Download Chapter 7: Linear Momentum and Collisions

Chapter 7: Linear Momentum and Collisions
Answers and Solutions
1.
Picture the Problem: A baseball pitcher gives the ball momentum when he throws it.
Strategy: Multiply the mass and the speed to calculate the magnitude of the momentum.
Solution: Calculate the momentum:
p = mv = ( 0.142 kg )( 45.1 m/s ) = 6.40 kg ⋅ m/s
Insight: This momentum will change slightly due to the effects of gravity, as we will shortly learn, but not by much
because the pitch will cross the plate less than half a second after leaving the pitcher’s hand!
2.
Picture the Problem: A dog has momentum as it races to meet its owner.
Strategy: Use the definition of momentum to find the dog’s speed.
p = mv
Solution: Solve the definition of momentum for v:
v=
p 37 kg ⋅ m/s
=
= 3.1 m/s
m
12 kg
Insight: The dog’s speed is equivalent to 6.9 mi/hr. Either the dog isn’t “racing” after all, or it’s a really small dog for
whom 6.9 mi/hr is a fast run!
3.
Picture the Problem: The speeds and masses of four objects are given.
Strategy: Multiply the mass and the speed of each object to find its momentum, and then use the results to rank the
momenta of the four objects.
pA = mA vA = (10 kg )(10 m/s ) = 100 kg ⋅ m/s
Solution: 1. Calculate the momenta:
pB = mB vB = (15 kg )( 4 m/s ) = 60 kg ⋅ m/s
pC = mC vC = (5 kg )( 20 m/s ) = 100 kg ⋅ m/s
pD = mD vD = ( 60 kg )(3 m/s ) = 180 kg ⋅ m/s
2. Compare the momenta to arrive at the ranking: pB < pA = pC < pD .
Insight: The momenta of objects A and C could be different even though their magnitudes are the same because
momentum is a vector, and the directions of these momentum vectors weren’t specified.
4.
Picture the Problem: This is a follow-up question to Guided Example 7.2. At a city park a person throws some bread
into a duck pond. Two 4.0-kg ducks and a 7.6-kg goose paddle rapidly toward the bread from opposite directions. The
ducks swim at 1.1 m/s and the goose swims with a speed of 1.3 m/s.
Strategy: The total momentum of the three birds points to the right, in the direction the goose is swimming. That
means the momentum of the goose is larger in magnitude than that of the two ducks combined. Use the definition of
momentum to find the reduced speed at which the goose must swim in order to make the total momentum equal to zero.
Solution: 1. The momentum of the goose is larger in magnitude than the two ducks combined. We conclude that the
speed of the goose should be decreased in order to make the total momentum of the three birds equal to zero.




p total = p d + p d + p g
2. Set the total momentum equal to zero:
0 = −2md vd + mg vg
2 ( 4.0 kg )(1.1 m/s )
2md vd
= vg =
= 1.2 m/s
7.6 kg
mg
3. Solve for the speed of the goose:
Insight: The goose need only slow down by a small amount because the total momentum of the system was already
close to zero.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–1
Chapter 7: Linear Momentum and Collisions
5.
Pearson Physics by James S. Walker
Picture the Problem: A car and a bicycle move toward each other, each having a momentum.
Strategy: The total momentum of the two vehicles is the vector sum of their individual momenta. Let the car’s
momentum point in the positive direction and the bicycle’s momentum point in the negative direction.



p total = p car + p bicycle
Solution: Add the two momenta as vectors:
(
= mcar vcar + − mbicycle vbicycle
)
= (1250 kg )(12.0 m/s ) − (81.6 kg )(12.0 m/s ) kg ⋅ m/s

p total = 14, 021 kg ⋅ m/s = 1.40 × 104 kg ⋅ m/s
Insight: The answer is written in scientific notation in order to emphasize that it has only three significant digits.
Sometimes this can be done with metric prefixes for the units, but in the case of momenta the results are often
awkward. For instance, we could give our answer as 14.0 Mg·m/s or 14.0 kg·km/s, but neither unit is well accepted.
6.
Picture the Problem: One car drives due north and an identical car drives due east with the same speed.
Strategy: The total momentum of the two vehicles is the vector sum of their individual momenta. Find the magnitude
of the momentum of each car and then add the two momentum vectors.
Solution: 1. Find the magnitude
of the momentum:
2. Add the components of the two
momenta:
p = mcar vcar = (1200 kg )(15 m/s ) = 18, 000 kg ⋅ m/s



p total = p car 1 + p car 2
ptotal, x = pcar 1, x + pcar 2, x = 0 + (18, 000 kg ⋅ m/s )
ptotal, y = pcar 1, y + pcar 2, y = (18, 000 kg ⋅ m/s ) + 0
3. Find the total momentum:
2
2
pcar
1, x + pcar 1, y
ptotal =
=
(18, 000 kg ⋅ m/s )2 + (18, 000 kg ⋅ m/s )2
= 25, 000 kg ⋅ m/s

p total = 2.5 × 104 kg ⋅ m/s at 45° north of east
Insight: The answer is written in scientific notation in order to emphasize that it has only two significant digits. The 45°
(
)
direction can either be inferred or calculated: θ = tan −1 ptotal, y ptotal, x = tan −1 (18, 000 kg ⋅ m/s 18, 000 kg ⋅ m/s ) = 45°.
7.
8.
9.
If the mass of an object doubles, its momentum also doubles if the speed remains the same, because p = mv.


Because the momentum vector is given by p = m v, and because mass is not a vector, then it is clear that the
momentum vector must point in the same direction as the velocity vector. The two sides of a vector equation such as


p = m v must be equal in both magnitude and direction.
If the speed of an object doubles, the magnitude of its momentum will double because p = m ( 2v ) = 2 m v. However, its
kinetic energy will quadruple because KE = 12 m ( 2v ) = 4 × 12 mv 2 . Therefore, the momentum doubles but the kinetic
2
energy quadruples.
10. The total momentum of two moving objects can be zero because momentum is a vector, and two equal and opposite
momenta will cancel each other. However, kinetic energy is a scalar quantity, and two moving objects will each have
nonzero, positive kinetic energy that cannot cancel. Therefore, it does not follow that if a system has zero momentum it
must have zero kinetic energy as well.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–2
Chapter 7: Linear Momentum and Collisions
Pearson Physics by James S. Walker
11. Picture the Problem: An apple moves in a straight line at a known speed.
Strategy: Multiply the mass and the speed to calculate the magnitude of the momentum.
Solution: Calculate the momentum:
p = mv = ( 0.16 kg )( 2.7 m/s ) = 0.43 kg ⋅ m/s
Insight: The momentum of the apple will not change unless a net force acts upon it.
12. Picture the Problem: A school bus and a baseball each have nonzero momentum.
Strategy: Set the momentum of the school bus equal to the momentum of the baseball and solve for the baseball’s
speed.
pbus = pball
Solution: Solve for the speed of the baseball:
mbus vbus = mball vball
mbus vbus
(18, 200 kg )(12.5 m/s )
= vball =
mball
(0.142 kg )
vball = 1, 600, 000 m/s = 1.60 × 106 m/s
Insight: The baseball’s speed would be 4,670 times faster than the speed of sound!
13. Picture the Problem: A dog has momentum as it runs.
Strategy: Use the definition of momentum to find the dog’s mass.
p 17 kg ⋅ m/s
=
= 8.9 kg
v
1.9 m/s
Insight: A dog of this mass would weigh 20 lb on Earth’s surface, so this is a medium-sized dog.
p = mv
Solution: Solve the definition of momentum for m:
m=
14. Picture the Problem: Two different carts approach each other on a frictionless track at different speeds.
Strategy: Add the momenta of the two carts, taking the momentum of cart 1 to be positive.



Solution: Add the two momenta:
p total = m1 v1 + m2 v 2
= ( 0.35 kg )(1.2 m/s ) + ( 0.61 kg )( − 0.85 m/s )

p total = 0.42 + ( − 0.52) kg ⋅ m/s = − 0.10 kg ⋅ m/s
Insight: If cart 2 were traveling at 0.69 m/s, its momentum would exactly cancel the momentum of cart 1 and the total
momentum of the system would be zero.
15. Picture the Problem: The force of a kick to a soccer ball delivers an impulse to the ball.
Strategy: Use the definition of impulse to calculate the magnitude of the force.
Solution: Apply the definition of impulse:
I = F Δt
F=
I
23 kg ⋅ m/s
=
= 92 N
Δt
0.25 s
Insight: This force is equivalent to about 21 lb. A 0.43-kg regulation soccer ball that is initially at rest would travel
v = Δ p m = ( 23 kg ⋅ m/s ) ( 0.43 kg ) = 53 m/s, or about 120 mi/h, after being kicked with this impulse!
16. Impulse is the product of force and time. In this problem, the gravitational force on the two apples is the same because
their masses are the same. However, apple 2 falls a greater distance and the unbalanced force of gravity acts on it for a
greater interval of time. Therefore, the impulse delivered by gravity to apple 1 is less than the impulse delivered by
gravity to apple 2.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–3
Chapter 7: Linear Momentum and Collisions
Pearson Physics by James S. Walker
17. Picture the Problem: The momentum of an object is changed by the impulse delivered to it.
Strategy: Use the definition of impulse to calculate the new momentum of the object.



p f = p i + Δp
Solution: The final momentum is the
initial momentum plus the impulse:
pf = 4.5 kg ⋅ m/s + 12.2 kg ⋅ m/s = 16.7 kg ⋅ m/s
Insight: If the impulse were pointed in the direction opposite to the object’s initial momentum, the final momentum
would be pf = 4.5 kg ⋅ m/s − 12.2 kg ⋅ m/s = −7.7 kg ⋅ m/s.
18. Picture the Problem: This is a follow-up question to Guided Example 7.6. Three friends push a stalled car and its
driver to a nearby service station. Assume that the car rolls without friction on a smooth and level road.
Strategy: Use the momentum-impulse theorem to find the final momentum of the car. The change in momentum equals
the force times the time. The car is initially at rest, so that its initial momentum is zero.
Solution: 1. (a) Find the impulse
delivered to the car:
I = F1 Δ t1 + F2 Δ t2
= (305 N )( 6.0 s ) + ( 215 N )( 6.0 s )
I = 3120 kg ⋅ m/s
2. Calculate the final momentum of
the car:
3. (b) Solve the momentum equation
for the mass of the car:
I = Δ p = pf − pi = pf − 0
pf = I = 3120 kg ⋅ m/s
vf =
pf = mvf
pf 3120 kg ⋅ m/s
=
= 2.29 m/s
m
1360 kg
Insight: As expected, the speed of the car is less than the 2.69 m/s it attained when it was pushed by all three students.
19. Picture the Problem: A baseball is bunted, or hit softly, back toward the pitcher.
Strategy: The impulse delivered by the bat changes the momentum of the ball. Here the directions of the momentum
and forces are important. Let the initial momentum of the ball be in the negative direction, and the impulse delivered
by the bat, as well as the final momentum of the ball, be in the positive direction.
Solution: 1. Find the impulse
delivered to the ball:
2. Calculate the final
momentum of the ball:
(
)
I = F Δ t = 6.50 × 103 N ( 0.00122 s ) = 7.93 kg ⋅ m/s
I = Δ p = pf − pi
pf = I + pi = I + mvi
= 7.93 kg ⋅ m/s + ( 0.144 kg )( − 43.0 m/s )
pf = 1.74 kg ⋅ m/s
3. Use the definition of momentum to
find the speed of the ball:
pf = mvf
vf =
pf 1.74 kg ⋅ m/s
=
= 12.1 m/s
m
0.144 kg
Insight: Despite being a “soft” hit, the average force exerted by the bat is 6500 N, or about 1460 lb.
20. Picture the Problem: The hand of a volleyball player exerts an impulse on the ball, imparting momentum.
Strategy: Use the definition of momentum together with the momentum-impulse theorem to find the mass of the ball.
Solution: Solve the momentum-impulse
theorem for the mass of the volleyball:
I = Δ p = m Δv
I
−9.3 kg ⋅ m/s
m=
=
= 0.33 kg
Δv −24 m/s − 4.2 m/s
Insight: A mass this large weighs about ¾ lb at sea level. This ball is 22% heavier than the 0.27 kg regulation ball.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–4
Chapter 7: Linear Momentum and Collisions
Pearson Physics by James S. Walker
21. Picture the Problem: An apple falls vertically downward due to gravity.
Strategy: The rate of change of the apple’s momentum is the net force acting on it, according to the momentumimpulse theorem. The only force acting on the apple is its weight mg.
Solution: 1. (a) Solve the momentumimpulse theorem for Δp Δt :
Δp
= F = 2.7 N
Δt
2. (b) Solve the same equation for Δp :
Δp = F Δt = ( 2.7 N )(1.4 s ) = 3.8 kg ⋅ m/s
Insight: We could also write the units of the answer to part (a) as 2.7 ( kg ⋅ m/s ) s.
22. Impulse is a vector that points in the same direction as the change in momentum and the net force, as defined by the
 

momentum-impulse theorem: I = FΔ t = Δ p .
 

