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  1. Preface
  2. Unit 1. Mechanics
    1. 1 Units and Measurement
      1. Introduction
      2. 1.1 The Scope and Scale of Physics
      3. 1.2 Units and Standards
      4. 1.3 Unit Conversion
      5. 1.4 Dimensional Analysis
      6. 1.5 Estimates and Fermi Calculations
      7. 1.6 Significant Figures
      8. 1.7 Solving Problems in Physics
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    2. 2 Vectors
      1. Introduction
      2. 2.1 Scalars and Vectors
      3. 2.2 Coordinate Systems and Components of a Vector
      4. 2.3 Algebra of Vectors
      5. 2.4 Products of Vectors
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    3. 3 Motion Along a Straight Line
      1. Introduction
      2. 3.1 Position, Displacement, and Average Velocity
      3. 3.2 Instantaneous Velocity and Speed
      4. 3.3 Average and Instantaneous Acceleration
      5. 3.4 Motion with Constant Acceleration
      6. 3.5 Free Fall
      7. 3.6 Finding Velocity and Displacement from Acceleration
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    4. 4 Motion in Two and Three Dimensions
      1. Introduction
      2. 4.1 Displacement and Velocity Vectors
      3. 4.2 Acceleration Vector
      4. 4.3 Projectile Motion
      5. 4.4 Uniform Circular Motion
      6. 4.5 Relative Motion in One and Two Dimensions
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    5. 5 Newton's Laws of Motion
      1. Introduction
      2. 5.1 Forces
      3. 5.2 Newton's First Law
      4. 5.3 Newton's Second Law
      5. 5.4 Mass and Weight
      6. 5.5 Newton’s Third Law
      7. 5.6 Common Forces
      8. 5.7 Drawing Free-Body Diagrams
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    6. 6 Applications of Newton's Laws
      1. Introduction
      2. 6.1 Solving Problems with Newton’s Laws
      3. 6.2 Friction
      4. 6.3 Centripetal Force
      5. 6.4 Drag Force and Terminal Speed
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    7. 7 Work and Kinetic Energy
      1. Introduction
      2. 7.1 Work
      3. 7.2 Kinetic Energy
      4. 7.3 Work-Energy Theorem
      5. 7.4 Power
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    8. 8 Potential Energy and Conservation of Energy
      1. Introduction
      2. 8.1 Potential Energy of a System
      3. 8.2 Conservative and Non-Conservative Forces
      4. 8.3 Conservation of Energy
      5. 8.4 Potential Energy Diagrams and Stability
      6. 8.5 Sources of Energy
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
    9. 9 Linear Momentum and Collisions
      1. Introduction
      2. 9.1 Linear Momentum
      3. 9.2 Impulse and Collisions
      4. 9.3 Conservation of Linear Momentum
      5. 9.4 Types of Collisions
      6. 9.5 Collisions in Multiple Dimensions
      7. 9.6 Center of Mass
      8. 9.7 Rocket Propulsion
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    10. 10 Fixed-Axis Rotation
      1. Introduction
      2. 10.1 Rotational Variables
      3. 10.2 Rotation with Constant Angular Acceleration
      4. 10.3 Relating Angular and Translational Quantities
      5. 10.4 Moment of Inertia and Rotational Kinetic Energy
      6. 10.5 Calculating Moments of Inertia
      7. 10.6 Torque
      8. 10.7 Newton’s Second Law for Rotation
      9. 10.8 Work and Power for Rotational Motion
      10. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    11. 11 Angular Momentum
      1. Introduction
      2. 11.1 Rolling Motion
      3. 11.2 Angular Momentum
      4. 11.3 Conservation of Angular Momentum
      5. 11.4 Precession of a Gyroscope
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    12. 12 Static Equilibrium and Elasticity
      1. Introduction
      2. 12.1 Conditions for Static Equilibrium
      3. 12.2 Examples of Static Equilibrium
      4. 12.3 Stress, Strain, and Elastic Modulus
      5. 12.4 Elasticity and Plasticity
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    13. 13 Gravitation
      1. Introduction
      2. 13.1 Newton's Law of Universal Gravitation
      3. 13.2 Gravitation Near Earth's Surface
      4. 13.3 Gravitational Potential Energy and Total Energy
      5. 13.4 Satellite Orbits and Energy
      6. 13.5 Kepler's Laws of Planetary Motion
      7. 13.6 Tidal Forces
      8. 13.7 Einstein's Theory of Gravity
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    14. 14 Fluid Mechanics
      1. Introduction
      2. 14.1 Fluids, Density, and Pressure
      3. 14.2 Measuring Pressure
      4. 14.3 Pascal's Principle and Hydraulics
      5. 14.4 Archimedes’ Principle and Buoyancy
      6. 14.5 Fluid Dynamics
      7. 14.6 Bernoulli’s Equation
      8. 14.7 Viscosity and Turbulence
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  3. Unit 2. Waves and Acoustics
    1. 15 Oscillations
      1. Introduction
      2. 15.1 Simple Harmonic Motion
      3. 15.2 Energy in Simple Harmonic Motion
      4. 15.3 Comparing Simple Harmonic Motion and Circular Motion
      5. 15.4 Pendulums
      6. 15.5 Damped Oscillations
      7. 15.6 Forced Oscillations
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    2. 16 Waves
      1. Introduction
      2. 16.1 Traveling Waves
      3. 16.2 Mathematics of Waves
      4. 16.3 Wave Speed on a Stretched String
      5. 16.4 Energy and Power of a Wave
      6. 16.5 Interference of Waves
      7. 16.6 Standing Waves and Resonance
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    3. 17 Sound
      1. Introduction
      2. 17.1 Sound Waves
      3. 17.2 Speed of Sound
      4. 17.3 Sound Intensity
      5. 17.4 Normal Modes of a Standing Sound Wave
      6. 17.5 Sources of Musical Sound
      7. 17.6 Beats
      8. 17.7 The Doppler Effect
      9. 17.8 Shock Waves
      10. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  4. A | Units
  5. B | Conversion Factors
  6. C | Fundamental Constants
  7. D | Astronomical Data
  8. E | Mathematical Formulas
  9. F | Chemistry
  10. G | The Greek Alphabet
  11. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
    8. Chapter 8
    9. Chapter 9
    10. Chapter 10
    11. Chapter 11
    12. Chapter 12
    13. Chapter 13
    14. Chapter 14
    15. Chapter 15
    16. Chapter 16
    17. Chapter 17
  12. Index

