9
Linear Momentum and Collisions
CHAPTER OUTLINE ANSWERS TO QUESTIONS
9.1 Linear Momentum and Its
Conservation
9.2 Impulse and Momentum
*Q9.1 (a) No. Impulse, F∆t, depends on the force and the
9.3 Collisions in One Dimension time for which it is applied.
9.4 Two-Dimensional Collisions
9.5 The Center of Mass (b) No. Work depends on the force and on the distance
9.6 Motion of a System of Particles
9.7 Deformable Systems over which it acts.
9.8 Rocket Propulsion
*Q9.2 The momentum magnitude is proportional to the speed and
the kinetic energy is proportional to the speed squared.
(i) The speed of the constant-mass object becomes
4 times larger and the kinetic energy 16 times
larger. Answer (a).
(ii) The speed and the momentum become two times
larger. Answer (d).
*Q9.3 (i) answer (c). For example, if one particle has 5 times larger mass, it will have 5 times smaller
speed and 5 times smaller kinetic energy.
(ii) answer (d). Momentum is a vector.
*Q9.4 (i) Equal net work inputs imply equal kinetic energies. Answer (c).
(ii) Imagine one particle has four times more mass. For equal kinetic energy it must have half the
speed. Then this more massive particle has 4(1Ⲑ2) = 2 times more momentum. Answer (a).
Q9.5 (a) It does not carry force, for if it did, it could accelerate itself.
(b) It cannot deliver more kinetic energy than it possesses. This would violate the law of energy
conservation.
(c) It can deliver more momentum in a collision than it possesses in its flight, by bouncing
from the object it strikes.
*Q9.6 Mutual gravitation brings the ball and the Earth together. As the ball moves downward, the
Earth moves upward, although with an acceleration on the order of 10 25 times smaller than that
of the ball. The two objects meet, rebound, and separate. Momentum of the ball-Earth system is
conserved. Answer (d).
Q9.7 (a) Linear momentum is conserved since there are no external forces acting on the system. The
fragments go off in different directions and their vector momenta add to zero.
(b) Kinetic energy is not conserved because the chemical potential energy initially in the
explosive is converted into kinetic energy of the pieces of the bomb.
209
,210 Chapter 9
Q9.8 Momentum conservation is not violated if we choose as our system the planet along with you.
When you receive an impulse forward, the Earth receives the same size impulse backwards. The
resulting acceleration of the Earth due to this impulse is much smaller than your acceleration
forward, but the planet’s backward momentum is equal in magnitude to your forward momentum.
*Q9.9 (i) During the short time the collision lasts, the total system momentum is constant. Whatever
momentum one loses the other gains. Answer (c).
(ii) When the car overtakes the manure spreader, the faster-moving one loses more energy than
the slower one gains. Answer (a).
Q9.10 The rifle has a much lower speed than the bullet and much less kinetic energy. Also, the butt
distributes the recoil force over an area much larger than that of the bullet.
*Q9.11 (i) answer (a). The ball gives more rightward momentum to the block when the ball reverses
its momentum.
(ii) answer (b). In case (a) there is no temperature increase because the collision is elastic.
Q9.12 His impact speed is determined by the acceleration of gravity and the distance of fall, in
v 2f = vi2 − 2 g ( 0 − yi ). The force exerted by the pad depends also on the unknown stiffness of the pad.
Q9.13 The sheet stretches and pulls the two students toward each other. These effects are larger for a
faster-moving egg. The time over which the egg stops is extended, more for a faster missile,
so that the force stopping it is never too large.
*Q9.14 Think about how much the vector momentum of the Frisbee changes in a horizontal plane. This
will be the same in magnitude as your momentum change. Since you start from rest, this quantity
directly controls your final speed. Thus f is largest and d is smallest. In between them, b is larger
than c and c is larger than g and g is larger than a. Also a is equal to e, because the ice can exert
a normal force to prevent you from recoiling straight down when you throw the Frisbee up. The
assembled answer is f > b > c > g > a = e > d.