23. The momentum-impulse theorem, I = FΔ t = Δ p , indicates that the impulse determines an object’s change in
momentum, not its momentum. Of course, if the object started with zero momentum, the change in momentum
and the momentum would be the same.
24. A helmet is supposed to reduce the force delivered to your head by increasing the time over which the impact occurs.
A thin steel helmet offers no soft cushion that can increase the time of impact, so the forces delivered to your head
would not be reduced very much by a thin steel helmet.
25. Picture the Problem: The forces and durations that produce four impulses are given.
Strategy: Multiply the force and the duration in each case to find the impulse, and then use the results to rank them.
I A = FA Δ tA = ( F )( Δ t ) = F Δ t
Solution: 1. Calculate the impulses:
I B = FB Δ tB = ( 2 F )( Δ t 3) = 23 F Δ t
I C = FC Δ tC = (5 F )( Δ t 10) = 12 F Δ t
I D = FD Δ tD = (10 F )( Δ t 100) =
1
10
F Δt
2. Compare the impulses to arrive at the ranking: ID < IC = IB < IA .
Insight: When you compare two car crashes (for example, into a concrete barrier and into bales of hay) with the same
initial speed, the impulses are the same (because the change in momentum is the same) but the forces are different. The
greatest force corresponds to the smallest time interval, which in this case would correspond to the concrete barrier.
26. (a) In each case the change in momentum of the car is the same because its initial speeds are the same and the final
speeds are each zero. Therefore, the impulse delivered in case 1 is equal to the impulse in case 2. (b) The interaction
time with the pile of plastic garbage bags is much larger than it is with the light pole. Therefore, the average force in
case 1 is greater than the average force in case 2.
27. Picture the Problem: A safety helmet extends the time of impact and reduces the force.
Strategy: Use the momentum-impulse theorem to calculate a ratio of the forces. The change in momentum of the
person’s head is the same in each case.
Solution: Write a ratio of the forces using
the momentum-impulse theorem:
Fnew Δ p Δ tnew
Δt
0.005 s 1
=
= old =
=
Fold
Δ p Δ told
Δ tnew 0.020 s 4
1
Fold the average force is reduced by a factor of 4 .
4
Insight: Neither the impulse nor the change in momentum can be changed in a collision, but increasing the duration of
the collision can reduce the force that is applied.
Fnew =
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–5
Chapter 7: Linear Momentum and Collisions
Pearson Physics by James S. Walker
28. Picture the Problem: A golf club exerts an impulse on a ball, imparting momentum.
Strategy: Find the change of momentum of the golf ball and use it to find the force exerted on it according to the
momentum-impulse theorem.
Solution: Use the general form
of Newton’s second law:
Fav =
Δp m ( vf − vi ) ( 0.045 kg )( 67 m/s − 0)
=
=
= 3.0 kN
Δt
Δt
0.0010 s
Insight: This is a force equivalent to 670 lb! A ball launched with a speed of 67 m/s launched at 45° will land 457 m
(500 yd) down range in the absence of air friction. A par 5 hole-in-one?
29. Picture the Problem: A croquet mallet exerts an impulse on a ball, imparting momentum.
Strategy: Find the change in momentum of the croquet ball and then use it to find Δt using the momentum-impulse
theorem.
Solution: Solve the momentumimpulse theorem for Δt :
Δt =
Δp m ( vf − vi ) ( 0.50 kg )(3.2 m/s − 0)
=
=
= 0.0070 s = 7.0 ms
Fav
Fav
230 N
Insight: The large force (52 lb) is exerted over a very brief time to give the ball its small velocity (7.2 mi/h).
30. Picture the Problem: This is a follow-up question to Guided Example 7.7. Two groups in canoes float motionless in
the middle of a lake. After a brief visit someone in canoe 1 pushes on canoe 2 with a force of 46 N. The force lasts for
1.20 s and moves the canoes in opposite directions.
Strategy: Guided Example 7.7 showed that the magnitude of the momentum of each canoe is 55 kg·m/s. Use this value
together with the definition of momentum to find the velocity of each canoe. Then test the results to see if the total
velocity of the canoes remains the same before and after the push.

p1 −55 kg ⋅ m/s



Solution: 1. Find the velocity of
=
= − 0.42 m/s
v1 =
p = mv
canoe 1:
130 kg
m1

p

55 kg ⋅ m/s


2. Find the velocity of canoe 2:
= 0.22 m/s
v2 = 2 =
p = mv
250 kg
m2


v total, before = v total, after
3. Check to determine whether
 
velocity is a conserved quantity:
0 = v1 + v 2
0 = − 0.42 + 0.22 m/s
0 ≠ − 0.20 m/s
No , velocity is not conserved
Insight: Note that in order for the momenta to be equal and opposite, the more massive canoe must have the lower
speed.
31. Picture the Problem: This is a follow-up question to Guided Example 7.7. Two groups in canoes float motionless in
the middle of a lake. After a brief visit someone in canoe 1 pushes on canoe 2 with a force of 56 N. The force lasts for
1.30 s and moves the canoes in opposite directions.
Strategy: Each canoe starts at rest, with zero initial momentum. Determine the change in momentum of each canoe

 
using the momentum-impulse theorem, FΔ t = p f − p i .





 
p1, f = FΔ t + p1, i
Solution: 1. Solve FΔ t = p f − pi for p f .
Substitute the numerical values for canoe 1,
= ( −56 N )(1.3 s ) + 0


including p1, i = 0 and F1 = −56 N (−x direction):
= −73 kg ⋅ m/s
( − x direction)
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–6
Chapter 7: Linear Momentum and Collisions
Pearson Physics by James S. Walker



p 2, f = FΔ t + p 2, i


 
2. Solve FΔ t = p f − pi for p f . Substitute the
numerical values for canoe 2, including


p 2, i = 0 and F2 = 56 N (+x direction):
= (56 N )(1.3 s ) + 0
= 73 kg ⋅ m/s
(+ x direction)
Insight: The total momentum of the system remains zero after the push because there is no net external force.
32. Picture the Problem: Two skaters push apart and move in opposite directions without friction.
Strategy: By applying the conservation of momentum we conclude that the total momentum of the two skaters after the
push is zero, just as it was before the push. Set the total momentum of the system to zero and solve for m2 . Let the


velocity v1 point in the negative direction, v 2 in the positive direction.
Solution: Set ptotal = 0 and solve for m2 :
p1x + p2 x = 0 = m1v1x + m2 v2 x
m2 =
− m1v1x − ( 45 kg )( − 0.62 m/s )
=
= 31 kg
v2 x
0.89 m/s
Insight: An alternative way to find the mass is to use the equations of kinematics, but conservation of momentum is
easier!
33. Picture the Problem: An astronaut and a satellite move in opposite directions after the astronaut pushes off. The
astronaut travels at constant speed a distance d before coming in contact with a space shuttle.
Strategy: As long as there is no friction the total momentum of the astronaut and the satellite must remain zero, as it
was before the astronaut pushed off. Use the conservation of momentum to determine the speed of the astronaut, and
then multiply the speed by the time to find the distance. Assume the satellite’s motion is in the negative x-direction.
Solution: 1. Find the speed of the astronaut using
conservation of momentum:
pa + ps = 0 = ma va + ms vs
va = −
ms vs
(1200 kg )( − 0.14 m/s) = −1.826 m/s
=−
ma
(92 kg )
ms vs
t = − (1.826 m/s )( 7.5 s ) = 14 m
ma
Insight: One of the tricky things about spacewalking is that whenever you push on a satellite or anything else, you
yourself get pushed because of Newton’s third law!
2. Find the distance to the space shuttle:
d = va t = −
34. If the total external force acting on a system is zero, then its total momentum cannot change. This is a statement of the
conservation of momentum and is also a consequence of Newton’s second law.
35. Internal forces cannot affect the total momentum of the system. Any change in momentum of one part of the system that
is due to an internal force is balanced by an equal and opposite change in momentum of another part of the system.
36. Momentum is conserved for the system of the two skaters. The internal forces are those that the two skaters exert on one
another; if skater A pushes on skater B, skater B pushes on A with an equal and opposite force, according to Newton’s
third law. The external forces on the system are the friction forces between the ice and the skates, but we assume these
to be zero so that the total momentum of the system is conserved.
37. (a) No, it is not possible for both objects to be at rest after the collision because that would require the total momentum
of the system to be zero after the collision when it was not zero before the collision. Friction could allow that to happen,
but in this case the surface is frictionless. (b) Yes, it is possible for one of the objects to be at rest after the collision.
You can try this on a pool table. If a cue ball strikes another (resting) ball head-on, the cue ball will come to rest and the
other ball will roll with the same velocity that the cue ball had initially.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–7
Chapter 7: Linear Momentum and Collisions
Pearson Physics by James S. Walker
38. The momentum of the keys is not conserved because the external force of gravity continually increases their
momentum. However, the momentum of the universe remains the same because both the Earth and the keys experience
equal and opposite gravitational forces that change their momenta by the same amount but in opposite directions.
39. Picture the Problem: Two canoes are pushed apart by the force exerted by the occupants.
Strategy: By applying the conservation of momentum we conclude that the total momentum of the two canoes after the
push is zero, just as it was before the push. Set the total momentum of the system to zero and solve for m2 . Let the


velocity v1 point in the negative direction, v 2 in the positive direction.
Solution: Set ptotal = 0 and solve for m2 :
p1x + p2 x = 0 = m1v1x + m2 v2 x
m2 =
− m1v1x − (320 kg )( − 0.58 m/s )
=
= 440 kg
0.42 m/s
v2 x
Insight: An alternative way to find the mass is to use the equations of kinematics, but conservation of momentum is
easier!
40. Picture the Problem: A lumberjack stands on a log that is floating at rest in a pond. He then starts jogging along the
length of the log while the log moves in the opposite direction.
Strategy: As long as there is no friction, the total momentum of the lumberjack and the log is zero, both before and
after the lumberjack started jogging. Use conservation of momentum to determine the log’s speed relative to the shore.