Check Your Understanding

9.1

To reach a final speed of vf=14(3.0×108m/s)vf=14(3.0×108m/s) at an acceleration of 10g, the time
required is
10g=vfΔtΔt=vf10g14(3.0×108m/s)10g=7.7×105s=8.9d10g=vfΔtΔt=vf10g14(3.0×108m/s)10g=7.7×105s=8.9d

9.2

If the phone bounces up with approximately the same initial speed as its impact speed, the change in momentum of the phone will be Δp=mΔv(mΔv)=2mΔvΔp=mΔv(mΔv)=2mΔv. This is twice the momentum change than when the phone does not bounce, so the impulse-momentum theorem tells us that more force must be applied to the phone.

9.3

If the smaller cart were rolling at 1.33 m/s to the left, then conservation of momentum gives
(m1+m2)vf=m1v1i^m2v2i^vf=(m1v1m2v2m1+m2)i^=[(0.675kg)(0.75m/s)(0.500kg)(1.33m/s)1.175kg]i^=(0.135m/s)i^(m1+m2)vf=m1v1i^m2v2i^vf=(m1v1m2v2m1+m2)i^=[(0.675kg)(0.75m/s)(0.500kg)(1.33m/s)1.175kg]i^=(0.135m/s)i^
Thus, the final velocity is 0.135 m/s to the left.

9.4

If the ball does not bounce, its final momentum p2p2 is zero, so
Δp=p2p1=(0)j^(−1.4kg·m/s)j^=+(1.4kg·m/s)j^Δp=p2p1=(0)j^(−1.4kg·m/s)j^=+(1.4kg·m/s)j^

9.5

Consider the impulse momentum theory, which is J=ΔpJ=Δp. If J=0J=0, we have the situation described in the example. If a force acts on the system, then J=FaveΔtJ=FaveΔt. Thus, instead of pf=pipf=pi, we have
FaveΔt=Δp=pfpiFaveΔt=Δp=pfpi
where FaveFave is the force due to friction.

