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Chapter 8
Conservation of Linear Momentum
Conceptual Problems
1 • [SSM] Show that if two particles have equal kinetic energies, the magnitudes of their momenta are equal only if they have the same mass.
Determine the Concept The kinetic energy of a particle, as a function of its
momentum, is given by K = p^2 2 m.
The kinetic energy of the particles is given by: (^1)
2 1 (^1 2) m
p K = and 2
2 2 (^2 2) m
p K =
Equate these kinetic energies to
obtain: (^2)
2 2 1
2 1 2 2 m
p m
p
Because the magnitudes of their
momenta are equal: (^12)
m m
= and m 1 (^) = m 2
2 • Particle A has twice the (magnitude) momentum and four times the kinetic energy of particle B. A also has four times the kinetic energy of B. What is the ratio of their masses (the mass of particle A to that of particle B)? Explain your reasoning.
Determine the Concept The kinetic energy of a particle, as a function of its
momentum, is given by K = p^2 2 m.
The kinetic energy of particle A is given by: (^) A
2 A A (^2) m
p K = ⇒ A
2 A A (^2) K
p m =
The kinetic energy of particle B is
given by: (^) B
2 B B 2 m
p K = ⇒ B
2 B B 2 K
p m =
Divide the first of these equations by the second and simplify to obtain: 2
B
A A
B 2 A B
2 B A
B
2 B
A
2 A
B
A
p
p K
K
K p
K p
K
p
K
p
m
m
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Because particle A has twice the (magnitude) momentum of particle B
and four times as much kinetic energy:
2
B
B B
B B
A =
p
p K
K
m
m
3 • Using SI units, show that the units of momentum squared divided by those of mass is equivalent to the joule.
Determine the Concept The SI units of momentum are kg⋅m/s.
Express the ratio of the square of the units of momentum to the units of
mass: (^) kg
s
m kg
2 ⎟ ⎠
Simplify to obtain:
m N m J s
m kg s
m kg kg
s
m kg
kg
s
m kg 2 2
2 2
2 2
2
⎟⋅ = ⋅^ =
= ⋅ =⎛^ ⋅
4 • True or false:
( a ) The total linear momentum of a system may be conserved even when the mechanical energy of the system is not. ( b ) For the total linear momentum of a system to be conserved, there must be no external forces acting on the system. ( c ) The velocity of the center of mass of a system changes only when there is a net external force on the system.
( a ) True. Consider the collision of two objects of equal mass traveling in opposite directions with the same speed. Assume that they collide inelastically. The
mechanical energy of the system is not conserved (it is transformed into other forms of energy), but the momentum of the system is the same after the collision
as before the collision; that is, zero. Therefore, for any inelastic collision, the
momentum of a system may be conserved even when mechanical energy is not.
( b ) False. The net external force must be zero if the linear momentum of the system is to be conserved.
( c ) True. This non-zero net force accelerates the center of mass. Hence its velocity
changes.
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relative to its center of mass. Because the bowling balls are identical and have the same velocity, the rolling ball has more kinetic energy. There is no problem here because the relationship K = p^22 m is between the center of mass kinetic energy of the ball and its linear momentum.
9 • A philosopher tells you, ″Changing motion of objects is impossible. Forces always come in equal but opposite pairs. Therefore, all forces cancel out. Because forces cancel, the momenta of objects can never be changed.″ Answer his argument.
Determine the Concept Think of someone pushing a box across a floor. Her push on the box is equal but opposite to the push of the box on her, but the action and reaction forces act on different objects. Newton’s second law is that the sum of the forces acting on the box equals the rate of change of momentum of the box. This sum does not include the force of the box on her.
10 • A moving object collides with a stationary object. Is it possible for both objects to be at rest immediately after the collision? (Assume any external forces acting on this two-object system are negligibly small.) Is it possible for one object to be at rest immediately after the collision? Explain.
Determine the Concept It’s not possible for both to remain at rest after the collision, as that wouldn't satisfy the requirement that momentum is conserved. It is possible for one to remain at rest. This is what happens for a one-dimensional collision of two identical particles colliding elastically.
