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Chapter 6 — Momentum
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Resource CD. They are organized by textbook chapter, and each animation comes within a shell that
provides information on how to use the animation, exploration activities, and a short quiz.
Answers to the Conceptual Questions
1. A physical quantity is conserved when its numerical value does not change.
2. If the system is “closed” so that nothing can enter or leave, the numerical value of its mass does not
change.
3. Toby is correct because it is the change in momentum that matters not the value it has at any particular
time.
4. Lee is correct because it is the change in momentum that matters not the value it has at any particular
time.
5. Supertankers have large inertia.
6. The child has greater momentum. Momentum is mass times velocity so no matter how large the mass,
if the speed is zero the momentum is zero.
7. The net force on an object is equal to the change in its momentum divided by the time required to make
the change. Or the change in momentum is equal to the impulse.
8. The momentum of every object is conserved unless acted upon by an unbalanced force.
9. An impulse (F∆t) is required to change your momentum. The padding and the air pockets lengthen the
time for your leg to stop thus decreasing the size of the force experienced by the leg.
10. Padded dashboards lengthen the time for the body to stop and therefore reduce the forces involved.
11. The impulses are the same because the change in momentum is the same in both cases.
12. In both cases the change in momentum is the same. Therefore, the impulses are also the same. But in
the deep powder, the interaction takes a longer time and the forces are therefore smaller.
13. Because the changes in momentum are the same, the impulses are equal.
14. The carpet lengthens the time for the egg to stop thus decreasing the sizes of the forces the egg
experiences. The carpet also spreads the forces out over a larger fraction of the egg's surface.
15. The initial momentum is 8 kilogram-meters per second down. It changes to zero, so the impulse is 8
kilogram-meters per second directed up.
16. The initial momentum is 8 kilogram-meters per second down. The final momentum is directed upward
so the change must be greater than 8 kilogram-meters per second.
17. Momentum is a vector. Both flowerpots have the same initial momentum. The change in momentum
is greater for the pot that bounces and so it must experience the greater impulse. Jeff’s head must apply
the greater force to the pot and therefore, by Newton’s third law, he must feel the greater force.
18. Momentum is a vector. Both balls have the same initial momentum. The change in momentum is
greater for the ball that bounces and so it must experience the greater impulse.
19. The large padded gloves used in amateur fights lengthen the interaction time between the fist and the
face, decreasing the forces involved.
20. The change in momentum and therefore the impulse is the same in both cases. In the second case the
time of impact is reduced, which increases the average force.
21. At each interaction there are equal and opposite momentum changes. This includes the interactions
involved in throwing and catching the ball as well as the interactions of the feet and the Earth. The
total momentum of the Earth-ball-people system remains constant at all times.
22. The ball experiences three interactions; one with your hand, one with the floor and one with the Earth
(gravity). The first two interactions produce impulses that cause large, rapid momentum changes for
the ball (both speed and direction changes). The steady impulse of gravity causes the ball to experience
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Chapter 6 — Momentum
a continual slow momentum change in the downward direction. However, if the Earth-ball-you-floor
system is considered, all of these impulses are equal and opposite and the total system momentum
remains constant during the motion.
3 N × 5 s = 15 N⋅s and 4 N × 4 s = 16 N⋅s. The second produces the larger impulse and therefore the
larger change in momentum.
3 N × 3 s = 9 N⋅s and 4 N × 2 s = 8 N⋅s. Therefore, the first case produces the larger impulse.
Conservation of linear momentum requires the bullet to go one way and the rifle the other.
You can reduce the recoil by reducing the momentum given to the bullet or by increasing the mass of
the rifle, but you can never eliminate the recoil. So-called recoilless rifles are really tubes for firing
rockets. The recoil momentum is taken up by the gases escaping from the rocket engine.