Solution: 1. Write out the conservation of
p = 0 = mL vL − mlog vlog
momentum with respect to the shore:
Σ
2. Solve the expression from step 1 for vlog :
mL vL = mlog vlog
mL vL
(94 kg )( 2.2 m/s) = 0.17 m/s
= vlog =
mlog
(1200 kg )
Insight: The log is moving slowly away from the shore as the lumberjack jogs toward the shore.
41. Picture the Problem: A hockey player tosses a helmet in one direction and recoils in the other.
Strategy: The total horizontal momentum of the hockey player mp and helmet mh is conserved because there is no
friction. The total horizontal momentum remains zero both before and after the helmet is tossed. Use this fact to find the
mass of the player. Let the helmet be tossed in the positive direction.
Solution: Set pi,x = pf,x and solve for mp :
Σp
x
= 0 = mp vp, x + mh vh, x
mp = −
mh vh, x
vp, x
=−
(1.3 kg )(6.5 m/s ) =
− 0.25 m/s
34 kg
Insight: This must be a junior player because a 34-kg player weighs only 75 lb. If a 70-kg adult tossed the helmet in the
same manner, the recoil velocity would be − 0.12 m/s.
42. Picture the Problem: This is a follow-up question to Guided Example
7.10. On a touchdown attempt a 95.0-kg running back runs toward the end
zone. A 111-kg linebacker moving at 4.10 m/s meets the runner in a headon collision.
Strategy: Guided Example 7.10 showed that the final velocity of the two
players is to the left. That implies that the total momentum of the twoplayer system is to the left, and the running back needs more momentum
toward the right in order to make the total momentum of the system equal
to zero.
Solution: The total momentum of the two-player system is toward the left, implying that the linebacker has a larger
momentum toward the left than the running back’s momentum toward the right. We conclude that the running back
should run faster in order to increase his momentum toward the right and make the total momentum equal to zero.
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7–8
Chapter 7: Linear Momentum and Collisions
Pearson Physics by James S. Walker
Insight: Because none of the momentum is lost in the collision, a total momentum of zero before the collision will
remain zero after the collision, and the final speed of the two players, stuck together, will be zero.
43. Picture the Problem: This is a follow-up question to Guided Example
7.10. On a touchdown attempt a 95.0-kg running back runs toward the end
zone. A 111-kg linebacker moving at 4.10 m/s meets the runner in a headon collision.
Strategy: Guided Example 7.10 showed that the final velocity of the two
players is to the left. That implies that the total momentum of the twoplayer system is to the left, and the running back needs more momentum
toward the right in order to make the total momentum of the system equal
to zero. Set the initial momentum to zero and solve for the speed of the
running back.




Solution: Set the initial momentum to
p1, i + p 2, i = 0 = m1 v1, i + m2 v 2, i


zero and solve for v1, i :
m2 v 2, i
(111 kg )( − 4.10 m/s) = 4.79 m/s

=−
v1,i = −
m1
(95.0 kg )
Insight: The result verifies the answer to the previous question that the running back must run faster than 3.75 m/s
in order for the total momentum to be zero. In fact, the less massive running back must have a higher speed than the
linebacker in order for the two momenta to be equal and opposite of each other.
44. Picture the Problem: A bullet strikes a wooden block and together they
move horizontally along the surface of a table.
Strategy: Use conservation of momentum to determine the horizontal
speed of the bullet and block after the collision. Let m = the mass of the
bullet, M = the mass of the block, and vi = the initial speed of the bullet.


Solution: 1. Set pi = p f
and solve for vf :
2. Substitute the numerical values:
mvi + M (0) = (m + M ) vf
æ m ö÷
vf = çç
v
çè m + M ø÷÷ i
æ 0.0105 kg ÷ö
÷÷(715 m/s) = 5.52 m/s
vf = ççç
è 0.0105 + 1.35 kg ø÷
Insight: If the block were twice as massive it would travel half as fast after the collision, and if it were to leave the
table, it would land on the floor about half as far away. In this way the landing spot is a measure of the mass of the
block, as long as the block is much more massive than the bullet.
45. Picture the Problem: A car is struck directly from behind by a truck, and the two vehicles stick together after the
collision.
Strategy: Use conservation of momentum to determine the speed of the two vehicles after the collision. Let m = the
mass of the car and M = the mass of the truck. Let each travel in the positive direction.


Solution: 1. Set pi = p f
mvcar, i + Mvtruck, i = ( m + M ) vf
and solve for vf :
mvcar, i + Mvtruck,i
vf =
m+M
2. Substitute the numerical values:
vf =
(1200 kg )( 2.5 m/s ) + ( 2600 kg )(6.2 m/s) =
1200 + 2600 kg
5.0 m/s
Insight: Note that the 5.0 m/s final speed is closer to the truck’s initial 6.2 m/s speed than the car’s initial 2.5 m/s speed.
That is because the final speed is the average of the two initial speeds weighted by the mass of the vehicles. The truck
therefore slows down a little bit whereas the car speeds up substantially.
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7–9
Chapter 7: Linear Momentum and Collisions
Pearson Physics by James S. Walker
46. Picture the Problem: This is a follow-up question to Guided Example
7.11. A 950-kg red car with a speed of 16 m/s approaches an intersection,
traveling in the positive x direction. A 1300-kg blue minivan traveling at
21 m/s is heading for the same intersection in the positive y direction. The
car and minivan collide and stick together, producing wreckage that moves
at an angle of θ= 61° relative to the positive x axis.
Strategy: Increasing the speed of the red car would increase the
momentum of the two-car system toward the right. Use this fact to answer
the question about the direction of motion of the two cars after they stick
together.
Solution: Increasing the speed of the red car would increase the
momentum of the two-car system toward the right. The cars will retain this
extra momentum toward the right after they stick together. We conclude
that the cars will move more toward the right, at an angle that is less than
61°, after the collision if the initial speed of the red car is increased.
Insight: Because none of the momentum is lost in the collision, any change of momentum of the system before the
collision will remain in the system after the collision.
47. Picture the Problem: This is a follow-up question to Guided Example
7.11. A 950-kg red car approaches an intersection, traveling in the positive
x direction. A 1300-kg blue minivan is heading for the same intersection
in the positive y direction. The car and minivan collide and stick together,
producing wreckage that moves at 12.5 m/s and at an angle of θ= 42°
relative to the positive x axis.
Strategy: The red car adds momentum in the positive x direction and the
blue minivan adds momentum in the positive y direction. These momenta
are the same before and after the collision. Use this principle to first find
the components of the system’s momentum after the collision, and use
these components to determine the speeds of the cars before the collision.
Solution: 1. Find the
momentum in the x
direction:
px , i = px , f = pf cos θ = ( m1 + m2 ) vf cos θ
2. Find the initial speed
of the red car:
px , i = m1v1, i = 20,900 kg ⋅ m/s
3. Find the momentum in
the y direction:
p y , i = p y , f = pf sin θ = ( m1 + m2 ) vf sin θ
= (950 + 1300 kg )(12.5 m/s ) cos 42°
px , i = 20,900 kg ⋅ m/s
v1, i =
20,900 kg ⋅ m/s 20,900 kg ⋅ m/s
=
= 22 m/s
m1
950 kg
= (950 + 1300 kg )(12.5 m/s ) sin 42°
p y , i = 18,800 kg ⋅ m/s
4. Find the initial speed
of the blue minivan:
p y , i = m2 v2, i = 18,800 kg ⋅ m/s
v2, i =
18,800 kg ⋅ m/s 18,800 kg ⋅ m/s
=
= 14 m/s
m2
1300 kg
Insight: The result verifies the answer to the previous question that if the red car has a higher initial speed (22 m/s
instead of 16 m/s) the resulting two-car combination would move at an angle that is less than 61°. However, it is also
true that a slower initial speed for the minivan would have the same effect because there would be less momentum in
the positive y direction. Indeed, the 14 m/s speed of the minivan is less than the 21 m/s it had in the original example.
Both of these effects combine to produce the 42° angle that is significantly less than the original 61° angle.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7 – 10
Chapter 7: Linear Momentum and Collisions
Pearson Physics by James S. Walker
48. Picture the Problem: An 82-kg football player is running in the positive
y direction with a speed of 3.1 m/s. A 96-kg player from the opposing team is
running in the negative x direction with a speed of 2.4 m/s when the tackle is
made.
θ
Strategy: The total momentum of the two players after they collide will be
identical to their total momentum before the collision. Use this fact to find the
momentum of the two players after the collision, and use their combined mass
to find their velocity after the collision.
Solution: 1. Find the momentum in the x direction:
2. Find the momentum in the
y direction:
y
12
2
x
1
px , i = px , f = m2 vx , 2 = (96 kg )( −2.4 m/s )
px , f = −230.4 kg ⋅ m/s
p y , i = p y , f = m1v y ,1 = (82 kg )(3.1 m/s )
p y , f = 254.2 kg ⋅ m/s
3. Find the magnitude of the
final momentum:
pf =
4. Divide by the total mass to
find the speed:
vf =
( −230.4 kg ⋅ m/s )2 + ( 254.2 kg ⋅ m/s) = 343.1 kg ⋅ m/s
px2, f + p y2, f =
pf
343.1 kg ⋅ m/s
=
= 1.9 m/s
m1 + m2
82 + 96 kg
Insight: By analyzing the components of the final momentum you can determine that after the collision the two players
æ p y , f ö÷
æ
ö
÷÷ = tan -1 çç 254.2 kg ⋅ m/s ÷÷ = 48° above the negative x axis.
move at θ = tan -1 ççç
÷
çè 230.4 kg ⋅ m/s ÷ø
èç px , f ø÷÷
49. If the external force (such as friction) acting on a system of colliding objects is zero, then its total momentum cannot
change. This is a statement of the conservation of momentum and is also a consequence of Newton’s second law.
50. No kinetic energy is lost in a perfectly elastic collision, so the initial and final kinetic energies are the same.
51. The kinetic energy that is lost in an inelastic collision is converted to other forms of energy, such as thermal energy,
sound energy, the work required to deform metal, etc.
52. A collision that results in the two objects sticking together is a completely inelastic collision.
53. In order for the total momentum in a system of colliding particles to be conserved, the total external force (such as
friction, gravity, normal forces, etc.) must be zero.
54. No. While elastic collisions result in the rebounding of the two objects, this is also true for partially inelastic collisions.
The only way to be certain that a collision is elastic is to check that the total kinetic energy before the collision is equal
to the total kinetic energy after the collision. Any collision between cars will be at least partially inelastic, due to
denting, sound production, heating, and other effects.
55. Picture the Problem: Two carts collide on a frictionless track and stick together.
Strategy: The collision is completely inelastic because the two carts stick together. Momentum is conserved during
the collision because the track has no friction. The two carts move as if they were a single object after the collision.
Use the conservation of momentum to find the final speed of the carts and final kinetic energy of the system.
Solution: 1. (a) Calculate the total
kinetic energy of the carts:
KEi = 12 m1v1,2 i + 12 m2 v2,2 i
=
1
2
(0.12 kg )(0.45 m/s )2 + 0 =
0.012 J
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7 – 11
Chapter 7: Linear Momentum and Collisions
2. (b) Conserve momentum to
find the final speed of the carts:
Pearson Physics by James S. Walker
pi = pf
m1v1, i + m2 (0) = ( m1 + m2 ) vf
m1
m
1
1
v1, i = vf = 1 v1, i = v1, i = ( 0.45 m/s ) = 0.225 m/s
2m1
2
2
m1 + m2
3. Calculate the final kinetic energy:
K f = 12 (2m1 )vf 2 =
1
2
( 2 × 0.12 kg )(0.225 m/s)2 =
0.0061 J
Insight: Half of the initial kinetic energy is gone, having been converted to heat, sound, and permanent deformation
of material during the inelastic collision.
56. Picture the Problem: A hammer strikes a rock with an elastic collision. The collision drives the rock in the forward
direction and slows down the speed of the hammer.
Strategy: This is a one-dimensional, elastic collision where one of the objects (the rock) is initially at rest. Therefore,
we can use the results of the analysis of elastic collisions to find the speed of the rock after being struck by the hammer.
In this case the subscript 1 refers to the hammer and the subscript 2 refers to the rock.
Solution: Use the results of the
analysis of elastic collisions to find
the final speed of the rock, v2,f :
é 2 (0.55 kg ) ù
æ 2m1 ö÷
ú
÷v = ê
v2,f = ççç
÷
ê 0.55 + 0.19 kg ú (4.5 m/s) = 6.7 m/s
çè m1 + m2 ÷ø
ë
û
Insight: The elastic collisions analysis can also be used to find the velocity of the hammer after the collision:
æ m - m2 ÷ö
é 0.55 - 0.19 kg ù
ú (4.5 m/s) = 2.2 m/s. Note that the hammer slows down but is still going in the
÷÷ v = ê
v1,f = ççç 1
ê 0.55 + 0.19 kg ú
çè m1 + m2 ÷ø
ë
û
forward direction after the collision. If the collision were instead considered to be inelastic (the hammer remains in
contact with the rock and they move together) the final velocity would be 3.3 m/s.
57. Picture the Problem: This is a follow-up question to Guided Example
7.11. A 950-kg red car approaches an intersection at 20.0 m/s, traveling
in the positive x direction. A 1300-kg blue minivan is heading for the same
intersection in the positive y direction. The two vehicles collide and stick
together, heading in a direction given by θ= 40.0° above the x axis after the
collision.
Strategy: If we ignore frictional forces, the total momentum of the two
vehicles before the crash equals the momentum after the collision. The
vehicles stick together and behave as a single object with a mass equal to
the sum of their masses. Use conservation of momentum to find the initial
speed v2 of the blue minivan.
Solution: 1. Conserve momentum
in the x direction:
2. Conserve momentum in the y direction:
3. Divide the equation from step 2 by the
equation for step 1 and solve for v2 :
Σp
Σp
x
= m1v1 + 0 = ( m1 + m2 ) vf cos θ
y
= m2 v2 + 0 = ( m1 + m2 ) vf sin θ
m2 v2 ( m1 + m2 ) vf sin θ
=
= tan θ
m1v1 ( m1 + m2 ) vf cos θ
v2 = v1
m1
(950 kg ) tan 40.0° = 12 m/s
tan θ = ( 20.0 m/s )
m2
(1300 kg )
Insight: In real life the assumption that there is no friction is quite incorrect unless the collision occurs on very slick,
icy pavement. Substantial deviations from the conservation of momentum will be observed if the pavement is dry.
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7 – 12
Chapter 7: Linear Momentum and Collisions
Pearson Physics by James S. Walker
58. Picture the Problem: Brittany stands on a skateboard as Dave tosses her a pumpkin. She and the skateboard begin to
roll backward when she catches the pumpkin.
Strategy: The total horizontal momentum of Brittany and the pumpkin is conserved because there is essentially no
friction acting on the system. Use conservation of momentum to find the initial speed of the pumpkin.
Solution: Set pi = pf and solve for vp :
pi = pf
(
(m
=
)
+ m )v
mp vp, i + 0 = mp + mB vf
vp, i
p
B
f
mp
=−
(3.7 + 61 kg )(0.16 m/s) =
3.7 kg
2.8 m/s
Insight: This is a pretty gentle pumpkin toss, equivalent to throwing an 8.1 lb pumpkin at 6.3 mi/h.
59. Yes. Momentum is the product of mass and velocity, so a moving baseball will have more momentum than a truck that
is at rest, which has zero momentum. However, a baseball can never have more inertia than a truck, because inertia is a
property of mass and the truck has far more mass than a baseball.
60. If a system of particles has zero kinetic energy, none of the particles can be moving. This is due to the fact that kinetic
energy is a scalar quantity, not a vector, so that the kinetic energy of one particle cannot be canceled out by the motion
of another. Because none of the particles are moving, we conclude that the total momentum of the system is also zero.
61. Momentum is a vector, so that it is possible that the momentum of one particle can cancel the momentum of another.
Kinetic energy is a scalar quantity, however, and such cancellation is not possible. That is why a system of particles
might have a total kinetic energy of 10,000 J but zero total momentum.
62. Momentum is the product of velocity and mass. Momentum is a conserved quantity, but velocity is not. The two are
similar because each is a vector that points in the direction of motion.
63. Picture the Problem: The speeds and masses of four objects are given.
Strategy: Multiply the mass and the speed of each object to find its momentum, and then use the results to rank the
momenta of the four objects.
pA = mA vA = (1 kg )(1 m/s ) = 1 kg ⋅ m/s
Solution: 1. Calculate the momenta:
pB = mB vB = ( 2 kg )(3 m/s ) = 6 kg ⋅ m/s
pC = mC vC = ( 0.1 kg )( 20 m/s ) = 2 kg ⋅ m/s
pD = mD vD = (5 kg )( 0.2 m/s ) = 1 kg ⋅ m/s
2. Compare the momenta to arrive at the ranking: pA = pD < pC < pB .
Insight: The momenta of objects A and D could be different even though their magnitudes are the same because
momentum is a vector, and the directions of these momentum vectors weren’t specified.
64. Picture the Problem: The speeds and momenta of four objects are given.
Strategy: Divide the momentum by the speed of each object to find its mass, and then use the results to rank the masses
of the four objects.
mA = pA vA = ( 4 kg ⋅ m/s ) (1 m/s ) = 4 kg
Solution: 1. Calculate the masses:
mB = pB vB = (5 kg ⋅ m/s ) ( 2 m/s ) = 2.5 kg
mC = pC vC = (1 kg ⋅ m/s ) ( 4 m/s ) = 0.25 kg
mD = pD vD = ( 2 kg ⋅ m/s ) ( 0.5 m/s ) = 4 kg
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7 – 13
Chapter 7: Linear Momentum and Collisions
Pearson Physics by James S. Walker
2. Compare the masses to arrive at the ranking: mC < mB < mA = mD .
Insight: A 4-kg object might also have a speed of 2 m/s and a momentum of 8 kg·m/s. There is no limit to the
momentum that a 4-kg object could have, but there is a limit to the speed: 3.0×108 m/s. The special theory of relativity
explains how objects that travel at nearly the speed of light can increase their momentum without changing their speed
by very much.
65. Picture the Problem: A baseball has momentum just before it lands on the ground.
Strategy: Use the definition of momentum to find the baseball’s speed.
p = mv
Solution: Solve the definition of momentum for v:
v=
p 0.78 kg ⋅ m/s
=
= 5.2 m/s
m
0.15 kg
Insight: The baseball’s speed is equivalent to only 12 mi/hr. It must have been tossed gently from a short distance.
66. Picture the Problem: A car has momentum as it moves.
Strategy: Use the definition of momentum to find the car’s mass.
p = mv
Solution: Solve the definition of momentum for m:
m=
p 32, 000 kg ⋅ m/s
=
= 1300 kg
v
25 m/s
Insight: This is a fairly small car, with a weight of 2800 lb or 1.4 tons.
67. Picture the Problem: A salmon has momentum as it swims upstream.
Strategy: Multiply the mass and the speed to calculate the magnitude of the momentum.
Solution: Calculate the momentum:
p = mv = (5.2 kg )( 2.3 m/s ) = 12 kg ⋅ m/s
Insight: A 5.2-kg salmon weighs 11 lb. If the fish swims twice as fast it will have double the momentum.
68. Picture the Problem: Two soccer players of known mass move with known velocities.
Strategy: The total momentum of the two soccer players is the vector sum of their individual momenta. Let velocities
to the right be positive and those to the left be negative.