11 • Several researchers in physics education claim that part of the cause of physical misconceptions amongst students comes from special effects they observe in cartoons and movies. Using the conservation of linear momentum, how would you explain to a class of high school physics students what is conceptually wrong with a superhero hovering at rest in midair while tossing massive objects such as cars at villains? Does this action violate conservation of energy as well? Explain.
Determine the Concept Hovering in midair while tossing objects violates the conservation of linear momentum! To throw something forward requires being pushed backward. Superheroes are not depicted as experiencing this backward motion that is predicted by conservation of linear momentum. This action does not violate the conservation of energy.
12 •• A struggling physics student asks ″If only external forces can cause the center of mass of a system of particles to accelerate, how can a car move? Doesn’t the car’s engine supply the force needed to accelerate the car? ″ Explain what external agent produces the force that accelerates the car, and explain how the engine makes that agent do so.
Conservation of Linear Momentum 725
Determine the Concept There is only one force which can cause the car to move forward−the friction of the road! The car’s engine causes the tires to rotate, but if the road were frictionless (as is closely approximated by icy conditions) the wheels would simply spin without the car moving anywhere. Because of friction, the car’s tire pushes backwards against the road and the frictional force acting on the tire pushes it forward. This may seem odd, as we tend to think of friction as being a retarding force only, but it is true.
13 •• When we push on the brake pedal to slow down a car, a brake pad is pressed against the rotor so that the friction of the pad slows the rotation of the rotor and thus the rotation of the wheel. However, the friction of the pad against the rotor can’t be the force that slows the car down, because it is an internal force—both the rotor and the wheel are parts of the car, so any forces between them are internal, not external, forces. What external agent exerts the force that slows down the car? Give a detailed explanation of how this force operates.
Determine the Concept The frictional force by the road on the tire causes the car to slow. Normally the wheel is rotating at just the right speed so both the road and the tread in contact with the road are moving backward at the same speed relative to the car. By stepping on the brake pedal, you slow the rotation rate of the wheel. The tread in contact with the road is no longer moving as fast, relative to the car, as the road. To oppose the tendency to skid, the tread exerts a forward frictional force on the road and the road exerts an equal and opposite force on the tread.
14 • Explain why a circus performer falling into a safety net can survive unharmed, while a circus performer falling from the same height onto the hard concrete floor suffers serious injury or death. Base your explanation on the impulse-momentum theorem.
Determine the Concept Because Δ p = F Δ t is constant, the safety net reduces the
force acting on the performer by increasing the time Δ t during which the slowing force acts.
15 •• [SSM] In Problem 14, estimate the ratio of the collision time with the safety net to the collision time with the concrete for the performer falling from a height of 25 m. Hint: Use the procedure outlined in Step 4 of the Problem- Solving Strategy located in Section 8-3.
Determine the Concept The stopping time for the performer is the ratio of the
distance traveled during stopping to the average speed during stopping.
Conservation of Linear Momentum 727
( d ) After a perfectly inelastic head-on collision along the east-west direction, the two objects are observed to be moving west. The initial total system momentum was, therefore, to the west.
( a ) False. Following a perfectly inelastic collision, the colliding bodies stick
together but may or may not continue moving, depending on the momentum each brings to the collision.
( b ) True. For a head-on elastic collision both kinetic energy and momentum are
conserved and the relative speeds of approach and recession are equal.
( c ) True. This is the definition of an inelastic collision.
( d ) True. The linear momentum of the system before the collision must be in the
same direction as the linear momentum of the system after the collision.
18 •• Under what conditions can all the initial kinetic energy of an isolated system consisting of two colliding objects be lost in a collision? Explain how this result can be, and yet the momentum of the system can be conserved.
Determine the Concept If the collision is perfectly inelastic, the bodies stick together and neither will be moving after the collision. Therefore, the final kinetic energy will be zero and all of it will have been lost (that is, transformed into some other form of energy). Momentum is conserved because in an isolated system the net external force is zero.