Only external forces can change a system’s momentum. With or without the plate, the exhaust gases
exert an external force on the rocket that should cause it to accelerate. With the plate, the exhaust gases
also exert an additional (backward) force on the plate. This yields a nearly zero net external force on
the rocket-plate system.
Assuming that the air leaves the board moving toward the sides, the momentum transferred by the fan
to the air molecules is canceled by that due to the air molecules colliding with the screen. If the
molecules have some recoil, there would be a small net forward force.
Momentum is only conserved for systems that have no external forces acting on them. The momentum
of the Earth-ball system is conserved but not the momentum of just the ball. The Earth’s large mass
allows it to acquire an equal and opposite momentum to the ball without a measurable velocity.
Momentum is only conserved for systems that have no external forces acting on them. There is an
external force for systems a and b, which changes their momenta. All forces in system c are internal so
its momentum is conserved.
Momentum is a vector. The initial momentum of the system is zero. The only way that the momentum
of the two stuck together can be zero is if they are at rest.
The initial momentum of the system is to the left, which means that after the collision the pair will be
moving to the left.
The initial momentum of the system is to the left, which means that after the collision the pair will be
moving to the left.
The initial momentum of the system is to the right, which means that after the collision the pair will be
moving to the right.
The magnitude of the total momentum stays constant at 250 kilogram-meters per second because the
forces between the teacher and the board are internal to the system.
The total momentum remains zero because the forces between your friend and the board are internal to
the system. Therefore, the board must have a momentum of 150 kilogram-meters per second to the left.
The momentum remains zero because the forces are internal.
Because the initial momentum is zero, the gliders must have equal and opposite momenta. Thus, glider
A with twice the mass will have half the velocity of glider B, namely 1 meter per second.
As they jump toward the shore, they give the rowboat momentum away from the dock.
As the astronaut pushes off the wall, the Shuttle experiences an impulse in the other direction,
increasing its momentum in the direction of the impulse.
The total momentum is conserved because the forces of the explosion are internal.
The total momentum of the fragments changes in the same way as if the rocket never exploded.
Conservation of linear momentum says that it is impossible for the astronaut to move anywhere unless
something is thrown in the opposite direction.
There are a variety of solutions. They all share the requirement that Al create a change in the
momentum of his body, moving it toward the spaceship. Some students suggest pushing off the bag of
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Chapter 6 — Momentum
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gold. A more creative solution is to throw coins away from the spaceship, one at a time.
The magnitude of the momentum of the northbound object is greater than the magnitude of the
momentum of the eastbound object. Therefore the final momentum must be more north than east,
which is path B.
The magnitude of the momentum of the northbound object is greater than the magnitude of the
momentum of the eastbound object. Therefore the final momentum must be more north than east,
which is path B.
The magnitude of the momentum of the eastbound object is greater than the magnitude of the
momentum of the northbound object. Therefore the final momentum must be more east than north,
which is path D.
The magnitude of the momentum of the eastbound object is equal to the magnitude of the momentum
of the northbound object. Therefore the final momentum must be northeast, which is path C.
Answers to the Exercises
1.
p = m v = (1200 kg ) ( 30 m s ) = 36,000 kg ⋅ m s
2.
pDE = m v = (120 kg ) ( 6 m s ) = 720 kg ⋅ m s
pRB = m v = (100 kg ) ( 8 m s ) = 800 kg ⋅ m s
Therefore, the running back has the larger momentum.
(10 g ) ( 900 m s ) = 62 m s
m v
3. vbb = bul bul =
145 g
mbb
4. v you =
m18 v18
=
m you
( 24,000 kg ) (1 mph )
80 kg
= 300 mph
5. F =
(1500 kg ) ( 30 m s ) = 5625 N
mv
∆p
=
=
8s
∆t
∆t
6. F =
( 60 kg ) ( 44.7 m s ) = 89.4 N
mv
∆p
=
=
30 s
∆t
∆t
F
F
89.4 N
89.4 N
=
=
=
= 0.15
2
W mg ( 60 kg ) (10 m/s ) 600 N
7. F ∆t = ∆ ( m v ) = 0 − (1400 kg ) ( 25 m s ) = − 35,000 N ⋅ s
8. F ∆t = ∆ ( m v ) = 0 − ( 0.5 kg ) ( 6 m s ) = − 3 N ⋅ s
If the ball had bounced, it would have experienced a greater change in momentum, which requires a
greater impulse.
