s. w ic ys ar ph sw .a w w w .c om ly ee b Pedagogical Color Chart Pedagogical Color Chart Mechanics and Thermodynamics S Linear ( p) S and angular (L) momentum vectors Displacement and position vectors Displacement and position component vectors Linear and angular momentum component vectors S Linear (v ) and angular (v) velocity vectors Velocity component vectors S Energy transfer arrows Weng Dimensional rotational arrow .c Schematic linear or rotational motion directions S Acceleration vectors ( a ) Acceleration component vectors ly S Force vectors (F) Force component vectors om Torque vectors (t) Torque component vectors ee b S Enlargement arrow Qh Springs s. w Qc Pulleys ic Process arrow w Resistors w Negative charges .a Positive charges Inductors (coils) sw Magnetic fields Magnetic field vectors Magnetic field component vectors Capacitors ar ph Electric fields Electric field vectors Electric field component vectors ys Electricity and Magnetism w Batteries and other DC power supplies   Voltmeters V Ammeters A AC Sources Lightbulbs Ground symbol   Current Switches Light and Optics Light ray Focal light ray Central light ray Mirror Curved mirror Objects Converging lens Diverging lens Images Some Physical Constants Quantity Symbol Valuea Atomic mass unit u 1.660 538 782 (83) 3 10227 kg 931.494 028 (23) MeV/c 2 Avogadro’s number NA 6.022 141 79 (30) 3 1023 particles/mol Bohr magneton mB 5 eU 2me 9.274 009 15 (23) 3 10224 J/T Bohr radius a0 5 U2 m e e 2k e 5.291 772 085 9 (36) 3 10211 m Boltzmann’s constant kB 5 Compton wavelength lC 5 h me c Coulomb constant ke 5 1 4pP0 Deuteron mass md Electron mass me Electron volt eV Elementary charge e om R NA 1.380 650 4 (24) 3 10223 J/K ly .c 2.426 310 217 5 (33) 3 10212 m ic s. w ee b 8.987 551 788 . . . 3 109 N ? m2/C 2 (exact) R ys Gas constant Gravitational constant Nuclear magneton sw .a Permittivity of free space w w Planck’s constant 8.314 472 (15) J/mol ? K mn 1.674 927 211 (84) 3 10227 kg 1.008 664 915 97 (43) u 939.565 346 (23) MeV/c 2 eU 2m p 4p 3 1027 T ? m/A (exact) m0 P0 5 h U5 5.050 783 24 (13) 3 10227 J/T 1 m 0c 2 h 2p 8.854 187 817 . . . 3 10212 C2/N ? m2 (exact) 6.626 068 96 (33) 3 10234 J ? s 1.054 571 628 (53) 3 10234 J ? s mp 1.672 621 637 (83) 3 10227 kg 1.007 276 466 77 (10) u 938.272 013 (23) MeV/c 2 Rydberg constant RH 1.097 373 156 852 7 (73) 3 107 m21 Speed of light in vacuum c 2.997 924 58 3 108 m/s (exact) w Proton mass 1.602 176 487 (40) 3 10219 C 6.674 28 (67) 3 10211 N ? m2/kg2 mn 5 Permeability of free space 1.602 176 487 (40) 3 10219 J G ar ph Neutron mass 3.343 583 20 (17) 3 10227 kg 2.013 553 212 724 (78) u 9.109 382 15 (45) 3 10231 kg 5.485 799 094 3 (23) 3 1024 u 0.510 998 910 (13) MeV/c 2 Note: These constants are the values recommended in 2006 by CODATA, based on a least-squares adjustment of data from different measurements. For a more complete list, see P. J. Mohr, B. N. Taylor, and D. B. Newell, “CODATA Recommended Values of the Fundamental Physical Constants: 2006.” Rev. Mod. Phys. 80:2, 633–730, 2008. aThe numbers in parentheses for the values represent the uncertainties of the last two digits. Solar System Data Mean Radius (m) 3.30 3 1023 4.87 3 1024 5.97 3 1024 6.42 3 1023 1.90 3 1027 5.68 3 1026 8.68 3 1025 1.02 3 1026 1.25 3 1022 7.35 3 1022 1.989 3 1030 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Plutoa Moon Sun Mean Distance from the Sun (m) Period (s) 2.44 3 106 6.05 3 106 6.37 3 106 3.39 3 106 6.99 3 107 5.82 3 107 2.54 3 107 2.46 3 107 1.20 3 106 1.74 3 106 6.96 3 108 7.60 3 106 1.94 3 107 3.156 3 107 5.94 3 107 3.74 3 108 9.29 3 108 2.65 3 109 5.18 3 109 7.82 3 109 — — a In 5.79 3 1010 1.08 3 1011 1.496 3 1011 2.28 3 1011 7.78 3 1011 1.43 3 1012 2.87 3 1012 4.50 3 1012 5.91 3 1012 — — om Mass (kg) .c Body ee b ly August 2006, the International Astronomical Union adopted a definition of a planet that separates Pluto from the other eight planets. Pluto is now defined as a “dwarf planet” (like the asteroid Ceres). Physical Data Often Used 3.84 3 108 m s. w Average Earth–Moon distance 1.496 3 1011 m Average Earth–Sun distance 6.37 3 106 m Density of air (208C and 1 atm) ys Density of air (0°C and 1 atm) ic Average radius of the Earth 1.20 kg/m3 1.29 kg/m3 1.00 3 103 kg/m3 Free-fall acceleration 9.80 m/s2 ar ph Density of water (208C and 1 atm) 5.97 3 1024 kg Mass of the Moon 7.35 3 1022 kg Mass of the Sun 1.99 3 1030 kg Standard atmospheric pressure 1.013 3 105 Pa sw Mass of the Earth w .a Note: These values are the ones used in the text. Some Prefixes for Powers of Ten 10221 10218 10215 Abbreviation Power Prefix Abbreviation yocto y 101 deka da zepto z 102 hecto h a 103 kilo k f 106 mega M w 10224 Prefix w Power atto femto 10212 pico p 109 giga G 1029 nano n 1012 tera T m 1015 peta P m 1018 exa E zetta Z yotta Y 1026 1023 micro milli 1022 centi c 1021 1021 deci d 1024 Physics Ninth Edition .c om for Scientists and Engineers with Modern Physics ly Raymond A. Serway ee b Emeritus, James Madison University John W. Jewett, Jr. ic ys ar ph With contributions from Vahé Peroomian, University of California at Los Angeles s. w Emeritus, California State Polytechnic University, Pomona w w w .a sw About the Cover The cover shows a view inside the new railway departures concourse opened in March 2012 at the Kings Cross Station in London. The wall of the older structure (completed in 1852) is visible at the left. The sweeping shell-like roof is claimed by the architect to be the largest single-span station structure in Europe. Many principles of physics are required to design and construct such an open semicircular roof with a radius of 74 meters and containing over 2 000 triangular panels. Other principles of physics are necessary to develop the lighting design, optimize the acoustics, and integrate the new structure with existing infrastructure, historic buildings, and railway platforms. © Ashley Cooper/Corbis Australia • Brazil • Japan • Korea • Mexico • Singapore • Spain • United Kingdom • United States Physics for Scientists and Engineers with Modern Physics, Ninth Edition Raymond A. Serway and John W. Jewett, Jr. Publisher, Physical Sciences: Mary Finch Publisher, Physics and Astronomy: Charlie Hartford Development Editor: Ed Dodd 2014, 2010, 2008 by Raymond A. Serway NO RIGHTS RESERVED. Any part of this work may be reproduced, transmitted, stored, or used in any form or by any means graphic, electronic, or mechanical, including but not limited to photocopying, recording, scanning, digitizing, taping, Web distribution, information networks, or information storage and retrieval systems, without the prior written permission of the publisher. Assistant Editor: Brandi Kirksey Editorial Assistant: Brendan Killion om Media Editor: Rebecca Berardy Schwartz Brand Manager: Nicole Hamm Senior Marketing Development Manager: Tom Ziolkowski ly .c Marketing Communications Manager: Linda Yip Library of Congress Control Number: 2012947242 Senior Art Director: Cate Barr ISBN-13: 978-1-133-95405-7 Manufacturing Planner: Sandee Milewski ISBN-10: 1-133-95405-7 s. w Rights Acquisition Specialist: Shalice Shah-Caldwell Production Service: Lachina Publishing Services Text and Cover Designer: Roy Neuhaus w w w .a sw ar ph Compositor: Lachina Publishing Services ic Cover Image Credit: © Ashley Cooper/Corbis Brooks/Cole 20 Channel Center Street Boston, MA 02210 USA ys Cover Image: The new Kings Cross railway station, London, UK ee b Content Project Manager: Alison Eigel Zade We dedicate this book to our wives, Elizabeth and Lisa, and all our children and grandchildren for their loving understanding when we spent time on writing instead of being with them. Printed in the United States of America 1 2 3 4 5 6 7 16 15 14 13 12 Brief Contents 1 p a r t p a r t 2 w w p a r t p a r t 3 Temperature 568 The First Law of Thermodynamics 590 The Kinetic Theory of Gases 626 Heat Engines, Entropy, and the Second Law of Thermodynamics 653 5 Light and Optics 1057 35 36 37 38 The Nature of Light and the Principles of Ray Optics 1058 Image Formation 1090 Wave Optics 1134 Diffraction Patterns and Polarization 1160 p a r t 533 Thermodynamics 567 19 20 21 22 .c ee b w Oscillatory Motion 450 Wave Motion 483 Sound Waves 507 Superposition and Standing Waves Electric Fields 690 Gauss’s Law 725 Electric Potential 746 Capacitance and Dielectrics 777 Current and Resistance 808 Direct-Current Circuits 833 Magnetic Fields 868 Sources of the Magnetic Field 904 Faraday’s Law 935 Inductance 970 Alternating-Current Circuits 998 Electromagnetic Waves 1030 s. w .a sw Oscillations and Mechanical Waves 449 15 16 17 18 23 24 25 26 27 28 29 30 31 32 33 34 ic ys 11 12 13 14 ar ph 7 8 9 10 Physics and Measurement 2 Motion in One Dimension 21 Vectors 59 Motion in Two Dimensions 78 The Laws of Motion 111 Circular Motion and Other Applications of Newton’s Laws 150 Energy of a System 177 Conservation of Energy 211 Linear Momentum and Collisions 247 Rotation of a Rigid Object About a Fixed Axis 293 Angular Momentum 335 Static Equilibrium and Elasticity 363 Universal Gravitation 388 Fluid Mechanics 417 om Electricity and Magnetism 689 Mechanics 1 1 2 3 4 5 6 4 ly p a r t 6 Modern Physics 1191 39 40 41 42 43 44 45 46 Relativity 1192 Introduction to Quantum Physics 1233 Quantum Mechanics 1267 Atomic Physics 1296 Molecules and Solids 1340 Nuclear Structure 1380 Applications of Nuclear Physics 1418 Particle Physics and Cosmology 1447 iii Contents 6 Circular Motion and Other Applications About the Authors viii 6.1 6.2 6.3 6.4 To the Student xxx p a r t 1 ly ee b s. w 10 21 ar ph sw 59 .a Coordinate Systems 59 Vector and Scalar Quantities 61 Some Properties of Vectors 62 Components of a Vector and Unit Vectors w 4 Motion in Two Dimensions 65 78 w w 4.1 The Position, Velocity, and Acceleration Vectors 78 4.2 Two-Dimensional Motion with Constant Acceleration 81 4.3 ​Projectile Motion 84 4.4 ​Analysis Model: Particle in Uniform Circular Motion 91 4.5 Tangential and Radial Acceleration 94 4.6 ​Relative Velocity and Relative Acceleration 96 5 The Laws of Motion 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 8 Conservation of Energy 8.1 8.2 8.3 8.4 8.5 ys Position, Velocity, and Speed 22 Instantaneous Velocity and Speed 25 Analysis Model: Particle Under Constant Velocity 28 Acceleration 31 Motion Diagrams 35 Analysis Model: Particle Under Constant Acceleration 36 Freely Falling Objects 40 Kinematic Equations Derived from Calculus 43 3 Vectors 3.1 3.2 3.3 3.4 2 Standards of Length, Mass, and Time 3 Matter and Model Building 6 Dimensional Analysis 7 Conversion of Units 9 Estimates and Order-of-Magnitude Calculations Significant Figures 11 2 Motion in One Dimension 150 177 7.1 Systems and Environments 178 7.2 Work Done by a Constant Force 178 7.3 The Scalar Product of Two Vectors 181 7.4 Work Done by a Varying Force 183 7.5 Kinetic Energy and the Work–Kinetic Energy Theorem 7.6 Potential Energy of a System 191 7.7 Conservative and Nonconservative Forces 196 7.8 Relationship Between Conservative Forces and Potential Energy 198 7.9 Energy Diagrams and Equilibrium of a System 199 ic 1 Physics and Measurement 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 Extending the Particle in Uniform Circular Motion Model Nonuniform Circular Motion 156 Motion in Accelerated Frames 158 Motion in the Presence of Resistive Forces 161 7 Energy of a System Mechanics 1 1.1 1.2 1.3 1.4 1.5 1.6 om of Newton’s Laws 150 ix .c Preface 111 The Concept of Force 111 Newton’s First Law and Inertial Frames 113 Mass 114 Newton’s Second Law 115 The Gravitational Force and Weight 117 Newton’s Third Law 118 Analysis Models Using Newton’s Second Law Forces of Friction 130 120 211 Analysis Model: Nonisolated System (Energy) 212 Analysis Model: Isolated System (Energy) 215 Situations Involving Kinetic Friction 222 Changes in Mechanical Energy for Nonconservative Forces 227 Power 232 9 Linear Momentum and Collisions 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 188 247 Linear Momentum 247 Analysis Model: Isolated System (Momentum) 250 Analysis Model: Nonisolated System (Momentum) 252 Collisions in One Dimension 256 Collisions in Two Dimensions 264 The Center of Mass 267 Systems of Many Particles 272 Deformable Systems 275 Rocket Propulsion 277 10 Rotation of a Rigid Object About a Fixed Axis 293 10.1 Angular Position, Velocity, and Acceleration 293 10.2 Analysis Model: Rigid Object Under Constant Angular Acceleration 296 10.3 Angular and Translational Quantities 298 10.4 Torque 300 10.5 Analysis Model: Rigid Object Under a Net Torque 302 10.6 Calculation of Moments of Inertia 307 10.7 Rotational Kinetic Energy 311 10.8 Energy Considerations in Rotational Motion 312 10.9 Rolling Motion of a Rigid Object 316 11 Angular Momentum 335 11.1 The Vector Product and Torque 335 11.2 Analysis Model: Nonisolated System (Angular Momentum) 338 iv v Contents 11.3 Angular Momentum of a Rotating Rigid Object 342 11.4 Analysis Model: Isolated System (Angular Momentum) 345 11.5 The Motion of Gyroscopes and Tops 350 12.1 12.2 12.3 12.4 Thermodynamics 567 363 Analysis Model: Rigid Object in Equilibrium 363 More on the Center of Gravity 365 Examples of Rigid Objects in Static Equilibrium 366 Elastic Properties of Solids 373 13 Universal Gravitation 19 Temperature 388 13.1 Newton’s Law of Universal Gravitation 389 13.2 Free-Fall Acceleration and the Gravitational Force 391 13.3 Analysis Model: Particle in a Field (Gravitational) 392 13.4 Kepler’s Laws and the Motion of Planets 394 13.5 Gravitational Potential Energy 400 13.6 Energy Considerations in Planetary and Satellite Motion 402 417 ee b 21 The Kinetic Theory of Gases 21.1 21.2 21.3 21.4 21.5 2 Oscillations and Mechanical Waves 449 ic 450 sw ar ph 15.1 Motion of an Object Attached to a Spring 450 15.2 Analysis Model: Particle in Simple Harmonic Motion 452 15.3 Energy of the Simple Harmonic Oscillator 458 15.4 Comparing Simple Harmonic Motion with Uniform Circular Motion 462 15.5 The Pendulum 464 15.6 Damped Oscillations 468 15.7 Forced Oscillations 469 483 .a 16 Wave Motion w w w 16.1 Propagation of a Disturbance 484 16.2 Analysis Model: Traveling Wave 487 16.3 The Speed of Waves on Strings 491 16.4 Reflection and Transmission 494 16.5 Rate of Energy Transfer by Sinusoidal Waves on Strings 16.6 The Linear Wave Equation 497 17 Sound Waves 17.1 17.2 17.3 17.4 507 Pressure Variations in Sound Waves 508 Speed of Sound Waves 510 Intensity of Periodic Sound Waves 512 The Doppler Effect 517 18 Superposition and Standing Waves 533 18.1 Analysis Model: Waves in Interference 534 18.2 Standing Waves 538 18.3 Analysis Model: Waves Under Boundary Conditions 18.4 Resonance 546 18.5 Standing Waves in Air Columns 546 18.6 Standing Waves in Rods and Membranes 550 18.7 Beats: Interference in Time 550 18.8 Nonsinusoidal Wave Patterns 553 541 626 Molecular Model of an Ideal Gas 627 Molar Specific Heat of an Ideal Gas 631 The Equipartition of Energy 635 Adiabatic Processes for an Ideal Gas 637 Distribution of Molecular Speeds 639 22 Heat Engines, Entropy, and the Second Law ys 15 Oscillatory Motion 590 20.1 Heat and Internal Energy 590 20.2 Specific Heat and Calorimetry 593 20.3 Latent Heat 597 20.4 Work and Heat in Thermodynamic Processes 601 20.5 The First Law of Thermodynamics 603 20.6 Some Applications of the First Law of Thermodynamics 604 20.7 Energy Transfer Mechanisms in Thermal Processes 608 ly Pressure 417 Variation of Pressure with Depth 419 Pressure Measurements 423 Buoyant Forces and Archimedes’s Principle 423 Fluid Dynamics 427 Bernoulli’s Equation 430 Other Applications of Fluid Dynamics 433 p a r t 20 The First Law of Thermodynamics s. w 14.1 14.2 14.3 14.4 14.5 14.6 14.7 568 19.1 Temperature and the Zeroth Law of Thermodynamics 568 19.2 Thermometers and the Celsius Temperature Scale 570 19.3 The Constant-Volume Gas Thermometer and the Absolute Temperature Scale 571 19.4 Thermal Expansion of Solids and Liquids 573 19.5 Macroscopic Description of an Ideal Gas 578 .c 14 Fluid Mechanics 3 om 12 Static Equilibrium and Elasticity p a r t 495 of Thermodynamics 653 22.1 Heat Engines and the Second Law of Thermodynamics 654 22.2 Heat Pumps and Refrigerators 656 22.3 Reversible and Irreversible Processes 659 22.4 The Carnot Engine 660 22.5 Gasoline and Diesel Engines 665 22.6 Entropy 667 22.7 Changes in Entropy for Thermodynamic Systems 671 22.8 Entropy and the Second Law 676 p a r t 4 Electricity and Magnetism 689 23 Electric Fields 690 23.1 Properties of Electric Charges 690 23.2 Charging Objects by Induction 692 23.3 Coulomb’s Law 694 23.4 Analysis Model: Particle in a Field (Electric) 699 23.5 Electric Field of a Continuous Charge Distribution 704 23.6 Electric Field Lines 708 23.7 Motion of a Charged Particle in a Uniform Electric Field 710 24 Gauss’s Law 725 24.1 Electric Flux 725 24.2 Gauss’s Law 728 24.3 Application of Gauss’s Law to Various Charge Distributions 731 24.4 Conductors in Electrostatic Equilibrium 735 25 Electric Potential 746 25.1 Electric Potential and Potential Difference 746 25.2 Potential Difference in a Uniform Electric Field 748 Contents 33 Alternating-Current Circuits 25.3 Electric Potential and Potential Energy Due to Point Charges 752 25.4 Obtaining the Value of the Electric Field from the Electric Potential 755 25.5 Electric Potential Due to Continuous Charge Distributions 756 25.6 Electric Potential Due to a Charged Conductor 761 25.7 The Millikan Oil-Drop Experiment 764 25.8 Applications of Electrostatics 765 om 808 p a r t 833 s. w 35 The Nature of Light and the Principles of Ray Optics 1058 852 ar ph sw 30 Sources of the Magnetic Field 904 w w .a 30.1 The Biot–Savart Law 904 30.2 The Magnetic Force Between Two Parallel Conductors 30.3 Ampère’s Law 911 30.4 The Magnetic Field of a Solenoid 915 30.5 Gauss’s Law in Magnetism 916 30.6 Magnetism in Matter 919 32 Inductance 32.1 32.2 32.3 32.4 32.5 32.6 935 Faraday’s Law of Induction 935 Motional emf 939 Lenz’s Law 944 Induced emf and Electric Fields 947 Generators and Motors 949 Eddy Currents 953 w 31.1 31.2 31.3 31.4 31.5 31.6 35.1 35.2 35.3 35.4 35.5 35.6 35.7 35.8 ys 868 29.1 Analysis Model: Particle in a Field (Magnetic) 869 29.2 Motion of a Charged Particle in a Uniform Magnetic Field 874 29.3 Applications Involving Charged Particles Moving in a Magnetic Field 879 29.4 Magnetic Force Acting on a Current-Carrying Conductor 882 29.5 Torque on a Current Loop in a Uniform Magnetic Field 885 29.6 The Hall Effect 890 31 Faraday’s Law 970 Self-Induction and Inductance 970 RL Circuits 972 Energy in a Magnetic Field 976 Mutual Inductance 978 Oscillations in an LC Circuit 980 The RLC Circuit 984 5 Light and Optics 1057 Electromotive Force 833 Resistors in Series and Parallel 836 Kirchhoff’s Rules 843 RC Circuits 846 Household Wiring and Electrical Safety 29 Magnetic Fields 1030 34.1 Displacement Current and the General Form of Ampère’s Law 1031 34.2 Maxwell’s Equations and Hertz’s Discoveries 1033 34.3 Plane Electromagnetic Waves 1035 34.4 Energy Carried by Electromagnetic Waves 1039 34.5 Momentum and Radiation Pressure 1042 34.6 Production of Electromagnetic Waves by an Antenna 1044 34.7 The Spectrum of Electromagnetic Waves 1045 Electric Current 808 Resistance 811 A Model for Electrical Conduction 816 Resistance and Temperature 819 Superconductors 819 Electrical Power 820 28 Direct-Current Circuits 28.1 28.2 28.3 28.4 28.5 34 Electromagnetic Waves 1015 .c 27 Current and Resistance 27.1 27.2 27.3 27.4 27.5 27.6 777 Definition of Capacitance 777 Calculating Capacitance 779 Combinations of Capacitors 782 Energy Stored in a Charged Capacitor 786 Capacitors with Dielectrics 790 Electric Dipole in an Electric Field 793 An Atomic Description of Dielectrics 795 998 The Nature of Light 1058 Measurements of the Speed of Light 1059 The Ray Approximation in Ray Optics 1061 Analysis Model: Wave Under Reflection 1061 Analysis Model: Wave Under Refraction 1065 Huygens’s Principle 1071 Dispersion 1072 Total Internal Reflection 1074 ic 26.1 26.2 26.3 26.4 26.5 26.6 26.7 AC Sources 998 Resistors in an AC Circuit 999 Inductors in an AC Circuit 1002 Capacitors in an AC Circuit 1004 The RLC Series Circuit 1007 Power in an AC Circuit 1011 Resonance in a Series RLC Circuit 1013 The Transformer and Power Transmission Rectifiers and Filters 1018 ly 26 Capacitance and Dielectrics 33.1 33.2 33.3 33.4 33.5 33.6 33.7 33.8 33.9 ee b vi 909 36 Image Formation 36.1 36.2 36.3 36.4 36.5 36.6 36.7 36.8 36.9 36.10 1090 Images Formed by Flat Mirrors 1090 Images Formed by Spherical Mirrors 1093 Images Formed by Refraction 1100 Images Formed by Thin Lenses 1104 Lens Aberrations 1112 The Camera 1113 The Eye 1115 The Simple Magnifier 1118 The Compound Microscope 1119 The Telescope 1120 37 Wave Optics 1134 37.1 Young’s Double-Slit Experiment 1134 37.2 Analysis Model: Waves in Interference 1137 37.3 Intensity Distribution of the Double-Slit Interference Pattern 1140 37.4 Change of Phase Due to Reflection 1143 37.5 Interference in Thin Films 1144 37.6 The Michelson Interferometer 1147 38 Diffraction Patterns and Polarization 38.1 38.2 38.3 38.4 38.5 38.6 Introduction to Diffraction Patterns 1160 Diffraction Patterns from Narrow Slits 1161 Resolution of Single-Slit and Circular Apertures 1166 The Diffraction Grating 1169 Diffraction of X-Rays by Crystals 1174 Polarization of Light Waves 1175 1160 vii Contents 6 44.5 The Decay Processes 1394 44.6 Natural Radioactivity 1404 44.7 Nuclear Reactions 1405 44.8 Nuclear Magnetic Resonance and Magnetic Resonance Imaging 1406 1192 39.1 The Principle of Galilean Relativity 1193 39.2 The Michelson–Morley Experiment 1196 39.3 Einstein’s Principle of Relativity 1198 39.4 Consequences of the Special Theory of Relativity 1199 39.5 The Lorentz Transformation Equations 1210 39.6 The Lorentz Velocity Transformation Equations 1212 39.7 Relativistic Linear Momentum 1214 39.8 Relativistic Energy 1216 39.9 The General Theory of Relativity 1220 40 Introduction to Quantum Physics 1233 41 Quantum Mechanics w .a sw 42.1 Atomic Spectra of Gases 1297 42.2 Early Models of the Atom 1299 42.3 Bohr’s Model of the Hydrogen Atom 1300 42.4 The Quantum Model of the Hydrogen Atom 1306 42.5 The Wave Functions for Hydrogen 1308 42.6 Physical Interpretation of the Quantum Numbers 1311 42.7 The Exclusion Principle and the Periodic Table 1318 42.8 More on Atomic Spectra: Visible and X-Ray 1322 42.9 Spontaneous and Stimulated Transitions 1325 42.10 Lasers 1326 w 1340 Molecular Bonds 1341 Energy States and Spectra of Molecules 1344 Bonding in Solids 1352 Free-Electron Theory of Metals 1355 Band Theory of Solids 1359 Electrical Conduction in Metals, Insulators, and Semiconductors 1361 43.7 Semiconductor Devices 1364 43.8 Superconductivity 1370 w 43.1 43.2 43.3 43.4 43.5 43.6 44 Nuclear Structure 44.1 44.2 44.3 44.4 1380 Some Properties of Nuclei 1381 Nuclear Binding Energy 1386 Nuclear Models 1387 Radioactivity 1390 The Fundamental Forces in Nature 1448 Positrons and Other Antiparticles 1449 Mesons and the Beginning of Particle Physics Classification of Particles 1454 Conservation Laws 1455 Strange Particles and Strangeness 1459 Finding Patterns in the Particles 1460 Quarks 1462 Multicolored Quarks 1465 The Standard Model 1467 The Cosmic Connection 1469 Problems and Perspectives 1474 1451 s. w ic 1281 1296 43 Molecules and Solids 1447 46.1 46.2 46.3 46.4 46.5 46.6 46.7 46.8 46.9 46.10 46.11 46.12 1418 Appendices ar ph 42 Atomic Physics 1234 1267 41.1 The Wave Function 1267 41.2 Analysis Model: Quantum Particle Under Boundary Conditions 1271 41.3 The Schrödinger Equation 1277 41.4 A Particle in a Well of Finite Height 1279 41.5 Tunneling Through a Potential Energy Barrier 41.6 Applications of Tunneling 1282 41.7 The Simple Harmonic Oscillator 1286 46 Particle Physics and Cosmology Interactions Involving Neutrons Nuclear Fission 1419 Nuclear Reactors 1421 Nuclear Fusion 1425 Radiation Damage 1432 Uses of Radiation 1434 ee b Blackbody Radiation and Planck’s Hypothesis The Photoelectric Effect 1240 The Compton Effect 1246 The Nature of Electromagnetic Waves 1249 The Wave Properties of Particles 1249 A New Model: The Quantum Particle 1252 The Double-Slit Experiment Revisited 1255 The Uncertainty Principle 1256 1418 45.1 45.2 45.3 45.4 45.5 45.6 ys 40.1 40.2 40.3 40.4 40.5 40.6 40.7 40.8 45 Applications of Nuclear Physics om 39 Relativity .c Modern Physics 1191 ly p a r t A Tables A-1 A.1 Conversion Factors A-1 A.2 Symbols, Dimensions, and Units of Physical Quantities A-2 B Mathematics Review B.1 B.2 B.3 B.4 B.5 B.6 B.7 B.8 A-4 Scientific Notation A-4 Algebra A-5 Geometry A-10 Trigonometry A-11 Series Expansions A-13 Differential Calculus A-13 Integral Calculus A-16 Propagation of Uncertainty A-20 C Periodic Table of the Elements D SI Units A-24 D.1 SI Units A-24 D.2 Some Derived SI Units A-22 A-24 Answers to Quick Quizzes and Odd-Numbered Problems A-25 Index I-1 About the Authors ys ic s. w ee b ly .c om Raymond A. Serway received his doctorate at Illinois Institute of Technology and is Professor Emeritus at James Madison University. In 2011, he was awarded with an honorary doctorate degree from his alma mater, Utica College. He received the 1990 Madison Scholar Award at James Madison University, where he taught for 17 years. Dr. Serway began his teaching career at Clarkson University, where he conducted research and taught from 1967 to 1980. He was the recipient of the Distinguished Teaching Award at Clarkson University in 1977 and the Alumni Achievement Award from Utica College in 1985. As Guest Scientist at the IBM Research Laboratory in Zurich, Switzerland, he worked with K. Alex Müller, 1987 Nobel Prize recipient. Dr. Serway also was a visiting scientist at Argonne National Laboratory, where he collaborated with his mentor and friend, the late Dr. Sam Marshall. Dr. Serway is the coauthor of College Physics, Ninth Edition; Principles of Physics, Fifth Edition; Essentials of College Physics; Modern Physics, Third Edition; and the high school textbook Physics, published by Holt McDougal. In addition, Dr. Serway has published more than 40 research papers in the field of condensed matter physics and has given more than 60 presentations at professional meetings. Dr. Serway and his wife, Elizabeth, enjoy traveling, playing golf, fishing, gardening, singing in the church choir, and especially spending quality time with their four children, ten grandchildren, and a recent great grandson. John W. Jewett, Jr. w w w .a sw ar ph earned his undergraduate degree in physics at Drexel University and his doctorate at Ohio State University, specializing in optical and magnetic properties of condensed matter. Dr. Jewett began his academic career at Richard Stockton College of New Jersey, where he taught from 1974 to 1984. He is currently Emeritus Professor of Physics at California State Polytechnic University, Pomona. Through his teaching career, Dr. Jewett has been active in promoting effective physics education. In addition to receiving four National Science Foundation grants in physics education, he helped found and direct the Southern California Area Modern Physics Institute (SCAMPI) and Science IMPACT (Institute for Modern Pedagogy and Creative Teaching). Dr. Jewett’s honors include the Stockton Merit Award at Richard Stockton College in 1980, selection as Outstanding Professor at California State Polytechnic University for 1991–1992, and the Excellence in Undergraduate Physics Teaching Award from the American Association of Physics Teachers (AAPT) in 1998. In 2010, he received an Alumni Lifetime Achievement Award from Drexel University in recognition of his contributions in physics education. He has given more than 100 presentations both domestically and abroad, including multiple presentations at national meetings of the AAPT. He has also published 25 research papers in condensed matter physics and physics education research. Dr. Jewett is the author of The World of Physics: Mysteries, Magic, and Myth, which provides many connections between physics and everyday experiences. In addition to his work as the coauthor for Physics for Scientists and Engineers, he is also the coauthor on Principles of Physics, Fifth Edition, as well as Global Issues, a four-volume set of instruction manuals in integrated science for high school. Dr. Jewett enjoys playing keyboard with his all-physicist band, traveling, underwater photography, learning foreign languages, and collecting antique quack medical devices that can be used as demonstration apparatus in physics lectures. Most importantly, he relishes spending time with his wife, Lisa, and their children and grandchildren. viii Preface ee b ly .c om In writing this Ninth Edition of Physics for Scientists and Engineers, we continue our ongoing efforts to improve the clarity of presentation and include new pedagogical features that help support the learning and teaching processes. Drawing on positive feedback from users of the Eighth Edition, data gathered from both professors and students who use Enhanced WebAssign, as well as reviewers’ suggestions, we have refined the text to better meet the needs of students and teachers. This textbook is intended for a course in introductory physics for students majoring in science or engineering. The entire contents of the book in its extended version could be covered in a three-semester course, but it is possible to use the material in shorter sequences with the omission of selected chapters and sections. The mathematical background of the student taking this course should ideally include one semester of calculus. If that is not possible, the student should be enrolled in a concurrent course in introductory calculus. s. w Content ar ph Objectives ys ic The material in this book covers fundamental topics in classical physics and provides an introduction to modern physics. The book is divided into six parts. Part 1 (Chapters 1 to 14) deals with the fundamentals of Newtonian mechanics and the physics of fluids; Part 2 (Chapters 15 to 18) covers oscillations, mechanical waves, and sound; Part 3 (Chapters 19 to 22) addresses heat and thermodynamics; Part 4 (Chapters 23 to 34) treats electricity and magnetism; Part 5 (Chapters 35 to 38) covers light and optics; and Part 6 (Chapters 39 to 46) deals with relativity and modern physics. .a sw This introductory physics textbook has three main objectives: to provide the student with a clear and logical presentation of the basic concepts and principles of physics, to strengthen an understanding of the concepts and principles through a broad range of interesting real-world applications, and to develop strong problem-solving skills through an effectively organized approach. To meet these objectives, we emphasize well-organized physical arguments and a focused problem-solving strategy. At the same time, we attempt to motivate the student through practical examples that demonstrate the role of physics in other disciplines, including engineering, chemistry, and medicine. w w Changes in the Ninth Edition w A large number of changes and improvements were made for the Ninth Edition of this text. Some of the new features are based on our experiences and on current trends in science education. Other changes were incorporated in response to comments and suggestions offered by users of the Eighth Edition and by reviewers of the manuscript. The features listed here represent the major changes in the Ninth Edition. Enhanced Integration of the Analysis Model Approach to Problem Solving. Students are faced with hundreds of problems during their physics courses. A relatively small number of fundamental principles form the basis of these problems. When faced with a new problem, a physicist forms a model of the problem that can be solved in a simple way by identifying the fundamental principle that is applicable in the problem. For example, many problems involve conservation of energy, Newton’s second law, or kinematic equations. Because the physicist has studied these principles and their applications extensively, he or she can apply this knowledge as a model for solving a new problem. Although it would be ideal for students to follow this same process, most students have difficulty becoming familiar with the entire palette of fundamental principles that are available. It is easier for students to identify a situation rather than a fundamental principle. ix x Preface s. w ee b ly .c om The Analysis Model approach we focus on in this revision lays out a standard set of situations that appear in most physics problems. These situations are based on an entity in one of four simplification models: particle, system, rigid object, and wave. Once the simplification model is identified, the student thinks about what the entity is doing or how it interacts with its environment. This leads the student to identify a particular Analysis Model for the problem. For example, if an object is falling, the object is recognized as a particle experiencing an acceleration due to gravity that is constant. The student has learned that the Analysis Model of a particle under constant acceleration describes this situation. Furthermore, this model has a small number of equations associated with it for use in starting problems, the kinematic equations presented in Chapter 2. Therefore, an understanding of the situation has led to an Analysis Model, which then identifies a very small number of equations to start the problem, rather than the myriad equations that students see in the text. In this way, the use of Analysis Models leads the student to identify the fundamental principle. As the student gains more experience, he or she will lean less on the Analysis Model approach and begin to identify fundamental principles directly. To better integrate the Analysis Model approach for this edition, Analysis Model descriptive boxes have been added at the end of any section that introduces a new Analysis Model. This feature recaps the Analysis Model introduced in the section and provides examples of the types of problems that a student could solve using the Analysis Model. These boxes function as a “refresher” before students see the Analysis Models in use in the worked examples for a given section. Worked examples in the text that utilize Analysis Models are now designated with an AM icon for ease of reference. The solutions of these examples integrate the Analysis Model approach to problem solving. The approach is further reinforced in the end-of-chapter summary under the heading Analysis Models for Problem Solving, and through the new Analysis Model Tutorials that are based on selected end-of-chapter problems and appear in Enhanced WebAssign. sw ar ph ys ic Analysis Model Tutorials. John Jewett developed 165 tutorials (indicated in each chapter’s problem set with an AMT icon) that strengthen students’ problem-solving skills by guiding them through the steps in the problem-solving process. Important first steps include making predictions and focusing on physics concepts before solving the problem quantitatively. A critical component of these tutorials is the selection of an appropriate Analysis Model to describe what is going on in the problem. This step allows students to make the important link between the situation in the problem and the mathematical representation of the situation. Analysis Model tutorials include meaningful feedback at each step to help students practice the problem-solving process and improve their skills. In addition, the feedback addresses student misconceptions and helps them to catch algebraic and other mathematical errors. Solutions are carried out symbolically as long as possible, with numerical values substituted at the end. This feature helps students understand the effects of changing the values of each variable in the problem, avoids unnecessary repetitive substitution of the same numbers, and eliminates round-off errors. Feedback at the end of the tutorial encourages students to compare the final answer with their original predictions. w .a Annotated Instructor’s Edition. New for this edition, the Annotated Instructor’s Edition provides instructors with teaching tips and other notes on how to utilize the textbook in the classroom, via cyan annotations. Additionally, the full complement of icons describing the various types of problems will be included in the questions/problems sets (the Student Edition contains only those icons needed by students). w w PreLecture Explorations. The Active Figure questions in WebAssign from the Eighth Edition have been completely revised. The simulations have been updated, with additional parameters to enhance investigation of a physical phenomenon. Students can make predictions, change the parameters, and then observe the results. Each new PreLecture Exploration comes with conceptual and analytical questions that guide students to a deeper understanding and help promote a robust physical intuition. New Master Its Added in Enhanced WebAssign. Approximately 50 new Master Its in Enhanced WebAssign have been added for this edition to the end-of-chapter problem sets. Chapter-by-Chapter Changes The list below highlights some of the major changes for the Ninth Edition. Preface Chapter 1 • Two new Master Its were added to the end-of-chapter problems set. • Three new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. • One new Master It was added to the end-of-chapter problems set. • Four new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. Chapter 8 Chapter 2 • Two Analysis Model descriptive boxes have been added, in Sections 8.1 and 8.2. • The problem-solving strategy in Section 8.2 has been reworded to account for a more general application to both isolated and nonisolated systems. • As a result of a suggestion from a PER team at University of Washington and Pennsylvania State University, Example 8.1 has been rewritten to demonstrate to students the effect of choosing different systems on the development of the solution. • All examples in the chapter have been rewritten to begin with Equation 8.2 directly rather than beginning with the format Ei 5 Ef . • Several examples have been modified so that numerical values are put in only at the end of the solution. • The problem-solving strategy in Section 8.4 has been deleted and the text material revised to incorporate these ideas on handling energy changes when nonconservative forces act. • Several textual sections have been revised to make more explicit references to analysis models. • One new Master It was added to the end-of-chapter problems set. • Four new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. .c om • A new introduction to the concept of Analysis Models has been included in Section 2.3. • Three Analysis Model descriptive boxes have been added, in Sections 2.3 and 2.6. • Several textual sections have been revised to make more explicit references to analysis models. • Three new Master Its were added to the end-of-chapter problems set. • Five new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. ly Chapter 3 ee b • Three new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. Chapter 4 ys ic s. w • An Analysis Model descriptive box has been added, in Section 4.6. • Several textual sections have been revised to make more explicit references to analysis models. • Three new Master Its were added to the end-of-chapter problems set. • Five new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. xi ar ph Chapter 5 w .a sw • Two Analysis Model descriptive boxes have been added, in Section 5.7. • Several examples have been modified so that numerical values are put in only at the end of the solution. • Several textual sections have been revised to make more explicit references to analysis models. • Four new Master Its were added to the end-of-chapter problems set. • Four new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. Chapter 6 w w • An Analysis Model descriptive box has been added, in Section 6.1. • Several examples have been modified so that numerical values are put in only at the end of the solution. • Four new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. Chapter 7 • The notation for work done on a system externally and internally within a system has been clarified. • The equations and discussions in several sections have been modified to more clearly show the comparisons of similar potential energy equations among different situations. Chapter 9 • Two Analysis Model descriptive boxes have been added, in Section 9.3. • Several examples have been modified so that numerical values are put in only at the end of the solution. • Five new Master Its were added to the end-of-chapter problems set. • Four new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. Chapter 10 • The order of four sections (10.4–10.7) has been modified so as to introduce moment of inertia through torque (rather than energy) and to place the two sections on energy together. The sections have been revised accordingly to account for the revised development of concepts. This revision makes the order of approach similar to the order of approach students have already seen in translational motion. • New introductory paragraphs have been added to several sections to show how the development of our analysis of rotational motion parallels that followed earlier for translational motion. • Two Analysis Model descriptive boxes have been added, in Sections 10.2 and 10.5. • Several textual sections have been revised to make more explicit references to analysis models. xii Preface • One new Master It was added to the end-of-chapter problems set. • Four new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. • Two new Master Its were added to the end-of-chapter problems set. • Four new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. Chapter 11 Chapter 15 • An Analysis Model descriptive box has been added, in Section 15.2. • Several textual sections have been revised to make more explicit references to Analysis Models. • Four new Master Its were added to the end-of-chapter problems set. • Four new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. om • Two Analysis Model descriptive boxes have been added, in Sections 11.2 and 11.4. • Angular momentum conservation equations have been revised so as to be presented as DL 5 (0 or tdt) in order to be consistent with the approach in Chapter 8 for energy conservation and Chapter 9 for linear momentum conservation. • Four new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. Chapter 16 Chapter 12 .c • A new Analysis Model descriptive box has been added, in Section 16.2. • Section 16.3, on the derivation of the speed of a wave on a string, has been completely rewritten to improve the logical development. • Four new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. ee b ly • One Analysis Model descriptive box has been added, in Section 12.1. • Several examples have been modified so that numerical values are put in only at the end of the solution. • Four new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. Chapter 17 • One new Master It was added to the end-of-chapter problems set. • Four new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. ic • Sections 13.3 and 13.4 have been interchanged to provide a better flow of concepts. • A new analysis model has been introduced: Particle in a Field (Gravitational). This model is introduced because it represents a physical situation that occurs often. In addition, the model is introduced to anticipate the importance of versions of this model later in electricity and magnetism, where it is even more critical. An Analysis Model descriptive box has been added in Section 13.3. In addition, a new summary flash card has been added at the end of the chapter, and textual material has been revised to make reference to the new model. • The description of the historical goals of the Cavendish experiment in 1798 has been revised to be more consistent with Cavendish’s original intent and the knowledge available at the time of the experiment. • Newly discovered Kuiper belt objects have been added, in Section 13.4. • Textual material has been modified to make a stronger tie-in to Analysis Models, especially in the energy sections 13.5 and 13.6. • All conservation equations have been revised so as to be presented with the change in the system on the left and the transfer across the boundary of the system on the right, in order to be consistent with the approach in earlier chapters for energy conservation, linear momentum conservation, and angular momentum conservation. • Four new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. s. w Chapter 13 w w w .a sw ar ph ys Chapter 18 Chapter 14 • Several textual sections have been revised to make more explicit references to Analysis Models. • Several examples have been modified so that numerical values are put in only at the end of the solution. • Two Analysis Model descriptive boxes have been added, in Sections 18.1 and 18.3. • Two new Master Its were added to the end-of-chapter problems set. • Four new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. Chapter 19 • Several examples have been modified so that numerical values are put in only at the end of the solution. • One new Master It was added to the end-of-chapter problems set. • Four new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. Chapter 20 • Section 20.3 was revised to emphasize the focus on systems. • Five new Master Its were added to the end-of-chapter problems set. • Four new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. Chapter 21 • A new introduction to Section 21.1 sets up the notion of structural models to be used in this chapter and future chapters for describing systems that are too large or too small to observe directly. • Fifteen new equations have been numbered, and all equations in the chapter have been renumbered. This    Preface xiii • Two new Master Its were added to the end-of-chapter problems set. • Four new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. Chapter 25 • Sections 25.1 and 25.2 have been significantly revised to make connections to the new particle in a field analysis models introduced in Chapters 13 and 23. • Example 25.4 has been moved so as to appear after the Problem-Solving Strategy in Section 25.5, allowing students to compare electric fields due to a small number of charges and a continuous charge distribution. • Two new Master Its were added to the end-of-chapter problems set. • Four new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. .c Chapter 26 ee b • The discussion of series and parallel capacitors in Section 26.3 has been revised for clarity. • The discussion of potential energy associated with an electric dipole in an electric field in Section 26.6 has been revised for clarity. • Four new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. s. w • In Section 22.4, the discussion of Carnot’s theorem has been rewritten and expanded, with a new figure added that is connected to the proof of the theorem. • The material in Sections 22.6, 22.7, and 22.8 has been completely reorganized, reordered, and rewritten. The notion of entropy as a measure of disorder has been removed in favor of more contemporary ideas from the physics education literature on entropy and its relationship to notions such as uncertainty, missing information, and energy spreading. • Two new Pitfall Preventions have been added in Section 22.6 to help students with their understanding of entropy. • There is a newly added argument for the equivalence of the entropy statement of the second law and the Clausius and Kelvin–Planck statements in Section 22.8. • Two new summary flashcards have been added relating to the revised entropy discussion. • Three new Master Its were added to the end-of-chapter problems set. • Four new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. om Chapter 22 ly new program of equation numbers allows easier and more efficient referencing to equations in the development of kinetic theory. • The order of Sections 21.3 and 21.4 has been reversed to provide a more continuous discussion of specific heats of gases. • One new Master It was added to the end-of-chapter problems set. • Four new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. ar ph ys ic Chapter 27 Chapter 23 w w w .a sw • A new analysis model has been introduced: Particle in a Field (Electrical). This model follows on the introduction of the Particle in a Field (Gravitational) model introduced in Chapter 13. An Analysis Model descriptive box has been added, in Section 23.4. In addition, a new summary flash card has been added at the end of the chapter, and textual material has been revised to make reference to the new model. • A new What If? has been added to Example 23.9 in order to make a connection to infinite planes of charge, to be further studied in later chapters. • Several textual sections and worked examples have been revised to make more explicit references to analysis models. • One new Master It was added to the end-of-chapter problems set. • Four new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. Chapter 24 • Section 24.1 has been significantly revised to clarify the geometry of area elements through which electric field lines pass to generate an electric flux. • Two new figures have been added to Example 24.5 to further explore the electric fields due to single and paired infinite planes of charge. • The discussion of the Drude model for electrical conduction in Section 27.3 has been revised to follow the outline of structural models introduced in Chapter 21. • Several textual sections have been revised to make more explicit references to analysis models. • Five new Master Its were added to the end-of-chapter problems set. • Four new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. Chapter 28 • The discussion of series and parallel resistors in Section 28.2 has been revised for clarity. • Time-varying charge, current, and voltage have been represented with lowercase letters for clarity in distinguishing them from constant values. • Five new Master Its were added to the end-of-chapter problems set. • Two new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. Chapter 29 • A new analysis model has been introduced: Particle in a Field (Magnetic). This model follows on the introduction of the Particle in a Field (Gravitational) model introduced in Chapter 13 and the Particle in a Field (Electrical) model in Chapter 23. An Analysis Model descriptive box has been added, in Section 29.1. In addition, a new summary flash card has been added at the end of the chapter, and textual material has been revised to make reference to the new model. xiv Preface been included, representing the first-ever direct optical image of a solar system beyond ours. • Five new Master Its were added to the end-of-chapter problems set. • Three new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. Chapter 32 Chapter 38 • Four new Master Its were added to the end-of-chapter problems set. • Three new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. Chapter 39 • Several textual sections have been revised to make more explicit references to analysis models. • Sections 39.8 and 39.9 from the Eighth Edition have been combined into one section. • Five new Master Its were added to the end-of-chapter problems set. • Four new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. ys ic • Several textual sections have been revised to make more explicit references to analysis models. • Time-varying charge, current, and voltage have been represented with lowercase letters for clarity in distinguishing them from constant values. • Two new Master Its were added to the end-of-chapter problems set. • Three new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. om • Several textual sections have been revised to make more explicit references to analysis models. • One new Master It was added to the end-of-chapter problems set. • Four new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. • An Analysis Model descriptive box has been added, in Section 37.2. • The discussion of the Laser Interferometer GravitationalWave Observatory (LIGO) in Section 37.6 has been updated. • Three new Master Its were added to the end-of-chapter problems set. • Four new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. .c Chapter 31 Chapter 37 ly • Several textual sections have been revised to make more explicit references to analysis models. • One new Master It was added to the end-of-chapter problems set. • Four new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. ee b Chapter 30 s. w • One new Master It was added to the end-of-chapter problems set. • Six new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. ar ph Chapter 33 sw • Phasor colors have been revised in many figures to improve clarity of presentation. • Three new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. Chapter 34 w Chapter 35 w w .a • Several textual sections have been revised to make more explicit references to analysis models. • The status of spacecraft related to solar sailing has been updated in Section 34.5. • Six new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. • Two new Analysis Model descriptive boxes have been added, in Sections 35.4 and 35.5. • Several textual sections and worked examples have been revised to make more explicit references to analysis models. • Five new Master Its were added to the end-of-chapter problems set. • Four new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. Chapter 36 • The discussion of the Keck Telescope in Section 36.10 has been updated, and a new figure from the Keck has Chapter 40 • The discussion of the Planck model for blackbody radiation in Section 40.1 has been revised to follow the outline of structural models introduced in Chapter 21. • The discussion of the Einstein model for the photoelectric effect in Section 40.2 has been revised to follow the outline of structural models introduced in Chapter 21. • Several textual sections have been revised to make more explicit references to analysis models. • Two new Master Its were added to the end-of-chapter problems set. • Two new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. Chapter 41 • An Analysis Model descriptive box has been added, in Section 41.2. • One new Analysis Model Tutorial was added for this chapter in Enhanced WebAssign. Chapter 42 • The discussion of the Bohr model for the hydrogen atom in Section 42.3 has been revised to follow the outline of structural models introduced in Chapter 21. • In Section 42.7, the tendency for atomic systems to drop to their lowest energy levels is related to the new discus-    Preface xv • Chapter 46 • A discussion of the ALICE (A Large Ion Collider Experiment) project searching for a quark–gluon plasma at the Large Hadron Collider (LHC) has been added to Section 46.9. • A discussion of the July 2012 announcement of the discovery of a Higgs-like particle from the ATLAS (A Toroidal LHC Apparatus) and CMS (Compact Muon Solenoid) projects at the Large Hadron Collider (LHC) has been added to Section 46.10. • A discussion of closures of colliders due to the beginning of operations at the Large Hadron Collider (LHC) has been added to Section 46.10. • A discussion of recent missions and the new Planck mission to study the cosmic background radiation has been added to Section 46.11. • Several textual sections have been revised to make more explicit references to analysis models. • One new Master It was added to the end-of-chapter problems set. • One new Analysis Model Tutorial was added for this chapter in Enhanced WebAssign. ys ic s. w • A new discussion of the contribution of carbon dioxide molecules in the atmosphere to global warming has been added to Section 43.2. A new figure has been added, showing the increasing concentration of carbon dioxide in the past decades. • A new discussion of graphene (Nobel Prize in Physics, 2010) and its properties has been added to Section 43.4. • The discussion of worldwide photovoltaic power plants in Section 43.7 has been updated. • The discussion of transistor density on microchips in Section 43.7 has been updated. • Several textual sections and worked examples have been revised to make more explicit references to analysis models. • One new Analysis Model Tutorial was added for this chapter in Enhanced WebAssign. om Chapter 43 .c • • Discussion of the March 2011 nuclear disaster after the earthquake and tsunami in Japan was added to Section 45.3. • The discussion of the International Thermonuclear Experimental Reactor (ITER) in Section 45.4 has been updated. • The discussion of the National Ignition Facility (NIF) in Section 45.4 has been updated. • The discussion of radiation dosage in Section 45.5 has been cast in terms of SI units grays and sieverts. • Section 45.6 from the Eighth Edition has been deleted. • Four new Master Its were added to the end-of-chapter problems set. • One new Analysis Model Tutorial was added for this chapter in Enhanced WebAssign. ly • Chapter 45 ee b • sion of the second law of thermodynamics appearing in Chapter 22. The discussion of the applications of lasers in Section 42.10 has been updated to include laser diodes, carbon dioxide lasers, and excimer lasers. Several textual sections have been revised to make more explicit references to analysis models. Five new Master Its were added to the end-of-chapter problems set. Three new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. Chapter 44 .a sw ar ph • Data for the helium-4 atom were added to Table 44.1. • Several textual sections have been revised to make more explicit references to analysis models. • Three new Master Its were added to the end-of-chapter problems set. • Two new Analysis Model Tutorials were added for this chapter in Enhanced WebAssign. w Text Features w w Most instructors believe that the textbook selected for a course should be the student’s primary guide for understanding and learning the subject matter. Furthermore, the textbook should be easily accessible and should be styled and written to facilitate instruction and learning. With these points in mind, we have included many pedagogical features, listed below, that are intended to enhance its usefulness to both students and instructors. Problem Solving and Conceptual Understanding General Problem-Solving Strategy. A general strategy outlined at the end of Chapter 2 (pages 45–47) provides students with a structured process for solving problems. In all remaining chapters, the strategy is employed explicitly in every example so that students learn how it is applied. Students are encouraged to follow this strategy when working end-of-chapter problems. Worked Examples. All in-text worked examples are presented in a two-column format to better reinforce physical concepts. The left column shows textual information xvi Preface ee b ly .c om that describes the steps for solving the problem. The right column shows the mathematical manipulations and results of taking these steps. This layout facilitates matching the concept with its mathematical execution and helps students organize their work. The examples closely follow the General Problem-­Solving Strategy introduced in Chapter 2 to reinforce effective problem-solving habits. All worked examples in the text may be assigned for homework in Enhanced WebAssign. A sample of a worked example can be found on the next page. Examples consist of two types. The first (and most common) example type presents a problem and numerical answer. The second type of example is conceptual in nature. To accommodate increased emphasis on understanding physical concepts, the many conceptual examples are labeled as such and are designed to help students focus on the physical situation in the problem. Worked examples in the text that utilize Analysis Models are now designated with an AM icon for ease of reference, and the solutions of these examples now more thoroughly integrate the Analysis Model approach to problem solving. Based on reviewer feedback from the Eighth Edition, we have made careful revisions to the worked examples so that the solutions are presented symbolically as far as possible, with numerical values substituted at the end. This approach will help students think symbolically when they solve problems instead of unnecessarily inserting numbers into intermediate equations. ys ic s. w What If? Approximately one-third of the worked examples in the text contain a What If? feature. At the completion of the example solution, a What If? question offers a variation on the situation posed in the text of the example. This feature encourages students to think about the results of the example, and it also assists in conceptual understanding of the principles. What If? questions also prepare students to encounter novel problems that may be included on exams. Some of the end-of-chapter problems also include this feature. w w w .a sw ar ph Quick Quizzes. Students are provided an opportunity to test their understanding of the physical concepts presented through Quick Quizzes. The questions require students to make decisions on the basis of sound reasoning, and some of the questions have been written to help students overcome common misconceptions. Quick Quizzes have been cast in an objective format, including multiple-choice, true–false, and ranking. Answers to all Quick Quiz questions are found at the end of the text. Many instructors choose to use such questions in a “peer instruction” teaching style or with the use of personal response system “clickers,” but they can be used in standard quiz format as well. An example of a Quick Quiz follows below. Pitfall Prevention 16.2 Two Kinds of Speed/Velocity Do not confuse v, the speed of the wave as it propagates along the string, with vy , the transverse velocity of a point on the string. The speed v is constant for a uniform medium, whereas vy varies sinusoidally. Q uick Quiz 7.5 ​A dart is inserted into a spring-loaded dart gun by pushing the spring in by a distance x. For the next loading, the spring is compressed a distance 2x. How much faster does the second dart leave the gun compared with the first? (a) four times as fast (b) two times as fast (c) the same (d) half as fast (e) one-fourth as fast Pitfall Preventions. More than two hundred Pitfall Preventions (such as the one to the left) are provided to help students avoid common mistakes and misunderstandings. These features, which are placed in the margins of the text, address both common student misconceptions and situations in which students often follow unproductive paths. Summaries. Each chapter contains a summary that reviews the important concepts and equations discussed in that chapter. The summary is divided into three sections: Definitions, Concepts and Principles, and Analysis Models for Problem Solving. In each section, flash card–type boxes focus on each separate definition, concept, principle, or analysis model. Preface xvii All worked examples are also available to be assigned as interactive examples in the Enhanced 1.1 First-Levelsystem. Head WebAssign homework management Example 3.2 A Vacation Trip A car travels 20.0 km due north and then 35.0 km in a direction 60.0° west of north as shown in Figure 3.11a. Find the magnitude and direction of the car’s resultant displacement. .c Figure 3.11 (Example 3.2) (a) Graphical method for finding the resul tant displacement vector (b) Adding the vectors in reverse order gives the same result for ly and drawn in Figure 3.11a help us conceptualize the problem. The resultant vector has also been drawn. We expect its magnitude to be a few tens of kilometers. The angle that the resultant vector makes with the axis is expected to be less than 60°, the angle that vector makes with the axis. om Conceptualize The vectors ee b Categorize We can categorize this example as a simple analysis problem in vector addition. The displacement is the resultant when the two individual displacements and are added. We can further categorize it as a problem about the analysis of triangles, so we appeal to our expertise in geometry and trigonometry. Analyze In this example, we show two ways to analyze the problem of finding the resultant of two vectors. The first way is ic s. w to solve the problem geometrically, using graph paper and a protractor to measure the magnitude of and its direction in Figure 3.11a. (In fact, even when you know you are going to be carrying out a calculation, you should sketch the vectors to check your results.) With an ordinary ruler and protractor, a large diagram typically gives answers to two-digit but not to three-digit precision. Try using these tools on in Figure 3.11a and compare to the trigonometric analysis below! The second way to solve the problem is to analyze it using algebra and trigonometry. The magnitude of can be obtained from the law of cosines as applied to the triangle in Figure 3.11a (see Appendix B.4). Use find cos cos from the law of cosines to Substitute numerical values, noting that 180° 60° 120°: ar ph Each step of the solution is detailed in a two-column format. The left column provides an explanation for each mathematical step in the right column, to better reinforce the physi cal concepts. Solution ys Each solution has been written to closely follow the General ProblemSolving Strategy as outlined on pages 45–47 in Chapter 2, so as to reinforce good problemsolving habits. 20.0 km 35.0 km 48.2 km sw Use the law of sines (Appendix B.4) to find the direction measured from the northerly direction: sin 20.0 km 2 1 35.0 km cos 120 sin sin b 5 sin u 5 35.0 km sin 1208 5 0.629 48.2 km 38.9° .a The resultant displacement of the car is 48.2 km in a direction 38.9° west of north. w w w Finalize Does the angle that we calculated agree with an estimate made by looking at Figure 3.11a or with an actual angle measured from the diagram using the graphical method? Is it reasonable that the magnitude of is larger than that of both and ? Are the units of correct? Although the head to tail method of adding vectors works well, it suffers from two disadvantages. First, some people find using the laws of cosines and sines to be awkward. Second, a triangle only results if you are adding two vectors. If you are adding three or more vectors, the resulting geometric shape is usually not a triangle. In Section 3.4, we explore a new method of adding vectors that will address both of these disadvantages. W h at Suppose the trip were taken with the two vectors in reverse order: 35.0 km at 60.0° west of north first and then 20.0 km due north. How would the magnitude and the direction of the resultant vector change? Answer They would not change. The commutative law for vector addition tells us that the order of vectors in an addition is irrelevant. Graphically, Figure 3.11b shows that the vectors added in the reverse order give us the same resultant vector. What If? statements appear in about one-third of the worked examples and offer a variation on the situation posed in the text of the example. For instance, this feature might explore the effects of changing the conditions of the situation, determine what happens when a quantity is taken to a particular limiting value, or question whether additional information can be determined about the problem situation. This feature encourages students to think about the results of the example and assists in conceptual understanding of the principles. xviii Preface Questions and Problems Sets. For the Ninth Edition, the authors reviewed each ques tion and problem and incorporated revisions designed to improve both readability and assignability. More than 10% of the problems are new to this edition. om Questions. The Questions section is divided into two sections: Objective Questions and Conceptual Questions. The instructor may select items to assign as homework or use in the classroom, possibly with “peer instruction” methods and possibly with personal response systems. More than 900 Objective and Conceptual Questions are included in this edition. Answers for selected questions are included in the Student Solutions Manual/Study Guide, and answers for all questions are found in the Instruc tor’s Solutions Manual. ee b ly .c Objective Questions are multiple-choice, 242 true–false, ranking, or other multiple Chapter Conservation of Energy guess–type questions. Some require calculations designed to facilitate students’ familiarity with the equations, the variables(a)used, concepts the variables After the the spring is compressed and therep popgun fired, to what height the conceptual projectile risein above resent, and the relationships between the concepts. Others aredoes more point ? (b) Draw four Objective energy bar charts for this situa nature and are designed to encourage conceptual thinking. Questions tion, analogous to those in Figures 8.6c–d. are also written with the personal response system user in mind, and most of the 57. As the driver steps on the gas pedal, a car of mass questions could easily be used in these systems. 1 160 kg accelerates from rest. During the first few seconds of motion, the car’s acceleration increases with s. w time according to the expression and essay-type questions Conceptual Questions are more traditional short-answer 1.16 0.210 0.240 that require students to think conceptually about a physical situation. 242 Chapter Conservation of Energy where is in seconds and is in m/s . (a) What is the change in kinetic energy of the car during the interval load a distance (4) /2 will move given time interv (a) Show that Arist the equation constant. (b) Show this part of Aristot particular, describe derive the equatio tions, and determin 61. A child’s pogo stick w w w .a sw ar ph ys ic stores energy in a sp Problems. An extensive set of problems is included the each chapter; in from 0 toatload 2.50aend s? (b)of What is the minimum aver /2, then (a) After the spring is compressed and the popgun distance /2 in time interval force constant of age power output of the engine over this time interval? all, fired, this edition contains more than 3 700 problems. Answers for odd-numbered to what height does the projectile rise above (4) /2 will move /2 the given distance in the N/m. At positio (c) Why is the value in part (b) described as the mini problems at thebar end of for thethis book. forinterval approximately 20% point are ? (b)provided Draw four energy charts situa Full solutions given time 0.100 m), the sp mum value? tion, analogous to those in Figuresin8.6c–d. is a maximu of the problems are included the Student Solutions(a)Manual/Study Guide, and soluare included pression Show that Aristotle’s proportions in 58. Review. Why is the following situation impossible? A new child is momentaril 57. As the driver steps on the gas pedal, a car of mass the equation bwd, where is a proportionality tions for all problems are found in the Instructor’s Solutions Manual. high-speed roller coaster is claimed be so of safe that includes position 0) 1 160 kg accelerates from rest. During the first few secconstant. (b) Show that ourtotheory motion the passengers do not need to wear seat belts or any case. is The end-of-chapter problems are organized by the sections in each chapter onds of motion, the car’s acceleration increases with this part of Aristotle’s theory as one special Inrelaxed and the ch other restraining device. The coaster is designed withit is true, ing upward. At posi timetwo-thirds according to the particular, describe situation in which (about of expression the problems are keyed to specific sections ofa the chapter). a vertical derive circularthe section over which the coaster trav- propor child is again mom equation representing Aristotle’s Within each section, the problems now “platform” to higher-order 1.16 0.210 0.240 els on thestudents inside of determine the circle the so that thethink passengers tions, and proportionality constant. rest at the top of the are upside down for asection short time interval. The radius ing where by presenting straightforward in the first, followed combined mass of is in secondsall andthe is in m/s . (a) What is problems the A child’ssection pogo stick (Fig. m, P8.61) of the61.circular is 12.0 and the coaster pogo stick is 25.0 kg change in kinetic energy of the car during interval by the intermediate problems. (Thethe problem numbers for straightforward prob stores energy in a spring with a the boy must lean from 0 to 2.50 s? (b) What is the minimum averenters the bottom of the circular section at a speed of constant of The 2.50 Additional lemsage are printed black; intermediate-level problems are in blue.) 22.0 m/s. force Assume the coaster moves without friction remain balanced, th power outputin of the engine over this time interval? N/m. At position on thekeyed track and the coaster as a particle. pogo stick is vertic Problems section contains problems not tomodel specific sections. At the (c) Why is the value in part (b) describedthat as theare mini 0.100 m), the spring com bend his legs during value?chapter is the Challenge Problems 59. Asection, horizontal springis attached tothe aand wall has adiffi force conend mum of each which most pression agathers maximum the energy of the child stant of child 850 isN/m. A block of mass 1.00 kg 58. Review. Why is the following situation impossible? A new momentarily athave rest. Atproblem cult problems for a given chapter in one place. (Challenge Problems gravitational and el is attached to the spring and rests on a frictionless, high-speed roller coaster is claimed to be so safe that position 0), the spring 0. (b) Determin numbers markeddoinnotred. horizontalis surface Figure the passengers need to wear seat belts or any andinthe child P8.59. is mov (a) The block relaxed as 0. (d) D child at 6.00 cm from is pulled to position There are several kinds problems featured in this text: At position other restraining device. Theof coaster is designed with inga upward. , the equilibrium The problem is identified in the Annotated Instructor’s Edition with a icon. Parts (a)–(c) of the problem ask for quantitative calculations. the kinetic energy o the elastic potential and released. a vertical circular section over which the coaster travchildFind is again momentarily at energy stored culate the child’s m in the spring when the block is 6.00 cm from equilibels on the inside of the circle so that the passengers rest Annotated at the top of theInstructor’s jump. The Quantitative/Conceptual problems (indicated in the Edi rium and when the block passes through equilibrium. 62. A 1.00-kg object sli are upside down for a short time interval. The radius combined mass of child and Figure (b) Find the speed block asconceptually. it passes through theP8.61 to the right on a s tion)of contain parts that isask students both quantitatively and the circular section 12.0 m, and to thethink coaster pogo stickofisthe 25.0 kg. Although equilibrium point. (c) What the speed face having a coe enters the bottom of the circular section at a speed of the boy must lean is forward to of the block An example of a Quantitative/Conceptual problem appears here: /2 3.00 cm? Why when it is remain at a position cient of kinetic frict 22.0 m/s. Assume the coaster moves without friction balanced, the angle is (d) small, so isn’t let’s assume the the answerpogo to part (c)ishalf the answer part (b)? 0.250 (Fig. P8.62 on the track and model the coaster as a particle. stick vertical. Also to assume the boy does not The object has a spe bend his legs during the motion. (a) Calculate the total 59. A horizontal spring attached to a wall has a force con3.00 m/s wh of energy of the child–stick–Earth system, taking both stant of 850 N/m. A block of mass 1.00 kg it makes contact w gravitational and elastic potential energies as zero for is attached to the spring and rests on a frictionless, a light spring ( 0. (b) Determine . (c) Calculate the speed of the horizontal surface as in Figure P8.59. (a) The block P8.62b) that has a fo 0. (d) Determine the value of for which child at 6.00 cm from equilibrium is pulled to a position constant of 50.0 N/ the kinetic energy of the system is a maximum. (e) Cal and released. Find the elastic potential energy stored The object comes culate the child’s maximum upward speed. in the spring when the block is 6.00 cm from equilibrest after the spr rium and when the block passes through equilibrium. 62. A 1.00-kgFigure objectP8.59 slides has been compres (b) Find the speed of the block as it passes through the to the right on a sur a distance ( equilibrium point. (c) What is the speed of the60. block having coeffi More thanface 2 300 years aago, the Greek teacher AristoP8.62c). The objec when it is at a position /2 3.00 cm? (d) Why isn’ttle wrote cient of kinetic frictionPhysics. Put into more the first book called then forced toward the answer to part (c) half the answer to part (b)? precise terminology, 0.250 (Fig.thisP8.62a). passage is from the end of its left by the spring ( Part (d) asks a conceptual The object has a speed Section Eta: P8.62d) and contin 3.00 m/sabout when of question situation. to move in that dir Let be the power of an agentthe causing motion; it makes contact with tion beyond the spr the load moved; , the distance covered; and a light spring (Fig. the object comes to , the time interval required. Then (1) a power P8.62b) that has a force unstretched spring ( equal to will in an interval of time equal to constant of 50.0 N/m. compression , (b) t move /2 a distance 2 or (2) it will move /2 The object comes to tion when the objec the given distance in the time interval /2. rest after the spring and (c) the distance Also, if (3) the given power moves the given has been compressed Figure P8.59 a distance (Fig. 60. More than 2 300 years ago, the Greek teacher AristoP8.62c). The object is tle wrote the first book called Physics. Put into more then forced toward the precise terminology, this passage is from the end of its ing force necessary to (a) shear a steel bolt 1.00 cm in thenumerical beam begins to tip. separation thea line of best fit. Express inwoman’s part (d)position dependwhen on the values given in diameter and (b) from punch 1.00-cm-diameter hole this in ascatter find the the appropriate analysis model for the beam as a 0.500 percentage. (e) In a short paragraph, state what (a) What thisisproblem, or is it true in general? Explain. steel plate cm thick. before it begins to tip? (b) Sketch a force diagram for the graph demonstrates and compare it with the the54. A puck of mass is tied 32. When water freezes, it expands by about 9.00%. What labeling the gravitational and normal forces oretical prediction. You will need to make reference the beam, to string andand allowed pressure increase would occur inside your automobile onathe beam placing the woman a distance to the quantities plotted on the axes, to the shape of acting to revolve in a circle of engine block if the water in it froze? (The bulk moduright of the first pivot, which is the origin. the graph line, to the data points, and to the results of to the radius a friction lus of ice is 2.00 10 N/m (c) Where is the on woman when the normal force is the parts (c) and (d). less,(d)horizontal What is table. when the beam is about to greatest? 33. A 200-kg load is hung on a wire of length 4.00 m, cross50. A basin surrounding a drain has the shape of a circular tip? (e)The end12.1 of the Use other Equation to find the value of when sectional area 0.200 10 , and Young’s modulus Preface cone opening upward, having everywhere an angle of the beam string through is passes about to tip. (f)a Using the result of part N/m . What is its increase in length? 8.00 10 35.0° with the horizontal. A 25.0-g ice cube is set slid (d) and small hole in the cen Equation 12.2, with torques computed around hotel lobby is supported at 34. A walkway ing suspended around theacross coneawithout friction in a horizontal ter of the find table, the second pivot, the and woman’s position when the Symbolic problems (indicated numerous in the Annotated Instructor’s Edition) ask students points along its edges by a vertical cable above is the answer to part (e) by circle of radius . (a) Find the speed the ice cube must beam isanabout object to of tip.mass (g) Check each point and vertical column steel P6.54). tied torques to it (Fig. havemanipulation. as a afunction of .Reviewers (b)underneath. Is any piece of data unnec Edi to solve a problem using only symbolic ofThe the Eighth computing around the first pivot Figure point. P6.54 cable isessary 1.27 cm in and is 5.75 m islong before for to thediameter solution? Suppose made two times for The suspended object tion (as well as the majority of respondents a large survey) asked specifically loading. The (c) aluminum a hollow cylinder remains in equilibrium while the puck on the tabletop larger. Will the column requiredisspeed increase, decrease, an increase in the number of symbolic problems in the text better revolves. Find symbolic expressions for (a) the tension in with anorinside diameter found of cm, anby outside diameter stay constant? If 16.14 it changes, what because factor? (d) it Will of their 16.24 cm, and an unloaded 3.25 m. When the time required forlength eachof solving revolution increase,prob the string, (b) the radial force acting on the puck, and reflects the way instructors want students to think when physics the walkway exerts astay load force of If 8 500 N on one thefactor? (c) the speed of the puck. (d) Qualitatively describe what decrease, or constant? it changes, by of what lems. An example of a Symbolicsupport problem appears here: points, how much to does the(c) point down? will happen in the motion of the puck if the value of (e) Do the answers parts andmove (d) seem contradic is increased by placing a small additional load on the tory? 35. Review. A Explain. 2.00-m-long cylindrical puck. (e) Qualitatively describe what will happen in the steel with ais cross-sectional 51.wire A truck moving with diam The problem is identified motion of the puck if the value of is instead decreased eter ofconstant 4.00 mm isacceleration placed over a light, in the Annotated by removing a part of the hanging load. frictionless Anmakes object of mass up apulley. hill that Instructor’s Edition with a 5.00 is hung with from the one end of 55. Because theFigure EarthP12.38 rotates about its axis, a point on ankgangle icon. and an object of mass the wire the equator experiences a centripetal acceleration of horizontal as in Figure 39. In exercise physiology studies, it only is sometimes impor from A thesmall othersphere end as shown 3.00 kgP6.51. The,figure shows 0.0337 m/s whereas a point at the poles experiences tant to determine the location of a person’s center in Figure P12.35. The objects are acceleration. no centripetal of mass is suspended symbolic quantities.If a person at the equator of mass. This determination can be done with the released andthe allowed has a mass of 75.0 kg, calculate (a) the gravitational from ceilingtoofmove the freely. arrangement shown in Figure P12.39. A light plank Compared length the Figure P12.35 force (true weight) on the person and (b) the normal truckwith by a its light cord.before If No numbers appear in Figure P6.51 rests on two scales, which read 380 N and objectsthe were attached, by how force (apparent weight) on the person. (c) Which force pendulum makes a much the problem statement. 320 N. A distance of 1.65 m separates the scales. How has theconstant wire stretched whilethe theperpendicular objects are in to motion? is greater? Assume the Earth is a uniform sphere and angle with the ceiling, far The fromanswer the woman’s is her center of mass? tom/s thefeet problem take 9.800 what is 36. Review. A 30.0-kg hammer, moving with speed 20.0 m/s, is purely symbolic. AMT strikes a steel spike in diameter. Thea hammer Galileo thought about whether acceleration should be 52. 51. The pilot an cm airplane executes loop-the-loop (cos tanof2.30 sin rebounds with speed 10.0 m/scircle. after The 0.110speed s. What is the defined as the rate of change of velocity over time or as maneuver in a vertical of the airplane averageis strain in the during impact? the rate of change in velocity over distance. He chose 300 mi/h at spike the top of thethe loop and 450 mi/h at the the former, so let’s use the name “vroomosity” for the bottom, and the radius of the circle is 1 200 ft. (a) What Guided Problems help students problems into steps. physics problem Additionalbreak Problems rate of change of velocity over distance. For motion of is the pilot’s apparent weight at theA lowest point if his typically asks for one physical a given context. Often, however, several a particle on a straight line with constant acceleration, trueofin weight is50.0 160 m lb?and (b) mass What is his10 apparent kg isweight 37. quantity A bridge length 8.00 gives its velocity as a function at the point? What If? as Describe how the that the equation onhighest a smooth pier (c) atare each end shown in supported concepts must be used and a number of calculations required to obtain of time. Similarly, for a particle’s linear motion with pilot could experience weightlessness if both the Figure P12.37. A truck of mass 3.00 10 kg is located final answer. Many students are notradius accustomed tocan this level of complexity Figure P12.39 , the equation gives andend. the What speed be forces varied. apparent and constant vroomosity are the onNote: the His bridge 15.0 m from one weight of isProblem equal to the magnitude of the force exerted into the velocity as a function of the position if the parti often don’t know where to start.atAthe Guided breaks a standard problem points support? 40. The lintel of prestressed reinforced concrete in Figcle’s speed is at 0. (a) Find the law describing the by the seat his concepts body. smaller steps, enabling students to grasp allon the and strategies required ure P12.40 is 1.50 m long. The concrete encloses total force acting on this object of mass . (b) Describe 53. Review. While learning to drive, you are in a 1 200-kg one steel reinforcing rod with cross-sectional area to arrive at a correct solution. Unlike standard physics problems, guidance is often an example such a motion or explain why it is unre car moving at 20.0 m/s across a large, vacant, level 1.50 cm . The rodofjoins two strong end plates. The alistic. Consider (c) the possibility of positive and built into the problem statement. Guided of heading how a crossstu sectional parking Problems lot. Suddenlyare youreminiscent realize you are area of the concrete perpendicular to (d) the possibility of toward the brick sidewall of a large (there supermardent might interact with a professor straight in an office visit. These problems is the onerod is 50.0 cm . Young’snegative. modulus for the concrete 57. Figure P6.57. After shows ket and are in danger of running into it. The pavement is 30.0 10 N/m the concrete cures and the in every chapter of the text) help train students to break down complex problems can exert a maximum horizontal force of 7 000 N on original AMT a photo tensionof ainswing the rod is released, the con into a series of simpler problems, anthe essential problem-solving of isride amusement car. (a) Explain why you shouldskill. expectAn theexample force to crete to at bean under comprespark.8.00 The 10 structure have a well-defined maximum value. (b) Suppose you sive stress a Guided Problem appears here: N/m . of a horizon apply the brakes do not turn the steering wheel. (a) Byconsists what distance will the Figureand P12.37 tal, rotating, Find the minimum distance you must be from the wall rod compress thecircular concrete Problems 383 of diameter to avoid a collision. If pivots you dohas nota brake 38. A uniform beam resting on(c) two lengthbut instead when platform the original tension in which(b) seats andpivot turnunder the steering 6.00 mmaintain and massconstant 90.0speed kg. The the left wheel, the rod isfrom released? What Figure P12.40 of mass are sus whataisnormal the minimum distance you and mustthe besec from the end exerts force on the beam, Evaluate Young’s modulus for the material whose pended at the end walllocated to avoid a collision? (d) twothe methods in ond pivot a distance 4.00Ofmthe from left stress–strain curve is shown in Figure 12.12. of massless chains partsa(b) and (c), which better for avoiding a colliend exerts normal force . Ais woman of mass Assume if the shear stress in steel exceeds about 4.00 The goal of the problem of length . When should youend use of both brakes the steerstepsOr onto the left thethe beam andand begins 55.0 kgsion? N/m , the steel ruptures. Determine the shearis identified. Figure P6.57 the system rotates at ing wheel, or neither? Explain. (e) Does the conclusion walking to the right as in Figure P12.38. The goal is to a steel 1.00 cm in ing force necessary to (a) Theshear problem is bolt identified find the woman’s position when the beam begins to tip. diameter and (b) punch a 1.00-cm-diameter hole in a (a) What is the appropriate analysis model for the beam with a icon. steel plate 0.500 cm thick. Analysis begins by identifying before it begins to tip? (b) Sketch a force diagram for When water freezes, it expands by about 9.00%. What the appropriate analysis model. the beam, labeling the gravitational and normal forces pressure increase would occur inside your automobile acting on the beam and placing the woman a distance engine block if the water in it froze? (The bulk moduto the right of the first pivot, which is the origin. lus of ice is 2.00 10 N/m (c) Where is the woman when the normal force is the Students are provided when the beam is about to greatest? (d) What is A 200-kg load is hung on a wire of length 4.00 m, crosswith suggestions for steps tip? (e) Use Equation 12.1 to find the value of when sectional area 0.200 10 , and Young’s modulus the beam is about to tip. (f) Using the result of part N/m . What is its increase in length? 8.00 10 to solve the problem. (d) and Equation 12.2, with torques computed around A walkway suspended across a hotel lobby is supported at the second pivot, find the woman’s position when the numerous points along its edges by a vertical cable above The calculation beam is about to tip. (g) Check the answer to part (e) by each point and a vertical column underneath. The steel associated with the computing torques around the first pivot point. cable is 1.27 cm in diameter and is 5.75 m long before goal is requested. loading. The aluminum column is a hollow cylinder 33. 34. with an inside diameter of 16.14 cm, an outside diameter of 16.24 cm, and an unloaded length of 3.25 m. When the walkway exerts a load force of 8 500 N on one of the support points, how much does the point move down? 35. Review. A 2.00-m-long cylindrical steel wire with a cross-sectional diam eter of 4.00 mm is placed over a light, frictionless pulley. An object of mass 5.00 kg is hung from one end of the wire and an object of mass 3.00 kg from the other end as shown in Figure P12.35. The objects are released and allowed to move freely. Compared with its length before the Figure P12.35 objects were attached, by how much has the wire stretched while the objects are in motion? 36. Review. A 30.0-kg hammer, moving with speed 20.0 m/s, AMT strikes a steel spike 2.30 cm in diameter. The hammer Figure P12.38 39. In exercise physiology studies, it is sometimes impor tant to determine the location of a person’s center of mass. This determination can be done with the arrangement shown in Figure P12.39. A light plank rests on two scales, which read 380 N and 320 N. A distance of 1.65 m separates the scales. How far from the woman’s feet is her center of mass? Stuart Gregory/Getty Images sw .a w 32. w 31. w 30. ar ph ys ic s. w ee b ly .c om xix Problems it enters a parabolic path with a velocity of 143 m/s nose high at 45.0° and exits with velocity 143 m/s at 45.0° nose low. During this portion of the flight, the aircraft and objects inside its padded cabin are in free fall; astronauts and equipment floatPreface freely as if there were no gravity. What are the aircraft’s (a) speed and (b) altitude at the top of the maneuver? (c) What is the time interval spent in microgravity? leave simultaneously and drive for 2.50 h in the directions shown. Car 1 has a speed of 90.0 km/h. If the cars arrive simultaneously at the lake, what is the speed of car 2? Impossibility problems. Physics education research has focused heavily on the problem-solving skills of students. Although most problems in this text are struc tured in the form of providing data and asking for a result of computation, two problems in each chapter, on average, are structured as impossibility problems. They begin with the phrase Why is the following situation impossible? That is followed by the description of a situation. The striking aspect of these problems is that no question is asked of the students, other than that in the initial italics. The student must determine what questions need to be asked and what calculations need to be performed. Based on the results of these calculations, the student must determine why the situation described is not possible. This determination may require infor Figureexperience, P4.64 mation from personal common sense, Internet or print research, mea surement, mathematical skills, knowledge of human norms, or scientific thinking. 65. A catapult launches a rocket at an angle of 53.0° above the horizontal with an of 100to m/s. The critical thinking skills in students. These problems caninitial be speed assigned build rocket engine immediately starts a burn, and for 3.00 s They are also fun, having the aspect of physics “mysteries” to be solved by students the rocket moves along its initial line of motion with individually or inof groups. example anand impossibility problem appears here: an acceleration 30.0 m/s .An Then its engineof fails, ly ee b s. w 68. As some molten metal splashes, one droplet flies off to the east with initial velocity at angle above the hor izontal, and another droplet flies off to the west with the same speed at the same angle above the horizontal as shown in Figure P4.68. In terms of and , find the distance between the two droplets as a function of time. Paired problems. These problems are otherwise A situation is described. No question is asked. The student must determine what needs to be calculated and why the situation is impossible. identical, one asking for a numeri cal solution and one asking for a symbolic derivation. There are now three pairs of these problems in most chapters, indicated in the Annotated Instructor’s Edition by cyan shading in the end-of-chapter problems set. sw 63. A flea is at point on a horizontal turntable, 10.0 cm from the center. The turntable is rotating at 33.3 rev/min in the clockwise direction. The flea jumps straight up to a height of 5.00 cm. At takeoff, it gives itself no horizontal velocity relative to the turntable. The flea lands on the turntable at point . Choose the origin of coor dinates to be at the center of the turntable and the posi tive axis passing through at the moment of takeoff. Then the original position of the flea is 10.0 cm. when the flea lands. (a) Find the position of point (b) Find the position of point when the flea lands. 