19 •• Consider a perfectly inelastic collision of two objects of equal mass. ( a ) Is the loss of kinetic energy greater if the two objects are moving in opposite directions, each moving at speed v /2, or if one of the two objects is initially at rest and the other has an initial speed of v? ( b ) In which of these situations is the percentage loss in kinetic energy the greatest?
Determine the Concept We can find the loss of kinetic energy in these two collisions by finding the initial and final kinetic energies. We’ll use conservation of momentum to find the final velocities of the two masses in each perfectly elastic collision.
( a ) Letting V represent the velocity of the masses after their perfectly inelastic collision, use conservation of momentum to determine V :
p before (^) = p after or mv − mv = 2 mV ⇒ V = 0
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Express the loss of kinetic energy for the case in which the two objects have oppositely directed velocities of magnitude v /2:
2 4 1
2 2
1 f i 2
mv
v K K K m
Letting V represent the velocity of the masses after their perfectly inelastic collision, use conservation of momentum to determine V :
p before (^) = p after or mv = 2 mV ⇒ V = 21 v
Express the loss of kinetic energy for the case in which the one object is initially at rest and the other has an initial velocity v :
2 2 1
f i
mv mv v m
K K K
⎟ − =^ −
The loss of kinetic energy is the same in both cases.
( b ) Express the percentage loss for the case in which the two objects have oppositely directed velocities of magnitude v /2:
4
1
2 4
1
before
mv
mv K
K
Express the percentage loss for the case in which the one object is initially at rest and the other has an initial velocity v :
2 1
2 4
1
before
mv
mv K
K
The percentage loss is greatest for the case in which the two objects have oppositely directed velocities of magnitude v /2.
20 •• A double-barreled pea shooter is shown in Figure 8-41. Air is blown into the left end of the pea shooter, and identical peas A and B are positioned inside each straw as shown. If the pea shooter is held horizontally while the peas are shot off, which pea, A or B, will travel farther after leaving the straw? Explain. Base your explanation on the impulse–momentum theorem.
Determine the Concept Pea A will travel farther. Both peas are acted on by the same force, but pea A is acted on by that force for a longer time. By the impulse- momentum theorem, its momentum (and, hence, speed) will be higher than pea B’s speed on leaving the shooter.
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Solve this equation for m 2 to determine its value for maximum
energy transfer:
m (^) 2 = m 1
( ) b is correct because all of particle 1’s kinetic energy is transferred to
particle 2 when m 2 (^) = m 1.
22 •• Suppose you are in charge of an accident-reconstruction team which has reconstructed an accident in which a car was ″rear-ended″ causing the two cars to lock bumpers and skid to a halt. During the trial, you are on the stand as an expert witness for the prosecution and the defense lawyer claims that you wrongly neglected friction and the force of gravity during the fraction of a second while the cars collided. Defend your report. Why were you correct in ignoring these forces? You did not ignore these two forces in your skid analysis both before and after the collision. Can you explain to the jury why you did not ignore these two forces during the pre- and post-collision skids?
Determine the Concept You only used conservation of linear momentum for a fraction of a second of actual contact between the cars. Over that short time, friction and other external forces can be neglected. In the long run, over the duration of the accident, they cannot.
23 •• Nozzles for a garden hose are often made with a right-angle shape as shown in Figure 8-42. If you open the nozzle and spray water out, you will find that the nozzle presses against your hand with a pretty strong force—much stronger than if you used a nozzle not bent into a right angle. Why is this situation true?
Determine the Concept The water is changing direction when it rounds the corner in the nozzle. Therefore, the nozzle must exert a force on the stream of water to change its momentum, and from Newton’s third law, the water exerts an equal but opposite force on the nozzle. This requires a net force in the direction of the momentum change.
Conceptual Problems from Optional Sections
24 •• Describe a perfectly inelastic head-on collision between two stunt cars as viewed in the center-of-mass reference frame.
Determine the Concept In the center-of-mass reference frame the two objects
approach with equal but opposite momenta and remain at rest after the collision.
Conservation of Linear Momentum 731
25 •• One air-hockey puck is initially at rest. An identical air-hockey puck collides with it, striking it with a glancing blow. Assume the collision was elastic and neglect any rotational motion of the pucks. Describe the collision in the center-of-mass frame of the pucks.