Q9.15 As one finger slides towards the center, the normal force exerted by the sliding finger on the ruler
increases. At some point, this normal force will increase enough so that static friction between the
sliding finger and the ruler will stop their relative motion. At this moment the other finger starts
sliding along the ruler towards the center. This process repeats until the fingers meet at the center
of the ruler.
Next step: Try a rod with a nonuniform mass distribution.
Next step: Wear a piece of sandpaper as a ring on one finger to change its coefficient of friction.
*Q9.16 (a) No: mechanical energy turns into internal energy in the coupling process.
(b) No: the Earth feeds momentum into the boxcar during the downhill rolling process.
(c) Yes: total energy is constant as it turns from gravitational into kinetic.
(d) Yes: If the boxcar starts moving north the Earth, very slowly, starts moving south.
(e) No: internal energy appears.
(f) Yes: Only forces internal to the two-car system act.
, Linear Momentum and Collisions 211
Q9.17 The center of mass of the balls is in free fall, moving up and then down with the acceleration due
to gravity, during the 40% of the time when the juggler’s hands are empty. During the 60% of the
time when the juggler is engaged in catching and tossing, the center of mass must accelerate up
with a somewhat smaller average acceleration. The center of mass moves around in a little closed
loop with a parabolic top and likely a circular bottom, making three revolutions for every one
revolution that one ball makes. Letting T represent the time for one cycle and Fg the weight of
one ball, we have FJ 0.60T = 3FgT and FJ = 5Fg. The average force exerted by the juggler is five
times the weight of one ball.
Q9.18 In empty space, the center of mass of a rocket-plus-fuel system does not accelerate during a burn,
because no outside force acts on this system. The rocket body does accelerate as it blows exhaust
containing momentum out the back.
According to the text’s ‘basic expression for rocket propulsion,’ the change in speed of the
rocket body will be larger than the speed of the exhaust relative to the rocket, if the final mass is
less than 37% of the original mass.
Q9.19 To generalize broadly, around 1740 the English favored position (a), the Germans position (b),
and the French position (c). But in France Emilie de Chatelet translated Newton’s Principia and
argued for a more inclusive view. A Frenchman, Jean D’Alembert, is most responsible for showing
that each theory is consistent with the others. All the theories are equally correct. Each is useful for
giving a mathematically simple and conceptually clear solution for some problems. There is another
comprehensive mechanical theory, the angular impulse—angular momentum theorem, which we
will glimpse in Chapter 11. It identifies the product of the torque of a force and the time it acts as
the cause of a change in motion, and change in angular momentum as the effect. We have here an
example of how scientific theories are different from what people call a theory in everyday life.
People who think that different theories are mutually exclusive should bring their thinking up to
date to around 1750.
SOLUTIONS TO PROBLEMS
Section 9.1 Linear Momentum and Its Conservation
P9.1 m = 3.00 kg,
( )
v = 3.00 ˆi − 4.00 ˆj m s
(a)
( )
p = mv = 9.00 ˆi − 12.0 ˆj kg ⋅ m s
Thus,
px = 9.00 kg ⋅ m s
and
py = −12.0 kg ⋅ m s
p= px2 + py2 = ( 9.00 ) + (12.0 ) = 15.0 kg ⋅ m s
2 2
(b)
⎛ py ⎞
θ = tan −1 ⎜ ⎟ = tan −1 ( −1.33) = 307°
⎝ px ⎠
*P9.2 (a) Whomever we consider the aggressor, brother and sister exert equal-magnitude oppositely-
directed forces on each other, to give each other equal magnitudes of momentum. We take
the eastward component of the equation
total original momentum = total final momentum for the two-sibling system
0 = 65 kg (−2.9 m Ⲑs) + 40 kg v v = 4.71 m Ⲑs, meaning she moves at 4.71 m Ⲑs east
continued on next page
, 212 Chapter 9
(b) original chemical energy in girl’s body = total final kinetic energy
Uchemical = (1Ⲑ2)(65 kg)(2.9 m Ⲑs)2 + (1Ⲑ2)(40 kg)(4.71 m Ⲑs)2 = 717 J
(c) System momentum is conserved with the value zero. The net forces on the two siblings
are of equal magnitude in opposite directions. Their impulses add to zero. Their final
momenta are of equal magnitude in opposite directions, to add as vectors to zero.