9. F ∆t = ∆ ( m v ) = 0 − (1500 kg ) ( 30 m s ) = − 45,000 N ⋅ s
F =
10. F =
∆ (m v )
− 45,000 N ⋅ s
=
= − 5625 N
8s
∆t
( 0.145 kg ) ( 50 m s ) = 181 N
mv
∆p
=
=
0.04 s
∆t
∆t
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Chapter 6 — Momentum
11. F =
( 0.6 kg ) (8 m s ) − ( 0.6 kg ) ( −8 m s ) = 240 N (upward)
∆p
=
0.04 s
∆t
12. F =
− ( 0.2 kg ) (8 m s ) − ( 0.2 kg ) (12 m s )
∆p
=
= − 100 N; magnitude would be 100 N
0.04 s
∆t
13. F =
14. vgun
∆p m∆v ( 0.01 kg )( 900 m/s − 0 )
=
=
= 900 N
0.001 s
∆t
∆t
⎛ 0.01 kg ⎞
m
= vbullet bullet = ( 800 m s ) ⎜
⎟ = 2m s
mgun
⎝ 4 kg ⎠
15. v father =
mson vson
=
m father
( 40 kg ) ( 3 m s )
80 kg
= 1.5 m s
17.
⎛
⎞
50 kg
m1
= (6 m s) ⎜
⎟ = 3.75 m s
m2
⎝ 50 kg + 30 kg ⎠
pi = p f = m1 v1 + m2 v2 = ( 3 kg ) ( 4 m s ) + ( 4 kg ) ( −3 m s ) = 0
18.
pi = p f = m1 v1 + m2 v2 = ( 4 kg ) ( 4 m s ) + ( 5 kg ) ( −2 m s ) = + 6 kg ⋅ m s
19.
pi = p f = m1 v1 + m2 v2 = (1200 kg ) (14 m s ) + ( 2000 kg ) ( 25 m s )
16. v2 = v1
= 66,800 kg ⋅ m s
20. v f =
pf
m1 + m2
=
( right )
( north )
66,800 kg ⋅ m s
= 20.9 m s
1200 kg + 2000 kg
21. v final =
m1 v1 + m2 v2
m v − m1 v1
= 1 1
= 0
m1 + m2
m1 + m2
22. v final =
m1 v1 + m2 v2
mv + 0
v
10 m s
=
=
=
= 2.5 m s
m1 + m2
m + 3m
4
4
Answers to the Problems in Problem Solving
1.
2.
3.
4.
I = p f − pi = m ( v f − vi ) = ( 2000 kg )(10 m/s − 20 m/s ) = −20,000 kg ⋅ m/s (+ is forward)
I = p f − pi = m ( v f − vi ) = ( 6000 kg )( 30 m/s − 25 m/s ) = +30,000 kg ⋅ m/s (+ is forward)
I = p f − pi = m ( v f − vi ) = ( 0.145 kg )( −60 m/s − 40 m/s ) = −14.5 kg ⋅ m/s (+ is forward)
I = p f − pi = m ( v f − vi ) = ( 5 kg ) (1 m/s − ( −3 m/s ) ) = +20 kg ⋅ m/s (+ is upward)
5. ∆t =
6. ∆t =
∆ (m v )
(1600 kg ) ( 20 m s ) = 40 s
=
F
800 N
∆ (m v)
F
=
(1200 kg ) ( 30 m s )
2000 N
= 18 s
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Chapter 6 — Momentum
∆ (m v )
( 70 kg ) ( 24 m s ) = 8400 N
=
∆t
0.2 s
7. F =
F
F
8400 N
=
=
= 12.2
W
mg
( 70 kg ) ( 9.8 m s2 )
8.