Determine the Concept In the center-of-mass frame the two velocities are equal and opposite, both before and after the collision. In addition, the speed of each puck is the same before and after the collision. The direction of the velocity of each puck changes by some angle during the collision.
26 •• A baton with one end more massive than the other is tossed at an angle into the air. ( a ) Describe the trajectory of the center of mass of the baton in the reference frame of the ground. ( b ) Describe the motion of the two ends of the baton in the center-of-mass frame of the baton.
Determine the Concept ( a ) In the center-of-mass frame of the ground, the center of mass moves in a parabolic arc.
( b ) Relative to the center of mass, each end of the baton would describe a circular path. The more massive end of the baton would travel in the circle with the smaller radius because it is closer to the location of the center of mass.
27 •• Describe the forces acting on a descending Lunar lander as it fires its retrorockets to slow it down for a safe landing. (Assume the lander’s mass loss during the rocket firing is not negligible.)
Determine the Concept The forces acting on a descending Lunar lander are the downward force of lunar gravity and the upward thrust provided by the rocket engines.
28 •• A railroad car rolling along by itself is passing by a grain elevator, which is dumping grain into it at a constant rate. ( a ) Does momentum conservation imply that the railroad car should be slowing down as it passes the grain elevator? Assume that the track is frictionless and perfectly level and that the grain is falling vertically. ( b ) If the car is slowing down, this situation implies that there is some external force acting on the car to slow it down. Where does this force come from? ( c ) After passing the elevator, the railroad car springs a leak, and grain starts leaking out of a vertical hole in its floor at a constant rate. Should the car speed up as it loses mass?
Determine the Concept We can apply conservation of linear momentum and Newton’s laws of motion to each of these scenarios.
( a ) Yes, the car should slow down. An easy way of seeing this is to imagine a "packet" of grain being dumped into the car all at once: This is a completely inelastic collision, with the packet having an initial horizontal velocity of 0. After
Conservation of Linear Momentum 733
( a ) Relate the stopping time to the
assumption that the center of the car travels halfway to the wall with
constant deceleration:
av
4 car 1
av
2 car 1 2 1
av
stopping v
L
v
L
v
d Δ t = = = (1)
Because a is constant, the average speed of the car is given by: 2
i f av
v v v
Substitute numerical values and evaluate v av:
- 5 m/s
km
1000 m 3600 s
1 h h
km 0 90 av =
+ × ×
v =
Substitute numerical values in
equation (1) and evaluate Δ t :
- 120 s 0. 12 s 12.5m/s
6.0m Δ 4
1 t = = =
( b ) Relate the average force exerted by the wall on the car to the car’s change in momentum:
4. 2 10 N
0.120s
km
1000 m 3600 s
1 h h
km 2000 kg 90
Δ
av = ×
× ×
t
p F
31 •• In hand-pumped railcar races, a speed of 32.0 km/h has been achieved by teams of four people. A car that has a mass equal to 350 kg is moving at that speed toward a river when Carlos, the chief pumper, notices that the bridge ahead is out. All four people (each with a mass of 75.0 kg) simultaneously jump backward off the car with a velocity that has a horizontal component of 4.00 m/s relative to the car. The car proceeds off the bank and falls into the water a horizontal distance of 25.0 m from the bank. ( a ) Estimate the time of the fall of the railcar. ( b ) What is the horizontal component of the velocity of the pumpers when they hit the ground?
Picture the Problem Let the direction the railcar is moving be the + x direction
and the system include Earth, the pumpers, and the railcar. We’ll also denote the
railcar with the letter c and the pumpers with the letter p. We’ll use conservation of linear momentum to relate the center of mass frame velocities of the car and
the pumpers and then transform to the Earth frame of reference to find the time of fall of the car.