9.6

The impulse is the change in momentum multiplied by the time required for the change to occur. By conservation of momentum, the changes in momentum of the probe and the comment are of the same magnitude, but in opposite directions, and the interaction time for each is also the same. Therefore, the impulse each receives is of the same magnitude, but in opposite directions. Because they act in opposite directions, the impulses are not the same. As for the impulse, the force on each body acts in opposite directions, so the forces on each are not equal. However, the change in kinetic energy differs for each, because the collision is not elastic.

9.7

This solution represents the case in which no interaction takes place: the first puck misses the second puck and continues on with a velocity of 2.5 m/s to the left. This case offers no meaningful physical insights.

9.8

If zero friction acts on the car, then it will continue to slide indefinitely (dd), so we cannot use the work-kinetic-energy theorem as is done in the example. Thus, we could not solve the problem from the information given.

9.9

Were the initial velocities not at right angles, then one or both of the velocities would have to be expressed in component form. The mathematical analysis of the problem would be slightly more involved, but the physical result would not change.

9.10

The volume of a scuba tank is about 11 L. Assuming air is an ideal gas, the number of gas molecules in the tank is
PV=NRTN=PVRT=(2500psi)(0.011m3)(8.31J/mol·K)(300K)(6894.8Pa1psi)=7.59×101molPV=NRTN=PVRT=(2500psi)(0.011m3)(8.31J/mol·K)(300K)(6894.8Pa1psi)=7.59×101mol
The average molecular mass of air is 29 g/mol, so the mass of air contained in the tank is about 2.2 kg. This is about 10 times less than the mass of the tank, so it is safe to neglect it. Also, the initial force of the air pressure is roughly proportional to the surface area of each piece, which is in turn proportional to the mass of each piece (assuming uniform thickness). Thus, the initial acceleration of each piece would change very little if we explicitly consider the air.

9.11

The average radius of Earth’s orbit around the Sun is 1.496×109m1.496×109m. Taking the Sun to be the origin, and noting that the mass of the Sun is approximately the same as the masses of the Sun, Earth, and Moon combined, the center of mass of the Earth + Moon system and the Sun is
RCM=mSunRSun+memRemmSun=(1.989×1030kg)(0)+(5.97×1024kg+7.36×1022kg)(1.496×109m)1.989×1030kg=4.6kmRCM=mSunRSun+memRemmSun=(1.989×1030kg)(0)+(5.97×1024kg+7.36×1022kg)(1.496×109m)1.989×1030kg=4.6km
Thus, the center of mass of the Sun, Earth, Moon system is 4.6 km from the center of the Sun.

9.12

On a macroscopic scale, the size of a unit cell is negligible and the crystal mass may be considered to be distributed homogeneously throughout the crystal. Thus,
rCM=1Mj=1Nmjrj=1Mj=1Nmrj=mMj=1Nrj=NmMj=1NrjNrCM=1Mj=1Nmjrj=1Mj=1Nmrj=mMj=1Nrj=NmMj=1NrjN
where we sum over the number N of unit cells in the crystal and m is the mass of a unit cell. Because Nm = M, we can write
rCM=mMj=1Nrj=NmMj=1NrjN=1Nj=1Nrj.rCM=mMj=1Nrj=NmMj=1NrjN=1Nj=1Nrj.
This is the definition of the geometric center of the crystal, so the center of mass is at the same point as the geometric center.

9.13

The explosions would essentially be spherically symmetric, because gravity would not act to distort the trajectories of the expanding projectiles.