 


p total = p1 + p 2 = m1 v1 + m2 v 2
Solution: Sum the two momenta as vectors:
= ( 45 kg )(1.4 m/s ) + (32 kg )( −2.1 m/s )

p total = − 4.2 kg ⋅ m/s (to the left)
Insight: The momentum of player 1 is 63.0 kg·m/s (to the right), and that of player 2 is −67.2 kg·m/s (to the left).
69. Picture the Problem: Two soccer players of known mass move in opposite directions.
Strategy: Set the total momentum of the two players equal to the vector sum of their individual momenta, and solve the
resulting expression for the velocity of player 2. Let velocities to the right be positive and those to the left be negative.

 


p total = p1 + p 2 = m1 v1 + m2 v 2
Solution: Solve for the velocity of player 2:



p total − m1 v1 = m2 v 2


2.2 kg ⋅ m/s − ( 47 kg )(1.1 m/s )
p total − m1 v1 
= v2 =
38 kg
m2

v 2 = − 1.3 m/s (to the left)
Insight: The momentum of player 1 is 51.7 kg·m/s (to the right), and that of player 2 is −49.4 kg·m/s (to the left).
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7 – 14
Chapter 7: Linear Momentum and Collisions
Pearson Physics by James S. Walker
70. Picture the Problem: Two carts approach each other on a frictionless track at different speeds.
Strategy: Add the momenta of the two carts and set it equal to zero. Solve the resulting expression for v2 .




Solution: Set
p = 0 and solve for v2 :
p = m1 v1 + m2 v 2 = 0
Σ
Σ
v2 = −
m1v1 ( 0.35 kg )(1.2 m/s )
=
= 0.69 m/s
0.61 kg
m2
Insight: If cart 1 is traveling in the positive x direction, then its momentum is 0.42 kg·m/s and the momentum of cart 2
is −0.42 kg·m/s.
71. Picture the Problem: A ball falls vertically downward, landing with a speed of 2.5 m/s and rebounding upward with a
speed of 2.0 m/s.
Strategy: Use the definition of momentum to find the change in momentum of the ball when it rebounds. Let upward be
the positive direction.


Solution: 1. (a) Use the definition

pi = m v i = ( 0.28 kg )( −2.5 m/s ) = − 0.70 kg ⋅ m/s
of momentum to find p i :
2. (b) Use the definition of

momentum to find p f :


p f = m v f = ( 0.28 kg )( 2.0 m/s ) = 0.56 kg ⋅ m/s

3. (c) Subtract the momenta to find Δp :
  
Δp = p f − p i = 0.56 kg ⋅ m/s − ( − 0.70 kg ⋅ m/s ) = 1.26 kg ⋅ m/s

Insight: If the ball were to rebound at 2.5 m/s upward we would find Δp = 2mv = 1.1 kg ⋅ m/s and pf − pi = 0 . Such a
collision with the floor would be called elastic because there would be no change in the kinetic energy of the ball.
72. Picture the Problem: A cat runs eastward and a dog runs northward.
Strategy: Sum the momenta of the dog and cat using the component method. Use the known components of the total
momentum to find its magnitude and direction. Let north be in the +y direction, east in the +x direction.