67. Why is the following situation impossible? Albert Pujols hits a home run so that the baseball just clears the top row of bleachers, 24.0 m high, located 130 m from home plate. The ball is hit at 41.7 m/s at an angle of 35.0° to the horizontal, and air resistance is negligible. ic 62. A boy throws a stone horizontally from the top of a cliff of height toward the ocean below. The stone strikes the ocean at distance from the base of the cliff. In terms of h, d, and , find expressions for (a) the time at which the stone lands in the ocean, (b) the initial speed of the stone, (c) the speed of the stone immediately before it reaches the ocean, and (d) the direction of the stone’s velocity immediately before it reaches the ocean. 66. A cannon with a muzzle speed of 1 000 m/s is used to an impossibility problem. start an avalanche on a mountain slope. The target is 2 000 m from the cannon horizontally and 800 m above the cannon. At what angle, above the horizontal, should the cannon be fired? ys 61. Lisa in her Lamborghini accelerates at the rate of 3.00 2.00 m/s , while Jill in her Jaguar acceler ates at 1.00 3.00 m/s . They both start from rest at the origin of an coordinate system. After 5.00 s, (a) what is Lisa’s speed with respect to Jill, (b) how far apart are they, and (c) what is Lisa’s acceleration relative to Jill? the rocket proceeds to move in free fall. Find (a) the maximum altitude reached by the rocket, (b) its total timeThe of flight, (c) itsinhorizontal range. initialand phrase italics signals ar ph Figure P4.60 .c om 60. A basketball player is standing on the floor 10.0 m from the basket as in Figure P4.60. The height of the basket is 3.05 m, and he shoots the ball at a 40.0 angle with the horizontal from a height of 2.00 m. (a) What is the acceleration of the basketball at the highest point in its trajectory? (b) At what speed must the player throw the basketball so that the ball goes through the hoop without striking the backboard? 107 .a Biomedical problems.Figure These problems (indicated in the Annotated Instructor’s Edi P4.68 tion with a icon) highlight the relevance of physics principles to those students 69. An astronaut on the surface of the Moon fires a cantaking course are majoring in one of the life sciences. non this to launch an who experiment package, which leaves w w w 64. Towns A and B in Figure P4.64 are 80.0 km apart. A couple arranges to drive from town A and meet a couple driving from town B at the lake, L. The two couples the barrel moving horizontally. Assume the free-fall acceleration on the Moon is one-sixth of that on the Review problems. Many chapters include review problems requiring the student to combine concepts covered in the chapter with those discussed in previous chapters. These problems (marked Review) reflect the cohesive nature of the principles in the text and verify that physics is not a scattered set of ideas. When facing a realworld issue such as global warming or nuclear weapons, it may be necessary to call on ideas in physics from several parts of a textbook such as this one. “Fermi problems.” One or more problems in most chapters ask the student to reason in order-of-magnitude terms. Design problems. Several chapters contain problems that ask the student to deter mine design parameters for a practical device so that it can function as required. Calculus-based problems. Every chapter contains at least one problem applying ideas and methods from differential calculus and one problem using integral calculus. Preface .c om Integration with Enhanced WebAssign. The textbook’s tight integration with Enhanced WebAssign content facilitates an online learning environment that helps students improve their problem-solving skills and gives them a variety of tools to meet their individual learning styles. Extensive user data gathered by WebAssign were used to ensure that the problems most often assigned were retained for this new edition. In each chapter’s problems set, the top quartile of problems assigned in Enhanced WebAssign have cyan-shaded problem numbers in the Annotated Instructor’s Edi tion for easy identification, allowing professors to quickly and easily find the most popular problems assigned in Enhanced WebAssign. New Analysis Model tutorials added for this edition have already been discussed (see page x). Master It tutorials help students solve problems by having them work through a stepped-out solution. Problems with Master It tutorials are indicated in each chapter’s problem set with a icon. In addition, Watch It solution videos are indicated in each chapter’s prob lem set with a icon and explain fundamental problem-solving strategies to help students step through the problem. s. w ee b ly Artwork. Every piece of artwork in the Ninth Edition is in a modern style that helps express the physics principles at work in a clear and precise fashion. Focus pointers are included with many figures in the text; these either point out important aspects of a figure or guide students through a process illustrated by the artwork or photo. This format helps those students who are more visual learners. An example of a figure with a focus pointer appears below. at ic sw ar ph Direction of . ys line tangent to the curve at Figure 4.2 As a particle moves between two points, its average velocity is in the direction of the displacement vector . By definition, the instantaneous velocity at is directed along the line tangent to the curve at     corresponding time intervals become smaller and smaller. w w w .a  Math Appendix. The math appendix (Appendix B), a valuable tool for students, shows the math tools in a physics context. This resource is ideal for students who need a quick review on topics such as algebra, trigonometry, and calculus. Helpful Features Style. To facilitate rapid comprehension, we have written the book in a clear, logi cal, and engaging style. We have chosen a writing style that is somewhat informal and relaxed so that students will find the text appealing and enjoyable to read. New terms are carefully defined, and we have avoided the use of jargon. xxi Preface Important Definitions and Equations. Most important definitions are set in boldface or are highlighted with a background screen for added emphasis and ease of review. Similarly, important equations are also highlighted with a background screen to facilitate location. Marginal Notes. Comments and notes appearing in the margin with a  icon can be used to locate important statements, equations, and concepts in the text. om Pedagogical Use of Color. Readers should consult the pedagogical color chart (inside the front cover) for a listing of the color-coded symbols used in the text diagrams. This system is followed consistently throughout the text. ee b ly .c Mathematical Level. We have introduced calculus gradually, keeping in mind that students often take introductory courses in calculus and physics concurrently. Most steps are shown when basic equations are developed, and reference is often made to mathematical appendices near the end of the textbook. Although vectors are discussed in detail in Chapter 3, vector products are introduced later in the text, where they are needed in physical applications. The dot product is introduced in Chapter 7, which addresses energy of a system; the cross product is introduced in Chapter 11, which deals with angular momentum. ys ic s. w Significant Figures. In both worked examples and end-of-chapter problems, significant figures have been handled with care. Most numerical examples are worked to either two or three significant figures, depending on the precision of the data provided. End-of-chapter problems regularly state data and answers to three-digit precision. When carrying out estimation calculations, we shall typically work with a single significant figure. (More discussion of significant figures can be found in Chapter 1, pages 11–13.) ar ph Units. The international system of units (SI) is used throughout the text. The U.S. customary system of units is used only to a limited extent in the chapters on mechanics and thermodynamics. sw Appendices and Endpapers. Several appendices are provided near the end of the textbook. Most of the appendix material represents a review of mathematical concepts and techniques used in the text, including scientific notation, algebra, geometry, trigonometry, differential calculus, and integral calculus. Reference to these appendices is made throughout the text. Most mathematical review sections in the appendices include worked examples and exercises with answers. In addition to the mathematical reviews, the appendices contain tables of physical data, conversion factors, and the SI units of physical quantities as well as a periodic table of the elements. Other useful information—fundamental constants and physical data, planetary data, a list of standard prefixes, mathematical symbols, the Greek alphabet, and standard abbreviations of units of measure—appears on the endpapers. .a w w w xxii  engageCompose Options for Physics for C Scientists and Engineers Would you like to easily create your own personalized text, selecting the elements that meet your specific learning objectives? CengageCompose puts the power of the vast Cengage Learning library of learning content at your fingertips to create exactly the text you need. The all-new, Webbased CengageCompose site lets you quickly scan content and review materials to pick what you need for your text. Site tools let you easily assemble the modular learning units into the order you want and immediately provide you with an online copy for review. Add enrichment content like case studies, exercises, and lab materials to Preface further build your ideal learning materials. Even choose from hundreds of vivid, artrich, customizable, full-color covers. Cengage Learning offers the fastest and easiest way to create unique customized learning materials delivered the way you want. For more information about custom publishing options, visit www.cengage.com/custom or contact your local Cengage Learning representative. Course Solutions That Fit Your Teaching Goals and Your Students’ Learning Needs ee b ly .c om Recent advances in educational technology have made homework management sys tems and audience response systems powerful and affordable tools to enhance the way you teach your course. Whether you offer a more traditional text-based course, are interested in using or are currently using an online homework management sys tem such as Enhanced WebAssign, or are ready to turn your lecture into an interac tive learning environment with JoinIn, you can be confident that the text’s proven content provides the foundation for each and every component of our technology and ancillary package. Homework Management Systems ys ic s. w Enhanced WebAssign for Physics for Scientists and Engineers, Ninth Edition. Exclu sively from Cengage Learning, Enhanced WebAssign offers an extensive online program for physics to encourage the practice that’s so critical for concept mastery. The meticulously crafted pedagogy and exercises in our proven texts become even more effective in Enhanced WebAssign. Enhanced WebAssign includes the Cen gage YouBook, a highly customizable, interactive eBook. WebAssign includes: .a sw ar ph All of the quantitative end-of-chapter problems Selected problems enhanced with targeted feedback. An example of targeted feedback appears below: w w w Selected problems include feedback to address common mistakes that students make. This feedback was developed by professors with years of classroom experience. Master It tutorials (indicated in the text by a icon), to help students work through the problem one step at a time. An example of a Master It tutorial appears on page xxiv: xxiii Preface om Master It tutorials help students organize what they need to solve a problem with Conceptualize and Categorize sections before they work through each step. ly .c Master It tutorials help students work through each step of the problem. Watch It solution videos help students visualize the steps needed to solve a problem. w w .a sw ar ph ys ic s. w ee b Watch It solution videos (indicated in the text by a icon) that explain fundamental problem-solving strategies, to help students step through the problem. In addition, instructors can choose to include video hints of problemsolving strategies. A screen shot from a Watch It solution video appears below: w xxiv Concept Checks PhET simulations Most worked examples, enhanced with hints and feedback, to help strengthen students’ problem-solving skills Every Quick Quiz, giving your students ample opportunity to test their conceptual understanding PreLecture Explorations. The Active Figure questions in WebAssign have been completely revised. The simulations have been updated, with additional parameters to enhance investigation of a physical phenomenon. Students can make predictions, change the parameters, and then observe the results. Each new PreLecture Exploration comes with conceptual and analytical questions, which guide students to a deeper understanding and help promote a robust physical intuition. Analysis Model tutorials. John Jewett developed 165 tutorials (indicated in each chapter’s problem set with an AMT icon) that strengthen students’ problemsolving skills by guiding them through the steps in the problem-solving process.    Preface xxv w w w .a sw ar ph ys ic s. w ee b ly .c om Important first steps include making predictions and focusing strategy on physics concepts before starting to solve the problem quantitatively. A critical component of these tutorials is the selection of an appropriate Analysis Model to describe what is going on in the problem. This step allows students to make the important link between the situation in the problem and the mathematical representation of the situation. Analysis Model tutorials include meaningful feedback at each step to help students practice the problem-solving process and improve their skills. In addition, the feedback addresses student misconceptions and helps them to catch algebraic and other mathematical errors. Solutions are carried out symbolically as long as possible, with numerical values substituted at the end. This feature helps students to understand the effects of changing the values of each variable in the problem, avoids unnecessary repetitive substitution of the same numbers, and eliminates round-off errors. Feedback at the end of the tutorial encourages students to think about how the final answer compares to their original predictions. • Personalized Study Plan. The Personal Study Plan in Enhanced WebAssign provides chapter and section assessments that show students what material they know and what areas require more work. For items that they answer incorrectly, students can click on links to related study resources such as videos, tutorials, or reading materials. Color-coded progress indicators let them see how well they are doing on different topics. You decide what chapters and sections to include—and whether to include the plan as part of the final grade or as a study guide with no scoring involved. • The Cengage YouBook. WebAssign has a customizable and interactive eBook, the Cengage YouBook, that lets you tailor the textbook to fit your course and connect with your students. You can remove and rearrange chapters in the table of contents and tailor assigned readings that match your syllabus exactly. Powerful editing tools let you change as much as you’d like—or leave it just like it is. You can highlight key passages or add sticky notes to pages to comment on a concept in the reading, and then share any of these individual notes and highlights with your students, or keep them personal. You can also edit narrative content in the textbook by adding a text box or striking out text. With a handy link tool, you can drop in an icon at any point in the eBook that lets you link to your own lecture notes, audio summaries, video lectures, or other files on a personal Web site or anywhere on the Web. A simple YouTube widget lets you easily find and embed videos from YouTube directly into eBook pages. The Cengage YouBook helps students go beyond just reading the textbook. Students can also highlight the text, add their own notes, and bookmark the text. Animations play right on the page at the point of learning so that they’re not speed bumps to reading but true enhancements. Please visit www.webassign.net/brookscole to view an interactive demonstration of Enhanced WebAssign. • Offered exclusively in WebAssign, Quick Prep for physics is algebra and trigonometry math remediation within the context of physics applications and principles. Quick Prep helps students succeed by using narratives illustrated throughout with video examples. The Master It tutorial problems allow students to assess and retune their understanding of the material. The Practice Problems that go along with each tutorial allow both the student and the instructor to test the student’s understanding of the material. Quick Prep includes the following features: • • • • 67 interactive tutorials 67 additional practice problems A thorough overview of each topic, including video examples Can be taken before the semester begins or during the first few weeks of the course • Can also be assigned alongside each chapter for “ just in time” remediation Preface Topics include units, scientific notation, and significant figures; the motion of objects along a line; functions; approximation and graphing; probability and error; vectors, displacement, and velocity; spheres; force and vector projections. MindTap™: The Personal Learning Experience ly .c om MindTap for Serway and Jewett Physics for Scientists and Engineers is a personalized, fully online digital learning platform of authoritative textbook content, assignments, and services that engages your students with interactivity while also offering you choice in the configuration of coursework and enhancement of the curriculum via complimentary Web-apps known as MindApps. MindApps range from ReadSpeaker (which reads the text out loud to students), to Kaltura (allowing you to insert inline video and audio into your curriculum), to ConnectYard (allowing you to create digital “yards” through social media—all without “friending” your students). MindTap is well beyond an eBook, a homework solution or digital supplement, a resource center Web site, a course delivery platform, or a Learning Management System. It is the first in a new category—the Personal Learning Experience. ee b CengageBrain.com s. w On CengageBrain.com students will be able to save up to 60% on their course materials through our full spectrum of options. Students will have the option to rent their textbooks, purchase print textbooks, e-textbooks, or individual e-chapters and audio books all for substantial savings over average retail prices. CengageBrain.com also includes access to Cengage Learning’s broad range of homework and study tools and features a selection of free content. ic Lecture Presentation Resources sw ar ph ys PowerLecture with ExamView® and JoinIn for Physics for Scientists and Engineers, Ninth ­Edition. Bringing physics principles and concepts to life in your lectures has never been easier! The full-featured, two-volume PowerLecture Instructor’s Resource DVD-ROM (Volume 1: Chapters 1–22; Volume 2: Chapters 23–46) provides everything you need for Physics for Scientists and Engineers, Ninth Edition. Key content includes the Instructor’s Solutions Manual, art and images from the text, premade chapter-specific PowerPoint lectures, ExamView test generator software with pre-loaded test questions, JoinIn response-system “clickers,” Active Figures animations, and a physics movie library. w w .a JoinIn. Assessing to Learn in the Classroom questions developed at the University of w xxvi ­ assachusetts Amherst. This collection of 250 advanced conceptual questions has M been tested in the classroom for more than ten years and takes peer learning to a new level. JoinIn helps you turn your lectures into an interactive learning environment that ­promotes conceptual understanding. Available exclusively for higher education from our partnership with Turning Technologies, JoinIn™ is the easiest way to turn your lecture hall into a personal, fully interactive experience for your students! Assessment and Course Preparation Resources A number of resources listed below will assist with your assessment and preparation processes. Instructor’s Solutions Manual by Vahé Peroomian (University of California at Los ­ ngeles). Thoroughly revised for this edition, the Instructor’s Solutions Manual contains A ­complete worked solutions to all end-of-chapter problems in the textbook as well as answers to the even-­numbered problems and all the questions. The solutions to problems new to the Ninth Edition are marked for easy identification. Volume 1 contains Chapters 1 through 22; Volume 2 contains Chapters 23 through 46. Electronic files of the Instructor’s Solutions Manual are available on the PowerLecture™ DVD-ROM.    Preface xxvii Test Bank by Ed Oberhofer (University of North Carolina at Charlotte and Lake Sumter Community College). The test bank is available on the two-volume PowerLecture™ DVD-ROM via the ExamView® test software. This two-volume test bank contains approximately 2 000 multiple-choice questions. Instructors may print and duplicate pages for distribution to students. Volume 1 contains Chapters 1 through 22, and Volume 2 contains Chapters 23 through 46. WebCT and Blackboard versions of the test bank are available on the instructor’s companion site at www.CengageBrain.com. Instructor’s Companion Web Site. Consult the instructor’s site by pointing your browser om to www.CengageBrain.com for a problem correlation guide, ­PowerPoint lectures, and JoinIn audience response content. Instructors adopting the Ninth Edition of Physics for Scientists and Engineers may download these materials after securing the appropriate password from their local sales representative. .c Supporting Materials for the Instructor s. w • request a desk copy • locate your local representative • download electronic files of select support materials ee b ly Supporting instructor materials are available to qualified adopters. Please consult your local Cengage Learning, Brooks/Cole representative for details. Visit www.CengageBrain.com to Student Resources ys ic Visit the Physics for Scientists and Engineers Web site at www.cengagebrain.com/ shop/ISBN/9781133954156 to see samples of select student supplements. Go to CengageBrain.com to purchase and access this product at Cengage Learning’s preferred online store. sw ar ph Student Solutions Manual/Study Guide by John R. Gordon, Vahé Peroomian, ­R aymond A. Serway, and John W. Jewett, Jr. This two-volume manual features detailed solutions to 20% of the end-of-chapter problems from the text. The manual also features a list of important equations, concepts, and notes from key sections of the text in addition to answers to selected end-of-chapter questions. Volume 1 contains Chapters 1 through 22; and Volume 2 contains Chapters 23 through 46. Physics Laboratory Manual, Third Edition by David Loyd (Angelo State University) w w w .a supplements the learning of basic physical principles while introducing laboratory procedures and equipment. Each chapter includes a prelaboratory assignment, objectives, an equipment list, the theory behind the experiment, experimental procedures, graphing exercises, and questions. A laboratory report form is included with each experiment so that the student can record data, calculations, and experimental results. Students are encouraged to apply statistical analysis to their data. A complete Instructor’s Manual is also available to facilitate use of this lab manual. Physics Laboratory Experiments, Seventh Edition by Jerry D. Wilson (Lander College) and Cecilia A. Hernández (American River College). This market-leading manual for the first-year physics laboratory course offers a wide range of class-tested experiments designed specifically for use in small to midsize lab programs. A series of integrated experiments emphasizes the use of computerized instrumentation and includes a set of “computer-assisted experiments” to allow students and instructors to gain experience with modern equipment. This option also enables instructors to determine the appropriate balance between traditional and computer-based experiments for their courses. By analyzing data through two different methods, students gain a greater understanding of the concepts behind the experiments. The Seventh Edition is updated with the latest information and techniques involving state-of-the-art equipment and a new Guided Learning feature addresses Preface the growing interest in guided-inquiry pedagogy. Fourteen additional experiments are also available through custom printing. Teaching Options 31.6 Eddy Currents 33.9 Rectifiers and Filters 34.6 Production of Electromagnetic Waves by an Antenna 36.5 Lens Aberrations 36.6 The Camera 36.7 The Eye 36.8 The Simple Magnifier 36.9 The Compound Microscope 36.10 The Telescope 38.5 Diffraction of X-Rays by Crystals 39.9 The General Theory of Relativity 41.6 Applications of Tunneling 42.9 Spontaneous and Stimulated Transitions 42.10 Lasers 43.7 Semiconductor Devices 43.8 Superconductivity 44.8 Nuclear Magnetic Resonance and Magnetic Resonance Imaging 45.5 Radiation Damage 45.6 Uses of Radiation w w .a sw ar ph ys ic s. w 2.8 Kinematic Equations Derived from Calculus 4.6 Relative Velocity and Relative Acceleration 6.3 Motion in Accelerated Frames 6.4 Motion in the Presence of Resistive Forces 7.9 Energy Diagrams and Equilibrium of a System 9.9 Rocket Propulsion 11.5 The Motion of Gyroscopes and Tops 14.7 Other Applications of Fluid Dynamics 15.6 Damped Oscillations 15.7 Forced Oscillations 18.6 Standing Waves in Rods and Membranes 18.8 Nonsinusoidal Wave Patterns 25.7 The Millikan Oil-Drop Experiment 25.8 Applications of Electrostatics 26.7 An Atomic Description of Dielectrics 27.5 Superconductors 28.5 Household Wiring and Electrical Safety 29.3 Applications Involving Charged Particles Moving in a Magnetic Field 29.6 The Hall Effect 30.6 Magnetism in Matter ee b ly .c om The topics in this textbook are presented in the following sequence: classical mechanics, oscillations and mechanical waves, and heat and thermodynamics, followed by electricity and magnetism, electromagnetic waves, optics, relativity, and modern physics. This presentation represents a traditional sequence, with the subject of mechanical waves being presented before electricity and magnetism. Some instructors may prefer to discuss both mechanical and electromagnetic waves together after completing electricity and magnetism. In this case, Chapters 16 through 18 could be covered along with Chapter 34. The chapter on relativity is placed near the end of the text because this topic often is treated as an introduction to the era of “modern physics.” If time permits, instructors may choose to cover Chapter 39 after completing Chapter 13 as a conclusion to the material on Newtonian mechanics. For those instructors teaching a two-semester sequence, some sections and chapters could be deleted without any loss of continuity. The following sections can be considered optional for this purpose: w xxviii Acknowledgments This Ninth Edition of Physics for Scientists and Engineers was prepared with the guidance and assistance of many professors who reviewed selections of the manuscript, the prerevision text, or both. We wish to acknowledge the following scholars and express our sincere appreciation for their suggestions, criticisms, and encouragement: Benjamin C. Bromley, University of Utah; Elena Flitsiyan, University of Central Florida; Yuankun Lin, University of North Texas; Allen Mincer, New York University; Yibin Pan, University of Wisconsin–Madison; N. M. Ravindra, New Jersey Institute of Technology; Masao Sako, University of Pennsylvania; Charles Stone, Colorado School of Mines; Robert Weidman, Michigan Technological University; Michael Winokur, University of Wisconsin–Madison Prior to our work on this revision, we conducted a survey of professors; their feedback and suggestions helped shape this revision, and so we would like to thank the survey participants:    Preface xxix ys ic s. w ee b ly .c om Elise Adamson, Wayland Baptist University; Saul Adelman, The Citadel; Yiyan Bai, Houston Community College; Philip Blanco, Grossmont College; Ken Bolland, Ohio State University; Michael Butros, Victor Valley College; Brian Carter, Grossmont College; Jennifer Cash, South Carolina State University; Soumitra Chattopadhyay, Georgia Highlands College; John Cooper, Brazosport College; Gregory Dolise, Harrisburg Area Community College; Mike Durren, Lake Michigan College; Tim Farris, Volunteer State Community College; Mirela Fetea, University of Richmond; Susan Foreman, Danville Area Community College; Richard Gottfried, Frederick Community College; Christopher Gould, University of Southern California; Benjamin Grinstein, University of California, San Diego; Wayne Guinn, Lon Morris College; Joshua Guttman, Bergen Community College; Carlos Handy, Texas Southern University; David Heskett, University of Rhode Island; Ed Hungerford, University of Houston; Matthew Hyre, Northwestern College; Charles Johnson, South Georgia College; Lynne Lawson, Providence College; Byron Leles, Northeast Alabama Community College; Rizwan Mahmood, Slippery Rock University; Virginia Makepeace, Kankakee Community College; David Marasco, Foothill College; Richard McCorkle, University of Rhode Island; Brian Moudry, Davis & Elkins College; Charles Nickles, University of Massachusetts Dartmouth; Terrence O’Neill, Riverside Community College; Grant O’Rielly, University of Massachusetts Dartmouth; Michael Ottinger, Missouri Western State University; Michael Panunto, Butte College; Eugenia Peterson, Richard J. Daley College; Robert Pompi, Binghamton University, State University of New York; Ralph Popp, Mercer County Community College; Craig Rabatin, West Virginia University at Parkersburg; Marilyn Rands, Lawrence Technological University; Christina Reeves-Shull, Cedar Valley College; John Rollino, Rutgers University, Newark; Rich Schelp, Erskine College; Mark Semon, Bates College; Walther Spjeldvik, Weber State University; Mark Spraker, North Georgia College and State University; Julie Talbot, University of West Georgia; James Tressel, Massasoit Community College; Bruce Unger, Wenatchee Valley College; Joan Vogtman, Potomac State College w w w .a sw ar ph This title was carefully checked for accuracy by Grant Hart, Brigham Young University; James E. Rutledge, University of California at Irvine; and Som Tyagi, Drexel University. We thank them for their diligent efforts under schedule pressure. Belal Abas, Zinoviy Akkerman, Eric Boyd, Hal Falk, Melanie Martin, Steve McCauley, and Glenn Stracher made corrections to problems taken from previous editions. Harvey Leff provided invaluable guidance on the restructuring of the discussion of entropy in Chapter 22. We are grateful to authors John R. Gordon and Vahé Peroomian for preparing the Student Solutions Manual/Study Guide and to Vahé Peroomian for preparing an excellent Instructor’s Solutions Manual. Susan English carefully edited and improved the test bank. Linnea Cookson provided an excellent accuracy check of the Analysis Model tutorials. Special thanks and recognition go to the professional staff at the Brooks/Cole Publishing Company—in particular, Charles Hartford, Ed Dodd, Stephanie VanCamp, Rebecca Berardy Schwartz, Tom Ziolkowski, Alison Eigel Zade, Cate Barr, and Brendan Killion (who managed the ancillary program)—for their fine work during the development, production, and promotion of this textbook. We recognize the skilled production service and excellent artwork provided by the staff at Lachina Publishing Services and the dedicated photo research efforts of Christopher Arena at the Bill Smith Group. Finally, we are deeply indebted to our wives, children, and grandchildren for their love, support, and long-term sacrifices. Raymond A. Serway St. Petersburg, Florida John W. Jewett, Jr. Anaheim, California To the Student om It is appropriate to offer some words of advice that should be of benefit to you, the student. Before doing so, we assume you have read the Preface, which describes the various features of the text and support materials that will help you through the course. How to Study s. w ee b ly .c Instructors are often asked, “How should I study physics and prepare for examinations?” There is no simple answer to this question, but we can offer some suggestions based on our own experiences in learning and teaching over the years. First and foremost, maintain a positive attitude toward the subject matter, keeping in mind that physics is the most fundamental of all natural sciences. Other science courses that follow will use the same physical principles, so it is important that you understand and are able to apply the various concepts and theories discussed in the text. ic Concepts and Principles w w w .a sw ar ph ys It is essential that you understand the basic concepts and principles before attempting to solve assigned problems. You can best accomplish this goal by carefully reading the textbook before you attend your lecture on the covered material. When reading the text, you should jot down those points that are not clear to you. Also be sure to make a diligent attempt at answering the questions in the Quick Quizzes as you come to them in your reading. We have worked hard to prepare questions that help you judge for yourself how well you understand the material. Study the What If? features that appear in many of the worked examples carefully. They will help you extend your understanding beyond the simple act of arriving at a numerical result. The Pitfall Preventions will also help guide you away from common misunderstandings about physics. During class, take careful notes and ask questions about those ideas that are unclear to you. Keep in mind that few people are able to absorb the full meaning of scientific material after only one reading; several readings of the text and your notes may be necessary. Your lectures and laboratory work supplement the textbook and should clarify some of the more difficult material. You should minimize your memorization of material. Successful memorization of passages from the text, equations, and derivations does not necessarily indicate that you understand the material. Your understanding of the material will be enhanced through a combination of efficient study habits, discussions with other students and with instructors, and your ability to solve the problems presented in the textbook. Ask questions whenever you believe that clarification of a concept is necessary. Study Schedule It is important that you set up a regular study schedule, preferably a daily one. Make sure that you read the syllabus for the course and adhere to the schedule set by your instructor. The lectures will make much more sense if you read the corresponding text material before attending them. As a general rule, you should devote about two hours of study time for each hour you are in class. If you are having trouble with the xxx    To the Student xxxi om course, seek the advice of the instructor or other students who have taken the course. You may find it necessary to seek further instruction from experienced students. Very often, instructors offer review sessions in addition to regular class periods. Avoid the practice of delaying study until a day or two before an exam. More often than not, this approach has disastrous results. Rather than undertake an all-night study session before a test, briefly review the basic concepts and equations, and then get a good night’s rest. If you believe that you need additional help in understanding the concepts, in preparing for exams, or in problem solving, we suggest that you acquire a copy of the Student Solutions Manual/Study Guide that accompanies this textbook. Visit the Physics for Scientists and Engineers Web site at www.cengagebrain.com/ shop/ISBN/9781133954156 to see samples of select student supplements. You can purchase any Cengage Learning product at your local college store or at our preferred online store CengageBrain.com. .c Use the Features sw ar ph ys ic s. w ee b ly You should make full use of the various features of the text discussed in the Preface. For example, marginal notes are useful for locating and describing important equations and concepts, and boldface indicates important definitions. Many useful tables are contained in the appendices, but most are incorporated in the text where they are most often referenced. Appendix B is a convenient review of mathematical tools used in the text. Answers to Quick Quizzes and odd-numbered problems are given at the end of the textbook, and solutions to selected end-of-chapter questions and problems are provided in the Student Solutions Manual/Study Guide. The table of contents provides an overview of the entire text, and the index enables you to locate specific material quickly. Footnotes are sometimes used to supplement the text or to cite other references on the subject discussed. After reading a chapter, you should be able to define any new quantities introduced in that chapter and discuss the principles and assumptions that were used to arrive at certain key relations. The chapter summaries and the review sections of the Student Solutions Manual/Study Guide should help you in this regard. In some cases, you may find it necessary to refer to the textbook’s index to locate certain topics. You should be able to associate with each physical quantity the correct symbol used to represent that quantity and the unit in which the quantity is specified. Furthermore, you should be able to express each important equation in concise and accurate prose. .a Problem Solving w w w R. P. Feynman, Nobel laureate in physics, once said, “You do not know anything until you have practiced.” In keeping with this statement, we strongly advise you to develop the skills necessary to solve a wide range of problems. Your ability to solve problems will be one of the main tests of your knowledge of physics; therefore, you should try to solve as many problems as possible. It is essential that you understand basic concepts and principles before attempting to solve problems. It is good practice to try to find alternate solutions to the same problem. For example, you can solve problems in mechanics using Newton’s laws, but very often an alternative method that draws on energy considerations is more direct. You should not deceive yourself into thinking that you understand a problem merely because you have seen it solved in class. You must be able to solve the problem and similar problems on your own. The approach to solving problems should be carefully planned. A systematic plan is especially important when a problem involves several concepts. First, read the problem several times until you are confident you understand what is being asked. Look for any key words that will help you interpret the problem and perhaps allow you to make certain assumptions. Your ability to interpret a question properly is To the Student ic s. w ee b ly .c om an integral part of problem solving. Second, you should acquire the habit of writing down the information given in a problem and those quantities that need to be found; for example, you might construct a table listing both the quantities given and the quantities to be found. This procedure is sometimes used in the worked examples of the textbook. Finally, after you have decided on the method you believe is appropriate for a given problem, proceed with your solution. The General ProblemSolving Strategy will guide you through complex problems. If you follow the steps of this procedure (Conceptualize, Categorize, Analyze, Finalize), you will find it easier to come up with a solution and gain more from your efforts. This strategy, located at the end of Chapter 2 (pages 45–47), is used in all worked examples in the remaining chapters so that you can learn how to apply it. Specific problem-solving strategies for certain types of situations are included in the text and appear with a special heading. These specific strategies follow the outline of the General Problem-Solving Strategy. Often, students fail to recognize the limitations of certain equations or physical laws in a particular situation. It is very important that you understand and remember the assumptions that underlie a particular theory or formalism. For example, certain equations in kinematics apply only to a particle moving with constant acceleration. These equations are not valid for describing motion whose acceleration is not constant, such as the motion of an object connected to a spring or the motion of an object through a fluid. Study the Analysis Models for Problem Solving in the chapter summaries carefully so that you know how each model can be applied to a specific situation. The analysis models provide you with a logical structure for solving problems and help you develop your thinking skills to become more like those of a physicist. Use the analysis model approach to save you hours of looking for the correct equation and to make you a faster and more efficient problem solver. ys Experiments sw ar ph Physics is a science based on experimental observations. Therefore, we recommend that you try to supplement the text by performing various types of “hands-on” experiments either at home or in the laboratory. These experiments can be used to test ideas and models discussed in class or in the textbook. For example, the common Slinky toy is excellent for studying traveling waves, a ball swinging on the end of a long string can be used to investigate pendulum motion, various masses attached to the end of a vertical spring or rubber band can be used to determine its elastic nature, an old pair of polarized sunglasses and some discarded lenses and a magnifying glass are the components of various experiments in optics, and an approximate measure of the free-fall acceleration can be determined simply by measuring with a stopwatch the time interval required for a ball to drop from a known height. The list of such experiments is endless. When physical models are not available, be imaginative and try to develop models of your own. .a w w w xxxii New Media If available, we strongly encourage you to use the Enhanced WebAssign product that is available with this textbook. It is far easier to understand physics if you see it in action, and the materials available in Enhanced WebAssign will enable you to become a part of that action. It is our sincere hope that you will find physics an exciting and enjoyable experience and that you will benefit from this experience, regardless of your chosen profession. Welcome to the exciting world of physics! The scientist does not study nature because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful. If nature were not beautiful, it would not be worth knowing, and if nature were not worth knowing, life would not be worth living. —Henri Poincaré Mechanics p a r t 1 s. w ee b ly .c om The Honda FCX Clarity, a fuel-cellpowered automobile available to the public, albeit in limited quantities. A fuel cell converts hydrogen fuel into electricity to drive the motor attached to the wheels of the car. Automobiles, whether powered by fuel cells, gasoline engines, or batteries, use many of the concepts and principles of mechanics that we will study in this first part of the book. Quantities that we can use to describe the operation of vehicles include position, velocity, acceleration, force, energy, and momentum. ys ic (PRNewsFoto/American Honda) sw ar ph Physics, the most fundamental physical science, is concerned with the fundamental principles of the Universe. It is the foundation upon which the other sciences—astronomy, biology, chemistry, and geology—are based. It is also the basis of a large number of engineering applications. The beauty of physics lies in the simplicity of its fundamental principles and in the manner in which just a small number of concepts and models can alter and expand our view of the world around us. The study of physics can be divided into six main areas: w w w .a 1. c lassical mechanics, concerning the motion of objects that are large relative to atoms and move at speeds much slower than the speed of light 2. relativity, a theory describing objects moving at any speed, even speeds approaching the speed of light 3. thermodynamics, dealing with heat, work, temperature, and the statistical behavior of systems with large numbers of particles 4. electromagnetism, concerning electricity, magnetism, and electromagnetic fields 5. optics, the study of the behavior of light and its interaction with materials 6. quantum mechanics, a collection of theories connecting the behavior of matter at the submicroscopic level to macroscopic observations The disciplines of mechanics and electromagnetism are basic to all other branches of classical physics (developed before 1900) and modern physics (c. 1900–present). The first part of this textbook deals with classical mechanics, sometimes referred to as Newtonian mechanics or simply mechanics. Many principles and models used to understand mechanical systems retain their importance in the theories of other areas of physics and can later be used to describe many natural phenomena. Therefore, classical mechanics is of vital importance to students from all disciplines. ■ 1 c h a p t e r 1 om Physics and Measurement 1.1 Standards of Length, Mass, and Time .c 1.2 Matter and Model Building ly 1.3 Dimensional Analysis ee b 1.4 Conversion of Units 1.5 Estimates and Order-ofMagnitude Calculations Like all other sciences, physics is based on experimental observations and quantitative measurements. The main objectives of physics are to identify a limited number of fundamental laws that govern natural phenomena and use them to develop theories that can predict the results of future experiments. The fundamental laws used in developing theories are expressed in the language of mathematics, the tool that provides a bridge between theory and experiment. When there is a discrepancy between the prediction of a theory and experimental results, new or modified theories must be formulated to remove the discrepancy. Many times a theory is satisfactory only under limited conditions; a more general theory might be satisfactory without such limitations. For example, the laws of motion discovered by Isaac Newton (1642–1727) accurately describe the motion of objects moving at normal speeds but do not apply to objects moving at speeds comparable to the speed of light. In contrast, the special theory of relativity developed later by Albert Einstein (1879–1955) gives the same results as Newton’s laws at low speeds but also correctly describes the motion of objects at speeds approaching the speed of light. Hence, Einstein’s special theory of relativity is a more general theory of motion than that formed from Newton’s laws. Classical physics includes the principles of classical mechanics, thermodynamics, optics, and electromagnetism developed before 1900. Important contributions to classical physics w w w .a sw Stonehenge, in southern England, was built thousands of years ago. Various theories have been proposed about its function, including a burial ground, a healing site, and a place for ancestor worship. One of the more intriguing theories suggests that Stonehenge was an observatory, allowing measurements of some of the quantities discussed in this chapter, such as position of objects in space and time intervals between repeating celestial events. ar ph ys ic s. w 1.6 Significant Figures (Stephen Inglis/Shutterstock.com) Interactive content from this and other chapters may be assigned online in Enhanced WebAssign. 2 1.1 Standards of Length, Mass, and Time ar ph ys ic s. w ee b ly .c om were provided by Newton, who was also one of the originators of calculus as a mathematical tool. Major developments in mechanics continued in the 18th century, but the fields of thermodynamics and electromagnetism were not developed until the latter part of the 19th century, principally because before that time the apparatus for controlled experiments in these disciplines was either too crude or unavailable. A major revolution in physics, usually referred to as modern physics, began near the end of the 19th century. Modern physics developed mainly because many physical phenomena could not be explained by classical physics. The two most important developments in this modern era were the theories of relativity and quantum mechanics. Einstein’s special theory of relativity not only correctly describes the motion of objects moving at speeds comparable to the speed of light; it also completely modifies the traditional concepts of space, time, and energy. The theory also shows that the speed of light is the upper limit of the speed of an object and that mass and energy are related. Quantum mechanics was formulated by a number of distinguished scientists to provide descriptions of physical phenomena at the atomic level. Many practical devices have been developed using the principles of quantum mechanics. Scientists continually work at improving our understanding of fundamental laws. Numerous technological advances in recent times are the result of the efforts of many scientists, engineers, and technicians, such as unmanned planetary explorations, a variety of developments and potential applications in nanotechnology, microcircuitry and high-speed computers, sophisticated imaging techniques used in scientific research and medicine, and several remarkable results in genetic engineering. The effects of such developments and discoveries on our society have indeed been great, and it is very likely that future discoveries and developments will be exciting, challenging, and of great benefit to humanity. 1.1 Standards of Length, Mass, and Time w w w .a sw To describe natural phenomena, we must make measurements of various aspects of nature. Each measurement is associated with a physical quantity, such as the length of an object. The laws of physics are expressed as mathematical relationships among physical quantities that we will introduce and discuss throughout the book. In mechanics, the three fundamental quantities are length, mass, and time. All other quantities in mechanics can be expressed in terms of these three. If we are to report the results of a measurement to someone who wishes to reproduce this measurement, a standard must be defined. It would be meaningless if a visitor from another planet were to talk to us about a length of 8 “glitches” if we do not know the meaning of the unit glitch. On the other hand, if someone familiar with our system of measurement reports that a wall is 2 meters high and our unit of length is defined to be 1 meter, we know that the height of the wall is twice our basic length unit. Whatever is chosen as a standard must be readily accessible and must possess some property that can be measured reliably. Measurement standards used by different people in different places—throughout the Universe—must yield the same result. In addition, standards used for measurements must not change with time. In 1960, an international committee established a set of standards for the fundamental quantities of science. It is called the SI (Système International), and its fundamental units of length, mass, and time are the meter, kilogram, and second, respectively. Other standards for SI fundamental units established by the committee are those for temperature (the kelvin), electric current (the ampere), luminous intensity (the candela), and the amount of substance (the mole). 3 4 Chapter 1 Physics and Measurement Length .c om We can identify length as the distance between two points in space. In 1120, the king of England decreed that the standard of length in his country would be named the yard and would be precisely equal to the distance from the tip of his nose to the end of his outstretched arm. Similarly, the original standard for the foot adopted by the French was the length of the royal foot of King Louis XIV. Neither of these standards is constant in time; when a new king took the throne, length measurements changed! The French standard prevailed until 1799, when the legal standard of length in France became the meter (m), defined as one ten-millionth of the distance from the equator to the North Pole along one particular longitudinal line that passes through Paris. Notice that this value is an Earth-based standard that does not satisfy the requirement that it can be used throughout the Universe. As recently as 1960, the length of the meter was defined as the distance between two lines on a specific platinum–iridium bar stored under controlled conditions in France. Current requirements of science and technology, however, necessitate more accuracy than that with which the separation between the lines on the bar can be determined. In the 1960s and 1970s, the meter was defined as 1 650 763.73 wavelengths1 of orange-red light emitted from a krypton-86 lamp. In October 1983, however, the meter was redefined as the distance traveled by light in vacuum during a time of 1/299 792 458 second. In effect, this latest definition establishes that the speed of light in vacuum is precisely 299 792 458 meters per second. This definition of the meter is valid throughout the Universe based on our assumption that light is the same everywhere. Table 1.1 lists approximate values of some measured lengths. You should study this table as well as the next two tables and begin to generate an intuition for what is meant by, for example, a length of 20 centimeters, a mass of 100 kilograms, or a time interval of 3.2 3 107 seconds. ly Pitfall Prevention 1.1 Mass ys ic s. w ee b Reasonable Values Generating intuition about typical values of quantities when solving problems is important because you must think about your end result and determine if it seems reasonable. For example, if you are calculating the mass of a housefly and arrive at a value of 100 kg, this answer is unreasonable and there is an error somewhere. sw ar ph The SI fundamental unit of mass, the kilogram (kg), is defined as the mass of a specific platinum–iridium alloy cylinder kept at the International Bureau of Weights and Measures at Sèvres, France. This mass standard was established in 1887 and Table 1.1 Distance from the Earth to the most remote known quasar Distance from the Earth to the most remote normal galaxies Distance from the Earth to the nearest large galaxy (Andromeda) Distance from the Sun to the nearest star (Proxima Centauri) One light-year Mean orbit radius of the Earth about the Sun Mean distance from the Earth to the Moon Distance from the equator to the North Pole Mean radius of the Earth Typical altitude (above the surface) of a satellite orbiting the Earth Length of a football field Length of a housefly Size of smallest dust particles Size of cells of most living organisms Diameter of a hydrogen atom Diameter of an atomic nucleus Diameter of a proton .a w w w Approximate Values of Some Measured Lengths Length (m) 1.4 3 1026 9 3 1025 2 3 1022 4 3 1016 9.46 3 1015 1.50 3 1011 3.84 3 108 1.00 3 107 6.37 3 106 2 3 105 9.1 3 101 5 3 1023 , 1024 , 1025 , 10210 , 10214 , 10215 1We will use the standard international notation for numbers with more than three digits, in which groups of three digits are separated by spaces rather than commas. Therefore, 10 000 is the same as the common American notation of 10,000. Similarly, p 5 3.14159265 is written as 3.141 592 65. 1.1 Standards of Length, Mass, and Time Table 1.3 Approximate Values of Some Time Intervals Time Interval (s) 8 3 1021 , 1023 , 1026 , 10213 , 10215 , 10222 a , 10224 ee b , 1042 1.99 3 1030 5.98 3 1024 7.36 3 1022 , 103 , 102 , 1021 , 1025 , 1 3 10215 1.67 3 10227 9.11 3 10231 om , 1052 4 3 1017 1.3 3 1017 6.3 3 108 3.2 3 107 8.6 3 104 3.0 3 103 .c Age of the Universe Age of the Earth Average age of a college student One year One day One class period Time interval between normal heartbeats Period of audible sound waves Period of typical radio waves Period of vibration of an atom in a solid Period of visible light waves Duration of a nuclear collision Time interval for light to cross a proton ly Mass (kg) Observable Universe Milky Way galaxy Sun Earth Moon Shark Human Frog Mosquito Bacterium Hydrogen atom Electron Reproduced with permission of the BIPM, which retains full internationally protected copyright. Table 1.2 Approximate Masses of Various Objects 5 Time w w w .a sw ar ph ys Before 1967, the standard of time was defined in terms of the mean solar day. (A solar day is the time interval between successive appearances of the Sun at the highest point it reaches in the sky each day.) The fundamental unit of a second (s) was defined as 1 1 1 1 60 2 1 60 2 1 24 2 of a mean solar day. This definition is based on the rotation of one planet, the Earth. Therefore, this motion does not provide a time standard that is universal. In 1967, the second was redefined to take advantage of the high precision attainable in a device known as an atomic clock (Fig. 1.1b), which measures vibrations of cesium atoms. One second is now defined as 9 192 631 770 times the period of vibration of radiation from the cesium-133 atom.2 Approximate values of time intervals are presented in Table 1.3. In addition to SI, another system of units, the U.S. customary system, is still used in the United States despite acceptance of SI by the rest of the world. In this system, the units of length, mass, and time are the foot (ft), slug, and second, respectively. In this book, we shall use SI units because they are almost universally accepted in science and industry. We shall make some limited use of U.S. customary units in the study of classical mechanics. In addition to the fundamental SI units of meter, kilogram, and second, we can also use other units, such as millimeters and nanoseconds, where the prefixes milliand nano- denote multipliers of the basic units based on various powers of ten. Prefixes for the various powers of ten and their abbreviations are listed in Table 1.4 (page 6). For example, 1023 m is equivalent to 1 millimeter (mm), and 103 m corresponds to 1 kilometer (km). Likewise, 1 kilogram (kg) is 103 grams (g), and 1 mega volt (MV) is 106 volts (V). The variables length, time, and mass are examples of fundamental quantities. Most other variables are derived quantities, those that can be expressed as a mathematical combination of fundamental quantities. Common examples are area (a product of two lengths) and speed (a ratio of a length to a time interval). 2Period is defined as the time interval needed for one complete vibration. AP Photo/Focke Strangmann ic s. w has not been changed since that time because platinum–iridium is an unusually stable alloy. A duplicate of the Sèvres cylinder is kept at the National Institute of Standards and Technology (NIST) in Gaithersburg, Maryland (Fig. 1.1a). Table 1.2 lists approximate values of the masses of various objects. b Figure 1.1 (a) The National Standard Kilogram No. 20, an accurate copy of the International Standard Kilogram kept at Sèvres, France, is housed under a double bell jar in a vault at the National Institute of Standards and Technology. (b) A cesium fountain atomic clock. The clock will neither gain nor lose a second in 20 million years. 6 Chapter 1 Physics and Measurement Table 1.4 Prefixes for Powers of Ten Prefix Abbreviation Power Prefix Abbreviation 10224 10221 10218 10215 10212 1029   1026   1023   1022   1021   yocto zepto atto femto pico nano micro milli centi deci y z a f p n m m c d 103   106   109   1012 1015 1018 1021 1024 kilo mega giga tera peta exa zetta yotta k M G T P E Z Y om Power Another example of a derived quantity is density. The density r (Greek letter rho) of any substance is defined as its mass per unit volume: .c A table of the letters in the  Greek alphabet is provided on the back endpaper of this book. Don Farrall/Photodisc/ Getty Images s. w ee b ly m (1.1) V In terms of fundamental quantities, density is a ratio of a mass to a product of three lengths. Aluminum, for example, has a density of 2.70 3 103 kg/m3, and iron has a density of 7.86 3 103 kg/m3. An extreme difference in density can be imagined by thinking about holding a 10-centimeter (cm) cube of Styrofoam in one hand and a 10-cm cube of lead in the other. See Table 14.1 in Chapter 14 for densities of several materials. r; ar ph ys ic Q uick Quiz 1.1 In a machine shop, two cams are produced, one of aluminum and one of iron. Both cams have the same mass. Which cam is larger? (a) The aluminum cam is larger. (b) The iron cam is larger. (c) Both cams have the same size. A piece of gold consists of gold atoms. 1.2 Matter and Model Building sw w w w Inside the nucleus are protons (orange) and neutrons (gray). Protons and neutrons are composed of quarks. The quark composition of a proton is shown here. Figure 1.2 in matter. If physicists cannot interact with some phenomenon directly, they often imagine a model for a physical system that is related to the phenomenon. For example, we cannot interact directly with atoms because they are too small. Therefore, we build a mental model of an atom based on a system of a nucleus and one or more electrons outside the nucleus. Once we have identified the physical components of the model, we make predictions about its behavior based on the interactions among the components of the system or the interaction between the system and the environment outside the system. As an example, consider the behavior of matter. A sample of solid gold is shown at the top of Figure 1.2. Is this sample nothing but wall-to-wall gold, with no empty space? If the sample is cut in half, the two pieces still retain their chemical identity as solid gold. What if the pieces are cut again and again, indefinitely? Will the smaller and smaller pieces always be gold? Such questions can be traced to early Greek philosophers. Two of them—Leucippus and his student Democritus—could not accept the idea that such cuttings could go on forever. They developed a model for matter by speculating that the process ultimately must end when it produces a particle that can no longer be cut. In Greek, atomos means “not sliceable.” From this Greek term comes our English word atom. The Greek model of the structure of matter was that all ordinary matter consists of atoms, as suggested in the middle of Figure 1.2. Beyond that, no additional structure was specified in the model; atoms acted as small particles that interacted with one another, but internal structure of the atom was not a part of the model. .a At the center of each atom is a nucleus. p u u d Levels of organization 1.3 Dimensional Analysis ar ph ys ic s. w ee b ly .c om In 1897, J. J. Thomson identified the electron as a charged particle and as a constituent of the atom. This led to the first atomic model that contained internal structure. We shall discuss this model in Chapter 42. Following the discovery of the nucleus in 1911, an atomic model was developed in which each atom is made up of electrons surrounding a central nucleus. A nucleus of gold is shown in Figure 1.2. This model leads, however, to a new question: Does the nucleus have structure? That is, is the nucleus a single particle or a collection of particles? By the early 1930s, a model evolved that described two basic entities in the nucleus: protons and neutrons. The proton carries a positive electric charge, and a specific chemical element is identified by the number of protons in its nucleus. This number is called the atomic number of the element. For instance, the nucleus of a hydrogen atom contains one proton (so the atomic number of hydrogen is 1), the nucleus of a helium atom contains two protons (atomic number 2), and the nucleus of a uranium atom contains 92 protons (atomic number 92). In addition to atomic number, a second number—mass number, defined as the number of protons plus neutrons in a nucleus—characterizes atoms. The atomic number of a specific element never varies (i.e., the number of protons does not vary), but the mass number can vary (i.e., the number of neutrons varies). Is that, however, where the process of breaking down stops? Protons, neutrons, and a host of other exotic particles are now known to be composed of six different varieties of particles called quarks, which have been given the names of up, down, strange, charmed, bottom, and top. The up, charmed, and top quarks have electric charges of 123 that of the proton, whereas the down, strange, and bottom quarks have charges of 213 that of the proton. The proton consists of two up quarks and one down quark as shown at the bottom of Figure 1.2 and labeled u and d. This structure predicts the correct charge for the proton. Likewise, the neutron consists of two down quarks and one up quark, giving a net charge of zero. You should develop a process of building models as you study physics. In this study, you will be challenged with many mathematical problems to solve. One of the most important problem-solving techniques is to build a model for the problem: identify a system of physical components for the problem and make predictions of the behavior of the system based on the interactions among its components or the interaction between the system and its surrounding environment. sw 1.3 Dimensional Analysis w w w .a In physics, the word dimension denotes the physical nature of a quantity. The distance between two points, for example, can be measured in feet, meters, or furlongs, which are all different ways of expressing the dimension of length. The symbols we use in this book to specify the dimensions of length, mass, and time are L, M, and T, respectively.3 We shall often use brackets [ ] to denote the dimensions of a physical quantity. For example, the symbol we use for speed in this book is v, and in our notation, the dimensions of speed are written [v] 5 L/T. As another example, the dimensions of area A are [A] 5 L2. The dimensions and units of area, volume, speed, and acceleration are listed in Table 1.5. The dimensions of other quantities, such as force and energy, will be described as they are introduced in the text. Table 1.5 Dimensions and Units of Four Derived Quantities Quantity Dimensions SI units U.S. customary units 3The Area (A) Volume (V ) Speed (v) Acceleration (a) L2 L3 m2 ft 2 m3 ft 3 L/T m/s ft/s L/T2 m/s2 ft/s2 dimensions of a quantity will be symbolized by a capitalized, nonitalic letter such as L or T. The algebraic symbol for the quantity itself will be an italicized letter such as L for the length of an object or t for time. 7 8 Chapter 1 Physics and Measurement In many situations, you may have to check a specific equation to see if it matches your expectations. A useful procedure for doing that, called dimensional analysis, can be used because dimensions can be treated as algebraic quantities. For example, quantities can be added or subtracted only if they have the same dimensions. Furthermore, the terms on both sides of an equation must have the same dimensions. By following these simple rules, you can use dimensional analysis to determine whether an expression has the correct form. Any relationship can be correct only if the dimensions on both sides of the equation are the same. To illustrate this procedure, suppose you are interested in an equation for the position x of a car at a time t if the car starts from rest at x 5 0 and moves with constant acceleration a. The correct expression for this situation is x 5 12 at 2 as we show in Chapter 2. The quantity x on the left side has the dimension of length. For the equation to be dimensionally correct, the quantity on the right side must also have the dimension of length. We can perform a dimensional check by substituting the dimensions for acceleration, L/T2 (Table 1.5), and time, T, into the equation. That is, the dimensional form of the equation x 5 12 at 2 is Pitfall Prevention 1.2 Symbols for Quantities Some L # 2 T 5L T2 ee b L5 ly .c om quantities have a small number of symbols that represent them. For example, the symbol for time is almost always t. Other quantities might have various symbols depending on the usage. Length may be described with symbols such as x, y, and z (for position); r (for radius); a, b, and c (for the legs of a right triangle); , (for the length of an object); d (for a distance); h (for a height); and so forth. s. w The dimensions of time cancel as shown, leaving the dimension of length on the right-hand side to match that on the left. A more general procedure using dimensional analysis is to set up an expression of the form x ~ an t m ar ph ys ic where n and m are exponents that must be determined and the symbol ~ indicates a proportionality. This relationship is correct only if the dimensions of both sides are the same. Because the dimension of the left side is length, the dimension of the right side must also be length. That is, 3 ant m 4 5 L 5 L1T0 sw Because the dimensions of acceleration are L/T2 and the dimension of time is T, we have 1 L/T2 2 n Tm 5 L1T0 S 1 Ln Tm22n 2 5 L1T0 Q uick Quiz 1.2 True or False: Dimensional analysis can give you the numerical value of constants of proportionality that may appear in an algebraic expression. w w w .a The exponents of L and T must be the same on both sides of the equation. From the exponents of L, we see immediately that n 5 1. From the exponents of T, we see that m 2 2n 5 0, which, once we substitute for n, gives us m 5 2. Returning to our original expression x ~ ant m , we conclude that x ~ at 2 . Example 1.1    Analysis of an Equation Show that the expression v 5 at, where v represents speed, a acceleration, and t an instant of time, is dimensionally correct. Solution Identify the dimensions of v from Table 1.5: 3v 4 5 L T 1.4 Conversion of Units 9 ▸ 1.1 c o n t i n u e d 3 at 4 5 Identify the dimensions of a from Table 1.5 and multiply by the dimensions of t : L L T 5 T T2 Therefore, v 5 at is dimensionally correct because we have the same dimensions on both sides. (If the expression were given as v 5 at 2, it would be dimensionally incorrect. Try it and see!)    Analysis of a Power Law om Example 1.2 ly .c Suppose we are told that the acceleration a of a particle moving with uniform speed v in a circle of radius r is proportional to some power of r, say r n , and some power of v, say v m . Determine the values of n and m and write the simplest form of an equation for the acceleration. Substitute the dimensions of a, r, and v: ar ph Write the acceleration expression: ys Solve the two equations for n: L L m Ln1m 5 Ln a b 5 m 2 T T T n 1 m 5 1 and m 5 2 ic Equate the exponents of L and T so that the dimensional equation is balanced: a 5 kr n v m s. w Write an expression for a with a dimensionless constant of proportionality k: ee b Solution n 5 21 a 5 kr21 v 2 5 k v2 r sw In Section 4.4 on uniform circular motion, we show that k 5 1 if a consistent set of units is used. The constant k would not equal 1 if, for example, v were in km/h and you wanted a in m/s2 . Pitfall Prevention 1.3 1.4 Conversion of Units .a Always Include Units When per- w w Sometimes it is necessary to convert units from one measurement system to another or convert within a system (for example, from kilometers to meters). Conversion factors between SI and U.S. customary units of length are as follows: w 1 mile 5 1 609 m 5 1.609 km 1 m 5 39.37 in. 5 3.281 ft 1 ft 5 0.304 8 m 5 30.48 cm 1 in. 5 0.025 4 m 5 2.54 cm (exactly) A more complete list of conversion factors can be found in Appendix A. Like dimensions, units can be treated as algebraic quantities that can cancel each other. For example, suppose we wish to convert 15.0 in. to centimeters. Because 1 in. is defined as exactly 2.54 cm, we find that 15.0 in. 5 1 15.0 in. 2 a 2.54 cm b 5 38.1 cm 1 in. where the ratio in parentheses is equal to 1. We express 1 as 2.54 cm/1 in. (rather than 1 in./2.54 cm) so that the unit “inch” in the denominator cancels with the unit in the original quantity. The remaining unit is the centimeter, our desired result. forming calculations with numerical values, include the units for every quantity and carry the units through the entire calculation. Avoid the temptation to drop the units early and then attach the expected units once you have an answer. By including the units in every step, you can detect errors if the units for the answer turn out to be incorrect. 10 Chapter 1 Physics and Measurement Q uick Quiz 1.3 The distance between two cities is 100 mi. What is the number of kilometers between the two cities? (a) smaller than 100 (b) larger than 100 (c) equal to 100 Example 1.3    Is He Speeding? On an interstate highway in a rural region of Wyoming, a car is traveling at a speed of 38.0 m/s. Is the driver exceeding the speed limit of 75.0 mi/h? 1 mi b 5 2.36 3 1022 mi/s 1 609 m 1 2.36 3 1022 mi/s 2 a 60 s 60 min b a b 5 85.0 mi/h 1 min 1h The driver is indeed exceeding the speed limit and should slow down. ee b ly Convert seconds to hours: .c 1 38.0 m/s 2 a Convert meters in the speed to miles: om Solution s. w W h at If ? What if the driver were from outside the United States and is familiar with speeds measured in kilometers per hour? What is the speed of the car in km/h? Answer We can convert our final answer to the appropriate units: © Cengage Learning/Ed Dodd ic 1.609 km b 5 137 km/h 1 mi ys 1 85.0 mi/h 2 a sw ar ph Figure 1.3 shows an automobile speedometer displaying speeds in both mi/h and km/h. Can you check the conversion we just performed using this photograph? Figure 1.3 The speedometer of a vehicle that shows speeds in both miles per hour and kilometers per hour. w w w .a 1.5 Estimates and Order-of-Magnitude Calculations Suppose someone asks you the number of bits of data on a typical musical compact disc. In response, it is not generally expected that you would provide the exact number but rather an estimate, which may be expressed in scientific notation. The estimate may be made even more approximate by expressing it as an order of magnitude, which is a power of ten determined as follows: 1. Express the number in scientific notation, with the multiplier of the power of ten between 1 and 10 and a unit. 2. If the multiplier is less than 3.162 (the square root of 10), the order of magnitude of the number is the power of 10 in the scientific notation. If the multiplier is greater than 3.162, the order of magnitude is one larger than the power of 10 in the scientific notation. We use the symbol , for “is on the order of.” Use the procedure above to verify the orders of magnitude for the following lengths: 0.008 6 m , 1022 m   0.002 1 m , 1023 m   720 m , 103 m 1.6 Significant Figures 11    Breaths in a Lifetime .c Example 1.4 om Usually, when an order-of-magnitude estimate is made, the results are reliable to within about a factor of 10. If a quantity increases in value by three orders of magnitude, its value increases by a factor of about 103 5 1 000. Inaccuracies caused by guessing too low for one number are often canceled by other guesses that are too high. You will find that with practice your guesstimates become better and better. Estimation problems can be fun to work because you freely drop digits, venture reasonable approximations for unknown numbers, make simplifying assumptions, and turn the question around into something you can answer in your head or with minimal mathematical manipulation on paper. Because of the simplicity of these types of calculations, they can be performed on a small scrap of paper and are often called “back-of-the-envelope calculations.” ly Estimate the number of breaths taken during an average human lifetime. Solution Find the approximate number of minutes in a year: s. w ee b We start by guessing that the typical human lifetime is about 70 years. Think about the average number of breaths that a person takes in 1 min. This number varies depending on whether the person is exercising, sleeping, angry, serene, and so forth. To the nearest order of magnitude, we shall choose 10 breaths per minute as our estimate. (This estimate is certainly closer to the true average value than an estimate of 1 breath per minute or 100 breaths per minute.) ys ar ph Find the approximate number of breaths in a lifetime: 400 days 1 yr b a 25 h 60 min b a b 5 6 3 105 min 1 day 1h number of minutes 5 (70 yr)(6 3 105 min/yr) ic Find the approximate number of minutes in a 70-year lifetime: 1 yr a 5 4 3 107 min number of breaths 5 (10 breaths/min)(4 3 107 min) 5 4 3 108 breaths Therefore, a person takes on the order of 109 breaths in a lifetime. Notice how much simpler it is in the first calculation above to multiply 400 3 25 than it is to work with the more accurate 365 3 24. What if the average lifetime were estimated as 80 years instead of 70? Would that change our final estimate? sw W h at If ? Answer We could claim that (80 yr)(6 3 105 min/yr) 5 5 3 107 min, so our final estimate should be 5 3 108 breaths. w .a This answer is still on the order of 109 breaths, so an order-of-magnitude estimate would be unchanged. w 1.6 Significant Figures w When certain quantities are measured, the measured values are known only to within the limits of the experimental uncertainty. The value of this uncertainty can depend on various factors, such as the quality of the apparatus, the skill of the experimenter, and the number of measurements performed. The number of significant figures in a measurement can be used to express something about the uncertainty. The number of significant figures is related to the number of numerical digits used to express the measurement, as we discuss below. As an example of significant figures, suppose we are asked to measure the radius of a compact disc using a meterstick as a measuring instrument. Let us assume the accuracy to which we can measure the radius of the disc is 60.1 cm. Because of the uncertainty of 60.1 cm, if the radius is measured to be 6.0 cm, we can claim only that its radius lies somewhere between 5.9 cm and 6.1 cm. In this case, we say that the measured value of 6.0 cm has two significant figures. Note that the 12 Chapter 1 Physics and Measurement ee b ly .c om significant figures include the first estimated digit. Therefore, we could write the radius as (6.0 6 0.1) cm. Zeros may or may not be significant figures. Those used to position the decimal point in such numbers as 0.03 and 0.007 5 are not significant. Therefore, there are one and two significant figures, respectively, in these two values. When the zeros come after other digits, however, there is the possibility of misinterpretation. For example, suppose the mass of an object is given as 1 500 g. This value is ambiguous because we do not know whether the last two zeros are being used to locate the decimal point or whether they represent significant figures in the measurement. To remove this ambiguity, it is common to use scientific notation to indicate the number of significant figures. In this case, we would express the mass as 1.5 3 103 g if there are two significant figures in the measured value, 1.50 3 103 g if there are three significant figures, and 1.500 3 103 g if there are four. The same rule holds for numbers less than 1, so 2.3 3 1024 has two significant figures (and therefore could be written 0.000 23) and 2.30 3 1024 has three significant figures (also written as 0.000 230). In problem solving, we often combine quantities mathematically through multiplication, division, addition, subtraction, and so forth. When doing so, you must make sure that the result has the appropriate number of significant figures. A good rule of thumb to use in determining the number of significant figures that can be claimed in a multiplication or a division is as follows: s. w When multiplying several quantities, the number of significant figures in the final answer is the same as the number of significant figures in the quantity having the smallest number of significant figures. The same rule applies to division. ic Let’s apply this rule to find the area of the compact disc whose radius we measured above. Using the equation for the area of a circle, ys A 5 pr 2 5 p 1 6.0 cm 2 2 5 1.1 3 102 cm2 Pitfall Prevention 1.4 When numbers are added or subtracted, the number of decimal places in the result should equal the smallest number of decimal places of any term in the sum or difference. w w w .a Read Carefully Notice that the rule for addition and subtraction is different from that for multiplication and division. For addition and subtraction, the important consideration is the number of decimal places, not the number of significant figures. sw ar ph If you perform this calculation on your calculator, you will likely see 113.097 335 5. It should be clear that you don’t want to keep all of these digits, but you might be tempted to report the result as 113 cm2. This result is not justified because it has three significant figures, whereas the radius only has two. Therefore, we must report the result with only two significant figures as shown above. For addition and subtraction, you must consider the number of decimal places when you are determining how many significant figures to report: As an example of this rule, consider the sum 23.2 1 5.174 5 28.4 Notice that we do not report the answer as 28.374 because the lowest number of decimal places is one, for 23.2. Therefore, our answer must have only one decimal place. The rule for addition and subtraction can often result in answers that have a different number of significant figures than the quantities with which you start. For example, consider these operations that satisfy the rule: 1.000 1 1 0.000 3 5 1.000 4 1.002 2 0.998 5 0.004 In the first example, the result has five significant figures even though one of the terms, 0.000 3, has only one significant figure. Similarly, in the second calculation, the result has only one significant figure even though the numbers being subtracted have four and three, respectively. 13 Summary WW Significant figure guidelines used in this book In this book, most of the numerical examples and end-of-chapter problems will yield answers having three significant figures. When carrying out estimation calculations, we shall typically work with a single significant figure. Pitfall Prevention 1.5 Symbolic Solutions When solving problems, it is very useful to perform the solution completely in algebraic form and wait until the very end to enter numerical values into the final symbolic expression. This method will save many calculator keystrokes, especially if some quantities cancel so that you never have to enter their values into your calculator! In addition, you will only need to round once, on the final result.    Installing a Carpet ic Example 1.5 s. w ee b ly .c om If the number of significant figures in the result of a calculation must be reduced, there is a general rule for rounding numbers: the last digit retained is increased by 1 if the last digit dropped is greater than 5. (For example, 1.346 becomes 1.35.) If the last digit dropped is less than 5, the last digit retained remains as it is. (For example, 1.343 becomes 1.34.) If the last digit dropped is equal to 5, the remaining digit should be rounded to the nearest even number. (This rule helps avoid accumulation of errors in long arithmetic processes.) A technique for avoiding error accumulation is to delay the rounding of numbers in a long calculation until you have the final result. Wait until you are ready to copy the final answer from your calculator before rounding to the correct number of significant figures. In this book, we display numerical values rounded off to two or three significant figures. This occasionally makes some mathematical manipulations look odd or incorrect. For instance, looking ahead to Example 3.5 on page 69, you will see the operation 217.7 km 1 34.6 km 5 17.0 km. This looks like an incorrect subtraction, but that is only because we have rounded the numbers 17.7 km and 34.6 km for display. If all digits in these two intermediate numbers are retained and the rounding is only performed on the final number, the correct three-digit result of 17.0 km is obtained. ys A carpet is to be installed in a rectangular room whose length is measured to be 12.71 m and whose width is measured to be 3.46 m. Find the area of the room. ar ph Solution w .a sw If you multiply 12.71 m by 3.46 m on your calculator, you will see an answer of 43.976 6 m2. How many of these numbers should you claim? Our rule of thumb for multiplication tells us that you can claim only the number of significant figures in your answer as are present in the measured quantity having the lowest number of significant figures. In this example, the lowest number of significant figures is three in 3.46 m, so we should express our final answer as 44.0 m2. w w Summary Definitions The three fundamental physical quantities of mechanics are length, mass, and time, which in the SI system have the units meter (m), kilogram (kg), and second (s), respectively. These fundamental quantities cannot be defined in terms of more basic quantities. The density of a substance is defined as its mass per unit volume: m (1.1) r; V continued 14 Chapter 1 Physics and Measurement Concepts and Principles When you compute a result from several measured numbers, each of which has a certain accuracy, you should give the result with the correct number of significant figures. The method of dimensional analysis is very powerful in solving physics problems. Dimensions can be treated as algebraic quantities. By making estimates and performing order-of-magnitude calculations, you should be able to approximate the answer to a problem when there is not enough information available to specify an exact solution completely. om When multiplying several quantities, the number of significant ­f igures in the final answer is the same as the number of significant figures in the quantity having the smallest number of significant figures. The same rule applies to division. 1. denotes answer available in Student Solutions Manual/Study Guide give them equal rank in your list. (a) 0.032 kg (b) 15 g (c) 2.7 3 105 mg (d) 4.1 3 1028 Gg (e) 2.7 3 108 mg 6. What is the sum of the measured values 21.4 s 1 15 s 1 17.17 s 1 4.00 3 s? (a) 57.573 s (b) 57.57 s (c) 57.6 s (d) 58 s (e) 60 s ic 1. One student uses a meterstick to measure the thickness of a textbook and obtains 4.3 cm 6 0.1 cm. Other students measure the thickness with vernier calipers and obtain four different measurements: (a) 4.32 cm 6 0.01 cm, (b) 4.31 cm 6 0.01 cm, (c) 4.24 cm 6 0.01 cm, and (d) 4.43 cm 6 0.01 cm. Which of these four measurements, if any, agree with that obtained by the first student? s. w Objective Questions ee b ly .c When numbers are added or subtracted, the number of decimal places in the result should equal the smallest number of decimal places of any term in the sum or difference. ys 7. Which of the following is the best estimate for the mass of all the people living on the Earth? (a) 2 3 108 kg (b) 1 3 109 kg (c) 2 3 1010 kg (d) 3 3 1011 kg (e) 4 3 1012 kg ar ph 2. A house is advertised as having 1 420 square feet under its roof. What is its area in square meters? (a) 4 660 m2 (b) 432 m2 (c) 158 m2 (d) 132 m2 (e) 40.2 m2 sw 3. Answer each question yes or no. Must two quantities have the same dimensions (a) if you are adding them? (b) If you are multiplying them? (c) If you are subtracting them? (d) If you are dividing them? (e) If you are equating them? w w w .a 4. The price of gasoline at a particular station is 1.5 euros per liter. An American student can use 33 euros to buy gasoline. Knowing that 4 quarts make a gallon and that 1 liter is close to 1 quart, she quickly reasons that she can buy how many gallons of gasoline? (a) less than 1 gallon (b) about 5 gallons (c) about 8 gallons (d) more than 10 gallons 5. Rank the following five quantities in order from the largest to the smallest. If two of the quantities are equal, Conceptual Questions 8. (a) If an equation is dimensionally correct, does that mean that the equation must be true? (b) If an equation is not dimensionally correct, does that mean that the equation cannot be true? 9. Newton’s second law of motion (Chapter 5) says that the mass of an object times its acceleration is equal to the net force on the object. Which of the following gives the correct units for force? (a) kg ? m/s2 (b) kg ? m2/s2 (c) kg/m ? s2 (d) kg ? m2/s (e) none of those answers 10. A calculator displays a result as 1.365 248 0 3 107 kg. The estimated uncertainty in the result is 62%. How many digits should be included as significant when the result is written down? (a) zero (b) one (c) two (d) three (e) four 1. denotes answer available in Student Solutions Manual/Study Guide 1. Suppose the three fundamental standards of the metric system were length, density, and time rather than length, mass, and time. The standard of density in this system is to be defined as that of water. What considerations about water would you need to address to make sure that the standard of density is as accurate as possible? 2. Why is the metric system of units considered superior to most other systems of units? 3. What natural phenomena could serve as alternative time standards? 4. Express the following quantities using the prefixes given in Table 1.4. (a) 3 3 1024 m (b) 5 3 1025 s (c) 72 3 102 g 15 Problems Problems The problems found in this chapter may be assigned online in Enhanced WebAssign AMT   Analysis Model tutorial available in Enhanced WebAssign GP Guided Problem 1. straightforward; 2. intermediate; 3. challenging M   Master It tutorial available in Enhanced 1. full solution available in the Student W   Watch It video solution available in Solutions Manual/Study Guide WebAssign BIO Q/C Enhanced WebAssign om S Section 1.1 Standards of Length, Mass, and Time .c Note: Consult the endpapers, appendices, and tables in the text whenever necessary in solving problems. For this chapter, Table 14.1 and Appendix B.3 may be particularly useful. Answers to odd-numbered problems appear in the back of the book. ly L d ee b a 1. (a) Use information on the endpapers of this book to Q/C calculate the average density of the Earth. (b) Where s. w does the value fit among those listed in Table 14.1 in Chapter 14? Look up the density of a typical surface rock like granite in another source and compare it with the density of the Earth. w w .a sw ar ph ys ic 2. The standard kilogram (Fig. 1.1a) is a platinum–iridium W cylinder 39.0 mm in height and 39.0 mm in diameter. What is the density of the material? 3. An automobile company displays a die-cast model of its first car, made from 9.35 kg of iron. To celebrate its hundredth year in business, a worker will recast the model in solid gold from the original dies. What mass of gold is needed to make the new model? 4. A proton, which is the nucleus of a hydrogen atom, can Q/C be modeled as a sphere with a diameter of 2.4 fm and a mass of 1.67 3 10227 kg. (a) Determine the density of the proton. (b) State how your answer to part (a) compares with the density of osmium, given in Table 14.1 in Chapter 14. 5. Two spheres are cut from a certain uniform rock. One W has radius 4.50 cm. The mass of the other is five times greater. Find its radius. w 6. What mass of a material with density r is required to S make a hollow spherical shell having inner radius r 1 and outer radius r 2? Section 1.2 Matter and Model Building 7. A crystalline solid consists of atoms stacked up in a repeating lattice structure. Consider a crystal as shown in Figure P1.7a. The atoms reside at the corners of cubes of side L 5 0.200 nm. One piece of evidence for the regular arrangement of atoms comes from the flat surfaces along which a crystal separates, or cleaves, when it is broken. Suppose this crystal cleaves along a face diagonal as shown in Figure P1.7b. Calculate the spacing d between two adjacent atomic planes that separate when the crystal cleaves. b Figure P1.7 8. The mass of a copper atom is 1.06 3 10225 kg, and the density of copper is 8 920 kg/m3 . (a) Determine the number of atoms in 1 cm3 of copper. (b) Visualize the one cubic centimeter as formed by stacking up identical cubes, with one copper atom at the center of each. Determine the volume of each cube. (c) Find the edge dimension of each cube, which represents an estimate for the spacing between atoms. Section 1.3 Dimensional Analysis 9. Which of the following equations are dimensionally correct? (a) vf 5 vi 1 ax (b) y 5 (2 m) cos (kx), where k 5 2 m21 10. Figure P1.10 shows a frustum r1 W of a cone. Match each of the expressions (a) p(r 1 1 r 2)[h 2 1 (r 2 2 r 1)2]1/2, (b) 2p(r 1 1 r 2), and (c) ph(r 12 1 r 1r 2 1 r 22)/3 h r2 with the quantity it describes: (d) the total circumference of Figure P1.10 the flat circular faces, (e) the volume, or (f) the area of the curved surface. 11. Kinetic energy K (Chapter 7) has dimensions kg ? m2/s2. It can be written in terms of the momentum p (Chapter 9) and mass m as K5 p2 2m 16 Chapter 1 Physics and Measurement Adam Sylvester/Photo Researchers, Inc. (a) Determine the proper units for momentum using dimensional analysis. (b) The unit of force is the newton N, where 1 N 5 1 kg ? m/s2 . What are the units of momentum p in terms of a newton and another fundamental SI unit? 12. Newton’s law of universal gravitation is represented by GMm r2 14. (a) Assume the equation x 5 At 3 1 Bt describes the motion of a particular object, with x having the dimension of length and t having the dimension of time. Determine the dimensions of the constants A and B. (b) Determine the dimensions of the derivative dx/dt 5 3At 2 1 B. om 22. Assume it takes 7.00 min to fill a 30.0-gal gasoline tank. W (a) Calculate the rate at which the tank is filled in gallons per second. (b) Calculate the rate at which the tank is filled in cubic meters per second. (c) Determine the time interval, in hours, required to fill a 1.00-m3 volume at the same rate. (1 U.S. gal 5 231 in.3) 23. A section of land has an area of 1 square mile and contains 640 acres. Determine the number of square meters in 1 acre. 24. A house is 50.0 ft long and 26 ft wide and has 8.0-ftM high ceilings. What is the volume of the interior of the house in cubic meters and in cubic centimeters? 