P9.3 I have mass 85.0 kg and can jump to raise my center of gravity 25.0 cm. I leave the ground with
speed given by
( )
v 2f − vi2 = 2a x f − xi : 0 − vi2 = 2 ( −9.80 m s 2 ) ( 0.250 m )
vi = 2.20 m s
Total momentum of the system of the Earth and me is conserved as I push the planet down and
myself up:
0 = ( 5.98 × 10 24 kg ) (− ve ) + (85.0 kg ) ( 2.20 m s )
ve ~ 10 −23 m s
*P9.4 (a) For the system of two blocks ∆p = 0 ,
or
pi = p f
Therefore,
0 = M vm + ( 3 M ) ( 2.00 m s )
Solving gives vm = −6.00 m s (motion toward the left).
1 2 1 1
(b) kx = M v M2 + ( 3 M ) v32M = 8.40 J FIG. P9.4
2 2 2
(c) The original energy is in the spring. A force had to be exerted over a distance to com-
press the spring, transferring energy into it by work. The cord exerts force, but over no
distance.
(d) System momentum is conserved with the value zero. The forces on the two blocks are of
equal magnitude in opposite directions. Their impulses add to zero. The final momenta
of the two blocks are of equal magnitude in opposite directions.
p
P9.5 (a) The momentum is p = mv, so v = and the kinetic energy is
m
2
p2
mv 2 = m ⎛ ⎞ =
1 1 p
K=
2 2 ⎝ m⎠ 2m
1 2K 2K
(b) K= mv 2 implies v = , so p = mv = m = 2 mK
2 m m
Linear Momentum and Collisions
CHAPTER OUTLINE ANSWERS TO QUESTIONS
9.1 Linear Momentum and Its
Conservation
9.2 Impulse and Momentum
*Q9.1 (a) No. Impulse, F∆t, depends on the force and the
9.3 Collisions in One Dimension time for which it is applied.
9.4 Two-Dimensional Collisions
9.5 The Center of Mass (b) No. Work depends on the force and on the distance
9.6 Motion of a System of Particles
9.7 Deformable Systems over which it acts.
9.8 Rocket Propulsion
*Q9.2 The momentum magnitude is proportional to the speed and
the kinetic energy is proportional to the speed squared.
(i) The speed of the constant-mass object becomes
4 times larger and the kinetic energy 16 times
larger. Answer (a).
(ii) The speed and the momentum become two times
larger. Answer (d).
*Q9.3 (i) answer (c). For example, if one particle has 5 times larger mass, it will have 5 times smaller
speed and 5 times smaller kinetic energy.
(ii) answer (d). Momentum is a vector.
*Q9.4 (i) Equal net work inputs imply equal kinetic energies. Answer (c).
(ii) Imagine one particle has four times more mass. For equal kinetic energy it must have half the
speed. Then this more massive particle has 4(1Ⲑ2) = 2 times more momentum. Answer (a).
Q9.5 (a) It does not carry force, for if it did, it could accelerate itself.
(b) It cannot deliver more kinetic energy than it possesses. This would violate the law of energy
conservation.
(c) It can deliver more momentum in a collision than it possesses in its flight, by bouncing
from the object it strikes.
*Q9.6 Mutual gravitation brings the ball and the Earth together. As the ball moves downward, the
Earth moves upward, although with an acceleration on the order of 10 25 times smaller than that
of the ball. The two objects meet, rebound, and separate. Momentum of the ball-Earth system is
conserved. Answer (d).
Q9.7 (a) Linear momentum is conserved since there are no external forces acting on the system. The
fragments go off in different directions and their vector momenta add to zero.