p total = p cat + p dog = mcat v cat + mdog v dog
Solution: 1. Use the component
method of vector addition to
= (5.3 kg )(3.0 m/s ) (east) + ( 26 kg )( 2.7 m/s ) (north)
find the total momentum:

p total = (15.9 kg ⋅ m/s ) (east) + ( 70.2 kg ⋅ m/s ) (north)
2. Find the magnitude of the momentum:
ptotal =
px2 + p y2 =
(15.9 kg ⋅ m/s)2 + (70.2 kg ⋅ m/s )2
= 72 kg ⋅ m/s
3. Use the known components to find the
direction of the momentum:
æ py ö
æ 70.2 kg ⋅ m/s ö÷
q = tan -1 ççç ÷÷÷ = tan -1 ççç
÷÷ = 77 north of east
è 15.9 kg ⋅ m/s ø÷
èç px ø÷
Insight: The dog makes the larger contribution to the momentum of the system because its mass is so much larger than
that of the cat.
73. An impulse is a force multiplied by a time, and it is equivalent to the change in momentum of the object that received
the impulse. A force is equal to the rate at which momentum changes, whereas an impulse is equal to the total change in
momentum.
74. Yes. A force is equal to the rate at which momentum changes, so a small force changes the momentum of an object at a
slower rate than does a large force. However, if it acts on the object for a much longer time than does the large force, it
can deliver a larger impulse.
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7 – 15
Chapter 7: Linear Momentum and Collisions
Pearson Physics by James S. Walker
75. In each case the school bus experienced the same change in momentum as its velocity slowed from its initial value to
zero. Because the change in momentum is the same, the magnitude of the impulse during the sudden stop is equal
to the magnitude of the impulse during the gradual stop.
76. The change in momentum of a driver during a crash is the same whether she hits the steering wheel or an air bag.
Therefore, the air bag does not change the impulse delivered to the driver. Instead, it makes the duration of the collision
longer, thereby reducing the force that is exerted on the driver. The reduced force is responsible for preventing injury.
77. Picture the Problem: A net force of 200 N acts on a 100-kg boulder, and a force of the same magnitude acts on a
100-g pebble.
Strategy: Use Newton’s second law in terms of momentum to answer the conceptual question.
Solution: 1. (a) One way to write Newton’s second law is to equate the net force with the change in momentum per
time. Both the boulder and the pebble have the same rate of momentum change because the same force acts on both
objects. We conclude that the change of the boulder’s momentum in one second is equal to the change of the pebble’s
momentum in the same time period.
2. (b) The best explanation is C. Equal force means equal change in momentum for a given time. Statement A is false,
and statement B is partly true (the pebble’s speed is larger) but the pebble’s change in momentum is the same because
it has a smaller mass than the boulder.
Insight: Newton originally formulated his second law by asserting that the net force is equal to the rate of change of
momentum. Later textbooks introduced the more familiar F = ma equation.
78. Picture the Problem: A net force of 200 N acts on a 100-kg boulder, and a force of the same magnitude acts on a
100-g pebble.
Strategy: Use Newton’s second law in terms of momentum to answer the conceptual question.
Solution: 1. (a) Both the boulder and the pebble have the same rate of momentum change because the same force acts
on both objects. As a result, the rate of change in velocity for the less massive object (the pebble) must be greater than
it is for the more massive object (the boulder). We conclude that the change of the boulder’s speed in one second is
less than the change of the pebble’s speed in the same time period.
2. (b) The best explanation is A. The larger mass of the boulder results in a smaller acceleration. Statement B is false
(the same force results in the same change in momentum, not speed), and statement C is true but irrelevant.
Insight: We know that the acceleration (rate of change of velocity) of an object is proportional to the force acting on
it and inversely proportional to its mass. These objects experience the same force, and therefore the less massive object
has the greater acceleration.
79. Picture the Problem: The forces and durations that produce four impulses are given.
Strategy: Multiply the force and the duration in each case to find the impulse, and then use the results to rank them.
I A = FA Δ tA = (1 N )(1 s ) = 1 kg ⋅ m/s
Solution: 1. Calculate the impulses:
I B = FB Δ tB = ( 2 N )( 0.3 s ) = 0.6 kg ⋅ m/s
I C = FC Δ tC = (5 N )( 0.1 s ) = 0.5 kg ⋅ m/s
I D = FD Δ tD = (10 N )( 0.07 s ) = 0.7 kg ⋅ m/s
2. Compare the impulses to arrive at the ranking: IC < IB < ID < IA .
Insight: When you compare two car crashes (for example, into a concrete barrier and into bales of hay) with the same
initial speed, the impulses are the same (because the change in momentum is the same) but the forces are different. The
greatest force corresponds to the smallest time interval, which in this case would correspond to the concrete barrier.
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7 – 16
Chapter 7: Linear Momentum and Collisions
Pearson Physics by James S. Walker
80. Picture the Problem: A baseball has momentum just before it lands on the ground after being dropped from rest.
Strategy: Use the definition of impulse to find the baseball’s time of flight.
I = Δ p = F Δt
Δ p Δ p 0.78 kg ⋅ m/s
Δt =
=
=
= 5.6 s
F
mg
0.14 N
Solution: Solve the definition of impulse for Δt :
Insight: The baseball’s momentum just before landing, together with its mass of
m = W g = ( 0.14 N ) 9.81 m/s 2 = 0.014 kg, can be used to determine that it lands with a speed of 56 m/s.
(
)
81. Picture the Problem: The force of a kick acting over a period of time delivers an impulse to a soccer ball.
Strategy: Multiply the force by the time of impact to find the impulse.
(
)
I = Fav Δ t = (1250 N ) 5.95 × 10−3 s = 7.44 kg ⋅ m/s
Solution: Apply the definition of impulse:
Insight: The 1250-N force is equivalent to 281 lb! The same impulse could be delivered to the ball with 7.44 N
(1.67 lb) of force acting over a period of one second.
82. Picture the Problem: A golf club delivers an impulse to a ball, imparting momentum.
Strategy: Use the momentum-impulse theorem to find the change of momentum of the ball and then its velocity.
Solution: 1. Apply the momentumimpulse theorem to find Δp :
Δp = Fav Δt = (2200 N)(0.0013 s) = 2.86 kg ⋅ m/s
2. Use the fact that vi = 0
Δp = m(vf − vi ) = mvf ⇒ vf =
vf :
Δp 2.86 kg ⋅ m/s
=
= 65 m/s
m
0.044 kg
Insight: The final speed is equivalent to 120 mi/h and will result in a downfield range of 373 m if the ball is launched at
a 30° angle above level ground. A pretty long tee shot!
83. Picture the Problem: A croquet mallet delivers an impulse to a ball, imparting momentum.
Strategy: Find the change in momentum of the croquet ball and then find vf using the momentum-impulse theorem.
Solution: Solve the momentumimpulse theorem for vf:
vf =
(
)
−3
Δ p F Δ t (190 N ) 7.2 × 10 s
=
=
= 2.6 m/s
m
m
0.52 kg
Insight: The large force (43 lb) is exerted over a very brief time to give the ball its small velocity (5.8 mi/h).
84. Picture the Problem: A thick gel delivers an impulse to a bullet, bringing it to rest.
Strategy: Find the change in momentum of the bullet and then find Δt using the momentum-impulse theorem.
Solution: Solve the momentumimpulse theorem for Δt :
Δt =
Δ p m ( vf − vi ) ( 0.0055 kg )( 0 − 325 m/s )
=
=
= 0.0033 s = 3.3 ms
−550 N
Fav
Fav
Insight: The large force (124 lb) is exerted over a very brief time to stop the bullet. However, the high speed of the
bullet means that the distance it travels through the gel is Δ x = v Δ t = (325 m/s )(0.0033 s ) = 1.1 m (3.5 ft).
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7 – 17
Chapter 7: Linear Momentum and Collisions
Pearson Physics by James S. Walker
85. Picture the Problem: The foot of a soccer player delivers an impulse to a ball, imparting momentum.
Strategy: Use the definition of impulse, together with the known mass and change of velocity of the ball, to determine
the impulse delivered to the ball.
Solution: Solve the definition of impulse:
I = Δ p = mΔv
= (0.425 kg)[(4.5 m/s) − ( −0.89 m/s)] = 2.3 kg ⋅ m/s
Insight: A impulse is in the same direction as the momentum change, forward in this case.
86. Picture the Problem: A marble drops straight down from rest and rebounds from the floor.
Strategy: Use the velocity-position equation to find the speed of the marble just before it hits the floor. Use the same
equation and the known rebound height to find the rebound speed. Then use the definition of impulse to find the
impulse delivered to the marble by the floor. Let upward be the positive direction, so that the marble hits the floor
with speed −vi and rebounds upward with speed vf .
Solution: 1. (a) Find vi
using the velocity-position equation:
vi2 = v02 − 2 g Δy
2. Find vf using the same equation:
v 2 = vf2 − 2 g Δy
(
)
vi = − v02 − 2 g Δy = − 02 − 2 9.81 m/s 2 ( −1.4 m ) = −5.24 m/s
(
)
vf = v 2 + 2 g Δy = 02 + 2 9.81 m/s 2 ( 0.64 m ) = 3.54 m/s
3. Use the definition of impulse to find I:
I = mΔv = m ( vf − vi )
= ( 0.015 kg ) 3.54
(
5.24) m/s
0.13 kg m/s upward
4. (b) If the marble had bounced to a greater height, its rebound speed would have been larger and the impulse would
have been greater than the impulse found in part (a).
Insight: By Newton’s third law the marble also delivers a downward impulse on the floor. The Earth theoretically
moves in response to this impulse, but its change of velocity is tiny (2.2×10-26 m/s) due to its enormous mass.
87. The total momentum of a system can only be changed by an external force, not an internal force.
88. No. Internal forces in a system can change the individual momenta of the objects in the system without changing the
total momentum.
89. If a 2.0-kg cart traveling at 3.0 m/s collides elastically with a 4.0-kg cart that is at rest, the 2.0-kg cart will rebound with
a velocity of −1.0 m/s and the 4.0-kg cart will move forward at 2.0 m/s. In this case the total momentum is +6.0 kg·m/s
before and after the collision, but the total velocity is +3.0 m/s before the collision and +1.0 m/s after the collision.
90. From the standpoint of Newton’s third law, a rocket exerts a force on the exhaust particles, and the exhaust particles
exert an equal and opposite force on the rocket. From the standpoint of conservation of momentum, the large backward
momentum of the exhaust particles is balanced by the forward momentum of the rocket, so that the total momentum of
the system remains unchanged by the internal forces between the rocket and the exhaust particles. In either case the
rocket is propelled forward without needing anything to push against.
91. An astronaut that is floating near the space shuttle can propel himself by throwing a large wrench directly away from
the space shuttle. The forward momentum of the wrench will be balanced by the backward momentum of the astronaut,
allowing the astronaut to move toward the shuttle.
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7 – 18
Chapter 7: Linear Momentum and Collisions
Pearson Physics by James S. Walker
92. Momentum increases linearly with speed, but kinetic energy increases with the square of speed. Therefore, the small
bullet with the large velocity will have a much greater kinetic energy than the more massive gun with the small recoil
velocity. The bullet’s ability to deliver a large kinetic energy over a small area is responsible for damage that it can
cause.
93. Picture the Problem: Two ice skaters at rest then push apart and move in opposite directions without friction.
Strategy: By applying the conservation of momentum we conclude that the total momentum of the two skaters after the
push is zero, just as it was before the push. Set the total momentum of the system to zero and solve for v2 . Let the


velocity v1 point in the negative direction, v 2 in the positive direction.
 



p1 + p 2 = 0 = m1 v1 + m2 v 2 = − m1v1 + m2 v2
Solution: Set p total = 0 and solve for v2 :
m v ( 54 kg )( 0.95 m/s )
v2 = 1 1 =
= 0.84 m/s
m2
61 kg
Insight: The more-massive skater 2 has the slower speed to ensure that the momenta of the two skaters are equal and
opposite.
94. Picture the Problem: Two people on roller blades push apart and move in opposite directions without friction.
Strategy: By applying the conservation of momentum we conclude that the total momentum of the two people after
the push is zero, just as it was before the push. Set the total momentum of the system to zero and solve for the speed


of the second person (Tracey). Let the velocity v Jonas point in the negative direction, v Tracey in the positive direction.





p Jonas + p Tracey = 0 = mJonas v Jonas + mTracey v Tracey
Solution: Set p total = 0 and solve for vTracey :