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( a ) Relate the time of fall of the railcar to the horizontal distance it travels and its horizontal speed as it leaves the bank:
v c
x t
Use conservation of momentum to find the speed of the car relative to the speed of its center of mass: (^0)
or c c p p
i f
mu m u
p p
r r
Relate u c to u p and solve for u c: u p − u c=− 4. 00 m/s and u p = u c− 4. 00 m/s
Substitute for u p to obtain: m c u c+^ m p( u^ c−^4.^00 m/s)^ =^0
Solving for u c yields:
p
c
c 1
- 00 m/s
m
m
u
Substitute numerical values and evaluate u c:
- 846 m/s
4 75.0kg
350 kg 1
- 00 m/s c =
u =
Relate the speed of the car to its speed relative to the center of mass of the system:
v c (^) = u c+ v cm
Substitute numerical values and evaluate v c:
- 73 m/s km
1000 m 3600 s
1 h h
km
s
m c 1 .846 ⎟= ⎠
v = +
Substitute numerical values in equation (1) and evaluate Δ t :
2.33s 10.73m/s
25.0m Δ t = =
( b ) The horizontal velocity of the pumpers when they hit the ground is:
6.7m/s
p c p 10.73m/s 4.00m/s
v = v − u = −
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Relate the time of flight Δ t to L
and v rel : v rel
L
Δ t =
Substitute and simplify to find the distance Δ s moved by the glider in time Δ t :
L
m m
m
v m
m m
L
v m
m v
L
v m
m s v t ⎟⎟ ⎠
p b
b
b p
p b
b p
b rel
b p
b Δ pΔ
Conservation of Linear Momentum
33 • [SSM] Tyrone, an 85-kg teenager, runs off the end of a horizontal pier and lands on a free-floating 150-kg raft that was initially at rest. After he lands on the raft, the raft, with him on it, moves away from the pier at 2.0 m/s. What was Tyrone’s speed as he ran off the end of the pier?
Picture the Problem Let the system include Earth, the raft, and Tyrone and apply conservation of linear momentum to find Tyrone’s speed when he ran off the end of the pier.
Apply conservation of linear momentum to the system consisting of the raft and Tyrone to obtain:
Δ p system (^) = Δ p Tyrone+Δ p raft= 0
r r r
or, because the motion is one- dimensional, p f,Tyrone − p i,Tyrone+ p f,raft− p i,raft= 0
Because the raft is initially at rest: p f,Tyrone −^ p i,Tyrone+ p f,raft=^0
Use the definition of linear momentum to obtain:
m Tyrone v f,Tyrone− m Tyrone v i,Tyrone+ m raft v f,raft= 0
Solve for v i, (^) Tyroneto obtain: f,raft f,Tyrone Tyrone
raft i,Tyrone (^) m v v
m v = +
Letting v represent the common final speed of the raft and Tyrone yields: (^) m v
m v ⎟
Tyrone
raft i,Tyrone 1
Conservation of Linear Momentum 737
Substitute numerical values and evaluate v i, (^) Tyrone:
5 .5 m/s
- 0 m/s 85 kg
150 kg i,Tyrone^1
=
v = +
34 •• A 55-kg woman contestant on a reality television show is at rest at the south end of a horizontal 150-kg raft that is floating in crocodile-infested waters. She and the raft are initially at rest. She needs to jump from the raft to a platform that is several meters off the north end of the raft. She takes a running start. When she reaches the north end of the raft she is running at 5.0 m/s relative to the raft. At that instant, what is her velocity relative to the water?
Picture the Problem Let the system include the woman, the raft, and Earth. Then the net external force is zero and linear momentum is conserved as she jumps off the raft. Let the direction the woman is running be the + x direction.
Apply conservation of linear momentum to the system:
∑ i v^ i= woman v woman+ raft v raft=^0
r r r m m m
Solving for v raft
r
yields: raft
woman woman raft m
m v v
r r =−
Substituting numerical values gives: raft woman 150 woman
150 kg
55 kg v v v
r r r ⎟ ⎠
It is given that: v r woman − v rraft=( 5. 0 m/s) i ˆ
Substituting for v raft
r yields:
v v ( 5. 0 m/s) i ˆ
woman ⎟ woman= ⎠
r r
Solve for v woman r to obtain:
v i ˆ ( 3.7m/s) i ˆ
- 0 m/s woman = ⎟⎟
r
35 • A 5.0-kg object and a 10-kg object, both resting on a frictionless table, are connected by a massless compressed spring. The spring is released and the objects fly off in opposite directions. The 5.0-kg object has a velocity of 8.0 m/s to the left. What is the velocity of the 10-kg object?