25. One cubic meter (1.00 m3) of aluminum has a mass of 3 M 2.70 3 10 kg, and the same volume of iron has a mass of 7.86 3 103 kg. Find the radius of a solid aluminum sphere that will balance a solid iron sphere of radius 2.00 cm on an equal-arm balance. ic Section 1.4 Conversion of Units Problems 20 and 21. .c 13. The position of a particle moving under uniform acceleration is some function of time and the acceleration. Suppose we write this position as x 5 kamtn , where k is a dimensionless constant. Show by dimensional analysis that this expression is satisfied if m 5 1 and n 5 2. Can this analysis give the value of k? Figure P1.20 ly where F is the magnitude of the gravitational force exerted by one small object on another, M and m are the masses of the objects, and r is a distance. Force has the SI units kg ? m/s2. What are the SI units of the proportionality constant G? ee b F5 s. w W 26. Let rAl represent the density of aluminum and rFe that S of iron. Find the radius of a solid aluminum sphere that balances a solid iron sphere of radius r Fe on an equal-arm balance. ys 15. A solid piece of lead has a mass of 23.94 g and a volume W of 2.10 cm3. From these data, calculate the density of lead in SI units (kilograms per cubic meter). ar ph 16. An ore loader moves 1 200 tons/h from a mine to the surface. Convert this rate to pounds per second, using 1 ton 5 2 000 lb. sw 17. A rectangular building lot has a width of 75.0 ft and a length of 125 ft. Determine the area of this lot in square meters. w w .a 18. Suppose your hair grows at the rate 1/32 in. per day. W Find the rate at which it grows in nanometers per second. Because the distance between atoms in a molecule is on the order of 0.1 nm, your answer suggests how rapidly layers of atoms are assembled in this protein synthesis. w 19. Why is the following situation impossible? A student’s dormitory room measures 3.8 m by 3.6 m, and its ceiling is 2.5 m high. After the student completes his physics course, he displays his dedication by completely wallpapering the walls of the room with the pages from his copy of volume 1 (Chapters 1–22) of this textbook. He even covers the door and window. 20. A pyramid has a height of 481 ft, and its base covers an W area of 13.0 acres (Fig. P1.20). The volume of a pyramid is given by the expression V 5 13 Bh, where B is the area of the base and h is the height. Find the volume of this pyramid in cubic meters. (1 acre 5 43 560 ft2) 21. The pyramid described in Problem 20 contains approximately 2 million stone blocks that average 2.50 tons each. Find the weight of this pyramid in pounds. 27. One gallon of paint (volume 5 3.78 3 10 –3 m3) covers 2 M an area of 25.0 m . What is the thickness of the fresh paint on the wall? 28. An auditorium measures 40.0 m 3 20.0 m 3 12.0 m. 3 W The density of air is 1.20 kg/m . What are (a) the volume of the room in cubic feet and (b) the weight of air in the room in pounds? 29. (a) At the time of this book’s printing, the U.S. M national debt is about $16 trillion. If payments were made at the rate of $1 000 per second, how many years would it take to pay off the debt, assuming no interest were charged? (b) A dollar bill is about 15.5 cm long. How many dollar bills attached end to end would it take to reach the Moon? The front endpapers give the Earth–Moon distance. Note: Before doing these calculations, try to guess at the answers. You may be very surprised. 30. A hydrogen atom has a diameter of 1.06 3 10210 m. The nucleus of the hydrogen atom has a diameter of approximately 2.40 3 10215 m. (a) For a scale model, represent the diameter of the hydrogen atom by the playing length of an American football field (100 yards 5 300 ft) and determine the diameter of the nucleus in millimeters. (b) Find the ratio of the volume of the hydrogen atom to the volume of its nucleus. 17 Problems Section 1.5 Estimates and Order-of-Magnitude Calculations 34. An automobile tire is rated to last for 50 000 miles. To an order of magnitude, through how many revolutions will it turn over its lifetime? Note: Appendix B.8 on propagation of uncertainty may be useful in solving some problems in this section. θ 9.00 m 46. Review. Prove that one solution of the equation 6.00 m φ Figure P1.45 2.00x 4 2 3.00x 3 1 5.00x 5 70.0 is x 5 22.22. 47. Review. A pet lamb grows rapidly, with its mass proM portional to the cube of its length. When the lamb’s length changes by 15.8%, its mass increases by 17.3 kg. Find the lamb’s mass at the end of this process. s. w Section 1.6 Significant Figures 45. Review. For the right triM angle shown in Figure P1.45, what are (a) the length of the unknown side, (b) the tangent of u, and (c) the sine of f? om 33. To an order of magnitude, how many piano tuners reside in New York City? The physicist Enrico Fermi was famous for asking questions like this one on oral Ph.D. qualifying examinations. 44. Review. Find every angle u between 0 and 360° for which the ratio of sin u to cos u is 23.00. .c 32. (a) Compute the order of magnitude of the mass of a bathtub half full of water. (b) Compute the order of magnitude of the mass of a bathtub half full of copper coins. ly 31. Find the order of magnitude of the number of tabletennis balls that would fit into a typical-size room (without being crushed). 43. Review. The ratio of the number of sparrows visiting a bird feeder to the number of more interesting birds is 2.25. On a morning when altogether 91 birds visit the feeder, what is the number of sparrows? ee b Note: In your solutions to Problems 31 through 34, state the quantities you measure or estimate and the values you take for them. that of Uranus is 1.19. The ratio of the radius of Neptune to that of Uranus is 0.969. Find the average density of Neptune. 48. Review. A highway curve forms a section of a circle. A car goes around the curve as shown in the helicopter view of Figure P1.48. Its dashboard compass shows that the car is initially heading due east. After it travels d 5 840 m, it is heading u 5 35.0° south of east. Find the radius of curvature of its path. Suggestion: You may find it useful to learn a geometric theorem stated in Appendix B.3. ic 35. A rectangular plate has a length of (21.3 6 0.2) cm and a width of (9.8 6 0.1) cm. Calculate the area of the plate, including its uncertainty. ys 36. How many significant figures are in the following num9 26 W bers? (a) 78.9 6 0.2 (b) 3.788 3 10 (c) 2.46 3 10 (d) 0.005 3 ar ph 37. The tropical year, the time interval from one vernal equinox to the next vernal equinox, is the basis for our calendar. It contains 365.242 199 days. Find the number of seconds in a tropical year. d sw 38. Carry out the arithmetic operations (a) the sum of the W measured values 756, 37.2, 0.83, and 2; (b) the product 0.003 2 3 356.3; and (c) the product 5.620 3 p. N W Note: The next 13 problems call on mathematical skills from your prior education that will be useful throughout this course. 40. Review. While you are on a trip to Europe, you must purchase hazelnut chocolate bars for your grandmother. Eating just one square each day, she makes each large bar last for one and one-third months. How many bars will constitute a year’s supply for her? 41. Review. A child is surprised that because of sales tax she must pay $1.36 for a toy marked $1.25. What is the effective tax rate on this purchase, expressed as a percentage? 42. Review. The average density of the planet Uranus is 1.27 3 103 kg/m3. The ratio of the mass of Neptune to u S .a w w w 39. Review. In a community college parking lot, the number of ordinary cars is larger than the number of sport utility vehicles by 94.7%. The difference between the number of cars and the number of SUVs is 18. Find the number of SUVs in the lot. E Figure P1.48 49.   Review. From the set of equations S p 5 3q pr 5 qs 1 2 2 pr 1 1 2 2 qs 5 12 qt 2 involving the unknowns p, q, r, s, and t, find the value of the ratio of t to r. 50. Review. Figure P1.50 on page 18 shows students study- Q/C ing the thermal conduction of energy into cylindrical S blocks of ice. As we will see in Chapter 20, this process is described by the equation Q Dt 5 k pd 2 1 Th 2 Tc 2 4L For experimental control, in one set of trials all quantities except d and Dt are constant. (a) If d is made three 18 Chapter 1 Physics and Measurement times larger, does the equation predict that Dt will get larger or get smaller? By what factor? (b) What pattern of proportionality of Dt to d does the equation predict? (c) To display this proportionality as a straight line on a graph, what quantities should you plot on the horizontal and vertical axes? (d) What expression represents the theoretical slope of this graph? If the sidewalk is to measure (1.00 6 0.01) m wide by (9.0 6 0.1) cm thick, what volume of concrete is needed and what is the approximate uncertainty of this volume? Additional Problems 54. Collectible coins are sometimes plated with gold to Q/C enhance their beauty and value. Consider a commemo- ly .c Alexandra Héder om rative quarter-dollar advertised for sale at $4.98. It has a diameter of 24.1 mm and a thickness of 1.78 mm, and it is completely covered with a layer of pure gold 0.180 mm thick. The volume of the plating is equal to the thickness of the layer multiplied by the area to which it is applied. The patterns on the faces of the coin and the grooves on its edge have a negligible effect on its area. Assume the price of gold is $25.0 per gram. (a) Find the cost of the gold added to the coin. (b) Does the cost of the gold significantly enhance the value of the coin? Explain your answer. ar ph ys ic ance. He cuts out various shapes in various sizes, calculates their areas, measures their masses, and prepares the graph of Figure P1.51. (a) Consider the fourth experimental point from the top. How far is it from the best-fit straight line? Express your answer as a difference in vertical-axis coordinate. (b) Express your answer as a percentage. (c) Calculate the slope of the line. (d) State what the graph demonstrates, referring to the shape of the graph and the results of parts (b) and (c). (e) Describe whether this result should be expected theoretically. (f) Describe the physical meaning of the slope. ee b 51. Review. A student is supplied with a stack of copy Q/C paper, ruler, compass, scissors, and a sensitive bal- 55. In a situation in which data are known to three significant digits, we write 6.379 m 5 6.38 m and 6.374 m 5 6.37 m. When a number ends in 5, we arbitrarily choose to write 6.375 m 5 6.38 m. We could equally well write 6.375 m 5 6.37 m, “rounding down” instead of “rounding up,” because we would change the number 6.375 by equal increments in both cases. Now consider an orderof-magnitude estimate, in which factors of change rather than increments are important. We write 500 m , 103 m because 500 differs from 100 by a factor of 5 while it differs from 1 000 by only a factor of 2. We write 437 m , 103 m and 305 m , 102 m. What distance differs from 100 m and from 1 000 m by equal factors so that we could equally well choose to represent its order of magnitude as , 102 m or as , 103 m? s. w Figure P1.50 sw Dependence of mass on area for paper shapes .a 0.2 w 0.1 w 0 200 400 Area (cm2) Rectangles Squares Circles tinal volume and assume 1% of it is occupied by bacteria. (b) Does the number of bacteria suggest whether the bacteria are beneficial, dangerous, or neutral for the human body? What functions could they serve? 57. The diameter of our disk-shaped galaxy, the Milky Way, is about 1.0 3 105 light-years (ly). The distance to the Andromeda galaxy (Fig. P1.57), which is the spiral galaxy nearest to the Milky Way, is about 2.0 million ly. If a scale model represents the Milky Way and Andromeda w Mass (g) 0.3 56. (a) What is the order of magnitude of the number of BIO microorganisms in the human intestinal tract? A typiQ/C cal bacterial length scale is 1026 m. Estimate the intes- 600 Triangles Best fit 52. The radius of a uniform solid sphere is measured to be (6.50 6 0.20) cm, and its mass is measured to be (1.85 6 0.02) kg. Determine the density of the sphere in kilograms per cubic meter and the uncertainty in the density. 53. A sidewalk is to be constructed around a swimming pool that measures (10.0 6 0.1) m by (17.0 6 0.1) m. Robert Gendler/NASA Figure P1.51 Figure P1.57 The Andromeda galaxy. Problems galaxies as dinner plates 25 cm in diameter, determine the distance between the centers of the two plates. 58. Why is the following situation impossible? In an effort to boost interest in a television game show, each weekly winner is offered an additional $1 million bonus prize if he or she can personally count out that exact amount from a supply of one-dollar bills. The winner must do this task under supervision by television show executives and within one 40-hour work week. To the dismay of the show’s producers, most contestants succeed at the challenge. 19 a disk of diameter , 1021 m and thickness , 1019 m. Find the order of magnitude of the number of stars in the Milky Way. Assume the distance between the Sun and our nearest neighbor is typical. 63. Assume there are 100 million passenger cars in the AMT United States and the average fuel efficiency is 20 mi/gal M of gasoline. If the average distance traveled by each car is 10 000 mi/yr, how much gasoline would be saved per year if the average fuel efficiency could be increased to 25 mi/gal? 64. A spherical shell has an outside radius of 2.60 cm and Q/C an inside radius of a. The shell wall has uniform thick- 60. A water fountain is at the center of a circular pool S as shown in Figure P1.59. A student walks around the pool and measures its circumference C. Next, he stands at the edge of the pool and uses a protractor to measure the angle of elevation f of his sightline to the top of the water jet. How high is the fountain? BIO ground, in water, and in the air. One micron (1026 m) om 59. A high fountain of water AMT is located at the center M of a circular pool as shown in Figure P1.59. A student walks around f the pool and measures its circumference to be 15.0 m. Next, the student stands at the edge Figure P1.59 of the pool and uses a Problems 59 and 60. protractor to gauge the angle of elevation of the top of the fountain to be f 5 55.0°. How high is the fountain? ee b ly .c ness and is made of a material with density 4.70 g/cm3. The space inside the shell is filled with a liquid having a density of 1.23 g/cm3. (a) Find the mass m of the sphere, including its contents, as a function of a. (b) For what value of the variable a does m have its maximum possible value? (c) What is this maximum mass? (d) Explain whether the value from part (c) agrees with the result of a direct calculation of the mass of a solid sphere of uniform density made of the same material as the shell. (e) What If? Would the answer to part (a) change if the inner wall were not concentric with the outer wall? ys ic s. w 65. Bacteria and other prokaryotes are found deep under- ar ph 61. The data in the following table represent measurements Q/C of the masses and dimensions of solid cylinders of alu- sw minum, copper, brass, tin, and iron. (a) Use these data to calculate the densities of these substances. (b) State how your results compare with those given in Table 14.1. Mass (g) Substance w w .a Aluminum  51.5 Copper  56.3 Brass  94.4 Tin  69.1 Iron 216.1 Diameter (cm) Length (cm) 2.52 1.23 1.54 1.75 1.89 3.75 5.06 5.69 3.74 9.77 is a typical length scale associated with these microbes. (a) Estimate the total number of bacteria and other prokaryotes on the Earth. (b) Estimate the total mass of all such microbes. 66. Air is blown into a spherical balloon so that, when its Q/C radius is 6.50 cm, its radius is increasing at the rate 0.900 cm/s. (a) Find the rate at which the volume of the balloon is increasing. (b) If this volume flow rate of air entering the balloon is constant, at what rate will the radius be increasing when the radius is 13.0 cm? (c) Explain physically why the answer to part (b) is larger or smaller than 0.9 cm/s, if it is different. 67. A rod extending between x 5 0 and x 5 14.0 cm has uniform cross-sectional area A 5 9.00 cm2. Its density increases steadily between its ends from 2.70 g/cm3 to 19.3 g/cm3. (a) Identify the constants B and C required in the expression r 5 B 1 Cx to describe the variable density. (b) The mass of the rod is given by 2 3 r dV 5 3 rA dx 5 3 1 B 1 Cx 2 1 9.00 cm 2 dx 14.0 cm w 62. The distance from the Sun to the nearest star is about 4 3 1016 m. The Milky Way galaxy (Fig. P1.62) is roughly m5 all material all x 0 Carry out the integration to find the mass of the rod. 68. In physics, it is important to use mathematical approximations. (a) Demonstrate that for small angles (, 20°) Richard Payne/NASA tan a < sin a < a 5 par 1808 where a is in radians and a9 is in degrees. (b) Use a calculator to find the largest angle for which tan a may be approximated by a with an error less than 10.0%. Figure P1.62 The Milky Way galaxy. 69. The consumption of natural gas by a company satisfies 2 M the empirical equation V 5 1.50t 1 0.008 00t , where V 20 Chapter 1 Physics and Measurement om s. w section of the bottle is circular, but the diameters of the circles have different values. You pour the brightly colored shampoo into the bottle at a constant rate of 16.5 cm3/s. At what rate is its level in the bottle rising (a) at a point where the diameter of the bottle is 6.30 cm and (b) at a point where the diameter is 1.35 cm? 73. You stand in a flat meadow and observe two cows (Fig. P1.73). Cow A is due north of you and 15.0 m from your position. Cow B is 25.0 m from your position. From your point of view, the angle between cow A and cow B is 20.0°, with cow B appearing to the right of cow A. (a) How far apart are cow A and cow B? (b) Consider the view seen by cow A. According to this cow, what is the angle between you and cow B? (c) Consider the view seen by cow B. According to this cow, what is the angle between you and cow A? Hint: What does the situation look like to a hummingbird hovering above the meadow? (d) Two stars in the sky appear to be 20.0° apart. Star A is 15.0 ly from the Earth, and star B, appearing to the right of star A, is 25.0 ly from the Earth. To an inhabitant of a planet orbiting star A, what is the angle in the sky between star B and our Sun? .c 71. A child loves to watch as you fill a transparent plastic AMT bottle with shampoo (Fig P1.71). Every horizontal cross 72. A woman stands at a horizontal distance x from a S mountain and measures the angle of elevation of the mountaintop above the horizontal as u. After walking a distance d closer to the mountain on level ground, she finds the angle to be f. Find a general equation for the height y of the mountain in terms of d, f, and u, neglecting the height of her eyes above the ground. ly 70. A woman wishing to know the height of a mountain GP measures the angle of elevation of the mountaintop as 12.0°. After walking 1.00 km closer to the mountain on level ground, she finds the angle to be 14.0°. (a) Draw a picture of the problem, neglecting the height of the woman’s eyes above the ground. Hint: Use two triangles. (b) Using the symbol y to represent the mountain height and the symbol x to represent the woman’s original distance from the mountain, label the picture. (c) Using the labeled picture, write two trigonometric equations relating the two selected variables. (d) Find the height y. Challenge Problems ee b is the volume of gas in millions of cubic feet and t is the time in months. Express this equation in units of cubic feet and seconds. Assume a month is 30.0 days. w w w ar ph .a sw Figure P1.71 ys 6.30 cm 1.35 cm Cow B ic Cow A Figure P1.73 Your view of two cows in a meadow. Cow A is due north of you. You must rotate your eyes through an angle of 20.0° to look from cow A to cow B. c h a p t e r 2 om Motion in One Dimension 2.1 Position, Velocity, and Speed .c 2.2 Instantaneous Velocity and Speed 2.4 Acceleration 2.5 Motion Diagrams 2.6 Analysis Model: Particle Under Constant Acceleration 2.7 Freely Falling Objects 2.8 Kinematic Equations Derived from Calculus General Problem-Solving Strategy sw ar ph ys ic s. w ee b ly 2.3 Analysis Model: Particle Under Constant Velocity In drag racing, a driver wants as large an acceleration as possible. In a distance of one-quarter mile, a vehicle reaches speeds of more than 320 mi/h, covering the entire distance in under 5 s. (George Lepp/ Stone/Getty Images) w w w .a As a first step in studying classical mechanics, we describe the motion of an object while ignoring the interactions with external agents that might be affecting or modifying that motion. This portion of classical mechanics is called kinematics. (The word kinematics has the same root as cinema.) In this chapter, we consider only motion in one dimension, that is, motion of an object along a straight line. From everyday experience, we recognize that motion of an object represents a continuous change in the object’s position. In physics, we can categorize motion into three types: translational, rotational, and vibrational. A car traveling on a highway is an example of translational motion, the Earth’s spin on its axis is an example of rotational motion, and the back-and-forth movement of a pendulum is an example of vibrational motion. In this and the next few chapters, we are concerned only with translational motion. (Later in the book we shall discuss rotational and vibrational motions.) In our study of translational motion, we use what is called the particle model and describe the moving object as a particle regardless of its size. Remember our discussion of making models for physical situations in Section 1.2. In general, a particle is a point-like object, that is, an object that has mass but is of infinitesimal size. For example, if we wish to describe the motion of the Earth around the Sun, we can treat the Earth as a particle and 21 22 Chapter 2 Motion in One Dimension obtain reasonably accurate data about its orbit. This approximation is justified because the radius of the Earth’s orbit is large compared with the dimensions of the Earth and the Sun. As an example on a much smaller scale, it is possible to explain the pressure exerted by a gas on the walls of a container by treating the gas molecules as particles, without regard for the internal structure of the molecules. 2.1 Position, Velocity, and Speed A particle’s position x is the location of the particle with respect to a chosen reference point that we can consider to be the origin of a coordinate system. The motion of a particle is completely known if the particle’s position in space is known at all times. Consider a car moving back and forth along the x axis as in Figure 2.1a. When we begin collecting position data, the car is 30 m to the right of the reference position x 5 0. We will use the particle model by identifying some point on the car, perhaps the front door handle, as a particle representing the entire car. We start our clock, and once every 10 s we note the car’s position. As you can see from Table 2.1, the car moves to the right (which we have defined as the positive direction) during the first 10 s of motion, from position A to position B. After B, the position values begin to decrease, suggesting the car is backing up from position B through position F. In fact, at D, 30 s after we start measuring, the car is at the origin of coordinates (see Fig. 2.1a). It continues moving to the left and is more than 50 m to the left of x 5 0 when we stop recording information after our sixth data point. A graphical representation of this information is presented in Figure 2.1b. Such a plot is called a position–time graph. Notice the alternative representations of information that we have used for the motion of the car. Figure 2.1a is a pictorial representation, whereas Figure 2.1b is a graphical representation. Table 2.1 is a tabular representation of the same information. Using an alternative representation is often an excellent strategy for understanding the situation in a given problem. The ultimate goal in many problems is a math- ee b ly .c om Position  30 52 38 0 237 253 ic x (m) ys t (s) A  0 B 10 C 20 D 30 E 40 F 50 ar ph Position s. w Table 2.1 Position of the Car at Various Times w .a sw The car moves to the right between positions A and B. E w F w 60 50 40 30 20 10 60 50 40 30 20 10 0 10 20 A 30 D 0 x (m) 60 B 40 50 60 x (m) 10 20 The car moves to the left between positions C and F. 30 40 x 40 A 50 60 x (m) C t 20 C B D 0 20 E 40 60 F 0 10 20 30 40 50 t (s) b a Figure 2.1 A car moves back and forth along a straight line. Because we are interested only in the car’s translational motion, we can model it as a particle. Several representations of the information about the motion of the car can be used. Table 2.1 is a tabular representation of the information. (a) A pictorial representation of the motion of the car. (b) A graphical representation (position–time graph) of the motion of the car. 2.1 Position, Velocity, and Speed 23 ematical representation, which can be analyzed to solve for some requested piece of information. Given the data in Table 2.1, we can easily determine the change in position of the car for various time intervals. The displacement Dx of a particle is defined as its change in position in some time interval. As the particle moves from an initial position xi to a final position xf , its displacement is given by Dx ; xf 2 xi WW Displacement Figure 2.2 On this basketball court, players run back and forth for the entire game. The distance that the players run over the duration of the game is nonzero. The displacement of the players over the duration of the game is approximately zero because they keep returning to the same point over and over again. w w w .a sw ar ph ys ic s. w ee b ly .c om We use the capital Greek letter delta (D) to denote the change in a quantity. From this definition, we see that Dx is positive if xf is greater than xi and negative if xf is less than xi . It is very important to recognize the difference between displacement and distance traveled. Distance is the length of a path followed by a particle. Consider, for example, the basketball players in Figure 2.2. If a player runs from his own team’s basket down the court to the other team’s basket and then returns to his own basket, the displacement of the player during this time interval is zero because he ended up at the same point as he started: xf 5 xi , so Dx 5 0. During this time interval, however, he moved through a distance of twice the length of the basketball court. Distance is always represented as a positive number, whereas displacement can be either positive or negative. Displacement is an example of a vector quantity. Many other physical quantities, including position, velocity, and acceleration, also are vectors. In general, a vector quantity requires the specification of both direction and magnitude. By contrast, a scalar quantity has a numerical value and no direction. In this chapter, we use positive (1) and negative (2) signs to indicate vector direction. For example, for horizontal motion let us arbitrarily specify to the right as being the positive direction. It follows that any object always moving to the right undergoes a positive displacement Dx . 0, and any object moving to the left undergoes a negative displacement so that Dx , 0. We shall treat vector quantities in greater detail in Chapter 3. One very important point has not yet been mentioned. Notice that the data in Table 2.1 result only in the six data points in the graph in Figure 2.1b. Therefore, the motion of the particle is not completely known because we don’t know its position at all times. The smooth curve drawn through the six points in the graph is only a possibility of the actual motion of the car. We only have information about six instants of time; we have no idea what happened between the data points. The smooth curve is a guess as to what happened, but keep in mind that it is only a guess. If the smooth curve does represent the actual motion of the car, the graph contains complete information about the entire 50-s interval during which we watch the car move. It is much easier to see changes in position from the graph than from a verbal description or even a table of numbers. For example, it is clear that the car covers more ground during the middle of the 50-s interval than at the end. Between positions C and D, the car travels almost 40 m, but during the last 10 s, between positions E and F, it moves less than half that far. A common way of comparing these different motions is to divide the displacement Dx that occurs between two clock readings by the value of that particular time interval Dt. The result turns out to be a very useful ratio, one that we shall use many times. This ratio has been given a special name: the average velocity. The average velocity vx,avg of a particle is defined as the particle’s displacement Dx divided by the time interval Dt during which that displacement occurs: Brian Drake/Time Life Pictures/Getty Images (2.1) v x,avg ; Dx Dt (2.2) where the subscript x indicates motion along the x axis. From this definition we see that average velocity has dimensions of length divided by time (L/T), or meters per second in SI units. WW Average velocity 24 Chapter 2 Motion in One Dimension s. w ee b ly .c om The average velocity of a particle moving in one dimension can be positive or negative, depending on the sign of the displacement. (The time interval Dt is always positive.) If the coordinate of the particle increases in time (that is, if xf . xi ), Dx is positive and vx,avg 5 Dx/Dt is positive. This case corresponds to a particle moving in the positive x direction, that is, toward larger values of x. If the coordinate decreases in time (that is, if xf , xi ), Dx is negative and hence vx,avg is negative. This case corresponds to a particle moving in the negative x direction. We can interpret average velocity geometrically by drawing a straight line between any two points on the position–time graph in Figure 2.1b. This line forms the hypotenuse of a right triangle of height Dx and base Dt. The slope of this line is the ratio Dx/Dt, which is what we have defined as average velocity in Equation 2.2. For example, the line between positions A and B in Figure 2.1b has a slope equal to the average velocity of the car between those two times, (52 m 2 30 m)/(10 s 2 0) 5 2.2 m/s. In everyday usage, the terms speed and velocity are interchangeable. In physics, however, there is a clear distinction between these two quantities. Consider a marathon runner who runs a distance d of more than 40 km and yet ends up at her starting point. Her total displacement is zero, so her average velocity is zero! Nonetheless, we need to be able to quantify how fast she was running. A slightly different ratio accomplishes that for us. The average speed vavg of a particle, a scalar quantity, is defined as the total distance d traveled divided by the total time interval required to travel that distance: v avg ; Average speed  d Dt (2.3) ic The SI unit of average speed is the same as the unit of average velocity: meters per second. Unlike average velocity, however, average speed has no direction and is always expressed as a positive number. Notice the clear distinction between the definitions of average velocity and average speed: average velocity (Eq. 2.2) is the displacement divided by the time interval, whereas average speed (Eq. 2.3) is the distance divided by the time interval. Knowledge of the average velocity or average speed of a particle does not provide information about the details of the trip. For example, suppose it takes you 45.0 s to travel 100 m down a long, straight hallway toward your departure gate at an airport. At the 100-m mark, you realize you missed the restroom, and you return back 25.0 m along the same hallway, taking 10.0 s to make the return trip. The magnitude of your average velocity is 175.0 m/55.0 s 5 11.36 m/s. The average speed for your trip is 125 m/55.0 s 5 2.27 m/s. You may have traveled at various speeds during the walk and, of course, you changed direction. Neither average velocity nor average speed provides information about these details. ys Pitfall Prevention 2.1 w w sw w .a average velocity is not the average speed. For example, consider the marathon runner discussed before Equation 2.3. The magnitude of her average velocity is zero, but her average speed is clearly not zero. ar ph Average Speed and Average Velocity The magnitude of the Example 2.1 Q uick Quiz 2.1 Under which of the following conditions is the magnitude of the average velocity of a particle moving in one dimension smaller than the average speed over some time interval? (a) A particle moves in the 1x direction without reversing. (b) A particle moves in the 2x direction without reversing. (c) A particle moves in the 1x direction and then reverses the direction of its motion. (d) There are no conditions for which this is true.   Calculating the Average Velocity and Speed Find the displacement, average velocity, and average speed of the car in Figure 2.1a between positions A and F. 2.2 Instantaneous Velocity and Speed 25 ▸ 2.1 c o n t i n u e d Solution Consult Figure 2.1 to form a mental image of the car and its motion. We model the car as a particle. From the position– time graph given in Figure 2.1b, notice that x A 5 30 m at t A 5 0 s and that x F 5 253 m at t F 5 50 s. Dx 5 x F 2 x A 5 253 m 2 30 m 5 283 m Use Equation 2.1 to find the displacement of the car: v x,avg 5 tF 2 tA 253 m 2 30 m 283 m 5 5 21.7 m/s 50 s 2 0 s 50 s ly 5 xF 2 xA .c Use Equation 2.2 to find the car’s average velocity: om This result means that the car ends up 83 m in the negative direction (to the left, in this case) from where it started. This number has the correct units and is of the same order of magnitude as the supplied data. A quick look at Figure 2.1a indicates that it is the correct answer. v avg 5 127 m 5 2.5 m/s 50 s s. w Use Equation 2.3 to find the car’s average speed: ee b We cannot unambiguously find the average speed of the car from the data in Table 2.1 because we do not have information about the positions of the car between the data points. If we adopt the assumption that the details of the car’s position are described by the curve in Figure 2.1b, the distance traveled is 22 m (from A to B) plus 105 m (from B to F), for a total of 127 m. ar ph ys ic Notice that the average speed is positive, as it must be. Suppose the red-brown curve in Figure 2.1b were different so that between 0 s and 10 s it went from A up to 100 m and then came back down to B. The average speed of the car would change because the distance is different, but the average velocity would not change. 2.2 Instantaneous Velocity and Speed w w w .a sw Often we need to know the velocity of a particle at a particular instant in time t rather than the average velocity over a finite time interval Dt. In other words, you would like to be able to specify your velocity just as precisely as you can specify your position by noting what is happening at a specific clock reading, that is, at some specific instant. What does it mean to talk about how quickly something is moving if we “freeze time” and talk only about an individual instant? In the late 1600s, with the invention of calculus, scientists began to understand how to describe an object’s motion at any moment in time. To see how that is done, consider Figure 2.3a (page 26), which is a reproduction of the graph in Figure 2.1b. What is the particle’s velocity at t 5 0? We have already discussed the average velocity for the interval during which the car moved from position A to position B (given by the slope of the blue line) and for the interval during which it moved from A to F (represented by the slope of the longer blue line and calculated in Example 2.1). The car starts out by moving to the right, which we defined to be the positive direction. Therefore, being positive, the value of the average velocity during the interval from A to B is more representative of the initial velocity than is the value of the average velocity during the interval from A to F, which we determined to be negative in Example 2.1. Now let us focus on the short blue line and slide point B to the left along the curve, toward point A, as in Figure 2.3b. The line between the points becomes steeper and steeper, and as the two points become extremely close together, the line becomes a tangent line to the curve, indicated by the green line in Figure 2.3b. The slope of this tangent line 26 Chapter 2 Motion in One Dimension x (m) 60 60 B C 40 A B 20 D 0 The blue line between positions A and B approaches the green tangent line as point B is 40 20 E F 0 10 20 30 t (s) 50 40 a A moved closer to point A. om 40 60 B B B b Figure 2.3 (a) Graph representing the motion of the car in Figure 2.1. (b) An enlargement of the upper-left-hand corner of the graph. .c Pitfall Prevention 2.2 ly Slopes of Graphs In any graph of physical data, the slope represents the ratio of the change in the quantity represented on the vertical axis to the change in the quantity represented on the horizontal axis. Remember that a slope has units (unless both axes have the same units). The units of slope in Figures 2.1b and 2.3 are meters per second, the units of velocity. ee b represents the velocity of the car at point A. What we have done is determine the instantaneous velocity at that moment. In other words, the instantaneous velocity vx equals the limiting value of the ratio Dx/Dt as Dt approaches zero:1 v x ; lim S Dx Dt (2.4) ic In calculus notation, this limit is called the derivative of x with respect to t, written dx/dt: Dx dx v x ; lim 5 (2.5) S Dt 0 Dt dt The instantaneous velocity can be positive, negative, or zero. When the slope of the position–time graph is positive, such as at any time during the first 10 s in Figure 2.3, vx is positive and the car is moving toward larger values of x. After point B, vx is negative because the slope is negative and the car is moving toward smaller values of x. At point B, the slope and the instantaneous velocity are zero and the car is momentarily at rest. From here on, we use the word velocity to designate instantaneous velocity. When we are interested in average velocity, we shall always use the adjective average. The instantaneous speed of a particle is defined as the magnitude of its instantaneous velocity. As with average speed, instantaneous speed has no direction associated with it. For example, if one particle has an instantaneous velocity of 125 m/s along a given line and another particle has an instantaneous velocity of 225 m/s along the same line, both have a speed2 of 25 m/s. ys Instantaneous velocity  Pitfall Prevention 2.3 w w w .a sw ar ph Instantaneous Speed and Instantaneous Velocity In Pitfall Prevention 2.1, we argued that the magnitude of the average velocity is not the average speed. The magnitude of the instantaneous velocity, however, is the instantaneous speed. In an infinitesimal time interval, the magnitude of the displacement is equal to the distance traveled by the particle. 0 s. w Dt Conceptual Example 2.2 Q uick Quiz 2.2 Are members of the highway patrol more interested in (a) your average speed or (b) your instantaneous speed as you drive?    The Velocity of Different Objects Consider the following one-dimensional motions: (A) a ball thrown directly upward rises to a highest point and falls back into the thrower’s hand; (B) a race car starts from rest and speeds up to 100 m/s; and (C) a spacecraft drifts through space at constant velocity. Are there any points in the motion of these objects at which the instantaneous velocity has the same value as the average velocity over the entire motion? If so, identify the point(s). 1Notice that the displacement Dx also approaches zero as Dt approaches zero, so the ratio looks like 0/0. While this ratio may appear to be difficult to evaluate, the ratio does have a specific value. As Dx and Dt become smaller and smaller, the ratio Dx/Dt approaches a value equal to the slope of the line tangent to the x-versus-t curve. 2 As with velocity, we drop the adjective for instantaneous speed. Speed means “instantaneous speed.” 27 2.2 Instantaneous Velocity and Speed ▸ 2.2 c o n t i n u e d Solution (A) The average velocity for the thrown ball is zero because the ball returns to the starting point; therefore, its displacement is zero. There is one point at which the instantaneous velocity is zero: at the top of the motion. (B) The car’s average velocity cannot be evaluated unambiguously with the information given, but it must have some value between 0 and 100 m/s. Because the car will have every instantaneous velocity between 0 and 100 m/s at some time during the interval, there must be some instant at which the instantaneous velocity is equal to the average velocity over the entire motion. ly    Average and Instantaneous Velocity ee b Example 2.3 .c om (C) Because the spacecraft’s instantaneous velocity is constant, its instantaneous velocity at any time and its average velocity over any time interval are the same. ic s. w A particle moves along the x axis. Its position varies with time according to the expression x 5 24t 1 2t 2, where x is in meters and t is in seconds.3 The position–time graph for this motion is shown in Figure 2.4a. Because the position of the particle is given by a mathematical function, the motion of the particle is completely known, unlike that of the car in Figure 2.1. Notice that the particle moves in the negative x direction for the first second of motion, is momentarily at rest at the moment t 5 1 s, and moves in the positive x direction at times t . 1 s. ar ph ys (A) Determine the displacement of the particle in the time intervals t 5 0 to t 5 1 s and t 5 1 s to t 5 3 s. Solution 8 Slope  4 m/s 6 4 2 0 C A 2 t (s) B 4 0 a 4 D Slope  2 m/s 1 2 B A 2 0 3 C 2 4 D 4 6 8 x b Figure 2.4 (Example 2.3) (a) ­Position– time graph for a particle having an x coordinate that varies in time according to the expression x 5 24t 1 2t 2. (b) The particle moves in one dimension along the x axis. w w w .a sw From the graph in Figure 2.4a, form a mental representation of the particle’s motion. Keep in mind that the particle does not move in a curved path in space such as that shown by the red-brown curve in the graphical representation. The particle moves only along the x axis in one dimension as shown in Figure 2.4b. At t 5 0, is it moving to the right or to the left? During the first time interval, the slope is negative and hence the average velocity is negative. Therefore, we know that the displacement between A and B must be a negative number having units of meters. Similarly, we expect the displacement between B and D to be positive. x (m) 10 In the first time interval, set ti 5 t A 5 0 and tf 5 t B 5 1 s and use Equation 2.1 to find the displacement: Dx ASB 5 xf 2 xi 5 x B 2 x A For the second time interval (t 5 1 s to t 5 3 s), set ti 5 t B 5 1 s and tf 5 t D 5 3 s: Dx BSD 5 xf 2 xi 5 x D 2 x B 5 [24(1) 1 2(1)2] 2 [24(0) 1 2(0)2] 5 22 m 5 [24(3) 1 2(3)2] 2 [24(1) 1 2(1)2] 5 18 m These displacements can also be read directly from the position–time graph. (B) Calculate the average velocity during these two time intervals. continued 3 Simply to make it easier to read, we write the expression as x 5 24t 1 2t 2 rather than as x 5 (24.00 m/s)t 1 (2.00 m/s2)t 2.00. When an equation summarizes measurements, consider its coefficients and exponents to have as many significant figures as other data quoted in a problem. Consider its coefficients to have the units required for dimensional consistency. When we start our clocks at t 5 0, we usually do not mean to limit the precision to a single digit. Consider any zero value in this book to have as many significant figures as you need. 28 Chapter 2 Motion in One Dimension ▸ 2.3 c o n t i n u e d Solution In the first time interval, use Equation 2.2 with Dt 5 tf 2 ti 5 t B 2 t A 5 1 s: v x,avg 1A S B 2 5 In the second time interval, Dt 5 2 s: v x,avg 1B S D 2 5 Dx A S B Dt Dx B S D Dt 5 22 m 5 22 m/s 1s 5 8m 5 14 m/s 2s These values are the same as the slopes of the blue lines joining these points in Figure 2.4a. om (C) Find the instantaneous velocity of the particle at t 5 2.5 s. Solution 10 m 2 1 24 m 2 5 16 m/s 3.8 s 2 1.5 s ly .c vx 5 Measure the slope of the green line at t 5 2.5 s (point C) in Figure 2.4a: s. w ee b Notice that this instantaneous velocity is on the same order of magnitude as our previous results, that is, a few meters per second. Is that what you would have expected? 2.3 Analysis Model: Particle Under Constant Velocity ic In Section 1.2 we discussed the importance of making models. A particularly important model used in the solution to physics problems is an analysis model. An analysis model is a common situation that occurs time and again when solving physics problems. Because it represents a common situation, it also represents a common type of problem that we have solved before. When you identify an analysis model in a new problem, the solution to the new problem can be modeled after that of the previously-solved problem. Analysis models help us to recognize those common situations and guide us toward a solution to the problem. The form that an analysis model takes is a description of either (1) the behavior of some physical entity or (2) the interaction between that entity and the environment. When you encounter a new problem, you should identify the fundamental details of the problem and attempt to recognize which of the situations you have already seen that might be used as a model for the new problem. For example, suppose an automobile is moving along a straight freeway at a constant speed. Is it important that it is an automobile? Is it important that it is a freeway? If the answers to both questions are no, but the car moves in a straight line at constant speed, we model the automobile as a particle under constant velocity, which we will discuss in this section. Once the problem has been modeled, it is no longer about an automobile. It is about a particle undergoing a certain type of motion, a motion that we have studied before. This method is somewhat similar to the common practice in the legal profession of finding “legal precedents.” If a previously resolved case can be found that is very similar legally to the current one, it is used as a model and an argument is made in court to link them logically. The finding in the previous case can then be used to sway the finding in the current case. We will do something similar in physics. For a given problem, we search for a “physics precedent,” a model with which we are already familiar and that can be applied to the current problem. All of the analysis models that we will develop are based on four fundamental simplification models. The first of the four is the particle model discussed in the introduction to this chapter. We will look at a particle under various behaviors and environmental interactions. Further analysis models are introduced in later chapters based on simplification models of a system, a rigid object, and a wave. Once w w w .a sw ar ph ys Analysis model  29 2.3 Analysis Model: Particle Under Constant Velocity x Slope  xi x  vx t t Figure 2.5 Position–time graph for a particle under constant velocity. The value of the constant velocity is the slope of the line. ee b ly .c om we have introduced these analysis models, we shall see that they appear again and again in different problem situations. When solving a problem, you should avoid browsing through the chapter looking for an equation that contains the unknown variable that is requested in the problem. In many cases, the equation you find may have nothing to do with the problem you are attempting to solve. It is much better to take this first step: Identify the analysis model that is appropriate for the problem. To do so, think carefully about what is going on in the problem and match it to a situation you have seen before. Once the analysis model is identified, there are a small number of equations from which to choose that are appropriate for that model, sometimes only one equation. Therefore, the model tells you which equation(s) to use for the mathematical representation. Let us use Equation 2.2 to build our first analysis model for solving problems. We imagine a particle moving with a constant velocity. The model of a particle under constant velocity can be applied in any situation in which an entity that can be modeled as a particle is moving with constant velocity. This situation occurs frequently, so this model is important. If the velocity of a particle is constant, its instantaneous velocity at any instant during a time interval is the same as the average velocity over the interval. That is, vx 5 vx,avg. Therefore, Equation 2.2 gives us an equation to be used in the mathematical representation of this situation: Dx Dt Remembering that Dx 5 xf 2 xi , we see that vx 5 (xf 2 xi)/Dt, or xf 5 xi 1 vx  Dt (2.6) s. w vx 5 ys ic This equation tells us that the position of the particle is given by the sum of its original position xi at time t 5 0 plus the displacement vx Dt that occurs during the time interval Dt. In practice, we usually choose the time at the beginning of the interval to be ti 5 0 and the time at the end of the interval to be tf 5 t, so our equation becomes (2.7) ar ph xf 5 xi 1 vxt (for constant vx) w w w .a sw Equations 2.6 and 2.7 are the primary equations used in the model of a particle under constant velocity. Whenever you have identified the analysis model in a problem to be the particle under constant velocity, you can immediately turn to these equations. Figure 2.5 is a graphical representation of the particle under constant velocity. On this position–time graph, the slope of the line representing the motion is constant and equal to the magnitude of the velocity. Equation 2.7, which is the equation of a straight line, is the mathematical representation of the particle under constant velocity model. The slope of the straight line is vx and the y intercept is xi in both representations. Example 2.4 below shows an application of the particle under constant velocity model. Notice the analysis model icon AM , which will be used to identify examples in which analysis models are employed in the solution. Because of the widespread benefits of using the analysis model approach, you will notice that a large number of the examples in the book will carry such an icon. WW Position as a function of time for the particle under ­constant velocity model Example 2.4    Modeling a Runner as a Particle AM A kinesiologist is studying the biomechanics of the human body. (Kinesiology is the study of the movement of the human body. Notice the connection to the word kinematics.) She determines the velocity of an experimental subject while he runs along a straight line at a constant rate. The kinesiologist starts the stopwatch at the moment the runner passes a given point and stops it after the runner has passed another point 20 m away. The time interval indicated on the stopwatch is 4.0 s. (A) What is the runner’s velocity? continued 30 Chapter 2 Motion in One Dimension ▸ 2.4 c o n t i n u e d Solution We model the moving runner as a particle because the size of the runner and the movement of arms and legs are unnecessary details. Because the problem states that the subject runs at a constant rate, we can model him as a particle under constant velocity. Having identified the model, we can use Equation 2.6 to find the constant velocity of the runner: vx 5 xf 2 xi Dx 20 m 2 0 5 5 5 5.0 m/s Dt Dt 4.0 s om (B) If the runner continues his motion after the stopwatch is stopped, what is his position after 10 s have passed? Solution xf 5 xi 1 vxt 5 0 1 (5.0 m/s)(10 s) 5 50 m ly .c Use Equation 2.7 and the velocity found in part (A) to find the position of the particle at time t 5 10 s: s. w ee b Is the result for part (A) a reasonable speed for a human? How does it compare to world-record speeds in 100-m and 200-m sprints? Notice the value in part (B) is more than twice that of the 20-m position at which the stopwatch was stopped. Is this value consistent with the time of 10 s being more than twice the time of 4.0 s? sw ar ph ys ic The mathematical manipulations for the particle under constant velocity stem from Equation 2.6 and its descendent, Equation 2.7. These equations can be used to solve for any variable in the equations that happens to be unknown if the other variables are known. For example, in part (B) of Example 2.4, we find the position when the velocity and the time are known. Similarly, if we know the velocity and the final position, we could use Equation 2.7 to find the time at which the runner is at this position. A particle under constant velocity moves with a constant speed along a straight line. Now consider a particle moving with a constant speed through a distance d along a curved path. This situation can be represented with the model of a particle under constant speed. The primary equation for this model is Equation 2.3, with the average speed v avg replaced by the constant speed v: d (2.8) Dt As an example, imagine a particle moving at a constant speed in a circular path. If the speed is 5.00 m/s and the radius of the path is 10.0 m, we can calculate the time interval required to complete one trip around the circle: w w .a v5 w v5 d Dt S Dt 5 2p 1 10.0 m 2 d 2pr 5 5 5 12.6 s v v 5.00 m/s Analysis Model   Particle Under Constant Velocity Imagine a moving object that can be modeled as a particle. If it moves at a constant speed through a displacement Dx in a straight line in a time interval Dt, its constant velocity is Dx (2.6) Dt The position of the particle as a function of time is given by vx 5 xf 5 xi 1 vxt v (2.7) Examples: • a meteoroid traveling through gravity-free space • a car traveling at a constant speed on a straight highway • a runner traveling at constant speed on a perfectly straight path • an object moving at terminal speed through a viscous medium (Chapter 6) 2.4 Acceleration 31 Analysis Model   Particle Under Constant Speed Examples: Imagine a moving object that can be modeled as a particle. If it moves at a constant speed through a distance d along a straight line or a curved path in a time interval Dt, its constant speed is v5 d Dt • a planet traveling around a perfectly circular orbit • a car traveling at a constant speed on a curved racetrack • a runner traveling at constant speed on a curved path • a charged particle moving through a uniform magnetic field (Chapter 29) (2.8) .c om v ee b ly 2.4 Acceleration v xf 2 v xi Dv x 5 Dt tf 2 ti (2.9) ar ph a x,avg ; ys ic s. w In Example 2.3, we worked with a common situation in which the velocity of a particle changes while the particle is moving. When the velocity of a particle changes with time, the particle is said to be accelerating. For example, the magnitude of a car’s velocity increases when you step on the gas and decreases when you apply the brakes. Let us see how to quantify acceleration. Suppose an object that can be modeled as a particle moving along the x axis has an initial velocity vxi at time ti at position A and a final velocity vxf at time tf at position B as in Figure 2.6a. The red-brown curve in Figure 2.6b shows how the velocity varies with time. The average acceleration ax,avg of the particle is defined as the change in velocity Dvx divided by the time interval Dt during which that change occurs: WW Average acceleration w w w .a sw As with velocity, when the motion being analyzed is one dimensional, we can use positive and negative signs to indicate the direction of the acceleration. Because the dimensions of velocity are L/T and the dimension of time is T, acceleration has dimensions of length divided by time squared, or L/T2. The SI unit of acceleration is meters per second squared (m/s2). It might be easier to interpret these units if you think of them as meters per second per second. For example, suppose an object has an acceleration of 12 m/s2. You can interpret this value by forming a mental image of the object having a velocity that is along a straight line and is increasing by 2 m/s during every time interval of 1 s. If the object starts from rest, The slope of the green line is the instantaneous acceleration of the car at point B (Eq. 2.10). vx The car moves with different velocities at points A and B. A ti v  vxi a B vxf B vxi vx A t x tf v  vxf ti b tf t The slope of the blue line connecting A and B is the average acceleration of the car during the time interval t  tf  ti (Eq. 2.9). Figure 2.6 (a) A car, modeled as a particle, moving along the x axis from A to B, has velocity vxi at t 5 ti and velocity vxf at t 5 tf . (b) Velocity–time graph (redbrown) for the particle moving in a straight line. 32 Chapter 2 Motion in One Dimension you should be able to picture it moving at a velocity of 12 m/s after 1 s, at 14 m/s after 2 s, and so on. In some situations, the value of the average acceleration may be different over different time intervals. It is therefore useful to define the instantaneous acceleration as the limit of the average acceleration as Dt approaches zero. This concept is analogous to the definition of instantaneous velocity discussed in Section 2.2. If we imagine that point A is brought closer and closer to point B in Figure 2.6a and we take the limit of Dvx /Dt as Dt approaches zero, we obtain the instantaneous acceleration at point B: Dv x dv x a x ; lim (2.10) 5 Dt S 0 Dt dt om Instantaneous acceleration  ee b ly .c That is, the instantaneous acceleration equals the derivative of the velocity with respect to time, which by definition is the slope of the velocity–time graph. The slope of the green line in Figure 2.6b is equal to the instantaneous acceleration at point B. Notice that Figure 2.6b is a velocity–time graph, not a position–time graph like Figures 2.1b, 2.3, 2.4, and 2.5. Therefore, we see that just as the velocity of a moving particle is the slope at a point on the particle’s x–t graph, the acceleration of a particle is the slope at a point on the particle’s vx –t graph. One can interpret the derivative of the velocity with respect to time as the time rate of change of velocity. If ax is positive, the acceleration is in the positive x direction; if ax is negative, the acceleration is in the negative x direction. Figure 2.7 illustrates how an acceleration–time graph is related to a velocity– time graph. The acceleration at any time is the slope of the velocity–time graph at that time. Positive values of acceleration correspond to those points in Figure 2.7a where the velocity is increasing in the positive x direction. The acceleration reaches a maximum at time t A, when the slope of the velocity–time graph is a maximum. The acceleration then goes to zero at time t B, when the velocity is a maximum (that is, when the slope of the vx –t graph is zero). The acceleration is negative when the velocity is decreasing in the positive x direction, and it reaches its most negative value at time t C. tB t tC ic tA s. w vx ys a ar ph The acceleration at any time equals the slope of the line tangent to the curve of vx versus t at that time. Q uick Quiz 2.3 Make a velocity–time graph for the car in Figure 2.1a. Suppose the speed limit for the road on which the car is driving is 30 km/h. True or False? The car exceeds the speed limit at some time within the time interval 0 2 50 s. ax t tB For the case of motion in a straight line, the direction of the velocity of an object and the direction of its acceleration are related as follows. When the object’s velocity and acceleration are in the same direction, the object is speeding up. On the other hand, when the object’s velocity and acceleration are in opposite directions, the object is slowing down. To help with this discussion of the signs of velocity and acceleration, we can relate the acceleration of an object to the total force exerted on the object. In Chapter 5, we formally establish that the force on an object is proportional to the acceleration of the object: F x ~ ax (2.11) .a b Figure 2.7 sw tC tA w w w (a) The velocity–time graph for a particle moving along the x axis. (b) The instantaneous acceleration can be obtained from the velocity–time graph. This proportionality indicates that acceleration is caused by force. Furthermore, force and acceleration are both vectors, and the vectors are in the same direction. Therefore, let us think about the signs of velocity and acceleration by imagining a force applied to an object and causing it to accelerate. Let us assume the velocity and acceleration are in the same direction. This situation corresponds to an object that experiences a force acting in the same direction as its velocity. In this case, the object speeds up! Now suppose the velocity and acceleration are in opposite directions. In this situation, the object moves in some direction and experiences a force acting in the opposite direction. Therefore, the object slows 2.4 Acceleration down! It is very useful to equate the direction of the acceleration to the direction of a force because it is easier from our everyday experience to think about what effect a force will have on an object than to think only in terms of the direction of the acceleration. Q uick Quiz 2.4 If a car is traveling eastward and slowing down, what is the direction of the force on the car that causes it to slow down? (a) eastward (b) westward (c) neither eastward nor westward 33 Pitfall Prevention 2.4 Negative Acceleration Keep in mind that negative acceleration does not necessarily mean that an object is slowing down. If the acceleration is negative and the velocity is negative, the object is speeding up! Pitfall Prevention 2.5 Deceleration The word deceleration has the common popular connotation of slowing down. We will not use this word in this book because it confuses the definition we have given for negative acceleration. .c om From now on, we shall use the term acceleration to mean instantaneous acceleration. When we mean average acceleration, we shall always use the adjective average. Because vx 5 dx/dt, the acceleration can also be written as dv x d dx d 2x ax 5 5 a b5 2 (2.12) dt dt dt dt Conceptual Example 2.5 ee b ly That is, in one-dimensional motion, the acceleration equals the second derivative of x with respect to time.   Graphical Relationships Between x, v x , and ax s. w The position of an object moving along the x axis varies with time as in Figure 2.8a. Graph the velocity versus time and the acceleration versus time for the object. ic Solution x w w w .a sw ar ph ys The velocity at any instant is the slope of the tangent to the x–t graph at that instant. Between t 5 0 and t 5 t A, the slope of the x–t graph increases uniformly, so the velocity increases linearly as shown in Figure 2.8b. Between t A and t B, the slope of the x–t graph is constant, so the velocity remains constant. Between t B and t tA tB tC tD tE tF t D, the slope of the x–t graph decreases, so the value of a the velocity in the vx –t graph decreases. At t D, the slope vx of the x–t graph is zero, so the velocity is zero at that instant. Between t D and t E, the slope of the x–t graph and therefore the velocity are negative and decrease unit tA tB tC tD tE tF formly in this interval. In the interval t E to t F, the slope b of the x–t graph is still negative, and at t F it goes to zero. Finally, after t F, the slope of the x–t graph is zero, meanax ing that the object is at rest for t . t F. The acceleration at any instant is the slope of the tangent to the vx –t graph at that instant. The graph of accelt tB tE tF tA eration versus time for this object is shown in Figure 2.8c. The acceleration is constant and positive between 0 and c t A, where the slope of the vx –t graph is positive. It is zero Figure 2.8 (Conceptual Example 2.5) (a) Position–time graph between t A and t B and for t . t F because the slope of the for an object moving along the x axis. (b) The velocity–time graph vx –t graph is zero at these times. It is negative between for the object is obtained by measuring the slope of the position– t B and t E because the slope of the vx –t graph is negative time graph at each instant. (c) The acceleration–time graph for during this interval. Between t E and t F, the acceleration the object is obtained by measuring the slope of the velocity–time graph at each instant. is positive like it is between 0 and t A, but higher in value because the slope of the vx –t graph is steeper. Notice that the sudden changes in acceleration shown in Figure 2.8c are unphysical. Such instantaneous changes cannot occur in reality. Example 2.6 Motion in One Dimension    Average and Instantaneous Acceleration The acceleration at B is equal to the slope of the green tangent line at t  2 s, which is 20 m/s2. The velocity of a particle moving along the x axis varies according to the expression vx 5 40 2 5t 2, where vx is in meters per second and t is in seconds. vx (m/s) 40 (A) Find the average acceleration in the time interval t 5 0 to t 5 2.0 s. 30 Solution 20 Think about what the particle is doing from the mathematical representation. Is it moving at t 5 0? In which direction? Does it speed up or slow down? Figure 2.9 is a vx –t graph that was created from the velocity versus time expression given in the problem statement. Because the slope of the entire vx –t curve is negative, we expect the acceleration to be negative. 10 A B 0 10 20 .c (Example 2.6) The velocity–time graph for a particle moving along the x axis according to the expression vx 5 40 2 5t 2. 30 0 1 ly Figure 2.9 om 34 Chapter 2 2 3 t (s) 4 ee b vx A 5 40 2 5t A2 5 40 2 5(0)2 5 140 m/s Find the velocities at ti 5 t A 5 0 and tf 5 t B 5 2.0 s by substituting these values of t into the expression for the velocity: vx B 5 40 2 5t B2 5 40 2 5(2.0)2 5 120 m/s a x,avg 5 v xf 2 v xi 5 s. w Find the average acceleration in the specified time interval Dt 5 t B 2 t A 5 2.0 s: tf 2 ti vx B 2 vx A tB 2 tA 5 20 m/s 2 40 m/s 2.0 s 2 0 s ic 5 210 m/s2 ar ph (B) Determine the acceleration at t 5 2.0 s. ys The negative sign is consistent with our expectations: the average acceleration, represented by the slope of the blue line joining the initial and final points on the velocity–time graph, is negative. Solution sw Knowing that the initial velocity at any time t is vxi 5 40 2 5t 2, find the velocity at any later time t 1 Dt: .a Find the change in velocity over the time interval Dt: vxf 5 40 2 5(t 1 Dt)2 5 40 2 5t 2 2 10t Dt 2 5(Dt)2 Dvx 5 vxf 2 vxi 5 210t Dt 2 5(Dt)2 Dv x 1 210t 2 5 Dt 2 5 210t 5 lim Dt S 0 Dt a x 5 lim S Substitute t 5 2.0 s: ax 5 (210)(2.0) m/s2 5 220 m/s2 Dt 0 w w w To find the acceleration at any time t, divide this expression by Dt and take the limit of the result as Dt approaches zero: Because the velocity of the particle is positive and the acceleration is negative at this instant, the particle is slowing down. Notice that the answers to parts (A) and (B) are different. The average acceleration in part (A) is the slope of the blue line in Figure 2.9 connecting points A and B. The instantaneous acceleration in part (B) is the slope of the green line tangent to the curve at point B. Notice also that the acceleration is not constant in this example. Situations involving constant acceleration are treated in Section 2.6. So far, we have evaluated the derivatives of a function by starting with the definition of the function and then taking the limit of a specific ratio. If you are familiar with calculus, you should recognize that there are specific rules for taking 2.5 Motion Diagrams 35 derivatives. These rules, which are listed in Appendix B.6, enable us to evaluate derivatives quickly. For instance, one rule tells us that the derivative of any constant is zero. As another example, suppose x is proportional to some power of t such as in the expression x 5 At n where A and n are constants. (This expression is a very common functional form.) The derivative of x with respect to t is dx 5 nAt n21 dt om Applying this rule to Example 2.6, in which vx 5 40 2 5t 2, we quickly find that the acceleration is ax 5 dvx /dt 5 210t, as we found in part (B) of the example. .c 2.5 Motion Diagrams w w w .a sw ar ph ys ic s. w ee b ly The concepts of velocity and acceleration are often confused with each other, but in fact they are quite different quantities. In forming a mental representation of a moving object, a pictorial representation called a motion diagram is sometimes useful to describe the velocity and acceleration while an object is in motion. A motion diagram can be formed by imagining a stroboscopic photograph of a moving object, which shows several images of the object taken as the strobe light flashes at a constant rate. Figure 2.1a is a motion diagram for the car studied in Section 2.1. Figure 2.10 represents three sets of strobe photographs of cars moving along a straight roadway in a single direction, from left to right. The time intervals between flashes of the stroboscope are equal in each part of the diagram. So as to not confuse the two vector quantities, we use red arrows for velocity and purple arrows for acceleration in Figure 2.10. The arrows are shown at several instants during the motion of the object. Let us describe the motion of the car in each diagram. In Figure 2.10a, the images of the car are equally spaced, showing us that the car moves through the same displacement in each time interval. This equal spacing is consistent with the car moving with constant positive velocity and zero acceleration. We could model the car as a particle and describe it with the particle under constant velocity model. In Figure 2.10b, the images become farther apart as time progresses. In this case, the velocity arrow increases in length with time because the car’s displacement between adjacent positions increases in time. These features suggest the car is moving with a positive velocity and a positive acceleration. The velocity and acceleration are in the same direction. In terms of our earlier force discussion, imagine a force pulling on the car in the same direction it is moving: it speeds up. This car moves at constant velocity (zero acceleration). a This car has a constant acceleration in the direction of its velocity. b This car has a constant acceleration in the direction opposite its velocity. c v v a v a Figure 2.10 Motion diagrams of a car moving along a straight roadway in a single direction. The velocity at each instant is indicated by a red arrow, and the constant acceleration is indicated by a purple arrow. 36 Chapter 2 Motion in One Dimension om In Figure 2.10c, we can tell that the car slows as it moves to the right because its displacement between adjacent images decreases with time. This case suggests the car moves to the right with a negative acceleration. The length of the velocity arrow decreases in time and eventually reaches zero. From this diagram, we see that the acceleration and velocity arrows are not in the same direction. The car is moving with a positive velocity, but with a negative acceleration. (This type of motion is exhibited by a car that skids to a stop after its brakes are applied.) The velocity and acceleration are in opposite directions. In terms of our earlier force discussion, imagine a force pulling on the car opposite to the direction it is moving: it slows down. Each purple acceleration arrow in parts (b) and (c) of Figure 2.10 is the same length. Therefore, these diagrams represent motion of a particle under constant acceleration. This important analysis model will be discussed in the next section. ee b ly .c Q uick Quiz 2.5 Which one of the following statements is true? (a) If a car is traveling eastward, its acceleration must be eastward. (b) If a car is slowing down, its acceleration must be negative. (c) A particle with constant acceleration can never stop and stay stopped. xi Slope  vxi t t ys a ar ph vx Slope  ax axt vx i vx f t t or w w .a b sw vxi ax If the acceleration of a particle varies in time, its motion can be complex and difficult to analyze. A very common and simple type of one-dimensional motion, however, is that in which the acceleration is constant. In such a case, the average acceleration ax,avg over any time interval is numerically equal to the instantaneous acceleration ax at any instant within the interval, and the velocity changes at the same rate throughout the motion. This situation occurs often enough that we identify it as an analysis model: the particle under constant acceleration. In the discussion that follows, we generate several equations that describe the motion of a particle for this model. If we replace ax,avg by ax in Equation 2.9 and take ti 5 0 and tf to be any later time t, we find that v xf 2 v xi ax 5 t20 ic Slope  vxf s. w 2.6 Analysis Model: Particle Under Constant Acceleration x w Slope  0 ax t c Figure 2.11 A particle under constant acceleration ax moving along the x axis: (a) the position– time graph, (b) the velocity–time graph, and (c) the acceleration– time graph. t vxf 5 vxi 1 axt (for constant ax) (2.13) This powerful expression enables us to determine an object’s velocity at any time t if we know the object’s initial velocity vxi and its (constant) acceleration ax . A velocity–time graph for this constant-acceleration motion is shown in Figure 2.11b. The graph is a straight line, the slope of which is the acceleration ax; the (constant) slope is consistent with ax 5 dvx/dt being a constant. Notice that the slope is positive, which indicates a positive acceleration. If the acceleration were negative, the slope of the line in Figure 2.11b would be negative. When the acceleration is constant, the graph of acceleration versus time (Fig. 2.11c) is a straight line having a slope of zero. Because velocity at constant acceleration varies linearly in time according to Equation 2.13, we can express the average velocity in any time interval as the arithmetic mean of the initial velocity vxi and the final velocity vxf : v x,avg 5 v xi 1 v xf 2 1 for constant a x 2 (2.14) 2.6 Analysis Model: Particle Under Constant Acceleration 37 Notice that this expression for average velocity applies only in situations in which the acceleration is constant. We can now use Equations 2.1, 2.2, and 2.14 to obtain the position of an object as a function of time. Recalling that Dx in Equation 2.2 represents xf 2 xi and recognizing that Dt 5 tf 2 ti 5 t 2 0 5 t, we find that x f 2 x i 5 v x,avg t 5 12 1 v xi 1 v xf 2 t x f 5 x i 1 12 1 v xi 1 v xf 2 t 1 for constant a x 2 (2.15) om This equation provides the final position of the particle at time t in terms of the initial and final velocities. We can obtain another useful expression for the position of a particle under constant acceleration by substituting Equation 2.13 into Equation 2.15: .c x f 5 x i 1 12 3 v xi 1 1 v xi 1 a xt 2 4 t 1 for constant a x 2 (2.16) ly x f 5 x i 1 v xit 1 12a xt 2 WW Position as a function of velocity and time for the particle under constant acceleration model b 5 xi 1 v xf 2 2 v xi 2 ys v xf 2 v xi ax ar ph x f 5 x i 1 12 1 v xi 1 v xf 2 a ic s. w ee b This equation provides the final position of the particle at time t in terms of the initial position, the initial velocity, and the constant acceleration. The position–time graph for motion at constant (positive) acceleration shown in Figure 2.11a is obtained from Equation 2.16. Notice that the curve is a parabola. The slope of the tangent line to this curve at t 5 0 equals the initial velocity vxi , and the slope of the tangent line at any later time t equals the velocity vxf at that time. Finally, we can obtain an expression for the final velocity that does not contain time as a variable by substituting the value of t from Equation 2.13 into Equation 2.15: WW Position as a function of time for the particle under constant acceleration model vxf 2 5 vxi2 1 2ax(xf 2 xi ) 2a x (for constant ax ) (2.17) sw This equation provides the final velocity in terms of the initial velocity, the constant acceleration, and the position of the particle. For motion at zero acceleration, we see from Equations 2.13 and 2.16 that .a v xf 5 v xi 5 v x f x f 5 x i 1 v xt when a x 5 0 w w w That is, when the acceleration of a particle is zero, its velocity is constant and its position changes linearly with time. In terms of models, when the acceleration of a particle is zero, the particle under constant acceleration model reduces to the particle under constant velocity model (Section 2.3). Equations 2.13 through 2.17 are kinematic equations that may be used to solve any problem involving a particle under constant acceleration in one dimension. These equations are listed together for convenience on page 38. The choice of which equation you use in a given situation depends on what you know beforehand. Sometimes it is necessary to use two of these equations to solve for two unknowns. You should recognize that the quantities that vary during the motion are position xf , velocity vxf , and time t. You will gain a great deal of experience in the use of these equations by solving a number of exercises and problems. Many times you will discover that more than one method can be used to obtain a solution. Remember that these equations of kinematics cannot be used in a situation in which the acceleration varies with time. They can be used only when the acceleration is constant. WW Velocity as a function of position for the particle under constant acceleration model 38 Chapter 2 Motion in One Dimension Q uick Quiz 2.6 In Figure 2.12, match each vx –t graph on the top with the ax –t graph on the bottom that best describes the motion. vx vx t a b t t t ly e f ee b Analysis Model   Particle Under Constant Acceleration c ax .c ax d t t ax Examples ic s. w Imagine a moving object that can be modeled as a particle. If it begins from position xi and initial velocity vxi and moves in a straight line with a constant acceleration ax , its subsequent position and velocity are described by the following kinematic equations: vx om Figure 2.12 (Quick Quiz 2.6) Parts (a), (b), and (c) are vx–t graphs of objects in one-dimensional motion. The possible accelerations of each object as a function of time are shown in scrambled order in (d), (e), and (f). (2.13) vxi 1 vxf 2 (2.14) ar ph vx,avg 5 ys vxf 5 vxi 1 axt xf 5 xi 1 12 1 vxi 1 vxf 2 t x f 5 x i 1 v xi t 1 12a x t 2 (2.16) vxf 2 5 vxi21 2ax(xf 2 xi ) (2.17) sw (2.15) v w w .a a • a car accelerating at a constant rate along a straight freeway • a dropped object in the absence of air resistance (Section 2.7) • an object on which a constant net force acts (Chapter 5) • a charged particle in a uniform electric field (Chapter 23)   Carrier Landing AM w Example 2.7 A jet lands on an aircraft carrier at a speed of 140 mi/h (< 63 m/s). (A) What is its acceleration (assumed constant) if it stops in 2.0 s due to an arresting cable that snags the jet and brings it to a stop? Solution You might have seen movies or television shows in which a jet lands on an aircraft carrier and is brought to rest surprisingly fast by an arresting cable. A careful reading of the problem reveals that in addition to being given the initial speed of 63 m/s, we also know that the final speed is zero. Because the acceleration of the jet is assumed constant, we model it as a particle under constant acceleration. We define our x axis as the direction of motion of the jet. Notice that we have no information about the change in position of the jet while it is slowing down. 39 2.6 Analysis Model: Particle Under Constant Acceleration ▸ 2.7 c o n t i n u e d Equation 2.13 is the only equation in the particle under constant acceleration model that does not involve position, so we use it to find the acceleration of the jet, modeled as a particle: ax 5 v xf 2 v xi t < 0 2 63 m/s 2.0 s 5 232 m/s2 (B) If the jet touches down at position xi 5 0, what is its final position? Solution x f 5 x i 1 12 1 v xi 1 v xf 2 t 5 0 1 12 1 63 m/s 1 0 2 1 2.0 s 2 5 63 m om Use Equation 2.15 to solve for the final position: .c Given the size of aircraft carriers, a length of 63 m seems reasonable for stopping the jet. The idea of using arresting cables to slow down landing aircraft and enable them to land safely on ships originated at about the time of World War I. The cables are still a vital part of the operation of modern aircraft carriers. W h at If ? ly Suppose the jet lands on the deck of the aircraft carrier with a speed faster than 63 m/s but has the same acceleration due to the cable as that calculated in part (A). How will that change the answer to part (B)? Example 2.8 s. w ee b Answer If the jet is traveling faster at the beginning, it will stop farther away from its starting point, so the answer to part (B) should be larger. Mathematically, we see in Equation 2.15 that if vxi is larger, then xf will be larger. AM    Watch Out for the Speed Limit! t A  1.00 s tB  0 tC  ? A B C ar ph ys ic A car traveling at a constant speed of 45.0 m/s passes a trooper on a motorcycle hidden behind a billboard. One second after the speeding car passes the billboard, the trooper sets out from the billboard to catch the car, accelerating at a constant rate of 3.00 m/s2. How long does it take the trooper to overtake the car? Solution w w w .a sw A pictorial representation (Fig. 2.13) helps clarify the Figure 2.13 (Example 2.8) A speeding car passes a hidsequence of events. The car is modeled as a particle under conden trooper. stant velocity, and the trooper is modeled as a particle under constant acceleration. First, we write expressions for the position of each vehicle as a function of time. It is convenient to choose the position of the billboard as the origin and to set t B 5 0 as the time the trooper begins moving. At that instant, the car has already traveled a distance of 45.0 m from the billboard because it has traveled at a constant speed of vx 5 45.0 m/s for 1 s. Therefore, the initial position of the speeding car is x B 5 45.0 m. Using the particle under constant velocity model, apply Equation 2.7 to give the car’s position at any time t : x car 5 x B 1 vx car t A quick check shows that at t 5 0, this expression gives the car’s correct initial position when the trooper begins to move: x car 5 x B 5 45.0 m. The trooper starts from rest at t B 5 0 and accelerates at ax 5 3.00 m/s2 away from the origin. Use Equation 2.16 to give her position at any time t : x f 5 x i 1 v xit 1 12a x t 2 x trooper 5 0 1 1 0 2 t 1 12a x t 2 5 12a x t 2 Set the positions of the car and trooper equal to represent the trooper overtaking the car at position C: x trooper 5 x car 1 2 2a x t 5 x B 1 v x cart continued 40 Chapter 2 Motion in One Dimension ▸ 2.8 c o n t i n u e d Rearrange to give a quadratic equation: 1 2 2a x t Solve the quadratic equation for the time at which the trooper catches the car (for help in solving quadratic equations, see Appendix B.2.): t5 Evaluate the solution, choosing the positive root because that is the only choice consistent with a time t . 0: t5 v x car 6 "v x2car 1 2a xx B ax 2x B v x2 car v x car 6 1 ax ax Å a x2 1 45.0 m/s 2 2 2 1 45.0 m 2 45.0 m/s 1 1 5 31.0 s 2 2 2 Å 1 3.00 m/s 2 3.00 m/s 3.00 m/s2 om (1) t 5 2 v x cart 2 x B 5 0 ly .c Why didn’t we choose t 5 0 as the time at which the car passes the trooper? If we did so, we would not be able to use the particle under constant acceleration model for the trooper. Her acceleration would be zero for the first second and then 3.00 m/s2 for the remaining time. By defining the time t 5 0 as when the trooper begins moving, we can use the particle under constant acceleration model for her movement for all positive times. ee b W h at If ? What if the trooper had a more powerful motorcycle with a larger acceleration? How would that change the time at which the trooper catches the car? Answer If the motorcycle has a larger acceleration, the trooper should catch up to the car sooner, so the answer for ar ph ys ic s. w the time should be less than 31 s. Because all terms on the right side of Equation (1) have the acceleration ax in the denominator, we see symbolically that increasing the acceleration will decrease the time at which the trooper catches the car. 2.7 Freely Falling Objects w Galileo Galilei w w .a Georgios Kollidas/Shutterstock.com sw It is well known that, in the absence of air resistance, all objects dropped near the Earth’s surface fall toward the Earth with the same constant acceleration under the influence of the Earth’s gravity. It was not until about 1600 that this conclusion was accepted. Before that time, the teachings of the Greek philosopher Aristotle (384–322 BC) had held that heavier objects fall faster than lighter ones. The Italian Galileo Galilei (1564–1642) originated our present-day ideas concerning falling objects. There is a legend that he demonstrated the behavior of falling objects by observing that two different weights dropped simultaneously from the Leaning Tower of Pisa hit the ground at approximately the same time. Although there is some doubt that he carried out this particular experiment, it is well established that Galileo performed many experiments on objects moving on inclined planes. In his experiments, he rolled balls down a slight incline and measured the distances they covered in successive time intervals. The purpose of the incline was to reduce the acceleration, which made it possible for him to make accurate measurements of the time intervals. By gradually increasing the slope of the incline, he was finally able to draw conclusions about freely falling objects because a freely falling ball is equivalent to a ball moving down a vertical incline. You might want to try the following experiment. Simultaneously drop a coin and a crumpled-up piece of paper from the same height. If the effects of air resistance are negligible, both will have the same motion and will hit the floor at the same time. In the idealized case, in which air resistance is absent, such motion is referred Italian physicist and astronomer (1564–1642) Galileo formulated the laws that govern the motion of objects in free fall and made many other significant discoveries in physics and astronomy. Galileo publicly defended Nicolaus Copernicus’s assertion that the Sun is at the center of the Universe (the heliocentric system). He published Dialogue Concerning Two New World Systems to support the Copernican model, a view that the Catholic Church declared to be heretical. 2.7 Freely Falling Objects Pitfall Prevention 2.6 g and g Be sure not to confuse the italic symbol g for free-fall acceleration with the nonitalic symbol g used as the abbreviation for the unit gram. Pitfall Prevention 2.7 The Sign of g Keep in mind that g is a positive number. It is tempting to substitute 29.80 m/s2 for g, but resist the temptation. Downward gravitational acceleration is indicated explicitly by stating the acceleration as ay 5 2g. om to as free-fall motion. If this same experiment could be conducted in a vacuum, in which air resistance is truly negligible, the paper and the coin would fall with the same acceleration even when the paper is not crumpled. On August 2, 1971, astronaut David Scott conducted such a demonstration on the Moon. He simultaneously released a hammer and a feather, and the two objects fell together to the lunar surface. This simple demonstration surely would have pleased Galileo! When we use the expression freely falling object, we do not necessarily refer to an object dropped from rest. A freely falling object is any object moving freely under the influence of gravity alone, regardless of its initial motion. Objects thrown upward or downward and those released from rest are all falling freely once they are released. Any freely falling object experiences an acceleration directed downward, regardless of its initial motion. We shall denote the magnitude of the free-fall acceleration, also called the acceleration due to gravity, by the symbol g. The value of g decreases with increasing altitude above the Earth’s surface. Furthermore, slight variations in g occur with changes in latitude. At the Earth’s surface, the value of g is approximately 9.80 m/s2. Unless stated otherwise, we shall use this value for g when performing calculations. For making quick estimates, use g 5 10 m/s2. If we neglect air resistance and assume the free-fall acceleration does not vary with altitude over short vertical distances, the motion of a freely falling object moving vertically is equivalent to the motion of a particle under constant acceleration in one dimension. Therefore, the equations developed in Section 2.6 for the particle under constant acceleration model can be applied. The only modification for freely falling objects that we need to make in these equations is to note that the motion is in the vertical direction (the y direction) rather than in the horizontal direction (x) and that the acceleration is downward and has a magnitude of 9.80 m/s2. Therefore, we choose ay 5 2g 5 29.80 m/s2, where the negative sign means that the acceleration of a freely falling object is downward. In Chapter 13, we shall study how to deal with variations in g with altitude. 41 Acceleration at the Top of the Motion A common misconcep- tion is that the acceleration of a projectile at the top of its trajectory is zero. Although the velocity at the top of the motion of an object thrown upward momentarily goes to zero, the acceleration is still that due to gravity at this point. If the velocity and acceleration were both zero, the projectile would stay at the top. ys ic s. w ee b ly .c Pitfall Prevention 2.8 sw ar ph Q uick Quiz 2.7 Consider the following choices: (a) increases, (b) decreases, (c) increases and then decreases, (d) decreases and then increases, (e) remains the same. From these choices, select what happens to (i) the acceleration and (ii) the speed of a ball after it is thrown upward into the air.    The Daring Skydivers w .a Conceptual Example 2.9 w w A skydiver jumps out of a hovering helicopter. A few seconds later, another skydiver jumps out, and they both fall along the same vertical line. Ignore air resistance so that both skydivers fall with the same acceleration. Does the difference in their speeds stay the same throughout the fall? Does the vertical distance between them stay the same throughout the fall? Solution At any given instant, the speeds of the skydivers are different because one had a head start. In any time interval Dt after this instant, however, the two skydivers increase their speeds by the same amount because they have the same acceleration. Therefore, the difference in their speeds remains the same throughout the fall. The first jumper always has a greater speed than the second. Therefore, in a given time interval, the first skydiver covers a greater distance than the second. Consequently, the separation distance between them increases. 