∆ (m v)
F =
∆t
12.
F =
∆ (m v)
∆t
∆ (m v)
∆t
14. vboy =
16.
= 2800 N
=
=
( 0.145 kg ) ( 50 m s ) − ( 0.145 kg ) ( − 40 m s )
0.05 s
( 5 kg ) (1 m s ) − ( 5 kg ) ( − 3 m s )
0.1 s
= 261 N
= 200 N
( 0.2 kg ) ( 3 m s ) = 0.15 m s
mb vb
=
4 kg
mc
13. vc =
15. v =
0.6 s
∆ ( mv )
50 m s
mv
v
=
=
=
= 0.51 s
2
10 m g
10 g
F max
98 m s
10. ∆t ≥
F =
( 70 kg ) ( 24 m s )
∆ (m v)
30 m s
mv
v
=
=
=
= 3.06 s
9.8 m s2
Fmax
mg
g
9. ∆t ≥
11.
=
mb vb
( 0.5 kg ) ( 25 m s ) = 0.313 m s
=
mboy
40 kg
(80 kg ) ( 5 m s ) = 4.76 m s
p
mv + 0
=
=
80 kg + 4 kg
M
M
pb = ptot − pw = ( 60 kg ) ( 5 m s ) − ( 50 kg ) ( 3 m s ) = 150 kg ⋅ m s
vb =
pb
150 kg ⋅ m s
=
= 15 m s
mb
10 kg
( + 2 m s ) m − (3 m s ) m = − 1 m s
ptot − p2
=
( + is right )
m
m
( 4 m s ) m − (3 m s ) m = + 1 m s
p − pr
18. vb = tot
=
( + is right )
m
m
The red ball cannot pass the blue ball.
19. ptotal = m1 v1 + m2 v2 = ( 2 kg ) ( 4 m s ) + ( 4 kg ) ( − 2 m s ) = 0
17. v1 =
v4 =
20.
( + is right )
ptotal = m1 v1 + m2 v2 = ( 2 kg ) ( 6 m s ) + ( 4 kg ) ( − 4 m s ) = − 4 kg ⋅ m s
v4 =
21.
0 − ( 2 kg ) ( − 2 m s )
ptotal − p2
=
= +1m s
m4
4 kg
pf =
− 4 kg ⋅ m s − ( 2 kg ) ( − 1 m s )
ptotal − p2
=
= − 0.5 m s
m4
4 kg
p12 + p12 =
( + is right )
⎣⎡ (1200 kg ) ( 25 m s ) ⎤⎦ + ⎣⎡ (1600 kg ) ( 35 m s ) ⎤⎦ = 63,500 kg ⋅ m s
2
2
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Chapter 6 — Momentum
vf =
22.
pf
mtot
=
63,500 kg ⋅ m s
= 22.7 m s
2800 kg
⎡⎣ (1200 kg ) ( 25 m s ) ⎤⎦ + ⎡⎣ (1500 kg ) ( 20 m s ) ⎤⎦
= 42, 400 kg ⋅ m s
southwest
pf =
2
pc2 + pt2 =
2
23. To end up traveling directly northeast after the collision, the two cars must have had the same initial
momenta. Therefore,
(1000 kg ) ( 30 m s ) = 21.4 m s
m v
ve = n n =
1400 kg
me
24.
⎡⎣ ( 50 kg ) ( 6 m s ) ⎤⎦ + ⎡⎣ (100 kg ) ( 4 m s ) ⎤⎦
∆p
500 kg ⋅ m s
=
= 250 N
F =
∆t
2s
pi =
p12 + p22 =
2
2
= 500 kg ⋅ m s