− mJonas v Jonas − ( 45 kg )( − 0.55 m/s )

v Tracey =
=
= 0.77 m/s
mTracey
32 kg
Insight: The less-massive rollerblader Tracey has the greater speed to ensure that the momentum of the two
rollerbladers are equal and opposite.
95. Picture the Problem: The locations of the canoeist and the canoe relative
to shore in each configuration are depicted in the figure at right.
Strategy: Because there are no external forces, the momentum of the
canoe-person system remains zero, both before and after the canoeist
begins walking. Use the conservation of momentum to find the speed of
the canoe relative to the shore.
Solution: 1. (a) Use conservation of
momentum to find the speed of the canoe.
Let the positive direction be to the right
(the direction the canoeist is walking):
vcanoe
0.95 m/s


p f = pi
mperson vperson − mcanoe vcanoe = 0
vcanoe =
=
mperson vperson
mcanoe
(63 kg )(0.95 m/s) = 1.9 m/s
32 kg
2. (b) If the mass of the canoe is increased, the speed found in part (a) will be less than before because a smaller canoe
speed is needed to ensure its backward momentum matches the forward momentum of the canoeist.
3. Repeat the calculation of part (a)
with the larger canoe mass:
vcanoe =
mperson vperson
mcanoe
=
(63 kg )(0.95 m/s) = 1.5 m/s
39 kg
Insight: If the canoe were very light, its speed would be high compared to the speed of the canoeist, and the canoeist
would hardly move at all toward the shore before she reached the end of the canoe. Boat recoil similar to this example
has led to many a boater falling in the water when they attempt to step off the boat and onto a dock!
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7 – 19
Chapter 7: Linear Momentum and Collisions
Pearson Physics by James S. Walker
96. Picture the Problem: An astronaut and a box move in opposite directions after the astronaut pushes on the box. The
box travels at constant speed a certain distance before reaching a second astronaut.
Strategy: As long as there is no friction the total momentum of the first astronaut and the box must remain zero, as it
was before the astronaut pushed it. Use the conservation of momentum to determine the speed of the box, and then
divide the distance by the speed to find the time. Assume the box’s motion is in the negative x-direction.
pa + ps = 0 = ma va + mbox vbox
Solution: 1. Find the speed of the box
using conservation of momentum:
vbox = −
ma va
(86 kg)( −0.26 m/s)
=−
= 1.86 m/s
mbox
12 kg
d
4.5 m
=
= 2.4 s
vbox 1.86 m/s
Insight: One of the tricky things about spacewalking is that whenever you push on a box or anything else, because of
Newton’s third law, you yourself get pushed! But conservation of momentum makes it easy to predict your speed.
t=
2. Find the time to reach the second astronaut:
97. Picture the Problem: A lumberjack moves to the right while the log he stands
upon moves to the left.
Strategy: As long as there is no friction the total momentum of the lumberjack
and the log remains zero, as it was before the lumberjack started trotting. Use the
conservation of momentum to find speed of log vlog relative to the shore. Let vL
be the speed of the lumberjack relative to the shore.

p total = 0 = mL vL − mlog vlog
Solution: 1. (a) Write out the
conservation of momentum with
mv
(85 kg )( 2.7 m/s )
respect to the shore and solve for
vlog = L L =
mlog
380 kg
the speed of the log:
= 0.60 m/s
2. (b) If the mass of the log had been greater, its speed relative to the shore would have been less than that found in part
(a), because the log would still have a momentum that is equal and opposite to the lumberjack, but a larger mass.
3. (c) Use the expression from step 1
to find the new speed of the log:
vlog =
mL vL (85 kg )( 2.7 m/s )
=
= 0.51 m/s
450 kg
mlog
Insight: Taking the argument in (b) to its extreme, if the mass of the log equaled the mass of the Earth, the speed of the
log would essentially be zero, and the lumberjack’s speed would be exactly 2.7 m/s relative to the shore (and the log). If
the mass of the log were the same as the mass of the lumberjack, their speeds relative to the shore would be equal.
98. A completely inelastic collision is the only situation in which the two objects stick together after the collision and move
as if they were a single object.
99. No. If you consider each billiard ball as a separate system, then the force applied to each ball by the collision with the
other ball is considered an external force that changes the ball’s momentum. If you regard the pair of balls as the
system, then the collision forces are internal forces that do not change the momentum of the system.
100. As long as there are no external forces, both the total momentum and the total kinetic energy are conserved in an elastic
collision.
101. Energy is lost in an inelastic collision, so the total kinetic energy of the system is not conserved.
102. No. Kinetic energy is a scalar quantity, not a vector quantity, so its value cannot be divided up into x and y components.
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7 – 20
Chapter 7: Linear Momentum and Collisions
Pearson Physics by James S. Walker
103. Yes. For example, if two objects have momenta of equal magnitude and collide head-on and stick together, they will be
at rest after the collision, and the system will have zero kinetic energy.
104. Picture the Problem: Two carts collide on a frictionless track and stick together.
Strategy: The collision is completely inelastic because the two carts stick together. Momentum is conserved during
the collision because the track has no friction. The two carts move as if they were a single object after the collision.
Use the conservation of momentum to find the final speed of the carts and final kinetic energy of the system.
pi = pf
Solution: 1. Conserve
momentum to find the final
speed of the carts:
m1v1, i + m2 (0) = ( m1 + m2 ) vf
2. Calculate the final kinetic energy:
K f = 12 (m1 + m2 )vf 2 =
m1
0.10 kg
v1, i = vf =
(0.66 m/s) = 0.22 m/s
0.10 + 0.20 kg
m1 + m2
2
Insight: The initial kinetic energy was KEi = 12 m1v1,i
+0=
1
2
1
2
(0.10 + 0.20 kg )(0.22 m/s )2 =
0.0073 J
(0.10 kg )(0.66 m/s)2 = 0.0218 J . Two-thirds of the initial
kinetic energy is gone, having been converted to heat, sound, and permanent deformation of material during the
inelastic collision.
105. Picture the Problem: A train car collides with another, and the two cars stick together after the collision.
Strategy: Use conservation of momentum to determine the horizontal speed of the two train cars after the collision.
Let each car travel in the positive direction. Find the total kinetic energy of the system before and after the collision
to determine the energy that is lost in the collision.


Solution: 1. (a) Set pi = p f and solve for vf :
mv1, i + mv2,i = ( m + m ) vf
vf =
=
2m
2.50 + 1.00 m/s
= 1.75 m/s
2
(
)
2. (b) Find the total kinetic
energy before the collision:
KEi = mv + mv2,2 i = 12 m v1,2 i + v2,2 i
3. Find the total kinetic energy
after the collision:
KEf = 12 mv1,2 f + 12 mv2,2 f =
4. Find the energy lost during the collision:
Δ KE = KEf − KEf = 90, 625 − 76,563 J = 14,100 J
1
2
=
1
2
2
1,i
m v1, i + m v2, i
1
2
( 25, 000 kg ) ( 2.50 m/s) + (1.00 m/s )2
2
= 90, 625 J
( 2m) vf2
2
= 12 ( 2 × 25, 000 kg )(1.75 m/s ) = 76,563 J
1
2
Insight: Note that the 1.75 m/s final speed is the average of the two initial speeds weighted by the mass of the cars.
Because the two cars have equal mass, the final speed is just the average of 2.50 m/s and 1.00 m/s.
106. Picture the Problem: A truck strikes a car from behind. The collision sends the car lurching forward and slows down
the speed of the truck.
Strategy: This is a one-dimensional, elastic collision where one of the objects (the car) is initially at rest. Therefore,
the equations for the velocities after an elastic collision apply and can be used to find the final speeds of the vehicles.
Let m1 be the mass of the truck, m2 be the mass of the car, and v be the initial speed of the truck.
Solution: 1. Use the elastic
collision equation to find v1,f :
æ m - m2 ÷ö
æ1720 - 732 kg ÷ö
÷÷ v = çç
÷÷(15.5 m/s) = 6.25 m/s = vtruck
v1,f = ççç 1
ç
çè m1 + m2 ø÷
è1720 + 732 kg ø÷
2. Use the elastic collision
equation to find v2,f :
é 2 (1720 kg ) ù
æ 2m1 ö÷
ú
÷÷ v = ê
v2,f = ççç
ê1720 + 732 kg ú (15.5 m/s) = 21.7 m/s = vcar
÷
çè m1 + m2 ø
ë
û
Insight: The elastic collision produces a bigger jolt for the car. If the collision were instead inelastic and the two
vehicles stuck together, the final speed of the car (and the truck) would be 10.9 m/s.
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7 – 21
Chapter 7: Linear Momentum and Collisions
Pearson Physics by James S. Walker
107. Picture the Problem: This is a similar question to Guided Example 7.10.
Two rugby players collide in midair. (The figure shows American football
players, but the basic idea is similar.) Each player has zero velocity after
the collision.
m1 = 92 kg v1 = 7.5 m/s m2 = 112 kg v2 = ?
Strategy: The total momentum of the two players is conserved, so that it
must be zero both before and after the collision. Use this fact to determine
the initial speed and momentum of the 112-kg player, then compare the
initial and final momentum of the players to compare the change in
momentum for each.




p1, i + p 2, i = 0 = m1 v1, i + m2 v 2,i

m1 v1, i
(92 kg )(7.5 m/s ) = − 6.16 m/s

v 2, i = −
=−
m2
(112 kg )
Solution: 1. (a) Find the initial velocity
of the 112-kg player by applying the
conservation of momentum:
2. This velocity can be used to label the above diagram of the collision. The remaining question parts analyze the
change in momentum of each player.


3. (b) Find the initial momentum
p1, i = m1 v1,i = (92 kg )( 7.5 m/s ) = 690 kg ⋅ m/s
of the 92-kg player:



4. (c) Find the change in momentum of
Δp1 = p1, f − p1, i = 0 − 690 kg ⋅ m/s = − 690 kg ⋅ m/s
the 92-kg player due to the collision:


5. (d) The initial momentum of the 112-kg player is p 2, i = m2 v 2, i = (112 kg )( − 6.16 m/s ) = − 690 kg ⋅ m/s. Because this
player also has zero momentum after the collision, the change in momentum of the 112-kg player due to the collision is
+690 kg m/s. We conclude that the two changes in momentum are equal in magnitude and opposite in direction.
6. (e) The initial speed of the 112-kg player was found above to be 6.16 m/s.
Insight: Note that the more massive player has a smaller speed in order for the two momenta to have equal magnitudes.
108. Picture the Problem: The initial and final momentum vectors for this
collision between two hockey players are depicted at right.
y

p 2i
Strategy: Assuming there is no friction between the players’ skates and
the ice, we can use conservation of momentum together with the fact that
the players stick together after the collision to find the final velocity. Let
the motion of player 1 be in the positive x-direction and the motion of
player 2 be at an angle of 90° measured counterclockwise from the
positive x-axis.
Solution: 1. Write out the
conservation of momentum along
the x direction and solve for vx , f :
p1x , i + p2x , i = px , f
2. Write out the conservation of
momentum along the y direction
and solve for v y , f :
p1 y , i + p2y , i = p y , f

