## SOLUTION MANUAL PHYSICS FOR SCIENTISTS AND ENGINEERS 9TH EDITION SERWAY

9

Linear Momentum and Collisions

CHAPTER OUTLINE

9.1 Linear Momentum

9.2 Analysis Model: Isolated System (Momentum)

9.3 Analysis Model: Nonisolated System (Momentum)

9.4 Collisions in One Dimension

9.5 Collisions in Two Dimensions

9.6 The Center of Mass

9.7 Systems of Many Particles

9.8 Deformable Systems

9.9 Rocket Propulsion

* An asterisk indicates a question or problem new to this edition.

ANSWERS TO OBJECTIVE QUESTIONS

OQ9.1 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 (b) is largest and (c) is smallest. In

between them, (e) is larger than (a) and (a) is larger than (c). Also (a) is

equal to (d), 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 b > e > a = d > c.

OQ9.2 (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.

Chapter 9 439

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(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.

OQ9.3 (i) Answer (c). During the short time the collision lasts, the total

system momentum is constant. Whatever momentum one loses

the other gains.

(ii) Answer (a). The problem implies that the tractor’s momentum is

negligible compared to the car’s momentum before the collision.

It also implies that the car carries most of the kinetic energy of the

system. The collision slows down the car and speeds up the

tractor, so that they have the same final speed. The faster-moving

car loses more energy than the slower tractor gains because a lot

of the car’s original kinetic energy is converted into internal

energy.

OQ9.4 Answer (a). We have m1 = 2 kg, v1i = 4 m/s; m2 = 1 kg, and v1i = 0. We

find the velocity of the 1-kg mass using the equation derived in Section

9.4 for an elastic collision:

v2 f = 2m1

m1 + m2

⎛

⎝

⎜

⎞

⎠

⎟ v1i +

m1 − m2

m1 + m2

⎛

⎝

⎜

⎞

⎠

⎟ v2i

v2 f = 4 kg

3 kg

⎛

⎝

⎜

⎞

⎠

⎟(4 m/s) +

1 kg

3 kg

⎛

⎝

⎜

⎞

⎠

⎟(0) = 5.33 m/s

OQ9.5 Answer (c). We choose the original direction of motion of the cart as

the positive direction. Then, vi = 6 m/s and vf = −2 m/s. The change in

the momentum of the cart is

Δp = mvf − mvi = m vf − v ( i) = (5 kg)(−2 m/s − 6 m/s)

= −40 kg⋅m/s.

OQ9.6 Answer (c). The impulse given to the ball is I = FavgΔt = mvf − mvi

.

Choosing the direction of the final velocity of the ball as the positive

direction, this gives

Favg = m(vf − vi)

Δt = 57.0 × 10−3 ( kg)[25.0 m/s − (−21.0 m/s)]

0.060 s

= 43.7 kg ⋅m/s2 = 43.7 N

440 Linear Momentum and Collisions

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OQ9.7 Answer (a). The magnitude of momentum is proportional to speed and

the kinetic energy is proportional to speed squared. The speed of the

rocket becomes 4 times larger, so the kinetic energy becomes 16 times

larger.

OQ9.8 Answer (d). The magnitude of momentum is proportional to speed

and the kinetic energy is proportional to speed squared. The speed of

the rocket becomes 2 times larger, so the magnitude of the momentum

becomes 2 times larger.

OQ9.9 Answer (c). The kinetic energy of a particle may be written as

KE = mv2

2 = m2

v2

2m = (mv)

2

2m = p2

2m

The ratio of the kinetic energies of two particles is then

(KE)2

(KE)1

= p2

2 2m2

p1

2 2m1

= p2

p1

⎛

⎝

⎜

⎞

⎠

⎟

2

m1

m2

⎛

⎝

⎜

⎞

⎠

⎟

We see that, if the magnitudes of the momenta are equal (p2 = p1), the

kinetic energies will be equal only if the masses are also equal. The

correct response is then (c).