(b) Kinetic energy is not conserved because the chemical potential energy initially in the
explosive is converted into kinetic energy of the pieces of the bomb.
209
,210 Chapter 9
Q9.8 Momentum conservation is not violated if we choose as our system the planet along with you.
When you receive an impulse forward, the Earth receives the same size impulse backwards. The
resulting acceleration of the Earth due to this impulse is much smaller than your acceleration
forward, but the planet’s backward momentum is equal in magnitude to your forward momentum.
*Q9.9 (i) During the short time the collision lasts, the total system momentum is constant. Whatever
momentum one loses the other gains. Answer (c).
(ii) When the car overtakes the manure spreader, the faster-moving one loses more energy than
the slower one gains. Answer (a).
Q9.10 The rifle has a much lower speed than the bullet and much less kinetic energy. Also, the butt
distributes the recoil force over an area much larger than that of the bullet.
*Q9.11 (i) answer (a). The ball gives more rightward momentum to the block when the ball reverses
its momentum.
(ii) answer (b). In case (a) there is no temperature increase because the collision is elastic.
Q9.12 His impact speed is determined by the acceleration of gravity and the distance of fall, in
v 2f = vi2 − 2 g ( 0 − yi ). The force exerted by the pad depends also on the unknown stiffness of the pad.
Q9.13 The sheet stretches and pulls the two students toward each other. These effects are larger for a
faster-moving egg. The time over which the egg stops is extended, more for a faster missile,
so that the force stopping it is never too large.
*Q9.14 Think about how much the vector momentum of the Frisbee changes in a horizontal plane. This
will be the same in magnitude as your momentum change. Since you start from rest, this quantity
directly controls your final speed. Thus f is largest and d is smallest. In between them, b is larger
than c and c is larger than g and g is larger than a. Also a is equal to e, because the ice can exert
a normal force to prevent you from recoiling straight down when you throw the Frisbee up. The
assembled answer is f > b > c > g > a = e > d.
Q9.15 As one finger slides towards the center, the normal force exerted by the sliding finger on the ruler
increases. At some point, this normal force will increase enough so that static friction between the
sliding finger and the ruler will stop their relative motion. At this moment the other finger starts
sliding along the ruler towards the center. This process repeats until the fingers meet at the center
of the ruler.
Next step: Try a rod with a nonuniform mass distribution.
Next step: Wear a piece of sandpaper as a ring on one finger to change its coefficient of friction.
*Q9.16 (a) No: mechanical energy turns into internal energy in the coupling process.
(b) No: the Earth feeds momentum into the boxcar during the downhill rolling process.
(c) Yes: total energy is constant as it turns from gravitational into kinetic.
(d) Yes: If the boxcar starts moving north the Earth, very slowly, starts moving south.
(e) No: internal energy appears.
(f) Yes: Only forces internal to the two-car system act.
, Linear Momentum and Collisions 211
Q9.17 The center of mass of the balls is in free fall, moving up and then down with the acceleration due
to gravity, during the 40% of the time when the juggler’s hands are empty. During the 60% of the
time when the juggler is engaged in catching and tossing, the center of mass must accelerate up
with a somewhat smaller average acceleration. The center of mass moves around in a little closed
loop with a parabolic top and likely a circular bottom, making three revolutions for every one
revolution that one ball makes. Letting T represent the time for one cycle and Fg the weight of
one ball, we have FJ 0.60T = 3FgT and FJ = 5Fg. The average force exerted by the juggler is five
times the weight of one ball.
Q9.18 In empty space, the center of mass of a rocket-plus-fuel system does not accelerate during a burn,
because no outside force acts on this system. The rocket body does accelerate as it blows exhaust
containing momentum out the back.
According to the text’s ‘basic expression for rocket propulsion,’ the change in speed of the
rocket body will be larger than the speed of the exhaust relative to the rocket, if the final mass is
less than 37% of the original mass.