3. Determine the magnitude of v f :
vf = vx2, f + v y2, f =

pf

p1i
x
m v + 0 = 2 m vx , f
1
2
v = vx , f =
1
2
(5.45 m/s) = 2.725 m/s
1
2
(5.45 m/s) = 2.725 m/s
0 + mv = 2mv y , f
1
2
v = vy, f =
( 2.725 m/s)2 + ( 2.725 m/s)2
= 3.85 m/s
Insight: The two players slide away from the collision at 45.0° above the positive x axis. Note that because the mass
cancels out, in this particular case only (the case of two identical masses colliding) the final velocity is the average of






the two initial velocities: mv1i + mv 2i = 2mv f
v f = 12 ( v1i + v 2i ) .
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7 – 22
Chapter 7: Linear Momentum and Collisions
Pearson Physics by James S. Walker
109. Picture the Problem: Two curling stones collide dead center on an ice rink. Stone 1 moves toward the north and
collides with stone 2, which is initially at rest.
Strategy: This is a one-dimensional, elastic collision where one of the objects (stone 2) is initially at rest. Therefore, the
equations for the velocities after an elastic collision apply and can be used to find the final speeds of the two stones. Let
m1 be the mass of stone 1, m2 be the mass of stone 2, and v be the initial speed of stone 1.
Solution: 1. (a) Use the elastic
collision equation to find v2,f :
é 2 (21 kg ) ù
æ 2m1 ö÷
ú (1.7 m/s) = 1.9 m/s
÷÷ v = ê
v2,f = ççç
ê
ú
çè m1 + m2 ø÷
ë 21 + 16 kg û
2. (b) Use the elastic collision
equation to find v1,f :
æ m - m2 ÷ö
æ 21-16 kg ÷ö
÷÷ v = çç
÷(1.7 m/s) = 0.23 m/s
v1,f = ççç 1
çè 21 + 16 kg ÷÷ø
èç m1 + m2 ø÷
Insight: If the two stones had equal mass, stone 1 would completely come to rest and stone 2 would have a final
velocity of 1.7 m/s toward the north. One interesting fact about elastic collisions is that the velocity difference
before and after the collision is the same, even if the masses are not. In this case, the initial velocity difference of
1.70 – 0 m/s = 1.70 m/s is the same as the final velocity difference of 1.93 – 0.23 m/s = 1.70 m/s.
110. Picture the Problem: Two curling stones collide dead center on an ice rink. Stone 1 moves toward the north and
collides with stone 2, which is initially at rest.
Strategy: This is a one-dimensional, elastic collision where one of the objects (stone 2) is initially at rest. Therefore,
the equations for the velocities after an elastic collision apply and can be used to find the mass of stone 2 and the final
speed of stone 1. Let m1 be the mass of stone 1, m2 be the mass of stone 2, and v be the initial speed of stone 1.
Solution: 1. (a) Use the elastic
collision equation to find v2,f :
æ 2m1 ö÷
÷v
v2,f = ççç
çè m1 + m2 ø÷÷
2. Multiply both sides by m1 + m2 and
divide both sides by v2, f v :
m1 + m2 =
3. Subtract m1 from both sides and
substitute numerical values to find m2:
m2 =
4. (b) Use the elastic collision equation to find v1,f :
æ m - m2 ÷ö
æ16 - 54 kg ÷ö
÷÷ v = çç
÷÷(1.5 m/s) = - 0.81 m/s
v1,f = ççç 1
ç
çè m1 + m2 ø÷
è16 + 54 kg ø÷
v2,f
v
=
2m1
m1 + m2
2m1
v2,f v
v
1.5 m/s
( 2m1 ) − m1 = 0.69 m/s ( 2 × 16 kg ) − 16 kg = 54 kg
v2,f
Insight: The less massive stone 1 rebounds backward after colliding with stone 2. One interesting fact about elastic
collisions is that the velocity difference before and after the collision is the same, even if the masses are not. In this
case, the initial velocity difference of 1.50 – 0 m/s = 1.50 m/s is the same as the final velocity difference of
0.69 − (− 0.81) m/s = 1.50 m/s.
111. Picture the Problem: A bullet collides with a block, passing completely through it and continuing in a straight line,
but at a slower speed than initially. Afterward the block moves with constant speed in the same direction as the bullet.
Strategy: Use conservation of momentum to find the speed of the bullet after the collision. Because this is an inelastic
collision we expect a loss of kinetic energy. Calculate the initial and final kinetic energies and confirm the energy loss.
Let the subscripts b and B denote the bullet and the block, respectively.


mb vb i + 0 = mb vb f + mB vB f
Solution: 1. (a) Let pi = p f and solve for vb f :
vb f =
=
mb vb i − mB vB f
mb
(0.0040 kg )(650 m/s) − (0.095 kg )( 23 m/s)
0.0040 kg
= 104 m/s = 1.0 × 102 m/s
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7 – 23
Chapter 7: Linear Momentum and Collisions
Pearson Physics by James S. Walker
2. (b) The final kinetic energy is less than the initial kinetic energy because energy is lost to the heating and
deformation of the bullet and block.
(0.0040 kg )(650 m/s)2 =
3. (c) Use its definition to find KEi :
KEi = 12 mb vb i 2 =
4. Use its definition to find KEf :
KEf = 12 mb vb f 2 + 12 mB vB f 2
=
1
2
1
2
850 J
(0.0040 kg )(104 m/s)2 + 12 (0.095 kg )( 23 m/s)2 =
47 J
Insight: In this collision 94% of the bullet’s initial kinetic energy is converted to heat, sound, and permanent
deformation of materials.
112. Picture the Problem: A wad of putty is thrown horizontally, strikes the side of a block, and sticks to it. The putty and
the block move together in the horizontal direction immediately after the collision, compressing a spring.
Strategy: Use conservation of momentum to find the speed of the putty-block conglomerate immediately after the
collision, then find the total kinetic energy. Use conservation of energy (convert the kinetic energy into the potential
energy stored in the spring) to find the maximum compression of the spring after the collision.


Solution: 1. Set pi = p f and solve for vf :
0 + mp vp = (mb + mp ) vf
æ mp ö÷
æ
ö÷
0.0500 kg
÷÷ vp = çç
÷÷(2.30 m/s) = 0.240 m/s
vf = ççç
ç
÷
è 0.430 + 0.0500 kg ø÷
èç mb + mp ø÷
2. Set Eafter = Erest after the collision and
solve the resulting expression for xmax :
1
2
(m
b
K after + 0 = 0 + U rest
)
2
+ mp vf2 = 12 kxmax
mb + mp
k
3. Substitute the numerical values:
xmax =
vf2 = xmax
0.430 + 0.0500 kg
(0.240 m/s)2 = 0.0372 m = 3.72 cm
20.0 N/m
Insight: The putty-block conglomerate will compress the spring even farther if vp is larger or if mp is larger.
113. Picture the Problem: The cart 4m collides with the cart 2m, which is
given kinetic energy as a result and later collides with the cart m.
v
Strategy: In each case a moving cart collides with a cart that is at rest,
so that the elastic collision equations given in the textbook will yield the
final velocities of all the carts. First apply the equations to the collision
between carts 4m and 2m, then to the collision between 2m and m. Let
the 4m cart be called cart 4, the 2m cart be called cart 2, and the m cart
be called cart 1:
æ m - m2 ö÷
æ 4m - 2m ö÷
÷v = ç
v4,f = ççç 4
÷ v = 13 v
çè m4 + m2 ø÷÷ 4,i çèç 4m + 2m ø÷
Solution: 1. (a) Apply the elastic collision
equations to the first collision:
é 2 ( 4m) ù
æ 2m4 ö÷
4
ú
÷÷ v4,i = ê
v2,f = ççç
ê 4m + 2m ú v = 3 v
÷
çè m4 + m2 ø
ë
û
2. Apply the elastic collision equation to the
second collision. In this case cart 2 has an
initial speed of 43 v :
æ m - m1 ö÷
æ 2m - m ÷ö 4
÷v = ç
v2,f = ççç 2
÷( v ) =
çè m2 + m1 ÷÷ø 2,i ççè 2m + m ÷ø 3
4
9
v
é 2 ( 2m) ù
æ 2m2 ö÷
ú 4
÷ v2,i = ê
v1,f = ççç
÷
ê 2m + m ú ( 3 v ) =
çè m2 + m1 ÷ø
ë
û
16
9
v
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7 – 24
Chapter 7: Linear Momentum and Collisions
Pearson Physics by James S. Walker
3. (b) Verify that K i = K f by applying the definition of
kinetic energy and dividing both sides by mv02 :
1
2
?
( 4m) v 2 = 12 ( 4m) ( 13 v )
2=
2
+ 12 ( 2m ) ( 94 v ) + 12 ( m ) ( 169 v )
2
2
2 16 256 36 32 256 324
+ +
=
+
+
=
=2
9 81 162 162 162 162 162
Insight: Note that due to the transfer of kinetic energy via collisions, the cart with the smallest mass ends up with the
largest speed and the largest kinetic energy.
114. Picture the Problem: Three objects have different masses but the same momentum.
Strategy: Write the kinetic energy of an object in terms of its momentum: K = 12 mv 2 = m 2 v 2 2m = p 2 2m . Use this
equation to determine the ranking of the kinetic energies.
Solution: The kinetic energy is inversely proportional to the mass when the momentum is constant. The smallest mass
will therefore have the highest kinetic energy. We arrive at the ranking KEB < KEA < KEC .
Insight: The smaller mass will have a higher speed in order to have the same momentum as a larger mass. This will
give the smaller mass a larger kinetic energy because K is proportional to the square of the speed.
115. Picture the Problem: Three objects have different masses but the same kinetic energy.
Strategy: Write the momentum of an object in terms of its kinetic energy: p 2 2m = K
p = 2 m K . Use this
equation to determine the ranking of the momenta.
Solution: The momentum is proportional to the square root of the mass when the kinetic energy is constant. The
smallest mass will therefore have the smallest momentum. We therefore arrive at the ranking pC < pA < pB .
Insight: Object A has four times more mass and twice the momentum of object C.
116. Picture the Problem: A block of wood is struck by a bullet. The bullet either embeds itself in the wood or it rebounds.
Strategy: Consider the change in the momentum of the bullet for the two scenarios. The larger the change
in momentum of the bullet, the larger the impulse the wood must deliver to the bullet, and by Newton’s
third law the larger the impulse the bullet delivers to the wood.
Solution: 1. (a) If the bullet bounces off the wood then its change in momentum is greater than if it embeds itself into
the wood and comes to rest. We conclude that block is more likely to be knocked over if the bullet is rubber and
bounces off the wood.
2. (b) The best explanation is A. The change in momentum when a bullet rebounds is twice as much as when it is
brought to rest. Statement B is true, but irrelevant, and statement C is false.
Insight: If the bullet were to bounce elastically, with the same rebound speed as its initial speed, the impulse delivered
to the wood is exactly twice the impulse delivered if the bullet embeds itself in the wood.
117. Picture the Problem: A small navigational rocket delivers an impulse to the Voyager II spacecraft, changing its
momentum and its speed.
Strategy: Find the change in momentum of the spacecraft and then find Δt using the momentum-impulse theorem.
Solution: Solve the momentumimpulse theorem for Δt :
Δt =
Δ p m ( vf − vi ) ( 722 kg )(1.0 m/s )
=
=
= 29 s
Fav
Fav
25 N
Insight: A rocket converts chemical energy into kinetic energy of the exhaust particles, and the momentum of the
particles gives the spacecraft an equal and opposite momentum.
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7 – 25
Chapter 7: Linear Momentum and Collisions
Pearson Physics by James S. Walker
118. Picture the Problem: A moving car collides with a stationary car that has half the mass of the first car.
Strategy: Conserve momentum in order to find the speed of the second car. Then calculate the initial and final kinetic
energies of the entire system to determine if any energy is lost. If energy is lost the collision is inelastic.