42 Chapter 2 Motion in One Dimension Example 2.10    Not a Bad Throw for a Rookie! AM B A stone thrown from the top of a building is given an initial velocity of 20.0 m/s straight upward. The stone is launched 50.0 m above the ground, and the stone just misses the edge of the roof on its way down as shown in Figure 2.14. tA  0 yA  0 vy A  20.0 m/s ay A  9.80 m/s2 (A) Using t A 5 0 as the time the stone leaves the thrower’s hand at position A, determine the time at which the stone reaches its maximum height. C A .c ly ee b D ic s. w 50.0 m ar ph ys t E  5.83 s y E  50.0 m E vy E  37.1 m/s2 ay E  9.80 m/s v yf 5 v yi 1 a yt t 5 tB 5 t D  5.00 s y D  22.5 m vy D  29.0 m/s ay D  9.80 m/s2 S t5 v yf 2 v yi ay 0 2 20.0 m/s 5 2.04 s 29.80 m/s2 .a Substitute numerical values: sw Use Equation 2.13 to calculate the time at which the stone reaches its maximum height: t C  4.08 s yC  0 vy C  20.0 m/s ay C  9.80 m/s2 om Solution You most likely have experience Figure 2.14 (Example 2.10) Position, velocity, and acceleration values at with dropping objects or throwvarious times for a freely falling stone ing them upward and watching thrown initially upward with a velocity them fall, so this problem should vyi 5 20.0 m/s. Many of the quantities describe a familiar experience. in the labels for points in the motion To simulate this situation, toss a of the stone are calculated in the example. Can you verify the other valsmall object upward and notice ues that are not? the time interval required for it to fall to the floor. Now imagine throwing that object upward from the roof of a building. Because the stone is in free fall, it is modeled as a particle under constant acceleration due to gravity. Recognize that the initial velocity is positive because the stone is launched upward. The velocity will change sign after the stone reaches its highest point, but the acceleration of the stone will always be downward so that it will always have a negative value. Choose an initial point just after the stone leaves the person’s hand and a final point at the top of its flight. t B  2.04 s y B  20.4 m vy B  0 ay B  9.80 m/s2 Solution w (B) Find the maximum height of the stone. w As in part (A), choose the initial and final points at the beginning and the end of the upward flight. w Set y A 5 0 and substitute the time from part (A) into Equation 2.16 to find the maximum height: ymax 5 y B 5 y A 1 v x A t 1 12a yt 2 y B 5 0 1 1 20.0 m/s 2 1 2.04 s 2 1 12 1 29.80 m/s2 2 1 2.04 s 2 2 5 20.4 m (C) Determine the velocity of the stone when it returns to the height from which it was thrown. Solution Choose the initial point where the stone is launched and the final point when it passes this position coming down. Substitute known values into Equation 2.17: vyC2 5 vyA2 1 2ay(y C 2 y A) vyC2 5 (20.0 m/s)2 1 2(29.80 m/s2)(0 2 0) 5 400 m2/s2 vyC 5 220.0 m/s 2.8 Kinematic Equations Derived from Calculus 43 ▸ 2.10 c o n t i n u e d When taking the square root, we could choose either a positive or a negative root. We choose the negative root because we know that the stone is moving downward at point C. The velocity of the stone when it arrives back at its original height is equal in magnitude to its initial velocity but is opposite in direction. (D) Find the velocity and position of the stone at t 5 5.00 s. Solution Choose the initial point just after the throw and the final point 5.00 s later. vyD5 vyA 1 ayt 5 20.0 m/s 1 (29.80 m/s2)(5.00 s) 5 229.0 m/s Use Equation 2.16 to find the position of the stone at t D 5 5.00 s: y D 5 yA 1 v y A t 1 12a yt 2 om Calculate the velocity at D from Equation 2.13: 5 0 1 (20.0 m/s)(5.00 s) 1 12(29.80 m/s2)(5.00 s)2 .c 5 222.5 m ee b ly The choice of the time defined as t 5 0 is arbitrary and up to you to select as the problem solver. As an example of this arbitrariness, choose t 5 0 as the time at which the stone is at the highest point in its motion. Then solve parts (C) and (D) again using this new initial instant and notice that your answers are the same as those above. W h at If ? What if the throw were from 30.0 m above the ground instead of 50.0 m? Which answers in parts (A) to (D) would change? ys ic s. w Answer None of the answers would change. All the motion takes place in the air during the first 5.00 s. (Notice that even for a throw from 30.0 m, the stone is above the ground at t 5 5.00 s.) Therefore, the height of the throw is not an issue. Mathematically, if we look back over our calculations, we see that we never entered the height of the throw into any equation. ar ph 2.8 Kinematic Equations Derived from Calculus w w w .a sw This section assumes the reader is familiar with the techniques of integral calculus. If you have not yet studied integration in your calculus course, you should skip this section or cover it after you become familiar with integration. The velocity of a particle moving in a straight line can be obtained if its position as a function of time is known. Mathematically, the velocity equals the derivative of the position with respect to time. It is also possible to find the position of a particle if its velocity is known as a function of time. In calculus, the procedure used to perform this task is referred to either as integration or as finding the antiderivative. Graphically, it is equivalent to finding the area under a curve. Suppose the vx –t graph for a particle moving along the x axis is as shown in Figure 2.15 on page 44. Let us divide the time interval tf 2 ti into many small intervals, each of duration Dtn . From the definition of average velocity, we see that the displacement of the particle during any small interval, such as the one shaded in Figure 2.15, is given by Dxn 5 vxn,avg Dtn , where vxn,avg is the average velocity in that interval. Therefore, the displacement during this small interval is simply the area of the shaded rectangle in Figure 2.15. The total displacement for the interval tf 2 ti is the sum of the areas of all the rectangles from ti to tf : Dx 5 a v xn,avg Dtn n where the symbol o (uppercase Greek sigma) signifies a sum over all terms, that is, over all values of n. Now, as the intervals are made smaller and smaller, the number of terms in the sum increases and the sum approaches a value equal to the area 44 Chapter 2 Motion in One Dimension Figure 2.15 Velocity versus time for a particle moving along the x axis. The total area under the curve is the total displacement of the particle. The area of the shaded rectangle is equal to the displacement in the time interval tn. vx vxn,avg ti tf t om t n ly Dx 5 lim v xn,avg Dtn Dtn S 0 a .c under the curve in the velocity–time graph. Therefore, in the limit n S `, or Dtn S 0, the displacement is n (2.18) s. w ee b If we know the vx –t graph for motion along a straight line, we can obtain the displacement during any time interval by measuring the area under the curve corresponding to that time interval. The limit of the sum shown in Equation 2.18 is called a definite integral and is written lim v xn,avg Dtn 5 3 v x 1 t 2 dt Dtn S 0 a Definite integral  t vxi where vx(t) denotes the velocity at any time t. If the explicit functional form of vx(t) is known and the limits are given, the integral can be evaluated. Sometimes the vx –t graph for a moving particle has a shape much simpler than that shown in Figure 2.15. For example, suppose an object is described with the particle under constant velocity model. In this case, the vx –t graph is a horizontal line as in Figure 2.16 and the displacement of the particle during the time interval Dt is simply the area of the shaded rectangle: ys vx  vxi  constant vxi tf Dx 5 vxi Dt (when vx 5 vxi 5 constant) Kinematic Equations w w w .a Figure 2.16 ​The velocity–time curve for a particle moving with constant velocity vxi . The displacement of the particle during the time interval tf 2 ti is equal to the area of the shaded rectangle. t sw ti ar ph vx (2.19) ti ic n tf We now use the defining equations for acceleration and velocity to derive two of our kinematic equations, Equations 2.13 and 2.16. The defining equation for acceleration (Eq. 2.10), dv x dt may be written as dvx 5 ax dt or, in terms of an integral (or antiderivative), as ax 5 v xf 2 v xi 5 3 a x dt t 0 For the special case in which the acceleration is constant, ax can be removed from the integral to give v xf 2 v xi 5 a x 3 dt 5 a x 1 t 2 0 2 5 a xt t 0 which is Equation 2.13 in the particle under constant acceleration model. Now let us consider the defining equation for velocity (Eq. 2.5): dx vx 5 dt (2.20) General Problem-Solving Strategy 45 We can write this equation as dx 5 vx dt or in integral form as x f 2 x i 5 3 v x dt t 0 Because vx 5 vxf 5 vxi 1 axt, this expression becomes t2    x f 2 x i 5 3 1 v xi 1 a xt 2 dt 5 3 v xi dt 1 a x 3 t dt 5 v xi 1 t 2 0 2 1 a x a 2 0b 2 t 0 x f 2 x i 5 v xit 1 t t 0 0 1 2 2 a xt om which is Equation 2.16 in the particle under constant acceleration model. ee b ly .c Besides what you might expect to learn about physics concepts, a very valuable skill you should hope to take away from your physics course is the ability to solve complicated problems. The way physicists approach complex situations and break them into manageable pieces is extremely useful. The following is a general problem-solving strategy to guide you through the steps. To help you remember the steps of the strategy, they are Conceptualize, Categorize, Analyze, and Finalize. s. w General Problem-Solving Strategy the details that are not important to the solution. For example, model a moving object as a particle. If appropriate, ignore air resistance or friction between a sliding object and a surface. ic Conceptualize ar ph ys • The first things to do when approaching a problem are to think about and understand the situation. Study carefully any representations of the information (for example, diagrams, graphs, tables, or photographs) that accompany the problem. Imagine a movie, running in your mind, of what happens in the problem. sw • If a pictorial representation is not provided, you should almost always make a quick drawing of the situation. Indicate any known values, perhaps in a table or directly on your sketch. w w .a • Now focus on what algebraic or numerical information is given in the problem. Carefully read the problem statement, looking for key phrases such as “starts from rest” (vi 5 0), “stops” (vf 5 0), or “falls freely” (ay 5 2g 5 29.80 m/s2). w • Now focus on the expected result of solving the problem. Exactly what is the question asking? Will the final result be numerical or algebraic? Do you know what units to expect? • Don’t forget to incorporate information from your own experiences and common sense. What should a reasonable answer look like? For example, you wouldn’t expect to calculate the speed of an automobile to be 5 3 106 m/s. Categorize • Once you have a good idea of what the problem is about, you need to simplify the problem. Remove • Once the problem is simplified, it is important to categorize the problem. Is it a simple substitution problem such that numbers can be substituted into a simple equation or a definition? If so, the problem is likely to be finished when this substitution is done. If not, you face what we call an analysis problem: the situation must be analyzed more deeply to generate an appropriate equation and reach a solution. • If it is an analysis problem, it needs to be categorized further. Have you seen this type of problem before? Does it fall into the growing list of types of problems that you have solved previously? If so, identify any analysis model(s) appropriate for the problem to prepare for the Analyze step below. We saw the first three analysis models in this chapter: the particle under constant velocity, the particle under constant speed, and the particle under constant acceleration. Being able to classify a problem with an analysis model can make it much easier to lay out a plan to solve it. For example, if your simplification shows that the problem can be treated as a particle under constant acceleration and you have already solved such a problem (such as the examples in Section 2.6), the solution to the present problem follows a similar pattern. continued 46 Chapter 2 Motion in One Dimension Analyze • Think about how this problem compared with others you have solved. How was it similar? In what critical ways did it differ? Why was this problem assigned? Can you figure out what you have learned by doing it? If it is a new category of problem, be sure you understand it so that you can use it as a model for solving similar problems in the future. • Now you must analyze the problem and strive for a mathematical solution. Because you have already categorized the problem and identified an analysis model, it should not be too difficult to select relevant equations that apply to the type of situation in the problem. For example, if the problem involves a particle under constant acceleration, Equations 2.13 to 2.17 are relevant. ly .c om • Use algebra (and calculus, if necessary) to solve symbolically for the unknown variable in terms of what is given. Finally, substitute in the appropriate numbers, calculate the result, and round it to the proper number of significant figures. ar ph ys ic s. w ee b Finalize • Examine your numerical answer. Does it have the correct units? Does it meet your expectations from your conceptualization of the problem? What about the algebraic form of the result? Does it make sense? Examine the variables in the problem to see whether the answer would change in a physically meaningful way if the variables were drastically increased or decreased or even became zero. Looking at limiting cases to see whether they yield expected values is a very useful way to make sure that you are obtaining reasonable results. When solving complex problems, you may need to identify a series of subproblems and apply the problemsolving strategy to each. For simple problems, you probably don’t need this strategy. When you are trying to solve a problem and you don’t know what to do next, however, remember the steps in the strategy and use them as a guide. For practice, it would be useful for you to revisit the worked examples in this chapter and identify the Conceptualize, Categorize, Analyze, and Finalize steps. In the rest of this book, we will label these steps explicitly in the worked examples. Many chapters in this book include a section labeled Problem-Solving Strategy that should help you through the rough spots. These sections are organized according to the General ProblemSolving Strategy outlined above and are tailored to the specific types of problems addressed in that chapter. To clarify how this Strategy works, we repeat Example 2.7 below with the particular steps of the Strategy identified. Simplify the problem. Remove the details that are not important to the solution. Then Categorize the problem. Is it a simple substitution problem such that numbers can be substituted into a simple equation or a definition? If not, you face an analysis problem. In this case, identify the appropriate analysis model. w w .a sw When you Conceptualize a problem, try to understand the situation that is presented in the problem statement. Study carefully any representations of the information (for example, diagrams, graphs, tables, or photographs) that accompany the problem. Imagine a movie, running in your mind, of what happens in the problem.   Carrier Landing AM w Example 2.7 A jet lands on an aircraft carrier at a speed of 140 mi/h (< 63 m/s). (A) What is its acceleration (assumed constant) if it stops in 2.0 s due to an arresting cable that snags the jet and brings it to a stop? Solution Conceptualize You might have seen movies or television shows in which a jet lands on an aircraft carrier and is brought to rest surprisingly fast by an arresting cable. A careful reading of the problem reveals that in addition to being given the initial speed of 63 m/s, we also know that the final speed is zero. Categorize Because the acceleration of the jet is assumed constant, we model it as a particle under constant acceleration.    Summary 47 ▸ 2.7 c o n t i n u e d Analyze We define our x axis as the direction of motion of the jet. Notice that we have no information about the change in position of the jet while it is slowing down. v xf 2 v xi 0 2 63 m/s Equation 2.13 is the only equation in the particle under   a x 5 5 t 2.0 s constant acceleration model that does not involve position, so we use it to find the acceleration of the jet, modeled as 5 232 m/s2 a particle: om (B) If the jet touches down at position xi 5 0, what is its final position? .c Solution x f 5 x i 1 12 1 v xi 1 v xf 2 t 5 0 1 12 1 63 m/s 1 0 2 1 2.0 s 2 5 63 m Use Equation 2.15 to solve for the final position: ee b ly Finalize Given the size of aircraft carriers, a length of 63 m seems reasonable for stopping the jet. The idea of using arresting cables to slow down landing aircraft and enable them to land safely on ships originated at about the time of World War I. The cables are still a vital part of the operation of modern aircraft carriers. Suppose the jet lands on the deck of the aircraft carrier with a speed higher than 63 m/s but has the same acceleration due to the cable as that calculated in part (A). How will that change the answer to part (B)? s. w W h at I f ? ys ic Answer If the jet is traveling faster at the beginning, it will stop farther away from its starting point, so the answer to part (B) should be larger. Mathematically, we see in Equation 2.15 that if vxi is larger, xf will be larger. Finalize the problem. Examine the numerical answer. Does it have the correct units? Does it meet your expectations from your conceptualization of the problem? Does the answer make sense? What about the algebraic form of the result? Examine the variables in the problem to see whether the answer would change in a physically meaningful way if the variables were drastically increased or decreased or even became zero. What If? questions will appear in many examples in the text, and offer a variation on the situation just explored. This feature encourages you to think about the results of the example and assists in conceptual understanding of the principles. .a sw ar ph Now Analyze the problem. Select relevant equations from the analysis model. Solve symbolically for the unknown variable in terms of what is given. Substitute in the appropriate numbers, calculate the result, and round it to the proper number of significant figures. w w Summary w Definitions When a particle moves along the x axis from some initial position xi to some final position xf , its displacement is Dx ; xf 2 xi (2.1) The average velocity of a particle during some time interval is the displacement Dx divided by the time interval Dt during which that displacement occurs: Dx (2.2) Dt The average speed of a particle is equal to the ratio of the total distance it travels to the total time interval during which it travels that distance: v x,avg ; v avg ; d Dt (2.3) continued 48 Chapter 2 Motion in One Dimension v x ; lim S Dt 0 Dx dx 5 Dt dt (2.5) The instantaneous speed of a particle is equal to the magnitude of its instantaneous velocity. The average acceleration of a particle is defined as the ratio of the change in its velocity Dvx divided by the time interval Dt during which that change occurs: v xf 2 v xi Dv x a x,avg ; 5 (2.9) Dt tf 2 ti The instantaneous acceleration is equal to the limit of the ratio Dvx /Dt as Dt approaches 0. By definition, this limit equals the derivative of vx with respect to t, or the time rate of change of the velocity: a x ; lim S Dt Dv x dv x 5 Dt dt (2.10) .c Concepts and Principles 0 om The instantaneous velocity of a particle is defined as the limit of the ratio Dx/Dt as Dt approaches zero. By definition, this limit equals the derivative of x with respect to t, or the time rate of change of the position: ly ee b ys ic An important aid to problem solving is the use of analysis models. Analysis models are situations that we have seen in previous problems. Each analysis model has one or more equations associated with it. When solving a new problem, identify the analysis model that corresponds to the problem. The model will tell you which equations to use. The first three analysis models introduced in this chapter are summarized below. ar ph Complicated problems are best approached in an organized manner. Recall and apply the Conceptualize, Categorize, Analyze, and Finalize steps of the General ProblemSolving Strategy when you need them. An object falling freely in the presence of the Earth’s gravity experiences free-fall acceleration directed toward the center of the Earth. If air resistance is neglected, if the motion occurs near the surface of the Earth, and if the range of the motion is small compared with the Earth’s radius, the free-fall acceleration ay 5 2g is constant over the range of motion, where g is equal to 9.80 m/s2. s. w When an object’s velocity and acceleration are in the same direction, the object is speeding up. On the other hand, when the object’s velocity and acceleration are in opposite directions, the object is slowing down. Remembering that Fx ~ ax is a useful way to identify the direction of the acceleration by associating it with a force. sw Analysis Models for Problem-Solving w .a Particle Under Constant Velocity. If a particle moves in a straight line with a constant speed vx , its constant velocity is given by w vx 5 Dx Dt Particle Under Constant Speed. If a particle moves a distance d along a curved or straight path with a constant speed, its constant speed is given by v5 (2.6) d Dt (2.8) w and its position is given by xf 5 xi 1 vxt (2.7) v v x f 5 x i 1 12 1 v xi 1 v xf 2 t Particle Under Constant Acceleration. If a particle moves in a straight line with a constant acceleration ax , its motion is described by the kinematic equations: vxf 5 vxi 1 axt v x,avg 5 v xi 1 v xf 2 (2.13) (2.14) v a (2.15) x f 5 x i 1 v xi t 1 12a x t 2 (2.16) vxf 2 5 vxi21 2ax(xf 2 xi ) (2.17) Objective Questions Objective Questions 49 1. denotes answer available in Student Solutions Manual/Study Guide but loses speed more and more slowly forever. (e) It never stops but continues to speed up in the forward direction. 8. A rock is thrown downward from the top of a 40.0-m-tall tower with an initial speed of 12 m/s. Assuming negligible air resistance, what is the speed of the rock just before hitting the ground? (a) 28 m/s (b) 30 m/s (c) 56 m/s (d) 784 m/s (e) More information is needed. 9. A skateboarder starts from rest and moves down a hill with constant acceleration in a straight line, traveling for 6 s. In a second trial, he starts from rest and moves along the same straight line with the same acceleration for only 2 s. How does his displacement from his starting point in this second trial compare with that from the first trial? (a) one-third as large (b) three times larger (c) one-ninth as large (d) nine times larger (e) 1/ !3 times as large 10. On another planet, a marble is released from rest at the top of a high cliff. It falls 4.00 m in the first 1 s of its motion. Through what additional distance does it fall in the next 1 s? (a) 4.00 m (b) 8.00 m (c) 12.0 m (d) 16.0 m (e) 20.0 m s. w 2. A racing car starts from rest at t 5 0 and reaches a final speed v at time t. If the acceleration of the car is constant during this time, which of the following statements are true? (a) The car travels a distance vt. (b) The average speed of the car is v/2. (c) The magnitude of the acceleration of the car is v/t. (d) The velocity of the car remains constant. (e) None of statements (a) through (d) is true. ly Figure OQ2.1 ee b 600 m .c om 1. One drop of oil falls straight down onto the road from the engine of a moving car every 5 s. Figure OQ2.1 shows the pattern of the drops left behind on the pavement. What is the average speed of the car over this section of its motion? (a) 20 m/s (b) 24 m/s (c) 30 m/s (d) 100 m/s (e) 120 m/s 11. As an object moves along the x axis, many measurements are made of its position, enough to generate a smooth, accurate graph of x versus t. Which of the following quantities for the object cannot be obtained from this graph alone? (a) the velocity at any instant (b) the acceleration at any instant (c) the displacement during some time interval (d) the average velocity during some time interval (e) the speed at any instant ar ph ys ic 3. A juggler throws a bowling pin straight up in the air. After the pin leaves his hand and while it is in the air, which statement is true? (a) The velocity of the pin is always in the same direction as its acceleration. (b) The velocity of the pin is never in the same direction as its acceleration. (c) The acceleration of the pin is zero. (d) The velocity of the pin is opposite its acceleration on the way up. (e) The velocity of the pin is in the same direction as its acceleration on the way up. w .a sw 4. When applying the equations of kinematics for an object moving in one dimension, which of the following statements must be true? (a) The velocity of the object must remain constant. (b) The acceleration of the object must remain constant. (c) The velocity of the object must increase with time. (d) The position of the object must increase with time. (e) The velocity of the object must always be in the same direction as its acceleration. w w 5. A cannon shell is fired straight up from the ground at an initial speed of 225 m/s. After how much time is the shell at a height of 6.20 3 102 m above the ground and moving downward? (a) 2.96 s (b) 17.3 s (c) 25.4 s (d) 33.6 s (e) 43.0 s 6. An arrow is shot straight up in the air at an initial speed of 15.0 m/s. After how much time is the arrow moving downward at a speed of 8.00 m/s? (a) 0.714 s (b) 1.24 s (c) 1.87 s (d) 2.35 s (e) 3.22 s 7. When the pilot reverses the propeller in a boat moving north, the boat moves with an acceleration directed south. Assume the acceleration of the boat remains constant in magnitude and direction. What happens to the boat? (a) It eventually stops and remains stopped. (b) It eventually stops and then speeds up in the forward direction. (c) It eventually stops and then speeds up in the reverse direction. (d) It never stops 12. A pebble is dropped from rest from the top of a tall cliff and falls 4.9 m after 1.0 s has elapsed. How much farther does it drop in the next 2.0 s? (a) 9.8 m (b) 19.6 m (c) 39 m (d) 44 m (e) none of the above 13. A student at the top of a building of height h throws one ball upward with a speed of vi and then throws a second ball downward with the same initial speed vi . Just before it reaches the ground, is the final speed of the ball thrown upward (a) larger, (b) smaller, or (c) the same in magnitude, compared with the final speed of the ball thrown downward? 14. You drop a ball from a window located on an upper floor of a building. It strikes the ground with speed v. You now repeat the drop, but your friend down on the ground throws another ball upward at the same speed v, releasing her ball at the same moment that you drop yours from the window. At some location, the balls pass each other. Is this location (a) at the halfway point between window and ground, (b) above this point, or (c) below this point? 15. A pebble is released from rest at a certain height and falls freely, reaching an impact speed of 4 m/s at the floor. Next, the pebble is thrown down with an initial speed of 3 m/s from the same height. What is its speed at the floor? (a) 4 m/s (b) 5 m/s (c) 6 m/s (d) 7 m/s (e) 8 m/s Motion in One Dimension om b c Figure OQ2.18 ys ar ph Conceptual Questions © Cengage Learning/Charles D. Winters ly .c a ic 17. A hard rubber ball, B not affected by air resistance in its moE tion, is tossed upward C from shoulder height, A falls to the sidewalk, rebounds to a smaller maximum height, and D is caught on its way down again. This moFigure OQ2.17 tion is represented in Figure OQ2.17, where the successive positions of the ball A through E are not equally spaced in time. At point D the center of the ball is at its lowest point in the motion. The motion of the ball is along a straight, vertical line, but the diagram shows successive positions offset to the right to avoid overlapping. Choose the positive y direction to be upward. (a) Rank the situations A through E according to the speed of the ball uvy u at each point, with the largest speed first. (b) Rank the same situations according to the acceleration ay of the ball at each point. (In both rankings, remember that zero is greater than a negative value. If two values are equal, show that they are equal in your ranking.) 18. Each of the strobe photographs (a), (b), and (c) in Figure OQ2.18 was taken of a single disk moving toward the right, which we take as the positive direction. Within each photograph, the time interval between images is constant. (i) Which photograph shows motion with zero acceleration? (ii) Which photograph shows motion with positive acceleration? (iii) Which photograph shows motion with negative acceleration? ee b 16. A ball is thrown straight up in the air. For which situation are both the instantaneous velocity and the acceleration zero? (a) on the way up (b) at the top of its flight path (c) on the way down (d) halfway up and halfway down (e) none of the above s. w 50 Chapter 2 Objective Question 18 and Problem 23. 1. denotes answer available in Student Solutions Manual/Study Guide sw 1. If the average velocity of an object is zero in some time interval, what can you say about the displacement of the object for that interval? w w w .a 2. Try the following experiment away from traffic where you can do it safely. With the car you are driving moving slowly on a straight, level road, shift the transmission into neutral and let the car coast. At the moment the car comes to a complete stop, step hard on the brake and notice what you feel. Now repeat the same experiment on a fairly gentle, uphill slope. Explain the difference in what a person riding in the car feels in the two cases. (Brian Popp suggested the idea for this question.) 3. If a car is traveling eastward, can its acceleration be westward? Explain. 4. If the velocity of a particle is zero, can the particle’s acceleration be zero? Explain. 5. If the velocity of a particle is nonzero, can the particle’s acceleration be zero? Explain. 6. You throw a ball vertically upward so that it leaves the ground with velocity 15.00 m/s. (a) What is its velocity when it reaches its maximum altitude? (b) What is its acceleration at this point? (c) What is the velocity with which it returns to ground level? (d) What is its acceleration at this point? 7. (a) Can the equations of kinematics (Eqs. 2.13–2.17) be used in a situation in which the acceleration varies in time? (b) Can they be used when the acceleration is zero? 8. (a) Can the velocity of an object at an instant of time be greater in magnitude than the average velocity over a time interval containing the instant? (b) Can it be less? 9. Two cars are moving in the same direction in parallel lanes along a highway. At some instant, the velocity of car A exceeds the velocity of car B. Does that mean that the acceleration of car A is greater than that of car B? Explain.    Problems 51 Problems The problems found in this chapter may be assigned online in Enhanced WebAssign AMT   Analysis Model tutorial available in Enhanced WebAssign GP Guided Problem 1. straightforward; 2. intermediate; 3. challenging M   Master It tutorial available in Enhanced 1. full solution available in the Student W   Watch It video solution available in WebAssign BIO Q/C Enhanced WebAssign Solutions Manual/Study Guide age velocity in the time intervals (a) 0 to 2 s, (b) 0 to 4 s, (c) 2 s to 4 s, (d) 4 s to 7 s, and (e) 0 to 8 s. x (m) 10 8 6 4 2 0 t (s) –2 1 2 3 4 5 6 7 8 –4 –6 ee b x (m) 12 10 Problems 1 and 9. s. w 8 ic Figure P2.1 .c 1. The position versus time for a certain particle moving W along the x axis is shown in Figure P2.1. Find the aver- 7. A position–time graph for a particle moving along the x axis is shown in Figure P2.7. (a) Find the average velocity in the time interval t 5 1.50 s to t 5 4.00 s. (b) Determine the instantaneous velocity at t 5 2.00 s by measuring the slope of the tangent line shown in the graph. (c) At what value of t is the velocity zero? ly Section 2.1 Position, Velocity, and Speed om S ys 2. The speed of a nerve impulse in the human body is BIO about 100 m/s. If you accidentally stub your toe in the ar ph dark, estimate the time it takes the nerve impulse to travel to your brain. 3. A person walks first at a constant speed of 5.00 m/s M along a straight line from point A to point B and then sw back along the line from B to A at a constant speed of 3.00 m/s. (a) What is her average speed over the entire trip? (b) What is her average velocity over the entire trip? 4. A particle moves according to the equation x 5 10t 2, .a W where x is in meters and t is in seconds. (a) Find the w average velocity for the time interval from 2.00 s to 3.00 s. (b) Find the average velocity for the time interval from 2.00 to 2.10 s. w w 5. The position of a pinewood derby car was observed at various times; the results are summarized in the following table. Find the average velocity of the car for (a) the first second, (b) the last 3 s, and (c) the entire period of observation. t (s) x (m) 0 0 1.0 2.3 2.0 9.2 3.0 20.7 4.0 36.8 5.0 57.5 Section 2.2 Instantaneous Velocity and Speed 6. The position of a particle moving along the x axis varies in time according to the expression x 5 3t 2, where x is in meters and t is in seconds. Evaluate its position (a) at t 5 3.00 s and (b) at 3.00 s 1 Dt. (c) Evaluate the limit of Dx/Dt as Dt approaches zero to find the velocity at t 5 3.00 s. 6 4 2 0 1 2 3 4 5 6 t (s) Figure P2.7 8. An athlete leaves one end of a pool of length L at t 5 0 and arrives at the other end at time t 1. She swims back and arrives at the starting position at time t 2. If she is swimming initially in the positive x direction, determine her average velocities symbolically in (a) the first half of the swim, (b) the second half of the swim, and (c) the round trip. (d) What is her average speed for the round trip? S 9. Find the instantaneous velocity of the particle W described in Figure P2.1 at the following times: (a) t 5 1.0 s, (b) t 5 3.0 s, (c) t 5 4.5 s, and (d) t 5 7.5 s. Section 2.3 Analysis Model: Particle Under Constant Velocity 10. Review. The North American and European plates of the Earth’s crust are drifting apart with a relative speed of about 25 mm/yr. Take the speed as constant and find when the rift between them started to open, to reach a current width of 2.9 3 103 mi. 11. A hare and a tortoise compete in a race over a straight course 1.00 km long. The tortoise crawls at a speed of 0.200 m/s toward the finish line. The hare runs at a speed of 8.00 m/s toward the finish line for 0.800 km and then stops to tease the slow-moving tortoise as the tortoise eventually passes by. The hare waits for a while after the tortoise passes and then runs toward the finish line again at 8.00 m/s. Both the hare and the tortoise cross the finish line at the exact same instant. Assume both animals, when moving, move steadily at Motion in One Dimension Section 2.4 Acceleration 14. Review. A 50.0-g Super Ball traveling at 25.0 m/s bounces W off a brick wall and rebounds at 22.0 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 3.50 ms, what is the magnitude of the average acceleration of the ball during this time interval? 20 cm 0 40 cm –2 om 17. Figure P2.17 shows a graph of vx versus t for the motion of a motorcyclist as he starts from rest and moves along the road in a straight line. (a) Find the average acceleration for the time interval t 5 0 to t 5 6.00 s. (b) Estimate the time at which the acceleration has its greatest positive value and the value of the acceleration at that instant. (c) When is the acceleration zero? (d) Estimate the maximum negative value of the acceleration and the time at which it occurs. vx (m/s) 10 8 6 4 2 0 ys 5 10 –4 15 20 w .a –6 –8 t (s) sw 8 6 4 2 0 ar ph vx (m/s) w Figure P2.15 w 16. A child rolls a marble on a bent track that is 100 cm long as shown in Figure P2.16. We use x to represent the position of the marble along the track. On the horizontal sections from x 5 0 to x 5 20 cm and from x 5 40 cm to x 5 60 cm, the marble rolls with constant speed. On the sloping sections, the marble’s speed changes steadily. At the places where the slope changes, the marble stays on the track and does not undergo any sudden changes in speed. The child gives the marble some initial speed at x 5 0 and t 5 0 and then watches it roll to x 5 90 cm, where it turns around, eventually returning to x 5 0 with the same speed with which the child released it. Prepare graphs of x versus t, vx versus t, and ax versus t, vertically aligned with their time axes identical, to show the motion of the marble. You will not be able to place numbers other than zero on the 60 cm Figure P2.16 ic 15. A velocity–time graph for an object moving along the x axis is shown in Figure P2.15. (a) Plot a graph of the acceleration versus time. Determine the average acceleration of the object (b) in the time interval t 5 5.00 s to t 5 15.0 s and (c) in the time interval t 5 0 to t 5 20.0 s. v .c 13. A person takes a trip, driving with a constant speed of M 89.5 km/h, except for a 22.0-min rest stop. If the person’s average speed is 77.8 km/h, (a) how much time is spent on the trip and (b) how far does the person travel? 100 cm S ly 12. A car travels along a straight line at a constant speed of AMT 60.0 mi/h for a distance d and then another distance d in the same direction at another constant speed. The average velocity for the entire trip is 30.0 mi/h. (a) What is the constant speed with which the car moved during the second distance d ? (b) What If? Suppose the second distance d were traveled in the opposite direction; you forgot something and had to return home at the same constant speed as found in part (a). What is the average velocity for this trip? (c) What is the average speed for this new trip? horizontal axis or on the velocity or acceleration axes, but show the correct graph shapes. ee b their respective speeds. (a) How far is the tortoise from the finish line when the hare resumes the race? (b) For how long in time was the hare stationary? s. w 52 Chapter 2 2 4 6 8 10 t (s) 12 Figure P2.17 18. (a) Use the data in Problem 5 to construct a smooth graph of position versus time. (b) By constructing tangents to the x(t) curve, find the instantaneous velocity of the car at several instants. (c) Plot the instantaneous velocity versus time and, from this information, determine the average acceleration of the car. (d) What was the initial velocity of the car? 19. A particle starts from rest W and accelerates as shown in Figure P2.19. Determine (a) the particle’s speed at t 5 10.0 s and at t 5 20.0 s, and (b) the distance traveled in the first 20.0 s. ax (m/s2) 2 1 0 1 5 10 15 t (s) 20 2 3 20. An object moves along Figure P2.19 W the x axis according to 2 the equation x 5 3.00t 2 2.00t 1 3.00, where x is in meters and t is in seconds. Determine (a) the average speed between t 5 2.00 s and t 5 3.00 s, (b) the instantaneous speed at t 5 2.00 s and at t 5 3.00 s, (c) the average acceleration between t 5 2.00 s and t 5 3.00 s, and (d) the instantaneous acceleration at t 5 2.00 s and t 5 3.00 s. (e) At what time is the object at rest? 21. A particle moves along the x axis according to the M equation x 5 2.00 1 3.00t 2 1.00t 2, where x is in meters and t is in seconds. At t 5 3.00 s, find (a) the position of the particle, (b) its velocity, and (c) its acceleration.    Problems 53 Section 2.5 Motion Diagrams of 5.00 m/s2 as it comes to rest. (a) From the instant the jet touches the runway, what is the minimum time interval needed before it can come to rest? (b) Can this jet land at a small tropical island airport where the runway is 0.800 km long? (c) Explain your answer. 22. Draw motion diagrams for (a) an object moving to the Q/C right at constant speed, (b) an object moving to the om .c Figure P2.31 (left) Col. John Stapp and his rocket sled are brought to rest in a very short time interval. (right) Stapp’s face is contorted by the stress of rapid negative acceleration. s. w Section 2.6 Analysis Model: Particle Under Constant Acceleration ly the right, which we take as the positive direction. Within each photograph the time interval between images is constant. For each photograph, prepare graphs of x versus t, vx versus t, and ax versus t, vertically aligned with their time axes identical, to show the motion of the disk. You will not be able to place numbers other than zero on the axes, but show the correct shapes for the graph lines. ee b 23. Each of the strobe photographs (a), (b), and (c) in Fig- Q/C ure OQ2.18 was taken of a single disk moving toward 31. Review. Colonel John P. Stapp, USAF, participated in M studying whether a jet pilot could survive emergency BIO ejection. On March 19, 1954, he rode a rocket-propelled sled that moved down a track at a speed of 632 mi/h. He and the sled were safely brought to rest in 1.40 s (Fig. P2.31). Determine (a) the negative acceleration he experienced and (b) the distance he traveled during this negative acceleration. left, Courtesy U.S. Air Force; right, NASA/Photo Researchers, Inc. right and speeding up at a constant rate, (c) an object moving to the right and slowing down at a constant rate, (d) an object moving to the left and speeding up at a constant rate, and (e) an object moving to the left and slowing down at a constant rate. (f) How would your drawings change if the changes in speed were not uniform, that is, if the speed were not changing at a constant rate? 32. Solve Example 2.8 by a graphical method. On the same graph, plot position versus time for the car and the trooper. From the intersection of the two curves, read the time at which the trooper overtakes the car. 33. A truck on a straight road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 20.0 m/s. Then the truck travels for 20.0 s at constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional 5.00 s. (a) How long is the truck in motion? (b) What is the average velocity of the truck for the motion described? 34. Why is the following situation impossible? Starting from rest, a charging rhinoceros moves 50.0 m in a straight line in 10.0 s. Her acceleration is constant during the entire motion, and her final speed is 8.00 m/s. ic 24. The minimum distance required to stop a car moving at 35.0 mi/h is 40.0 ft. What is the minimum stopping distance for the same car moving at 70.0 mi/h, assuming the same rate of acceleration? ar ph ys 25. An electron in a cathode-ray tube accelerates uniformly from 2.00 3 104 m/s to 6.00 3 106 m/s over 1.50 cm. (a) In what time interval does the electron travel this 1.50 cm? (b) What is its acceleration? sw 26. A speedboat moving at 30.0 m/s approaches a no-wake buoy marker 100 m ahead. The pilot slows the boat with a constant acceleration of 23.50 m/s2 by reducing the throttle. (a) How long does it take the boat to reach the buoy? (b) What is the velocity of the boat when it reaches the buoy? .a 27. A parcel of air moving in a straight tube with a constant 2 Q/C acceleration of 24.00 m/s has a velocity of 13.0 m/s at w w w 10:05:00 a.m. (a) What is its velocity at 10:05:01 a.m.? (b) At 10:05:04 a.m.? (c) At 10:04:59 a.m.? (d) Describe the shape of a graph of velocity versus time for this parcel of air. (e) Argue for or against the following statement: “Knowing the single value of an object’s constant acceleration is like knowing a whole list of values for its velocity.” 28. A truck covers 40.0 m in 8.50 s while smoothly slowing W down to a final speed of 2.80 m/s. (a) Find its original speed. (b) Find its acceleration. 29. An object moving with uniform acceleration has a M velocity of 12.0 cm/s in the positive x direction when its x coordinate is 3.00 cm. If its x coordinate 2.00 s later is 25.00 cm, what is its acceleration? 30. In Example 2.7, we investigated a jet landing on an M aircraft carrier. In a later maneuver, the jet comes in Q/C for a landing on solid ground with a speed of 100 m/s, and its acceleration can have a maximum magnitude 35. The driver of a car slams on the brakes when he sees AMT a tree blocking the road. The car slows uniformly W with an acceleration of 25.60 m/s2 for 4.20 s, making straight skid marks 62.4 m long, all the way to the tree. With what speed does the car then strike the tree? 36. In the particle under constant acceleration model, S we identify the variables and parameters vxi , vxf , ax , t, and xf 2 xi. Of the equations in the model, Equations 2.13–2.17, the first does not involve xf 2 xi , the second and third do not contain ax , the fourth omits vxf , and the last leaves out t. So, to complete the set, there should be an equation not involving vxi . (a) Derive it from the others. (b) Use the equation in part (a) to solve Problem 35 in one step. 37. A speedboat travels in a straight line and increases in AMT speed uniformly from vi 5 20.0 m/s to vf 5 30.0 m/s in GP a displacement ∆x of 200 m. We wish to find the time interval required for the boat to move through this 54 Chapter 2 Motion in One Dimension displacement. (a) Draw a coordinate system for this situation. (b) What analysis model is most appropriate for describing this situation? (c) From the analysis model, what equation is most appropriate for finding the acceleration of the speedboat? (d) Solve the equation selected in part (c) symbolically for the boat’s acceleration in terms of vi , vf , and ∆x. (e) Substitute numerical values to obtain the acceleration numerically. (f) Find the time interval mentioned above. the time interval? (d) What is the total distance it travels during the interval in part (c)? 42. At t 5 0, one toy car is set rolling on a straight track Q/C with initial position 15.0 cm, initial velocity 23.50 cm/s, and constant acceleration 2.40 cm/s2. At the same moment, another toy car is set rolling on an adjacent track with initial position 10.0 cm, initial velocity 15.50 cm/s, and constant acceleration zero. (a) At what time, if any, do the two cars have equal speeds? (b) What are their speeds at that time? (c) At what time(s), if any, do the cars pass each other? (d) What are their locations at that time? (e) Explain the difference between question (a) and question (c) as clearly as possible. om 38. A particle moves along the x axis. Its position is given 2 W by the equation x 5 2 1 3t 2 4t , with x in meters and t in seconds. Determine (a) its position when it changes direction and (b) its velocity when it returns to the position it had at t 5 0. 43. Figure P2.43 represents part vx (m/s) a b 50 of the performance data 40 of a car owned by a proud 30 physics student. (a) Calculate the total distance trav20 eled by computing the area 10 c t (s) under the red-brown graph 0 10 20 30 40 50 line. (b) What distance does the car travel between the Figure P2.43 times t 5 10 s and t 5 40 s? (c) Draw a graph of its acceleration versus time between t 5 0 and t 5 50 s. (d) Write an equation for x as a function of time for each phase of the motion, represented by the segments 0a, ab, and bc. (e) What is the average velocity of the car between t 5 0 and t 5 50 s? 39. A glider of length , moves through a stationary pho- .c Q/C togate on an air track. A photogate (Fig. P2.39) is Figure P2.39 Problems 39 and 40. .a 40. A glider of length 12.4 cm moves on an air track with Q/C constant acceleration (Fig P2.39). A time interval of w w w 0.628 s elapses between the moment when its front end passes a fixed point A along the track and the moment when its back end passes this point. Next, a time interval of 1.39 s elapses between the moment when the back end of the glider passes the point A and the moment when the front end of the glider passes a second point B farther down the track. After that, an additional 0.431 s elapses until the back end of the glider passes point B. (a) Find the average speed of the glider as it passes point A. (b) Find the acceleration of the glider. (c) Explain how you can compute the acceleration without knowing the distance between points A and B. 41. An object moves with constant acceleration 4.00 m/s2 and over a time interval reaches a final velocity of 12.0 m/s. (a) If its initial velocity is 6.00 m/s, what is its displacement during the time interval? (b) What is the distance it travels during this interval? (c) If its initial velocity is 26.00 m/s, what is its displacement during Ralph McGrew sw ar ph ys ic s. w ee b ly a device that measures the time interval Dtd during which the glider blocks a beam of infrared light passing across the photogate. The ratio vd 5 ,/Dtd is the average velocity of the glider over this part of its motion. Suppose the glider moves with constant acceleration. (a) Argue for or against the idea that vd is equal to the instantaneous velocity of the glider when it is halfway through the photogate in space. (b) Argue for or against the idea that vd is equal to the instantaneous velocity of the glider when it is halfway through the photogate in time. 44. A hockey player is standing on his skates on a frozen M pond when an opposing player, moving with a uniform speed of 12.0 m/s, skates by with the puck. After 3.00 s, the first player makes up his mind to chase his opponent. If he accelerates uniformly at 4.00 m/s2, (a) how long does it take him to catch his opponent and (b) how far has he traveled in that time? (Assume the player with the puck remains in motion at constant speed.) Section 2.7 Freely Falling Objects Note: In all problems in this section, ignore the effects of air resistance. 45. In Chapter 9, we will define the center of mass of an object and prove that its motion is described by the particle under constant acceleration model when constant forces act on the object. A gymnast jumps straight up, with her center of mass moving at 2.80 m/s as she leaves the ground. How high above this point is her center of mass (a) 0.100 s, (b) 0.200 s, (c) 0.300 s, and (d) 0.500 s thereafter? 46. An attacker at the base of a castle wall 3.65 m high Q/C throws a rock straight up with speed 7.40 m/s from a height of 1.55 m above the ground. (a) Will the rock reach the top of the wall? (b) If so, what is its speed at the top? If not, what initial speed must it have to reach the top? (c) Find the change in speed of a rock thrown straight down from the top of the wall at an initial speed of 7.40 m/s and moving between the same two    Problems 55 56. A package is dropped at time t 5 0 from a helicopter S that is descending steadily at a speed vi . (a) What is the speed of the package in terms of vi , g, and t? (b) What vertical distance d is it from the helicopter in terms of g and t ? (c) What are the answers to parts (a) and (b) if the helicopter is rising steadily at the same speed? om .c 58. A student drives a vx (m/s) 8 moped along a straight road as described 4 by the ­ velocity–time t (s) graph in Figure P2.58. 0 2 4 6 8 10 Sketch this graph in the middle of a 4 sheet of graph paper. 8 (a) Directly above your Figure P2.58 graph, sketch a graph of the position versus time, aligning the time coordinates of the two graphs. (b) Sketch a graph of the acceleration versus time directly below the velocity–time graph, again aligning the time coordinates. On each graph, show the numerical values of x and ax for all points of inflection. (c) What is the acceleration at t 5 6.00 s? (d) Find the position (relative to the starting point) at t 5 6.00 s. (e) What is the moped’s final position at t 5 9.00 s? ic s. w 49. It is possible to shoot an arrow at a speed as high as 100 m/s. (a) If friction can be ignored, how high would an arrow launched at this speed rise if shot straight up? (b) How long would the arrow be in the air? 57. Automotive engineers refer to the time rate of change S of acceleration as the “ jerk.” Assume an object moves in one dimension such that its jerk J is constant. (a) Determine expressions for its acceleration ax(t), velocity vx(t), and position x(t), given that its initial acceleration, velocity, and position are axi , vxi , and xi , respectively. (b) Show that ax 2 5 axi 2 1 2J(vx 2 vxi ). ly 48. A baseball is hit so that it travels straight upward after W being struck by the bat. A fan observes that it takes 3.00 s for the ball to reach its maximum height. Find (a) the ball’s initial velocity and (b) the height it reaches. Section 2.8 Kinematic Equations Derived from Calculus ee b 47. Why is the following situation impossible? Emily challenges David to catch a $1 bill as follows. She holds the bill vertically as shown in Figure P2.47, with the center of the bill between but not touching David’s index finger and thumb. Without warning, Figure P2.47 Emily releases the bill. David catches the bill without moving his hand downward. David’s reaction time is equal to the average human reaction time. © Cengage Learning/George Semple points. (d) Does the change in speed of the downwardmoving rock agree with the magnitude of the speed change of the rock moving upward between the same elevations? (e) Explain physically why it does or does not agree. ar ph ys 50. The height of a helicopter above the ground is given by h 5 3.00t 3, where h is in meters and t is in seconds. At t 5 2.00 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground? 51. A ball is thrown directly downward with an initial W speed of 8.00 m/s from a height of 30.0 m. After what time interval does it strike the ground? .a sw 52. A ball is thrown upward from the ground with an iniM tial speed of 25 m/s; at the same instant, another ball is dropped from a building 15 m high. After how long will the balls be at the same height above the ground? w w w 53. A student throws a set of keys vertically upward to her M sorority sister, who is in a window 4.00 m above. The second student catches the keys 1.50 s later. (a) With what initial velocity were the keys thrown? (b) What was the velocity of the keys just before they were caught? 54. At time t 5 0, a student throws a set of keys vertically S upward to her sorority sister, who is in a window at distance h above. The second student catches the keys at time t. (a) With what initial velocity were the keys thrown? (b) What was the velocity of the keys just before they were caught? 55. A daring ranch hand sitting on a tree limb wishes AMT to drop vertically onto a horse galloping under the tree. The constant speed of the horse is 10.0 m/s, and the distance from the limb to the level of the saddle is 3.00 m. (a) What must be the horizontal distance between the saddle and limb when the ranch hand makes his move? (b) For what time interval is he in the air? 59. The speed of a bullet as it travels down the barrel of a rifle toward the opening is given by v 5 (25.00 3 107)t 2 1 (3.00 3 105)t where v is in meters per second and t is in seconds. The acceleration of the bullet just as it leaves the barrel is zero. (a) Determine the acceleration and position of the bullet as functions of time when the bullet is in the barrel. (b) Determine the time interval over which the bullet is accelerated. (c) Find the speed at which the bullet leaves the barrel. (d) What is the length of the barrel? Additional Problems 60. A certain automobile manufacturer claims that its deluxe sports car will accelerate from rest to a speed of 42.0 m/s in 8.00 s. (a) Determine the average acceleration of the car. (b) Assume that the car moves with constant acceleration. Find the distance the car travels in the first 8.00 s. (c) What is the speed of the car 10.0 s after it begins its motion if it can continue to move with the same acceleration? 61. The froghopper Philaenus spumarius is supposedly the BIO best jumper in the animal kingdom. To start a jump, this insect can accelerate at 4.00 km/s2 over a distance of 2.00 mm as it straightens its specially adapted Motion in One Dimension 10 5 10 t (s) 15 68. Why is the following situation impossible? A freight train is lumbering along at a constant speed of 16.0 m/s. Behind the freight train on the same track is a passenger train traveling in the same direction at 40.0 m/s. When the front of the passenger train is 58.5 m from the back of the freight train, the engineer on the passenger train recognizes the danger and hits the brakes of his train, causing the train to move with acceleration 23.00 m/s2. Because of the engineer’s action, the trains do not collide. s. w 0 top of the elevator car passes a bolt loosely attached to the wall of the elevator shaft, the bolt falls from rest. (a) At what time does the bolt hit the top of the stilldescending elevator? (b) In what way is this problem similar to Example 2.8? (c) Estimate the highest floor from which the bolt can fall if the elevator reaches the ground floor before the bolt hits the top of the elevator. ee b vx (m/s) 20 67. An elevator moves downward in a tall building at a Q/C constant speed of 5.00 m/s. Exactly 5.00 s after the om 62. An object is at x 5 0 at t 5 0 and moves along the x axis according to the velocity–time graph in Figure P2.62. (a) What is the object’s acceleration between 0 and 4.0 s? (b) What is the object’s acceleration between 4.0 s and 9.0 s? (c) What is the object’s acceleration between 13.0 s and 18.0 s? (d) At what time(s) is the object moving with the lowest speed? (e) At what time is the object farthest from x 5 0? (f) What is the final position x of the object at t 5 18.0 s? (g) Through what total distance has the object moved between t 5 0 and t 5 18.0 s? (a) the speed of the woman just before she collided with the ventilator and (b) her average acceleration while in contact with the box. (c) Modeling her acceleration as constant, calculate the time interval it took to crush the box. .c “ jumping legs.” Assume the acceleration is constant. (a) Find the upward velocity with which the insect takes off. (b) In what time interval does it reach this velocity? (c) How high would the insect jump if air resistance were negligible? The actual height it reaches is about 70 cm, so air resistance must be a noticeable force on the leaping froghopper. ly 56 Chapter 2 69. The Acela is an electric train on the Washington–New 10 Q/C York–Boston run, carrying passengers at 170 mi/h. A velocity–time graph for the Acela is shown in Figure P2.69. (a) Describe the train’s motion in each successive time interval. (b) Find the train’s peak positive acceleration in the motion graphed. (c) Find the train’s displacement in miles between t 5 0 and t 5 200 s. ic Figure P2.62 sw ar ph ys 63. An inquisitive physics student and mountain climber M climbs a 50.0-m-high cliff that overhangs a calm pool of water. He throws two stones vertically downward, 1.00 s apart, and observes that they cause a single splash. The first stone has an initial speed of 2.00 m/s. (a) How long after release of the first stone do the two stones hit the water? (b) What initial velocity must the second stone have if the two stones are to hit the water simultaneously? (c) What is the speed of each stone at the instant the two stones hit the water? w w w .a 64. In Figure 2.11b, the area under the velocity–time Q/C graph and between the vertical axis and time t (verS tical dashed line) represents the displacement. As shown, this area consists of a rectangle and a triangle. (a) Compute their areas. (b) Explain how the sum of the two areas compares with the expression on the right-hand side of Equation 2.16. 65. A ball starts from rest and accelerates at 0.500 m/s2 while moving down an inclined plane 9.00 m long. When it reaches the bottom, the ball rolls up another plane, where it comes to rest after moving 15.0 m on that plane. (a) What is the speed of the ball at the bottom of the first plane? (b) During what time interval does the ball roll down the first plane? (c) What is the acceleration along the second plane? (d) What is the ball’s speed 8.00 m along the second plane? 66. A woman is reported to have fallen 144 ft from the 17th floor of a building, landing on a metal ventilator box that she crushed to a depth of 18.0 in. She suffered only minor injuries. Ignoring air resistance, calculate v (mi/h) 200 150 100 50 0 –50 –50 0 50 100 150 200 250 300 350 400 t (s) –100 Figure P2.69 Velocity–time graph for the Acela. 70. Two objects move with initial velocity 28.00 m/s, final velocity 16.0 m/s, and constant accelerations. (a) The first object has displacement 20.0 m. Find its acceleration. (b) The second object travels a total distance of 22.0 m. Find its acceleration. 71. At t 5 0, one athlete in a race running on a long, Q/C straight track with a constant speed v1 is a distance d1 S behind a second athlete running with a constant speed v 2. (a) Under what circumstances is the first athlete able to overtake the second athlete? (b) Find the time t at which the first athlete overtakes the second athlete, in terms of d1, v1, and v 2. (c) At what minimum distance d 2 from the leading athlete must the finish line    Problems 57 Height (m) 2.75 3.00 3.25 3.50 3.75 4.00 4.25 4.50 4.75 5.00 7.62 7.25 6.77 6.20 5.52 4.73 3.85 2.86 1.77 0.58 .c 77. A motorist drives along a straight road at a constant speed of 15.0 m/s. Just as she passes a parked motorcycle police officer, the officer starts to accelerate at 2.00 m/s2 to overtake her. Assuming that the officer maintains this acceleration, (a) determine the time interval required for the police officer to reach the motorist. Find (b) the speed and (c) the total displacement of the officer as he overtakes the motorist. 78. A commuter train travels between two downtown stations. Because the stations are only 1.00 km apart, the train never reaches its maximum possible cruising speed. During rush hour the engineer minimizes the time interval Δt between two stations by accelerating at a rate a1 5 0.100 m/s2 for a time interval Dt 1 and then immediately braking with acceleration a 2 5 20.500 m/s2 for a time interval Dt 2. Find the minimum time interval of travel Dt and the time interval Dt 1. s. w moves with a constant acceleration of 3.50 m/s2, while Kathy maintains an acceleration of 4.90 m/s2. Find (a) the time at which Kathy overtakes Stan, (b) the distance she travels before she catches him, and (c) the speeds of both cars at the instant Kathy overtakes Stan. Time (s) 5.00 5.75 6.40 6.94 7.38 7.72 7.96 8.10 8.13 8.07 7.90 ly 73. Kathy tests her new sports car by racing with Stan, AMT an experienced racer. Both start from rest, but Kathy M leaves the starting line 1.00 s after Stan does. Stan Height (m) 0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 ee b 72. A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.0 m/s at ground level. The engines then fire, and the rocket accelerates upward at 4.00 m/s2 until it reaches an altitude of 1 000 m. At that point, its engines fail and the rocket goes into free fall, with an acceleration of 29.80 m/s2. (a) For what time interval is the rocket in motion above the ground? (b) What is its maximum altitude? (c) What is its velocity just before it hits the ground? (You will need to consider the motion while the engine is operating and the free-fall motion separately.) Time (s) om be located so that the trailing athlete can at least tie for first place? Express d 2 in terms of d1, v1, and v 2 by using the result of part (b). ar ph ys ic 74. Two students are on a balcony a distance h above the S street. One student throws a ball vertically downward at a speed vi ; at the same time, the other student throws a ball vertically upward at the same speed. Answer the following symbolically in terms of vi , g, h, and t. (a) What is the time interval between when the first ball strikes the ground and the second ball strikes the ground? (b) Find the velocity of each ball as it strikes the ground. (c) How far apart are the balls at a time t after they are thrown and before they strike the ground? w w w .a sw y 75. Two objects, A and B, are conQ/C nected by hinges to a rigid S rod that has a length L. The x B objects slide along perpenL dicular guide rails as shown in y S v Figure P2.75. Assume object A u slides to the left with a constant A O speed v. (a) Find the velocity v B x of object B as a function of the Figure P2.75 angle u. (b) Describe v B relative to v. Is v B always smaller than v, larger than v, or the same as v, or does it have some other relationship? 76. Astronauts on a distant planet toss a rock into the Q/C air. With the aid of a camera that takes pictures at a steady rate, they record the rock’s height as a function of time as given in the following table. (a) Find the rock’s average velocity in the time interval between each measurement and the next. (b) Using these average velocities to approximate instantaneous velocities at the midpoints of the time intervals, make a graph of velocity as a function of time. (c) Does the rock move with constant acceleration? If so, plot a straight line of best fit on the graph and calculate its slope to find the acceleration. 79. Liz rushes down onto a subway platform to find her train already departing. She stops and watches the cars go by. Each car is 8.60 m long. The first moves past her in 1.50 s and the second in 1.10 s. Find the constant acceleration of the train. 80. A hard rubber ball, released at chest height, falls to the Q/C pavement and bounces back to nearly the same height. When it is in contact with the pavement, the lower side of the ball is temporarily flattened. Suppose the maximum depth of the dent is on the order of 1 cm. Find the order of magnitude of the maximum acceleration of the ball while it is in contact with the pavement. State your assumptions, the quantities you estimate, and the values you estimate for them. Challenge Problems 81. A blue car of length 4.52 m is moving north on a roadway that intersects another perpendicular roadway (Fig. P2.81, page 58). The width of the intersection from near edge to far edge is 28.0 m. The blue car has a constant acceleration of magnitude 2.10 m/s2 directed south. The time interval required for the nose of the blue car to move from the near (south) edge of the intersection to the north edge of the intersection is 3.10 s. (a) How far is the nose of the blue car from the south edge of the intersection when it stops? (b) For what time interval is any part of the blue car within the boundaries of the intersection? (c) A red car is at rest on the perpendicular intersecting roadway. As the nose of the blue car Motion in One Dimension enters the intersection, the red car starts from rest and accelerates east at 5.60 m/s2. What is the minimum distance from the near (west) edge of the intersection at which the nose of the red car can begin its motion if it is to enter the intersection after the blue car has entirely left the intersection? (d) If the red car begins its motion at the position given by the answer to part (c), with what speed does it enter the intersection? N E om S 28.0 m aR 84. Two thin rods are fastened A to the inside of a circular ring as shown in Figure P2.84. One rod of length D is vertical, and the other of D length L makes an angle u with the horizontal. The two B L rods and the ring lie in a veru tical plane. Two small beads are free to slide without fricC tion along the rods. (a) If the Figure P2.84 two beads are released from rest simultaneously from the positions shown, use your intuition and guess which bead reaches the bottom first. (b) Find an expression for the time interval required for the red bead to fall from point A to point C in terms of g and D. (c) Find an expression for the time interval required for the blue bead to slide from point B to point C in terms of g, L, and u. (d) Show that the two time intervals found in parts (b) and (c) are equal. Hint: What is the angle between the chords of the circle A B and B C? (e) Do these results surprise you? Was your intuitive guess in part (a) correct? This problem was inspired by an article by Thomas B. Greenslade, Jr., “Galileo’s Paradox,” Phys. Teach. 46, 294 (May 2008). .c vB ly aB Figure P2.81 ic 82. Review. As soon as a traffic light turns green, a car speeds up from rest to 50.0 mi/h with constant acceleration 9.00 mi/h/s. In the adjoining bicycle lane, a cyclist speeds up from rest to 20.0 mi/h with constant acceleration 13.0 mi/h/s. Each vehicle maintains constant velocity after reaching its cruising speed. (a) For what time interval is the bicycle ahead of the car? (b) By what maximum distance does the bicycle lead the car? ee b W ahead at the 6.00-s mark, and by how much? (d) What is the maximum distance by which Healan is behind Laura, and at what time does that occur? s. w 58 Chapter 2 85. A man drops a rock into a well. (a) The man hears the sound of the splash 2.40 s after he releases the rock from rest. The speed of sound in air (at the ambient temperature) is 336 m/s. How far below the top of the well is the surface of the water? (b) What If? If the travel time for the sound is ignored, what percentage error is introduced when the depth of the well is calculated? ys w w w .a sw ar ph 83. In a women’s 100-m race, accelerating uniformly, Laura takes 2.00 s and Healan 3.00 s to attain their maximum speeds, which they each maintain for the rest of the race. They cross the finish line simultaneously, both setting a world record of 10.4 s. (a) What is the acceleration of each sprinter? (b) What are their respective maximum speeds? (c) Which sprinter is c h a p t e r 3 Vectors 3.1 Coordinate Systems 3.2 Vector and Scalar Quantities om 3.3 Some Properties of Vectors ar ph ys ic s. w ee b ly .c 3.4 Components of a Vector and Unit Vectors A signpost in Saint Petersburg, Florida, shows the distance and direction to several cities. Quantities that are defined by both a magnitude and a direction are called vector quantities. (Raymond A. Serway) w w .a sw In our study of physics, we often need to work with physical quantities that have both numerical and directional properties. As noted in Section 2.1, quantities of this nature are vector quantities. This chapter is primarily concerned with general properties of vector quantities. We discuss the addition and subtraction of vector quantities, together with some common applications to physical situations. Vector quantities are used throughout this text. Therefore, it is imperative that you master the techniques discussed in this chapter. w 3.1 Coordinate Systems y Many aspects of physics involve a description of a location in space. In Chapter 2, for 10 (x, y) example, we saw that the mathematical description of an object’s motion requires a method for describing the object’s position at various times. In two dimensions, Q 5 this description is accomplished with the use of the Cartesian coordinate system, P (5, 3) (–3, 4) in which perpendicular axes intersect at a point defined as the origin O (Fig. 3.1). Cartesian coordinates are also called rectangular coordinates. x Sometimes it is more convenient to represent a point in a plane by its plane polar 5 O 10 coordinates (r, u) as shown in Figure 3.2a (page 60). In this polar coordinate system, r is Figure 3.1 ​Designation of points the distance from the origin to the point having Cartesian coordinates (x, y) and u in a Cartesian coordinate system. is the angle between a fixed axis and a line drawn from the origin to the point. The Every point is labeled with coordifixed axis is often the positive x axis, and u is usually measured counterclockwise nates (x, y). 59 (x, y) r u 60 Chapter 3 x O Vectors a y (x, y) y sin u = r r r cos u = xr u x O tan u = y x y u x b a (a) The plane polar coordinates of a point are represented by the distance r and the angle u, where u is measured counterclockwise from the positive x axis. (b) The right triangle used to relate (x, y) to (r, u). y sin u = r r y cos u = xr .c om Figure 3.2 ee b ly from it. From the right triangle in Figure 3.2b, we find that sin u 5 y/r and that cos u 5 x/r. (A reviewy of trigonometric functions is given in Appendix B.4.) Therefore, u tan u = starting with the xplane polar coordinates of any point, we can obtain the Cartesian x coordinates by using the equations b Cartesian coordinates  in terms of polar coordinates (3.1) (3.2) x 5 r cos u y 5 r sin u s. w Furthermore, if we know the Cartesian coordinates, the definitions of trigonometry tell us that y tan u 5 (3.3) x ys ic Polar coordinates in terms  of Cartesian coordinates r 5 "x 2 1 y 2 (3.4)   Polar Coordinates w Example 3.1 w .a sw ar ph Equation 3.4 is the familiar Pythagorean theorem. These four expressions relating the coordinates (x, y) to the coordinates (r, u) apply only when u is defined as shown in Figure 3.2a—in other words, when positive u is an angle measured counterclockwise from the positive x axis. (Some scientific calculators perform conversions between Cartesian and polar coordinates based on these standard conventions.) If the reference axis for the polar angle u is chosen to be one other than the positive x axis or if the sense of increasing u is chosen differently, the expressions relating the two sets of coordinates will change. w The Cartesian coordinates of a point in the xy plane are (x, y) 5 (23.50, 22.50) m as shown in Figure 3.3. Find the polar coordinates of this point. Solution Conceptualize ​The drawing in Figure 3.3 helps us conceptualize the problem. We wish to find r and u. We expect r to be a few meters and u to be larger than 180°. y (m) Categorize Based on the statement of the problem and the Conceptualize step, we recognize that we are simply converting from Cartesian coordinates to polar coordinates. We therefore categorize this example as a substitution problem. Substitution problems generally do not have an extensive Analyze step other than the substitution of numbers into a given equation. Similarly, the Finalize step u x (m) Figure 3.3 (Example 3.1) Finding polar coordinates when Cartesian coordinates are given. r (–3.50, –2.50) 3.2 Vector and Scalar Quantities 61 ▸ 3.1 c o n t i n u e d consists primarily of checking the units and making sure that the answer is reasonable and consistent with our expectations. Therefore, for substitution problems, we will not label Analyze or Finalize steps. r 5 "x 2 1 y 2 5 " 1 23.50 m 2 2 1 1 22.50 m 2 2 5 4.30 m Use Equation 3.4 to find r : tan u 5 Use Equation 3.3 to find u: y 22.50 m 5 5 0.714 x 23.50 m om u 5 2168 ly .c Notice that you must use the signs of x and y to find that the point lies in the third quadrant of the coordinate system. That is, u 5 216°, not 35.5°, whose tangent is also 0.714. Both answers agree with our expectations in the Conceptualize step. ee b 3.2 Vector and Scalar Quantities s. w We now formally describe the difference between scalar quantities and vector quantities. When you want to know the temperature outside so that you will know how to dress, the only information you need is a number and the unit “degrees C” or “degrees F.” Temperature is therefore an example of a scalar quantity: ys ic A scalar quantity is completely specified by a single value with an appropriate unit and has no direction. sw ar ph Other examples of scalar quantities are volume, mass, speed, time, and time intervals. Some scalars are always positive, such as mass and speed. Others, such as temperature, can have either positive or negative values. The rules of ordinary arithmetic are used to manipulate scalar quantities. If you are preparing to pilot a small plane and need to know the wind velocity, you must know both the speed of the wind and its direction. Because direction is important for its complete specification, velocity is a vector quantity: w .a A vector quantity is completely specified by a number with an appropriate unit (the magnitude of the vector) plus a direction. B w w Another example of a vector quantity is displacement, as you know from Chapter 2. Suppose a particle moves from some point A to some point B along a straight path as shown in Figure 3.4. We represent this displacement by drawing an arrow from A to B, with the tip of the arrow pointing away from the starting point. The direction of the arrowhead represents the direction of the displacement, and the length of the arrow represents the magnitude of the displacement. If the particle travels along some other path from A to B such as shown by the broken line in Figure 3.4, its displacement is still the arrow drawn from A to B. Displacement depends only on the initial and final positions, so the displacement vector is independent of the path taken by the particle between these two points. S In this text, we use a boldface letter with an arrow over the letter, such as A , to represent a vector. Another common notation for vectors with which you should be S familiar is a simple boldface character: A. The magnitude of the vector A is writS ten either A or 0 A 0 . The magnitude of a vector has physical units, such as meters for displacement or meters per second for velocity. The magnitude of a vector is always a positive number. A Figure 3.4 ​A s a particle moves from A to B along an arbitrary path represented by the broken line, its displacement is a vector quantity shown by the arrow drawn from A to B. 62 Chapter 3 Vectors Q uick Quiz 3.1 ​Which of the following are vector quantities and which are scalar quantities? (a) your age (b) acceleration (c) velocity (d) speed (e) mass y 3.3 Some Properties of Vectors x In this section, we shall investigate general properties of vectors representing physical quantities. We also discuss how to add and subtract vectors using both algebraic and geometric methods. Figure 3.5 ​T hese four vectors are equal because they have equal lengths and point in the same direction. Equality of Two Vectors S om O S ee b ly .c For many purposes, two vectors A and B may be defined to be equal if they have S S the same magnitude and if they point in the same direction. That is, A 5 B only if S S A 5 B and if A and B point in the same direction along parallel lines. For example, all the vectors in Figure 3.5 are equal even though they have different starting points. This property allows us to move a vector to a position parallel to itself in a diagram without affecting the vector. Adding Vectors Pitfall Prevention 3.1 sw ar ph ys ic A 1 B 5 C is very different from A 1 B 5 C. The first equation is a vector sum, which must be handled carefully, such as with the graphical method. The second equation is a simple algebraic addition of numbers that is handled with the normal rules of arithmetic. S S S S (3.5) A 1 B 5 B 1 A .a Commutative law of addition  The rules for adding vectors are conveniently S described by a graphical method. S S To add vector B to vector A , first draw vector A on graph paper, with its magniS tude represented by a convenient length scale, and then draw vector B to the same S scale, with its tail starting from the tip of A , as shown in Figure 3.6. The resultant S S S S S vector R 5 A 1 B is the vector drawn from the tail of A to the tip of B . A geometric construction can also be used to add more than Stwo vectors as S S shown in Figure 3.7 for the case of four vectors. The resultant vector R 5 A 1 B 1 S S S C 1 D is the vector that completes the polygon. In other words, R is the vector drawn from the tail of the first vector to the tip of the last vector. This technique for adding vectors is often called the “head to tail method.” When two vectors are added, the sum is independent of the order of the addition. (This fact may seem trivial, but as you will see in Chapter 11, the order is important when vectors are multiplied. Procedures for multiplying vectors are discussed in Chapters 7 and 11.) This property, which can be seen from the geometric construction in Figure 3.8, is known as the commutative law of addition: s. w Vector Addition Versus Scalar Addition Notice that S S S S S S A Figure 3.6 A5 S S S S B S A S When vector B is S S added to vector A , the resultant R is the vector that runs from the tail of S S A to the tip of B . R5 S B1 S C 1 B 1 S A 1 S B B C S R5 S S S D S B R 5 S S A S D B A1 S S A1 S B w w w Draw B , S then add A . Figure 3.7 Geometric construction for summing four vectors. The S resultant vector R is by definition the one that completes the polygon. S A S Draw A , S then add B . Figure 3.8S ThisSconstruction S S shows that A 1 B 5 B 1 A or, in other words, that vector addition is commutative. 3.3 Some Properties of Vectors S S S S S C S B) 1 C S C C) (B S 1 B; Add A and S then add C to the result. 1 S S (A S 1 B1C A S Figure 3.9 ​G eometric constructions for verifying the associative law of addition. S S Add B and C ; then add the S result to A. 63 S S A1B S S B B S S A A S S S S S .c om When three or more vectors are added, their sum is independent of the way in which the individual vectors are grouped together. A geometric proof of this rule for three vectors is given in Figure 3.9. This property is called the associative law of addition: S A 1 1B 1 C2 5 1A 1 B2 1 C ly (3.6) WW Associative law of addition ys Negative of a Vector ic s. w ee b In summary, a vector quantity has both magnitude and direction and also obeys the laws of vector addition as described in Figures 3.6 to 3.9. When two or more vectors are added together, they must all have the same units and they must all be the same type of quantity. It would be meaningless to add a velocity vector (for example, 60 km/h to the east) to a displacement vector (for example, 200 km to the north) because these vectors represent different physical quantities. The same rule also applies to scalars. For example, it would be meaningless to add time intervals to temperatures. S S Subtracting Vectors ar ph The negative of the vector A is defined as the vector that when added Sto A gives S S S zero for the vector sum. That is, A 1 1 2 A 2 5 0. The vectors A and 2 A have the same magnitude but point in opposite directions. sw The operation of vector subtraction makes use of theSdefinition of the negative of a S S S vector. We define the operation A 2 B as vector 2 B added to vector A : .a S S S S A 2 B 5 A 1 12 B 2 (3.7) w w w The geometric construction for subtracting two vectors in this way is illustrated in Figure 3.10a. Another way of looking atSvectorSsubtraction is to notice that the difference S S A 2 B between two vectors A and B is what you have to add to the second vector S We would draw B here if we were S adding it to A. S S Vector C 5 A 2 B is the vector we must S S add to B to obtain A. S S B S A S S S A2B B 2B S S Adding 2B to A is equivalent to S subtracting B S from A. a S A b S S C5A2B S S FigureS3.10 ​(a) Subtracting S vector B from vector A . The vecS tor 2 B is equal in magnitude to S vector B and points in the opposite direction. (b) A second way of looking at vector subtraction. 64 Chapter 3 Vectors S S to obtain the first. In this case, as Figure 3.10b shows, the vector A 2 B points from the tip of the second vector to the tip of the first. Multiplying a Vector by a Scalar S S If vector A is multiplied by a positive scalar quantity m, the product m A is a vector S S that has the same direction as A and magnitude mA. If vector A is multiplied by S S a negative scalarSquantity 2m, the product 2m A is directed opposite A . For examS S ple, the vector S 5 A is five times as long as A and points in the same direction as A ; S 1 the vector 23 A is one-third the length of A and points in the direction oppoS site A . S S .c S om Q uick Quiz 3.2 The magnitudes of two vectors A and B are A 5 12 units and B 5 8 units. Which pair of numbers represents theSlargest andSsmallest possible S values for the magnitude of the resultant vector R 5 A 1 B ? (a) 14.4 units, 4 units (b) 12 units, 8 units (c) 20 units, 4 units (d) none of these answers S Example 3.2 s. w ee b ly Q uick Quiz 3.3 If vector B is added to vector A , which two of the following S choices must be true for the resultant vector toSbe equal to zero? (a) A and S S B are parallel and in the same S direction. (b) A and B are parallel and S in S S opposite directions. (c) A and B have the same magnitude. (d) A and B are perpendicular.    A Vacation Trip S Conceptualize ​The 40 S ys B ar ph Solution ic y (km) A car travels 20.0 km due north and then 35.0 km in a direction 60.0° west of north as shown in Figure 3.11a. Find the magnitude and direction of the car’s resultant displacement. S R 60.0 sw .a w W 40 E S 20 u S R S A 20 S S b A S vectors A and B drawn in Figure 3.11a help us conceptualize the problem. S The resultant vector R has also been drawn. We expect its magnitude to be a few tens of kilometers. The angle b that the resultant vector makes with the y axis is expected to be less than 60°, the S angle that vector B makes with the y axis. y (km) N 0 220 x (km) a B b 220 0 x (km) b Figure 3.11 ​(Example 3.2) (a) Graphical method for finding the resulS S S tant displacement vector R 5 A 1 B . (b) Adding the vectors in reverse S S S order 1 B 1 A 2 gives the same result for R . S Categorize ​We can categorize this example as a simple analysis problem in vector addition. The displacement R is the S S w w resultant when the two individual displacements A and B are added. We can further categorize it as a problem about the analysis of triangles, so we appeal to our expertise in geometry and trigonometry. Analyze ​In this example, we show two ways to analyze the problem of finding the resultant of two vectors. The first way is S to solve the problem geometrically, using graph paper and a protractor to measure the magnitude of R and its direction in Figure 3.11a. (In fact, even when you know you are going to be carrying out a calculation, you should sketch the vectors to check your results.) With an ordinary ruler and protractor, a large diagram typically gives answers to two-digit but not to S three-digit precision. Try using these tools on R in Figure 3.11a and compare to the trigonometric analysis below! S The second way to solve the problem is to analyze it using algebra and trigonometry. The magnitude of R can be obtained from the law of cosines as applied to the triangle in Figure 3.11a (see Appendix B.4). Use R 2 5 A 2 1 B 2 2 2AB cos u from the law of cosines to find R: Substitute numerical values, noting that u 5 180° 2 60° 5 120°: R 5 "A2 1 B 2 2 2AB cos u R 5 " 1 20.0 km 2 2 1 1 35.0 km 2 2 2 2 1 20.0 km 2 1 35.0 km 2 cos 1208 5 48.2 km 65 3.4 Components of a Vector and Unit Vectors ▸ 3.2 c o n t i n u e d Use the law of sines (Appendix B.4) to find the direction S of R measured from the northerly direction: sin b sin u 5 B R B 35.0 km sin b 5 sin u 5 sin 1208 5 0.629 R 48.2 km b 5 38.9° The resultant displacement of the car is 48.2 km in a direction 38.9° west of north. W h at I f ? om people find using the laws of cosines and sines to be awkward. Second, a triangle only results if you are adding two vectors. If you are adding three or more vectors, the resulting geometric shape is usually not a triangle. In Section 3.4, we explore a new method of adding vectors that will address both of these disadvantages. .c estimate made by looking at Figure 3.11a or with an actual angle measured from the diagram using the graphical S method? Is it reasonable that the magnitude of R is larger S S S than that of both A and B ? Are the units of R correct? Although the head to tail method of adding vectors works well, it suffers from two disadvantages. First, some ly Finalize Does the angle b that we calculated agree with an ee b Suppose the trip were taken with the two vectors in reverse order: 35.0 km at 60.0° west of north first and then 20.0 km due north. How would the magnitude and the direction of the resultant vector change? Answer They would not change. The commutative law for vector addition tells us that the order of vectors in an ic s. w addition is irrelevant. Graphically, Figure 3.11b shows that the vectors added in the reverse order give us the same resultant vector. ys 3.4 Components of a Vector and Unit Vectors w w w .a sw ar ph The graphical method of adding vectors is not recommended whenever high accuracy is required or in three-dimensional problems. In this section, we describe a method of adding vectors that makes use of the projections of vectors along coordinate axes. These projections are called the components of the vector or its rectangular components. Any vector can be completely described by its components. S Consider a vector A lying in the xy plane and making an arbitrary angle u with the positive x axis as shown in S Figure 3.12a. This vector can be expressed as the S sum of two other component vectors A x , which is parallel to the x axis, and A y , which is parallel to the y axis. S From SFigure 3.12b, we see that the three vectors form a S right triangle and that A 5 A 1 A x y. We shall often refer to the “components S of a vector A ,” written Ax and Ay (without the boldface notation). The compoS nent Ax represents the projection of A along the x axis, and the component Ay S represents the projection of A along the y axis. These components can be positive S or negative. The component Ax is positiveSif the component vector A x points in the positive x direction and is negative if A x points in the negative x direction. A similar statement is made for the component Ay. y y Figure 3.12 S S S A Ay u O a S Ax A x u O b S Ax S Ay x S (​ a) A vector A lying in the xy plane can be represented by its component vectors S S A x and A y . (b) The y component S vector A y can be moved to the S right so that it adds to A x. The vector sum of the component S vectors is A . These three vectors form a right triangle. Vectors From Figure 3.12 and the definition of sine andScosine, we see that cos u 5 Ax /A and that sin u 5 Ay /A. Hence, the components of A are Pitfall Prevention 3.2 x and y Components Equations 3.8 and 3.9 associate the cosine of the angle with the x component and the sine of the angle with the y component. This association is true only because we measured the angle u with respect to the x axis, so do not memorize these equations. If u is measured with respect to the y axis (as in some problems), these equations will be incorrect. Think about which side of the triangle containing the components is adjacent to the angle and which side is opposite and then assign the cosine and sine accordingly. Ax 5 A cos u (3.8) A y 5 A sin u (3.9) The magnitudes of these components are the lengths of the two sides of a right triS angle with a hypotenuse of length A. Therefore, the magnitude and direction of A are related to its components through the expressions A 5 "Ax2 1 Ay2 Ay Ax b om u 5 tan21 a (3.10) (3.11) Ax points left and is 2 Ax points right and is 1 Ay points up and is 1 Ay points up and is 1 Ax points left and is 2 x Ax points right and is 1 Ay points down and is 2 Ay points down and is 2 ic Figure 3.13 ​The signs Sof the c­ omponents of a vector A depend on the quadrant in which the vector is located. ee b y ly .c Notice that the signs of the components Ax and A y depend on the angle u. For example, if u 5 120°, Ax is negative and Ay is positive. If u 5 225°, both Ax and A y are S negative. Figure 3.13 summarizes the signs of the components when A lies in the various quadrants. S When solving problems, you can specify a vector A either with its components Ax and A y or with its magnitude and direction A and u. Suppose you are working a physics problem that requires resolving a vector into its components. In many applications, it is convenient to express the components in a coordinate system having axes that are not horizontal and vertical but that are still perpendicular to each other. For example, we will consider the motion of objects sliding down inclined planes. For these examples, it is often convenient to orient the x axis parallel to the plane and the y axis perpendicular to the plane. s. w 66 Chapter 3 ar ph ys Q uick Quiz 3.4 ​Choose the correct response to make the sentence true: A component of a vector is (a) always, (b) never, or (c) sometimes larger than the magnitude of the vector. Unit Vectors z w w w .a sw Vector quantities often are expressed in terms ofy unit vectors. A unit vector is a dimensionless vector having a magnitude of exactly 1. Unit vectors are used to specify a given direction and have no other physical significance. They are used solely as a bookkeeping convenience in describing a direction in space. We shall use the symbols ^i, ^j, and k^ to represent unit vectors pointing in the positive x, y, and z x directions, respectively. (The “hats,” or circumflexes, on the symbols are a standard ĵ î notation for unit vectors.) The unit vectors ^i, ^j, and k^ form a set of mutually perpendicular vectors in a right-handed coordinate system as shown in Figure 3.14a. The magnitude of each unit vector equals 1; that is, 0 i^ 0 5k̂ 0 j^ 0 5 0 k^ 0 5 1. S product Consider a vector A lying in the xy plane as shown in Figure 3.14b. The S of the component Ax and the unit vector ^i is the component vector A x 5 A x i^ , y a y x ĵ î S k̂ Figure 3.14 (a) The unit vectors ^i, ^j, and k^ are directed along the x, y, and z axes, respectively. (b) VecS tor A 5 Ax i^ 1 Ay ^j lying in the xy plane has components Ax and Ay . A y ĵ z A x î b a y S A x 67 3.4 Components of a Vector and Unit Vectors S S y which lies on the x axis and has magnitude 0 Ax 0 . Likewise, A y 5 A y j is the com­ 0 Ay 0 lying on the y axis. Therefore, the unit-vector ponent vector of magnitude S ­notation for the vector A is S A 5 Ax i^ 1 A y ^j (x, y) S r (3.12) For example, consider a point lying in the xy plane and having Cartesian coordinates (x, y) as in Figure 3.15. The point can be specified by the position vector S r, which in unit-vector form is given by S r 5 x i^ 1 y ^j (3.13) S x î x O Figure 3.15 ​The point whose Cartesian coordinates are (x, y) can be represented by the position vector S r 5 x ^i 1 y ^j. .c om This notation tells us that the components of r are the coordinates x and y. Now let us see how to use components to add vectors when the graphical method S S is not sufficiently accurate. Suppose we wish to add vector to vector in EquaB A S tion 3.12, where vector B has components Bx and By. Because of the bookkeeping convenience of the unit vectors, all we do is add the x and y components separately. S S S The resultant vector R 5 A 1 B is y ĵ or S R 5 1 Ax 1 Bx 2 i^ 1 1 Ay 1 By 2 ^j ee b R 5 1 Ax i^ 1 Ay j^ 2 1 1 Bx i^ 1 By j^ 2 ly S y (3.14) R x 5 A x 1 Bx R y 5 A y 1 By s. w S Because R 5 R x i^ 1 R y ^j, we see that the components of the resultant vector are (3.15) ar ph ys ic Therefore, we see that in the component method of adding vectors, we add all the x components together to find the x component of the resultant vector and use the same process for the y components. We can check this addition by components with a geometric construction as shown in Figure 3.16. S The magnitude of R and the angle it makes with the x axis are obtained from its components using the relationships sw R 5 "R x2 1 R y2 5 " 1 Ax 1 Bx 2 2 1 1 Ay 1 By 2 2 tan u 5 Ry Rx 5 Ay 1 By Ax 1 B x (3.16) (3.17) w w w .a At times, we need to consider situations involving motion in three component directions. The extension of our methods to three-dimensional vectors is straightS S forward. If A and B both have x, y, and z components, they can be expressed in the form S (3.18) A 5 Ax i^ 1 Ay j^ 1 Az k^ S S B 5 Bx i^ 1 By j^ 1 Bz k^ (3.19) R 5 1 Ax 1 Bx 2 i^ 1 1 Ay 1 By 2 j^ 1 1 Az 1 Bz 2 k^ (3.20) S The sum of A and B is S Notice that Equation 3.20 differs from Equation 3.14: in Equation 3.20, the resulS tant vector also has a z component R z 5 Az 1 Bz . If a vector R has x, y, and z components, the magnitude of the vector is R 5 !R x2 1 R y2 1 R z2 . The angle ux S that R makes with the x axis is found from the expression cos ux 5 R x /R, with similar expressions for the angles with respect to the y and z axes. The extension of our method to adding more than two vectors is also straightS S S forward. For example, A 1 B 1 C 5 1 Ax 1 Bx 1 Cx 2 i^ 1 1 Ay 1 By 1 Cy 2 j^ 1 1 Az 1 Bz 1 Cz 2 k^ . We have described adding displacement vectors in this section because these types of vectors are easy to visualize. We can also add other types of Ry S By R Ay S B S A x Bx Ax Rx Figure 3.16 This geometric construction for the sum of two vectors shows the relationship between the components of the S resultant R and the components of the individual vectors. Pitfall Prevention 3.3 Tangents on Calculators Equation 3.17 involves the calculation of an angle by means of a tangent function. Generally, the inverse tangent function on calculators provides an angle between 290° and 190°. As a consequence, if the vector you are studying lies in the second or third quadrant, the angle measured from the positive x axis will be the angle your calculator returns plus 180°. 