9.14

The notation mgmg stands for the mass of the fuel and m stands for the mass of the rocket plus the initial mass of the fuel. Note that mgmg changes with time, so we write it as mg(t)mg(t). Using mRmR as the mass of the rocket with no fuel, the total mass of the rocket plus fuel is m=mR+mg(t)m=mR+mg(t). Differentiation with respect to time gives
dmdt=dmRdt+dmg(t)dt=dmg(t)dtdmdt=dmRdt+dmg(t)dt=dmg(t)dt
where we used dmRdt=0dmRdt=0 because the mass of the rocket does not change. Thus, time rate of change of the mass of the rocket is the same as that of the fuel.

Conceptual Questions

1.

Since K=p2/2mK=p2/2m, then if the momentum is fixed, the object with smaller mass has more kinetic energy.

3.

Yes; impulse is the force applied multiplied by the time during which it is applied (J=FΔtJ=FΔt), so if a small force acts for a long time, it may result in a larger impulse than a large force acting for a small time.

5.

By friction, the road exerts a horizontal force on the tires of the car, which changes the momentum of the car.

7.

Momentum is conserved when the mass of the system of interest remains constant during the interaction in question and when no net external force acts on the system during the interaction.

9.

To accelerate air molecules in the direction of motion of the car, the car must exert a force on these molecules by Newton’s second law F=dp/dtF=dp/dt. By Newton’s third law, the air molecules exert a force of equal magnitude but in the opposite direction on the car. This force acts in the direction opposite the motion of the car and constitutes the force due to air resistance.

11.

No, he is not a closed system because a net nonzero external force acts on him in the form of the starting blocks pushing on his feet.

13.

Yes, all the kinetic energy can be lost if the two masses come to rest due to the collision (i.e., they stick together).

15.

The angle between the directions must be 90°. Any system that has zero net external force in one direction and nonzero net external force in a perpendicular direction will satisfy these conditions.

17.

Yes, the rocket speed can exceed the exhaust speed of the gases it ejects. The thrust of the rocket does not depend on the relative speeds of the gases and rocket, it simply depends on conservation of momentum.

Problems

19.

a. magnitude: 25kg·m/s;25kg·m/s; b. same as a.

21.

1.78×1029kg·m/s1.78×1029kg·m/s

23.

1.3×109kg·m/s1.3×109kg·m/s

25.

a. 1.50×106N1.50×106N; b. 1.00×105N1.00×105N

27.

4.69×105N4.69×105N

29.

2.10×103N2.10×103N

31.

p(t)=(10i^+20tj^)kg·m/sp(t)=(10i^+20tj^)kg·m/s;F=(20N)j^F=(20N)j^

33.

Let the positive x-axis be in the direction of the original momentum. Then px=1.5kg·m/spx=1.5kg·m/s and py=7.5kg·m/spy=7.5kg·m/s

35.

(0.122m/s)i^(0.122m/s)i^

37.

a. 47 m/s in the bullet to block direction; b.70.6N·s70.6N·s, toward the bullet; c. 70.6N·s70.6N·s, toward the block; d. magnitude is 2.35×104N2.35×104N

39.

2:5

41.

5.9 m/s

43.

a. 6.80 m/s, 5.33°; b. yes (calculate the ratio of the initial and final kinetic energies)

45.

2.5 cm

47.

the speed of the leading bumper car is 6.00 m/s and that of the trailing bumper car is 5.60 m/s

49.

6.6%

51.

1.8 m/s

53.

22.1 m/s at 32.2°32.2° below the horizontal

55.

a. 33 m/s and 110 m/s; b. 57 m; c. 480 m

57.

(732m/s)i^+(−79.6m/s)j^(732m/s)i^+(−79.6m/s)j^

59.

(0.21m/s)i^+(0.25m/s)j^(0.21m/s)i^+(0.25m/s)j^

61.

341 m/s at 86.8° with respect to the i^i^ axis.

63.

With the origin defined to be at the position of the 150-g mass, xCM=−1.23cmxCM=−1.23cm and yCM=0.69cmyCM=0.69cm

65.

yCM={h214gt2,t<Th12gt214gT2+12gtT,tTyCM={h214gt2,t<Th12gt214gT2+12gtT,tT

67.

a. R1=4mR1=4m, R2=2mR2=2m; b. XCM=m1x1+m2x2m1+m2,YCM=m1y1+m2y2m1+m2XCM=m1x1+m2x2m1+m2,YCM=m1y1+m2y2m1+m2; c. yes, with R=1m1+m216m12+4m22R=1m1+m216m12+4m22

69.

xcm=34L(ρ1+ρ0ρ1+2ρ0)xcm=34L(ρ1+ρ0ρ1+2ρ0)

71.