Q9.19 To generalize broadly, around 1740 the English favored position (a), the Germans position (b),
and the French position (c). But in France Emilie de Chatelet translated Newton’s Principia and
argued for a more inclusive view. A Frenchman, Jean D’Alembert, is most responsible for showing
that each theory is consistent with the others. All the theories are equally correct. Each is useful for
giving a mathematically simple and conceptually clear solution for some problems. There is another
comprehensive mechanical theory, the angular impulse—angular momentum theorem, which we
will glimpse in Chapter 11. It identifies the product of the torque of a force and the time it acts as
the cause of a change in motion, and change in angular momentum as the effect. We have here an
example of how scientific theories are different from what people call a theory in everyday life.
People who think that different theories are mutually exclusive should bring their thinking up to
date to around 1750.
SOLUTIONS TO PROBLEMS
Section 9.1 Linear Momentum and Its Conservation
P9.1 m = 3.00 kg,
( )
v = 3.00 ˆi − 4.00 ˆj m s
(a)
( )
p = mv = 9.00 ˆi − 12.0 ˆj kg ⋅ m s
Thus,
px = 9.00 kg ⋅ m s
and
py = −12.0 kg ⋅ m s
p= px2 + py2 = ( 9.00 ) + (12.0 ) = 15.0 kg ⋅ m s
2 2
(b)
⎛ py ⎞
θ = tan −1 ⎜ ⎟ = tan −1 ( −1.33) = 307°
⎝ px ⎠
*P9.2 (a) Whomever we consider the aggressor, brother and sister exert equal-magnitude oppositely-
directed forces on each other, to give each other equal magnitudes of momentum. We take
the eastward component of the equation
total original momentum = total final momentum for the two-sibling system
0 = 65 kg (−2.9 m Ⲑs) + 40 kg v v = 4.71 m Ⲑs, meaning she moves at 4.71 m Ⲑs east
continued on next page
, 212 Chapter 9
(b) original chemical energy in girl’s body = total final kinetic energy
Uchemical = (1Ⲑ2)(65 kg)(2.9 m Ⲑs)2 + (1Ⲑ2)(40 kg)(4.71 m Ⲑs)2 = 717 J
(c) System momentum is conserved with the value zero. The net forces on the two siblings
are of equal magnitude in opposite directions. Their impulses add to zero. Their final
momenta are of equal magnitude in opposite directions, to add as vectors to zero.
P9.3 I have mass 85.0 kg and can jump to raise my center of gravity 25.0 cm. I leave the ground with
speed given by
( )
v 2f − vi2 = 2a x f − xi : 0 − vi2 = 2 ( −9.80 m s 2 ) ( 0.250 m )
vi = 2.20 m s
Total momentum of the system of the Earth and me is conserved as I push the planet down and
myself up:
0 = ( 5.98 × 10 24 kg ) (− ve ) + (85.0 kg ) ( 2.20 m s )
ve ~ 10 −23 m s
*P9.4 (a) For the system of two blocks ∆p = 0 ,
or
pi = p f
Therefore,
0 = M vm + ( 3 M ) ( 2.00 m s )
Solving gives vm = −6.00 m s (motion toward the left).
1 2 1 1
(b) kx = M v M2 + ( 3 M ) v32M = 8.40 J FIG. P9.4
2 2 2
(c) The original energy is in the spring. A force had to be exerted over a distance to com-
press the spring, transferring energy into it by work. The cord exerts force, but over no
distance.
(d) System momentum is conserved with the value zero. The forces on the two blocks are of
equal magnitude in opposite directions. Their impulses add to zero. The final momenta
of the two blocks are of equal magnitude in opposite directions.
p
P9.5 (a) The momentum is p = mv, so v = and the kinetic energy is
m
2
p2
mv 2 = m ⎛ ⎞ =
1 1 p
K=
2 2 ⎝ m⎠ 2m
1 2K 2K
(b) K= mv 2 implies v = , so p = mv = m = 2 mK
2 m m