68 Chapter 3 Vectors vectors, such as velocity, force, and electric field vectors, which we will do in later chapters. Q uick Quiz 3.5 ​For which of the following vectors isSthe magnitude of the ­vector equal to one of theScomponents of the vector? (a) A 5 2 i^ 1 5 ^j S (b) B 5 23 ^j (c) C 5 15 k^ Example 3.3    The Sum of Two Vectors S S Find the sum of two displacement vectors A and B lying in the xy plane and given by A 5 1 2.0 i^ 1 2.0 j^ 2 m and B 5 1 2.0 i^ 2 4.0 j^ 2 m .c Solution S om S ly Conceptualize You can conceptualize the situation by drawing the vectors on graph paper. Draw an approximation of the expected resultant vector. S ee b Categorize We categorize this example as a simple substitution problem. Comparing this expression for A with S the general expression A 5 Ax i^ 1 Ay j^ 1 Az k^ , we see that Ax 5 2.0 m, A y 5 2.0 m, and Az 5 0. Likewise, Bx 5 2.0 m, By 5 24.0 m, and Bz 5 0. We can use a two-dimensional approach because there are no z components. S S S S R 5 A 1 B 5 1 2.0 1 2.0 2 i^ m 1 1 2.0 2 4.0 2 j^ m s. w Use Equation 3.14 to obtain the resultant vector R : S Evaluate the components of R : Rx 5 4.0 m ic S tan u 5 Ry Rx 5 22.0 m 5 20.50 4.0 m ar ph Find the direction of R from Equation 3.17: R 5 "Rx2 1 Ry2 5 " 1 4.0 m 2 2 1 1 22.0 m 2 2 5 "20 m 5 4.5 m ys S Use Equation 3.16 to find the magnitude of R : Ry 5 22.0 m    The Resultant Displacement .a Example 3.4 sw Your calculator likely gives the answer 227° for u 5 tan21(20.50). This answer is correct if we interpret it to mean 27° clockwise from the x axis. Our standard form has been to quote the angles measured counterclockwise from the 1x axis, and that angle for this vector is u 5 333° . w w A particle undergoes three consecutive displacements: DS r 1 5 1 15 i^ 1 30 j^ 1 12 k^ 2 cm, DS r 2 5 1 23 i^ 2 14 j^ 2 5.0 k^ 2 cm, S and D r 3 5 1 213 i^ 1 15 j^ 2 cm. Find unit-vector notation for the resultant displacement and its magnitude. w Solution Conceptualize Although x is sufficient to locate a point S in one dimension, we need a vector r to locate a point in S two or three dimensions. The notation D r is a generalization of the one-dimensional displacement Dx in Equation 2.1. Three-dimensional displacements are more difficult to conceptualize than those in two dimensions because they cannot be drawn on paper like the latter. For this problem, let us imagine that you start with your pencil at the origin of a piece of graph paper on which you have drawn x and y axes. Move your pencil 15 cm to the right along the x axis, then 30 cm upward along the y axis, and then 12 cm perpendicularly toward you away from the graph paper. This procedure provides the displacement described by DS r 1. From this point, move your pencil 23 cm to the right parallel to the x axis, then 14 cm parallel to the graph paper in the 2y direction, and then 5.0 cm perpendicularly away from you toward the graph paper. You are now at the displacement from the origin described by DS r 1 1 DS r 2. From this point, move your pencil 13 cm to the left in the 2x direction, and (finally!) 15 cm parallel to the graph paper along the y axis. Your final position is at a displacement DS r 1 1 DS r 2 1 DS r3 from the origin. 69 3.4 Components of a Vector and Unit Vectors ▸ 3.4 c o n t i n u e d Categorize Despite the difficulty in conceptualizing in three dimensions, we can categorize this problem as a substitution problem because of the careful bookkeeping methods that we have developed for vectors. The mathematical manipulation keeps track of this motion along the three perpendicular axes in an organized, compact way, as we see below. DS r 5 DS r 1 1 DS r 2 1 DS r3 5 1 15 1 23 2 13 2 i^ cm 1 1 30 2 14 1 15 2 j^ cm 1 1 12 2 5.0 1 0 2 k^ cm 5 1 25 i^ 1 31 j^ 1 7.0 k^ 2 cm To find the resultant displacement, add the three vectors: .c 5 " 1 25 cm 2 2 1 1 31 cm 2 2 1 1 7.0 cm 2 2 5 40 cm    Taking a Hike ly Example 3.5 om R 5 "R x2 1 R y2 1 R z2 Find the magnitude of the resultant vector: N y (km) (A) D ​ etermine the components of the hiker’s displacement for each day. 20 s. w ee b A hiker begins a trip by first walking 25.0 km southeast from her car. She stops and sets up her tent for the night. On the second day, she walks 40.0 km in a direction 60.0° north of east, at which point she discovers a forest ranger’s tower. Solution S ys ic 3.17. If we denote the displacement vectors on the first and second days by A and S B , respectively, and use the car as the origin of coordinates, we obtain the vectors shown in Figure 3.17. The sketch allows us to estimate the resultant vector as shown. ar ph S S Categorize Having drawn the resultant R , we can now categorize this problem 0 Car 210 220 Tower S R 10 Conceptualize ​We conceptualize the problem by drawing a sketch as in Figure E W 45.0 20 S A S B 30 40 x (km) 60.0 Tent Figure 3.17 ​(Example 3.5) The sw total displacement of the hiker is as one we’ve solved before: an addition of two vectors. You should now have a S S S the vector R 5 A 1 B . hint of the power of categorization in that many new problems are very similar to problems we have already solved if we are careful to conceptualize them. Once we have drawn the displacement vectors and categorized the problem, this problem is no longer about a hiker, a walk, a car, a tent, or a tower. It is a problem about vector addition, one that we have already solved. S .a Analyze ​Displacement A has a magnitude of 25.0 km and is directed 45.0° below the positive x axis. S w Find the components of A using Equations 3.8 and 3.9: Ax 5 A cos 1 245.08 2 5 1 25.0 km 2 1 0.707 2 5 17.7 km w Ay 5 A sin 1 245.08 2 5 1 25.0 km 2 1 20.707 2 5 217.7 km w The negative value of Ay indicates that the hiker walks in the negative y direction on the first day. The signs of Ax and Ay also are evident from Figure 3.17. S Find the components of B using Equations 3.8 and 3.9: Bx 5 B cos 60.08 5 1 40.0 km 2 1 0.500 2 5 20.0 km By 5 B sin 60.08 5 1 40.0 km 2 1 0.866 2 5 34.6 km S S (B) ​Determine the components of the hiker’s resultant displacement R for the trip. Find an expression for R in terms of unit vectors. Solution Use Equation 3.15 to find the components of the resulS S S tant displacement R 5 A 1 B : Rx 5 Ax 1 Bx 5 17.7 km 1 20.0 km 5 37.7 km Ry 5 Ay 1 By 5 217.7 km 1 34.6 km 5 17.0 km continued 70 Chapter 3 Vectors ▸ 3.5 c o n t i n u e d S R 5 1 37.7 i^ 1 17.0 j^ 2 km Write the total displacement in unit-vector form: Finalize ​Looking at the graphical representation in Figure 3.17, we estimate the position of the tower to be about S (38 km, 17 km), which is consistent with the components of R in our result for the final position of the hiker. Also, S both components of R are positive, putting the final position in the first quadrant of the coordinate system, which is also consistent with Figure 3.17. After reaching the tower, the hiker wishes to return to her car along a single straight line. What are the components of the vector representing this hike? What should the direction of the hike be? W h at I f ? S S S R car 5 2 R 5 1 237.7 i^ 2 17.0 j^ 2 km om Answer ​The desired vector R car is the negative of vector R : S R car,y R car,x 5 217.0 km 5 0.450 237.7 km s. w ee b which gives an angle of u 5 204.2°, or 24.2° south of west. ly tan u 5 .c The direction is found by calculating the angle that the vector makes with the x axis: Definitions sw Concepts and Principles Vector quantities have both magnitude and direction and obey the laws of vector addition. The magnitude of a vector is always a positive number. ar ph Scalar quantities are those that have only a numerical value and no associated direction. ys ic Summary w w w .a When two or more vectors are added together, they must all have the same units and they all must be the S same type of quantity. We can add two vectors A and S B graphically. In this method (Fig. 3.6), the resultant S S S S vector R 5 A 1 B runs from the tail of A to the S tip of B . S If a vector A has an x component Ax and a y component Ay, the vector can be expressed in unit-vector form S as A 5 Ax i^ 1 Ay ^j. In this notation, ^i is a unit vector pointing in the positive x direction and ^j is a unit vector pointing in the positive y direction. Because ^i and ^j are unit vectors, 0 i^ 0 5 0 j^ 0 5 1. A second method of adding vectors involves com­ ponents of the vectors. The x component Ax of the S S ­vector A is equal to the projection of A along the x axis of a coordinate system, where Ax 5 A cos u. S S The y component Ay of A is the projection of A along the y axis, where Ay 5 A sin u. We can find the resultant of two or more vectors by resolving all vectors into their x and y components, adding their resultant x and y components, and then using the Pythagorean theorem to find the magnitude of the resultant vector. We can find the angle that the resultant vector makes with respect to the x axis by using a suitable trigonometric function. Conceptual Questions Objective Questions 71 1. denotes answer available in Student Solutions Manual/Study Guide S S of S the 3. Figure OQ3.3 shows two vectors D 1 and D 2. Which S possibilities (a) through (d) is the vector D 2 2 2 D 1, or (e) is it none of them? S must be in which quadrant, (a) the first, (b) the second, (c) the third, or (d) the fourth, or (e) is more than one answer possible? 7. Yes or no: Is each of the following quantities a vector? (a) force (b) temperature (c) the volume of water in a can (d) the ratings of a TV show (e) the height of a building (f) the velocity of a sports car (g) the age of the Universe 8. What is the y component of the vector 1 3 i^ 2 8 k^ 2 m/s? (a) 3 m/s (b) 28 m/s (c) 0 (d) 8 m/s (e) none of those answers om 1. What is the magnitude of the vector 1 10 i^ 2 10 k^ 2 m/s? (a) 0 (b) 10 m/s (c) 210 m/s (d) 10 (e) 14.1 m/s 2. A vector lying in the xy plane has components of opposite sign. The vector must lie in which quadrant? (a) the first quadrant (b) the second quadrant (c) the third quadrant (d) the fourth quadrant (e) either the second or the fourth quadrant 9. What is the x component of the vector shown in Figure OQ3.9? (a) 3 cm (b) 6 cm (c) 24 cm (d) 26 cm (e) none of those answers .c D1 S ly D2 a b c ee b y (cm) d Figure OQ3.3 s. w Figure OQ3.9 d e Figure OQ3.4 S w .a 5. TheSmagnitude of vector A is 8 km, and the magnitude following are possible valof B is 6 km. Which of the S S ues for the magnitude of A 1 B ? Choose all possible answers. (a) 10 km (b) 8 km (c) 2 km (d) 0 (e) 22 km S w w 6. Let vector A point from the originSinto the second the quadrant of the xy plane and vector B point from S S origin into the fourth quadrant. The vector B 2 A Conceptual Questions 2 x (cm) 22 Objective Questions 9 and 10. 10. What is the y component of the vector shown in Figure OQ3.9? (a) 3 cm (b) 6 cm (c) 24 cm (d) 26 cm (e) none of those answers ic ys ar ph c sw b 0 24 22 4. The cutting tool on a lathe is given two displacements, one of magnitude 4 cm and one of magnitude 3 cm, in each one of five situations (a) through (e) diagrammed in Figure OQ3.4. Rank these situations according to the magnitude of the total displacement of the tool, putting the situation with the greatest resultant magnitude first. If the total displacement is the same size in two situations, give those letters equal ranks. a 2 S 11. Vector A lies in the xy plane. Both of its components will be negative if it points from the origin into which quadrant? (a) the first quadrant (b) the second quadrant (c) the third quadrant (d) the fourth quadrant (e) the second or fourth quadrants 12. A submarine dives from the water surface at an angle of 30° below the horizontal, following a straight path 50 m long. How far is the submarine then below the water surface? (a) 50 m (b) (50 m)/sin 30° (c) (50 m) sin 30° (d) (50 m) cos 30° (e) none of those answers 13. A vector points from the origin into the second quadrant of the xy plane. What can you conclude about its components? (a) Both components are positive. (b) The x component is positive, and the y component is negative. (c) The x component is negative, and the y component is positive. (d) Both components are negative. (e) More than one answer is possible. 1. denotes answer available in Student Solutions Manual/Study Guide 1. Is it possible to add a vector quantity to a scalar quantity? Explain. 2. Can the magnitude of a vector have a negative value? Explain. 3. A book is moved once around the perimeter of a tabletop with the dimensions 1.0 m by 2.0 m. The book ends up at its initial position. (a) What is its displacement? (b) What is the distance traveled? S 4. If the component of vector A along the direction of vector S B is zero, what can you conclude about the two vectors? 5. On a certain calculator, the inverse tangent function returns a value between 290° and 190°. In what cases will this value correctly state the direction of a vector in the xy plane, by giving its angle measured counterclockwise from the positive x axis? In what cases will it be incorrect? 72 Chapter 3 Vectors Problems AMT   Analysis Model tutorial available in The problems found in this chapter may be assigned online in Enhanced WebAssign Enhanced WebAssign GP Guided Problem 1. straightforward; 2. intermediate; 3. challenging M   Master It tutorial available in Enhanced 1. full solution available in the Student W   Watch It video solution available in WebAssign BIO Q/C Enhanced WebAssign Solutions Manual/Study Guide Section 3.1 Coordinate Systems S 3. Two points in the xy plane have Cartesian coordinates (2.00, 24.00) m and (23.00, 3.00) m. Determine (a) the distance between these points and (b) their polar coordinates. 4. Two points in a plane have polar coordinates (2.50 m, W 30.0°) and (3.80 m, 120.0°). Determine (a) the Carte- .c ly S 10. A force F 1 of magnitude 6.00 units acts on an object at the origin in a direction u 5 30.0° above the positive x axis (Fig. P3.10). A S second force F 2 of magnitude 5.00 units acts on the object in the direction of the positive y axis. Find graphically the magnitude andSdirection of the resulS tant force F 1 1 F 2. S F2 S F1 ic sian coordinates of these points and (b) the distance between them. 9. Why is the following situation impossible? A skater glides along a circular path. She defines a certain point on the circle as her origin. Later on, she passes through a point at which the distance she has traveled along the path from the origin is smaller than the magnitude of her displacement vector from the origin. ee b 2. The rectangular coordinates of a point are given by (2, y), and its polar coordinates are (r, 30°). Determine (a) the value of y and (b) the value of r. S the resultant vector A 1 B points in the negative y direction with a magnitude of 14 units. Find the magS nitude and direction of B . s. w 1. The polar coordinates of a point are r 5 5.50 m and W u 5 240°. What are the Cartesian coordinates of this point? om S ys u ar ph 5. The polar coordinates of a certain point are (r 5 4.30 cm, u 5 214°). (a) Find its Cartesian coordinates x and y. Find the polar coordinates of the points with Cartesian coordinates (b) (2x, y), (c) (22x, 22y), and (d) (3x, 23y). sw 6. Let the polar coordinates of the point (x, y) be (r, u). S Determine the polar coordinates for the points (a) (2x, y), (b) (22x, 22y), and (c) (3x, 23y). Section 3.2 Vector and Scalar Quantities .a Section 3.3 Some Properties of Vectors w 7. A surveyor measures the distance across a straight river W by the following method (Fig. P3.7). Starting directly w w across from a tree on the opposite bank, she walks d 5 100 m along the riverbank to establish a baseline. Then she sights across to the tree. The angle from her baseline to the tree is u 5 35.0°. How wide is the river? u d Figure P3.7 S 8. Vector A has a magnitude of 29 units and points Sin S the positive y direction. When vector B is added to A , Figure P3.10 S 11. The S displacement vectors A M and B shown in Figure P3.11 both have magnitudes of 3.00 Sm. The direction of vec5 30.0°. Find gra­ tor A is u S S S S A 1 B , (b)S A 2 S B, phically (a) S S (c) B 2 A , and (d) A 2 2 B . (Report all angles counterclockwise from the positive x axis.) S y S B S A u O x Figure P3.11 12. Three displacements are A 5 Problems 11 and 22. S B 5 250 m Q/C 200 m due south, S due west, and C 5 150 m at 30.0° east of north. (a) Construct a separate diagram for each of the following posS S S S 5 A 1 B 1 C; sible ways of adding these vectors: R1 S S S S S S S S R 2 5 B 1 C 1 A ; R 3 5 C 1 B 1 A . (b) Explain what you can conclude from comparing the diagrams. 13. A roller-coaster car moves 200 ft horizontally and then rises 135 ft at an angle of 30.0° above the horizontal. It next travels 135 ft at an angle of 40.0° downward. What is its displacement from its starting point? Use graphical techniques. 14. A plane flies from base camp to Lake A, 280 km away in the direction 20.0° north of east. After dropping off supplies, it flies to Lake B, which is 190 km at 30.0° west of north from Lake A. Graphically determine the distance and direction from Lake B to the base camp. 73 Problems Section 3.4 Components of a Vector and Unit Vectors this information to find the displacement from Dallas to Chicago. 15. A vector has an x component of 225.0 units and a y W component of 40.0 units. Find the magnitude and direction of this vector. S 16. Vector A has a magnitude of 35.0 units and points in the direction 325° counterclockwise from the positive x axis. Calculate the x and y components of this vector. 730 mi Dallas Atlanta om 5.00 Figure P3.24 .c 25. Your dog is running around the grass in your back M yard. He undergoes successive displacements 3.50 m south, 8.20 m northeast, and 15.0 m west. What is the resultant displacement? S S B S 5 26. Given the vectors A 5 2.00 i^ 1 6.00 ^j and S S sum C 5 A 1 B W 3.00 i^ 2 2.00 j^ , (a) draw theS vector S S and theSvector dif­ ference D 5 A 2 B . (b) Calculate S S C and D , in terms of unit vectors. (c) Calculate C and S D in terms of polar coordinates, with angles measured with respect to the positive x axis. s. w 19. Obtain expressions in component form for the posiM tion vectors having the polar coordinates (a) 12.8 m, 150°; (b) 3.30 cm, 60.0°; and (c) 22.0 in., 215°. 560 mi ly 18. A person walks 25.0° north of east for 3.10 km. How far would she have to walk due north and due east to arrive at the same location? 21.0 ee b 17. A minivan travels straight north in the right lane of a Q/C divided highway at 28.0 m/s. A camper passes the minivan and then changes from the left lane into the right lane. As it does so, the camper’s path on the road is a straight displacement at 8.50° east of north. To avoid cutting off the minivan, the north–south distance between the camper’s back bumper and the minivan’s front bumper should not decrease. (a) Can the camper be driven to satisfy this requirement? (b) Explain your answer. Chicago 27. A novice golfer on the green takes three strokes to sink the ball. The successive displacements of the ball are 4.00 m to the north, 2.00 m northeast, and 1.00 m at 30.0° west of south (Fig. P3.27). Starting at the same initial point, an expert golfer could make the hole in what single displacement? N ic 20. A girl delivering newspapers covers her route by traveling 3.00 blocks west, 4.00 blocks north, and then 6.00 blocks east. (a) What is her resultant displacement? (b) What is the total distance she travels? sw ar ph ys 21. While exploring a cave, a spelunker starts at the entrance and moves the following distances in a horizontal plane. She goes 75.0 m north, 250 m east, 125 m at an angle u 5 30.0° north of east, and 150 m south. Find her resultant displacement from the cave entrance. Figure P3.21 suggests the situation but is not drawn to scale. u .a N W w w Cave entrance E S Final position Figure P3.21 S w 22. Use S the component method to add the vectors A vectors have magand B shown in Figure P3.11. Both S of nitudes of 3.00 m and vector A makes an angle S S u 5 30.0° with the x axis. Express the resultant A 1 B in unit-vector notation. S S ^i 2 23. Consider the two vectors A 5 3Si^ 2 S2 ^j and BS5 2S S S ^ 0 (a) A 1 B , (b) A 2 B , (c)S A S 1 B 0, M 4 j. Calculate S S 0 0 A 1 B 2 B and (e) the directions of and A , (d) S S A 2 B. 24. A map suggests that Atlanta is 730 miles in a direction of 5.00° north of east from Dallas. The same map shows that Chicago is 560 miles in a direction of 21.0° west of north from Atlanta. Figure P3.24 shows the locations of these three cities. Modeling the Earth as flat, use W 2.00 m 1.00 m E S 30.0 4.00 m Figure P3.27 28. A snow-covered ski slope makes an angle of 35.0° with the horizontal. When a ski jumper plummets onto the hill, a parcel of splashed snow is thrown up to a maximum displacement of 1.50 m at 16.0° from the vertical in the uphill direction as shown in Figure P3.28. Find the components of its maximum displacement (a) parallel to the surface and (b) perpendicular to the surface. 16.0 35.0 Figure P3.28 29. The helicopter view in Fig. P3.29 (page 74) shows two mule. The person on W people pulling on a stubborn S the right pulls with a force F 1 of magnitude 120 N Vectors and direction of u1 5 60.0°. The person onSthe left pulls with a force F 2 of magnitude 80.0 N and direction of u2 5 75.0°. Find (a) the single force that is equivalent to the two forces shown and (b) the force that a third person would have to exert on the mule to make the resultant force equal to zero. The forces are measured in units of newtons (symbolized N). student has learned that a single equation cannot be solved to determine values for more than one unknown in it. How would you explain to him that both a and b can be determined from the single equation used in part (a)? y S F2 u2 S 38. Three displacement vectors of a cro­quet ball are S shown in Figure 0 0 5 20.0 units, A P3.38, where S S 0 B 0 5 40.0 units, and 0 C 0 5 30.0 units. Find (a) the resultant in unitvector notation and (b) the magnitude and direction of the resultant displacement. F1 u1 x y S B S A 45.0 O x 45.0 S om 74 Chapter 3 C 39. A man pushing a mop across a floor Figure P3.38 M causes it to undergo two displacements. The first has a magnitude of 150 cm and makes an angle of 120° with the positive x axis. The resultant displacement has a magnitude of 140 cm and is directed at an angle of 35.0° to the positive x axis. Find the magnitude and direction of the second displacement. 31. Consider theS three displacement vectors A 5 S W 1 3 i^ 2 3 j^ 2 m, B 5 1 i^ 2 4 j^ 2 m, and C 5 1 22 i^ 1 5 j^ 2 m. Use the component methodSto determine (a) the S S S magnitude and direction of D 5 A 1SB 1 C Sand (b) the magnitude and direction of E 5 2 A 2 S S B 1 C. BIO and female (f) anatomies. The displacements d 1m and ly ee b 40. Figure P3.40 illustrates typical proportions of male (m) S S s. w d 1f from the soles of the feet to the navel have magnitudes of 104 cm and 84.0 cm, respectively. The disS S placements d 2m and d 2f from the navel to outstretched fingertips have magnitudes of 100 cm and 86.0 cm, respectively. Find the vector sum of these displacements S S S d 3 5 d 1 1 d 2 for both people. ic S .c 30. In a game of American footFigure P3.29 ball, a quarterback takes the ball from the line of scrimmage, runs backward a distance of 10.0 yards, and then runs sideways parallel to the line of scrimmage for 15.0 yards. At this point, he throws a forward pass downfield 50.0 yards perpendicular to the line of scrimmage. What is the magnitude of the football’s resultant displacement? S ar ph ys 32. Vector A has x and y components of 28.70 cm and S W 15.0 cm, respectively; vector B has x and y components of 13.2 cm and 26.60 cm, respectively. S S S S If A 2 B 1 3 C 5 0, what are the components of C ? S .a sw 33. The vector A has x, y, and z components of 8.00, units, respectively. (a) Write a vector M 12.0, and 24.00 S (b) Obtain a expression for A in unit-vector notation. S unit-vector S expression for a vector B one-fourth the S length of A pointing in the same direction as AS. (c) Obtain a unit-vector S expression for a vector C three times the length of SA pointing in the direction opposite the direction of A . S S w w 34. Vector B has x, y, and z components of 4.00, 6.00, and 3.00 units, respectively. Calculate (a) the magnitude of S S B and (b) the angle that B makes with each coordinate axis. w 35. Vector A has a negative x component 3.00 units in 2.00 units in length. M length and a positive y component S (a) Determine an expression for A in unit-vector notaS tion. (b) Determine the magnitude and direction of A . S S (c) What vector B when added to A gives a resultant vector with no x component and a negative y component 4.00 units in length? S 36. GivenSthe displacement vectors A 5 1 3 i^ 2 4 j^ 1 4 k^ 2 m W and B 5 1 2 i^ 1 3 j^ 2 7 k^ 2 m, find the magnitudes of the following vectors and express each inSterms of its S S S C 5 A 1 B (b) D 5 rectangular components. (a) S S 2A 2 B S S 37. (a) Taking A 5 1 6.00 i^ 2 8.00 j^ 2 units, B 5 1 28.00 i^ 1 S Q/C 3.00 j^ 2 units, and C 5 1 26.0 i^ 1 19.0 j^ 2 units, deterS S S mine a and b such that a A 1 b B 1 C 5 0. (b) A S d2m S d2f 23.0 28.0 S S d1m d1f Figure P3.40 41. Express in unit-vector notation the following vectors, S each of which has magnitude 17.0 cm. (a) Vector E is directed 27.0° S counterclockwise from the positive x axis. (b) Vector F is directed 27.0°Scounterclockwise from the positive y axis. (c) Vector G is directed 27.0° clockwise from the negative y axis. 42. A radar station locates a sinking ship at range 17.3 km and bearing 136° clockwise from north. From the same station, a rescue plane is at horizontal range 19.6 km, 153° clockwise from north, with elevation 2.20 km. (a) Write the position vector for the ship relative to the plane, letting ^i represent east, ^j north, and k^ up. (b) How far apart are the plane and ship? 43. Review. As it passes over Grand Bahama Island, the AMT eye of a hurricane is moving in a direction 60.08 north GP of west with a speed of 41.0 km/h. (a) What is the unit- vector expression for the velocity of the hurricane? Problems 45. Review. You are standing on the ground at the origin Ea st Additional Problems .c om 48. A fly lands on one wall Figure P3.47 W of a room. The lowerleft corner of the wall is selected as the origin of a two-dimensional Cartesian coordinate system. If the fly is located at the point having coordinates (2.00, 1.00) m, (a) how far is it from the origin? (b) What is its location in polar coordinates? 49. As she picks up her riders, a bus driver traverses four successive displacements represented by the expression 1 26.30 b 2 i^ 2 1 4.00 b cos 408 2 i^ 2 1 4.00 b sin 408 2 ^j AMT of a coordinate system. An airplane flies over you with 1 1 3.00 b cos 508 2 i^ 2 1 3.00 b sin 508 2 j^ 2 1 5.00 b 2 ^j Here b represents one city block, a convenient unit of distance of uniform size; ^i is east; and ^j is north. The displacements at 40° and 50° represent travel on roadways in the city that are at these angles to the main east–west and north–south streets. (a) Draw a map of the successive displacements. (b) What total distance did she travel? (c) Compute the magnitude and direction of her total displacement. The logical structure of this problem and of several problems in later chapters was suggested by Alan Van Heuvelen and David Maloney, American Journal of Physics 67(3) 252–256, March 1999. S P0 S ar ph ys ic s. w constant velocity parallel to the x axis and at a fixed height of 7.60 3 103 m. At time t 5 0, the airplane is directlySabove you so that the vector leading from you to it is P 0 5 7.60 3 103 j^ m. At t 5 30.0 s, the position vector leading from you to the airplane is S P 30 5 1 8.04 3 103 i^ 1 7.60 3 103 j^ 2 m as suggested in Figure P3.45. Determine the magnitude and orientation of the airplane’s position vector at t 5 45.0 s. rth No ly 44. Why is the following situation impossible? A shopper pushing a cart through a market follows directions to the canned goods and moves through a displacement 8.00 ^i m down one aisle. He then makes a 90.0° turn and moves 3.00 m along the y axis. He then makes another 90.0° turn and moves 4.00 m along the x axis. Every shopper who follows these directions correctly ends up 5.00 m from the starting point. circle of radius 3.70 cm that lies in a north– south vertical plane. Find (a) the magnitude of the total displacement of the object and (b) the angle the total displacement makes with the vertical. ee b It maintains this velocity for 3.00 h, at which time the course of the hurricane suddenly shifts due north, and its speed slows to a constant 25.0 km/h. This new velocity is maintained for 1.50 h. (b) What is the unitvector expression for the new velocity of the hurricane? (c) What is the unit-vector expression for the displacement of the hurricane during the first 3.00 h? (d) What is the unit-vector expression for the displacement of the hurricane during the latter 1.50 h? (e) How far from Grand Bahama is the eye 4.50 h after it passes over the island? 75 sw P30 Figure P3.45 51. A person going for a walk follows the path shown in M Figure P3.51. The total trip consists of four straightline paths. At the end of the walk, what is the person’s resultant displacement measured from the starting point? y w w w .a y 46. In Figure P3.46, the line seg(16, 12) Q/C ment represents a path from the point with position vector 1 5 i^ 1 3 j^ 2 m to the point with A location (16 i^ 1 12 ^j) m. Point (5, 3) x A is along this path, a fraction f O of the way to the destination. (a) Find the position vector of Figure P3.46 Point point A in terms of f. (b) Evalu- A is a fraction f of the ate the expression from part distance from the ini(a) for f 5 0. (c) Explain whether tial point (5, 3) to the the result in part (b) is reason- final point (16, 12). able. (d) Evaluate the expression for f 5 1. (e) Explain whether the result in part (d) is reasonable. 50. A jet airliner, moving initially at 300 mi/h to the east, suddenly enters a region where the wind is blowing at 100 mi/h toward the direction 30.0° north of east. What are the new speed and direction of the aircraft relative to the ground? 47. In an assembly operation illustrated in Figure P3.47, a robot moves an object first straight upward and then also to the east, around an arc forming one-quarter of a circle of radius 4.80 cm that lies in an east–west vertical plane. The robot then moves the object upward and to the north, through one-quarter of a Start 100 m x 300 m End 200 m 60.0 30.0 150 m Figure P3.51 52. Find the horizontal and vertical components of the 100-m displacement of a superhero who flies from the Vectors 53. Review. The biggest 100 m animal in the world is a snake 420 m long, constructed by Norwegian children. Suppose the snake is laid Figure P3.52 out in a park as shown in Figure P3.53, forming two straight sides of a 105° angle, with one side 240 m long. Olaf and Inge run a race they invent. Inge runs directly from the tail of the snake to its head, and Olaf starts from the same place at the same moment but runs along the snake. Figure P3.53 (a) If both children run steadily at 12.0 km/h, Inge reaches the head of the snake how much earlier than Olaf? (b) If Inge runs the race again at a constant speed of 12.0 km/h, at what constant speed must Olaf run to reach the end of the snake at the same time as Inge? AMT stuffed S S equal mag59. Two vectors A and B haveSprecisely S nitudes. For the magnitude of AS 1 BS to be 100 times larger than the magnitude of A 2 B , what must be the angle between them? S S equal magni60. Two vectors A and B haveSprecisely S to be larger than A 1 B of S tudes. For the magnitude S S the magnitude of A 2 B by the factor n, what must be the angle between them? S 61. Let A 5 60.0 cm at 270° measured from the horiS Q/C zontal. Let B 5 80.0 cm at some angle u. (a) Find the S S the magnitude of A 1 B as a function of u. (b) From S S answer to part (a), for what value of u does 0 A 1 B 0 take on its maximum value? What is this maximum value? (c) From the answer to part (a), for what value S S of u does 0 A 1 B 0 take on its minimum value? What is this minimum value? (d) Without reference to the answer to part (a), argue that the answers to each of parts (b) and (c) do or do not make sense. ar ph ys ic 54. An air-traffic controller observes two aircraft on his radar screen. The first is at altitude 800 m, horizontal distance 19.2 km, and 25.0° south of west. The second aircraft is at altitude 1 100 m, horizontal distance 17.6 km, and 20.0° south of west. What is the distance between the two aircraft? (Place the x axis west, the y axis south, and the z axis vertical.) 58. A ferry transports tourists between three islands. It sails from the first island to the second island, 4.76 km away, in a direction 37.0° north of east. It then sails from the second island to the third island in a direction 69.0° west of north. Finally it returns to the first island, sailing in a direction 28.0° east of south. Calculate the distance between (a) the second and third islands and (b) the first and third islands. om x 30.0 S 57. A vector is given by R 5 2 i^ 1 j^ 1 3 k^ . Find (a) the magnitudesSof the x, y, and z components; (b) the magS nitude of R ; and (c) the angles between R and the x, y, and z axes. .c y ly top of a tall building following the path shown in Figure P3.52. ee b Chapter 3 s. w 76 w w w .a sw 55. In Figure P3.55, a spider is x resting after starting to spin y its web. The gravitational Tx force on the spider makes it Ty exert a downward force of 0.150 N on the junction of the three strands of silk. The junction is supported by difFigure P3.55 ferent tension forces in the two strands above it so that the resultant force on the junction is zero. The two sloping strands are perpendicular, and we have chosen the x and y directions to be along them. The tension Tx is 0.127 N. Find (a) the tension Ty, (b) the angle the x axis makes with the horizontal, and (c) the angle the y axis makes with the horizontal. 56. The rectangle shown in Figure P3.56 has sides parallel to the x and y axes. The position vectors S A 5 10.0 m of two corners are S at 50.0° and B 5 12.0 m at 30.0°. (a) Find the perimeter of the rectangle. (b) Find the magnitude and direction of the vector from the origin to the upper-right corner of the rectangle. y S A S B Figure P3.56 x 62. After a ball rolls off the edge of a horizontal table at time t 5 0, its velocity as a function of time is given by S v 5 1.2 i^ 2 9.8t ^j S where v is in meters per second and t is in seconds. The ball’s displacement away from the edge of the table, during the time interval of 0.380 s for which the ball is in flight, is given by DS r 53 0.380 s S v dt 0 To perform the integral, you can use the calculus theorem 3 3 A 1 Bf 1 x 2 4 dx 5 3 A dx 1 B 3 f 1 x 2 dx You can think of the units and unit vectors as constants, represented by A and B. Perform the integration to calculate the displacement of the ball from the edge of the table at 0.380 s. 63. Review. The instantaneous position of an object is W specified by its position vector leading from a fixed Q/C origin to the location of the object, modeled as a particle. Suppose for a certain object the position vector is a function of time given by S r 5 4 i^ 1 3 j^ 2 2t k^ , where S r /dt. r is in meters and t is in seconds. (a) Evaluate d S r /dt represent about (b) What physical quantity does d S the object? 77 Problems 67. A pirate has buried his treasure on an island with five Q/C trees located at the points (30.0 m, 220.0 m), (60.0 m, z a S R2 om B E c ys S x y ic O s. w b .c 65. A rectangular parallelepiped has dimensions a, b, and S c as shown in Figure P3.65. (a) Obtain a vector expresS is the sion for the face diagonal vector R 1. (b) What S ^ , and R , c k magnitude of this vector? (c) Notice that 1 S R 2 make a right triangle. Obtain a vector expression S for the body diagonal vector R 2. 80.0 m), (210.0 m, 210.0 m), (40.0 m, 230.0 m), and (270.0 m, 60.0 m), all measured relative to some origin, as shown in Figure P3.67. His ship’s log instructs you to start at tree A and move toward tree B, but to cover only one-half the distance between A and B. Then move toward tree C, covering one-third the distance between your current location and C. Next move toward tree D, covering one-fourth the distance between where you are and D. Finally move toward tree E, covering one-fifth the distance between you and E, stop, and dig. (a) Assume you have correctly determined the order in which the pirate labeled the trees as A, B, C, D, and E as shown in the figure. What are the coordinates of the point where his treasure is buried? (b) What If? What if you do not really know the way the pirate labeled the trees? What would happen to the answer if you rearranged the order of the trees, for instance, to B (30 m, 220 m), A (60 m, 80 m), E (210 m, 210 m), C (40 m, 230 m), and D (270 m, 60 m)? State reasoning to show that the answer does not depend on the order in which the trees are labeled. ly den as latitude 0.002 43 degree south of the equator, longitude 75.642 38 degrees west. They wish to visit a tree at latitude 0.001 62 degree north, longitude 75.644 26 degrees west. (a) Determine the straightline distance and the direction in which they can walk to reach the tree as follows. First model the Earth as a sphere of radius 6.37 3 10 6 m to determine the westward and northward displacement components required, in meters. Then model the Earth as a flat surface to complete the calculation. (b) Explain why it is possible to use these two geometrical models together to solve the problem. Challenge Problem ee b 64. Ecotourists use their global positioning system indica- Q/C tor to determine their location inside a botanical gar- R1 ar ph y Figure P3.65 S S w w w .a sw 66. Vectors A and equal magnitudes of 5.00. B have S S vector 6.00 ^j. Determine The sum of A andS B is the S the angle between A and B . x C A D Figure P3.67 c h a p t e r 4 om Motion in Two Dimensions 4.1 The Position, Velocity, and Acceleration Vectors .c 4.2 Two-Dimensional Motion with Constant Acceleration ly 4.3 Projectile Motion ee b 4.4 Analysis Model: Particle in Uniform Circular Motion s. w 4.5 Tangential and Radial Acceleration In this chapter, we explore the kinematics of a particle moving in two dimensions. Knowing the basics of two-dimensional motion will allow us—in future chapters—to examine a variety of situations, ranging from the motion of satellites in orbit to the motion of electrons in a uniform electric field. We begin by studying in greater detail the vector nature of position, velocity, and acceleration. We then treat projectile motion and uniform circular motion as special cases of motion in two dimensions. We also discuss the concept of relative motion, which shows why observers in different frames of reference may measure different positions and velocities for a given particle. w .a sw Fireworks erupt from the Sydney Harbour Bridge in New South Wales, Australia. Notice the parabolic paths of embers projected into the air. All projectiles follow a parabolic path in the absence of air resistance. (Graham Monro/ ar ph ys ic 4.6 Relative Velocity and Relative Acceleration w w Photolibrary/Jupiter Images) 4.1 The Position, Velocity, and Acceleration Vectors In Chapter 2, we found that the motion of a particle along a straight line such as the x axis is completely known if its position is known as a function of time. Let us now extend this idea to two-dimensional motion of a particle in the xy plane. We begin by describing the position of the particle. In one dimension, a single numerical value describes a particle’s position, but in two dimensions, we indicate its position by its position vector S r, drawn from the origin of some coordinate system to the location of the particle in the xy plane as in Figure 4.1. At time ti , the particle is at point A, described by position vector S r i. At some later time tf , it is at point B, described by position vector S r f . The path followed by the particle from 78 79 4.1 The Position, Velocity, and Acceleration Vectors A to B is not necessarily a straight line. As the particle moves from A to B in the time interval Dt 5 tf 2 ti , its position vector changes from S r i to S r f . As we learned in Chapter 2, displacement is a vector, and the displacement of the particle is the difference between its final position and its initial position. We now define the displacement vector DS r for a particle such as the one in Figure 4.1 as being the difference between its final position vector and its initial position vector: r ;S rf 2 S ri DS (4.1) WWDisplacement vector S S v avg ; ee b ly .c om The direction of D r is indicated in Figure 4.1. As we see from the figure, the magr is less than the distance traveled along the curved path followed by the nitude of DS particle. As we saw in Chapter 2, it is often useful to quantify motion by looking at the displacement divided by the time interval during which that displacement occurs, which gives the rate of change of position. Two-dimensional (or three-dimensional) kinematics is similar to one-dimensional kinematics, but we must now use full vector notation rather than positive and negative signs to indicate the direction of motion. We define the average velocity S v avg of a particle during the time interval Dt as the displacement of the particle divided by the time interval: DS r Dt (4.2) ic s. w Multiplying or dividing a vector quantity by a positive scalar quantity such as Dt changes only the magnitude of the vector, not its direction. Because displacement is a vector quantity and the time interval is a positive scalar quantity, we conclude that the average velocity is a vector quantity directed along DS r . Compare Equation 4.2 with its one-dimensional counterpart, Equation 2.2. The average velocity between points is independent of the path taken. That is because average velocity is proportional to displacement, which depends only on the initial and final position vectors and not on the path taken. As with onedimensional motion, we conclude that if a particle starts its motion at some point and returns to this point via any path, its average velocity is zero for this trip because its displacement is zero. Consider again our basketball players on the court in Figure 2.2 (page 23). We previously considered only their one-dimensional motion back and forth between the baskets. In reality, however, they move over a two-dimensional surface, running back and forth between the baskets as well as left and right across the width of the court. Starting from one basket, a given player may follow a very complicated two-dimensional path. Upon returning to the original basket, however, a player’s average velocity is zero because the player’s displacement for the whole trip is zero. Consider again the motion of a particle between two points in the xy plane as shown in Figure 4.2 (page 80). The dashed curve shows the path of the particle. As the time interval over which we observe the motion becomes smaller and smaller— that is, as B is moved to B9 and then to B0 and so on—the direction of the displacement approaches that of the line tangent to the path at A. The instantaneous velocity S v is defined as the limit of the average velocity DS r /Dt as Dt approaches zero: WW Average velocity The displacement of the S particle is the vector r. ti A Sr B tf S ri S rf Path of particle O S DS r dS r 5 dt Dt S 0 Dt v ; lim (4.3) That is, the instantaneous velocity equals the derivative of the position vector with respect to time. The direction of the instantaneous velocity vector at any point in a particle’s path is along a line tangent to the path at that point and in the direction of motion. Compare Equation 4.3 with the corresponding one-dimensional version, Equation 2.5. The magnitude of the instantaneous velocity vector v 5 0 S v 0 of a particle is called the speed of the particle, which is a scalar quantity. x Figure 4.1 ​A particle moving in the xy plane is located with the position vector S r drawn from the origin to the particle. The displacement of the particle as it moves from A to B in the time interval Dt 5 tf 2 ti is equal to the vector DS r 5S rf 2 S r i. w w w .a sw ar ph ys y WW Instantaneous velocity 80 Chapter 4 Motion in Two Dimensions Figure 4.2 ​A s a particle moves between two points, its average velocity is in the direction of the r . By definidisplacement vector DS tion, the instantaneous velocity at A is directed along the line tangent to the curve at A. As the end point approaches A, t approaches zero and the direction S of r approaches that of the green line tangent to the curve at A. y S A Direction of v at A S S S As the end point of the path is moved from B to Bto B, the respective displacements and corresponding time intervals become smaller and smaller. .c B om r1 r2 r3 B ee b ly B x O ys ic s. w As a particle moves from one point to another along some path, its instantaneous velocity vector changes from S vi at time ti to S vf at time tf . Knowing the velocity at these points allows us to determine the average acceleration of the particle. The average acceleration S a avg of a particle is defined as the change in its instantaneous velocity vector DS v divided by the time interval Dt during which that change occurs: S a avg ; S vf 2 S vi DS v 5 Dt tf 2 ti (4.4) ar ph Average acceleration  w .a sw Because S a avg is the ratio of a vector quantity DS v and a positive scalar quantity Dt, we conclude that average acceleration is a vector quantity directed along DS v . As indicated in Figure 4.3, the direction of DS v is found by adding the vector 2S v i (the negative of S vi) to the vector S vf because, by definition, DS v 5S vf 2 S v i. Compare Equation 4.4 with Equation 2.9. When the average acceleration of a particle changes during different time intervals, it is useful to define its instantaneous acceleration. The instantaneous accela is defined as the limiting value of the ratio DS v /Dt as Dt approaches zero: eration S w w Instantaneous acceleration  S a ; lim Dt S 0 S Dv dS v 5 Dt dt In other words, the instantaneous acceleration equals the derivative of the velocity vector with respect to time. Compare Equation 4.5 with Equation 2.10. Various changes can occur when a particle accelerates. First, the magnitude of the velocity vector (the speed) may change with time as in straight-line (one-­ Pitfall Prevention 4.1 y Vector Addition Although the vector addition discussed in Chapter 3 involves displacement vectors, vector addition can be applied to any type of vector quantity. Figure 4.3, for example, shows the addition of velocity vectors using the graphical approach. (4.5) A vi S ri O vf v S B Figure 4.3 ​A particle moves from position A to position B. Its velocity vector changes from S v i to S vf . The vector diagrams at the upper right show two ways of determining the vector DS v from the initial and final velocities. S S S S rf –vi or vf S vi S v S vf x 4.2 Two-Dimensional Motion with Constant Acceleration 81 dimensional) motion. Second, the direction of the velocity vector may change with time even if its magnitude (speed) remains constant as in two-dimensional motion along a curved path. Finally, both the magnitude and the direction of the velocity vector may change simultaneously. Q uick Quiz 4.1 ​Consider the following controls in an automobile in motion: gas pedal, brake, steering wheel. What are the controls in this list that cause an acceleration of the car? (a) all three controls (b) the gas pedal and the brake (c) only the brake (d) only the gas pedal (e) only the steering wheel om 4.2 T  wo-Dimensional Motion with Constant Acceleration .a sw ar ph ys ic s. w ee b ly .c In Section 2.5, we investigated one-dimensional motion of a particle under constant acceleration and developed the particle under constant acceleration model. Let us now consider two-dimensional motion during which the acceleration of a particle remains constant in both magnitude and direction. As we shall see, this approach is useful for analyzing some common types of motion. Before embarking on this investigation, we need to emphasize an important point regarding two-dimensional motion. Imagine an air hockey puck moving in a straight line along a perfectly level, friction-free surface of an air hockey table. Figure 4.4a shows a motion diagram from an overhead point of view of this puck. Recall that in Section 2.4 we related the acceleration of an object to a force on the object. Because there are no forces on the puck in the horizontal plane, it moves with constant velocity in the x direction. Now suppose you blow a puff of air on the puck as it passes your position, with the force from your puff of air exactly in the y direction. Because the force from this puff of air has no component in the x direction, it causes no acceleration in the x direction. It only causes a momentary acceleration in the y direction, causing the puck to have a constant y component of velocity once the force from the puff of air is removed. After your puff of air on the puck, its velocity component in the x direction is unchanged as shown in Figure 4.4b. The generalization of this simple experiment is that motion in two dimensions can be modeled as two independent motions in each of the two perpendicular directions associated with the x and y axes. That is, any influence in the y direction does not affect the motion in the x direction and vice versa. The position vector for a particle moving in the xy plane can be written S r 5 x i^ 1 y ^j (4.6) S w w w where x, y, and r change with time as the particle moves while the unit vectors ^i and ^j remain constant. If the position vector is known, the velocity of the particle can be obtained from Equations 4.3 and 4.6, which give The horizontal red vectors, representing the x component of the velocity, are the same length in both parts of the figure, which demonstrates that motion in two dimensions can be modeled as two independent motions in perpendicular directions. S v 5 dS r dx ^ dy ^ 5 i1 j 5 v x ^i 1 v y ^j dt dt dt (4.7) y x a y x b Figure 4.4 ​(a) A puck moves across a horizontal air hockey table at constant velocity in the x direction. (b) After a puff of air in the y direction is applied to the puck, the puck has gained a y component of velocity, but the x component is unaffected by the force in the perpendicular direction. 82 Chapter 4 Motion in Two Dimensions Because the acceleration S a of the particle is assumed constant in this discussion, its components ax and ay also are constants. Therefore, we can model the particle as a particle under constant acceleration independently in each of the two directions and apply the equations of kinematics separately to the x and y components of the velocity vector. Substituting, from Equation 2.13, vxf 5 vxi 1 axt and vyf 5 vyi 1 ayt into Equation 4.7 to determine the final velocity at any time t, we obtain S vf 5 1 v xi 1 a xt 2 i^ 1 1 v yi 1 a yt 2 j^ 5 1 v xi ^i 1 v yi ^j 2 1 1 a x ^i 1 a y ^j 2 t S vf 5 S vi 1 S at Velocity vector as  a function of time for a particle under constant acceleration in two dimensions (4.8) ly .c om This result states that the velocity of a particle at some time t equals the vector sum of its initial velocity S vi at time t 5 0 and the additional velocity S a t acquired at time t as a result of constant acceleration. Equation 4.8 is the vector version of Equation 2.13. Similarly, from Equation 2.16 we know that the x and y coordinates of a particle under constant acceleration are yf 5 yi 1 v yit 1 12a yt 2 ee b x f 5 x i 1 v xit 1 12a xt 2 Substituting these expressions into Equation 4.6 (and labeling the final position vector S r f ) gives S 5 1 x i ^i 1 yi ^j 2 1 1 v xi ^i 1 v yi ^j 2 t 1 12 1 a x ^i 1 a y ^j 2 t 2 rf 5 S ri 1 S v i t 1 12S at2 (4.9) which is the vector version of Equation 2.16. Equation 4.9 tells us that the position vector S r f of a particle is the vector sum of the original position S r i, a displacement S vi t arising from the initial velocity of the particle, and a displacement 12S a t 2 resulting from the constant acceleration of the particle. We can consider Equations 4.8 and 4.9 to be the mathematical representation of a two-dimensional version of the particle under constant acceleration model. Graphical representations of Equations 4.8 and 4.9 are shown in Figure 4.5. The components of the position and velocity vectors are also illustrated in the figure. Notice from Figure 4.5a that S vf is generally not along the direction of either S vi or S a because the relationship between these quantities is a vector expression. For the same reason, from Figure 4.5b we see that S r f is generally not along the direction of S S r i, vi, or S a . Finally, notice that S vf and S r f are generally not in the same direction. w .a sw ar ph ys Position vector as  a function of time for a particle under constant acceleration in two dimensions ic S s. w r f 5 1 x i 1 v xit 1 12a xt 2 2 i^ 1 1 yi 1 v yit 1 12a yt 2 2 ^j w w y y ayt S vf vyf vyi Vector representations and components of (a) the velocity and (b) the position of a particle under constant acceleration in two dimensions. at 1 a t2 2 y yf S vi x S vit yi S ri axt x vxit xi vxf a xf b 1S at 2 2 rf vyit S vxi Figure 4.5 S 1 a t2 2 x 4.2 Two-Dimensional Motion with Constant Acceleration Example 4.1 83    Motion in a Plane AM A particle moves in the xy plane, starting from the origin at t 5 0 with an initial velocity having an x component of 20 m/s and a y component of 215 m/s. The particle experiences an acceleration in the x direction, given by ax 5 4.0 m/s2. (A) Determine the total velocity vector at any time. Solution Conceptualize The components of the initial velocity tell y ee b ly .c om x us that the particle starts by moving toward the right and downward. The x component of velocity starts at 20 m/s and increases by 4.0 m/s every second. The y component of velocity never changes from its initial value of 215 m/s. We sketch a motion diagram of the situation in Figure 4.6. Because the particle is accelerating in the 1x direction, its velocity component in this direction increases and the path curves as shown in the diagram. Notice that the spacing between successive images increases as time goes on because the speed is increasing. The placement of the acceleration and velocity vectors in Figure 4.6 ​(Example 4.1) Motion diagram for the particle. Figure 4.6 helps us further conceptualize the situation. s. w Categorize Because the initial velocity has components in both the x and y directions, we categorize this problem ic as one involving a particle moving in two dimensions. Because the particle only has an x component of acceleration, we model it as a particle under constant acceleration in the x direction and a particle under constant velocity in the y direction. S vf 5 S vi 1 S a t 5 1 v xi 1 a xt 2 i^ 1 1 v yi 1 a yt 2 ^j ar ph Use Equation 4.8 for the velocity vector: ys Analyze To begin the mathematical analysis, we set vxi 5 20 m/s, vyi 5 215 m/s, ax 5 4.0 m/s2, and ay 5 0. S Substitute numerical values with the velocity in meters per second and the time in seconds: (1) vf 5 3 20 1 1 4.0 2 t 4 i^ 1 3 215 1 1 0 2 t 4 ^j S vf 5 3 1 20 1 4.0t 2 i^ 2 15 j^ 4 sw Finalize Notice that the x component of velocity increases in time while the y component remains constant; this result is consistent with our prediction. Analyze w Solution .a (B) Calculate the velocity and speed of the particle at t 5 5.0 s and the angle the velocity vector makes with the x axis. w w Evaluate the result from Equation (1) at t 5 5.0 s: S Determine the angle u that vf makes with the x axis at t 5 5.0 s: Evaluate the speed of the particle as the magnitude S of vf : S vf 5 3 1 20 1 4.0 1 5.0 2 2 i^ 2 15 j^ 4 5 1 40 i^ 2 15 j^ 2 m/s u 5 tan21 a v yf v xf b 5 tan21 a 215 m/s b 5 2218 40 m/s vf 5 0 S vf 0 5 "v x f 2 1 v yf 2 5 " 1 40 2 2 1 1 215 2 2 m/s 5 43 m/s Finalize The negative sign for the angle u indicates that the velocity vector is directed at an angle of 21° below the posiS tive x axis. Notice that if we calculate vi from the x and y components of vi, we find that vf . vi . Is that consistent with our prediction? (C) Determine the x and y coordinates of the particle at any time t and its position vector at this time. continued 84 Chapter 4 Motion in Two Dimensions ▸ 4.1 c o n t i n u e d Solution Analyze Use the components of Equation 4.9 with xi 5 yi 5 0 at t 5 0 and with x and y in meters and t in seconds: x f 5 v xi t 1 12 a x t 2 5 20t 1 2.0t 2 Express the position vector of the particle at any time t: S yf 5 v yit 5 215t rf 5 x f ^i 1 yf ^j 5 1 20t 1 2.0t 2 2 ^i 2 15t ^j om Finalize Let us now consider a limiting case for very large values of t. What if we wait a very long time and then observe the motion of the particle? How would we describe the motion of the particle for large values of the time? W h at I f ? s. w ee b ly .c Answer Looking at Figure 4.6, we see the path of the particle curving toward the x axis. There is no reason to assume this tendency will change, which suggests that the path will become more and more parallel to the x axis as time grows large. Mathematically, Equation (1) shows that the y component of the velocity remains constant while the x component grows linearly with t. Therefore, when t is very large, the x component of the velocity will be much larger than the y component, suggesting that the velocity vector becomes more and more parallel to the x axis. The magnitudes of both xf and yf continue to grow with time, although xf grows much faster. 4.3 ​Projectile Motion Pitfall Prevention 4.2 sw ar ph ys Anyone who has observed a baseball in motion has observed projectile motion. The ball moves in a curved path and returns to the ground. Projectile motion of an object is simple to analyze if we make two assumptions: (1) the free-fall acceleration is constant over the range of motion and is directed downward,1 and (2) the effect of air resistance is negligible.2 With these assumptions, we find that the path of a projectile, which we call its trajectory, is always a parabola as shown in Figure 4.7. We use these assumptions throughout this chapter. The expression for the position vector of the projectile as a function of time follows directly from Equation 4.9, with its acceleration being that due to gravity, S a 5S g: .a As discussed in Pitfall Prevention 2.8, many people claim that the acceleration of a projectile at the topmost point of its trajectory is zero. This mistake arises from confusion between zero vertical velocity and zero acceleration. If the projectile were to experience zero acceleration at the highest point, its velocity at that point would not change; rather, the projectile would move horizontally at constant speed from then on! That does not happen, however, because the acceleration is not zero anywhere along the trajectory. ic Acceleration at the Highest Point ri 1 S v i t 1 12S gt 2 rf 5 S (4.10) w w where the initial x and y components of the velocity of the projectile are Lester Lefkowitz/Taxi/Getty Images w A welder cuts holes through a heavy metal construction beam with a hot torch. The sparks generated in the process follow parabolic paths. S v xi 5 v i cos u i v yi 5 v i sin u i (4.11) The expression in Equation 4.10 is plotted in Figure 4.8 for a projectile launched from the origin, so that S r i 5 0. The final position of a particle can be considered to be the superposition of its initial position S r i ; the term S vi t, which is its displacement if no acceleration were present; and the term 12S g t 2 that arises from its acceleration due to gravity. In other words, if there were no gravitational acceleration, the particle would continue to move along a straight path in the direction of S vi. Therefore, the vertical distance 12S g t 2 through which the particle “falls” off the straight-line path is the same distance that an object dropped from rest would fall during the same time interval. 1This assumption is reasonable as long as the range of motion is small compared with the radius of the Earth (6.4 3 10 6 m). In effect, this assumption is equivalent to assuming the Earth is flat over the range of motion considered. 2 This assumption is often not justified, especially at high velocities. In addition, any spin imparted to a projectile, such as that applied when a pitcher throws a curve ball, can give rise to some very interesting effects associated with aerodynamic forces, which will be discussed in Chapter 14. 85 4.3 Projectile Motion Figure 4.7 The y component of velocity is zero at the peak of the path. S vy  0 vC S vB vy S vi vy i B S g C u vx i The x component of velocity remains constant because there is no acceleration in the x direction. vx i  u vy S v ui A  vx i vx i x ui The projectile is launched S with initial velocity vi . om y The parabolic path of a projectile that leaves the origin with a velocity S v i . The velocity vector S v changes with time in both magnitude and direction. This change is the result of acceleration S a 5S g in the negative y direction. S vy .c v y ee b ly In Section 4.2, we stated that two-dimensional motion with constant acceleration can be analyzed as a combination of two independent motions in the x and y directions, with accelerations ax and ay. Projectile motion can also be handled in this way, with acceleration ax 5 0 in the x direction and a constant acceleration ay 5 2g in the y direction. Therefore, when solving projectile motion problems, use two analysis models: (1) the particle under constant velocity in the horizontal direction (Eq. 2.7): Figure 4.8 ​T he position vector r f of a projectile launched from the origin whose initial velocity at the origin is S v i . The vector S vit would be the displacement of the projectile if gravity were absent, and the vector 12S g t 2 is its vertical displacement from a straight-line path due to its downward gravitational acceleration. S ic ys v yi 1 v yf ar ph yf 5 yi 1 x O v yf 5 v yi 2 gt 2 1 2 1 v yi v it rf and (2) the particle under constant acceleration in the vertical direction (Eqs. 2.13–2.17 with x changed to y and ay = –g): v y,avg 5 (x,y) S S s. w x f 5 x i 1 v xit 1S gt 2 2 1 v yf 2 t yf 5 yi 1 v yit 2 12gt 2 sw v yf 2 5 v yi2 2 2g 1 yf 2 yi 2 w .a The horizontal and vertical components of a projectile’s motion are completely independent of each other and can be handled separately, with time t as the common variable for both components. Q uick Quiz 4.2 ​(i) As a projectile thrown upward moves in its parabolic path (such as in Fig. 4.8), at what point along its path are the velocity and acceleration vectors for the projectile perpendicular to each other? (a) nowhere (b) the highest point (c) the launch point (ii) From the same choices, at what point are the velocity and acceleration vectors for the projectile parallel to each other? w y w vy A  0 S vi h Horizontal Range and Maximum Height of a Projectile Before embarking on some examples, let us consider a special case of projectile motion that occurs often. Assume a projectile is launched from the origin at ti 5 0 with a positive vyi component as shown in Figure 4.9 and returns to the same horizontal level. This situation is common in sports, where baseballs, footballs, and golf balls often land at the same level from which they were launched. Two points in this motion are especially interesting to analyze: the peak point A, which has Cartesian coordinates (R/2, h), and the point B, which has coordinates (R, 0). The distance R is called the horizontal range of the projectile, and the distance h is its maximum height. Let us find h and R mathematically in terms of vi , ui , and g. A ui B O x R Figure 4.9 A projectile launched over a flat surface from the origin at ti 5 0 with an initial velocity S v i . The maximum height of the projectile is h, and the horizontal range is R. At A, the peak of the trajectory, the particle has coordinates (R/2, h). 86 Chapter 4 Motion in Two Dimensions We can determine h by noting that at the peak vy A 5 0. Therefore, from the particle under constant acceleration model, we can use the y direction version of Equation 2.13 to determine the time t A at which the projectile reaches the peak: S 0 5 v i sin u i 2 gt A v yf 5 v yi 2 gt tA 5 v i sin u i g Substituting this expression for t A into the y direction version of Equation 2.16 and replacing y f 5 y A with h, we obtain an expression for h in terms of the magnitude and direction of the initial velocity vector: v i2 sin2 u i 2g (4.12) ly h5 v i sin u i 1 v i sin u i 2 2 2g a b g g .c h 5 1 v i sin u i 2 om yf 5 yi 1 vyi t 2 12g t 2 S ee b The range R is the horizontal position of the projectile at a time that is twice the time at which it reaches its peak, that is, at time t B 5 2t A. Using the particle under constant velocity model, noting that vxi 5 vxB 5 vi cos ui , and setting x B 5 R at t 5 2t A, we find that s. w xf 5 xi 1 vxit S R 5 v xit B 5 1 v i cos u i 2 2t A 2v i sin u i 2v i 2 sin u i cos u i 5 g g ic 5 1 v i cos u i 2 Using the identity sin 2u 5 2 sin u cos u (see Appendix B.4), we can write R in the more compact form ys Pitfall Prevention 4.3 R5 v i 2 sin 2u i g (4.13) sw The maximum value of R from Equation 4.13 is Rmax 5 v i 2 /g . This result makes sense because the maximum value of sin 2ui is 1, which occurs when 2ui 5 90°. Therefore, R is a maximum when ui 5 45°. Figure 4.10 illustrates various trajectories for a projectile having a given initial speed but launched at different angles. As you can see, the range is a maximum for ui 5 45°. In addition, for any ui other than 45°, a point having Cartesian coordinates (R, 0) can be reached by using either one of two complementary values of ui , such as 75° and 15°. Of course, the maximum height and time of flight for one of these values of ui are different from the maximum height and time of flight for the complementary value. w w w .a 4.13 is useful for calculating R only for a symmetric path as shown in Figure 4.10. If the path is not symmetric, do not use this equation. The particle under constant velocity and particle under constant acceleration models are the important starting points because they give the position and velocity components of any projectile moving with constant acceleration in two dimensions at any time t. ar ph The Range Equation Equation Q uick Quiz 4.3 Rank the launch angles for the five paths in Figure 4.10 with respect to time of flight from the shortest time of flight to the longest. y (m) 150 vi  50 m/s 75 100 Figure 4.10 A projectile launched over a flat surface from the origin with an initial speed of 50 m/s at various angles of projection. 60 45 50 Complementary values of the initial angle ui result in the same value of R. 30 15 50 100 150 200 250 x (m) 4.3 Projectile Motion 87 Problem-Solving Strategy   Projectile Motion We suggest you use the following approach when solving projectile motion problems. 1. Conceptualize. Think about what is going on physically in the problem. Establish 2. Categorize. Confirm that the problem involves a particle in free fall and that air resistance is neglected. Select a coordinate system with x in the horizontal direction and y in the vertical direction. Use the particle under constant velocity model for the x component of the motion. Use the particle under constant acceleration model for the y direction. In the special case of the projectile returning to the same level from which it was launched, use Equations 4.12 and 4.13. om the mental representation by imagining the projectile moving along its trajectory. .c 3. Analyze. If the initial velocity vector is given, resolve it into x and y components. Select the appropriate equation(s) from the particle under constant acceleration model for the vertical motion and use these along with Equation 2.7 for the horizontal motion to solve for the unknown(s). ly 4. Finalize. Once you have determined your result, check to see if your answers are    The Long Jump s. w Example 4.2 ee b consistent with the mental and pictorial representations and your results are realistic. ic A long jumper (Fig. 4.11) leaves the ground at an angle of 20.0° above the horizontal and at a speed of 11.0 m/s. Solution ar ph Conceptualize The arms and legs of a long jumper move in a complicated way, but we will ignore this motion. We conceptualize the motion of the long jumper as equivalent to that of a simple projectile. Categorize We categorize this example as a projectile motion problem. Sipa via AP Images ys (A) How far does he jump in the horizontal direction? Figure 4.11 (Example 4.2) Romain Barras of France competes in the men’s decathlon long jump at the 2008 Beijing Olympic Games. w Analyze w .a sw Because the initial speed and launch angle are given and because the final height is the same as the initial height, we further categorize this problem as satisfying the conditions for which Equations 4.12 and 4.13 can be used. This approach is the most direct way to analyze this problem, although the general methods that have been described will always give the correct answer. R5 w Use Equation 4.13 to find the range of the jumper: 1 11.0 m/s 2 2 sin 2 1 20.08 2 v i 2 sin 2u i 5 7.94 m 5 g 9.80 m/s2 (B) What is the maximum height reached? Solution Analyze Find the maximum height reached by using Equation 4.12: h5 1 11.0 m/s 2 2 1 sin 20.08 2 2 v i 2sin2u i 5 5 0.722 m 2g 2 1 9.80 m/s2 2 Finalize Find the answers to parts (A) and (B) using the general method. The results should agree. Treating the long jumper as a particle is an oversimplification. Nevertheless, the values obtained are consistent with experience in sports. We can model a complicated system such as a long jumper as a particle and still obtain reasonable results. 88 Chapter 4 Example 4.3 Motion in Two Dimensions    A Bull’s-Eye Every Time AM In a popular lecture demonstration, a projectile is fired at a target in such a way that the projectile leaves the gun at the same time the target is dropped from rest. Show that if the gun is initially aimed at the stationary target, the projectile hits the falling target as shown in Figure 4.12a. Solution y ly ee b L Gun 0 ui Point of collision 1 2 gt 2 x T tan ui yT x xT b ic a g f si o ine s. w © Cengage Learning/Charles D. Winters Target ht .c The velocity of the projectile (red arrows) changes in direction and magnitude, but its acceleration (purple arrows) remains constant. om Conceptualize We conceptualize the problem by studying Figure 4.12a. Notice that the problem does not ask for numerical values. The expected result must involve an algebraic argument. Figure 4.12 ar ph ys ​(Example 4.3) (a) Multiflash photograph of the projectile–target demonstration. If the gun is aimed directly at the target and is fired at the same instant the target begins to fall, the projectile will hit the target. (b) Schematic diagram of the projectile–target demonstration. sw Categorize Because both objects are subject only to gravity, we categorize this problem as one involving two objects in free fall, the target moving in one dimension and the projectile moving in two. The target T is modeled as a particle under constant acceleration in one dimension. The projectile P is modeled as a particle under constant acceleration in the y direction and a particle under constant velocity in the x direction. Analyze Figure 4.12b shows that the initial y coordinate yi T of the target is x T tan ui and its initial velocity is zero. It falls .a with acceleration ay 5 2g. w w Write an expression for the y coordinate of the target at any moment after release, noting that its initial velocity is zero: w Write an expression for the y coordinate of the projectile at any moment: Write an expression for the x coordinate of the projectile at any moment: Solve this expression for time as a function of the horizontal position of the projectile: Substitute this expression into Equation (2): (1) y T 5 yi T 1 1 0 2 t 2 12 gt 2 5 x T tan u i 2 12gt 2 (2) y P 5 yi P 1 v yi Pt 2 12g t 2 5 0 1 1 v i P sin u i 2 t 2 12g t 2 5 1 v i P sin u i 2 t 2 12gt 2 x P 5 x iP 1 v xi Pt 5 0 1 1 v i P cos u i 2 t 5 1 v iP cos u i 2 t t5 xP v i P cos u i (3) y P 5 1 v iP sin u i 2 a xP b 2 12g t 2 5 x P tan u i 2 12gt 2 v iP cos u i Finalize Compare Equations (1) and (3). We see that when the x coordinates of the projectile and target are the same—that is, when x T 5 x P—their y coordinates given by Equations (1) and (3) are the same and a collision results. 4.3 Projectile Motion Example 4.4 89    That’s Quite an Arm! AM A stone is thrown from the top of a building upward at an angle of 30.0° to the horizontal with an initial speed of 20.0 m/s as shown in Figure 4.13. The height from which the stone is thrown is 45.0 m above the ground. (A) How long does it take the stone to reach the ground? Solution y Conceptualize Study Figure 4.13, in which we have indi- O Categorize We categorize this problem as a projectile motion problem. The stone is modeled as a particle under constant acceleration in the y direction and a particle under constant velocity in the x direction. .c ly Figure 4.13 (Example 4.4) A stone is thrown from the top of a building. s. w v xi 5 v i cos u i 5 1 20.0 m/s 2 cos 30.08 5 17.3 m/s v yi 5 v i sin u i 5 1 20.0 m/s 2 sin 30.08 5 10.0 m/s yf 5 yi 1 v yi t 2 12g t 2 ar ph ys ic Express the vertical position of the stone from the particle under constant acceleration model: Solve the quadratic equation for t: ee b ay 5 2g, and vi 5 20.0 m/s (the numerical value of yf is negative because we have chosen the point of the throw as the origin). Substitute numerical values: x 45.0 m Analyze We have the information xi 5 yi 5 0, yf 5 245.0 m, Find the initial x and y components of the stone’s velocity: ui  30.0 om cated the trajectory and various parameters of the motion of the stone. v i  20.0 m/s 245.0 m 5 0 1 1 10.0 m/s 2 t 1 12 1 29.80 m/s2 2 t 2 t 5 4.22 s (B) What is the speed of the stone just before it strikes the ground? Solution sw Analyze Use the velocity equation in the particle v y f 5 v yi 2 g t .a under constant acceleration model to obtain the y component of the velocity of the stone just before it strikes the ground: w Substitute numerical values, using t 5 4.22 s: w w Use this component with the horizontal component vxf 5 vxi 5 17.3 m/s to find the speed of the stone at t 5 4.22 s: v y f 5 10.0 m/s 1 1 29.80 m/s2 2 1 4.22 s 2 5 231.3 m/s v f 5 "v xf 2 1 v yf 2 5 " 1 17.3 m/s 2 2 1 1 231.3 m/s 2 2 5 35.8 m/s Finalize Is it reasonable that the y component of the final velocity is negative? Is it reasonable that the final speed is larger than the initial speed of 20.0 m/s? What if a horizontal wind is blowing in the same direction as the stone is thrown and it causes the stone to have a horizontal acceleration component ax 5 0.500 m/s2? Which part of this example, (A) or (B), will have a different answer? W h at I f ? Answer Recall that the motions in the x and y directions are independent. Therefore, the horizontal wind cannot affect the vertical motion. The vertical motion determines the time of the projectile in the air, so the answer to part (A) does not change. The wind causes the horizontal velocity component to increase with time, so the final speed will be larger in part (B). Taking ax 5 0.500 m/s2, we find vxf 5 19.4 m/s and vf 5 36.9 m/s. 90 Chapter 4 Example 4.5 Motion in Two Dimensions    The End of the Ski Jump AM A ski jumper leaves the ski track moving in the horizontal direction with a speed of 25.0 m/s as shown in Figure 4.14. The landing incline below her falls off with a slope of 35.0°. Where does she land on the incline? Solution 25.0 m/s Conceptualize We can conceptualize this problem based on memories of observing winter Olympic ski competitions. We estimate the skier to be airborne for perhaps 4 s and to travel a distance of about 100 m horizontally. We should expect the value of d, the distance traveled along the incline, to be of the same order of magnitude. O ly x ee b Figure 4.14 (Example 4.5) A ski jumper leaves the track moving in a horizontal direction. (1) x f 5 v xi t s. w Express the coordinates of the jumper as a function of time, using the particle under constant velocity model for x and the position equation from the particle under constant acceleration model for y: d .c motion. As with other projectile motion problems, we use the particle under constant velocity model for the horizontal motion and the particle under constant acceleration model for the vertical motion. gin. The initial velocity components are vxi 5 25.0 m/s and vyi 5 0. From the right triangle in Figure 4.14, we see that the jumper’s x and y coordinates at the landing point are given by xf 5 d cos f and yf 5 2d sin f. om y Categorize We categorize the problem as one of a particle in projectile Analyze It is convenient to select the beginning of the jump as the ori- f  35.0 (2) yf 5 v yit 2 12g t 2 ic (3) d cos f 5 vxi t ys (4) 2d sin f 5 212g t 2 2d sin f 5 212g a ar ph Solve Equation (3) for t and substitute the result into Equation (4): 2v xi 2 sin f d cos f 2 b v xi 2 1 25.0 m/s 2 2 sin 35.08 5 109 m 1 9.80 m/s2 2 cos2 35.08 d5 Evaluate the x and y coordinates of the point at which the skier lands: x f 5 d cos f 5 1 109 m 2 cos 35.08 5 89.3 m sw Solve for d and substitute numerical values: g cos2 f 5 yf 5 2d sin f 5 2 1 109 m 2 sin 35.08 5 262.5 m w .a Finalize Let us compare these results with our expectations. We expected the horizontal distance to be on the order of 100 m, and our result of 89.3 m is indeed on this order of magnitude. It might be useful to calculate the time interval that the jumper is in the air and compare it with our estimate of about 4 s. w w W h at I f ? Suppose everything in this example is the same except the ski jump is curved so that the jumper is projected upward at an angle from the end of the track. Is this design better in terms of maximizing the length of the jump? Answer If the initial velocity has an upward component, the skier will be in the air longer and should therefore travel farther. Tilting the initial velocity vector upward, however, will reduce the horizontal component of the initial velocity. Therefore, angling the end of the ski track upward at a large angle may actually reduce the distance. Consider the extreme case: the skier is projected at 90° to the horizontal and simply goes up and comes back down at the end of the ski track! This argument suggests that there must be an optimal angle between 0° and 90° that represents a balance between making the flight time longer and the horizontal velocity component smaller. Let us find this optimal angle mathematically. We modify Equations (1) through (4) in the following way, assuming the skier is projected at an angle u with respect to the horizontal over a landing incline sloped with an arbitrary angle f: (1) and (3) S xf 5 (vi cos u)t 5 d cos f (2) and (4) S yf 5 (vi sin u)t 2 12 gt 2 5 2d sin f 4.4 Analysis Model: Particle In Uniform Circular Motion 91 ▸ 4.5 c o n t i n u e d By eliminating the time t between these equations and using differentiation to maximize d in terms of u, we arrive (after several steps; see Problem 88) at the following equation for the angle u that gives the maximum value of d: u 5 458 2 f 2 om For the slope angle in Figure 4.14, f 5 35.0°; this equation results in an optimal launch angle of u 5 27.5°. For a slope angle of f 5 0°, which represents a horizontal plane, this equation gives an optimal launch angle of u 5 45°, as we would expect (see Figure 4.10). ly .c 4.4 A ​ nalysis Model: Particle in Uniform Circular Motion Pitfall Prevention 4.4 Acceleration of a Particle in Uniform Circular Motion Remember that acceleration in physics is defined as a change in the velocity, not a change in the speed (contrary to the everyday interpretation). In circular motion, the velocity vector is always changing in direction, so there is indeed an acceleration. w w w .a sw ar ph ys ic s. w ee b Figure 4.15a shows a car moving in a circular path; we describe this motion by calling it circular motion. If the car is moving on this path with constant speed v, we call it uniform circular motion. Because it occurs so often, this type of motion is recognized as an analysis model called the particle in uniform circular motion. We discuss this model in this section. It is often surprising to students to find that even though an object moves at a constant speed in a circular path, it still has an acceleration. To see why, consider the defining equation for acceleration, S a 5 dS v /dt (Eq. 4.5). Notice that the acceleration depends on the change in the velocity. Because velocity is a vector quantity, an acceleration can occur in two ways as mentioned in Section 4.1: by a change in the magnitude of the velocity and by a change in the direction of the velocity. The latter situation occurs for an object moving with constant speed in a circular path. The constant-magnitude velocity vector is always tangent to the path of the object and perpendicular to the radius of the circular path. Therefore, the direction of the velocity vector is always changing. Let us first argue that the acceleration vector in uniform circular motion is always perpendicular to the path and always points toward the center of the circle. If that were not true, there would be a component of the acceleration parallel to the path and therefore parallel to the velocity vector. Such an acceleration component would lead to a change in the speed of the particle along the path. This situation, however, is inconsistent with our setup of the situation: the particle moves with constant speed along the path. Therefore, for uniform circular motion, the acceleration vector can only have a component perpendicular to the path, which is toward the center of the circle. Let us now find the magnitude of the acceleration of the particle. Consider the diagram of the position and velocity vectors in Figure 4.15b. The figure also shows the vector representing the change in position DS r for an arbitrary time interval. The particle follows a circular path of radius r, part of which is shown by the dashed S vi A B rS r S vf S vi S O Top view a v S ri b S u q rf u S vf c S v Figure 4.15 (a) A car moving along a circular path at constant speed experiences uniform circular motion. (b) As a particle moves along a portion of a circular path from A to B, its velocity vector changes from S v i to S v f . (c) The construction for determining the direction of the change in velocity S Dv , which is toward the center of the circle for small DrS. 92 Chapter 4 Motion in Two Dimensions .c ly 0 DS 0 DS v0 r0 5 v r om curve. The particle is at A at time ti , and its velocity at that time is S v i; it is at B at v i and S some later time tf , and its velocity at that time is S v f . Let us also assume S vf differ only in direction; their magnitudes are the same (that is, vi 5 vf 5 v because it is uniform circular motion). In Figure 4.15c, the velocity vectors in Figure 4.15b have been redrawn tail to tail. The vector DS v connects the tips of the vectors, representing the vector addition S vf 5 S v i 1 DS v . In both Figures 4.15b and 4.15c, we can identify triangles that help us analyze the motion. The angle Du between the two position vectors in Figure 4.15b is the same as the angle between the velocity vectors in Figure 4.15c because the velocity vector S v is always perpendicular to the position vector S r . Therefore, the two triangles are similar. (Two triangles are similar if the angle between any two sides is the same for both triangles and if the ratio of the lengths of these sides is the same.) We can now write a relationship between the lengths of the sides for the two triangles in Figures 4.15b and 4.15c: ee b where v 5 vi 5 vf and r 5 ri 5 rf . This equation can be solved for 0 DS v 0 , and the S S expression obtained can be substituted into Equation 4.4, a avg 5 D v /Dt , to give the magnitude of the average acceleration over the time interval for the particle to move from A to B: 0 DS v0 v 0 DS r0 5 0 Dt 0 r Dt s. w 0S a avg 0 5 ar ph ys ic Now imagine that points A and B in Figure 4.15b become extremely close together. As A and B approach each other, Dt approaches zero, 0 DS r 0 approaches r 0 /Dt the distance traveled by the particle along the circular path, and the ratio 0 DS approaches the speed v. In addition, the average acceleration becomes the instantaneous acceleration at point A. Hence, in the limit Dt S 0, the magnitude of the acceleration is v2 ac 5 (4.14) r An acceleration of this nature is called a centripetal acceleration (centripetal means center-seeking). The subscript on the acceleration symbol reminds us that the acceleration is centripetal. In many situations, it is convenient to describe the motion of a particle moving with constant speed in a circle of radius r in terms of the period T, which is defined as the time interval required for one complete revolution of the particle. In the time interval T, the particle moves a distance of 2pr, which is equal to the circumference of the particle’s circular path. Therefore, because its speed is equal to the circumference of the circular path divided by the period, or v 5 2pr/T, it follows that w w w .a sw Centripetal acceleration  for a particle in uniform circular motion Period of circular motion  for a particle in uniform circular motion T5 2pr v (4.15) The period of a particle in uniform circular motion is a measure of the number of seconds for one revolution of the particle around the circle. The inverse of the period is the rotation rate and is measured in revolutions per second. Because one full revolution of the particle around the circle corresponds to an angle of 2p radians, the product of 2p and the rotation rate gives the angular speed v of the particle, measured in radians/s or s21: v5 2p T (4.16) 93 4.4 Analysis Model: Particle In Uniform Circular Motion Combining this equation with Equation 4.15, we find a relationship between angular speed and the translational speed with which the particle travels in the circular path: v 5 2pa v v b5 r 2pr S (4.17) v 5 rv Pitfall Prevention 4.5 Centripetal Acceleration Is Not Constant We derived the magnitude of the centripetal acceleration vector and found it to be constant for uniform circular motion, but the centripetal acceleration vector is not constant. It always points toward the center of the circle, but it continuously changes direction as the object moves around the circular path. om Equation 4.17 demonstrates that, for a fixed angular speed, the translational speed becomes larger as the radial position becomes larger. Therefore, for example, if a merry-go-round rotates at a fixed angular speed v, a rider at an outer position at large r will be traveling through space faster than a rider at an inner position at smaller r. We will investigate Equations 4.16 and 4.17 more deeply in Chapter 10. We can express the centripetal acceleration of a particle in uniform circular motion in terms of angular speed by combining Equations 4.14 and 4.17: 1 rv 2 2 r a c 5 rv 2 ac 5 .c (4.18) ee b ly Equations 4.14–4.18 are to be used when the particle in uniform circular motion model is identified as appropriate for a given situation. s. w Q uick Quiz 4.4 ​A particle moves in a circular path of radius r with speed v. It then increases its speed to 2v while traveling along the same circular path. (i) The centripetal acceleration of the particle has changed by what factor? Choose one: (a) 0.25 (b) 0.5 (c) 2 (d) 4 (e) impossible to determine (ii) From the same choices, by what factor has the period of the particle changed? ic Analysis Model   Particle in Uniform Circular Motion ar ph ys Imagine a moving object that can be modeled as a particle. If it moves in a circular path of radius r at a constant speed v, the magnitude of its centripetal acceleration is v2 ac 5 (4.14) S S ac r v sw and the period of the particle’s motion is given by 2pr T5 (4.15) v The angular speed of the particle is .a 2p T (4.16) w w v5 r    The Centripetal Acceleration of the Earth • a rock twirled in a circle on a string of constant length • a planet traveling around a perfectly circular orbit (Chapter 13) • a charged particle moving in a uniform magnetic field (Chapter 29) • an electron in orbit around a nucleus in the Bohr model of the hydrogen atom (Chapter 42) AM w Example 4.6 Examples: (A) What is the centripetal acceleration of the Earth as it moves in its orbit around the Sun? Solution Conceptualize Think about a mental image of the Earth in a circular orbit around the Sun. We will model the Earth as a particle and approximate the Earth’s orbit as circular (it’s actually slightly elliptical, as we discuss in Chapter 13). Categorize The Conceptualize step allows us to categorize this problem as one of a particle in uniform circular motion. Analyze We do not know the orbital speed of the Earth to substitute into Equation 4.14. With the help of Equation 4.15, however, we can recast Equation 4.14 in terms of the period of the Earth’s orbit, which we know is one year, and the radius of the Earth’s orbit around the Sun, which is 1.496 3 1011 m. continued 94 Chapter 4 Motion in Two Dimensions ▸ 4.6 c o n t i n u e d 2pr 2 b T v2 4p2r ac 5 5 5 r r T2 a Combine Equations 4.14 and 4.15: Substitute numerical values: ac 5 2 1 yr 4p2 1 1.496 3 1011 m 2 a b 5 5.93 3 1023 m/s2 2 7 1 1 yr 2 3.156 3 10 s (B) What is the angular speed of the Earth in its orbit around the Sun? om Solution Analyze 1 yr 2p a b 5 1.99 3 1027 s21 1 yr 3.156 3 107 s ly v5 .c Substitute numerical values into Equation 4.16: s. w ee b Finalize The acceleration in part (A) is much smaller than the free-fall acceleration on the surface of the Earth. An important technique we learned here is replacing the speed v in Equation 4.14 in terms of the period T of the motion. In many problems, it is more likely that T is known rather than v. In part (B), we see that the angular speed of the Earth is very small, which is to be expected because the Earth takes an entire year to go around the circular path once. ic 4.5 Tangential and Radial Acceleration w .a sw ar ph ys Let us consider a more general motion than that presented in Section 4.4. A particle moves to the right along a curved path, and its velocity changes both in direction and in magnitude as described in Figure 4.16. In this situation, the velocity vector is always tangent to the path; the acceleration vector S a , however, is at some angle to the path. At each of three points A, B, and C in Figure 4.16, the dashed blue circles represent the curvature of the actual path at each point. The radius of each circle is equal to the path’s radius of curvature at each point. As the particle moves along the curved path in Figure 4.16, the direction of the total acceleration vector S a changes from point to point. At any instant, this vector can be resolved into two components based on an origin at the center of the dashed circle corresponding to that instant: a radial component ar along the radius of the circle and a tangential component at perpendicular to this radius. The total acceleration vector S a can be written as the vector sum of the component vectors: S a 5S ar 1 S at w w Total acceleration  The tangential acceleration component causes a change in the speed v of the particle. This component is parallel to the instantaneous velocity, and its magnitude is given by Tangential acceleration  at 5 ` Path of particle Figure 4.16 The motion of a particle along an arbitrary curved path lying in the xy plane. If the v (always tangent velocity vector S to the path) changes in direction and magnitude, the components a are a tanof the acceleration S gential component a t and a radial component a r . (4.19) a S A at a (4.20) B at S ar dv ` dt ar ar C S a at 95 4.5 Tangential and Radial Acceleration The radial acceleration component arises from a change in direction of the velocity vector and is given by a r 5 2a c 5 2 v2 r (4.21) WW Radial acceleration s. w ee b ly .c om where r is the radius of curvature of the path at the point in question. We recognize the magnitude of the radial component of the acceleration as the centripetal acceleration discussed in Section 4.4 with regard to the particle in uniform circular motion model. Even in situations in which a particle moves along a curved path with a varying speed, however, Equation 4.14 can be used for the centripetal acceleration. In this situation, the equation gives the instantaneous centripetal acceleration at any time. The negative sign in Equation 4.21 indicates that the direction of the centripetal acceleration is toward the center of the circle representing the radius of curvature. The direction is opposite that of the radial unit vector r^ , which always points away from the origin at the center of the circle. Because S a r and S a t are perpendicular component vectors of S a , it follows that S the magnitude of a is a 5 !a r2 1 a t2. At a given speed, ar is large when the radius of curvature is small (as at points A and B in Fig. 4.16) and small when r is large (as at point C). The direction of S a t is either in the same direction as S v (if v is S increasing) or opposite v (if v is decreasing, as at point B). In uniform circular motion, where v is constant, at 5 0 and the acceleration is always completely radial as described in Section 4.4. In other words, uniform circular motion is a special case of motion along a general curved path. Furthermore, if the direction of S v does not change, there is no radial acceleration and the motion is one dimensional (in this case, ar 5 0, but at may not be zero).    