OQ9.10 Answer (d). Expressing the kinetic energy as KE = p

2

/2m, we see that

the ratio of the magnitudes of the momenta of two particles is

p2

p1

= 2m2 (KE)2

2m1(KE)1

= m2

m1

⎛

⎝

⎜

⎞

⎠

⎟

(KE)2

(KE)1

Thus, we see that if the particles have equal kinetic energies [(KE)2 =

(KE)1], the magnitudes of their momenta are equal only if the masses

are also equal. However, momentum is a vector quantity and we can

say the two particles have equal momenta only it both the magnitudes

and directions are equal, making choice (d) the correct answer.

OQ9.11 Answer (b). Before collision, the bullet, mass m1 = 10.0 g, has speed

v1i = vb, and the block, mass m2 = 200 g, has speed v2i = 0. After collision,

the objects have a common speed (velocity) v1f = v2f = v. The collision of

the bullet with the block is completely inelastic:

m1v1i + m2v2 = m1v1f + m2v2f

m1vb = (m1 + m2)v , so vb = v

m1 + m2

m1

⎛

⎝

⎜

⎞

⎠

⎟

Chapter 9 441

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The kinetic friction, fk = µk

n, slows down the block with acceleration of

magnitude µk

g. The block slides to a stop through a distance d = 8.00 m.

Using vf

2 = vi

2 + 2a(xf − xi

), we find the speed of the block just after the

collision:

v = 2(0.400)(9.80 m/s2

)(8.00 m) = 7.92 m/s.

Using the results above, the speed of the bullet before collision is

vb = (7.92m/s) 10 + 200

10.0

⎛

⎝

⎜

⎞

⎠

⎟ = 166 m/s.

OQ9.12 Answer (c). The masses move through the same distance under the

same force. Equal net work inputs imply equal kinetic energies.

OQ9.13 Answer (a). The same force gives the larger mass a smaller

acceleration, so the larger mass takes a longer time interval to move

through the same distance; therefore, the impulse given to the larger

mass is larger, which means the larger mass will have a greater final

momentum.

OQ9.14 Answer (d). Momentum of the ball-Earth system is conserved. Mutual

gravitation brings the ball and the Earth together into one system. As

the ball moves downward, the Earth moves upward, although with an

acceleration on the order of 1025 times smaller than that of the ball. The

two objects meet, rebound, and separate.

OQ9.15 Answer (d). Momentum is the same before and after the collision.

Before the collision the momentum is

m1v1 + m2v2 = (3 kg)(+2 m/s) + (2 kg)(−4 m/s) = −2 kg ⋅m/s

OQ9.16 Answer (a). The ball gives more rightward momentum to the block

when the ball reverses its momentum.

OQ9.17 Answer (c). Assuming that the collision was head-on so that, after

impact, the wreckage moves in the original direction of the car’s

motion, conservation of momentum during the impact gives

(mc + mt )vf = mcv0c + mtv0t = mcv + mt(0)

or

vf = mc

mc + mt

⎛

⎝

⎜

⎞

⎠

⎟ v = m

m + 2m

⎛

⎝

⎜

⎞

⎠

⎟ v = v

3

442 Linear Momentum and Collisions

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OQ9.18 Answer (c). Billiard balls all have the same mass and collisions

between them may be considered to be elastic. The dual requirements

of conservation of kinetic energy and conservation of momentum in a

one-dimensional, elastic collision are summarized by the two relations:

m1v1i + m2v2i = m1v1 f + m2v2 f [1]

and

v1i − v2i = v1 f − v ( 2 f) [2]

In this case, m1 = m2 and the masses cancel out of the first equation.

Call the blue ball #1 and the red ball #2 so that v1i = −3v, v2i = +v,

v 1f = vblue, and v2f = vred. Then, the two equations become

−3v + v = vblue + vred or vblue + vred = v [1]

and

−3v − v = − vblue − v ( red ) or vblue − v ( red ) = 4v [2]

Adding the final versions of these equations yields 2vblue = 2v, or vblue =

v. Substituting this result into either [1] or [2] above then yields vred =

−3v.