(2a3,2b3)(2a3,2b3)

73.

(xCM,yCM,zCM)=(0,0,h/4)(xCM,yCM,zCM)=(0,0,h/4)

75.

(xCM,yCM,zCM)=(0,4R/(3π),0)(xCM,yCM,zCM)=(0,4R/(3π),0)

77.

(a) 0.413 m/s, (b) about 0.2 J

79.

1551 kg

81.

4.9 km/s

Additional Problems

84.

the elephant has a higher momentum

86.

Answers may vary. The first clause is true, but the second clause is not true in general because the velocity of an object with small mass may be large enough so that the momentum of the object is greater than that of a larger-mass object with a smaller velocity.

88.

4.5×103N4.5×103N

90.

J=0τ[mgmg(1ebt/m)]dt=m2bg(ebτ/m1)J=0τ[mgmg(1ebt/m)]dt=m2bg(ebτ/m1)

92.

a. (2.1×103kg·m/s)i^(2.1×103kg·m/s)i^, b. (24×103N)i^(24×103N)i^

94.

a. (1.1×103kg·m/s)i^(1.1×103kg·m/s)i^, b. (0.010kg·m/s)i^(0.010kg·m/s)i^, c. (0.00093m/s)i^(0.00093m/s)i^, d. (0.0012m/s)i^(0.0012m/s)i^

96.

(7.2m/s)i^(7.2m/s)i^

98.

v1,f=v1,im1m2m1+m2,v2,f=v1,i2m1m1+m2v1,f=v1,im1m2m1+m2,v2,f=v1,i2m1m1+m2

100.

2.8 m/s

102.

0.094 m/s

104.

final velocity of cue ball is (0.76m/s)i^(0.76m/s)i^, final velocities of the other two balls are 2.6 m/s at ±30° with respect to the initial velocity of the cue ball

106.

ball 1: (1.4m/s)i^(0.4m/s)j^(1.4m/s)i^(0.4m/s)j^, ball 2: (2.2m/s)i^+(2.4m/s)j^(2.2m/s)i^+(2.4m/s)j^

108.

ball 1: (1.4m/s)i^(1.7m/s)j^(1.4m/s)i^(1.7m/s)j^, ball 2: (2.8m/s)i^+(0.012m/s)j^(2.8m/s)i^+(0.012m/s)j^

110.

(r,θ)=(2R/3,π/8)(r,θ)=(2R/3,π/8)

112.

Answers may vary. The rocket is propelled forward not by the gasses pushing against the surface of Earth, but by conservation of momentum. The momentum of the gas being expelled out the back of the rocket must be compensated by an increase in the forward momentum of the rocket.

Challenge Problems

114.

a. 617N·s617N·s, 108°; b. Fx=2.91×104NFx=2.91×104N, Fy=2.6×105NFy=2.6×105N; c. Fx=5850NFx=5850N, Fy=5265NFy=5265N

116.

Conservation of momentum demands m1v1,i+m2v2,i=m1v1,f+m2v2,fm1v1,i+m2v2,i=m1v1,f+m2v2,f. We are given that m1=m2m1=m2, v1,i=v2,fv1,i=v2,f, and v2,i=v1,f=0v2,i=v1,f=0. Combining these equations with the equation given by conservation of momentum gives v1,i=v1,iv1,i=v1,i, which is true, so conservation of momentum is satisfied. Conservation of energy demands 12m1v1,i2+12m2v2,i2=12m1v1,f2+12m2v2,f212m1v1,i2+12m2v2,i2=12m1v1,f2+12m2v2,f2. Again combining this equation with the conditions given above give v1,i=v1,iv1,i=v1,i, so conservation of energy is satisfied.

118.

Assume origin on centerline and at floor, then (xCM,yCM)=(0,86cm)(xCM,yCM)=(0,86cm)

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