Over the Rise sw Example 4.7 ar ph ys ic Q uick Quiz 4.5 ​A particle moves along a path, and its speed increases with time. (i) In which of the following cases are its acceleration and velocity vectors parallel? (a) when the path is circular (b) when the path is straight (c) when the path is a parabola (d) never (ii) From the same choices, in which case are its acceleration and velocity vectors perpendicular everywhere along the path? at  0.300 m/s2 S at S v v  6.00 m/s a w w w .a A car leaves a stop sign and exhibits a constant acceleration of 0.300 m/s2 parallel to the roadway. The car passes over a rise in the roadway such that the top of the rise is shaped like an arc of a circle of radius 500 m. At the moment the car is at the top of the rise, its velocity vector is horizontal and has a magnitude of 6.00 m/s. What are the magnitude and direction of the total acceleration vector for the car at this instant? Solution S at Conceptualize Conceptualize the situation using Figure 4.17a and any experiences you have had in driving over rises on a roadway. Categorize Because the accelerating car is moving along a curved path, we categorize this problem as one involving a particle experiencing both tangential and radial acceleration. We recognize that it is a relatively simple substitution problem. f S a S ar b Figure 4.17 (Example 4.7) (a) A car passes over a rise that is shaped like an arc of a circle. (b) The total acceleration vector S a is the sum of the tangential and radial acceleration vectors S a t and S ar. The tangential acceleration vector has magnitude 0.300 m/s2 and is horizontal. The radial acceleration is given by Equation 4.21, with v 5 6.00 m/s and r 5 500 m. The radial acceleration vector is directed straight downward. continued 96 Chapter 4 Motion in Two Dimensions ▸ 4.7 c o n t i n u e d 1 6.00 m/s 2 2 v2 52 5 20.072 0 m/s2 r 500 m Evaluate the radial acceleration: ar 5 2 Find the magnitude of S a: "a r 2 1 a t2 5 " 1 20.072 0 m/s2 2 2 1 1 0.300 m/s2 2 2 5 0.309 m/s2 Find the angle f (see Fig. 4.17b) between S a and the horizontal: P 5 om A 4.6 ​Relative Velocity and Relative Acceleration x 0 +5 +5 +10 P ee b P xA s. w A B 0 In this section, we describe how observations made by different observers in different frames of reference are related to one another. A frame of reference can be described by a Cartesian coordinate system for which an observer is at rest with respect to the origin. Let us conceptualize a sample situation in which there will be different observations for different observers. Consider the two observers A and B along the number line in Figure 4.18a. Observer A is located 5 units to the right of observer B. Both observers measure the position of point P, which is located 5 units to the right of observer A. Suppose each observer decides that he is located at the origin of an x axis as in Figure 4.18b. Notice that the two observers disagree on the value of the position of point P. Observer A claims point P is located at a position with a value of x A 5 15, whereas observer B claims it is located at a position with a value of x B 5 110. Both observers are correct, even though they make different measurements. Their measurements differ because they are making the measurement from different frames of reference. Imagine now that observer B in Figure 4.18b is moving to the right along the x B axis. Now the two measurements are even more different. Observer A claims point P remains at rest at a position with a value of 15, whereas observer B claims the position of P continuously changes with time, even passing him and moving behind him! Again, both observers are correct, with the difference in their measurements arising from their different frames of reference. We explore this phenomenon further by considering two observers watching a man walking on a moving beltway at an airport in Figure 4.19. The woman standing on the moving beltway sees the man moving at a normal walking speed. The woman observing from the stationary floor sees the man moving with a higher speed because the beltway speed combines with his walking speed. Both observers look at the same man and arrive at different values for his speed. Both are correct; the difference in their measurements results from the relative velocity of their frames of reference. In a more general situation, consider a particle located at point P in Figure 4.20. Imagine that the motion of this particle is being described by two observers, observer A in a reference frame SA fixed relative to the Earth and a second observer B in a reference frame S B moving to the right relative to SA (and therefore relative to the Earth) with a constant velocity S v BA. In this discussion of relative velocity, we use a double-subscript notation; the first subscript represents what is being observed, and the second represents who is doing the observing. Therefore, the notation S v BA means the velocity of observer B (and the attached frame S B) as measured by observer A. With this notation, observer B measures A to be moving to the left with a velocity S v AB 5 2S v BA. For purposes of this discussion, let us place each observer at her or his respective origin. We define the time t 5 0 as the instant at which the origins of the two reference frames coincide in space. Therefore, at time t, the origins of the reference frames ly a –5 ar 20.072 0 m/s2 5 tan21 a b 5 213.58 at 0.300 m/s 2 .c B f 5 tan21 xB b .a sw ar ph ys ic Figure 4.18 Different observers make different measurements. (a) Observer A is located 5 units to the right of Observer B. Both observers measure the position of a particle at P. (b) If both observers see themselves at the origin of their own coordinate system, they disagree on the value of the position of the particle at P. w w w The woman standing on the beltway sees the man moving with a slower speed than does the woman observing the man from the stationary floor. Figure 4.19 Two observers measure the speed of a man walking on a moving beltway. 97 4.6 Relative Velocity and Relative Acceleration will be separated by a distance v BAt. We label the position P of the particle relative to observer A with the position vector S r P A and that relative to observer B with the position vector S r P B, both at time t. From Figure 4.20, we see that the vectors S r PA and S r P B are related to each other through the expression S r PA 5 S r PB 1 S v BAt (4.22) S By differentiating Equation 4.22 with respect to time, noting that v BA is constant, we obtain dS rPA d Sr P B S 5 1 v BA dt dt S uP A 5 S uP B 1 S v BA (4.23) om WW Galilean velocity transformation SA SB P s. w ee b ly .c where S u PA is the velocity of the particle at P measured by observer A and S u P B is its velocity measured by B. (We use the symbol S u for particle velocity rather than S v, which we have already used for the relative velocity of two reference frames.) Equations 4.22 and 4.23 are known as Galilean transformation equations. They relate the position and velocity of a particle as measured by observers in relative motion. Notice the pattern of the subscripts in Equation 4.23. When relative velocities are added, the inner subscripts (B) are the same and the outer ones (P, A) match the subscripts on the velocity on the left of the equation. Although observers in two frames measure different velocities for the particle, v BA is constant. We can verify that by taking they measure the same acceleration when S the time derivative of Equation 4.23: ic dS uP A dS uP B dS v BA 5 1 dt dt dt Example 4.8 ar ph ys Because S v BA is constant, d S v BA/dt 5 0. Therefore, we conclude that S aPA 5 S aPB S S S S because a P A 5 d u P A/dt and a P B 5 d u P B /dt. That is, the acceleration of the particle measured by an observer in one frame of reference is the same as that measured by any other observer moving with constant velocity relative to the first frame. S rPA S rPB B A x S vBAt S vBA Figure 4.20 ​A particle located at P is described by two observers, one in the fixed frame of reference SA and the other in the frame S B, which moves to the right with a constant velocity S v BA. The vector S r P A is the particle’s position vector relative to SA , and S r P B is its position vector relative to S B.    A Boat Crossing a River sw A boat crossing a wide river moves with a speed of 10.0 km/h relative to the water. The water in the river has a uniform speed of 5.00 km/h due east relative to the Earth. S .a (A) If the boat heads due north, determine the velocity of the boat relative to an observer standing on either bank. w tive to the river and the river relative to the Earth, we can categorize this problem as one involving relative velocities. S v bE v bE S v br w w Categorize Because of the combined velocities of you rela- vrE S Solution Conceptualize Imagine moving in a boat across a river while the current pushes you down the river. You will not be able to move directly across the river, but will end up downstream as suggested in Figure 4.21a. S v rE N vbr u E W S E W S a u N S b Figure 4.21 (Example 4.8) (a) A boat aims directly across a river and ends up downstream. (b) To move directly across the river, the boat must aim upstream. Analyze We know S v br, the velocity of the boat relative to the river, and S v rE, the velocity of the river relative to the Earth. What we must find is S v bE, the velocity of the boat relative to the Earth. The relationship between these three quantities is S v bE 5 S v br 1 S v rE. The terms in the equation must be manipulated as vector quantities; the vectors are shown in Figure 4.21a. The quantity S v br is due north; S v rE is due east; and the vector sum of the two, S v bE, is at an angle u as defined in Figure 4.21a. continued 98 Chapter 4 Motion in Two Dimensions ▸ 4.8 c o n t i n u e d v bE 5 "v br2 1 v rE2 5 " 1 10.0 km/h 2 2 1 1 5.00 km/h 2 2 Find the speed v bE of the boat relative to the Earth using the Pythagorean theorem: 5 11.2 km/h Find the direction of S v bE: u 5 tan21 a v rE 5.00 b 5 tan21 a b 5 26.68 v br 10.0 om Finalize The boat is moving at a speed of 11.2 km/h in the direction 26.6° east of north relative to the Earth. Notice that the speed of 11.2 km/h is faster than your boat speed of 10.0 km/h. The current velocity adds to yours to give you a higher speed. Notice in Figure 4.21a that your resultant velocity is at an angle to the direction straight across the river, so you will end up downstream, as we predicted. .c (B) If the boat travels with the same speed of 10.0 km/h relative to the river and is to travel due north as shown in Figure 4.21b, what should its heading be? ly Solution Conceptualize/Categorize This question is an extension of part (A), so we have already conceptualized and catego- ee b rized the problem. In this case, however, we must aim the boat upstream so as to go straight across the river. Analyze The analysis now involves the new triangle shown in Figure 4.21b. As in part (A), we know S v rE and the mag- Find the direction in which the boat is heading: v bE 5 "v br2 2 v rE2 5 " 1 10.0 km/h 2 2 2 1 5.00 km/h 2 2 5 8.66 km/h u 5 tan21 a v rE 5.00 b 5 tan21 a b 5 30.08 v bE 8.66 ic Use the Pythagorean theorem to find v bE: s. w nitude of the vector S v br, and we want S v bE to be directed across the river. Notice the difference between the triangle in Figure 4.21a and the one in Figure 4.21b: the hypotenuse in Figure 4.21b is no longer S v bE. ys Finalize The boat must head upstream so as to travel directly northward across the river. For the given situation, the boat must steer a course 30.0° west of north. For faster currents, the boat must be aimed upstream at larger angles. ar ph W h at I f ? Imagine that the two boats in parts (A) and (B) are racing across the river. Which boat arrives at the opposite bank first? w w Summary .a sw Answer In part (A), the velocity of 10 km/h is aimed directly across the river. In part (B), the velocity that is directed across the river has a magnitude of only 8.66 km/h. Therefore, the boat in part (A) has a larger velocity component directly across the river and arrives first. w Definitions The displacement vector DrS for a particle is the difference between its final position vector and its initial position vector: DS r ;S rf 2 S ri (4.1) The average velocity of a particle during the time interval Dt is defined as the displacement of the particle divided by the time interval: DS r S (4.2) v avg ; Dt The instantaneous velocity of a particle is defined as the limit of the average velocity as Dt approaches zero: S DS r dS r 5 dt Dt S 0 Dt v ; lim (4.3)    Summary 99 The average acceleration of a particle is defined as the change in its instantaneous velocity vector divided by the time interval Dt during which that change occurs: S a avg ; S vf 2 S vi DS v 5 Dt tf 2 ti Projectile motion is one type of twodimensional motion, exhibited by an object launched into the air near the Earth’s surface and experiencing free fall. This common motion can be analyzed by applying the particle under constant velocity model to the motion of the projectile in the x direction and the particle under constant acceleration model (ay 5 2g) in the y direction. A particle moving in a circular path with constant speed is exhibiting uniform circular motion. (4.4) The instantaneous acceleration of a particle is defined as the limiting value of the average acceleration as Dt approaches zero: a ; lim Dt S 0 DS v dS v 5 Dt dt (4.5) om S .c Concepts and Principles S ee b ly If a particle moves with constant acceleration S a and has velocity S v i and position S r i at t 5 0, its velocity and position vectors at some later time t are vf 5 S vi 1 S at S S S r f 5 r i 1 vi t 1 (4.8) 1 S 2 2 at (4.9) ic s. w For two-dimensional motion in the xy plane under constant acceleration, ea–ch of these vector expressions is equivalent to two component expressions: one for the motion in the x direction and one for the motion in the y direction. A particle in uniform circular motion undergoes a radial acceleration S a r because the direction of S v changes in time. This acceleration is called centripetal acceleration, and its direction is always toward the center of the circle. If a particle moves along a curved path in such a way that both the magnitude and the direction of S v change in time, the particle has an acceleration vector that can be described by two component vectors: (1) a radial component vector S a r that causes the change in direction of S v and (2) a tangential component vector S a t that causes the change in magnitude of S v. The magnitude of S a r is v 2/r, and the magnitude of S a t is |dv/dt|. The velocity S u PA of a particle measured in a fixed frame of reference SA can be related to the velocity S u P B of the same particle measured in a moving frame of reference S B by S uP A 5 S uP B 1 S v BA (4.23) S where v BA is the velocity of S B relative to SA . w w .a sw ar ph ys It is useful to think of projectile motion in terms of a combination of two analysis models: (1) the particle under constant velocity model in the x direction and (2) the particle under constant acceleration model in the vertical direction with a constant downward acceleration of magnitude g 5 9.80 m/s2. w Analysis Model for Problem Solving Particle in Uniform Circular Motion If a particle moves in a circular path of radius r with a constant speed v, the magnitude of its centripetal acceleration is given by v2 r and the period of the particle’s motion is given by ac 5 (4.14) T5 2pr v (4.15) v5 2p T (4.16) The angular speed of the particle is S ac r S v Motion in Two Dimensions Objective Questions 1. denotes answer available in Student Solutions Manual/Study Guide 1. Figure OQ4.1 shows a bird’s-eye view of a car going around a highway curve. As the car moves from point 1 to point 2, its speed doubles. Which of the vectors (a) through (e) shows the direction of the car’s average acceleration between these two points? (a) (c) 6. An astronaut hits a golf ball on the Moon. Which of the following quantities, if any, remain constant as a ball travels through the vacuum there? (a) speed (b) acceleration (c) horizontal component of velocity (d) vertical component of velocity (e) velocity Figure OQ4.1 7. A projectile is launched on the Earth with a certain initial velocity and moves without air resistance. Another projectile is launched with the same initial velocity on the Moon, where the acceleration due to gravity is onesixth as large. How does the range of the projectile on the Moon compare with that of the projectile on the Earth? (a) It is one-sixth as large. (b) It is the same. (c) It is !6 times larger. (d) It is 6 times larger. (e) It is 36 times larger. ic 2. Entering his dorm room, a student tosses his book bag to the right and upward at an angle of 45° with the horizontal (Fig. OQ4.2). Air resistance does not affect the bag. The bag moves through point A immediately after it leaves the student’s hand, through point B at the top of its flight, and through point C immediately before it lands on the top bunk bed. (i) Rank the following horizontal and vertical velocity components from the largest to the smallest. (a) v Ax (b) v Ay (c) v Bx (d) v By (e) v Cy. Note that zero is larger than a negative number. If two quantities are equal, show them as equal in your list. If any quantity is equal to zero, show that fact in your list. (ii) Similarly, rank the following acceleration components. (a) a Ax (b) a Ay (c) a Bx (d) a By (e) a Cy. ee b (e) ly .c (d) 1 5. Does a car moving around a circular track with constant speed have (a) zero acceleration, (b) an acceleration in the direction of its velocity, (c) an acceleration directed away from the center of its path, (d) an acceleration directed toward the center of its path, or (e) an acceleration with a direction that cannot be determined from the given information? om (b) 2 projectile on the Moon compare with that of the projectile on the Earth? (a) It is one-sixth as large. (b) It is the same. (c) It is !6 times larger. (d) It is 6 times larger. (e) It is 36 times larger. s. w 100 Chapter 4 B C sw 45 w w .a A ar ph ys 8. A girl, moving at 8 m/s on in-line skates, is overtaking a boy moving at 5 m/s as they both skate on a straight path. The boy tosses a ball backward toward the girl, giving it speed 12 m/s relative to him. What is the speed of the ball relative to the girl, who catches it? (a) (8 1 5 1 12) m/s (b) (8 2 5 2 12) m/s (c) (8 1 5 2 12) m/s (d) (8 2 5 1 12) m/s (e) (28 1 5 1 12) m/s w Figure OQ4.2 3. A student throws a heavy red ball horizontally from a balcony of a tall building with an initial speed vi. At the same time, a second student drops a lighter blue ball from the balcony. Neglecting air resistance, which statement is true? (a) The blue ball reaches the ground first. (b) The balls reach the ground at the same instant. (c) The red ball reaches the ground first. (d) Both balls hit the ground with the same speed. (e) None of statements (a) through (d) is true. 4. A projectile is launched on the Earth with a certain initial velocity and moves without air resistance. Another projectile is launched with the same initial velocity on the Moon, where the acceleration due to gravity is onesixth as large. How does the maximum altitude of the 9. A sailor drops a wrench from the top of a sailboat’s vertical mast while the boat is moving rapidly and steadily straight forward. Where will the wrench hit the deck? (a) ahead of the base of the mast (b) at the base of the mast (c) behind the base of the mast (d) on the windward side of the base of the mast (e) None of the choices (a) through (d) is true. 10. A baseball is thrown from the outfield toward the catcher. When the ball reaches its highest point, which statement is true? (a) Its velocity and its acceleration are both zero. (b) Its velocity is not zero, but its acceleration is zero. (c) Its velocity is perpendicular to its acceleration. (d) Its acceleration depends on the angle at which the ball was thrown. (e) None of statements (a) through (d) is true. 11. A set of keys on the end of a string is swung steadily in a horizontal circle. In one trial, it moves at speed v in a circle of radius r. In a second trial, it moves at a higher speed 4v in a circle of radius 4r. In the second trial, how does the period of its motion compare with its period in the first trial? (a) It is the same as in the first trial. (b) It is 4 times larger. (c) It is one-fourth as large. (d) It is 16 times larger. (e) It is one-sixteenth as large. Problems 12. A rubber stopper on the end of a string is swung steadily in a horizontal circle. In one trial, it moves at speed v in a circle of radius r. In a second trial, it moves at a higher speed 3v in a circle of radius 3r. In this second trial, is its acceleration (a) the same as in the first trial, (b) three times larger, (c) one-third as large, (d) nine times larger, or (e) one-ninth as large? 13. In which of the following situations is the moving object appropriately modeled as a projectile? Choose all correct answers. (a) A shoe is tossed in an arbitrary Conceptual Questions 101 direction. (b) A jet airplane crosses the sky with its engines thrusting the plane forward. (c) A rocket leaves the launch pad. (d) A rocket moves through the sky, at much less than the speed of sound, after its fuel has been used up. (e) A diver throws a stone under water. 14. A certain light truck can go around a curve having a radius of 150 m with a maximum speed of 32.0 m/s. To have the same acceleration, at what maximum speed can it go around a curve having a radius of 75.0 m? (a) 64 m/s (b) 45 m/s (c) 32 m/s (d) 23 m/s (e) 16 m/s om 1. denotes answer available in Student Solutions Manual/Study Guide 4. Describe how a driver can steer a car traveling at constant speed so that (a) the acceleration is zero or (b) the magnitude of the acceleration remains constant. .c 1. A spacecraft drifts through space at a constant velocity. Suddenly, a gas leak in the side of the spacecraft gives it a constant acceleration in a direction perpendicular to the initial velocity. The orientation of the spacecraft does not change, so the acceleration remains perpendicular to the original direction of the velocity. What is the shape of the path followed by the spacecraft in this situation? ee b ly 5. A projectile is launched at some angle to the horizontal with some initial speed vi , and air resistance is negligible. (a) Is the projectile a freely falling body? (b) What is its acceleration in the vertical direction? (c) What is its acceleration in the horizontal direction? 2. An ice skater is executing a figure eight, consisting of two identically shaped, tangent circular paths. Throughout the first loop she increases her speed uniformly, and during the second loop she moves at a constant speed. Draw a motion diagram showing her velocity and acceleration vectors at several points along the path of motion. s. w 6. Construct motion diagrams showing the velocity and acceleration of a projectile at several points along its path, assuming (a) the projectile is launched horizontally and (b) the projectile is launched at an angle u with the horizontal. ic 3. If you know the position vectors of a particle at two points along its path and also know the time interval during which it moved from one point to the other, can you determine the particle’s instantaneous velocity? Its average velocity? Explain. ar ph ys 7. Explain whether or not the following particles have an acceleration: (a) a particle moving in a straight line with constant speed and (b) a particle moving around a curve with constant speed. Problems sw The problems found in this chapter may be assigned online in Enhanced WebAssign AMT   Analysis Model tutorial available in Enhanced WebAssign GP Guided Problem M   Master It tutorial available in Enhanced 1. full solution available in the Student W   Watch It video solution available in .a 1. straightforward; 2. intermediate; 3. challenging Enhanced WebAssign w w Solutions Manual/Study Guide WebAssign w Section 4.1 The Position, Velocity, and Acceleration Vectors 1. A motorist drives south at 20.0 m/s for 3.00 min, then turns west and travels at 25.0 m/s for 2.00 min, and finally travels northwest at 30.0 m/s for 1.00 min. For this 6.00-min trip, find (a) the total vector displacement, (b) the average speed, and (c) the average velocity. Let the positive x axis point east. 2. When the Sun is directly overhead, a hawk dives toward the ground with a constant velocity of 5.00 m/s at 60.0° below the horizontal. Calculate the speed of its shadow on the level ground. 3. Suppose the position vector for a particle is given as S a function of time by r 1 t 2 5 x 1 t 2 ^i 1 y 1 t 2 ^j, with x(t) 5 at 1 b and y(t) 5 ct 2 1 d, where a 5 1.00 m/s, b 5 1.00 m, BIO Q/C S c 5 0.125 m/s2, and d 5 1.00 m. (a) Calculate the average velocity during the time interval from t 5 2.00 s to t 5 4.00 s. (b) Determine the velocity and the speed at t 5 2.00 s. 4. The coordinates of an object moving in the xy plane Q/C vary with time according to the equations x 5 25.00 sin vt and y 5 4.00 2 5.00 cos vt, where v is a constant, x and y are in meters, and t is in seconds. (a) Determine the components of velocity of the object at t 5 0. (b) Determine the components of acceleration of the object at t 5 0. (c) Write expressions for the position vector, the velocity vector, and the acceleration vector of the object at any time t . 0. (d) Describe the path of the object in an xy plot. Motion in Two Dimensions 5. A golf ball is hit off a tee at the edge of a cliff. Its x and y coordinates as functions of time are given by x 5 18.0t and y 5 4.00t 2 4.90t 2, where x and y are in meters and t is in seconds. (a) Write a vector expression for the ball’s position as a function of time, using the unit vectors ^i and ^j. By taking derivatives, obtain expressions for v as a function of time and (b) the velocity vector S a as a function of time. (c) the acceleration vector S (d) Next use unit-vector notation to write expressions for the position, the velocity, and the acceleration of the golf ball at t 5 3.00 s. Section 4.2 Two-Dimensional Motion with Constant Acceleration 6. A particle initially located at the origin has an accelS the x axis, moving with velocity 4.50 m/s at 40.0°. It moves with constant acceleration 1.90 m/s2 at 200°. After 5.00 s have elapsed, find (a) its velocity and (b) its position vector. Section 4.3 Projectile Motion Note: Ignore air resistance in all problems and take g 5 9.80 m/s2 at the Earth’s surface. 11. Mayan kings and many school sports teams are named BIO for the puma, cougar, or mountain lion—Felis concolor— the best jumper among animals. It can jump to a height of 12.0 ft when leaving the ground at an angle of 45.0°. With what speed, in SI units, does it leave the ground to make this leap? om 102 Chapter 4 12. An astronaut on a strange planet finds that she can jump a maximum horizontal distance of 15.0 m if her initial speed is 3.00 m/s. What is the free-fall acceleration on the planet? 7. The vector position of a particle varies in time accordS S 2 W ing to the expression r 5 3.00i^ 2 6.00t ^j, where r is in meters and t is in seconds. (a) Find an expression for the velocity of the particle as a function of time. (b) Determine the acceleration of the particle as a function of time. (c) Calculate the particle’s position and velocity at t 5 1.00 s. AMT the counter for a refill. The height of the counter is M 1.22 m. The mug slides off the counter and strikes the ly floor 1.40 m from the base of the counter. (a) With what velocity did the mug leave the counter? (b) What was the direction of the mug’s velocity just before it hit the floor? 14. In a local bar, a customer slides an empty beer mug S down the counter for a refill. The height of the counter is h. The mug slides off the counter and strikes the floor at distance d from the base of the counter. (a) With what velocity did the mug leave the counter? (b) What was the direction of the mug’s velocity just before it hit the floor? ic 8. It is not possible to see very small objects, such as 13. In a local bar, a customer slides an empty beer mug down ee b vi 5 5.00i^ m/s. Find (a) the vector position of the particle at any time t, (b) the velocity of the particle at any time t, (c) the coordinates of the particle at t 5 2.00 s, and (d) the speed of the particle at t 5 2.00 s. s. w S .c 2 W eration of a 5 3.00 ^j m/s and an initial velocity of ys BIO viruses, using an ordinary light microscope. An elec- w w w .a sw ar ph tron microscope, however, can view such objects using an electron beam instead of a light beam. Electron microscopy has proved invaluable for investigations of viruses, cell membranes and subcellular structures, bacterial surfaces, visual receptors, chloroplasts, and the contractile properties of muscles. The “lenses” of an electron microscope consist of electric and magnetic fields that control the electron beam. As an example of the manipulation of an electron beam, consider an electron traveling away from the origin along the S x axis in the xy plane with initial velocity vi 5 v i ^i. As it passes through the region x 5 0 to x 5 d, the electron S experiences acceleration a 5 a x ^i 1 a y ^j, where ax and ay are constants. For the case vi 5 1.80 3 107 m/s, ax 5 8.00 3 1014 m/s2, and ay 5 1.60 × 1015 m/s2, determine at x 5 d 5 0.010 0 m (a) the position of the electron, (b) the velocity of the electron, (c) the speed of the electron, and (d) the direction of travel of the electron (i.e., the angle between its velocity and the x axis). 9. A fish swimming in a horizontal plane has velocS AMT ity v i 5 1 4.00i^ 1 1.00 ^j 2 m/s at a point in the M ocean where the position relative to a certain rock After the fish swims is S r i 5 1 10.0 ^i 2 4.00 ^j 2 m. with constant acceleration for 20.0 s, its velocity is S v 5 1 20.0 ^i 2 5.00 ^j 2 m/s. (a) What are the components of the acceleration of the fish? (b) What is the direction of its acceleration with respect to unit vector ^i ? (c) If the fish maintains constant acceleration, where is it at t 5 25.0 s and in what direction is it moving? 10. Review. A snowmobile is originally at the point with position vector 29.0 m at 95.0° counterclockwise from 15. A projectile is fired in such a way that its horizontal range is equal to three times its maximum height. What is the angle of projection? 16. To start an avalanche on a mountain slope, an artillery W shell is fired with an initial velocity of 300 m/s at 55.0° above the horizontal. It explodes on the mountainside 42.0 s after firing. What are the x and y coordinates of the shell where it explodes, relative to its firing point? 17. Chinook salmon are able to move through water espe- BIO cially fast by jumping out of the water periodically. This behavior is called porpoising. Suppose a salmon swimming in still water jumps out of the water with velocity 6.26 m/s at 45.0° above the horizontal, sails through the air a distance L before returning to the water, and then swims the same distance L underwater in a straight, horizontal line with velocity 3.58 m/s before jumping out again. (a) Determine the average velocity of the fish for the entire process of jumping and swimming underwater. (b) Consider the time interval required to travel the entire distance of 2L. By what percentage is this time interval reduced by the jumping/swimming process compared with simply swimming underwater at 3.58 m/s? 18. A rock is thrown upward from level ground in such a Q/C way that the maximum height of its flight is equal to S its horizontal range R. (a) At what angle u is the rock thrown? (b) In terms of its original range R, what is the range R max the rock can attain if it is launched at 103 Problems of mass, which we will define in Chapter 9. His center of mass is at elevation 1.02 m when he leaves the floor. It reaches a maximum height of 1.85 m above the floor and is at elevation 0.900 m when he touches down again. Determine (a) his time of flight (his “hang time”), (b) his horizontal and (c) vertical velocity components at the instant of takeoff, and (d) his takeoff angle. (e) For comparison, determine the hang time of a whitetail deer making a jump (Fig. P4.24b) with center-of-mass elevations yi 5 1.20 m, ymax 5 2.50 m, and yf 5 0.700 m. d .c ee b b Figure P4.24 s. w 25. A playground is on the flat roof of a city school, 6.00 m above the street below (Fig. P4.25). The vertical wall of the building is h 5 7.00 m high, forming a 1-m-high railing around the playground. A ball has fallen to the street below, and a passerby returns it by launching it at an angle of u 5 53.0° above the horizontal at a point d 5 24.0 m from the base of the building wall. The ball takes 2.20 s to reach a point vertically above the wall. (a) Find the speed at which the ball was launched. (b) Find the vertical distance by which the ball clears the wall. (c) Find the horizontal distance from the wall to the point on the roof where the ball lands. ic ar ph ui ys h vi a ly 21. A firefighter, a distance d from a burning building, S directs a stream of water from a fire hose at angle ui above the horizontal as shown in Figure P4.21. If the initial speed of the stream is vi , at what height h does the water strike the building? S Figure P4.21 sw 22. A landscape architect is planning an artificial waterfall in a city park. Water flowing at 1.70 m/s will h leave the end of a horizontal channel at the top of a vertical wall h 5 2.35 m high, and from there it will fall into a pool (Fig. P4.22). (a) Will the space behind Figure P4.22 the waterfall be wide enough for a pedestrian walkway? (b) To sell her plan to the city council, the architect wants to build a model to standard scale, which is one-twelfth actual size. How fast should the water flow in the channel in the model? 23. A placekicker must kick a football from a point 36.0 m AMT (about 40 yards) from the goal. Half the crowd hopes M the ball will clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 20.0 m/s at an angle of 53.0° to the horizontal. (a) By how much does the ball clear or fall short of clearing the crossbar? (b) Does the ball approach the crossbar while still rising or while falling? 24. A basketball star covers 2.80 m horizontally in a jump to dunk the ball (Fig. P4.24a). His motion through space can be modeled precisely as that of a particle at his center .a h u w w w B. G. Smith/Shutterstock.com om © epa european pressphoto agency b.v./ Alamy the same speed but at the optimal angle for maximum range? (c) What If? Would your answer to part (a) be different if the rock is thrown with the same speed on a different planet? Explain. 19. The speed of a projectile when it reaches its maximum height is one-half its speed when it is at half its maximum height. What is the initial projection angle of the projectile? 20. A ball is tossed from an upper-story window of a buildW ing. The ball is given an initial velocity of 8.00 m/s at an angle of 20.0° below the horizontal. It strikes the ground 3.00 s later. (a) How far horizontally from the base of the building does the ball strike the ground? (b) Find the height from which the ball was thrown. (c) How long does it take the ball to reach a point 10.0 m below the level of launching? d Figure P4.25 26. The motion of a human body through space can be modeled as the motion of a particle at the body’s center of mass as we will study in Chapter 9. The components of the displacement of an athlete’s center of mass from the beginning to the end of a certain jump are described by the equations xf 5 0 1 (11.2 m/s)(cos 18.5°)t 0.360 m 5 0.840 m 1 1 11.2 m/s 2 1 sin 18.58 2 t 2 12 1 9.80 m/s2 2 t 2 where t is in seconds and is the time at which the athlete ends the jump. Identify (a) the athlete’s position and (b) his vector velocity at the takeoff point. (c) How far did he jump? 27. A soccer player kicks a rock horizontally off a W 40.0-m-high cliff into a pool of water. If the player Note: Problems 6 and 13 in Chapter 6 can also be assigned with this section. Adrian Dennis/AFP/Getty Images om 33. The athlete shown in Figure P4.33 rotates a 1.00-kg discus along a circular path of radius 1.06 m. The maximum speed of the discus is 20.0 m/s. Determine the magnitude of the maximum radial acceleration of the discus. Figure P4.33 34. In Example 4.6, we found the centripetal acceleration of the Earth as it revolves around the Sun. From information on the endpapers of this book, compute the centripetal acceleration of a point on the surface of the Earth at the equator caused by the rotation of the Earth about its axis. w 31. w w .a sw 30. Q/C ar ph ys ic GP Section 4.4 Analysis Model: Particle in Uniform Circular Motion .c 29. initial speed of the ball, (b) the time it takes the ball to reach the wall, and (c) the velocity components and the speed of the ball when it reaches the wall. (Assume the ball is hit at a height of 1.00 m above the ground.) ly S hears the sound of the splash 3.00 s later, what was the initial speed given to the rock? Assume the speed of sound in air is 343 m/s. A projectile is fired from the top of a cliff of height h above the ocean below. The projectile is fired at an angle u above the horizontal and with an initial speed vi . (a) Find a symbolic expression in terms of the variables vi , g, and u for the time at which the projectile reaches its maximum height. (b) Using the result of part (a), find an expression for the maximum height h max above the ocean attained by the projectile in terms of h, vi , g, and u. y A student stands at the S edge of a cliff and throws vi a stone horizontally over the edge with a speed of vi 5 18.0 m/s. The cliff is S h 5 50.0 m above a body g h of water as shown in Figure P4.29. (a) What are the coordinates of the initial position of the stone? (b) What are the compox nents of the initial velocity of the stone? (c) What is the S v appropriate analysis model Figure P4.29 for the vertical motion of the stone? (d) What is the appropriate analysis model for the horizontal motion of the stone? (e) Write symbolic equations for the x and y components of the velocity of the stone as a function of time. (f) Write symbolic equations for the position of the stone as a function of time. (g) How long after being released does the stone strike the water below the cliff? (h) With what speed and angle of impact does the stone land? The record distance in the sport of throwing cowpats is 81.1 m. This record toss was set by Steve Urner of the United States in 1981. Assuming the initial launch angle was 45° and neglecting air resistance, determine (a) the initial speed of the projectile and (b) the total time interval the projectile was in flight. (c) How would the answers change if the range were the same but the launch angle were greater than 45°? Explain. A boy stands on a diving board and tosses a stone into a swimming pool. The stone is thrown from a height of 2.50 m above the water surface with a velocity of 4.00 m/s at an angle of 60.0° above the horizontal. As the stone strikes the water surface, it immediately slows down to exactly half the speed it had when it struck the water and maintains that speed while in the water. After the stone enters the water, it moves in a straight line in the direction of the velocity it had when it struck the water. If the pool is 3.00 m deep, how much time elapses between when the stone is thrown and when it strikes the bottom of the pool? A home run is hit in such a way that the baseball just clears a wall 21.0 m high, located 130 m from home plate. The ball is hit at an angle of 35.0° to the horizontal, and air resistance is negligible. Find (a) the ee b 28. Motion in Two Dimensions s. w 104 Chapter 4 32. M 35. Casting molten metal is important in many industrial processes. Centrifugal casting is used for manufacturing pipes, bearings, and many other structures. A variety of sophisticated techniques have been invented, but the basic idea is as illustrated in Figure P4.35. A cylindrical enclosure is rotated rapidly and steadily about a horizontal axis. Molten metal is poured into the rotating cylinder and then cooled, forming the finished product. Turning the cylinder at a high rotation rate forces the solidifying metal strongly to the outside. Any bubbles are displaced toward the axis, so unwanted voids will not be present in the casting. Sometimes it is desirable to form a composite casting, such as for a bearing. Here a strong steel outer surface is poured and then inside it a lining of special low-friction metal. In some applications, a very strong metal is given a coating of corrosion-resistant metal. Centrifugal casting results in strong bonding between the layers. Preheated steel sheath Axis of rotation Molten metal Figure P4.35 Problems 105 Suppose a copper sleeve of inner radius 2.10 cm and outer radius 2.20 cm is to be cast. To eliminate bubbles and give high structural integrity, the centripetal acceleration of each bit of metal should be at least 100g. What rate of rotation is required? State the answer in revolutions per minute. 41. A train slows down as it rounds a sharp horizontal M turn, going from 90.0 km/h to 50.0 km/h in the 15.0 s it takes to round the bend. The radius of the curve is 150 m. Compute the acceleration at the moment the train speed reaches 50.0 km/h. Assume the train continues to slow down at this time at the same rate. 36. A tire 0.500 m in radius rotates at a constant rate of 200 rev/min. Find the speed and acceleration of a small stone lodged in the tread of the tire (on its outer edge). 42. A ball swings counterclockwise in a vertical circle at the end of a rope 1.50 m long. When the ball is 36.9° past the lowest point on its way up, its total acceleration is 1 222.5 i^ 1 20.2 j^ 2 m/s2. For that instant, (a) sketch a vector diagram showing the components of its acceleration, (b) determine the magnitude of its radial acceleration, and (c) determine the speed and velocity of the ball. 37. Review. The 20-g centrifuge at NASA’s Ames Research AMT Center in Mountain View, California, is a horizontal, om cylindrical tube 58.0 ft long and is represented in Figure P4.37. Assume an astronaut in training sits in a seat at one end, facing the axis of rotation 29.0 ft away. Determine the rotation rate, in revolutions per second, required to give the astronaut a centripetal acceleration of 20.0g. ly .c 43. (a) Can a particle moving with instantaneous speed 3.00 m/s on a path with radius of curvature 2.00 m have an acceleration of magnitude 6.00 m/s2? (b) Can it have an acceleration of magnitude 4.00 m/s2? In each case, if the answer is yes, explain how it can happen; if the answer is no, explain why not. ee b 29 ft Section 4.6 Relative Velocity and Relative Acceleration s. w 44. The pilot of an airplane notes that the compass indicates a heading due west. The airplane’s speed relative to the air is 150 km/h. The air is moving in a wind at 30.0 km/h toward the north. Find the velocity of the airplane relative to the ground. Figure P4.37 45. An airplane maintains a speed of 630 km/h relative to the air it is flying through as it makes a trip to a city 750 km away to the north. (a) What time interval is required for the trip if the plane flies through a headwind blowing at 35.0 km/h toward the south? (b) What time interval is required if there is a tailwind with the same speed? (c) What time interval is required if there is a crosswind blowing at 35.0 km/h to the east relative to the ground? ar ph ys ic 38. An athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the rate of 8.00 rev/s when the length of the chain is 0.600 m. When he increases the length to 0.900 m, he is able to rotate the ball only 6.00 rev/s. (a) Which rate of rotation gives the greater speed for the ball? (b) What is the centripetal acceleration of the ball at 8.00 rev/s? (c) What is the centripetal acceleration at 6.00 rev/s? w w Section 4.5 Tangential and Radial Acceleration 40. Figure P4.40 represents the W total acceleration of a particle moving clockwise in a circle of radius 2.50 m at a certain instant of time. For that instant, find (a) the radial acceleration of the particle, (b) the speed of the particle, and (c) its tangential acceleration. a  15.0 m/s2 S v S a 2.50 m 30.0 Figure P4.40 NASA w .a sw 39. The astronaut orbiting the Earth in Figure P4.39 is preparing to dock with a Westar VI satellite. The satellite is in a circular orbit 600 km above the Earth’s surface, where the free-fall acceleration is 8.21 m/s2. Take Figure P4.39 the radius of the Earth as 6 400 km. Determine the speed of the satellite and the time interval required to complete one orbit around the Earth, which is the period of the satellite. 46. A moving beltway at an airport has a speed v 1 and a S length L. A woman stands on the beltway as it moves from one end to the other, while a man in a hurry to reach his flight walks on the beltway with a speed of v 2 relative to the moving beltway. (a) What time interval is required for the woman to travel the distance L? (b) What time interval is required for the man to travel this distance? (c) A second beltway is located next to the first one. It is identical to the first one but moves in the opposite direction at speed v1. Just as the man steps onto the beginning of the beltway and begins to walk at speed v 2 relative to his beltway, a child steps on the other end of the adjacent beltway. The child stands at rest relative to this second beltway. How long after stepping on the beltway does the man pass the child? 47. A police car traveling at 95.0 km/h is traveling west, chasing a motorist traveling at 80.0 km/h. (a) What is the velocity of the motorist relative to the police car? (b) What is the velocity of the police car relative to the motorist? (c) If they are originally 250 m apart, in what time interval will the police car overtake the motorist? 48. A car travels due east with a speed of 50.0 km/h. RainM drops are falling at a constant speed vertically with Motion in Two Dimensions 51. A river flows with a steady speed v. A student swims Q/C upstream a distance d and then back to the starting S point. The student can swim at speed c in still water. (a) In terms of d, v, and c, what time interval is required for the round trip? (b) What time interval would be required if the water were still? (c) Which time interval is larger? Explain whether it is always larger. om 56. A ball is thrown with an initial speed vi at an angle ui S with the horizontal. The horizontal range of the ball is R, and the ball reaches a maximum height R/6. In terms of R and g, find (a) the time interval during which the ball is in motion, (b) the ball’s speed at the peak of its path, (c) the initial vertical component of its velocity, (d) its initial speed, and (e) the angle ui . (f) Suppose the ball is thrown at the same initial speed found in (d) but at the angle appropriate for reaching the greatest height that it can. Find this height. (g) Suppose the ball is thrown at the same initial speed but at the angle for greatest possible range. Find this maximum horizontal range. 57. Why is the following situation impossible? A normally proportioned adult walks briskly along a straight line in the 1x direction, standing straight up and holding his right arm vertical and next to his body so that the arm does not swing. His right hand holds a ball at his side a distance h above the floor. When the ball passes above a point marked as x 5 0 on the horizontal floor, he opens his fingers to release the ball from rest relative to his hand. The ball strikes the ground for the first time at position x 5 7.00h. 58. A particle starts from the origin with velocity 5i^ m/s at t 5 0 and moves in the xy plane with a varying accela is in meters per eration given by S a 5 1 6!t j^ 2 , where S second squared and t is in seconds. (a) Determine the velocity of the particle as a function of time. (b) Determine the position of the particle as a function of time. ar ph ys ic 52. A Coast Guard cutter detects an unidentified ship at a distance of 20.0 km in the direction 15.0° east of north. The ship is traveling at 26.0 km/h on a course at 40.0° east of north. The Coast Guard wishes to send a speedboat to intercept and investigate the vessel. If the speedboat travels at 50.0 km/h, in what direction should it head? Express the direction as a compass bearing with respect to due north. 55. A ball on the end of a string is whirled around in a horizontal circle of radius 0.300 m. The plane of the circle is 1.20 m above the ground. The string breaks and the ball lands 2.00 m (horizontally) away from the point on the ground directly beneath the ball’s location when the string breaks. Find the radial acceleration of the ball during its circular motion. .c 50. A river has a steady speed of 0.500 m/s. A student swims M upstream a distance of 1.00 km and swims back to the Q/C starting point. (a) If the student can swim at a speed of 1.20 m/s in still water, how long does the trip take? (b) How much time is required in still water for the same length swim? (c) Intuitively, why does the swim take longer when there is a current? Additional Problems ly 49. A bolt drops from the ceiling of a moving train car Q/C that is accelerating northward at a rate of 2.50 m/s2. (a) What is the acceleration of the bolt relative to the train car? (b) What is the acceleration of the bolt relative to the Earth? (c) Describe the trajectory of the bolt as seen by an observer inside the train car. (d) Describe the trajectory of the bolt as seen by an observer fixed on the Earth. can and catch it. In this observer’s frame of reference, (d) describe the shape of the can’s path and (e) determine the initial velocity of the can. ee b respect to the Earth. The traces of the rain on the side windows of the car make an angle of 60.0° with the vertical. Find the velocity of the rain with respect to (a) the car and (b) the Earth. s. w 106 Chapter 4 w .a sw 53. A science student is riding on a flatcar of a train travAMT eling along a straight, horizontal track at a constant M speed of 10.0 m/s. The student throws a ball into the air along a path that he judges to make an initial angle of 60.0° with the horizontal and to be in line with the track. The student’s professor, who is standing on the ground nearby, observes the ball to rise vertically. How high does she see the ball rise? w 54. A farm truck moves Q/C due east with a constant S v Altitude (ft) w velocity of 9.50 m/s on a limitless, horizontal stretch of road. A boy riding on the back of the truck Figure P4.54 throws a can of soda upward (Fig. P4.54) and catches the projectile at the same location on the truck bed, but 16.0 m farther down the road. (a) In the frame of reference of the truck, at what angle to the vertical does the boy throw the can? (b) What is the initial speed of the can relative to the truck? (c) What is the shape of the can’s trajectory as seen by the boy? An observer on the ground watches the boy throw the 59. The “Vomit Comet.” In microgravity astronaut training and equipment testing, NASA flies a KC135A aircraft along a parabolic flight path. As shown in Figure P4.59, the aircraft climbs from 24 000 ft to 31 000 ft, where 31 000 45 nose high 24 000 45 nose low Zero g 0 65 Maneuver time (s) Figure P4.59 107 Problems it enters a parabolic path with a velocity of 143 m/s nose high at 45.0° and exits with velocity 143 m/s at 45.0° nose low. During this portion of the flight, the aircraft and objects inside its padded cabin are in free fall; astronauts and equipment float freely as if there were no gravity. What are the aircraft’s (a) speed and (b) altitude at the top of the maneuver? (c) What is the time interval spent in microgravity? leave simultaneously and drive for 2.50 h in the directions shown. Car 1 has a speed of 90.0 km/h. If the cars arrive simultaneously at the lake, what is the speed of car 2? L 60. A basketball player is standing on the floor 10.0 m from the basket as in Figure P4.60. The height of the basket is 3.05 m, and he shoots the ball at a 40.0o angle with the horizontal from a height of 2.00 m. (a) What is the acceleration of the basketball at the highest point in its trajectory? (b) At what speed must the player throw the basketball so that the ball goes through the hoop without striking the backboard? 1 om 2 .c 40.0° 80.0 km B ly A ee b 40.0 Figure P4.64 3.05 m ic 10.0 m Figure P4.60 65. A catapult launches a rocket at an angle of 53.0° above the horizontal with an initial speed of 100 m/s. The rocket engine immediately starts a burn, and for 3.00 s the rocket moves along its initial line of motion with an acceleration of 30.0 m/s2. Then its engine fails, and the rocket proceeds to move in free fall. Find (a) the maximum altitude reached by the rocket, (b) its total time of flight, and (c) its horizontal range. s. w 2.00 m ar ph ys 61. Lisa in her Lamborghini accelerates at the rate of 1 3.00i^ 2 2.00 j^ 2 m/s2, while Jill in her Jaguar accelerates at 1 1.00i^ 1 3.00 j^ 2 m/s2. They both start from rest at the origin of an xy coordinate system. After 5.00 s, (a) what is Lisa’s speed with respect to Jill, (b) how far apart are they, and (c) what is Lisa’s acceleration relative to Jill? w w .a sw 62. A boy throws a stone horizontally from the top of a cliff S of height h toward the ocean below. The stone strikes the ocean at distance d from the base of the cliff. In terms of h, d, and g, find expressions for (a) the time t at which the stone lands in the ocean, (b) the initial speed of the stone, (c) the speed of the stone immediately before it reaches the ocean, and (d) the direction of the stone’s velocity immediately before it reaches the ocean. w 63. A flea is at point A on a horizontal turntable, 10.0 cm from the center. The turntable is rotating at 33.3 rev/min in the clockwise direction. The flea jumps straight up to a height of 5.00 cm. At takeoff, it gives itself no horizontal velocity relative to the turntable. The flea lands on the turntable at point B. Choose the origin of coordinates to be at the center of the turntable and the positive x axis passing through A at the moment of takeoff. Then the original position of the flea is 10.0 ^i cm. (a) Find the position of point A when the flea lands. (b) Find the position of point B when the flea lands. 64. Towns A and B in Figure P4.64 are 80.0 km apart. A M couple arranges to drive from town A and meet a couple driving from town B at the lake, L. The two couples 66. A cannon with a muzzle speed of 1 000 m/s is used to start an avalanche on a mountain slope. The target is 2 000 m from the cannon horizontally and 800 m above the cannon. At what angle, above the horizontal, should the cannon be fired? 67. Why is the following situation impossible? Albert Pujols hits a home run so that the baseball just clears the top row of bleachers, 24.0 m high, located 130 m from home plate. The ball is hit at 41.7 m/s at an angle of 35.0° to the horizontal, and air resistance is negligible. 68. As some molten metal splashes, one droplet flies off to S the east with initial velocity vi at angle ui above the horizontal, and another droplet flies off to the west with the same speed at the same angle above the horizontal as shown in Figure P4.68. In terms of vi and ui , find the distance between the two droplets as a function of time. S vi ui S vi ui Figure P4.68 69. An astronaut on the surface of the Moon fires a cannon to launch an experiment package, which leaves the barrel moving horizontally. Assume the free-fall acceleration on the Moon is one-sixth of that on the Motion in Two Dimensions om .c 45.0° θ D Figure P4.74 75. A World War II bomber flies horizontally over level terrain with a speed of 275 m/s relative to the ground and at an altitude of 3.00 km. The bombardier releases one bomb. (a) How far does the bomb travel horizontally between its release and its impact on the ground? Ignore the effects of air resistance. (b) The pilot maintains the plane’s original course, altitude, and speed through a storm of flak. Where is the plane when the bomb hits the ground? (c) The bomb hits the target seen in the telescopic bombsight at the moment of the bomb’s release. At what angle from the vertical was the bombsight set? ar ph ys ic 71. A hawk is flying horizontally at 10.0 m/s in a straight M line, 200 m above the ground. A mouse it has been carrying struggles free from its talons. The hawk continues on its path at the same speed for 2.00 s before attempting to retrieve its prey. To accomplish the retrieval, it dives in a straight line at constant speed and recaptures the mouse 3.00 m above the ground. (a) Assuming no air resistance acts on the mouse, find the diving speed of the hawk. (b) What angle did the hawk make with the horizontal during its descent? (c) For what time interval did the mouse experience free fall? θ ly 70. A pendulum with a cord of length r 5 1.00 m swings in a vertical plane (Fig. P4.70). When the pendulum is in the two horizontal positions u u 5 90.0° and u 5 270°, its r speed is 5.00 m/s. Find the S magnitude of (a) the radial g acceleration and (b) the ar tangential acceleration for f S a these positions. (c) Draw vector diagrams to deterat mine the direction of the Figure P4.70 total acceleration for these two positions. (d) Calculate the magnitude and direction of the total acceleration at these two positions. 74. An outfielder throws a baseball to his catcher in an attempt to throw out a runner at home plate. The ball bounces once before reaching the catcher. Assume the angle at which the bounced ball leaves the ground is the same as the angle at which the outfielder threw it as shown in Figure P4.74, but that the ball’s speed after the bounce is one-half of what it was before the bounce. (a) Assume the ball is always thrown with the same initial speed and ignore air resistance. At what angle u should the fielder throw the ball to make it go the same distance D with one bounce (blue path) as a ball thrown upward at 45.0° with no bounce (green path)? (b) Determine the ratio of the time interval for the one-bounce throw to the flight time for the no-bounce throw. ee b Earth. (a) What must the muzzle speed of the package be so that it travels completely around the Moon and returns to its original location? (b) What time interval does this trip around the Moon require? s. w 108 Chapter 4 72. A projectile is launched from the point (x 5 0, y 5 0), Q/C with velocity 1 12.0i^ 1 49.0 j^ 2 m/s, at t 5 0. (a) Make a w w w .a sw r | from the oritable listing the projectile’s distance |S gin at the end of each second thereafter, for 0 # t # 10 s. Tabulating the x and y coordinates and the components of velocity vx and vy will also be useful. (b) Notice that the projectile’s distance from its starting point increases with time, goes through a maximum, and starts to decrease. Prove that the distance is a maximum when the position vector is perpendicular to the velocv is not perpendicular to ity. Suggestion: Argue that if S S r , then |S r | must be increasing or decreasing. (c) Determine the magnitude of the maximum displacement. (d) Explain your method for solving part (c). 73. A spring cannon is located at the edge of a table that Q/C is 1.20 m above the floor. A steel ball is launched from the cannon with speed vi at 35.0° above the horizontal. (a) Find the horizontal position of the ball as a function of vi at the instant it lands on the floor. We write this function as x(vi ). Evaluate x for (b) vi 5 0.100 m/s and for (c) vi 5 100 m/s. (d) Assume vi is close to but not equal to zero. Show that one term in the answer to part (a) dominates so that the function x(vi ) reduces to a simpler form. (e) If vi is very large, what is the approximate form of x(vi )? (f) Describe the overall shape of the graph of the function x(vi ). 76. A truck loaded with cannonball watermelons stops suddenly to avoid running over the edge of a washed-out bridge (Fig. P4.76). The quick stop causes a number of melons to fly off the truck. One melon leaves the hood of the truck with an initial speed vi 5 10.0 m/s in the horizontal direction. A cross section of the bank has the shape of the bottom half of a parabola, with its vertex at the initial location of the projected watermelon and with the equation y 2 5 16x, where x and y are meay vi  10.0 m/s x Figure P4.76 The blue dashed curve shows the parabolic shape of the bank. Problems sured in meters. What are the x and y coordinates of the melon when it splatters on the bank? Challenge Problems 81. A skier leaves the ramp of a ski jump with a velocity of Q/C v 5 10.0 m/s at u 5 15.0° above the horizontal as shown om in Figure P4.81. The slope where she will land is inclined downward at f 5 50.0°, and air resistance is negligible. Find (a) the distance from the end of the ramp to where the jumper lands and (b) her velocity components just before the landing. (c) Explain how you think the results might be affected if air resistance were included. S u ly .c v s. w 78. An aging coyote cannot run BEEP fast enough to catch a roadBEEP runner. He purchases on eBay a set of jet-powered roller skates, which provide a constant horizontal acceleration of 15.0 m/s2 (Fig. P4.78). The coyote starts at rest 70.0 m from the edge of a cliff at the instant the roadrunner zips past in the Figure P4.78 direction of the cliff. (a) Determine the minimum constant speed the roadrunner must have to reach the cliff before the coyote. At the edge of the cliff, the roadrunner escapes by making a sudden turn, while the coyote continues straight ahead. The coyote’s skates remain horizontal and continue to operate while he is in flight, so his acceleration while in the air is 1 15.0 i^ 2 9.80 j^ 2 m/s2. (b) The cliff is 100 m above the flat floor of the desert. Determine how far from the base of the vertical cliff the coyote lands. (c) Determine the components of the coyote’s impact velocity. order-of-magnitude estimate of this acceleration, stating the quantities you measure or estimate and their values. ee b 77. A car is parked on a steep incline, making an angle of M 37.0° below the horizontal and overlooking the ocean, when its brakes fail and it begins to roll. Starting from rest at t 5 0, the car rolls down the incline with a constant acceleration of 4.00 m/s2, traveling 50.0 m to the edge of a vertical cliff. The cliff is 30.0 m above the ocean. Find (a) the speed of the car when it reaches the edge of the cliff, (b) the time interval elapsed when it arrives there, (c) the velocity of the car when it lands in the ocean, (d) the total time interval the car is in motion, and (e) the position of the car when it lands in the ocean, relative to the base of the cliff. 109 f Figure P4.81 ar ph ys ic 82. Two swimmers, Chris and Sarah, start together at the S same point on the bank of a wide stream that flows with a speed v. Both move at the same speed c (where c . v) relative to the water. Chris swims downstream a distance L and then upstream the same distance. Sarah swims so that her motion relative to the Earth is perpendicular to the banks of the stream. She swims the distance L and then back the same distance, with both swimmers returning to the starting point. In terms of L, c, and v, find the time intervals required (a) for Chris’s round trip and (b) for Sarah’s round trip. (c) Explain which swimmer returns first. w w w .a sw 79. A fisherman sets out upstream on a river. His small boat, powered by an outboard motor, travels at a constant speed v in still water. The water flows at a lower constant speed vw . The fisherman has traveled upstream for 2.00 km when his ice chest falls out of the boat. He notices that the chest is missing only after he has gone upstream for another 15.0 min. At that point, he turns around and heads back downstream, all the time traveling at the same speed relative to the water. He catches up with the floating ice chest just as he returns to his starting point. How fast is the river flowing? Solve this problem in two ways. (a) First, use the Earth as a reference frame. With respect to the Earth, the boat travels upstream at speed v 2 vw and downstream at v 1 vw. (b) A second much simpler and more elegant solution is obtained by using the water as the reference frame. This approach has important applications in many more complicated problems; examples are calculating the motion of rockets and satellites and analyzing the scattering of subatomic particles from massive targets. 80. Do not hurt yourself; do not strike your hand against Q/C anything. Within these limitations, describe what you do to give your hand a large acceleration. Compute an 83. The water in a river flows uniformly at a constant speed of 2.50 m/s between parallel banks 80.0 m apart. You are to deliver a package across the river, but you can swim only at 1.50 m/s. (a) If you choose to minimize the time you spend in the water, in what direction should you head? (b) How far downstream will you be carried? (c) If you choose to minimize the distance downstream that the river carries you, in what direction should you head? (d) How far downstream will you be carried? 84. A person standing at the top of a hemispherical rock of S radius R kicks a ball (initially at rest on the top of the v i as shown in Figrock) to give it horizontal velocity S ure P4.84. (a) What must be its minimum initial speed S vi R Figure P4.84 x Motion in Two Dimensions if the ball is never to hit the rock after it is kicked? (b) With this initial speed, how far from the base of the rock does the ball hit the ground? 85. A dive-bomber has a velocity of 280 m/s at an angle u below the horizontal. When the altitude of the aircraft is 2.15 km, it releases a bomb, which subsequently hits a target on the ground. The magnitude of the displacement from the point of release of the bomb to the target is 3.25 km. Find the angle u. 86. A projectile is fired up an incline (incline angle f) with an initial speed vi at an angle ui with respect to the horizontal (ui > f) as shown in Figure P4.86. (a) Show that the projectile travels a distance d up the incline, where 2v i cos ui sin 1 u i 2 f 2 g cos 2 f S θi where f is the angle the hill makes with the horizontal in Figure 4.14. Prove this claim by deriving the equation above. 89. An enemy ship is on the east side of a mountain island as shown in Figure P4.89. The enemy ship has maneuvered to within 2 500 m of the 1 800-m-high mountain peak and can shoot projectiles with an initial speed of 250 m/s. If the western shoreline is horizontally 300 m from the peak, what are the distances from the western shore at which a ship can be safe from the bombardment of the enemy ship? ee b Path of the projectile vi d φ S ys ic s. w Figure P4.86 uH 1 800 m ar ph v i  250 m/s vi uL w w .a sw 2 500 m w f 2 ly d5 u 5 458 2 .c 2 (b) For what value of ui is d a maximum, and what is that maximum value? 87. A fireworks rocket explodes at height h, the peak of S its vertical trajectory. It throws out burning fragments in all directions, but all at the same speed v. Pellets of solidified metal fall to the ground without air resistance. Find the smallest angle that the final velocity of an impacting fragment makes with the horizontal. 88. In the What If? section of Example 4.5, it was claimed S that the maximum range of a ski jumper occurs for a launch angle u given by om 110 Chapter 4 Figure P4.89 300 m c h a p t e r 5 The Laws of Motion 5.1 The Concept of Force om 5.2 Newton’s First Law and Inertial Frames 5.3 Mass .c 5.4 Newton’s Second Law ly 5.5 The Gravitational Force and Weight 5.7 Analysis Models Using Newton’s Second Law 5.8 Forces of Friction ar ph ys ic s. w ee b 5.6 Newton’s Third Law A person sculls on a calm waterway. The water exerts forces on the oars to accelerate the boat. (Tetra Images/ Getty Images) w w .a sw In Chapters 2 and 4, we described the motion of an object in terms of its position, velocity, and acceleration without considering what might influence that motion. Now we consider that influence: Why does the motion of an object change? What might cause one object to remain at rest and another object to accelerate? Why is it generally easier to move a small object than a large object? The two main factors we need to consider are the forces acting on an object and the mass of the object. In this chapter, we begin our study of dynamics by discussing the three basic laws of motion, which deal with forces and masses and were formulated more than three centuries ago by Isaac Newton. w 5.1 The Concept of Force Everyone has a basic understanding of the concept of force from everyday experience. When you push your empty dinner plate away, you exert a force on it. Similarly, you exert a force on a ball when you throw or kick it. In these examples, the word force refers to an interaction with an object by means of muscular activity and some change in the object’s velocity. Forces do not always cause motion, however. For example, when you are sitting, a gravitational force acts on your body and yet you remain stationary. As a second example, you can push (in other words, exert a force) on a large boulder and not be able to move it. What force (if any) causes the Moon to orbit the Earth? Newton answered this and related questions by stating that forces are what cause any change in the velocity of an object. The Moon’s velocity changes in direction as it moves in a nearly circular 111 112 Chapter 5 The Laws of Motion Figure 5.1 Some examples of applied forces. In each case, a force is exerted on the object within the boxed area. Some agent in the environment external to the boxed area exerts a force on the object. Contact forces a c b Field forces q M Iron Q e S f .c d N om m Isaac Newton w ar ph .a sw Bridgeman-Giraudon/Art Resource, NY ys ic s. w ee b ly orbit around the Earth. This change in velocity is caused by the gravitational force exerted by the Earth on the Moon. When a coiled spring is pulled, as in Figure 5.1a, the spring stretches. When a stationary cart is pulled, as in Figure 5.1b, the cart moves. When a football is kicked, as in Figure 5.1c, it is both deformed and set in motion. These situations are all examples of a class of forces called contact forces. That is, they involve physical contact between two objects. Other examples of contact forces are the force exerted by gas molecules on the walls of a container and the force exerted by your feet on the floor. Another class of forces, known as field forces, does not involve physical contact between two objects. These forces act through empty space. The gravitational force of attraction between two objects with mass, illustrated in Figure 5.1d, is an example of this class of force. The gravitational force keeps objects bound to the Earth and the planets in orbit around the Sun. Another common field force is the electric force that one electric charge exerts on another (Fig. 5.1e), such as the attractive electric force between an electron and a proton that form a hydrogen atom. A third example of a field force is the force a bar magnet exerts on a piece of iron (Fig. 5.1f). The distinction between contact forces and field forces is not as sharp as you may have been led to believe by the previous discussion. When examined at the atomic level, all the forces we classify as contact forces turn out to be caused by electric (field) forces of the type illustrated in Figure 5.1e. Nevertheless, in developing models for macroscopic phenomena, it is convenient to use both classifications of forces. The only known fundamental forces in nature are all field forces: (1) gravitational forces between objects, (2) electromagnetic forces between electric charges, (3) strong forces between subatomic particles, and (4) weak forces that arise in certain radioactive decay processes. In classical physics, we are concerned only with gravitational and electromagnetic forces. We will discuss strong and weak forces in Chapter 46. w English physicist and mathematician (1642–1727) w Isaac Newton was one of the most brilliant scientists in history. Before the age of 30, he formulated the basic concepts and laws of mechanics, discovered the law of universal gravitation, and invented the mathematical methods of calculus. As a consequence of his theories, Newton was able to explain the motions of the planets, the ebb and flow of the tides, and many special features of the motions of the Moon and the Earth. He also interpreted many fundamental observations concerning the nature of light. His contributions to physical theories dominated scientific thought for two centuries and remain important today. The Vector Nature of Force It is possible to use the deformation of a spring to measure force. Suppose a vertical force is applied to a spring scale that has a fixed upper end as shown in Figure 5.2a. The spring elongates when the force is applied, and a pointer on the scale reads the extension of the spring. We can calibrate the spring by defining a S reference force F 1 as the force that produces a pointer reading of 1.00 cm. If we S now apply a different downward force F whose magnitude is twice that of the ref2 S erence force F 1 as seen in Figure 5.2b, the pointer moves to 2.00 cm. Figure 5.2c shows that the combined effect of the two collinear forces is the sum of the effects of the individual forces. S Now suppose the two forces are applied simultaneously with F 1 downward and S F 2 horizontal as S illustrated in Figure 5.2d. In this case, the pointer reads 2.24 cm. The single force F that would produce this same reading is the sum of the two vecS S S tors F 1 and F 2 as described in Figure 5.2d. That is, 0 F 1 0 5 !F12 1 F22 5 2.24 units, 5.2 Newton’s First Law and Inertial Frames S A downward S force F2 elongates the spring 2.00 cm. S When F1 and F2 are applied together in the same direction, the spring elongates by 3.00 cm. S S When F1 is downward and F2 is horizontal, the combination of the two forces elongates the spring by 2.24 cm. 1 0 A downward S force F1 elongates the spring 1.00 cm. 113 2 0 1 2 3 4 4 S u S F1 om F2 S .c 0 1 2 3 4 3 0 1 2 3 4 S F1 F1 S F2 S F2 b c d Figure 5.2 The vector nature of a force is tested with a spring scale. s. w a ee b S ly F ys ic and its direction is u 5 tan21 (20.500) 5 226.6°. Because forces have been experimentally verified to behave as vectors, you must use the rules of vector addition to obtain the net force on an object. ar ph 5.2 Newton’s First Law and Inertial Frames w w w .a sw We begin our study of forces by imagining some physical situations involving a puck on a perfectly level air hockey table (Fig. 5.3). You expect that the puck will remain stationary when it is placed gently at rest on the table. Now imagine your air hockey table is located on a train moving with constant velocity along a perfectly smooth track. If the puck is placed on the table, the puck again remains where it is placed. If the train were to accelerate, however, the puck would start moving along the table opposite the direction of the train’s acceleration, just as a set of papers on your dashboard falls onto the floor of your car when you step on the accelerator. As we saw in Section 4.6, a moving object can be observed from any number of reference frames. Newton’s first law of motion, sometimes called the law of inertia, defines a special set of reference frames called inertial frames. This law can be stated as follows: If an object does not interact with other objects, it is possible to identify a reference frame in which the object has zero acceleration. Such a reference frame is called an inertial frame of reference. When the puck is on the air hockey table located on the ground, you are observing it from an inertial reference frame; there are no horizontal interactions of the puck with any other objects, and you observe it to have zero acceleration in that direction. When you are on the train moving at constant velocity, you are also observing the puck from an inertial reference frame. Any reference frame that moves with constant velocity relative to an inertial frame is itself an inertial frame. When you and the train accelerate, however, you are observing the puck from a noninertial reference frame because the train is accelerating relative to the inertial reference frame of the Earth’s surface. While the puck appears to be accelerating according to your observations, a reference frame can be identified in which the puck has zero acceleration. Airflow Electric blower Figure 5.3 On an air hockey table, air blown through holes in the surface allows the puck to move almost without friction. If the table is not accelerating, a puck placed on the table will remain at rest. WW Newton’s first law WW Inertial frame of reference The Laws of Motion For example, an observer standing outside the train on the ground sees the puck sliding relative to the table but always moving with the same velocity with respect to the ground as the train had before it started to accelerate (because there is almost no friction to “tie” the puck and the train together). Therefore, Newton’s first law is still satisfied even though your observations as a rider on the train show an apparent acceleration relative to you. A reference frame that moves with constant velocity relative to the distant stars is the best approximation of an inertial frame, and for our purposes we can consider the Earth as being such a frame. The Earth is not really an inertial frame because of its orbital motion around the Sun and its rotational motion about its own axis, both of which involve centripetal accelerations. These accelerations are small compared with g, however, and can often be neglected. For this reason, we model the Earth as an inertial frame, along with any other frame attached to it. Let us assume we are observing an object from an inertial reference frame. (We will return to observations made in noninertial reference frames in Section 6.3.) Before about 1600, scientists believed that the natural state of matter was the state of rest. Observations showed that moving objects eventually stopped moving. Galileo was the first to take a different approach to motion and the natural state of matter. He devised thought experiments and concluded that it is not the nature of an object to stop once set in motion: rather, it is its nature to resist changes in its motion. In his words, “Any velocity once imparted to a moving body will be rigidly maintained as long as the external causes of retardation are removed.” For example, a spacecraft drifting through empty space with its engine turned off will keep moving forever. It would not seek a “natural state” of rest. Given our discussion of observations made from inertial reference frames, we can pose a more practical statement of Newton’s first law of motion: om 114 Chapter 5 Pitfall Prevention 5.1 ly ee b s. w ic Another statement of  Newton’s first law ar ph ys In the absence of external forces and when viewed from an inertial reference frame, an object at rest remains at rest and an object in motion continues in motion with a constant velocity (that is, with a constant speed in a straight line). sw w Pitfall Prevention 5.2 In other words, when no force acts on an object, the acceleration of the object is zero. From the first law, we conclude that any isolated object (one that does not interact with its environment) is either at rest or moving with constant velocity. The tendency of an object to resist any attempt to change its velocity is called inertia. Given the statement of the first law above, we can conclude that an object that is accelerating must be experiencing a force. In turn, from the first law, we can define force as that which causes a change in motion of an object. .a Definition of force  w Force Is the Cause of Changes in Motion An object can have motion in the absence of forces as described in Newton’s first law. Therefore, don’t interpret force as the cause of motion. Force is the cause of changes in motion. w  .c Newton’s First Law Newton’s first law does not say what happens for an object with zero net force, that is, multiple forces that cancel; it says what happens in the absence of external forces. This subtle but important difference allows us to define force as that which causes a change in the motion. The description of an object under the effect of forces that balance is covered by Newton’s second law. Definition of mass  Q uick Quiz 5.1 Which of the following statements is correct? (a) It is possible for an object to have motion in the absence of forces on the object. (b) It is possible to have forces on an object in the absence of motion of the object. (c) Neither statement (a) nor statement (b) is correct. (d) Both statements (a) and (b) are correct. 5.3 Mass Imagine playing catch with either a basketball or a bowling ball. Which ball is more likely to keep moving when you try to catch it? Which ball requires more effort to throw it? The bowling ball requires more effort. In the language of physics, we say that the bowling ball is more resistant to changes in its velocity than the basketball. How can we quantify this concept? Mass is that property of an object that specifies how much resistance an object exhibits to changes in its velocity, and as we learned in Section 1.1, the SI unit of mass is the kilogram. Experiments show that the greater the mass of an object, the less that object accelerates under the action of a given applied force. To describe mass quantitatively, we conduct experiments in which we compare the accelerations a given force produces on different objects. Suppose a force act- 5.4 Newton’s Second Law ic ys 5.4 Newton’s Second Law WW Mass and weight are different quantities s. w ee b ly .c om ing on an object of mass m1 produces a change in motion of the object that we can quantify with the object’s acceleration S a 1, and the same force acting on an object of mass m 2 produces an acceleration S a 2. The ratio of the two masses is defined as the inverse ratio of the magnitudes of the accelerations produced by the force: a2 m1 ; (5.1) a1 m2 For example, if a given force acting on a 3-kg object produces an acceleration of 4 m/s2, the same force applied to a 6-kg object produces an acceleration of 2 m/s2. According to a huge number of similar observations, we conclude that the magnitude of the acceleration of an object is inversely proportional to its mass when acted on by a given force. If one object has a known mass, the mass of the other object can be obtained from acceleration measurements. Mass is an inherent property of an object and is independent of the object’s surroundings and of the method used to measure it. Also, mass is a scalar quantity and thus obeys the rules of ordinary arithmetic. For example, if you combine a 3-kg mass with a 5-kg mass, the total mass is 8 kg. This result can be verified experimentally by comparing the acceleration that a known force gives to several objects separately with the acceleration that the same force gives to the same objects combined as one unit. Mass should not be confused with weight. Mass and weight are two different quantities. The weight of an object is equal to the magnitude of the gravitational force exerted on the object and varies with location (see Section 5.5). For example, a person weighing 180 lb on the Earth weighs only about 30 lb on the Moon. On the other hand, the mass of an object is the same everywhere: an object having a mass of 2 kg on the Earth also has a mass of 2 kg on the Moon. 115 w w w .a sw ar ph Newton’s first law explains what happens to an object when no forces act on it: it maintains its original motion; it either remains at rest or moves in a straight line with constant speed. Newton’s second law answers the question of what happens to an object when one or more forces act on it. Imagine performing an experiment in which you push a block of massSm across a frictionless, horizontal surface. When you exert some horizontal force F on the block, it moves with some acceleration S a . If you apply a force twice as great on the same block, experimental results show that the acceleration of the block doubles; if S you increase the applied force to 3 F , the acceleration triples; and so on. From such observations, we concludeSthat the acceleration of an object is directly proportional to the force acting on it: F ~ S a . This idea was first introduced in Section 2.4 when we discussed the direction of the acceleration of an object. We also know from the preceding section that the magnitude of the acceleration of an object is inversely proportional to its mass: 0 S a 0 ~ 1/m. These experimental observations are summarized in Newton’s second law: When viewed from an inertial reference frame, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass: S aF S a ~ m If we choose a proportionality constant of 1, we can relate mass, acceleration, and force through the following mathematical statement of Newton’s second law:1 S a F 5 ma S 1Equation (5.2) 5.2 is valid only when the speed of the object is much less than the speed of light. We treat the relativistic situation in Chapter 39. Pitfall Prevention 5.3 S ma Is Not a Force Equation 5.2 does not say that the product mS a is a force. All forces on an object are added vectorially to generate the net force on the left side of the equation. This net force is then equated to the product of the mass of the object and the acceleration that results from the net force. Do not include an “mS a force” in your analysis of the forces on an object. WW Newton’s second law 116 Chapter 5 In both the textual and mathematical statements of Newton’s second law, we have S indicated that the acceleration is due to the net force g F acting on an object. The net force on an object is the vector sum of all forces acting on the object. (We sometimes refer to the net force as the total force, the resultant force, or the unbalanced force.) In solving a problem using Newton’s second law, it is imperative to determine the correct net force on an object. Many forces may be acting on an object, but there is only one acceleration. Equation 5.2 is a vector expression and hence is equivalent to three component equations: (5.3) a Fx 5 ma x    a Fy 5 ma y   a Fz 5 ma z Newton’s second law:  component form om Q uick Quiz 5.2 An object experiences no acceleration. Which of the following cannot be true for the object? (a) A single force acts on the object. (b) No forces act on the object. (c) Forces act on the object, but the forces cancel. ee b ly .c Q uick Quiz 5.3 You push an object, initially at rest, across a frictionless floor with a constant force for a time interval Dt, resulting in a final speed of v for the object. You then repeat the experiment, but with a force that is twice as large. What time interval is now required to reach the same final speed v? (a) 4 Dt (b) 2 Dt (c) Dt (d) Dt/2 (e) Dt/4 s. w The SI unit of force is the newton (N). A force of 1 N is the force that, when acting on an object of mass 1 kg, produces an acceleration of 1 m/s2. From this definition and Newton’s second law, we see that the newton can be expressed in terms of the following fundamental units of mass, length, and time: 1 N ; 1 kg ? m/s2 (5.4) ic Definition of the newton  ys In the U.S. customary system, the unit of force is the pound (lb). A force of 1 lb is the force that, when acting on a 1-slug mass, 2 produces an acceleration of 1 ft/s2: ar ph 1 lb ; 1 slug ? ft/s2 A convenient approximation is 1 N < 14 lb.    An Accelerating Hockey Puck AM y S sw Example 5.1 F2 F1 = 5.0 N F2 = 8.0 N w w .a A hockey puck having a mass of 0.30 kg slides on the frictionless, horizontal surface of an ice rink. Two hockey sticks strike the puck simultaneously, exerting the forces on the puck shown S in Figure 5.4. The force F1 has a magnitude of 5.0 N, and is S directed at u 5 20° below the x axis. The force F2 has a magnitude of 8.0 N and its direction is f 5 60° above the x axis. Determine both the magnitude and the direction of the puck’s acceleration. w  The Laws of Motion Solution Figure 5.4 (Example 5.1) A hockey puck moving on a frictionless surface is subject to two S S forces F 1 and F 2. 60 20 x S F1 Conceptualize Study Figure 5.4. Using your expertise in vector addition from Chapter 3, predict the approximate direction of the net force vector on the puck. The acceleration of the puck will be in the same direction. Categorize Because we can determine a net force and we want an acceleration, this problem is categorized as one that may be solved using Newton’s second law. In Section 5.7, we will formally introduce the particle under a net force analysis model to describe a situation such as this one. Analyze Find the component of the net force acting on the puck in the x direction: 2The a Fx 5 F1x 1 F2x 5 F1 cos u 1 F2 cos f slug is the unit of mass in the U.S. customary system and is that system’s counterpart of the SI unit the kilogram. Because most of the calculations in our study of classical mechanics are in SI units, the slug is seldom used in this text. 5.5 The Gravitational Force and Weight ▸ 5.1 c o n t i n u e d 117 a Fy 5 F1y 1 F2y 5 F1 sin u 1 F2 sin f Find the component of the net force acting on the puck in the y direction: Use Newton’s second law in component form (Eq. 5.3) to find the x and y components of the puck’s acceleration: ax 5 ay 5 F1 cos u 1 F2 cos f a Fx 5 m m F1 sin u 1 F2 sin f a Fy 5 m m 1 5.0 N 2 cos 1 2208 2 1 1 8.0 N 2 cos 1 608 2 5 29 m/s2 0.30 kg 1 5.0 N 2 sin 1 2208 2 1 1 8.0 N 2 sin 1 608 2 ay 5 5 17 m/s2 0.30 kg Substitute numerical values: .c om ax 5 u 5 tan21 a ay ax ly Find the direction of the acceleration relative to the positive x axis: a 5 " 1 29 m/s2 2 2 1 1 17 m/s2 2 2 5 34 m/s2 b 5 tan21 a ee b Find the magnitude of the acceleration: 17 b 5 318 29 s. w Finalize The vectors in Figure 5.4 can be added graphically to check the reasonableness of our answer. Because the acceleration vector is along the direction of the resultant force, a drawing showing the resultant force vector helps us check the validity of the answer. (Try it!) ic W h at I f ? Suppose three hockey sticks strike the puck simultaneously, with two of them exerting the forces shown in Figure 5.4. The result of the three forces is that the hockey puck shows no acceleration. What must be the components of the third force? ar ph ys Answer If there is zero acceleration, the net force acting on the puck must be zero. Therefore, the three forces must cancel. The components of the third force must be of equal magnitude and opposite sign compared to the components of the net force applied by the first two forces so that all the components add to zero. Therefore, F 3x 5 2a Fx 5 2 1 0.30 kg 2 1 29 m/s 2 2 5 28.7 N and F 3y 5 2a Fy 5 2 1 0.30 kg 2 1 17 m/s 2 2 5 25.2 N. sw 5.5 The Gravitational Force and Weight Pitfall Prevention 5.4 w w w .a All objects are attracted to the Earth. TheSattractive force exerted by the Earth on an object is called the gravitational force F g . This force is directed toward the center of the Earth,3 and its magnitude is called the weight of the object. S We saw in Section 2.6 that a freely falling object experiences an acceleration g S S acting toward the center of the Earth. Applying Newton’s second law g F 5 m a to S S a freely falling object of mass m, with S a 5S g and g F 5 F g , gives S g F g 5 mS (5.5) S Therefore, the weight of an object, being defined as the magnitude of F g , is given by Fg = mg (5.6) Because it depends on g, weight varies with geographic location. Because g decreases with increasing distance from the center of the Earth, objects weigh less at higher altitudes than at sea level. For example, a 1 000-kg pallet of bricks used in the construction of the Empire State Building in New York City weighed 9 800 N at street level, but weighed about 1 N less by the time it was lifted from sidewalk level to the top of the building. As another example, suppose a student has a mass 3This statement ignores that the mass distribution of the Earth is not perfectly spherical. “Weight of an Object” We are familiar with the everyday phrase, the “weight of an object.” Weight, however, is not an inherent property of an object; rather, it is a measure of the gravitational force between the object and the Earth (or other planet). Therefore, weight is a property of a system of items: the object and the Earth. Pitfall Prevention 5.5 Kilogram Is Not a Unit of Weight You may have seen the “conversion” 1 kg 5 2.2 lb. Despite popular statements of weights expressed in kilograms, the kilogram is not a unit of weight, it is a unit of mass. The conversion statement is not an equality; it is an equivalence that is valid only on the Earth’s surface. The Laws of Motion ee b ly .c The life-support unit strapped to the back of astronaut Harrison Schmitt weighed 300 lb on the Earth and had a mass of 136 kg. During his training, a 50-lb mockup with a mass of 23 kg was used. Although this strategy effectively simulated the reduced weight the unit would have on the Moon, it did not correctly mimic the unchanging mass. It was more difficult to accelerate the 136-kg unit (perhaps by jumping or twisting suddenly) on the Moon than it was to accelerate the 23-kg unit on the Earth. of 70.0 kg. The student’s weight in a location where g 5 9.80 m/s2 is 686 N (about 150 lb). At the top of a mountain, however, where g 5 9.77 m/s2, the student’s weight is only 684 N. Therefore, if you want to lose weight without going on a diet, climb a mountain or weigh yourself at 30 000 ft during an airplane flight! Equation 5.6 quantifies the gravitational force on the object, but notice that this equation does not require the object to be moving. Even for a stationary object or for an object on which several forces act, Equation 5.6 can be used to calculate the magnitude of the gravitational force. The result is a subtle shift in the interpretation of m in the equation. The mass m in Equation 5.6 determines the strength of the gravitational attraction between the object and the Earth. This role is completely different from that previously described for mass, that of measuring the resistance to changes in motion in response to an external force. In that role, mass is also called inertial mass. We call m in Equation 5.6 the gravitational mass. Even though this quantity is different in behavior from inertial mass, it is one of the experimental conclusions in Newtonian dynamics that gravitational mass and inertial mass have the same value. Although this discussion has focused on the gravitational force on an object due to the Earth, the concept is generally valid on any planet. The value of g will vary from one planet to the next, but the magnitude of the gravitational force will always be given by the value of mg. om NASA/Eugene Cernan 118 Chapter 5    How Much Do You Weigh in an Elevator? ar ph Conceptual Example 5.2 ys ic s. w Q uick Quiz 5.4 Suppose you are talking by interplanetary telephone to a friend who lives on the Moon. He tells you that he has just won a newton of gold in a contest. Excitedly, you tell him that you entered the Earth version of the same contest and also won a newton of gold! Who is richer? (a) You are. (b) Your friend is. (c) You are equally rich. sw You have most likely been in an elevator that accelerates upward as it moves toward a higher floor. In this case, you feel heavier. In fact, if you are standing on a bathroom scale at the time, the scale measures a force having a magnitude that is greater than your weight. Therefore, you have tactile and measured evidence that leads you to believe you are heavier in this situation. Are you heavier? Solution w w w .a No; your weight is unchanged. Your experiences are due to your being in a noninertial reference frame. To provide the acceleration upward, the floor or scale must exert on your feet an upward force that is greater in magnitude than your weight. It is this greater force you feel, which you interpret as feeling heavier. The scale reads this upward force, not your weight, and so its reading increases. 