ANSWERS TO CONCEPTUAL QUESTIONS

CQ9.1 The passenger must undergo a certain momentum change in the

collision. This means that a certain impulse must be exerted on the

passenger by the steering wheel, the window, an air bag, or something.

By increasing the distance over which the momentum change occurs,

the time interval during which this change occurs is also increased,

resulting in the force on the passenger being decreased.

CQ9.2 If the golfer does not “follow through,” the club is slowed down by the

golfer before it hits the ball, so the club has less momentum available

to transfer to the ball during the collision.

CQ9.3 Its speed decreases as its mass increases. There are no external

horizontal forces acting on the box, so its momentum cannot change as

it moves along the horizontal surface. As the box slowly fills with

water, its mass increases with time. Because the product mv must be

constant, and because m is increasing, the speed of the box must

decrease. Note that the vertically falling rain has no horizontal

momentum of its own, so the box must “share” its momentum with

the rain it catches.

Chapter 9 443

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CQ9.4 (a) It does not carry force, force requires another object on which to

act.

(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.

CQ9.5 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. If we choose you as the

system, momentum conservation is not violated because you are not

an isolated system.

CQ9.6 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.

CQ9.7 The time interval over which the egg is stopped by the sheet (more for

a faster missile) is much longer than the time interval over which the

egg is stopped by a wall. For the same change in momentum, the

longer the time interval, the smaller the force required to stop the egg.

The sheet increases the time interval so that the stopping force is never

too large.

CQ9.8 (a) Assuming that both hands are never in contact with a ball, and

one hand is in contact with any one ball 20% of the time, the total

contact time with the system of three balls is 3(20%) = 60% of the

time. 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.

(b) On average, in one cycle of the system, the center of mass of the

balls does not change position, so its average acceleration is zero

(i.e., the average net force on the system is zero). 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.

444 Linear Momentum and Collisions

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CQ9.9 (a) 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 itself does accelerate as it blows

exhaust containing momentum out the back.

(b) 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.

CQ9.10 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.

CQ9.11 No. Impulse,

FΔt, depends on the force and the time interval during

which it is applied.

CQ9.12 No. Work depends on the force and on the displacement over which it

acts.

CQ9.13 (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.

Chapter 9 445

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SOLUTIONS TO END-OF-CHAPTER PROBLEMS

Section 9.1 Linear Momentum

P9.1 (a) The momentum is p = mv, so v = p/m and the kinetic energy is

K = 1

2

mv2 = 1

2

m p

m

⎛

⎝

⎜

⎞

⎠

⎟

2

= p2

2m

(b) K = 1

2

mv2 implies v = 2K

m

so p = mv = m

2K

m = 2mK .

P9.2 K = p

2

/2m, and hence, p = 2mK. Thus,

m = p2

2 ⋅K = (25.0 kg ⋅m/s)

2

2(275 J) = 1.14 kg

and

v = p

m = 2m(K)

m = 2(K)

m = 2(275 J)

1.14 kg = 22.0 m/s

P9.3 We apply the impulse-momentum theorem to relate the change in the

horizontal momentum of the sled to the horizontal force acting on it:

Δpx = FxΔt → Fx = Δpx

Δt = mvxf − mvxi

Δt

Fx = −(17.5 kg)(3.50 m/s)

8.75 s

Fx = 7.00 N

*P9.4 We are given m = 3.00 kg and

v = 3.00ˆ

i − 4.00ˆ ( j) m/s.

(a) The vector momentum is then

p = m

v = (3.00 kg) 3.00ˆ

i − 4.00ˆ ⎡( j) m/s ⎣ ⎤

⎦

= 9.00ˆ

i − 12.0ˆ ( j) kg ⋅m/s

Thus, px = 9.00 kg ⋅m/s and py = −12.0 kg ⋅m/s .