Conservation of Linear Momentum 739
37 • A shell of mass m and speed v explodes into two identical fragments. If the shell was moving horizontally with respect to Earth, and one of the fragments is subsequently moving vertically with speed v , find the velocity
r v ′ of the other fragment immediately following the explosion.
Picture the Problem Choose the direction the shell is moving just before the explosion to be the positive x direction and apply conservation of momentum.
Use conservation of momentum to
relate the masses of the fragments to their velocities:
p i (^) p f r r = or i ˆ^21 ˆ j 21 v '
r mv = mv + m ⇒ v ' = 2 v i ˆ− v j ˆ
r
38 •• During this week’s physics lab, the experimental setup consists of two gliders on a horizontal frictionless air track (see Figure 8-45). Each glider supports a strong magnet centered on top of it, and the magnets are oriented so they attract each other. The mass of glider 1 and its magnet is 0.100 kg and the mass of glider 2 and its magnet is 0.200 kg. You and your lab partners take the origin to be at the left end of the track and to center glider 1 at x 1 = 0.100 m and glider 2 at x 2 = 1.600 m. Glider 1 is 10.0 cm long, while glider 2 is 20.0 cm long and each glider has its center of mass at its geometric center. When the two are released from rest, they will move toward each other and stick. ( a ) Predict the position of the center of each glider when they first touch. ( b ) Predict the velocity the two gliders will continue to move with after they stick. Explain the reasoning behind this prediction for your lab partners.
Picture the Problem Because no external forces act on either glider, the center of mass of the two-glider system can’t move. We can use the data concerning the masses and separation of the gliders initially to calculate its location and then apply the definition of the center of mass a second time to relate the positions x 1 and x 2 of the centers of the carts when they first touch. We can also use the separation of the centers of the gliders when they touch to obtain a second equation in x 1 and x 2 that we can solve simultaneously with the equation obtained from the location of the center of mass.
( a ) The x coordinate of the center of mass of the 2-glider system is given by: 1 2
1 1 2 2 cm m m
mx mx x
Substitute numerical values and evaluate x cm:
- 10 m 0.100kg 0.200kg
0.100kg 0.100m 0.200kg 1.600m cm =
x =
from the left end of the air track.
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Because the location of the center of mass has not moved when two gliders first touch: 1 2
- 10 m^1122 m m
mX mX
Substitute numerical values and simplify to obtain:
3 2 2 3 1
- 10 m=^1 X + X
Also, when they first touch, their centers are separated by half their combined lengths:
0.150 m
2 1 21 10.0cm 20.0cm
X − X = +
Thus we have: (^13) X 1 + 32 X 2 = 1. 10 m and X 2 − X 1 =0.150m
Solving these equations simultaneously yields:
X (^) 1 = 1. 00 m and X 2 = 1. 15 m
( b ) Because the momentum of the system was zero initially, it must be zero just before the collision and after the collision in which the gliders stick together. Hence their velocity after the collision must be 0.
39 •• Bored, a boy shoots his pellet gun at a piece of cheese that sits on a massive block of ice. On one particular shot, his 1.2 g pellet gets stuck in the cheese, causing it to slide 25 cm before coming to a stop. If the muzzle velocity of the gun is known to be 65 m/s, and the cheese has a mass of 120 g, what is the coefficient of friction between the cheese and ice?
Picture the Problem Let the system consist of the pellet and the cheese. Then we can apply the conservation of linear momentum and the conservation of energy with friction to this inelastic collision to find the coefficient of friction between the cheese and the ice.
Apply conservation of linear
momentum to the system to obtain:
Δ p system (^) = Δ p pellet+Δ p cheese= 0
r r r
or, because the motion is one- dimensional, p f,pellet − p i,pellet+ p f,cheese− p i,cheese= 0
Because the cheese is initially at rest:
p f,pellet − p i,pellet+ p f,cheeset= 0