5.6 Newton’s Third Law If you press against a corner of this textbook with your fingertip, the book pushes back and makes a small dent in your skin. If you push harder, the book does the same and the dent in your skin is a little larger. This simple activity illustrates that forces are interactions between two objects: when your finger pushes on the book, the book pushes back on your finger. This important principle is known as Newton’s third law: Newton’s third law  S If two objects interact, the force F 12 exerted by object 1 on object 2 is equal in S magnitude and opposite in direction to the force F 21 exerted by object 2 on object 1: S S F 12 5 2 F 21 (5.7) 119 5.6 Newton’s Third Law S S F12  F21 2 S F12 S F21 1 Figure 5.5S Newton’s third law. The force F 12 exerted by object 1 on object 2 is equal in magnitude and opposite in direction to S the force F 21 exerted by object 2 on object 1. w w w .a sw ar ph ys ic s. w ee b ly .c om When it is important to designate forces as interactions between two objects, we S will use this subscript notation, where F ab means “the force exerted by a on b.” The third law is illustrated in Figure 5.5. The force that object 1 exerts on object 2 is popularly called the action force, and the force of object 2 on object 1 is called the reaction force. These italicized terms are not scientific terms; furthermore, either force can be labeled the action or reaction force. We will use these terms for convenience. In all cases, the action and reaction forces act on different objects and must be of the same type (gravitational, electrical, etc.). For example, the force acting on a freely falling projectile is the gravitational force exerted by the Earth on S S the projectile F g 5 F Ep (E 5 Earth, p 5 projectile), and the magnitude of this force is mg. The reaction to this force is the gravitational force exerted by the proS S S jectile on the Earth F pE 5 2 F Ep. The reactionSforce F pE must accelerate the Earth toward the projectile just as the action force F Ep accelerates the projectile toward the Earth. Because the Earth has such a large mass, however, its acceleration due to this reaction force is negligibly small. Consider a computer monitorS at rest on a table as in Figure 5.6a. The gravitaS tional force on the monitor is F 5 F . The reaction to this force is the force g Em S S F mE 5 2 F Em exerted by the monitor on the Earth. The monitor does not accelerate becauseSit is held up by the table. The table exerts on the monitor an upward force S n 5 F tm, called the normal force. (Normal in this context means perpendicular.) In general, whenever an object is in contact with a surface, the surface exerts a normal force on the object. The normal force on the monitor can have any value needed, up to the point of breaking the table. Because the monitor has zero accelS eration, Newton’s second law applied to the monitor gives us g F 5 S n 1 mS g 5 0, ^ ^ so n j 2 mg j 5 0, or n 5 mg. The normal force balances the gravitational force on the monitor, so the net force on the monitor is zero. The reaction force to S n is the S S S force exerted by the monitor downward on the table, F 5 2 F 5 2n . mt tm S Notice that the forces acting on the monitor are F g and S n as shown in Figure 5.6b. S S The two forces F mE and F mt are exerted on objects other than the monitor. Figure 5.6 illustrates an extremely important step in solving problems involving forces. Figure 5.6a shows many of the forces in the situation: those acting on the monitor, one acting on the table, and one acting on the Earth. Figure 5.6b, by contrast, shows only the forces acting on one object, the monitor, and is called a force diagram or a diagram showing the forces on the object. The important pictorial representation in Figure 5.6c is called a free-body diagram. In a free-body diagram, the particle model is used by representing the object as a dot and showing the forces that act on the object as being applied to the dot. When analyzing an object subject to forces, we are interested in the net force acting on one object, which we will model as a particle. Therefore, a free-body diagram helps us isolate only those forces on the object and eliminate the other forces from our analysis. S S S S S S S S Fg  FEm S S S Fg  FEm S FmE a S n  Ftm Fg  FEm Fmt S n  Ftm n  Ftm b c Pitfall Prevention 5.6 n Does Not Always Equal mg In the situation shown in Figure 5.6 and in many others, we find that n 5 mg (the normal force has the same magnitude as the gravitational force). This result, however, is not generally true. If an object is on an incline, if there are applied forces with vertical components, or if there is a vertical acceleration of the system, then n ? mg. Always apply Newton’s second law to find the relationship between n and mg. Pitfall Prevention 5.7 Newton’s Third Law Remember that Newton’s third-law action and reaction forces act on different objects. For example, in Figure 5.6, S S S n 5 F tm 5 2mS g 5 2 F Em. The S S forces n and mg are equal in magnitude and opposite in direction, but they do not represent an action–reaction pair because both forces act on the same object, the monitor. Pitfall Prevention 5.8 Free-Body Diagrams The most important step in solving a problem using Newton’s laws is to draw a proper sketch, the free-body diagram. Be sure to draw only those forces that act on the object you are isolating. Be sure to draw all forces acting on the object, including any field forces, such as the gravitational force. Figure 5.6 (a) When a computer monitor is at rest on a table, the forces acting on the monitor are the normal force S n and S S the gravitational force F g . The reaction to S n is the force F mt S exerted by the monitor on the table. The reaction to F g is the S force F mE exerted by the monitor on the Earth. (b) A force diagram shows the forces on the monitor. (c) A free-body diagram shows the monitor as a black dot with the forces acting on it. 120 Chapter 5 The Laws of Motion Q uick Quiz 5.5 (i) If a fly collides with the windshield of a fast-moving bus, which experiences an impact force with a larger magnitude? (a) The fly. (b) The bus. (c) The same force is experienced by both. (ii) Which experiences the greater acceleration? (a) The fly. (b) The bus. (c) The same acceleration is experienced by both. Conceptual Example 5.3    You Push Me and I’ll Push You A large man and a small boy stand facing each other on frictionless ice. They put their hands together and push against each other so that they move apart. om (A) Who moves away with the higher speed? .c Solution ee b ly This situation is similar to what we saw in Quick Quiz 5.5. According to Newton’s third law, the force exerted by the man on the boy and the force exerted by the boy on the man are a third-law pair of forces, so they must be equal in magnitude. (A bathroom scale placed between their hands would read the same, regardless of which way it faced.) Therefore, the boy, having the smaller mass, experiences the greater acceleration. Both individuals accelerate for the same amount of time, but the greater acceleration of the boy over this time interval results in his moving away from the interaction with the higher speed. s. w (B) Who moves farther while their hands are in contact? Solution ys ic Because the boy has the greater acceleration and therefore the greater average velocity, he moves farther than the man during the time interval during which their hands are in contact. ar ph 5.7 Analysis Models Using Newton’s Second Law w w w .a sw In this section, we discuss two analysis models for solving problems in which objects are either in equilibrium 1 S a 5 0 2 or accelerating under the action of constant external forces. Remember that when Newton’s laws are applied to an object, we are interested only in external forces that act on the object. If the objects are modeled as particles, we need not worry about rotational motion. For now, we also neglect the effects of friction in those problems involving motion, which is equivalent to stating that the surfaces are frictionless. (The friction force is discussed in Section 5.8.) We usually neglect the mass of any ropes, strings, or cables involved. In this approximation, the magnitude of the force exerted by any element of the rope on the adjacent element is the same for all elements along the rope. In problem statements, the synonymous terms light and of negligible mass are used to indicate that a mass is to be ignored when you work the problems. When a rope attached to an object is pulling on the object, the rope exerts a force on the object in a direction away from the object, parallel to the rope. The magnitude T of that force is called the tension in the rope. Because it is the magnitude of a vector quantity, tension is a scalar quantity. Analysis Model: The Particle in Equilibrium If the acceleration of an object modeled as a particle is zero, the object is treated with the particle in equilibrium model. In this model, the net force on the object is zero: (5.8) a F 50 Consider a lamp suspended from a light chain fastened to the ceiling as in Figure 5.7a. The force diagram for the lamp (Fig. 5.7b) shows that the forces acting on the S 121 5.7 Analysis Models Using Newton’s Second Law S S lamp are the downward gravitational force F g and the upward force T exerted by the chain. Because there are no forces in the x direction, o Fx 5 0 provides no helpful information. The condition o F y 5 0 gives S S T o Fy 5 T 2 Fg 5 0 or T 5 Fg S Again, notice that T and F g are not an action–reaction pair because they act on S the same object, the lamp. The reaction force to T is a downward force exerted by the lamp on the chain. Example 5.4 (page 122) shows an application of the particle in equilibrium model. S Fg b a Figure 5.7 (a) A lamp suspended from a ceiling by a chain of negligible mass. (b) The forces acting on the lamp are the graviS S tational force F g and the force T exerted by the chain. om Analysis Model: The Particle Under a Net Force If an object experiences an acceleration, its motion can be analyzed with the particle under a net force model. The appropriate equation for this model is Newton’s second law, Equation 5.2: .c S a F 5 ma S ly (5.2) a ic s. w ee b Consider a crate being pulled to the right on a frictionless, horizontal floor as in Figure 5.8a. Of course, the floor directly under the boy must have friction; otherwise, his feet would simply slip when he tries to pull on the crate! Suppose you wish to find the acceleration of the crate and the force the floor exerts on it. The forces acting on the crate are illustrated in the free-body diagram in Figure 5.8b. Notice S that the horizontal force T being applied to the crate acts through the rope. The S S T , magnitude of T is equal to the tension in the rope. In addition to the force the S free-body diagram for the crate includes the gravitational force F g and the normal force S n exerted by the floor on the crate. We can now apply Newton’s second S law in component form to the crate. The only force acting in the x direction is T. Applying o Fx 5 max to the horizontal motion gives T a Fx 5 T 5 ma x or a x 5 m No acceleration occurs in the y direction because the crate moves only horizontally. Therefore, we use the particle in equilibrium model in the y direction. Applying the y component of Equation 5.8 yields S n ar ph ys y sw o F y 5 n 2 Fg 5 0 or n 5 Fg x S Fg b Figure 5.8 (a) A crate being pulled to the right on a frictionless floor. (b) The free-body diagram representing the external forces acting on the crate. w w .a That is, the normal force has the same magnitude as the gravitational force but acts in theSopposite direction. If T is a constant force, the acceleration ax 5 T/m also is constant. Hence, the crate is also modeled as a particle under constant acceleration in the x direction, and the equations of kinematics from Chapter 2 can be used to obtain the crate’s position x and velocity vx as functions of time. Notice from this discussion two concepts that will be important in future problem solving: (1) In a given problem, it is possible to have different analysis models applied in different directions. The crate in Figure 5.8 is a particle in equilibrium in the vertical direction and a particle under a net force in the horizontal direction. (2) It is possible to describe an object by multiple analysis models. The crate is a particle under a net force in the horizontal direction and is also a particle under constant acceleration in the same direction. In the situation just described, the magnitude of the normal force S n is equal S to the magnitude of F g , but that is not always the case, as noted in Pitfall Prevention 5.6. For example,Ssuppose a book is lying on a table and you push down on the book with a force F as in Figure 5.9. Because the book is at rest and therefore not accelerating, o F y 5 0, which gives n 2 Fg 2 F 5 0, or n 5 Fg 1 F 5 mg 1 F. In this situation, the normal force is greater than the gravitational force. Other examples in which n ? Fg are presented later. S T w S F Physics S Fg Figure 5.9 S n S When a force F pushes vertically downward on another object, the normal force S n on the object is greater than the gravitational force: n 5 Fg 1 F. 122 Chapter 5 The Laws of Motion Several examples below demonstrate the use of the particle under a net force model. Analysis Model   Particle in Equilibrium Examples Imagine an object that can be modeled as a particle. If it has several forces acting on it so that the forces all cancel, giving a net force of zero, the object will have an acceleration of zero. This condition is mathematically described as a F 50 S • a chandelier hanging over a dining room table • an object moving at terminal speed through a viscous medium (Chapter 6) • a steel beam in the frame of a building (Chapter 12) • a boat floating on a body of water (Chapter 14) om (5.8) S a0 m S ly .c F  0 ee b Analysis Model   Particle Under a Net Force Examples Imagine an object that can be modeled as a particle. If it has one or more forces acting on it so that there is a net force on the object, it will accelerate in the direction of the net force. The relationship between the net force and the acceleration is s. w S a F 5ma S (5.2) S a ic m • a crate pushed across a factory floor • a falling object acted upon by a gravitational force • a piston in an automobile engine pushed by hot gases (Chapter 22) • a charged particle in an electric field (Chapter 23) S ys F AM ar ph Example 5.4    A Traffic Light at Rest A traffic light weighing 122 N hangs from a cable tied to two other cables fastened to a support as in Figure 5.10a. The upper cables make angles of u1 5 37.0° and u2 5 53.0° with the horizontal. These upper cables are not as strong as the vertical cable and will break if the tension in them exceeds 100 N. Does the traffic light remain hanging in this situation, or will one of the cables break? S T3 sw u1 u2 T1 u1 w .a T3 S w Categorize If nothing is moving, no part of the system is accelerating. We can now model the light as a particle in equilibrium on which the net force is zero. Similarly, the net force on the knot (Fig. 5.10c) is zero, so it is also modeled as a particle in equilibrium. b a u2 x S Fg Conceptualize Inspect the drawing in Figure 5.10a. Let us assume the cables do not break and nothing is moving. T2 T2 T1 w Solution S y S T3 c Figure 5.10 (Example 5.4) (a) A traffic light suspended by cables. (b) The forces acting on the traffic light. (c) The free-body diagram for the knot where the three cables are joined. Analyze We construct a diagram of the forces acting on the traffic light, shown in Figure 5.10b, and a free-body diagram for the knot that holds the three cables together, shown in Figure 5.10c. This knot is a convenient object to choose because all the forces of interest act along lines passing through the knot. From the particle in equilibrium model, apply Equation 5.8 for the traffic light in the y direction: oF y 5 0 S T3 2 Fg 5 0 T3 5 Fg 123 5.7 Analysis Models Using Newton’s Second Law ▸ 5.4 c o n t i n u e d Choose the coordinate axes as shown in Figure 5.10c and resolve the forces acting on the knot into their components: Force S T1 T2 S T3 S (2) S oF oF y Component 2T1 cos u1 T2 cos u2 0 T1 sin u1 T2 sin u2 2Fg x 5 2T1 cos u1 1 T2 cos u2 5 0 y 5 T1 sin u1 1 T2 sin u2 1 (2Fg) 5 0 om (1) Apply the particle in equilibrium model to the knot: x Component S .c Equation (1) shows that the horizontal components of T1 and T2 must be equal in magnitude, and Equation (2) S S S shows that the sum of the vertical components of T1 and T2 must balance the downward force T3, which is equal in ­magnitude to the weight of the light. cos u 1 b cos u 2 (3) T2 5 T1 a Substitute this value for T2 into Equation (2): T1 sin u1 1 T1 a Solve for T1: T1 5 ly Solve Equation (1) for T2 in terms of T1: ee b cos u 1 b(sin u2 ) 2Fg 5 0 cos u 2 Fg s. w sin u 1 1 cos u 1 tan u 2 T1 5 122 N 5 73.4 N sin 37.08 1 cos 37.08 tan 53.08 ic Substitute numerical values: ys Using Equation (3), solve for T2: T2 5 1 73.4 N 2 a cos 37.08 b 5 97.4 N cos 53.08 ar ph Both values are less than 100 N ( just barely for T2), so the cables will not break . Finalize Let us finalize this problem by imagining a change in the system, as in the following What If? W h at I f ? Suppose the two angles in Figure 5.10a are equal. What would be the relationship between T1 and T2? Answer We can argue from the symmetry of the problem that the two tensions T1 and T2 would be equal to each other. .a sw Mathematically, if the equal angles are called u, Equation (3) becomes T2 5 T1 a cos u b 5 T1 cos u w w w which also tells us that the tensions are equal. Without knowing the specific value of u, we cannot find the values of T1 and T2. The tensions will be equal to each other, however, regardless of the value of u. Conceptual Example 5.5    Forces Between Cars in a Train Train cars are connected by couplers, which are under tension as the locomotive pulls the train. Imagine you are on a train speeding up with a constant acceleration. As you move through the train from the locomotive to the last car, measuring the tension in each set of couplers, does the tension increase, decrease, or stay the same? When the engineer applies the brakes, the couplers are under compression. How does this compression force vary from the locomotive to the last car? (Assume only the brakes on the wheels of the engine are applied.) Solution While the train is speeding up, tension decreases from the front of the train to the back. The coupler between the locomotive and the first car must apply enough force to accelerate the rest of the cars. As you move back along the continued 124 Chapter 5 The Laws of Motion ▸ 5.5 c o n t i n u e d train, each coupler is accelerating less mass behind it. The last coupler has to accelerate only the last car, and so it is under the least tension. When the brakes are applied, the force again decreases from front to back. The coupler connecting the locomotive to the first car must apply a large force to slow down the rest of the cars, but the final coupler must apply a force large enough to slow down only the last car.    The Runaway Car AM y om Example 5.6 A car of mass m is on an icy driveway inclined at an angle u as in Figure 5.11a. S n ly .c (A) Find the acceleration of the car, assuming the driveway is frictionless. ee b Solution u Categorize We categorize the car as a particle a mg cos u u x S S Fg = m g b Figure 5.11 (Example 5.6) (a) A car on a frictionless incline. (b) The freebody diagram for the car. The black dot represents the position of the center of mass of the car. We will learn about center of mass in Chapter 9. ic alize the situation. From everyday experience, we know that a car on an icy incline will accelerate down the incline. (The same thing happens to a car on a hill with its brakes not set.) x s. w Conceptualize Use Figure 5.11a to conceptu- mg sin u ys under a net force because it accelerates. Furthermore, this example belongs to a very common category of problems in which an object moves under the influence of gravity on an inclined plane. ar ph Analyze Figure 5.11b shows the free-body diagram for the car. The only forces acting on the car are the normal force S n exerted by the inclined plane, which acts perpendicular to the plane, and the gravitational force F g 5 mS g , which acts vertically downward. For problems involving inclined planes, it is convenient to choose the coordinate axes with x along the incline and y perpendicular to it as in Figure 5.11b. With these axes, we represent the gravitational force by a component of magnitude mg sin u along the positive x axis and one of magnitude mg cos u along the negative y axis. Our choice of axes results in the car being modeled as a particle under a net force in the x direction and a particle in equilibrium in the y direction. sw S w .a Apply these models to the car: (2) oF oF x 5 mg sin u 5 max y 5 n 2 mg cos u 5 0 (3) ax 5 g sin u w w Solve Equation (1) for ax: (1) Finalize Note that the acceleration component ax is independent of the mass of the car! It depends only on the angle of inclination and on g. S From Equation (2), we conclude that the component of F g perpendicular to the incline is balanced by the normal force; that is, n 5 mg cos u. This situation is a case in which the normal force is not equal in magnitude to the weight of the object (as discussed in Pitfall Prevention 5.6 on page 119). It is possible, although inconvenient, to solve the problem with “standard” horizontal and vertical axes. You may want to try it, just for practice. (B) Suppose the car is released from rest at the top of the incline and the distance from the car’s front bumper to the bottom of the incline is d. How long does it take the front bumper to reach the bottom of the hill, and what is the car’s speed as it arrives there? 125 5.7 Analysis Models Using Newton’s Second Law ▸ 5.6 c o n t i n u e d Solution Conceptualize Imagine the car is sliding down the hill and you use a stopwatch to measure the entire time interval until it reaches the bottom. Categorize This part of the problem belongs to kinematics rather than to dynamics, and Equation (3) shows that the acceleration ax is constant. Therefore, you should categorize the car in this part of the problem as a particle under constant acceleration. Solve for t: 2d 2d 5 Å ax Å g sin u ly (4) t 5 om as xi 5 0 and its final position as xf 5 d, and recognizing that vxi 5 0, choose Equation 2.16 from the particle under constant acceleration model, x f 5 x i 1 v xi t 1 12a x t 2 : d 5 12a x t 2 .c Analyze Defining the initial position of the front bumper vxf 2 5 2axd ee b Use Equation 2.17, with vxi 5 0, to find the final velocity of the car: (5) v xf 5 "2a x d 5 "2gd sin u Finalize We see from Equations (4) and (5) that the time t at which the car reaches the bottom and its final speed vxf What previously solved problem does this situation become if u 5 90°? ys W h at I f ? ic s. w are independent of the car’s mass, as was its acceleration. Notice that we have combined techniques from Chapter 2 with new techniques from this chapter in this example. As we learn more techniques in later chapters, this process of combining analysis models and information from several parts of the book will occur more often. In these cases, use the General Problem-Solving Strategy to help you identify what analysis models you will need. ar ph Answer Imagine u going to 90° in Figure 5.11. The inclined plane becomes vertical, and the car is an object in free fall! Equation (3) becomes ax 5 g sin u 5 g sin 90° 5 g .a sw which is indeed the free-fall acceleration. (We find ax 5 g rather than ax 5 2g because we have chosen positive x to be downward in Fig. 5.11.) Notice also that the condition n 5 mg cos u gives us n 5 mg cos 90° 5 0. That is consistent with the car falling downward next to the vertical plane, in which case there is no contact force between the car and the plane.    One Block Pushes Another AM S F w Example 5.7 w w Two blocks of masses m1 and m 2 , with m1 . m 2, are placed in contact with each other on a frictionless, horizontal surface as in Figure 5.12a. A S constant horizontal force F is applied to m1 as shown. (A) Find the magnitude of the acceleration of the system. Solution Categorize We categorize this problem as one involving a particle under a net force because a force is applied to a system of blocks and we are looking for the acceleration of the system. m2 a S n1 y S n2 S S S P21 P12 F x Conceptualize Conceptualize the situation by using Figure 5.12a and realize that both blocks must experience the same acceleration because they are in contact with each other and remain in contact throughout the motion. m1 S m 1g b S m 2g c Figure 5.12 (Example 5.7) (a) A force is applied to a block of mass m1, which pushes on a second block of mass m 2. (b) The forces acting on m1. (c) The forces acting on m 2. continued 126 Chapter 5 The Laws of Motion ▸ 5.7 c o n t i n u e d Analyze First model the combination of two blocks as a single particle under a net force. Apply Newton’s second law to the combination in the x direction to find the acceleration: oF x 5 F 5 (m1 1 m 2)ax F m1 1 m2 (1) ax 5 Finalize The acceleration given by Equation (1) is the same as that of a single object of mass m1 1 m 2 and subject to the same force. om (B) Determine the magnitude of the contact force between the two blocks. Solution Conceptualize The contact force is internal to the system of two blocks. Therefore, we cannot find this force by model- .c ing the whole system (the two blocks) as a single particle. ly Categorize Now consider each of the two blocks individually by categorizing each as a particle under a net force. S ee b Analyze We construct a diagram of forces acting on the object for each block as shown in Figures 5.12b and 5.12c, where the contact force is denoted by P. From Figure 5.12c, we see that the only horizontal force acting on m 2 is the S contact force P12 (the force exerted by m1 on m 2), which is directed to the right. (2) x 5 P 12 5 m 2ax (3) P 12 5 m 2ax 5 a m2 m1 1 m2 ic Substitute the value of the acceleration ax given by Equation (1) into Equation (2): oF s. w Apply Newton’s second law to m 2: bF sw Apply Newton’s second law to m1: ar ph ys Finalize This result shows that the contact force P 12 is less than the applied force F. The force required to accelerate block 2 alone must be less than the force required to produce the same acceleration for the two-block system. To finalize further, let us check this expression for P 12 by considering the forces acting on m1, shown in Figure 5.12b. S S The horizontal forces acting on m1 are the applied force F to the right and the contact force P 21 to the left (the S S force exerted by m 2 on m1). From Newton’s third law, P 21 is the reaction force to P 12, so P 21 5 P 12. .a Solve for P 12 and substitute the value of ax from Equation (1): w This result agrees with Equation (3), as it must. (4) oF x 5 F 2 P 21 5 F 2 P 12 5 m1ax P12 5 F 2 m 1a x 5 F 2 m 1 a m2 F b5a bF m1 1 m2 m1 1 m2 S Imagine that the force F in Figure 5.12 is applied toward the left on the right-hand block of mass m 2. S Is the magnitude of the force P 12 the same as it was when the force was applied toward the right on m1? w W h at I f ? Answer When the force is applied toward the left on m 2, the contact force must accelerate m1. In the original situation, w S the contact force accelerates m 2. Because m1 . m 2, more force is required, so the magnitude of P 12 is greater than in S the original situation. To see this mathematically, modify Equation (4) appropriately and solve for P 12. Example 5.8    Weighing a Fish in an Elevator AM A person weighs a fish of mass m on a spring scale attached to the ceiling of an elevator as illustrated in Figure 5.13. (A) Show that if the elevator accelerates either upward or downward, the spring scale gives a reading that is different from the weight of the fish. 127 5.7 Analysis Models Using Newton’s Second Law Solution Conceptualize The reading on the scale is related to the When the elevator accelerates downward, the spring scale reads a value less than the weight of the fish. When the elevator accelerates upward, the spring scale reads a value greater than the weight of the fish. extension of the spring in the scale, which is related to the force on the end of the spring as in Figure 5.2. Imagine that the fish is hanging on a string attached to the end of the spring. In this case, the magnitude of the force exerted on the spring is equal to the tension T in the string. ThereS fore, we are looking for T. The force T pulls down on the string and pulls up on the fish. S a S a S 8 7 6 54 T 2 3 8 7 9 0 1 6 54 2 3 S T ee b ly .c Categorize We can categorize this problem by identifying the fish as a particle in equilibrium if the elevator is not accelerating or as a particle under a net force if the elevator is accelerating. 9 0 1 om ▸ 5.8 c o n t i n u e d ar ph Apply Newton’s second law to the fish: ys ic s. w Analyze Inspect the diagrams of the forces acting on the S S mg mg fish in Figure 5.13 and notice that the external forces acting S a b on the fish are the downward gravitational force F g 5 mS g S and the force T exerted by the string. If the elevator is Figure 5.13 (Example 5.8) A fish is weighed on a spring scale in either at rest or moving at constant velocity, the fish is a par- an accelerating elevator car. ticle in equilibrium, so o F y 5 T 2 Fg 5 0 or T 5 Fg 5 mg. (Remember that the scalar mg is the weight of the fish.) Now suppose the elevator is moving with an acceleration S a relative to an observer standing outside the elevator in an inertial frame. The fish is now a particle under a net force. Solve for T : oF y 5 T 2 mg 5 may (1) T 5 ma y 1 mg 5 mg a ay g 1 1b 5 Fg a ay g 1 1b sw where we have chosen upward as the positive y direction. We conclude from Equation (1) that the scale reading T is greater than the fish’s weight mg if S a is upward, so ay is positive (Fig. 5.13a), and that the reading is less than mg if S a is downward, so ay is negative (Fig. 5.13b). w Solution .a (B) Evaluate the scale readings for a 40.0-N fish if the elevator moves with an acceleration ay 5 62.00 m/s2. w Evaluate the scale reading from Equation (1) if S a is upward: w Evaluate the scale reading from Equation (1) if S a is downward: T 5 1 40.0 N 2 a T 5 1 40.0 N 2 a 2.00 m/s2 1 1b 5 48.2 N 9.80 m/s2 22.00 m/s2 1 1b 5 31.8 N 9.80 m/s2 Finalize Take this advice: if you buy a fish in an elevator, make sure the fish is weighed while the elevator is either at rest or accelerating downward! Furthermore, notice that from the information given here, one cannot determine the direction of the velocity of the elevator. W h at I f ? Suppose the elevator cable breaks and the elevator and its contents are in free fall. What happens to the reading on the scale? Answer If the elevator falls freely, the fish’s acceleration is ay 5 2g. We see from Equation (1) that the scale reading T is zero in this case; that is, the fish appears to be weightless. 128 Chapter 5 Example 5.9 The Laws of Motion    The Atwood Machine AM When two objects of unequal mass are hung vertically over a frictionless pulley of negligible mass as in Figure 5.14a, the arrangement is called an Atwood machine. The device is sometimes used in the laboratory to determine the value of g. Determine the magnitude of the acceleration of the two objects and the tension in the lightweight string. S T + m1 Solution Conceptualize Imagine the situation pictured in Figure 5.14a in action: as one object moves upward, the other object moves downward. Because the objects are connected by an inextensible string, their accelerations must be of equal magnitude. m2 .c a om + S m 2g b Figure 5.14 (Example 5.9) The Atwood machine. (a) Two objects connected by a massless inextensible string over a frictionless pulley. (b) The free-body diagrams for the two objects. ee b Analyze The free-body diagrams for the two objects are shown in Figure 5.14b. S m 1g ly Categorize The objects in the Atwood machine are subject to the gravitational force as well as to the forces exerted by the strings connected to them. Therefore, we can categorize this problem as one involving two particles under a net force. m1 m2 S T S ar ph ys ic s. w Two forces act on each object: the upward force T exerted by the string and the downward gravitational force. In problems such as this one in which the pulley is modeled as massless and frictionless, the tension in the string on both sides of the pulley is the same. If the pulley has mass or is subject to friction, the tensions on either side are not the same and the situation requires techniques we will learn in Chapter 10. We must be very careful with signs in problems such as this one. In Figure 5.14a, notice that if object 1 accelerates upward, object 2 accelerates downward. Therefore, for consistency with signs, if we define the upward direction as positive for object 1, we must define the downward direction as positive for object 2. With this sign convention, both objects accelerate in the same direction as defined by the choice of sign. Furthermore, according to this sign convention, the y component of the net force exerted on object 1 is T 2 m1g, and the y component of the net force exerted on object 2 is m 2g 2 T. (1) oF y 5 T 2 m1g 5 m1ay Apply Newton’s second law to object 2: (2) oF y 5 m 2g 2 T 5 m 2ay sw From the particle under a net force model, apply Newton’s second law to object 1: 2 m1g 1 m 2g 5 m1ay 1 m 2ay Solve for the acceleration: (3) ay 5 a w .a Add Equation (2) to Equation (1), noticing that T cancels: w Substitute Equation (3) into Equation (1) to find T: m2 2 m1 m1 1 m2 bg (4) T 5 m1(g 1 ay) 5 a 2m 1m 2 m1 1 m2 bg w Finalize The acceleration given by Equation (3) can be interpreted as the ratio of the magnitude of the unbalanced force on the system (m 2 2 m1)g to the total mass of the system (m1 1 m 2), as expected from Newton’s second law. Notice that the sign of the acceleration depends on the relative masses of the two objects. W h at I f ? Describe the motion of the system if the objects have equal masses, that is, m1 5 m 2. Answer If we have the same mass on both sides, the system is balanced and should not accelerate. Mathematically, we see that if m1 5 m 2, Equation (3) gives us ay 5 0. W h at I f ? What if one of the masses is much larger than the other: m1 .. m 2? Answer In the case in which one mass is infinitely larger than the other, we can ignore the effect of the smaller mass. Therefore, the larger mass should simply fall as if the smaller mass were not there. We see that if m1 .. m 2, Equation (3) gives us ay 5 2g. 129 5.7 Analysis Models Using Newton’s Second Law AM Example 5.10    Acceleration of Two Objects Connected by a Cord A ball of mass m1 and a block of mass m 2 are attached by a lightweight cord that passes over a frictionless pulley of negligible mass as in Figure 5.15a. The block lies on a frictionless incline of angle u. Find the magnitude of the acceleration of the two objects and the tension in the cord. y S a S T m2 S x S m 1g m1 a u Solution Conceptualize Imagine the objects in Figure 5.15 in motion. If m 2 moves a b down the incline, then m1 moves upward. Because the objects are connected by a cord (which we assume does not stretch), their accelerations have the same magnitude. Notice the normal coordinate axes in Figure 5.15b for the ball and the “tilted” axes for the block in Figure 5.15c. om n S .c T looking for an acceleration, so we categorize the objects as particles under a net force. For the block, this model is only valid for the x9 direction. In the y9 direction, we apply the particle in equilibrium model because the block does not accelerate in that direction. ly Categorize We can identify forces on each of the two objects and we are y S Analyze Consider the free-body diagrams shown in Figures 5.15b and Figure 5.15 m 2g cos u m2g sin u u S m2g ee b s. w 5.15c. (1) oF y c (Example 5.10) (a) Two objects connected by a lightweight cord strung over a frictionless pulley. (b) The free-body diagram for the ball. (c) The free-body diagram for the block. (The incline is frictionless.) 5 T 2 m1g 5 m1ay 5 m1a ys ic Apply Newton’s second law in the y direction to the ball, choosing the upward direction as positive: x ar ph For the ball to accelerate upward, it is necessary that T . m1g. In Equation (1), we replaced ay with a because the acceleration has only a y component. For the block, we have chosen the x9 axis along the incline as in Figure 5.15c. For consistency with our choice for the ball, we choose the positive x9 direction to be down the incline. sw Apply the particle under a net force model to the block in the x9 direction and the particle in equilibrium model in the y9 direction: (2) (3) oF oF x9 5 m 2g sin u 2 T 5 m 2ax9 5 m 2a y9 5 n 2 m 2g cos u 5 0 .a In Equation (2), we replaced ax9 with a because the two objects have accelerations of equal magnitude a. w Solve Equation (1) for T : m 2g sin u 2 m1(g 1 a) 5 m 2a w w Substitute this expression for T into Equation (2): (4) T 5 m1(g 1 a) Solve for a: (5) a 5 a Substitute this expression for a into Equation (4) to find T : (6) T 5 a m 2 sin u 2 m 1 m1 1 m2 bg m 1m 2 1 sin u 1 1 2 m1 1 m2 bg Finalize The block accelerates down the incline only if m 2 sin u . m1. If m1 . m 2 sin u, the acceleration is up the incline for the block and downward for the ball. Also notice that the result for the acceleration, Equation (5), can be interpreted as the magnitude of the net external force acting on the ball–block system divided by the total mass of the system; this result is consistent with Newton’s second law. W h at I f ? What happens in this situation if u 5 90°? continued 130 Chapter 5 The Laws of Motion ▸ 5.10 c o n t i n u e d Answer If u 5 90°, the inclined plane becomes vertical and there is no interaction between its surface and m 2. Therefore, this problem becomes the Atwood machine of Example 5.9. Letting u S 90° in Equations (5) and (6) causes them to reduce to Equations (3) and (4) of Example 5.9! W h at I f ? What if m1 5 0? om Answer If m1 5 0, then m 2 is simply sliding down an inclined plane without interacting with m1 through the string. Therefore, this problem becomes the sliding car problem in Example 5.6. Letting m1 S 0 in Equation (5) causes it to reduce to Equation (3) of Example 5.6! .c 5.8 Forces of Friction ys ic s. w ee b ly When an object is in motion either on a surface or in a viscous medium such as air or water, there is resistance to the motion because the object interacts with its surroundings. We call such resistance a force of friction. Forces of friction are very important in our everyday lives. They allow us to walk or run and are necessary for the motion of wheeled vehicles. Imagine that you are working in your garden and have filled a trash can with yard clippings. You then try to drag the trash can across the surface of your concrete patio as in Figure 5.16a. This surface is real, not an idealized, frictionless surface. S If we apply an external horizontal force F to the trash can, acting to the right, S the trash canSremains stationary when F is small. The force on the trash can that counteracts F and keeps it from moving acts toward the left and is called the sw ar ph For small applied forces, the magnitude of the force of static friction equals the magnitude of the applied force. S S .a n n S w fs S fk F S F S mg a Motion S S w w When the magnitude of the applied force exceeds the magnitude of the maximum force of static friction, the trash can breaks free and accelerates to the right. mg b S |f | fs,max Figure 5.16 (a) and (b) When pulling on a trash can, the direcS tion of the force of friction f between the can and a rough surface is opposite the direction of S the applied force F. (c) A graph of friction force versus applied force. Notice that fs,max . f k . fs  F fk  mkn O c F Static region Kinetic region 5.8 Forces of Friction S WW Force of static friction om force of static friction fs. As long as the trash can is not moving, fs 5 F. Therefore, S S S S if F is increased, fs also increases. Likewise, if F decreases, fs also decreases. Experiments show that the friction force arises from the nature of the two surfaces: because of their roughness, contact is made only at a few locations where peaks of the material touch. At these locations, the friction force arises in part because one peak physically blocks the motion of a peak from the opposing surface and in part from chemical bonding (“spot welds”) of opposing peaks as they come into contact. Although the details of friction are quite complex at the atomic level, this force ultimately involves an electrical interaction between atoms or molecules. S If we increase the magnitude of F as in Figure 5.16b, the trash can eventually slips. When the trash can is on the verge of slipping, fs has its maximum value fs,max as shown in Figure 5.16c. When F exceeds fs,max, the trash can moves and accelerates to the right. We call the friction force for an object in motion the force of kinetic S friction fk . When the trash can is in motion, the force of kinetic friction on the can is less than fs ,max (Fig. 5.16c). The net force F 2 f k in the x direction produces an acceleration to the right, according to Newton’s second law. If F 5 f k , the acceleration is zeroSand the trash can moves to the right with constant speed. If the applied S force F is removed from the moving can, the friction force fk acting to the left provides an acceleration of the trash can in the 2x direction and eventually brings it to rest, again consistent with Newton’s second law. Experimentally, we find that, to a good approximation, both fs,max and f k are proportional to the magnitude of the normal force exerted on an object by the surface. The following descriptions of the force of friction are based on experimental observations and serve as the simplification model we shall use for forces of friction in problem solving: 131 ic s. w ee b ly .c WW Force of kinetic friction ys • The magnitude of the force of static friction between any two surfaces in contact can have the values (5.9) ar ph fs # msn .a sw where the dimensionless constant ms is called the coefficient of static friction and n is the magnitude of the normal force exerted by one surface on the other. The equality in Equation 5.9 holds when the surfaces are on the verge of slipping, that is, when fs 5 fs,max 5 msn. This situation is called impending motion. The inequality holds when the surfaces are not on the verge of slipping. • The magnitude of the force of kinetic friction acting between two surfaces is f k 5 mkn (5.10) w w w where mk is the coefficient of kinetic friction. Although the coefficient of kinetic friction can vary with speed, we shall usually neglect any such variations in this text. • The values of mk and ms depend on the nature of the surfaces, but mk is generally less than ms . Typical values range from around 0.03 to 1.0. Table 5.1 (page 132) lists some reported values. • The direction of the friction force on an object is parallel to the surface with which the object is in contact and opposite to the actual motion (kinetic friction) or the impending motion (static friction) of the object relative to the surface. • The coefficients of friction are nearly independent of the area of contact between the surfaces. We might expect that placing an object on the side having the most area might increase the friction force. Although this method provides more points in contact, the weight of the object is spread out over a larger area and the individual points are not pressed together as tightly. Because these effects approximately compensate for each other, the friction force is independent of the area. Pitfall Prevention 5.9 The Equal Sign Is Used in Limited Situations In Equation 5.9, the equal sign is used only in the case in which the surfaces are just about to break free and begin sliding. Do not fall into the common trap of using fs 5 msn in any static situation. Pitfall Prevention 5.10 Friction Equations Equations 5.9 and 5.10 are not vector equations. They are relationships between the magnitudes of the vectors representing the friction and normal forces. Because the friction and normal forces are perpendicular to each other, the vectors cannot be related by a multiplicative constant. Pitfall Prevention 5.11 The Direction of the Friction Force Sometimes, an incorrect statement about the friction force between an object and a surface is made—“the friction force on an object is opposite to its motion or impending motion”—rather than the correct phrasing, “the friction force on an object is opposite to its motion or impending motion relative to the surface.” 132 Chapter 5 The Laws of Motion Table 5.1 Coefficients of Friction Rubber on concrete Steel on steel Aluminum on steel Glass on glass Copper on steel Wood on wood Waxed wood on wet snow Waxed wood on dry snow Metal on metal (lubricated) Teflon on Teflon Ice on ice Synovial joints in humans 30 S F Ms Mk 1.0  0.74 0.61 0.94 0.53 0.25–0.5 0.14 — 0.15 0.04 0.1  0.01 0.8 0.57 0.47 0.4 0.36 0.2 0.1 0.04 0.06 0.04 0.03 0.003 Note: All values are approximate. In some cases, the coefficient of friction can exceed 1.0. ly .c a om Table 5.1 Q uick Quiz 5.6 You press your physics textbook flat against a vertical wall with your hand. What is the direction of the friction force exerted by the wall on the book? (a) downward (b) upward (c) out from the wall (d) into the wall S ee b F 30 Q uick Quiz 5.7 You are playing with your daughter in the snow. She sits on a sled and asks you to slide her across a flat, horizontal field. You have a choice of (a) pushing her from behind by applying a force downward on her shoulders at 30° below the horizontal (Fig. 5.17a) or (b) attaching a rope to the front of the sled and pulling with a force at 30° above the horizontal (Fig. 5.17b). Which would be easier for you and why? s. w b ys ic Figure 5.17 (Quick Quiz 5.7) A father slides his daughter on a sled either by (a) pushing down on her shoulders or (b) pulling up on a rope. AM ar ph Example 5.11    Experimental Determination of Ms and Mk y sw The following is a simple method of measuring coefficients of friction. Suppose a block is placed on a rough surface inclined relative to the horizontal as shown in Figure 5.18. The incline angle is increased until the block starts to move. Show that you can obtain ms by measuring the critical angle uc at which this slipping just occurs. n fs mg sin u mg cos u .a Solution S S u u w w w Conceptualize Consider Figure 5.18 and imagine that the block tends to slide down the incline due to the gravitational force. To simulate the situation, place a coin on this book’s cover and tilt the book until the coin begins to slide. Notice how this example differs from Example 5.6. When there is no friction on an incline, any angle of the incline will cause a stationary object to begin moving. When there is friction, however, there is no movement of the object for angles less than the critical angle. Categorize The block is subject to various forces. Because we are raising the plane to the angle at which the block is just ready to begin to move but is not moving, we categorize the block as a particle in equilibrium. S mg x Figure 5.18 (Example 5.11) The external forces exerted on a block lying on a rough incline are the gravS itational force mg , the normal force S S n , and the force of friction f s . For convenience, the gravitational force is resolved into a component mg sin u along the incline and a component mg cos u perpendicular to the incline. Analyze The diagram in Figure 5.18 shows the forces on the block: the gravitational force mS g , the normal force S n , and S the force of static friction fs . We choose x to be parallel to the plane and y perpendicular to it. From the particle in equilibrium model, apply Equation 5.8 to the block in both the x and y directions: (1) (2) oF oF x 5 mg sin u 2 fs 5 0 y 5 n 2 mg cos u 5 0 133 5.8 Forces of Friction ▸ 5.11 c o n t i n u e d (3) fs 5 mg sin u 5 a Substitute mg 5 n/cos u from Equation (2) into Equation (1): msn 5 n tan uc When the incline angle is increased until the block is on the verge of slipping, the force of static friction has reached its maximum value msn. The angle u in this situation is the critical angle uc . Make these substitutions in Equation (3): n b sin u 5 n tan u cos u ms 5 tan uc We have shown, as requested, that the coefficient of static friction is related only to the critical angle. For example, if the block just slips at uc 5 20.0°, we find that ms 5 tan 20.0° 5 0.364. om Finalize Once the block starts to move at u $ uc , it accelerates down the incline and the force of friction is f k 5 mkn. ee b ly .c If u is reduced to a value less than uc , however, it may be possible to find an angle u9c such that the block moves down the incline with constant speed as a particle in equilibrium again (ax 5 0). In this case, use Equations (1) and (2) with fs replaced by f k to find mk: mk 5 tan u9c , where u9c , uc . AM Example 5.12    The Sliding Hockey Puck A hockey puck on a frozen pond is given an initial speed of 20.0 m/s. If the puck always remains on the ice and slides 115 m before coming to rest, determine the coefficient of kinetic friction between the puck and ice. S s. w S fk ic Solution Motion n ys Conceptualize Imagine that the puck in Figure 5.19 slides to the right. The kinetic ar ph friction force acts to the left and slows the puck, which eventually comes to rest due to that force. Categorize The forces acting on the puck are identified in Figure 5.19, but the S mg Figure 5.19 (Example 5.12) After the puck is given an initial velocity to the right, the only external forces acting on it are the gravitational n , and force mS g , the normal force S S the force of kinetic friction fk. .a sw text of the problem provides kinematic variables. Therefore, we categorize the problem in several ways. First, it involves modeling the puck as a particle under a net force in the horizontal direction: kinetic friction causes the puck to accelerate. There is no acceleration of the puck in the vertical direction, so we use the particle in equilibrium model for that direction. Furthermore, because we model the force of kinetic friction as independent of speed, the acceleration of the puck is constant. So, we can also categorize this problem by modeling the puck as a particle under constant acceleration. w Analyze First, let’s find the acceleration algebraically in terms of the coefficient of kinetic friction, using Newton’s w w second law. Once we know the acceleration of the puck and the distance it travels, the equations of kinematics can be used to find the numerical value of the coefficient of kinetic friction. The diagram in Figure 5.19 shows the forces on the puck. Apply the particle under a net force model in the x direction to the puck: (1) oF Apply the particle in equilibrium model in the y direction to the puck: (2) oF Substitute n 5 mg from Equation (2) and f k 5 mkn into Equation (1): 2 mkn 5 2 mkmg 5 max x y 5 2 f k 5 max 5 n 2 mg 5 0 ax 5 2 mk g The negative sign means the acceleration is to the left in Figure 5.19. Because the velocity of the puck is to the right, the puck is slowing down. The acceleration is independent of the mass of the puck and is constant because we assume mk remains constant. continued 134 Chapter 5 The Laws of Motion ▸ 5.12 c o n t i n u e d Apply the particle under constant acceleration model to the puck, choosing Equation 2.17 from the model, vxf 2 5 vxi2 1 2ax(xf 2 xi ), with xi 5 0 and vxf 5 0: 0 5 vxi2 1 2axxf 5 vxi2 2 2mk gxf v xi 2 2gx f Solve for the coefficient of kinetic friction: mk 5 Substitute the numerical values: mk 5 1 20.0 m/s 2 2 5 0.177 2 1 9.80 m/s2 2 1 115 m 2 om Finalize Notice that mk is dimensionless, as it should be, and that it has a low value, consistent with an object sliding .c on ice. Solution S a u s. w F S T ic a a S ar ph the block. Assuming F is large enough to break the block free from static friction but not large enough to lift the block, the block slides to the right and the ball rises. F sin u n S T S F u S fk m 2 F cos u m1 S S m 2g m1g m1 S x S m2 Conceptualize Imagine what happens as F is applied to S AM y S b c ys A block of mass m 2 on a rough, horizontal surface is connected to a ball of mass m1 by a lightweight cord over a lightweight, frictionless pulley as shown in Figure 5.20a. A force of magnitude F at an angle u with the horizontal is applied to the block as shown, and the block slides to the right. The coefficient of kinetic friction between the block and surface is mk . Determine the magnitude of the acceleration of the two objects. ee b ly Example 5.13    Acceleration of Two Connected Objects When Friction Is Present S Figure 5.20 (Example 5.13) (a) The external force F applied as shown can cause the block to accelerate to the right. (b, c) Diagrams showing the forces on the two objects, assuming the block accelerates to the right and the ball accelerates upward. sw Categorize We can identify forces and we want an acceleration, so we categorize this problem as one involving two particles under a net force, the ball and the block. Because we assume that the block does not rise into the air due to the applied force, we model the block as a particle in equilibrium in the vertical direction. .a Analyze First draw force diagrams for the two objects as shown in Figures 5.20b and 5.20c. Notice that the string S w w exerts a force of magnitude T on both objects. The applied force F has x and y components F cos u and F sin u, respectively. Because the two objects are connected, we can equate the magnitudes of the x component of the acceleration of the block and the y component of the acceleration of the ball and call them both a. Let us assume the motion of the block is to the right. (1) oF Because the block moves only horizontally, apply the particle in equilibrium model to the block in the vertical direction: (2) oF y 5 n 1 F sin u 2 m 2g 5 0 Apply the particle under a net force model to the ball in the vertical direction: (3) oF y 5 T 2 m1g 5 m1ay 5 m1a Solve Equation (2) for n: n 5 m 2g 2 F sin u Substitute n into f k 5 mkn from Equation 5.10: (4) f k 5 mk(m 2g 2 F sin u) w Apply the particle under a net force model to the block in the horizontal direction: x 5 F cos u 2 f k 2 T 5 m 2ax 5 m 2a    Summary 135 ▸ 5.13 c o n t i n u e d Substitute Equation (4) and the value of T from Equation (3) into Equation (1): F cos u 2 mk(m 2g 2 F sin u) 2 m1(a 1 g) 5 m 2a Solve for a: (5) a 5 F 1 cos u 1 mk sin u 2 2 1 m 1 1 mkm 2 2 g m1 1 m2 ee b ly .c om Finalize The acceleration of the block can be either to the right or to the left depending on the sign of the numerator in Equation (5). If the velocity is to the left, we must reverse the sign of f k in Equation (1) because the force of kinetic friction must oppose the motion of the block relative to the surface. In this case, the value of a is the same as in Equation (5), with the two plus signs in the numerator changed to minus signs. S What does Equation (5) reduce to if the force F is removed and the surface becomes frictionless? Call this expression Equation (6). Does this algebraic expression match your intuition about the physical situation in this case? Now go back to Example 5.10 and let angle u go to zero in Equation (5) of that example. How does the resulting equation compare with your Equation (6) here in Example 5.13? Should the algebraic expressions compare in this way based on the physical situations? s. w Summary ic Definitions ar ph ys An inertial frame of reference is a frame in which an object that does not interact with other objects experiences zero acceleration. Any frame moving with constant velocity relative to an inertial frame is also an inertial frame. We define force as that which causes a change in motion of an object. Concepts and Principles .a sw Newton’s first law states that it is possible to find an inertial frame in which an object that does not interact with other objects experiences zero acceleration, or, equivalently, in the absence of an external force, when viewed from an inertial frame, an object at rest remains at rest and an object in uniform motion in a straight line maintains that motion. The gravitational force exerted on an object is equal to the product of its mass (a scalar quantity) and the freefall acceleration: w S w Newton’s second law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. w Newton’s third law states that if two objects interact, the force exerted by object 1 on object 2 is equal in magnitude and opposite in direction to the force exerted by object 2 on object 1. S The maximum force of static friction fs,max between an object and a surface is proportional to the normal force acting on the object. In general, fs # msn, where ms is the coefficient of static friction and n is the magnitude of the normal force. F g 5 mS g (5.5) The weight of an object is the magnitude of the gravitational force acting on the object: Fg 5 mg (5.6) When an object slides over a surface, the S magnitude of the force of kinetic friction fk is given by f k 5 mkn, where mk is the coefficient of kinetic friction. continued 136 Chapter 5 The Laws of Motion Analysis Models for Problem Solving Particle Under a Net Force If a particle of mass m experiences a nonzero net force, its acceleration is related to the net force by Newton’s second law: S a F 5ma S (5.2) a F 50 S S a (5.8) S S a0 m F om m Particle in Equilibrium If a particle maintains a constant velocity (so that S a 5 0), which could include a velocity of zero, the forces on the particle balance and Newton’s second law reduces to S 12 m/s 400 g a ys 12 m/s b 9 m/s ic 2. In Figure OQ5.2, a locomotive has broken through the wall of a train station. During the collision, what can be said about the force exerted by the locomotive on the wall? (a) The force exerted by the locomotive on the wall was larger than the force the wall could exert on the locomotive. (b) The force exerted by the locomotive on the wall was the same in magnitude as the force exerted by the wall on the locomotive. (c) The force exerted by the locomotive on the wall was less than the force exerted by the wall on the locomotive. (d) The wall cannot be said to “exert” a force; after all, it broke. 300 g ee b 1. The driver of a speeding empty truck slams on the brakes and skids to a stop through a distance d. On a second trial, the truck carries a load that doubles its mass. What will now be the truck’s “skidding distance”? (a) 4d (b) 2d (c) !2d (d) d (e) d/2 ly 1. denotes answer available in Student Solutions Manual/Study Guide s. w Objective Questions .c F  0 200 g 10 m/s 8 m/s 400 g e c 500 g d Figure OQ5.3 3. The third graders are on one side of a schoolyard, and the fourth graders are on the other. They are throwing snowballs at each other. Between them, snowballs of various masses are moving with different velocities as shown in Figure OQ5.3. Rank the snowballs (a) through (e) according to the magnitude of the total force exerted on each one. Ignore air resistance. If two snowballs rank together, make that fact clear. 6. The manager of a department store is pushing horizontally with a force of magnitude 200 N on a box of shirts. The box is sliding across the horizontal floor with a forward acceleration. Nothing else touches the box. What must be true about the magnitude of the force of kinetic friction acting on the box (choose one)? (a) It is greater than 200 N. (b) It is less than 200 N. (c) It is equal to 200 N. (d) None of those statements is necessarily true. w w w © ND/Roger Viollet/The Image Works .a sw ar ph Figure OQ5.2 4. The driver of a speeding truck slams on the brakes and skids to a stop through a distance d. On another trial, the initial speed of the truck is half as large. What now will be the truck’s skidding distance? (a) 2d (b) !2d (c) d (d) d/2 (e) d/4 5. An experiment is performed on a puck on a level air hockey table, where friction is negligible. A constant horizontal force is applied to the puck, and the puck’s acceleration is measured. Now the same puck is transported far into outer space, where both friction and gravity are negligible. The same constant force is applied to the puck (through a spring scale that stretches the same amount), and the puck’s acceleration (relative to the distant stars) is measured. What is the puck’s acceleration in outer space? (a) It is somewhat greater than its acceleration on the Earth. (b) It is the same as its acceleration on the Earth. (c) It is less than its acceleration on the Earth. (d) It is infinite because neither friction nor gravity constrains it. (e) It is very large because acceleration is inversely proportional to weight and the puck’s weight is very small but not zero. Conceptual Questions om a down 13. An object of mass m moves with acceleration S a rough incline. Which of the following forces should appear in a free-body diagram of the object? Choose all correct answers. (a) the gravitational force exerted a in the direction of motion (c) the by the planet (b) m S normal force exerted by the incline (d) the friction force exerted by the incline (e) the force exerted by the object on the incline ys ic s. w 10. A large crate of mass m is place on the flatbed of a truck but not tied down. As the truck accelerates forward with acceleration a, the crate remains at rest relative to the truck. What force causes the crate to accelerate? (a) the normal force (b) the gravitational 12. A crate remains stationary after it has been placed on a ramp inclined at an angle with the horizontal. Which of the following statements is or are correct about the magnitude of the friction force that acts on the crate? Choose all that are true. (a) It is larger than the weight of the crate. (b) It is equal to msn. (c) It is greater than the component of the gravitational force acting down the ramp. (d) It is equal to the component of the gravitational force acting down the ramp. (e) It is less than the component of the gravitational force acting down the ramp. .c 9. A truck loaded with sand accelerates along a highway. The driving force on the truck remains constant. What happens to the acceleration of the truck if its trailer leaks sand at a constant rate through a hole in its bottom? (a) It decreases at a steady rate. (b) It increases at a steady rate. (c) It increases and then decreases. (d) It decreases and then increases. (e) It remains constant. 11. If an object is in equilibrium, which of the following statements is not true? (a) The speed of the object remains constant. (b) The acceleration of the object is zero. (c) The net force acting on the object is zero. (d) The object must be at rest. (e) There are at least two forces acting on the object. ly 8. An object of mass m is sliding with speed vi at some instant across a level tabletop, with which its coefficient of kinetic friction is m. It then moves through a distance d and comes to rest. Which of the following equations for the speed vi is reasonable? (a) v i 5 !22mmgd (b) v i 5 !2mmgd (c) v i 5 !22mgd (d) v i 5 !2mgd (e) v i 5 !2md force (c) the friction force (d) the ma force exerted by the crate (e) No force is required. ee b 7. Two objects are connected by a string that passes over a frictionless pulley as in Figure 5.14a, where m1 , m 2 and a1 and a 2 are the magnitudes of the respective accelerations. Which mathematical statement is true regarding the magnitude of the acceleration a 2 of the mass m 2? (a) a 2 , g (b) a 2 . g (c) a 2 5 g (d) a 2 , a1 (e) a 2 . a1 137 Conceptual Questions ar ph 1. denotes answer available in Student Solutions Manual/Study Guide sw 1. If you hold a horizontal metal bar several centimeters above the ground and move it through grass, each leaf of grass bends out of the way. If you increase the speed of the bar, each leaf of grass will bend more quickly. How then does a rotary power lawn mower manage to cut grass? How can it exert enough force on a leaf of grass to shear it off? 7. w .a 2. Your hands are wet, and the restroom towel dispenser is empty. What do you do to get drops of water off your hands? How does the motion of the drops exemplify one of Newton’s laws? Which one? 8. w w 3. In the motion picture It Happened One Night (Columbia Pictures, 1934), Clark Gable is standing inside a stationary bus in front of Claudette Colbert, who is seated. The bus suddenly starts moving forward and Clark falls into Claudette’s lap. Why did this happen? 4. If a car is traveling due westward with a constant speed of 20 m/s, what is the resultant force acting on it? 5. A passenger sitting in the rear of a bus claims that she was injured when the driver slammed on the brakes, causing a suitcase to come flying toward her from the front of the bus. If you were the judge in this case, what disposition would you make? Why? 6. A child tosses a ball straight up. She says that the ball is moving away from her hand because the ball feels an upward “force of the throw” as well as the gravitational force. (a) Can the “force of the throw” exceed the 9. 10. gravitational force? How would the ball move if it did? (b) Can the “force of the throw” be equal in magnitude to the gravitational force? Explain. (c) What strength can accurately be attributed to the “force of the throw”? Explain. (d) Why does the ball move away from the child’s hand? A person holds a ball in her hand. (a) Identify all the external forces acting on the ball and the Newton’s third-law reaction force to each one. (b) If the ball is dropped, what force is exerted on it while it is falling? Identify the reaction force in this case. (Ignore air resistance.) A spherical rubber balloon inflated with air is held stationary, with its opening, on the west side, pinched shut. (a) Describe the forces exerted by the air inside and outside the balloon on sections of the rubber. (b) After the balloon is released, it takes off toward the east, gaining speed rapidly. Explain this motion in terms of the forces now acting on the rubber. (c) Account for the motion of a skyrocket taking off from its launch pad. A rubber ball is dropped onto the floor. What force causes the ball to bounce? Twenty people participate in a tug-of-war. The two teams of ten people are so evenly matched that neither team wins. After the game they notice that a car is stuck in the mud. They attach the tug-of-war rope to the bumper of the car, and all the people pull on the The Laws of Motion 14. An athlete grips a light rope that passes over a lowfriction pulley attached to the ceiling of a gym. A sack of sand precisely equal in weight to the athlete is tied to the other end of the rope. Both the sand and the athlete are initially at rest. The athlete climbs the rope, sometimes speeding up and slowing down as he does so. What happens to the sack of sand? Explain. 21. Identify action–reaction pairs in the following situations: (a) a man takes a step (b) a snowball hits a girl in the back (c) a baseball player catches a ball (d) a gust of wind strikes a window 22. As shown in Figure CQ5.22, student A, a 55-kg girl, sits on one chair with metal runners, at rest on a classroom floor. Student B, an 80-kg boy, sits on an identical chair. Both students keep their feet off the floor. A rope runs from student A’s hands around a light pulley and then over her shoulder to the hands of a teacher standing on the floor behind her. The low-friction axle of the pulley is attached to a second rope held by student B. All ropes run parallel to the chair runners. (a) If student A pulls on her end of the rope, will her chair or will B’s chair slide on the floor? Explain why. (b) If instead the teacher pulls on his rope end, which chair slides? Why this one? (c) If student B pulls on his rope, which chair slides? Why? (d) Now the teacher ties his end of the rope to student A’s chair. Student A pulls on the end of the rope in her hands. Which chair slides and why? ic 15. Suppose you are driving a classic car. Why should you avoid slamming on your brakes when you want to stop in the shortest possible distance? (Many modern cars have antilock brakes that avoid this problem.) om 13. A weightlifter stands on a bathroom scale. He pumps a barbell up and down. What happens to the reading on the scale as he does so? What If? What if he is strong enough to actually throw the barbell upward? How does the reading on the scale vary now? .c 12. When you push on a box with a 200-N force instead of a 50-N force, you can feel that you are making a greater effort. When a table exerts a 200-N normal force instead of one of smaller magnitude, is the table really doing anything differently? 20. Balancing carefully, three boys inch out onto a horizontal tree branch above a pond, each planning to dive in separately. The third boy in line notices that the branch is barely strong enough to support them. He decides to jump straight up and land back on the branch to break it, spilling all three into the pond. When he starts to carry out his plan, at what precise moment does the branch break? Explain. Suggestion: Pretend to be the third boy and imitate what he does in slow motion. If you are still unsure, stand on a bathroom scale and repeat the suggestion. ly 11. Can an object exert a force on itself? Argue for your answer. floor on the ball be different in magnitude from the force the ball exerts on the floor? ee b rope. The heavy car has just moved a couple of decimeters when the rope breaks. Why did the rope break in this situation when it did not break when the same twenty people pulled on it in a tug-of-war? s. w 138 Chapter 5 Student B w w w .a sw ar ph ys B A 16. In Figure CQ5.16, the light, taut, unstretchable cord B 2 1 joins block 1 and the largermass block 2. Cord A exerts Figure CQ5.16 a force on block 1 to make it accelerate forward. (a) How does the magnitude of the force exerted by cord A on block 1 compare with the magnitude of the force exerted by cord B on block 2? Is it larger, smaller, or equal? (b) How does the acceleration of block 1 compare with the acceleration (if any) of block 2? (c) Does cord B exert a force on block 1? If so, is it forward or backward? Is it larger, smaller, or equal in magnitude to the force exerted by cord B on block 2? 17. Describe two examples in which the force of friction exerted on an object is in the direction of motion of the object. 18. The mayor of a city reprimands some city employees because they will not remove the obvious sags from the cables that support the city traffic lights. What explanation can the employees give? How do you think the case will be settled in mediation? 19. Give reasons for the answers to each of the following questions: (a) Can a normal force be horizontal? (b) Can a normal force be directed vertically downward? (c) Consider a tennis ball in contact with a stationary floor and with nothing else. Can the normal force be different in magnitude from the gravitational force exerted on the ball? (d) Can the force exerted by the Student A Teacher Figure CQ5.22 23. A car is moving forward slowly and is speeding up. A student claims that “the car exerts a force on itself” or that “the car’s engine exerts a force on the car.” (a) Argue that this idea cannot be accurate and that friction exerted by the road is the propulsive force on the car. Make your evidence and reasoning as persuasive as possible. (b) Is it static or kinetic friction? Suggestions: Consider a road covered with light gravel. Consider a sharp print of the tire tread on an asphalt road, obtained by coating the tread with dust. Problems 139 Problems AMT   Analysis Model tutorial available in The problems found in this chapter may be assigned online in Enhanced WebAssign Enhanced WebAssign GP Guided Problem 1. straightforward; 2. intermediate; 3. challenging M   Master It tutorial available in Enhanced 1. full solution available in the Student W   Watch It video solution available in WebAssign Enhanced WebAssign Solutions Manual/Study Guide BIO Q/C om S Section 5.1 The Concept of Force Section 5.2 Newton’s First Law and Inertial Frames Section 5.3 Mass Section 5.4 Newton’s Second Law Section 5.5 The Gravitational Force and Weight Section 5.6 Newton’s Third Law opposite direction, what is the average acceleration of the molecule during this time interval? (b) What average force does the molecule exert on the wall? 1. A woman weighs 120 lb. Determine (a) her weight in newtons and (b) her mass in kilograms. s. w 2. If a man weighs 900 N on the Earth, what would he weigh on Jupiter, where the free-fall acceleration is 25.9 m/s2? ee b ly .c 7. The distinction between mass and weight was discovered after Jean Richer transported pendulum clocks from Paris, France, to Cayenne, French Guiana, in 1671. He found that they quite systematically ran slower in Cayenne than in Paris. The effect was reversed when the clocks returned to Paris. How much weight would a 90.0 kg person lose in traveling from Paris, where g 5 9.809 5 m/s2, to Cayenne, where g 5 9.780 8 m/s2? (We will consider how the free-fall acceleration influences the period of a pendulum in Section 15.5.) 8. (a) A car with a mass of 850 kg is moving to the right with a constant speed of 1.44 m/s. What is the total force on the car? (b) What is the total force on the car if it is moving to the left? 3. A 3.00-kg object undergoes an acceleration given by S 2 W a 5 1 2.00 i^ 1 5.00 j^ 2 m/s . Find (a) the resultant force ys ic acting on the object and (b) the magnitude of the resultant force. 4. A certain orthodontist uses a wire brace to align a BIO patient’s crooked tooth as in Figure P5.4. The tension 14° S S T w w .a T x 14° sw y ar ph in the wire is adjusted to have a magnitude of 18.0 N. Find the magnitude of the net force exerted by the wire on the crooked tooth. Figure P5.4 w 5. A toy rocket engine is securely fastened to a large puck M that can glide with negligible friction over a horizon- tal surface, taken as the xy plane. The 4.00-kg puck has a velocity of 3.00 ^i m/s at one instant. Eight seconds later, its velocity is 1 8.00 i^ 1 10.00 j^ 2 m/s. Assuming the rocket engine exerts a constant horizontal force, find (a) the components of the force and (b) its magnitude. 6. The average speed of a nitrogen molecule in air is about 6.70 3 102 m/s, and its mass is 4.68 3 10226 kg. (a) If it takes 3.00 3 10213 s for a nitrogen molecule to hit a wall and rebound with the same speed but moving in the 9. Review. The gravitational force exerted on a baseball is 2.21 N down. A pitcher throws the ball horizontally with velocity 18.0 m/s by uniformly accelerating it along a straight horizontal line for a time interval of 170 ms. The ball starts from rest. (a) Through what distance does it move before its release? (b) What are the magnitude and direction of the force the pitcher exerts on the ball? 10. Review. The gravitational force exerted on a baseball is S 2Fg j^. A pitcher throws the ball with velocity v ^i by uniformly accelerating it along a straight horizontal line for a time interval of Dt 5 t 2 0 5 t. (a) Starting from rest, through what distance does the ball move before its release? (b) What force does the pitcher exert on the ball? 11. Review. An electron of mass 9.11 3 10231 kg has an M initial speed of 3.00 3 105 m/s. It travels in a straight line, and its speed increases to 7.00 3 105 m/s in a distance of 5.00 cm. Assuming its acceleration is constant, (a) determine the magnitude of the force exerted on the electron and (b) compare this force with the weight of the electron, which we ignored. 12. Besides the gravitational force, a 2.80-kg object is subjected to one other constant force. The object starts from rest and in 1.20 s experiences a displacement of 1 4.20 i^ 2 3.30 j^ 2 m, where the direction of ^j is the upward vertical direction. Determine the other force. The Laws of Motion 16. The force exerted by the wind on the sails of a sailboat M is 390 N north. The water exerts a force of 180 N east. If the boat (including its crew) has a mass of 270 kg, what are the magnitude and direction of its acceleration? om .c 22. Review. Three forces acting onSan object are given by S W F 1 5S(22.00i^ 1 2.00 ^j ) N, and F 2 5(5.00i^ 2 3.00 ^j ) N, and F 3 5 1 245.0 i^ 2 N. The object experiences an acceleration of magnitude 3.75 m/s2. (a) What is the direction of the acceleration? (b) What is the mass of the object? (c) If the object is initially at rest, what is its speed after 10.0 s? (d) What are the velocity components of the object after 10.0 s? 23. A 1 000-kg car is pulling a 300-kg trailer. Together, the car and trailer move forward with an acceleration of 2.15 m/s2. Ignore any force of air drag on the car and all friction forces on the trailer. Determine (a) the net force on the car, (b) the net force on the trailer, (c) the force exerted by the trailer on the car, and (d) the resultant force exerted by the car on the road. sw S S w .a 18. A force F applied to an object of mass m1 produces 2 W an acceleration of 3.00 m/s . The same force applied to a second object of mass m 2 produces an acceleration of 1.00 m/s2. (a) What is the value of the ratio m1/m 2? find its (b) If m1 and m 2 are combined into one object, S acceleration under the action of the force F . S w w 19. Two forces F 1 and F 2 act on a 5.00-kg object. Taking M F 1 5 20.0 N and F 2 5 15.0 N, find the accelerations of the object for the configurations of forces shown in parts (a) and (b) of Figure P5.19. S F2 S F2 90.0 m S F1 60.0 m a b Figure P5.19 S F1 24. If a single constant force acts on an object that moves S on a straight line, the object’s velocity is a linear function of time. The equation v 5 vi 1 at gives its velocity v as a function of time, where a is its constant acceleration. What if velocity is instead a linear function of position? Assume that as a particular object moves through a resistive medium, its speed decreases as described by the equation v 5 vi 2 kx, where k is a constant coefficient and x is the position of the object. Find the law describing the total force acting on this object. Section 5.7 Analysis Models Using Newton’s Second Law 25. Review. Figure P5.25 shows a worker poling a boat—a very efficient mode of transportation—across a shallow lake. He pushes parallel to the length of the light pole, exerting a force of magnitude 240 N on the bottom of the lake. Assume the pole lies in the vertical plane containing the keel of the boat. At one moment, the pole Figure P5.25 makes an angle of 35.0° with the vertical and the water exerts a horizontal drag force of 47.5 N on the boat, opposite to its forward velocity of magnitude 0.857 m/s. The mass of the boat including its cargo and the worker is 370 kg. (a) The water exerts AP Photo/Rebecca Blackwell ar ph ys ic 17. An object of mass m is dropped at t 5 0 from the roof S of a building of height h. While the object is falling, a wind blowing parallel to the face of the building exerts a constant horizontal force F on the object. (a) At what time t does the object strike the ground? Express t in terms of g and h. (b) Find an expression in terms of m and F for the acceleration ax of the object in the horizontal direction (taken as the positive x direction). (c) How far is the object displaced horizontally before hitting the ground? Answer in terms of m, g, F, and h. (d) Find the magnitude of the object’s acceleration while it is falling, using the variables F, m, and g. 21. A 15.0-lb block rests on the floor. (a) What force does the floor exert on the block? (b) A rope is tied to the block and is run vertically over a pulley. The other end is attached to a free-hanging 10.0-lb object. What now is the force exerted by the floor on the 15.0-lb block? (c) If the 10.0-lb object in part (b) is replaced with a 20.0-lb object, what is the force exerted by the floor on the 15.0-lb block? ly 14. A brick of mass M has been placed on a rubber cushion of mass m. Together they are sliding to the right at constant velocity on an ice-covered parking lot. (a) Draw a free-body diagram of the brick and identify each force acting on it. (b) Draw a free-body diagram of the cushion and identify each force acting on it. (c) Identify all of the action–­reaction pairs of forces in the brick–cushion–planet system. S S 15. Two forces, F1 5 126.00 i^ 2 4.00 j^ 2 N and F2 5 123.00 i^ 1 7.00 j^ 2 N, act on a particle of mass 2.00 kg that is initially at rest at coordinates (22.00 m, 14.00 m). (a) What are the components of the particle’s velocity at t 5 10.0 s? (b) In what direction is the particle moving at t 5 10.0 s? (c) What displacement does the particle undergo during the first 10.0 s? (d) What are the coordinates of the particle at t 5 10.0 s? 20. You stand on the seat of a chair and then hop off. (a) During the time interval you are in flight down to the floor, the Earth moves toward you with an acceleration of what order of magnitude? In your solution, explain your logic. Model the Earth as a perfectly solid object. (b) The Earth moves toward you through a distance of what order of magnitude? ee b 13. One or more external forces, large enough to be easily measured, are exerted on each object enclosed in a dashed box shown in Figure 5.1. Identify the reaction to each of these forces. s. w 140 Chapter 5 141 Problems 1.0 kg 2.0 kg om Figure P5.29 .c 30. A block slides down a frictionless plane having an incliW nation of u 5 15.0°. The block starts from rest at the top, and the length of the incline is 2.00 m. (a) Draw a free-body diagram of the block. Find (b) the acceleration of the block and (c) its speed when it reaches the bottom of the incline. 31. The distance between two telephone poles is 50.0 m. When a 1.00-kg bird lands on the telephone wire midway between the poles, the wire sags 0.200 m. (a) Draw a free-body diagram of the bird. (b) How much tension does the bird produce in the wire? Ignore the weight of the wire. s. w Q/C 42 N 3.0 kg ly 27. 29. Assume the three blocks portrayed in Figure P5.29 M move on a frictionless surface and a 42-N force acts as shown on the 3.0-kg block. Determine (a) the acceleration given this system, (b) the tension in the cord connecting the 3.0-kg and the 1.0-kg blocks, and (c) the force exerted by the 1.0-kg block on the 2.0-kg block. ee b 26. a buoyant force vertically upward on the boat. Find the magnitude of this force. (b) Model the forces as constant over a short interval of time to find the velocity of the boat 0.450 s after the moment described. An iron bolt of mass 65.0 g hangs from a string 35.7 cm long. The top end of the string is fixed. Without touching it, a magnet attracts the bolt so that it remains stationary, but is displaced horizontally 28.0 cm to the right from the previously vertical line of the string. The magnet is located to the right of the bolt and on the same vertical level as the bolt in the final configuration. (a) Draw a free-body diagram of the bolt. (b) Find the tension in the string. (c) Find the magnetic force on the bolt. Figure P5.27 shows the S P horizontal forces acting on S u a sailboat moving north at n constant velocity, seen N from a point straight above its mast. At the particular W E speed of the sailboat, the 220 N S water exerts a 220-N drag force on its hull and u 5 Figure P5.27 40.0°. For each of the situations (a) and (b) described below, write two component equations representing Newton’s second law. Then solve the equations for P (the force exerted by the wind on the sail) and for n (the force exerted by the water on the keel). (a) Choose the x direction as east and the y direction as north. (b) Now choose the x direction as u 5 40.0° north of east and the y direction as u 5 40.0° west of north. (c) Compare your solutions to parts (a) and (b). Do the results agree? Is one method significantly easier? The systems shown in Figure P5.28 are in equilibrium. If the spring scales are calibrated in newtons, what do they read? Ignore the masses of the pulleys and strings and assume the pulleys and the incline in Figure P5.28d are frictionless. .a sw W 5.00 kg b w a 5.00 kg ure P5.33. Two of the wires make angles u1 5 60.0° and u2 5 40.0° with the horizontal. Assuming the system is in equilibrium, find the tensions T1, T2, and T3 in the wires. T1 5 5.00 kg 5.00 kg 30.0 5.00 kg c d Figure P5.28 u2 u1 T1 T2 T3 Fg 34. A bag of cement whose weight S is Fg hangs in equilibrium from Figure P5.33 three wires as shown in Figure Problems 33 and 34. P5.33. Two of the wires make angles u1 and u2 with the horizontal. Assuming the system is in equilibrium, show that the tension in the lefthand wire is w w 5.00 kg 33. A bag of cement weighing 325 N AMT hangs in equilibrium from W three wires as suggested in Fig- CEMENT 28. ar ph ys ic 32. A 3.00-kg object is moving in a plane, with its x and y W coordinates given by x 5 5t 2 2 1 and y 5 3t 3 1 2, where x and y are in meters and t is in seconds. Find the magnitude of the net force acting on this object at t 5 2.00 s. Fg cos u 2 sin 1 u 1 1 u 2 2 35. Two people pull as hard as they can on horizontal ropes attached to a boat that has a mass of 200 kg. If they pull in the same direction, the boat has an acceleration of 1.52 m/s2 to the right. If they pull in opposite directions, the boat has an acceleration of 0.518 m/s2 to the left. What is the magnitude of the force each person exerts on the boat? Disregard any other horizontal forces on the boat. 142 Chapter 5 The Laws of Motion 36. Figure P5.36 shows loads hanging from the ceiling of an elevator that is moving at constant velocity. Find the tension in each of the three strands of cord supporting each load. 60.0° 50.0° T1 T2 T2 30 T3 a b Figure P5.36 10 ar ph sw 70 .a 8.00 kg w Figure P5.38 w w 39. A simple accelerometer is constructed inside a car by suspending an object of mass m from a string of length L that is tied to the car’s ceiling. As the car accelerates the string–object system makes a constant angle of u with the vertical. (a) Assuming that the string mass is negligible compared with m, derive an expression for the car’s acceleration in terms of u and show that it is independent of the mass m and the length L. (b) Determine the acceleration of the car m1 when u 5 23.0°. 40. An object of mass m1 5 5.00 kg AMT placed on a frictionless, horizonW tal table is connected to a string that passes over a pulley and then is fastened to a hanging object of mass m 2 5 9.00 kg as shown in Figure P5.40. (a) Draw free-body m2 Figure P5.40 Problems 40, 63, and 87. 1.0 time (s) 2.0 1.5 Figure P5.41 s. w 42. Two objects are connected by a light string that passes over a frictionless pulley as shown in Figure P5.42. Assume the incline is frictionless and take m1 5 2.00 kg, m2 5 6.00 kg, and u 5 55.0°. (a) Draw free-body diagrams of both objects. Find (b) the magnitude of the acceleration of the objects, (c) the tension in the string, and (d) the speed of each object 2.00 s after it is released from rest. ys 38. A setup similar to the one shown in Figure P5.38 is often BIO used in hospitals to support and apply a horizontal traction force to an injured leg. (a) Determine the force of tension in the rope supporting the leg. (b) What is the traction force exerted to the right on the leg? 0.5 ee b 0 ic 37. An object of mass m 5 1.00 kg S S a is observed to have an accelF2 S 60.0 eration a with a magnitude of 10.0 m/s2 in a direction 60.0° east of north. Figure P5.37 m S F1 shows a view of the S object from above. The force F 2 actFigure P5.37 ing on the object has a magnitude of 5.00 N and is directed north. Determine the magnitude and direction of the S one other horizontal force F 1 acting on the object. 20 om 10.0 kg 5.00 kg .c T3 ly T1 41. Figure P5.41 shows the speed of a person’s body as he does a chin-up. Assume the motion is vertical and the mass of the person’s body is 64.0 kg. Determine the force exerted by the chin-up bar on his body at (a) t 5 0, (b) t 5 0.5 s, (c) t 5 1.1 s, and (d) t 5 1.6 s. speed (cm/s) 40.0° diagrams of both objects. Find (b) the magnitude of the acceleration of the objects and (c) the tension in the string. m1 m2 u Figure P5.42 43. Two blocks, each of mass m 5 3.50 kg, are hung from the ceiling of an elevator as in Figure P5.43. (a) If the elevator moves with an T1 a of magniupward acceler­ation S 2 tude 1.60 m/s , find the tensions S a m T1 and T2 in the upper and lower strings. (b) If the strings can T2 withstand a maximum tension of m 85.0 N, what maximum acceleration can the elevator have before a string breaks? Figure P5.43 44. Two blocks, each of mass m, are Problems 43 and 44. S hung from the ceiling of an elevator as in Figure P5.43. The elevator has an upward acceleration a. The strings have negligible mass. (a) Find the tensions T1 and T2 in the upper and lower strings in terms of m, a, and g. (b) Compare the two tensions and determine which string would break first if a is made sufficiently large. (c) What are the tensions if the cable supporting the elevator breaks? 45. In the system shown in Figure P5.45, a horizontal force S M F x acts on an object of mass m 2 5 8.00 kg. The hori- 143 Problems acts on the top pin. Draw a free-body diagram of the pin. Use the condition for equilibrium of the pin to translate the free-body diagram into equations. From the equations calculate the forces exerted by struts A and B. If you obtain a positive answer, you correctly guessed the direction of the force. A negative answer means that the direction should be reversed, but the absolute value correctly gives the magnitude of the force. If a strut pulls on a pin, it is in tension. If it pushes, the strut is in compression. Identify whether each strut is in tension or in compression. zontal surface is frictionless. Consider the acceleration of the sliding object as a function of Fx . (a) For what values of Fx does the object of mass m1 5 2.00 kg accelerate upward? (b) For what values of Fx is the tension in the cord zero? (c) Plot the acceleration of the m 2 object versus Fx . Include values of Fx from 2100 N to 1100 N. S m2 Fx m1 om B Figure P5.45 60.0° P1 ar ph ys m2 m1 sw Figure P5.46 w w .a 47. A block is given an initial velocity of 5.00 m/s up a frictionless incline of angle u 5 20.0° (Fig. P5.47). How far up the incline does the block slide before coming to rest? w .c ly ee b 49. Two blocks of mass 3.50 kg and 8.00 kg are connected by a massless string that passes over a frictionless pulley (Fig. P5.49). The inclines are frictionless. Find (a) the magnitude of the acceleration of each block and (b) the tension in the string. ic P2 θ Figure P5.47 48. A car is stuck in the mud. A tow truck pulls on the car with the arrangement shown in Fig. P5.48. The tow cable is under a tension of 2 500 N and pulls downward and to the left on the pin at its upper end. The light pin is held in equilibrium by forces exerted by the two bars A and B. Each bar is a strut; that is, each is a bar whose weight is small compared to the forces it exerts and which exerts forces only through hinge pins at its ends. Each strut exerts a force directed parallel to its length. Determine the force of tension or compression in each strut. Proceed as follows. Make a guess as to which way (pushing or pulling) each force 50.0° Figure P5.48 s. w 46. An object of mass m1 hangs from a string that passes S over a very light fixed pulley P1 as shown in Figure P5.46. The string connects to a second very light pulley P2. A second string passes around this pulley with one end attached to a wall and the other to an object of mass m 2 on a frictionless, horizontal table. (a) If a1 and a 2 are the accelerations of m1 and m 2, respectively, what is the relation between these accelerations? Find expressions for (b) the tensions in the strings and (c) the accelerations a1 and a 2 in terms of the masses m1 and m 2, and g. A 8.00 kg 3.50 kg 35.0° Figure P5.49 35.0° Problems 49 and 71. 50. In the Atwood machine discussed in Example 5.9 and shown in Figure 5.14a, m1 5 2.00 kg and m 2 5 7.00 kg. The masses of the pulley and string are negligible by comparison. The pulley turns without friction, and the string does not stretch. The lighter object is released with a sharp push that sets it into motion at vi 5 2.40 m/s downward. (a) How far will m1 descend below its initial level? (b) Find the velocity of m1 after 1.80 s. 51. In Example 5.8, we investigated the apparent weight of AMT a fish in an elevator. Now consider a 72.0-kg man standM ing on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.20 m/s in 0.800 s. It travels with this constant speed for the next 5.00 s. The elevator then undergoes a uniform acceleration in the negative y direction for 1.50 s and comes to rest. What does the spring scale register (a) before the elevator starts to move, (b) during the first 0.800 s, (c) while the elevator is traveling at constant speed, and (d) during the time interval it is slowing down? Section 5.8 Forces of Friction 52. Consider a large truck carrying a heavy load, such as steel beams. A significant hazard for the driver is that the load may slide forward, crushing the cab, if the truck stops suddenly in an accident or even in braking. Assume, for example, that a 10 000-kg load sits on the W 61. M ys ar ph 62. BIO w w w 63. Jamie Squire/Allsport/Getty Images 58. Q/C .a sw 57. W 64. Q/C Figure P5.58 om W ic 56. 60. .c 55. 