mv1, i + 0 = mv1, f + ( 12 m ) v2, f
Solution: 1. (a) Set pi = p f and solve for v2f :
v2, f =
(
m v1, i − v1, f
m2
) = 2 (v − v 3) =
2. (b) Use its definition to find KEi :
KEi = 12 mv 2
3. Use equation 7-6 to find K f :
K f = 12 mv1f2 + 12 ( 12 m ) v2f2
4
v
3
2
1
= 12 m ( 13 v ) + 12 ( 12 m ) ( 34 v ) = 12 mv 2 ( 19 + 16
18 ) = 2 mv
2
2
4. Because K i = K f we conclude the collision is elastic.
Insight: The cars bounce off each other and separate because the final speed of the second car, ( 4 3) v , is greater
than the final speed of the first car, v 3. You can verify that the first car transfers 8 9 of its initial kinetic energy
to the second car.
119. Picture the Problem: An apple falls vertically downward due to gravity.
Strategy: The rate of change of the apple’s momentum is the net force acting on it, according to Newton’s second law.
The only force acting on the apple is its weight mg. Calculate the change in momentum produced by gravity using the
momentum-impulse theorem.
Solution: 1. (a) Solve Newton’s
second law for Δp Δt :
Δp
= F = 2.7 N
Δt
2. (b) Solve the momentum-impulse theorem for Δp :
Δp = F Δt = ( 2.7 N )(1.4 s ) = 3.8 kg ⋅ m/s
Insight: We could also write the units of the answer to part (a) as 2.7 kg ⋅ m/s s. That is, a force in Newtons is the same
as a rate of change of momentum.
120. Picture the Problem: A hockey player tosses a helmet in one direction and recoils in the other.
Strategy: The total horizontal momentum of the hockey player mp and helmet mh is conserved because there is no
friction. The total horizontal momentum remains zero both before and after the helmet is tossed. Use this fact to find the
mass of the player. Let the helmet be tossed in the positive direction.
Solution: Set px , i = px , f and solve for mp :
Σp
x
= 0 = mp vp, x + mh vh, x
mp = −
mh vh, x
vp, x
=−
(1.3 kg )(6.5 m/s ) cos11° =
− 0.25 m/s
33 kg
Insight: This must be a junior player (a 33 kg player weighs 73 lb). If a 70-kg adult tossed the helmet in the same
manner, the recoil velocity would be − 0.118 m/s. Note that the momentum in the vertical direction is not conserved; the
normal force from the ice is an external force that allows the vertical momentum of the system to change from zero to
mh vh sin11° = 1.6 kg ⋅ m/s .
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7 – 26
Chapter 7: Linear Momentum and Collisions
Pearson Physics by James S. Walker
121. Picture the Problem: An apple falls from rest and then is caught.
Strategy: Use conservation of energy to find the speed of the apple after it has fallen from rest a known distance. Then
use the momentum-impulse theorem, in the form of Newton’s second law, to find the average force required to bring the
apple to rest. Note that two forces act on the apple while it is being brought to rest, the force of gravity and the force of
your hand.
0 + PEi = KEf + 0
Solution: 1. (a) Use conservation of
energy to find the apple’s speed:
mghi = 12 mvf2
(
)
2 ghi = vf = 2 9.81 m/s 2 (3.5 m ) = 8.3 m/s
2. (b) Apply the momentum-impulse
theorem in the form of Newton’s second
law. Let upward be the positive direction:
Σ=F
hand
− mg =
Fhand = mg +
Δp
Δt
m ( vf − vi )
Δt
(
)
= ( 0.22 kg ) 9.81 m/s 2 +
(0.22 kg )
0
(
8.3 m/s )
0.28 s
= 2.2 N + 6.5 N = 8.7 N
Insight: The force of your hand must do two things: it must support the weight of the apple and it must change the
apple’s momentum, bringing it to rest.
122. Picture the Problem: Falling raindrops during a storm are stopped by the upward force from the ground.
Strategy: Calculate the rate Δm Δt at which water is delivered to a square meter of the ground, and then use the
momentum-impulse theorem to estimate the force.
Solution: 1. Find the rate at which
the water fell over 1.0 m2:
Dm æç 0.79 m öæ
÷÷çç 1 h ö÷÷´(1.0 m 2 )æçç1000 kg ö÷÷ = 0.024 kg/s
=ç
ç
çè m3 ø÷
Dt è 9.0 h ÷øèç 3600 s ÷ø
2. Find the force from the
momentum-impulse theorem:
Fav =
D p çæ D m ÷ö
÷ v = (0.024 kg/s)(10 m/s) = 0.24 N
=ç
Dt çè Dt ÷ø÷
Insight: Because a Newton is about ¼ lb, this force is about 1 16 lb or about 1 oz on a square meter. However, if you
allow all 31 in of water to pool on top of the square meter it will weigh 7.7 kN = 1700 lb = 0.87 ton!
123. Picture the Problem: A car traveling west collides with a minivan that is traveling east. The two cars stick together and
move in the eastward direction after the collision.
Strategy: This is a completely inelastic collision because the two cars stick together. Use the conservation of
momentum to find the mass of the minivan. Let eastward be the positive direction.


pi = pf
Solution: 1. Write a statement of the



conservation of momentum:
mcar v car, i + mmv v mv, i = ( mcar + mmv ) v f




mcar v car, i − mcar v f = mmv v f − mmv v mv, i
2. Rearrange to solve for the mass


 
mmv of the minivan:
mcar v car, i − v f = mmv v f − v mv,i


v car,i − v f
( −12 m/s ) − ( +1.5 m/s)
mcar  
= mmv = (1540 kg )
v f − v mv, i
( +1.5 m/s ) − ( +9.4 m/s)
(
)
(
)
mmv = 2630 kg = 2.6 × 103 kg
Insight: The more massive minivan has an initial momentum of +24,740 kg·m/s (eastward) while the car has an initial
momentum of −18,480 kg·m/s (westward). The total momentum is thus 6,260 kg·m/s eastward and the two move
toward the east after they stick together.
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7 – 27
Chapter 7: Linear Momentum and Collisions
Pearson Physics by James S. Walker
124. Picture the Problem: A bullet strikes a wood block and together they move horizontally across a rough surface before
coming to rest.
Strategy: Use conservation of momentum to determine the horizontal speed of the bullet m and block M after the
collision. Find the kinetic energy of the bullet-block combination, and then apply the work-energy theorem to find the
friction force. Finally, use the weight of the bullet-block combination to find the coefficient of kinetic friction.


Solution: 1. Set pi = p f and
mvi + M (0) = (m + M ) vf
solve for vf :
æ 0.011 kg ö÷
æ m ö÷
÷÷(670 m/s) = 4.31 m/s
vf = çç
vi = ççç
÷
÷
çè m + M ø
è 0.011 + 1.7 kg ø÷
( m + M ) vf2 = 12 (0.011 + 1.7 kg )( 4.31 m/s)2
2. Find the kinetic energy of the
bullet-block combination:
KE =
3. Use the work-energy theorem
to find the friction force:
W = − f d = Δ KE
4. Use the relationship between friction
force and normal force to find the
coefficient of kinetic friction:
f = μk N = μk ( m + M ) g
1
2
= 15.9 J
f =−
μk =
KEf − KEi
0 − 15.9 J
=−
= 6.63 N
d
2.4 m
6.63 N
f
=
= 0.39
( m + M ) g (0.011 + 1.7 kg ) 9.81 m/s 2
(
)
Insight: If the coefficient of friction were smaller, the block would slide farther than 2.4 m. If the surface were
frictionless, the block would still be sliding, and would never stop!
125. Picture the Problem: A bullet strikes a wood block and together they are
launched horizontally off the edge of the table.
Strategy: Use conservation of momentum to determine the horizontal
speed of the bullet and block, then apply the equations that apply to a
horizontally-launched projectile to find the landing site of the bullet-block
combination. Let m = the mass of the bullet, M = the mass of the block,
vi = the initial speed of the bullet, and h = the height of the table.


Solution: 1. Set pi = p f and
solve for vf :
2. Find the time required for
the bullet-block combination
to drop a distance h. See the
position-time equations for
projectiles in Chapter 4:
mvi + M (0) = (m + M ) vf
æ 0.0105 kg ö÷
æ m ö÷
÷(715 m/s) = 5.52 m/s
vf = çç
÷ v = çç
èç m + M ø÷ i çè 0.0105 + 1.35 kg ÷ø÷
y = yi + v y , i t − 12 g t 2
0 = h + 0 t − 12 g t 2
1
2
g t2 = h
t=
2. Now find the horizontal
distance traveled during the
time the bullet-block
combination is falling:
2h
=
g
2 ( 0.782 m )
9.81 m/s 2
= 0.399 s
xf = xi + vx , i t
d = xf − xi = vx ,i t = vf t = (5.52 m/s )( 0.399 s ) = 2.20 m
Insight: If the block were twice as massive it would land about half as far, 1.11 m. In this way the landing spot is a
measure of the mass of the block, as long as the block is much more massive than the bullet.
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7 – 28
Chapter 7: Linear Momentum and Collisions
Pearson Physics by James S. Walker
126. Picture the Problem: The two moving carts collide with the
v
v
4m cart that is at rest.
Strategy: Use conservation of momentum to find the speed
of the carts before and after the two collisions. Because the
momentum always remains the same, it is not necessary to
calculate the intermediate speed of carts 2 and 3 before cart
1 collides with them. Finally, find the ratio of the final kinetic
energy to initial kinetic energy.


mv + 2mv + 0 = ( m + 2m + 4m ) vf
Solution: 1. (a) Set pi = p f and solve for vf :
3v = 7vf
2. (b) Calculate K f K i :
Kf
=
Ki
1
2
(7m) ( 73 v )
2
1
2 ( 3m ) v
vf =
2
=
3
7
v
3
7
Insight: If you calculate the intermediate speed of the 2m and 4m carts after they collide, you should get v 3 . In part
(b) we learn that 4 7 or 57% of the initial energy is dissipated as heat, sound, and permanent deformation of material.
127. Answers will vary. When a tennis racket strikes a tennis ball, the force the racket exerts on the ball is equal and opposite
to the force the ball exerts on the racket. The impulse delivered to the ball brings it to rest and then gives it a high speed
in the opposite direction. The impulse delivered to the racket slows it down a bit.
128. Momentum is a combination of velocity and inertia, a property of mass. For instance, a 0.050-kg bullet traveling at 648
m/s has the same momentum as a 7.2-kg bowling ball traveling at 4.5 m/s. On the other hand, a 25,000-kg railroad car
traveling at 5.0 m/s has a momentum of 125,000 kg·m/s, much larger than the 400 kg·m/s momentum of an 80-kg
bicycle rider that is riding at the same speed.
129. Picture the Problem: The spacecraft approaches the planet with speed
vi , interacts (“collides”) with it, and recedes in the opposite direction
with speed vf , as depicted in the figure at right.
Strategy: Subtract the initial velocities in order to find the speed of
approach relative to an observer on the planet.
Solution: The speed of approach can be found by subtracting the two
initial velocities. The calculated answer is choice A.

 
v app = u − v i
vapp = u − ( −vi ) = u + vi = vi + u
Insight: The spacecraft appears to be approaching at a high rate of speed due to the motion of the planet itself. If the
spacecraft were at rest relative to the Sun, it would still approach the observer at speed u.
130. Picture the Problem: The spacecraft approaches the planet with speed
vi , interacts (“collides”) with it, and recedes in the opposite direction
with speed vf , as depicted in the figure at right.
Strategy: Subtract the final velocities in order to find the speed of
departure relative to an observer on the planet.
Solution: The speed of departure can be found by subtracting the two
final velocities. The calculated answer is choice B.

 
vdep = vf − u = vf − u
v dep = v f − u
Insight: The spacecraft appears to be receding at a relatively low rate of speed due to the motion of the planet itself. If
the spacecraft were at rest relative to the Sun, it would actually approach the observer at speed u.
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7 – 29
Chapter 7: Linear Momentum and Collisions
Pearson Physics by James S. Walker
131. Picture the Problem: The spacecraft approaches the planet with speed
vi , interacts (“collides”) with it, and recedes in the opposite direction
with speed vf , as depicted in the figure at right.
Strategy: Set the speed of approach from the previous questions to the
speed of departure and solve for vf .
Solution: Set vapp = vdep
vapp = u + vi = vf − u = vdep
and solve for vf . The calculated
vf = vi + 2u
answer is choice C.
Insight: The spacecraft has gained velocity 2u at the expense of a tiny bit of the planet’s kinetic energy.
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