59. mum value of ms is necessary to achieve the record time? (b) Suppose the driver were able to increase his or her engine power, keeping other things equal. How would this change affect the elapsed time? To meet a U.S. Postal Service requirement, employees’ footwear must have a coefficient of static friction of 0.5 or more on a specified tile surface. A typical athletic shoe has a coefficient of static friction of 0.800. In an emergency, what is the minimum time interval in which a person starting from rest can move 3.00 m on the tile surface if she is wearing (a) footwear meeting the Postal Service minimum and (b) a typical athletic shoe? A woman at an airport is towing her 20.0-kg suitcase at constant speed by pulling on a strap at an angle u above the horizontal (Fig. P5.60). She pulls on the u strap with a 35.0-N force, and the friction force on the suitcase is 20.0 N. (a) Draw a freebody diagram of the suitcase. Figure P5.60 (b) What angle does the strap make with the horizontal? (c) What is the magnitude of the normal force that the ground exerts on the suitcase? Review. A 3.00-kg block starts from rest at the top of a 30.0° incline and slides a distance of 2.00 m down the incline in 1.50 s. Find (a) the magnitude of the acceleration of the block, (b) the coefficient of kinetic friction between block and plane, (c) the friction force acting on the block, and (d) the speed of the block after it has slid 2.00 m. The person in Figure P5.62 weighs 170 lb. As seen from the front, each light crutch makes an angle of 22.0° with the vertical. Half of the person’s weight is supported by the crutches. The other half is supported by the vertical forces of the ground on the person’s feet. Assum22.0 22.0 ing that the person is moving with constant velocity and the force exerted by the Figure P5.62 ground on the crutches acts along the crutches, determine (a) the smallest possible coefficient of friction between crutches and ground and (b) the magnitude of the compression force in each crutch. A 9.00-kg hanging object is connected by a light, inextensible cord over a light, frictionless pulley to a 5.00kg block that is sliding on a flat table (Fig. P5.40). Taking the coefficient of kinetic friction as 0.200, find the tension in the string. Three objects are connected on a table as shown in Figure P5.64. The coefficient of kinetic friction between the block of mass m 2 and the table is 0.350. The objects have masses of m1 5 4.00 kg, m 2 5 1.00 kg, and m 3 5 ly 54. flatbed of a 20 000-kg truck moving at 12.0 m/s. Assume that the load is not tied down to the truck, but has a coefficient of friction of 0.500 with the flatbed of the truck. (a) Calculate the minimum stopping distance for which the load will not slide forward relative to the truck. (b) Is any piece of data unnecessary for the solution? Review. A rifle bullet with a mass of 12.0 g traveling toward the right at 260 m/s strikes a large bag of sand and penetrates it to a depth of 23.0 cm. Determine the magnitude and direction of the friction force (assumed constant) that acts on the bullet. Review. A car is traveling at 50.0 mi/h on a horizontal highway. (a) If the coefficient of static friction between road and tires on a rainy day is 0.100, what is the minimum distance in which the car will stop? (b) What is the stopping distance when the surface is dry and ms 5 0.600? A 25.0-kg block is initially at rest on a horizontal surface. A horizontal force of 75.0 N is required to set the block in motion, after which a horizontal force of 60.0 N is required to keep the block moving with constant speed. Find (a) the coefficient of static friction and (b) the coefficient of kinetic friction between the block and the surface. Why is the following situation impossible? Your 3.80-kg physics book is placed next to you on the horizontal seat of your car. The coefficient of static friction between the book and the seat is 0.650, and the coefficient of kinetic friction is 0.550. You are traveling forward at 72.0 km/h and brake to a stop with constant acceleration over a distance of 30.0 m. Your physics book remains on the seat rather than sliding forward onto the floor. To determine the coefficients of friction between rubber and various surfaces, a student uses a rubber eraser and an incline. In one experiment, the eraser begins to slip down the incline when the angle of inclination is 36.0° and then moves down the incline with constant speed when the angle is reduced to 30.0°. From these data, determine the coefficients of static and kinetic friction for this experiment. Before 1960, people believed that the maximum attainable coefficient of static friction for an automobile tire on a roadway was ms 5 1. Around 1962, three companies independently developed racing tires with coefficients of 1.6. This problem shows that tires have improved further since then. The shortest time interval in which a piston-engine car initially at rest has covered a distance of one-quarter mile is about 4.43 s. (a) Assume the car’s rear wheels lift the front wheels off the pavement as shown in Figure P5.58. What mini- ee b 53. The Laws of Motion s. w 144 Chapter 5 Problems 2.00 kg, and the pulleys are frictionless. (a) Draw a freebody diagram of each object. (b) Determine the acceleration of each object, including its direction. (c) Determine the tensions in the two cords. What If? (d) If the tabletop were smooth, would the tensions increase, decrease, or remain the same? Explain. m2 m1 T m2 S F s. w zontal force (Fig. P5.65). Figure P5.65 Suppose F 5 68.0 N, m1 5 12.0 kg, m 2 5 18.0 kg, and the coefficient of kinetic friction between each block and the surface is 0.100. (a) Draw a free-body diagram for each block. Determine (b) the acceleration of the system and (c) the tension T in the rope. .c 65. Two blocks connected by a 70. A 5.00-kg block is placed on top of a 10.0-kg block (Fig. P5.70). A horizontal force of 45.0 N is applied to the 10-kg block, and the 5.00-kg block is tied to the wall. The coefficient of kinetic friction between all moving surfaces is 0.200. (a) Draw a free-body diagram for each block and identify the action–reaction forces between the blocks. (b) Determine the tension in the string and the magnitude of the acceleration of the 10.0-kg block. ly Figure P5.64 AMT rope of negligible mass are M being dragged by a hori- 69. Review. A magician pulls a tablecloth from under a 200-g mug located 30.0 cm from the edge of the cloth. The cloth exerts a friction force of 0.100 N on the mug, and the cloth is pulled with a constant acceleration of 3.00 m/s2. How far does the mug move relative to the horizontal tabletop before the cloth is completely out from under it? Note that the cloth must move more than 30 cm relative to the tabletop during the process. om m3 the fluid friction force disappears as soon as the fish’s head breaks the water surface and assume the force on its tail is constant. Model the gravitational force as suddenly switching full on when half the length of the fish is out of the water. Find the value of P. ee b m1 145 66. A block of mass 3.00 kg is pushed S Q/C up against a wall by a force P that sw ar ph ys ic makes an angle of u 5 50.0° with the horizontal as shown in Figure u P5.66. The coefficient of static fricS tion between the block and the wall P is 0.250. (a) Determine the possible S Figure P5.66 values for the magnitude of P that allow the block to remain station-S ary. (b) Describe what happens if 0 P 0 has a larger value and what happens if it is smaller. (c) Repeat parts (a) and (b), assuming the force makes an angle of u 5 13.0° with the horizontal. w w w .a 67. Review. One side of the roof of a house slopes up at 37.0°. A roofer kicks a round, flat rock that has been thrown onto the roof by a neighborhood child. The rock slides straight up the incline with an initial speed of 15.0 m/s. The coefficient of kinetic friction between the rock and the roof is 0.400. The rock slides 10.0 m up the roof to its peak. It crosses the ridge and goes into free fall, following a parabolic trajectory above the far side of the roof, with negligible air resistance. Determine the maximum height the rock reaches above the point where it was kicked. 68. Review. A Chinook salmon can swim underwater at BIO 3.58 m/s, and it can also jump vertically upward, leav- ing the water with a speed of 6.26 m/s. A record salmon has length 1.50 m and mass 61.0 kg. Consider the fish swimming straight upward in the water below the surface of a lake. The gravitational force exerted on it is very nearly canceled out by a buoyant force exerted by the water as we will study in Chapter 14. The fish experiences an upward force P exerted by the water on its threshing tail fin and a downward fluid friction force that we model as acting on its front end. Assume 5.00 kg 10.0 kg F = 45.0 N Figure P5.70 71. The system shown in Figure P5.49 has an acceleration of magnitude 1.50 m/s2. Assume that the coefficient of kinetic friction between block and incline is the same for both inclines. Find (a) the coefficient of kinetic friction and (b) the tension in the string. Additional Problems 72. A black aluminum glider floats on a film of air above a level aluminum air track. Aluminum feels essentially no force in a magnetic field, and air resistance is negligible. A strong magnet is attached to the top of the glider, forming a total mass of 240 g. A piece of scrap iron attached to one end stop on the track attracts the magnet with a force of 0.823 N when the iron and the magnet are separated by 2.50 cm. (a) Find the acceleration of the glider at this instant. (b) The scrap iron is now attached to another green glider, forming total mass 120 g. Find the acceleration of each glider when the gliders are simultaneously released at 2.50-cm separation. 73. A young woman buys an inexpensive used car for stock car racing. It can attain highway speed with an acceleration of 8.40 mi/h · s. By making changes to its engine, she can increase the net horizontal force on the car by 24.0%. With much less expense, she can remove material from the body of the car to decrease its mass by 24.0%. (a) Which of these two changes, if either, will result in the greater increase in the car’s acceleration? (b) If she makes both changes, what acceleration can she attain? 74. Why is the following situation impossible? A book sits on an inclined plane on the surface of the Earth. The angle The Laws of Motion y θ ar ph vx om 81. ys h0 x 80. Q/C ic z .c 76. A 1.00-kg glider on a horizontal air track is pulled by a string at an angle u. The taut string runs over a pulley and is attached to a hanging object of mass 0.500 kg as shown in Figure P5.76. (a) Show that the speed vx of the glider and the speed vy of the hanging object are related by vx 5 uvy, where u 5 z(z 2 2 h 02)–1/2. (b) The glider is released from rest. Show that at that instant the acceleration ax of the glider and the acceleration ay of the hanging object are related by ax 5 uay. (c) Find the tension in the string at the instant the glider is released for h 0 5 80.0 cm and u 5 30.0°. GP S ly 75. Review. A hockey puck struck by a hockey stick is given an S initial speed v in the positive x direction. The coefficient i of kinetic friction between the ice and the puck is mk . (a) Obtain an expression for the acceleration of the puck as it slides across the ice. (b) Use the result of part (a) to obtain an expression for the distance d the puck slides. The answer should be in terms of the variables vi , mk , and g only. 79. to the force on the block as the rope’s mass approaches zero? What can you state about the tension in a light cord joining a pair of moving objects? Two blocks of masses m1 and m 2 are placed on a table in contact with each other as discussed in Example 5.7 and shown in Figure 5.12a. The coefficient of kinetic friction between the block of mass m1 and the table is m1, and that between the block of mass m 2 and the table is m2. A horizontal force of magnitude F is applied to the block of mass m1. We wish to find P, the magnitude of the contact force between the blocks. (a) Draw diagrams showing the forces for each block. (b) What is the net force on the ­system of two blocks? (c) What is the net force acting on m1? (d) What is the net force acting on m2? (e) Write Newton’s second law in the x direction for each block. (f) Solve the two equations in two unknowns for the acceleration a of the blocks in terms of the masses, the applied force F, the coefficients of friction, and g. (g) Find the magnitude P of the contact force between the blocks in terms of the same quantities. On a single, light, vertical cable that does not stretch, a crane is lifting a 1 207-kg Ferrari and, below it, a­ 1 461-kg BMW Z8. The Ferrari is moving upward with speed 3.50 m/s and acceleration 1.25 m/s2. (a) How do the velocity and acceleration of the BMW compare with those of the Ferrari? (b) Find the tension in the cable between the BMW and the Ferrari. (c) Find the tension in the cable above the Ferrari. An inventive child named Nick wants to reach an apple in a tree without climbing the tree. Sitting in a chair connected to a rope that passes over a frictionless pulley (Fig. P5.81), Nick pulls on the loose end of the rope with such a force that the spring scale reads 250 N. Nick’s true weight is 320 N, and the chair weighs 160 N. Nick’s feet are not touching the ground. (a) Draw one pair of diagrams showing the forces for Nick and the chair considered as separate systems and another diagram for Nick and the chair considered as one system. (b) Show that the acceleration of the system is upward and find its magnitude. (c) Find the force Nick exerts on the chair. ee b of the plane with the horizontal is 60.0°. The coefficient of kinetic friction between the book and the plane is 0.300. At time t 5 0, the book is released from rest. The book then slides through a distance of 1.00 m, measured along the plane, in a time interval of 0.483 s. s. w 146 Chapter 5 m sw vy Figure P5.76 w w w .a 77. A frictionless plane is 10.0 m long and inclined at 35.0°. M A sled starts at the bottom with an initial speed of 5.00 m/s up the incline. When the sled reaches the point at which it momentarily stops, a second sled is released from the top of the incline with an initial speed vi . Both sleds reach the bottom of the incline at the same moment. (a) Determine the distance that the first sled traveled up the incline. (b) Determine the initial speed of the second sled. 78. A rope with mass mr is Q/C attached to a block with S mass m as in Figure P5.78. b mb mr S F The block rests on a frictionFigure P5.78 less, horizontal surface. The rope does not stretch. The free end of the rope is pulled to the right with a horiS zontal force F . (a) Draw force diagrams for the rope and the block, noting that the tension in the rope is not uniform. (b) Find the acceleration of the system in terms of mb, mr , and F. (c) Find the magnitude of the force the rope exerts on the block. (d) What happens Figure P5.81 Problems 81 and 82. 82. In the situation described in Problem 81 and Figure Q/C P5.81, the masses of the rope, spring balance, and pul- Problems ley are negligible. Nick’s feet are not touching the ground. (a) Assume Nick is momentarily at rest when he stops pulling down on the rope and passes the end of the rope to another child, of weight 440 N, who is standing on the ground next to him. The rope does not break. Describe the ensuing motion. (b) Instead, assume Nick is momentarily at rest when he ties the end of the rope to a strong hook projecting from the tree trunk. Explain why this action can make the rope break. or the inclined plane, are very simple. Some machines do not even look like machines. For example, your car is stuck in the mud and you can’t pull hard enough to get it out. You do, however, have a long cable that you connect taut between your front bumper and the trunk of a stout tree. You now pull sideways on the cable at its midpoint, exerting a force f. Each half of the cable is displaced through a small angle u from the straight line between the ends of the cable. (a) Deduce an expression for the force acting on the car. (b) Evaluate the cable tension for the case where u 5 7.00° and f 5 100 N. 83. In Example 5.7, we pushed on two blocks on a table. .c om 87. Objects with masses m1 5 10.0 kg and m 2 5 5.00 kg are M connected by a light string that passes over a frictionless pulley as in Figure P5.40. If, when the system starts from rest, m 2 falls 1.00 m in 1.20 s, determine the coefficient of kinetic friction between m1 and the table. ly 88. Consider the three connected objects shown in Figure P5.88. Assume first that the inclined plane is frictionless and that the system is in equilibrium. In terms of m, g, and u, find (a) the mass M and (b) the tensions T1 and T2. Now assume that the value of M is double the value found in part (a). Find (c) the acceleration of each object and (d) the tensions T1 and T2. Next, assume that the coefficient of static friction between m and 2m and the inclined plane is ms and that the system is in equilibrium. Find (e) the maximum value of M and (f) the minimum value of M. (g) Compare the values of T2 when M has its minimum and maximum values. S m1 m2 m3 ar ph F ys ic s. w on a frictionless, horizontal surface as shown in Figure S P5.83. A horizontal force F is applied to m1. Take m1 5 2.00 kg, m 2 5 3.00 kg, m 3 5 4.00 kg, and F 5 18.0 N. (a) Draw a separate free-body diagram for each block. (b) Determine the acceleration of the blocks. (c) Find the resultant force on each block. (d) Find the magnitudes of the contact forces between the blocks. (e) You are working on a construction project. A coworker is nailing up plasterboard on one side of a light partition, and you are on the opposite side, providing “backing” by leaning against the wall with your back pushing on it. Every hammer blow makes your back sting. The supervisor helps you put a heavy block of wood between the wall and your back. Using the situation analyzed in parts (a) through (d) as a model, explain how this change works to make your job more comfortable. ee b Q/C Suppose three blocks are in contact with one another 147 Figure P5.83 sw .a w w w 86. Any device that allows you to increase the force you exert is a kind of machine. Some machines, such as the prybar T1 m 2m M θ 84. An aluminum block of Aluminum Copper mass m1 5 2.00 kg and a m1 copper block of mass m 2 5 m2 6.00 kg are connected by a Steel u light string over a frictionless pulley. They sit on a Figure P5.84 steel surface as shown in Figure P5.84, where u 5 30.0°. (a) When they are released from rest, will they start to move? If they do, determine (b) their acceleration and (c) the tension in the string. If they do not move, determine (d) the sum of the magnitudes of the forces of friction T4 acting on the blocks. 85. An object of mass M S is held in place S by an applied force F and a pulley system as shown in Figure P5.85. The pulleys are massless and frictionless. (a) Draw diagrams showing the forces on each pulley. Find (b) the tension in each section of rope, T1, T2, T3, T4,Sand T5 and (c) the mag­ nitude of F . T2 Figure P5.88 89. AScrate of weight Fg is pushed by a force S P S P on a horizontal floor as shown in Figure P5.89. The coefficient of S static friction is ms, and P is directed at angle u below the horizontal. (a) Show that the minimum value of P Figure P5.89 that will move the crate is given by P5 ms Fg sec u 1 2 ms tan u (b) Find the condition on u in terms of ms for which motion of the crate is impossible for any value of P. 90. A student is asked to measure the acceleration of a Q/C glider on a frictionless, inclined plane, using an air T1 S F T2 T3 T5 M Figure P5.85 track, a stopwatch, and a meterstick. The top of the track is measured to be 1.774 cm higher than the bottom of the track, and the length of the track is d 5 127.1 cm. The cart is released from rest at the top of the incline, taken as x 5 0, and its position x along the incline is measured as a function of time. For x values of 10.0 cm, 20.0 cm, 35.0 cm, 50.0 cm, 75.0 cm, and 100 cm, the measured times at which these positions are reached (averaged over five runs) are 1.02 s, 1.53 s, The Laws of Motion om S .c 96. AS time-dependent force, F 5 1 8.00 i^ 2 4.00t j^ 2 , where F is in newtons and t is in seconds, is exerted on a 2.00-kg object initially at rest. (a) At what time will the object be moving with a speed of 15.0 m/s? (b) How far is the object from its initial position when its speed is 15.0 m/s? (c) Through what total displacement has the object traveled at this moment? 97. The board sandwiched between two other boards in Figure P5.97 weighs 95.5 N. If the coefficient of static friction between the boards is 0.663, what must be the magnitude of the compression forces (assumed horizontal) acting on both sides of the center board to keep it from slipping? w .a sw ar ph ys ic 92. In Figure P5.92, the pulleys Q/C and the cord are light, all surfaces are frictionless, and the m1 cord does not stretch. (a) How does the acceleration of block 1 compare with the acceleration of block 2? Explain your m2 reasoning. (b) The mass of block 2 is 1.30 kg. Find its acceleration as it depends on Figure P5.92 the mass m1 of block 1. (c) What If? What does the result of part (b) predict if m1 is very much less than 1.30 kg? (d) What does the result of part (b) predict if m1 approaches infinity? (e) In this last case, what is the tension in the cord? (f) Could you anticipate the answers to parts (c), (d), and (e) without first doing part (b)? Explain. Challenge Problems ly 91. A flat cushion of mass m is Cushion Q/C released from rest at the corWind ner of the roof of a building, force at height h. A wind blowing along the side of the building h exerts a constant horizontal force of magnitude F on the cushion as it drops as shown in Figure P5.91. The air exerts no vertical force. (a) Show that the path of the cushion is Figure P5.91 a straight line. (b) Does the cushion fall with constant velocity? Explain. (c) If m 5 1.20 kg, h 5 8.00 m, and F 5 2.40 N, how far from the building will the cushion hit the level ground? What If? (d) If the cushion is thrown downward with a nonzero speed at the top of the building, what will be the shape of its trajectory? Explain. 95. A car accelerates down a M hill (Fig. P5.95), going from rest to 30.0 m/s in 6.00 s. A u toy inside the car hangs by u a string from the car’s ceiling. The ball in the figure represents the toy, of mass Figure P5.95 0.100 kg. The acceleration is such that the string remains perpendicular to the ceiling. Determine (a) the angle u and (b) the tension in the string. ee b 2.01 s, 2.64 s, 3.30 s, and 3.75 s, respectively. (a) Construct a graph of x versus t 2, with a best-fit straight line to describe the data. (b) Determine the acceleration of the cart from the slope of this graph. (c) Explain how your answer to part (b) compares with the theoretical value you calculate using a 5 g sin u as derived in Example 5.6. s. w 148 Chapter 5 w w m2 93. What horizontal force must S be applied to a large block m1 of mass M shown in Figure S M F P5.93 so that the tan blocks remain stationary relative to M? Assume all surfaces Figure P5.93 and the pulley are frictionProblems 93 and 98. less. Notice that the force exerted by the string accelerates m 2. 94. An 8.40-kg object slides down a fixed, frictionless, Q/C inclined plane. Use a computer to determine and tabulate (a) the normal force exerted on the object and (b) its acceleration for a series of incline angles (measured from the horizontal) ranging from 0° to 90° in 5° increments. (c) Plot a graph of the normal force and the acceleration as functions of the incline angle. (d) In the limiting cases of 0° and 90°, are your results consistent with the known behavior? Figure P5.97 98. Initially, the system of objects shown in Figure P5.93 is S held motionless. The pulley and all surfaces and wheels S are frictionless. Let the force F be zero and assume that m1 can move only vertically. At the instant after the system of objects is released, find (a) the tension T in the string, (b) the acceleration of m 2, (c) the acceleration of M, and (d) the acceleration of m1. (Note: The pulley accelerates along with the cart.) 99. A block of mass 2.20 kg is accelQ/C erated across a rough surface by a light cord passing over a small pulley as shown in Figure P5.99. The tension T in the T cord is maintained at 10.0 N, x and the pulley is 0.100 m above M the top of the block. The coefficient of kinetic friction is 0.400. (a) Determine the accelFigure P5.99 eration of the block when x 5 0.400 m. (b) Describe the general behavior of the acceleration as the block slides from a location where x is large to x 5 0. (c) Find the maximum value of the acceleration and the position x for which it occurs. (d) Find the value of x for which the acceleration is zero. 149 Problems m u m F M om a S F m .c M ly b Figure P5.103 104. A mobile is formed by supporting four metal butterS flies of equal mass m from a string of length L. The points of support are evenly spaced a distance , apart as shown in Figure P5.104. The string forms an angle u1 with the ceiling at each endpoint. The center section of string is horizontal. (a) Find the tension in each section of string in terms of u1, m, and g. (b) In terms of u1, find the angle u2 that the sections of string between the outside butterflies and the inside butterflies form with the horizontal. (c) Show that the distance D between the endpoints of the string is ys ic H R Problems 101 and 102. ar ph Figure P5.101 L b2 cos u 1 1 2 cos 3 tan21 1 12 tan u 1 2 4 1 1r 5 D5 D   u2 u2  w w u1 u1 .a sw 102. In Figure P5.101, the incline has mass M and is fasS tened to the stationary horizontal tabletop. The block of mass m is placed near the bottom of the incline and is released with a quick push that sets it sliding upward. The block stops near the top of the incline as shown in the figure and then slides down again, always without friction. Find the force that the tabletop exerts on the incline throughout this motion in terms of m, M, g, and u. 103. A block of mass m 5 2.00 kg rests on the left edge of a block of mass M 5 8.00 kg. The coefficient of kinetic friction between the two blocks is 0.300, and the surface on which the 8.00-kg block rests is frictionless. A constant horizontal force of magnitude F 5 10.0 N is applied to the 2.00-kg block, setting it in motion as w L S s. w h shown in Figure P5.103a. If the distance L that the leading edge of the smaller block travels on the larger block is 3.00 m, (a) in what time interval will the smaller block make it to the right side of the 8.00-kg block as shown in Figure P5.103b? (Note: Both blocks S are set into motion when F is applied.) (b) How far does the 8.00-kg block move in the process? ee b 100. Why is the following situation impossible? A 1.30-kg toaster is not plugged in. The coefficient of static friction between the toaster and a horizontal countertop is 0.350. To make the toaster start moving, you carelessly pull on its electric cord. Unfortunately, the cord has become frayed from your previous similar actions and will break if the tension in the cord exceeds 4.00 N. By pulling on the cord at a particular angle, you successfully start the toaster moving without breaking the cord. 101. Review. A block of mass m 5 2.00 kg is released from rest at h 5 0.500 m above the surface of a table, at the top of a u 5 30.0° incline as shown in Figure P5.101. The frictionless incline is fixed on a table of height H 5 2.00 m. (a) Determine the acceleration of the block as it slides down the incline. (b) What is the velocity of the block as it leaves the incline? (c) How far from the table will the block hit the floor? (d) What time interval elapses between when the block is released and when it hits the floor? (e) Does the mass of the block affect any of the above calculations?   m m m L  5 m Figure P5.104 c h a p t e r 6 .c om Circular Motion and Other Applications of Newton’s Laws ee b ly 6.1 Extending the Particle in Uniform Circular Motion Model 6.2 Nonuniform Circular Motion 6.3 Motion in Accelerated Frames sw In the preceding chapter, we introduced Newton’s laws of motion and incorporated them into two analysis models involving linear motion. Now we discuss motion that is slightly more complicated. For example, we shall apply Newton’s laws to objects traveling in circular paths. We shall also discuss motion observed from an accelerating frame of reference and motion of an object through a viscous medium. For the most part, this chapter consists of a series of examples selected to illustrate the application of Newton’s laws to a variety of new circumstances. w w w .a Kyle Busch, driver of the #18 Snickers Toyota, leads Jeff Gordon, driver of the #24 Dupont Chevrolet, during the NASCAR Sprint Cup Series Kobalt Tools 500 at the Atlanta Motor Speedway on March 9, 2008, in Hampton, Georgia. The cars travel on a banked roadway to help them undergo circular motion on the turns. (Chris Graythen/Getty ar ph ys ic s. w 6.4 Motion in the Presence of Resistive Forces Images for NASCAR) 6.1 Extending the Particle in Uniform Circular Motion Model In Section 4.4, we discussed the analysis model of a particle in uniform circular motion, in which a particle moves with constant speed v in a circular path having a radius r. The particle experiences an acceleration that has a magnitude ac 5 150 v2 r 6.1 Extending the Particle in Uniform Circular Motion Model 151 S m S Fr r When the string breaks, the puck moves in the direction tangent to the circle. r S S Fr Figure 6.2 The string holding the puck in its circular path breaks. ee b An overhead view of a puck moving in a circular path in a horizontal plane. .c Figure 6.1 om v ly A force Fr , directed toward the center of the circle, keeps the puck moving in its circular path. sw ar ph ys ic s. w The acceleration is called centripetal acceleration because S a c is directed toward a c is always perpendicular to S v. (If there the center of the circle. Furthermore, S were a component of acceleration parallel to S v , the particle’s speed would be changing.) Let us now extend the particle in uniform circular motion model from Section 4.4 by incorporating the concept of force. Consider a puck of mass m that is tied to a string of length r and moves at constant speed in a horizontal, circular path as illustrated in Figure 6.1. Its weight is supported by a frictionless table, and the string is anchored to a peg at the center of the circular path of the puck. Why does the puck move in a circle? According to Newton’s first law, the puck would move in a straight line if there were no force on it; the string, however, prevents motion S along a straight line by exerting on the puck a radial force F r that makes it follow the circular path. This force is directed along the string toward the center of the circle as shown in Figure 6.1. If Newton’s second law is applied along the radial direction, the net force causing the centripetal acceleration can be related to the acceleration as follows: v2 r (6.1) .a a F 5 ma c 5 m w w w A force causing a centripetal acceleration acts toward the center of the circular path and causes a change in the direction of the velocity vector. If that force should vanish, the object would no longer move in its circular path; instead, it would move along a straight-line path tangent to the circle. This idea is illustrated in Figure 6.2 for the puck moving in a circular path at the end of a string in a horizontal plane. If the string breaks at some instant, the puck moves along the straight-line path that is tangent to the circle at the position of the puck at this instant. Q uick Quiz 6.1 You are riding on a Ferris wheel that is rotating with constant speed. The car in which you are riding always maintains its correct upward orientation; it does not invert. (i) What is the direction of the normal force on you from the seat when you are at the top of the wheel? (a) upward (b) downward (c) impossible to determine (ii) From the same choices, what is the direction of the net force on you when you are at the top of the wheel? WW Force causing centripetal acceleration Pitfall Prevention 6.1 Direction of Travel When the String Is Cut Study Figure 6.2 very carefully. Many students (wrongly) think that the puck will move radially away from the center of the circle when the string is cut. The velocity of the puck is tangent to the circle. By Newton’s first law, the puck continues to move in the same direction in which it is moving just as the force from the string disappears. 152 Chapter 6 Circular Motion and Other Applications of Newton’s Laws Analysis Model   Particle in Uniform Circular Motion (Extension)    The Conical Pendulum S v om • the tension in a string of constant length S ac acting on a rock twirled in a circle • the gravitational force acting on a planet r traveling around the Sun in a perfectly circular orbit (Chapter 13) • the magnetic force acting on a charged particle moving in a uniform magnetic field (Chapter 29) • the electric force acting on an electron in orbit around a nucleus in the Bohr model of the hydrogen atom (Chapter 42) AM ee b Example 6.1 (6.1) S F .c v2 F 5 ma 5 m c a r Examples ly Imagine a moving object that can be modeled as a particle. If it moves in a circular path of radius r at a constant speed v, it experiences a centripetal acceleration. Because the particle is accelerating, there must be a net force acting on the particle. That force is directed toward the center of the circular path and is given by A small ball of mass m is suspended from a string of length L. The ball revolves with constant speed v in a horizontal circle of radius r as shown in Figure 6.3. (Because the string sweeps out the surface of a cone, the system is known as a conical pendulum.) Find an expression for v in terms of the geometry in Figure 6.3. L T sin u ic ys S ar ph S mg mg a b Categorize The ball in Figure 6.3 does not accelerate vertically. Therefore, we model it as a particle in equilibrium in the vertical direction. It experiences a centripetal acceleration in the horizontal direction, so it is modeled as a particle in uniform circular motion in this direction. u r Conceptualize Imagine the motion of the ball in Figure 6.3a and convince yourself that the string sweeps out a cone and that the ball moves in a horizontal circle. T cos u T s. w S o l u ti o n u S Figure 6.3 (Example 6.1) (a) A conical pendulum. The path of the ball is a horizontal circle. (b) The forces acting on the ball. Analyze Let u represent the angle between the string and the vertical. In the diaS sw gram of forces acting on the ball in Figure 6.3b, the force T exerted by the string on the ball is resolved into a vertical component T cos u and a horizontal component T sin u acting toward the center of the circular path. w .a Apply the particle in equilibrium model in the vertical direction: w w Use Equation 6.1 from the particle in uniform circular motion model in the horizontal direction: oF y 5 T cos u 2 mg 5 0 (1) T cos u 5 mg mv 2 (2) a Fx 5 T sin u 5 ma c 5 r v2 rg Divide Equation (2) by Equation (1) and use sin u/cos u 5 tan u: tan u 5 Solve for v: v 5 "rg tan u Incorporate r 5 L sin u from the geometry in Figure 6.3a: v 5 "Lg sin u tan u Finalize Notice that the speed is independent of the mass of the ball. Consider what happens when u goes to 908 so that the string is horizontal. Because the tangent of 908 is infinite, the speed v is infinite, which tells us the string canS not possibly be horizontal. If it were, there would be no vertical component of the force T to balance the gravitational force on the ball. That is why we mentioned in regard to Figure 6.1 that the puck’s weight in the figure is supported by a frictionless table. 153 6.1 Extending the Particle in Uniform Circular Motion Model Example 6.2 AM    How Fast Can It Spin? A puck of mass 0.500 kg is attached to the end of a cord 1.50 m long. The puck moves in a horizontal circle as shown in Figure 6.1. If the cord can withstand a maximum tension of 50.0 N, what is the maximum speed at which the puck can move before the cord breaks? Assume the string remains horizontal during the motion. S o l u ti o n Conceptualize It makes sense that the stronger the cord, the faster the puck can move before the cord breaks. Also, we expect a more massive puck to break the cord at a lower speed. (Imagine whirling a bowling ball on the cord!) ation into Newton’s second law as described by Equation 6.1: v2 r Solve for v: (1) v 5 Find the maximum speed the puck can have, which corresponds to the maximum tension the string can withstand: v max 5 Tr Åm .c T5m ly Analyze Incorporate the tension and the centripetal acceler- om Categorize Because the puck moves in a circular path, we model it as a particle in uniform circular motion. ee b 1 50.0 N 2 1 1.50 m 2 Tmaxr 5 5 12.2 m/s Å m Å 0.500 kg Finalize Equation (1) shows that v increases with T and decreases with larger m, as we expected from our conceptual- s. w ization of the problem. Suppose the puck moves in a circle of larger radius at the same speed v. Is the cord more likely or less likely to break? ic W h at I f ? Example 6.3 ar ph ys Answer The larger radius means that the change in the direction of the velocity vector will be smaller in a given time interval. Therefore, the acceleration is smaller and the required tension in the string is smaller. As a result, the string is less likely to break when the puck travels in a circle of larger radius.    What Is the Maximum Speed of the Car? AM S fs w .a sw A 1 500-kg car moving on a flat, horizontal road negotiates a curve as shown in Figure 6.4a. If the radius of the curve is 35.0 m and the coefficient of static friction between the tires and dry pavement is 0.523, find the maximum speed the car can have and still make the turn successfully. S o l u ti o n w w Conceptualize Imagine that the curved roadway is part of a large circle so that the car is moving in a circular path. a S n Categorize Based on the Conceptualize step of the problem, we model the car as a particle in uniform circular motion in the horizontal direction. The car is not accelerating vertically, so it is modeled as a particle in equilibrium in the vertical direction. S fs Analyze Figure 6.4b shows the forces on the car. The force that enables the car to remain in its circular path is the force of static friction. (It is static because no slipping occurs at the point of contact between road and tires. If this force of static friction were zero—for example, if the car were on an icy road—the car would continue in a straight line and slide off the curved road.) The maximum speed v max the car can have around the curve is the speed at which it is on the verge of skidding outward. At this point, the friction force has its maximum value fs,max 5 msn. S mg b Figure 6.4 (Example 6.3) (a) The force of static friction directed toward the center of the curve keeps the car moving in a circular path. (b) The forces acting on the car. continued 154 Chapter 6 Circular Motion and Other Applications of Newton’s Laws ▸ 6.3 c o n t i n u e d 2 v max r Apply Equation 6.1 from the particle in uniform circular motion model in the radial direction for the maximum speed condition: (1) fs,max 5 msn 5 m Apply the particle in equilibrium model to the car in the vertical direction: oF Solve Equation (1) for the maximum speed and substitute for n: (2) v max 5 Substitute numerical values: v max 5 " 1 0.523 2 1 9.80 m/s2 2 1 35.0 m 2 5 13.4 m/s 5 0 S n 2 mg 5 0 S n 5 mg msmgr msnr 5 5 "ms gr Å m Å m om y .c Finalize This speed is equivalent to 30.0 mi/h. Therefore, if the speed limit on this roadway is higher than 30 mi/h, this roadway could benefit greatly from some banking, as in the next example! Notice that the maximum speed does not depend on the mass of the car, which is why curved highways do not need multiple speed limits to cover the various masses of vehicles using the road. ee b ly W h at I f ? Suppose a car travels this curve on a wet day and begins to skid on the curve when its speed reaches only 8.00 m/s. What can we say about the coefficient of static friction in this case? ms 5 2 1 8.00 m/s 2 2 v max 5 0.187 5 gr 1 9.80 m/s2 2 1 35.0 m 2 ys ms 5 2 v max gr ic Substituting the numerical values gives s. w Answer The coefficient of static friction between the tires and a wet road should be smaller than that between the tires and a dry road. This expectation is consistent with experience with driving because a skid is more likely on a wet road than a dry road. To check our suspicion, we can solve Equation (2) for the coefficient of static friction:    The Banked Roadway AM sw Example 6.4 ar ph which is indeed smaller than the coefficient of 0.523 for the dry road. S n u ny w w w .a A civil engineer wishes to redesign the curved roadway in Example 6.3 in such a way that a car will not have to rely on friction to round the curve without skidding. In other words, a car moving at the designated speed can negotiate the curve even when the road is covered with ice. Such a road is usually banked, which means that the roadway is tilted toward the inside of the curve as seen in the opening photograph for this chapter. Suppose the designated speed for the road is to be 13.4 m/s (30.0 mi/h) and the radius of the curve is 35.0 m. At what angle should the curve be banked? nx S o l u ti o n Conceptualize The difference between this example and Example 6.3 is that the car is no longer moving on a flat roadway. Figure 6.5 shows the banked roadway, with the center of the circular path of the car far to the left of the figure. Notice that the horizontal component of the normal force participates in causing the car’s centripetal acceleration. Categorize As in Example 6.3, the car is modeled as a particle in equilibrium in the vertical direction and a particle in uniform circular motion in the horizontal direction. Analyze On a level (unbanked) road, the force that causes the centripetal acceleration is the force of static friction between tires and the road as we saw in the preceding example. If the road is banked at an angle u as in Figure 6.5, however, the u Figure 6.5 S Fg (Example 6.4) A car moves into the page and is rounding a curve on a road banked at an angle u to the horizontal. When friction is neglected, the force that causes the centripetal acceleration and keeps the car moving in its circular path is the horizontal component of the normal force. 155 6.1 Extending the Particle in Uniform Circular Motion Model ▸ 6.4 c o n t i n u e d normal force S n has a horizontal component toward the center of the curve. Because the road is to be designed so that the force of static friction is zero, the component nx 5 n sin u is the only force that causes the centripetal acceleration. Write Newton’s second law for the car in the radial direction, which is the x direction: mv 2 (1) a Fr 5 n sin u 5 r oF 5 n cos u 2 mg 5 0 Apply the particle in equilibrium model to the car in the vertical direction: (2) n cos u 5 mg Divide Equation (1) by Equation (2): (3) tan u 5 Solve for the angle u: u 5 tan21 c om v2 rg 1 13.4 m/s 2 2 d 5 27.68 1 35.0 m 2 1 9.80 m/s2 2 .c y ly Finalize Equation (3) shows that the banking angle is independent of the mass of the vehicle negotiating the curve. If a s. w ee b car rounds the curve at a speed less than 13.4 m/s, the centripetal acceleration decreases. Therefore, the normal force, which is unchanged, is sufficient to cause two accelerations: the lower centripetal acceleration and an acceleration of the car down the inclined roadway. Consequently, an additional friction force parallel to the roadway and upward is needed to keep the car from sliding down the bank (to the left in Fig. 6.5). Similarly, a driver attempting to negotiate the curve at a speed greater than 13.4 m/s has to depend on friction to keep from sliding up the bank (to the right in Fig. 6.5). Imagine that this same roadway were built on Mars in the future to connect different colony centers. Could it be traveled at the same speed? W h at I f ? Example 6.5 sw ar ph ys ic Answer The reduced gravitational force on Mars would mean that the car is not pressed as tightly to the roadway. The reduced normal force results in a smaller component of the normal force toward the center of the circle. This smaller component would not be sufficient to provide the centripetal acceleration associated with the original speed. The centripetal acceleration must be reduced, which can be done by reducing the speed v. Mathematically, notice that Equation (3) shows that the speed v is proportional to the square root of g for a roadway of fixed radius r banked at a fixed angle u. Therefore, if g is smaller, as it is on Mars, the speed v with which the roadway can be safely traveled is also smaller.    Riding the Ferris Wheel AM Top S v A child of mass m rides on a Ferris wheel as shown in Figure 6.6a. The child moves in a vertical circle of radius 10.0 m at a constant speed of 3.00 m/s. .a S nbot S w ntop R w w (A) Determine the force exerted by the seat on the child at the bottom of the ride. Express your answer in terms of the weight of the child, mg. S o l u ti o n Conceptualize Look carefully at Figure 6.6a. Based S v S mg S mg on experiences you may have had on a Ferris wheel or Bottom c driving over small hills on a roadway, you would expect a b to feel lighter at the top of the path. Similarly, you Figure 6.6 (Example 6.5) (a) A child rides on a Ferris wheel. would expect to feel heavier at the bottom of the path. (b) The forces acting on the child at the bottom of the path. At both the bottom of the path and the top, the nor(c) The forces acting on the child at the top of the path. mal and gravitational forces on the child act in opposite directions. The vector sum of these two forces gives a force of constant magnitude that keeps the child moving in a circular path at a constant speed. To yield net force vectors with the same magnitude, the normal force at the bottom must be greater than that at the top. continued 156 Chapter 6 Circular Motion and Other Applications of Newton’s Laws ▸ 6.5 c o n t i n u e d Categorize Because the speed of the child is constant, we can categorize this problem as one involving a particle (the child) in uniform circular motion, complicated by the gravitational force acting at all times on the child. Analyze We draw a diagram of forces acting on the child at the bottom of the ride as shown in Figure 6.6b. The only S Substitute numerical values given for the speed and radius: n bot 5 mg c 1 1 v2 v2 5 mg a1 1 b r rg 1 3.00 m/s 2 2 d 1 10.0 m 2 1 9.80 m/s2 2 .c n bot 5 mg 1 m ly Solve for the force exerted by the seat on the child: om forces acting on him are the downward gravitational force F g 5 mS n bot exerted by the seat. g and the upward force S The net upward force on the child that provides his centripetal acceleration has a magnitude n bot 2 mg. v2 F 5 n 2 mg 5 m Using the particle in uniform circular motion model, bot a r apply Newton’s second law to the child in the radial direction when he is at the bottom of the ride: ee b 5 1.09 mg n bot exerted by the seat on the child is greater than the weight of the child by a factor Hence, the magnitude of the force S of 1.09. So, the child experiences an apparent weight that is greater than his true weight by a factor of 1.09. s. w (B) Determine the force exerted by the seat on the child at the top of the ride. S o l u ti o n ic Analyze The diagram of forces acting on the child at the top of the ride is shown in Figure 6.6c. The net downward force that provides the centripetal acceleration has a magnitude mg 2 n top. v2 a F 5 mg 2 n top 5 m r ys ar ph Apply Newton’s second law to the child at this position: n top 5 mg 2 m Substitute numerical values: n top 5 mg c 1 2 sw Solve for the force exerted by the seat on the child: v2 v2 5 mg a1 2 b r rg 1 3.00 m/s 2 2 d 1 10.0 m 2 1 9.80 m/s2 2 5 0.908 mg .a In this case, the magnitude of the force exerted by the seat on the child is less than his true weight by a factor of 0.908, and the child feels lighter. w Finalize The variations in the normal force are consistent with our prediction in the Conceptualize step of the problem. w w Suppose a defect in the Ferris wheel mechanism causes the speed of the child to increase to 10.0 m/s. What does the child experience at the top of the ride in this case? W h at I f ? Answer If the calculation above is performed with v 5 10.0 m/s, the magnitude of the normal force at the top of the ride is negative, which is impossible. We interpret it to mean that the required centripetal acceleration of the child is larger than that due to gravity. As a result, the child will lose contact with the seat and will only stay in his circular path if there is a safety bar or a seat belt that provides a downward force on him to keep him in his seat. At the bottom of the ride, the normal force is 2.02 mg, which would be uncomfortable. 6.2 Nonuniform Circular Motion In Chapter 4, we found that if a particle moves with varying speed in a circular path, there is, in addition to the radial component of acceleration, a tangential component having magnitude udv/dtu. Therefore, the force acting on the particle 157 6.2 Nonuniform Circular Motion The net force exerted on the particle is the vector sum of the radial force and the tangential force. Figure 6.7 When the net force acting on a particle moving in a circular path has a tangential component o F t , the particle’s speed changes. S F S om  Fr S  Ft ee b ly .c must also have a tangential and a radial component. Because the totalSacceleraS S tion is S a 5S ar 1 S a t , the total force exerted on the particle is g F 5 g Fr 1 g Ft as shown in Figure 6.7. (We express the radial and tangential forces as net forces with the summation notation Sbecause each force could consist of multiple forces that combine.) The vector g Fr is directed toward the center of the circle and is S responsible for the centripetal acceleration. The vector g Ft tangent to the circle is responsible for the tangential acceleration, which represents a change in the particle’s speed with time. s. w A B ic Q uick Quiz 6.2 A bead slides at constant speed along a curved wire lying on a horizontal surface as shown in Figure 6.8. (a) Draw the vectors representing the force exerted by the wire on the bead at points A, B, and C. (b) Suppose the bead in Figure 6.8 speeds up with constant tangential acceleration as it moves toward the right. Draw the vectors representing the force on the bead at points A, B, and C. Example 6.6 ar ph ys C    Keep Your Eye on the Ball Figure 6.8 (Quick Quiz 6.2) A bead slides along a curved wire. AM sw A small sphere of mass m is attached to the end of a cord of length R and set into motion in a vertical circle about a fixed point O as illustrated in Figure 6.9. Determine the tangential acceleration of the sphere and the tension in the cord at any instant when the speed of the sphere is v and the cord makes an angle u with the vertical. S vtop S .a Ttop S mg w R w S o l u ti o n O w Conceptualize Compare the motion of the sphere in Figure 6.9 with that of the child in Figure 6.6a associated with Example 6.5. Both objects travel in a circular path. Unlike the child in Example 6.5, however, the speed of the sphere is not uniform in this example because, at most points along the path, a tangential component of acceleration arises from the gravitational force exerted on the sphere. Categorize We model the sphere as a particle under a net force and moving in a circular path, but it is not a particle in uniform circular motion. We need to use the techniques discussed in this section on nonuniform circular motion. Analyze From the force diagram in Figure 6.9, we see that the only forces acting on the sphere are the gravitational force S T mg cos u u u S Tbot S vbot mg sin u S mg S mg Figure 6.9 (Example 6.6) The forces acting on a sphere of mass m connected to a cord of length R and rotating in a vertical circle centered at O. Forces acting on the sphere are shown when the sphere is at the top and bottom of the circle and at an arbitrary location. 158 Chapter 6 Circular Motion and Other Applications of Newton’s Laws ▸ 6.6 c o n t i n u e d S S S Fg 5 mS g exerted by the Earth and the force T exerted by the cord. We resolve Fg into a tangential component mg sin u and a radial component mg cos u. o F 5 mg sin u 5 ma From the particle under a net force model, apply Newton’s second law to the sphere in the tangential direction: t t at 5 g sin u a Fr 5 T 2 mg cos u 5 v2 1 cos ub Rg Ttop 5 mg a 2 v top Rg 2 1b Tbot 5 mg a 2 v bot 1 1b Rg ly Finalize Let us evaluate this result at the top and bottom of the circular path (Fig. 6.9): om T 5 mg a mv 2 R .c Apply Newton’s second law to the forces acting on the sphere S in the radial direction, noting that both T and S a r are directed toward O. As noted in Section 4.5, we can use Equation 4.14 for the centripetal acceleration of a particle even when it moves in a circular path in nonuniform motion: s. w What if the ball is set in motion with a slower speed? ic W h at I f ? ee b These results have similar mathematical forms as those for the normal forces n top and n bot on the child in Example 6.5, which is consistent with the normal force on the child playing a similar physical role in Example 6.5 as the tension in the string plays in this example. Keep in mind, however, that the normal force S n on the child in Example 6.5 S is always upward, whereas the force T in this example changes direction because it must always point inward along the string. Also note that v in the expressions above varies for different positions of the sphere, as indicated by the subscripts, whereas v in Example 6.5 is constant. ys (A) What speed would the ball have as it passes over the top of the circle if the tension in the cord goes to zero instantaneously at this point? ar ph Answer Let us set the tension equal to zero in the expression for Ttop: 0 5 mg a 2 v top Rg 2 1b S v top 5 "gR (B) What if the ball is set in motion such that the speed at the top is less than this value? What happens? w w w .a sw Answer In this case, the ball never reaches the top of the circle. At some point on the way up, the tension in the string goes to zero and the ball becomes a projectile. It follows a segment of a parabolic path over the top of its motion, rejoining the circular path on the other side when the tension becomes nonzero again. 6.3 Motion in Accelerated Frames Newton’s laws of motion, which we introduced in Chapter 5, describe observations that are made in an inertial frame of reference. In this section, we analyze how Newton’s laws are applied by an observer in a noninertial frame of reference, that is, one that is accelerating. For example, recall the discussion of the air hockey table on a train in Section 5.2. The train moving at constant velocity represents an inertial frame. An observer on the train sees the puck at rest remain at rest, and Newton’s first law appears to be obeyed. The accelerating train is not an inertial frame. According to you as the observer on this train, there appears to be no force on the puck, yet it accelerates from rest toward the back of the train, appearing to violate Newton’s first law. This property is a general property of observations made in noninertial frames: there appear to be unexplained accelerations of objects that are not “fastened” to the frame. Newton’s first law is not violated, of course. It only appears to be violated because of observations made from a noninertial frame. On the accelerating train, as you watch the puck accelerating toward the back of the train, you might conclude based on your belief in Newton’s second law that a 6.3 Motion in Accelerated Frames a From the passenger’s frame of reference, a force appears to push her toward the right door, but it is a fictitious force. Fictitious force b Relative to the reference frame of the Earth, the car seat applies a real force (friction) toward the left on the passenger, causing her to change direction along with the rest of the car. w w w .a sw ar ph ys ic s. w ee b ly .c om force has acted on the puck to cause it to accelerate. We call an apparent force such as this one a fictitious force because it is not a real force and is due only to observations made in an accelerated reference frame. A fictitious force appears to act on an object in the same way as a real force. Real forces are always interactions between two objects, however, and you cannot identify a second object for a fictitious force. (What second object is interacting with the puck to cause it to accelerate?) In general, simple fictitious forces appear to act in the direction opposite that of the acceleration of the noninertial frame. For example, the train accelerates forward and there appears to be a fictitious force causing the puck to slide toward the back of the train. The train example describes a fictitious force due to a change in the train’s speed. Another fictitious force is due to the change in the direction of the velocity vector. To understand the motion of a system that is noninertial because of a change in direction, consider a car traveling along a highway at a high speed and approaching a curved exit ramp on the left as shown in Figure 6.10a. As the car takes the sharp left turn on the ramp, a person sitting in the passenger seat leans or slides to the right and hits the door. At that point the force exerted by the door on the passenger keeps her from being ejected from the car. What causes her to move toward the door? A popular but incorrect explanation is that a force acting toward the right in Figure 6.10b pushes the passenger outward from the center of the circular path. Although often called the “centrifugal force,” it is a fictitious force. The car represents a noninertial reference frame that has a centripetal ­acceleration toward the center of its circular path. As a result, the passenger feels an apparent force which is outward from the center of the circular path, or to the right in Figure 6.10b, in the direction opposite that of the acceleration. Let us address this phenomenon in terms of Newton’s laws. Before the car enters the ramp, the passenger is moving in a straight-line path. As the car enters the ramp and travels a curved path, the passenger tends to move along the original straight-line path, which is in accordance with Newton’s first law: the natural tendency of an object is to continue moving in a straight line. If a sufficiently large force (toward the center of curvature) acts on the passenger as in Figure 6.10c, however, she moves in a curved path along with the car. This force is the force of friction between her and the car seat. If this friction force is not large enough, the seat follows a curved path while the passenger tends to continue in the straight-line path of the car before the car began the turn. Therefore, from the point of view of an observer in the car, the passenger leans or slides to the right relative to the seat. Eventually, she encounters the door, which provides a force large enough to enable her to follow the same curved path as the car. Another interesting fictitious force is the “Coriolis force.” It is an apparent force caused by changing the radial position of an object in a rotating coordinate system. For example, suppose you and a friend are on opposite sides of a rotating circular platform and you decide to throw a baseball to your friend. Figure 6.11a on page 160 represents what an observer would see if the ball is viewed while the observer is hovering at rest above the rotating platform. According to this observer, who is in an inertial frame, the ball follows a straight line as it must according to Newton’s first law. At t 5 0 you throw the ball toward your friend, but by the time tf when the ball has crossed the platform, your friend has moved to a new position and can’t catch the ball. Now, however, consider the situation from your friend’s viewpoint. Your friend is in a noninertial reference frame because he is undergoing a centripetal acceleration relative to the inertial frame of the Earth’s surface. He starts off seeing the baseball coming toward him, but as it crosses the platform, it veers to one side as shown in Figure 6.11b. Therefore, your friend on the rotating platform states that the ball does not obey Newton’s first law and claims that a sideways force is causing the ball to follow a curved path. This fictitious force is called the Coriolis force. Fictitious forces may not be real forces, but they can have real effects. An object on your dashboard really slides off if you press the accelerator of your car. As you ride on a merry-go-round, you feel pushed toward the outside as if due to the fictitious “centrifugal force.” You are likely to fall over and injure yourself due to the 159 Real force c Figure 6.10 (a) A car approaching a curved exit ramp. What causes a passenger in the front seat to move toward the righthand door? (b) Passenger’s frame of reference. (c) Reference frame of the Earth. 160 Chapter 6 Circular Motion and Other Applications of Newton’s Laws By the time tf that the ball arrives at the other side of the platform, your friend is no longer there to catch it. According to this observer, the ball follows a straight-line path, consistent with Newton’s laws. Friend at t0 From your friend’s point of view, the ball veers to one side during its flight. Your friend introduces a fictitious force to explain this deviation from the expected path. Ball at t  tf You at t  tf ly .c om Friend at t  tf Ball at t0 a ee b You at t0 b Figure 6.11 ic s. w You and your friend stand at the edge of a rotating circular platform. You throw the ball at t 5 0 in the direction of your friend. (a) Overhead view observed by someone in an inertial reference frame attached to the Earth. The ground appears stationary, and the platform rotates clockwise. (b) Overhead view observed by someone in an inertial reference frame attached to the platform. The platform appears stationary, and the ground rotates counterclockwise. Pitfall Prevention 6.2 ar ph ys Coriolis force if you walk along a radial line while a merry-go-round rotates. (One of the authors did so and suffered a separation of the ligaments from his ribs when he fell over.) The Coriolis force due to the rotation of the Earth is responsible for rotations of hurricanes and for large-scale ocean currents. sw Q uick Quiz 6.3 Consider the passenger in the car making a left turn in Figure 6.10. Which of the following is correct about forces in the horizontal direction if she is making contact with the right-hand door? (a) The passenger is in equilibrium between real forces acting to the right and real forces acting to the left. (b) The passenger is subject only to real forces acting to the right. (c) The passenger is subject only to real forces acting to the left. (d) None of those statements is true.    Fictitious Forces in Linear Motion AM w Example 6.7 w .a Centrifugal Force The commonly heard phrase “centrifugal force” is described as a force pulling outward on an object moving in a circular path. If you are feeling a “centrifugal force” on a rotating carnival ride, what is the other object with which you are interacting? You cannot identify another object because it is a fictitious force that occurs when you are in a noninertial reference frame. w A small sphere of mass m hangs by a cord from the ceiling of a boxcar that is accelerating to the right as shown in Figure 6.12. Both the inertial observer on the ground in Figure 6.12a and the noninertial observer on the train in Figure 6.12b agree that the cord makes an angle u with respect to the vertical. The noninertial observer claims that a force, which we know to be fictitious, causes the observed deviation of the cord from the vertical. How is the magnitude of this force related to the boxcar’s acceleration measured by the inertial observer in Figure 6.12a? S o l u ti o n Conceptualize Place yourself in the role of each of the two observers in Figure 6.12. As the inertial observer on the ground, you see the boxcar accelerating and know that the deviation of the cord is due to this acceleration. As the noninertial observer on the boxcar, imagine that you ignore any effects of the car’s motion so that you are not aware of its acceleration. Because you are unaware of this acceleration, you claim that a force is pushing sideways on the sphere to cause the deviation of the cord from the vertical. To make the conceptualization more real, try running from rest while holding a hanging object on a string and notice that the string is at an angle to the vertical while you are accelerating, as if a force is pushing the object backward. 6.4 Motion in the Presence of Resistive Forces 161 ▸ 6.7 c o n t i n u e d A noninertial observer riding in in thethe carcar says that thethe netnet A noninertial observer riding says that force onon thethe sphere is zero and that thethe deflection of of thethe force sphere is zero and that deflection S S cord from thethe vertical is due to to a fictitious force F cord from vertical is due a fictitious force F fictitious fictitious S S that balances thethe horizontal component of of T. T. that balances horizontal component AnAn inertial observer at at rest outside thethe carcar claims that thethe inertial observer rest outside claims that acceleration ofS thethe sphere is provided byby thethe horizontal acceleration of sphere is provided horizontal S component of of T. T. component S S a a Inertial Inertial observer observer S S mg mg a a T Tu u S S mgmg .c b b (Example 6.7) A small sphere suspended from the ceiling of a boxcar accelerating to the right is deflected as shown. ly Figure 6.12 Noninertial Noninertial observer observer S S S S Ffictitious Ffictitious om S S T Tu u ee b Categorize For the inertial observer, we model the sphere as a particle under a net force in the horizontal direction and a particle in equilibrium in the vertical direction. For the noninertial observer, the sphere is modeled as a particle in equilibrium in both directions. S s. w Analyze According to the inertial observer at rest (Fig. 6.12a), the forces on the sphere are the force T exerted by the Inertial observer ys For this observer, apply the particle under a net force and particle in equilibrium models: 789 ic cord and the gravitational force. The inertial observer concludes that the sphere’s acceleration is the same as that of S the boxcar and that this acceleration is provided by the horizontal component of T. (1) (2) oF oF x 5 T sin u 5 ma y 5 T cos u 2 mg 5 0 sw Apply the particle in equilibrium model for this observer in both directions: Noninertial observer 789 ar ph According to the noninertial observer riding in the car (Fig. 6.12b), the cord also makes an angle u with the vertical; to that observer, however, the sphere is at rest and so its acceleration is zero. Therefore, the noninertial observer introduces a force (which we know to be fictitious) in the horizontal direction to balance the horizontal component of S T and claims that the net force on the sphere is zero. a F xr 5T sin u 2 Ffictitious 5 0 a F yr 5T cos u 2 mg 5 0 w .a These expressions are equivalent to Equations (1) and (2) if F fictitious 5 ma , where a is the acceleration according to the inertial observer. w Finalize If we make this substitution in the equation for o F xr above, we obtain the same mathematical results as the inertial observer. The physical interpretation of the cord’s deflection, however, differs in the two frames of reference. w W h at I f ? Suppose the inertial observer wants to measure the acceleration of the train by means of the pendulum (the sphere hanging from the cord). How could she do so? Answer Our intuition tells us that the angle u the cord makes with the vertical should increase as the acceleration increases. By solving Equations (1) and (2) simultaneously for a, we find that a 5 g tan u. Therefore, the inertial observer can determine the magnitude of the car’s acceleration by measuring the angle u and using that relationship. Because the deflection of the cord from the vertical serves as a measure of acceleration, a simple pendulum can be used as an accelerometer. 6.4 Motion in the Presence of Resistive Forces In Chapter 5, we described the force of kinetic friction exerted on an object moving on some surface. We completely ignored any interaction between the object and the medium through which it moves. Now consider the effect of that medium, which 162 Chapter 6 Circular Motion and Other Applications of Newton’s Laws S s. w ee b ly .c om can be either a liquid or a gas. The medium exerts a resistive force R on the object moving through it. Some examples are the air resistance associated with moving vehicles (sometimes called air drag) and S the viscous forces that act on objects moving through a liquid. The magnitude of R depends on factors such as the speed of S the object, and the direction of R is always opposite the direction of the object’s motion relative to the medium. This direction may or may not be in the direction opposite the object’s velocity according to the observer. For example, if a marble is dropped into a bottle of shampoo, the marble moves downward and the resistive force is upward, resisting the falling of the marble. In contrast, imagine the moment at which there is no wind and you are looking at a flag hanging limply on a flagpole. When a breeze begins to blow toward the right, the flag moves toward the right. In this case, the drag force on the flag from the moving air is to the right and the motion of the flag in response is also to the right, the same direction as the drag force. Because the air moves toward the right with respect to the flag, the flag moves to the left relative to the air. Therefore, the direction of the drag force is indeed opposite to the direction of the motion of the flag with respect to the air! The magnitude of the resistive force can depend on speed in a complex way, and here we consider only two simplified models. In the first model, we assume the resistive force is proportional to the velocity of the moving object; this model is valid for objects falling slowly through a liquid and for very small objects, such as dust particles, moving through air. In the second model, we assume a resistive force that is proportional to the square of the speed of the moving object; large objects, such as skydivers moving through air in free fall, experience such a force. Model 1: Resistive Force Proportional to Object Velocity ys ic If we model the resistive force acting on an object moving through a liquid or gas as proportional to the object’s velocity, the resistive force can be expressed as S sw ar ph R 5 2bS v (6.2) where b is a constant whose value depends on the properties of the medium and on the shape and dimensions of the object and S v is the velocity of the object relative to S the medium. The negative sign indicates that R is in the opposite direction to S v. Consider a small sphere of mass m released from rest in a liquid as in Figure 6.13a. S Assuming the only forces acting on the sphere are the resistive force R 5 2bS v and S the gravitational force Fg , let us describe its motion.1 We model the sphere as a parv0 ag w .a The sphere approaches a maximum (or terminal) speed vT . w v vT S w R S v v  vT a0 S mg Figure 6.13 (a) A small sphere falling through a liquid. (b) A motion diagram of the sphere as it falls. Velocity vectors (red) and acceleration vectors (violet) are shown for each image after the first one. (c) A speed–time graph for the sphere. 0.632vT t a 1A b c t The time constant t is the time at which the sphere reaches a speed of 0.632vT . buoyant force is also acting on the submerged object. This force is constant, and its magnitude is equal to the weight of the displaced liquid. This force can be modeled by changing the apparent weight of the sphere by a constant factor, so we will ignore the force here. We will discuss buoyant forces in Chapter 14. 6.4 Motion in the Presence of Resistive Forces 163 ticle under a net force. Applying Newton’s second law to the vertical motion of the sphere and choosing the downward direction to be positive, we obtain oF y (6.3) 5 ma S mg 2 bv 5 ma om where the acceleration of the sphere is downward. Noting that the acceleration a is equal to dv/dt gives dv b 5g2 v (6.4) m dt This equation is called a differential equation, and the methods of solving it may not be familiar to you as yet. Notice, however, that initially when v 5 0, the magnitude of the resistive force is also zero and the acceleration of the sphere is simply g. As t increases, the magnitude of the resistive force increases and the acceleration decreases. The acceleration approaches zero when the magnitude of the resistive force approaches the sphere’s weight so that the net force on the sphere is zero. In this situation, the speed of the sphere approaches its terminal speed v T . The terminal speed is obtained from Equation 6.4 by setting dv/dt 5 0, which gives mg mg 2 bv T 5 0 or v T 5 (6.5) b Because you may not be familiar with differential equations yet, we won’t show the details of the process that gives the expression for v for all times t. If v 5 0 at t 5 0, this expression is mg 1 1 2 e2bt/m 2 5 v T 1 1 2 e2t/t 2 v5 (6.6) b This function is plotted in Figure 6.13c. The symbol e represents the base of the natural logarithm and is also called Euler’s number: e 5 2.718 28. The time constant t 5 m/b (Greek letter tau) is the time at which the sphere released from rest at t 5 0 reaches 63.2% of its terminal speed; when t 5 t, Equation 6.6 yields v 5 0.632vT . (The number 0.632 is 1 2 e21.) We can check that Equation 6.6 is a solution to Equation 6.4 by direct differentiation: ar ph ys ic s. w ee b ly .c WW Terminal speed mg dv d mg b 1 1 2 e2bt/m 2 d 5 5 c a0 1 e2bt/m b 5 ge2bt/m m dt dt b b w .a sw (See Appendix Table B.4 for the derivative of e raised to some power.) Substituting into Equation 6.4 both this expression for dv/dt and the expression for v given by Equation 6.6 shows that our solution satisfies the differential equation.    Sphere Falling in Oil AM w Example 6.8 w A small sphere of mass 2.00 g is released from rest in a large vessel filled with oil, where it experiences a resistive force proportional to its speed. The sphere reaches a terminal speed of 5.00 cm/s. Determine the time constant t and the time at which the sphere reaches 90.0% of its terminal speed. S o l u ti o n Conceptualize With the help of Figure 6.13, imagine dropping the sphere into the oil and watching it sink to the bottom of the vessel. If you have some thick shampoo in a clear container, drop a marble in it and observe the motion of the marble. Categorize We model the sphere as a particle under a net force, with one of the forces being a resistive force that depends on the speed of the sphere. This model leads to the result in Equation 6.5. Analyze From Equation 6.5, evaluate the coefficient b: b5 mg vT continued 164 Chapter 6 Circular Motion and Other Applications of Newton’s Laws ▸ 6.8 c o n t i n u e d vt vt m 5 ma b 5 mg g b Evaluate the time constant t: t5 Substitute numerical values: t5 Find the time t at which the sphere reaches a speed of 0.900v T by setting v 5 0.900v T in Equation 6.6 and solving for t : 0.900v T 5 v T (1 2 e2t/t) 5.00 cm/s 5 5.10 3 1023 s 980 cm/s2 1 2 e2t/t 5 0.900 t 2 5 ln 1 0.100 2 5 22.30 t om e2t/t 5 0.100 .c t 5 2.30t 5 2.30(5.10 3 1023 s) 5 11.7 3 1023 s ly 5 11.7 ms s. w ee b Finalize The sphere reaches 90.0% of its terminal speed in a very short time interval. You should have also seen this behavior if you performed the activity with the marble and the shampoo. Because of the short time interval required to reach terminal velocity, you may not have noticed the time interval at all. The marble may have appeared to immediately begin moving through the shampoo at a constant velocity. Model 2: Resistive Force Proportional to Object Speed Squared ar ph ys ic For objects moving at high speeds through air, such as airplanes, skydivers, cars, and baseballs, the resistive force is reasonably well modeled as proportional to the square of the speed. In these situations, the magnitude of the resistive force can be expressed as R 5 12 DrAv 2 (6.7) .a sw where D is a dimensionless empirical quantity called the drag coefficient, r is the density of air, and A is the cross-sectional area of the moving object measured in a plane perpendicular to its velocity. The drag coefficient has a value of about 0.5 for spherical objects but can have a value as great as 2 for irregularly shaped objects. Let us analyze the motion of a falling object subject to an upward air resistive force of magnitude R 5 12 DrAv 2. Suppose an object of mass m is released from rest. 2 the downward As Figure 6.14 shows, the object experiences two external forces: S S S gravitational force Fg 5 m g and the upward resistive force R . Hence, the magnitude of the net force is S S w S w S R w R v S S mg a vT mg b Figure 6.14 (a) An object falling through air experiences S a resistive force R and a graviS tational force Fg 5 mS g . (b) The object reaches terminal speed when the net force acting on it is S S zero, that is, when R 5 2 Fg or R 5 mg. 1 2 a F 5 mg 2 2 DrAv (6.8) where we have taken downward to be the positive vertical direction. Modeling the object as a particle under a net force, with the net force given by Equation 6.8, we find that the object has a downward acceleration of magnitude DrA 2 (6.9) a5g2a bv 2m We can calculate the terminal speed v T by noticing that when the gravitational force is balanced by the resistive force, the net force on the object is zero and therefore its acceleration is zero. Setting a 5 0 in Equation 6.9 gives g2a 2 As DrA bv 2 5 0 2m T with Model 1, there is also an upward buoyant force that we neglect. 6.4 Motion in the Presence of Resistive Forces Table 6.1 165 Terminal Speed for Various Objects Falling Through Air Mass (kg) Cross-Sectional Area (m 2) vT (m/s) 75 0.145 0.046 4.8 3 1024 3.4 3 1025 0.70 4.2 3 1023 1.4 3 1023 7.9 3 1025 1.3 3 1025 60 43 44 14 9.0 Object Skydiver Baseball (radius 3.7 cm) Golf ball (radius 2.1 cm) Hailstone (radius 0.50 cm) Raindrop (radius 0.20 cm) vT 5 2mg om so (6.10) Å DrA .c Table 6.1 lists the terminal speeds for several objects falling through air. Conceptual Example 6.9   The Skysurfer s. w ee b ly Q uick Quiz 6.4 A baseball and a basketball, having the same mass, are dropped through air from rest such that their bottoms are initially at the same height above the ground, on the order of 1 m or more. Which one strikes the ground first? (a) The baseball strikes the ground first. (b) The basketball strikes the ground first. (c) Both strike the ground at the same time. ys ic Consider a skysurfer (Fig. 6.15) who jumps from a plane with his feet attached firmly to his surfboard, does some tricks, and then opens his parachute. Describe the forces acting on him during these maneuvers. ar ph S o l u ti o n w w w .a Oliver Furrer/Jupiter Images sw When the surfer first steps out of the plane, he has no vertical velocity. The downward gravitational force causes him to accelerate toward the ground. As his downward speed increases, so does the upward resistive force exerted by the air on his body and the board. This upward force reduces their acceleration, and so their speed increases more slowly. Eventually, they are going so fast that the upward resistive force matches the downward gravitational force. Now the net force is zero and they no longer accelerate, but instead reach their terminal speed. At some point after reaching terminal speed, he opens his parachute, resulting in a drastic increase in the upward resistive force. The net force (and therefore the acceleration) is now upward, in the direction opposite the direction of the velocity. The downward velocity therefore decreases rapidly, and the Figure 6.15 (Conceptual Example resistive force on the parachute also decreases. Eventually, the upward resistive 6.9) A skysurfer. force and the downward gravitational force balance each other again and a much smaller terminal speed is reached, permitting a safe landing. (Contrary to popular belief, the velocity vector of a skydiver never points upward. You may have seen a video in which a skydiver appears to “rocket” upward once the parachute opens. In fact, what happens is that the skydiver slows down but the person holding the camera continues falling at high speed.) Example 6.10    Falling Coffee Filters AM The dependence of resistive force on the square of the speed is a simplification model. Let’s test the model for a specific situation. Imagine an experiment in which we drop a series of bowl-shaped, pleated coffee filters and measure their terminal speeds. Table 6.2 on page 166 presents typical terminal speed data from a real experiment using these coffee filters as continued 166 Chapter 6 Circular Motion and Other Applications of Newton’s Laws ▸ 6.10 c o n t i n u e d they fall through the air. The time constant t is small, so a dropped filter quickly reaches terminal speed. Each filter has a mass of 1.64 g. When the filters are nested together, they combine in such a way that the front-facing surface area does not increase. Determine the relationship between the resistive force exerted by the air and the speed of the falling filters. S o l u ti o n Conceptualize Imagine dropping the coffee filters through the air. (If you have some coffee filters, try dropping them.) Because of the relatively small mass of the coffee filter, you probably won’t notice the time interval during which there is an acceleration. The filters will appear to fall at constant velocity immediately upon leaving your hand. om Categorize Because a filter moves at constant velocity, we model it as a particle in equilibrium. Analyze At terminal speed, the upward resistive force on the filter balances the downward gravitational force so that R 5 mg. .c 1 000 g b 1 9.80 m/s2 2 5 0.016 1 N Table 6.2 ee b Terminal Speed and Resistive Force for Nested Coffee Filters Number of Filters vT (m/s)a R (N) 1.01 1.40 1.63 2.00 2.25 2.40 2.57 2.80 3.05 3.22 0.016 1 0.032 2 0.048 3 0.064 4 0.080 5 0.096 6 0.112 7 0.128 8 0.144 9 0.161 0 ys ic s. w Likewise, two filters nested together experience 0.032 2 N of resistive force, and so forth. These values of resistive force are shown in the far right column of Table 6.2. A graph of the resistive force on the filters as a function of terminal speed is shown in Figure 6.16a. A straight line is not a good fit, indicating that the resistive force is not proportional to the speed. The behavior is more clearly seen in Figure 6.16b, in which the resistive force is plotted as a function of the square of the terminal speed. This graph indicates that the resistive force is proportional to the square of the speed as suggested by Equation 6.7. 1 kg ly R 5 mg 5 1 1.64 g 2 a Evaluate the magnitude of the resistive force: sw w .a Resistive force (N) The data points do not lie along a straight line, but instead suggest a curve. w 0.18 0.16 0.14 0.12 0.10 0.08 0.06 0.04 0.02 0.00 w Resistive force (N) ar ph Finalize Here is a good opportunity for you to take some actual data at home on real coffee filters and see if you can reproduce the results shown in Figure 6.16. If you have shampoo and a marble as mentioned in Example 6.8, take data on that system too and see if the resistive force is appropriately modeled as being proportional to the speed. 0 1 2 3 4 0.18 0.16 0.14 0.12 0.10 0.08 0.06 0.04 0.02 0.00 a All values of vT are approximate. The fit of the straight line to the data points indicates that the resistive force is proportional to the terminal speed squared. 0 2 4 6 8 10 12 Terminal speed squared (m/s)2 Terminal speed (m/s) a   1   2   3   4   5   6   7   8   9 10 b Figure 6.16 (Example 6.10) (a) Relationship between the resistive force acting on falling coffee filters and their terminal speed. (b) Graph relating the resistive force to the square of the terminal speed. Example 6.11    Resistive Force Exerted on a Baseball AM A pitcher hurls a 0.145-kg baseball past a batter at 40.2 m/s (5 90 mi/h). Find the resistive force acting on the ball at this speed.    Summary 167 ▸ 6.11 c o n t i n u e d S o l u ti o n Conceptualize This example is different from the previous ones in that the object is now moving horizontally through the air instead of moving vertically under the influence of gravity and the resistive force. The resistive force causes the ball to slow down, and gravity causes its trajectory to curve downward. We simplify the situation by assuming the velocity vector is exactly horizontal at the instant it is traveling at 40.2 m/s. Categorize In general, the ball is a particle under a net force. Because we are considering only one instant of time, however, we are not concerned about acceleration, so the problem involves only finding the value of one of the forces. that we drop the baseball and allow it to reach terminal speed. Solve Equation 6.10 for D: 2mg om D5 v T2rA .c Analyze To determine the drag coefficient D, imagine 1 2mg v 2 a 2 brAv 2 5 mg a b vt 2 v t rA R 5 12DrAv 2 5 Substitute numerical values, using the terminal speed from Table 6.1: R 5 1 0.145 kg 2 1 9.80 m/s2 2 a ee b ly Use this expression for D in Equation 6.7 to find an expression for the magnitude of the resistive force: 40.2 m/s 2 b 5 1.2 N 43 m/s s. w Finalize The magnitude of the resistive force is similar in magnitude to the weight of the baseball, which is about ys ic 1.4 N. Therefore, air resistance plays a major role in the motion of the ball, as evidenced by the variety of curve balls, floaters, sinkers, and the like thrown by baseball pitchers. ar ph Summary sw Concepts and Principles An object moving through a liquid or gas experiences a speed-dependent resistive force. This resistive force is in a direction opposite that of the velocity of the object relative to the medium and generally increases with speed. The magnitude of the resistive force depends on the object’s size and shape and on the properties of the medium through which the object is moving. In the limiting case for a falling object, when the magnitude of the resistive force equals the object’s weight, the object reaches its terminal speed. w .a A particle moving in uniform circular motion has a centripetal acceleration; this acceleration must be provided by a net force directed toward the center of the circular path. w w An observer in a noninertial (accelerating) frame of reference introduces fictitious forces when applying Newton’s second law in that frame. Analysis Model for Problem-Solving Particle in Uniform Circular Motion (Extension) With our new knowledge of forces, we can extend the model of a particle in uniform circular motion, first introduced in Chapter 4. Newton’s second law applied to a particle moving in uniform circular motion states that the net force causing the particle to undergo a centripetal acceleration (Eq. 4.14) is related to the acceleration according to v a F 5 ma c 5 m r 2 (6.1) S F S ac r S v 168 Chapter 6 Circular Motion and Other Applications of Newton’s Laws 1. denotes answer available in Student Solutions Manual/Study Guide 2. Consider a skydiver who has stepped from a helicopter and is falling through air. Before she reaches terminal speed and long before she opens her parachute, does her speed (a) increase, (b) decrease, or (c) stay constant? om .c ly 5. As a raindrop falls through the atmosphere, its speed initially changes as it falls toward the Earth. Before the raindrop reaches its terminal speed, does the magnitude of its acceleration (a) increase, (b) decrease, (c) stay constant at zero, (d) stay constant at 9.80 m/s2, or (e) stay constant at some other value? 6. An office door is given a sharp push and swings open against a pneumatic device that slows the door down and then reverses its motion. At the moment the door is open the widest, (a) does the doorknob have a centripetal acceleration? (b) Does it have a tangential acceleration? ic 3. A door in a hospital has a pneumatic closer that pulls the door shut such that the doorknob moves with constant speed over most of its path. In this part of its motion, (a) does the doorknob experience a centripetal acceleration? (b) Does it experience a tangential acceleration? direction of its total acceleration at this point? (b) Of these points, is there a point where the bob has nonzero tangential acceleration and zero radial acceleration? If so, which point? What is the direction of its total acceleraA tion at this point? (c) Is there a B C point where the bob has no acceleration? If so, which point? (d) Is Figure OQ6.4 there a point where the bob has both nonzero tangential and radial acceleration? If so, which point? What is the direction of its total acceleration at this point? ee b 1. A child is practicing for a BMX race. His B speed remains conN stant as he goes counA C terclockwise around W E a level track with two S straight sections and D E two nearly semicircular sections as shown in Figure OQ6.1 the aerial view of Figure OQ6.1. (a) Rank the magnitudes of his acceleration at the points A, B, C, D, and E from largest to smallest. If his acceleration is the same size at two points, display that fact in your ranking. If his acceleration is zero, display that fact. (b) What are the directions of his velocity at points A, B, and C ? For each point, choose one: north, south, east, west, or nonexistent. (c) What are the directions of his acceleration at points A, B, and C ? s. w Objective Questions ar ph ys 7. Before takeoff on an airplane, an inquisitive student on the plane dangles an iPod by its earphone wire. It hangs straight down as the plane is at rest waiting to take off. The plane then gains speed rapidly as it moves down the runway. (i) Relative to the student’s hand, does the iPod (a) shift toward the front of the plane, (b) continue to hang straight down, or (c) shift toward the back of the plane? (ii) The speed of the plane increases at a constant rate over a time interval of several seconds. During this interval, does the angle the earphone wire makes with the vertical (a) increase, (b) stay constant, or (c) decrease? w .a sw 4. A pendulum consists of a small object called a bob hanging from a light cord of fixed length, with the top end of the cord fixed, as represented in Figure OQ6.4. The bob moves without friction, swinging equally high on both sides. It moves from its turning point A through point B and reaches its maximum speed at point C. (a) Of these points, is there a point where the bob has nonzero radial acceleration and zero tangential acceleration? If so, which point? What is the w Conceptual Questions 1. denotes answer available in Student Solutions Manual/Study Guide tion is constant in magnitude at all times and parallel to the velocity. 2. A falling skydiver reaches terminal speed with her parachute closed. After the parachute is opened, what parameters change to decrease this terminal speed? 5. The observer in the accelerating elevator of Example 5.8 would claim that the “weight” of the fish is T, the scale reading, but this answer is obviously wrong. Why does this observation differ from that of a person outside the elevator, at rest with respect to the Earth? 6. If someone told you that astronauts are weightless in orbit because they are beyond the pull of gravity, would you accept the statement? Explain. 7. It has been suggested that rotating cylinders about 20 km in length and 8 km in diameter be placed in w 1. What forces cause (a) an automobile, (b) a propellerdriven airplane, and (c) a rowboat to move? 3. An object executes circular motion with constant speed whenever a net force of constant magnitude acts perpendicular to the velocity. What happens to the speed if the force is not perpendicular to the velocity? 4. Describe the path of a moving body in the event that (a) its acceleration is constant in magnitude at all times and perpendicular to the velocity, and (b) its accelera- Problems 10. A pail of water can be whirled in a vertical path such that no water is spilled. Why does the water stay in the pail, even when the pail is above your head? 11. “If the current position and velocity of every particle in the Universe were known, together with the laws describing the forces that particles exert on one another, the whole future of the Universe could be calculated. The future is determinate and preordained. Free will is an illusion.” Do you agree with this thesis? Argue for or against it. om space and used as colonies. The purpose of the rotation is to simulate gravity for the inhabitants. Explain this concept for producing an effective imitation of gravity. 8. Consider a small raindrop and a large raindrop falling through the atmosphere. (a) Compare their terminal speeds. (b) What are their accelerations when they reach terminal speed? 9. Why does a pilot tend to black out when pulling out of a steep dive? Problems AMT   Analysis Model tutorial available in .c Enhanced WebAssign GP Guided Problem ly The problems found in this chapter may be assigned online in Enhanced WebAssign 1. straightforward; 2. intermediate; 3. challenging M   Master It tutorial available in Enhanced 1. full solution available in the Student W   Watch It video solution available in WebAssign BIO ee b Solutions Manual/Study Guide 169 Enhanced WebAssign Q/C Section 6.1 Extending the Particle in Uniform Circular Motion Model What is the total horizontal force on the driver if the speed on the same curve is 18.0 m/s instead? 5. In a cyclotron (one type of particle accelerator), a deuteron (of mass 2.00 u) reaches a final speed of 10.0% of the speed of light while moving in a circular path of radius 0.480 m. What magnitude of magnetic force is required to maintain the deuteron in a circular path? ic 1. A light string can s. w S m ys AMT support a stationM ary hanging load r .a sw ar ph of 25.0 kg before breaking. An object of mass m 5 3.00 kg attached to the string rotates on a frictionless, horizontal table in a circle of radius r 5 0.800 m, and Figure P6.1 the other end of the string is held fixed as in Figure P6.1. What range of speeds can the object have before the string breaks? w w w 2. Whenever two Apollo astronauts were on the surface of the Moon, a third astronaut orbited the Moon. Assume the orbit to be circular and 100 km above the surface of the Moon, where the acceleration due to gravity is 1.52 m/s2. The radius of the Moon is 1.70 3 106 m. Determine (a) the astronaut’s orbital speed and (b) the period of the orbit. 3. In the Bohr model of the hydrogen atom, an electron moves in a circular path around a proton. The speed of the electron is approximately 2.20 3 10 6 m/s. Find (a) the force acting on the electron as it revolves in a circular orbit of radius 0.529 3 10210 m and (b) the centripetal acceleration of the electron. 4. A curve in a road forms part of a horizontal circle. As a car goes around it at constant speed 14.0 m/s, the total horizontal force on the driver has magnitude 130 N. y 6. A car initially traveling eastward turns north by W traveling in a circular path at uniform speed x as shown in Figure P6.6. 35.0 C O The length of the arc ABC is 235 m, and the B car completes the turn in 36.0 s. (a) What is the A acceleration when the car is at B located at an Figure P6.6 angle of 35.08? Express your answer in term