(b) p = px

2 + py

2 = (9.00 kg ⋅m/s)

2

+ (12.0 kg ⋅m/s)

2

= 15.0 kg ⋅m/s

446 Linear Momentum and Collisions

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at an angle of

θ = tan−1 py

px

⎛

⎝

⎜ ⎞

⎠

⎟ = tan−1 (−1.33) = 307°

P9.5 We apply the impulse-momentum theorem to find the average force

the bat exerts on the baseball:

Δ

p =

FΔt →

F = Δ

p

Δt = m

v f −

vi

Δt

⎛

⎝

⎜ ⎞

⎠

⎟

Choosing the direction toward home plate as the positive x direction,

we have

vi = (45.0 m/s)ˆ

i,

v f = (55.0 m/s)ˆ

j, and Δt = 2.00 ms:

Fon ball = m

v f −

vi

Δt = (0.145 kg)

(55.0 m/s)ˆ

j − (45.0 m/s)ˆ

i

2.00 × 10−3

s

Fon ball = −3.26ˆ

i + 3.99ˆ ( j) N

By Newton’s third law,

Fon bat = −

Fon ball so

Fon bat = +3.26ˆ

i − 3.99ˆ ( j) N

Section 9.2 Analysis Model: Isolated system (Momentum)

P9.6 (a) The girl-plank system is isolated, so horizontal momentum is

conserved.

We measure momentum relative to the ice:

pgi +

ppi =

pgf +

ppf .

The motion is in one dimension, so we can write,

vgi

ˆ

i = v

gp

ˆ

i + vpi

ˆ

i → vgi = v

gp

+ vpi

where vgi denotes the velocity of the girl relative to the ice, vgp the

velocity of the girl relative to the plank, and vpi the velocity of the

plank relative to the ice. The momentum equation becomes

0 = mg vgi

ˆ

i + mpvpi

ˆ

i → 0 = mg vgi + mpvpi

0 = mg vgp + v ( pi) + mpvpi

0 = mg vgp + (mg + mp )vpi → vpi = − mg

mg + mp

⎛

⎝

⎜

⎞

⎠

⎟ vgp

Chapter 9 447

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solving for the velocity of the plank gives

vpi = − mg

mg + mp

⎛

⎝

⎜

⎞

⎠

⎟ vgp = − 45.0 kg

45.0 kg + 150 kg

⎛

⎝

⎜

⎞

⎠

⎟(1.50 m/s)

vpi = −0.346 m/s

(b) Using our result above, we find that

vgi = vgp + vpi = (1.50 m/s) + (−0.346 m/s)

vgi = 1.15 m/s

P9.7 (a) The girl-plank system is isolated, so horizontal momentum is

conserved.

We measure momentum relative to the ice:

pgi +

ppi =

pgf +

ppf .

The motion is in one dimension, so we can write

vgi

ˆ

i = vgpˆ

i + vpi

ˆ

i → vgi = vgp + vpi

where vgi denotes the velocity of the girl relative to the ice, vgp the

velocity of the girl relative to the plank, and vpi the velocity of the

plank relative to the ice. The momentum equation becomes

0 = mg vgi

ˆ

i + mpvpi

ˆ

i → 0 = mg vgi + mpvpi

0 = mg vgp + v ( pi) + mpvpi

0 = mg vgp + (mg + mp )vpi

solving for the velocity of the plank gives

vpi = − mg

mg + mp

⎛

⎝

⎜

⎞

⎠

⎟ vgp

(b) Using our result above, we find that

vgi = vgp + vpi = vgp

(mg + mp )

mg + mp

− mg

mg + mp

vgp

vgi = (mg + mp )vgp − mg vgp

mg + mp

vgi = mg vgp + mpvgp − mg vgp

mg + mp

448 Linear Momentum and Collisions

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vgi = mp

mg + mp

⎛

⎝

⎜

⎞

⎠

⎟ vgp

P9.8 (a) Brother and sister exert equal-magnitude oppositely-directed

forces on each other for the same time interval; therefore, the

impulses acting on them are equal and opposite. Taking east as

the positive direction, we have

impulse on boy: I = FΔt = Δp = (65.0 kg)(−2.90 m/s) = −189 N⋅ s

impulse on girl: I = −FΔt = −Δp = +189 N⋅ s = mvf

Her speed is then

vf = I

m = 189 N⋅ s

40.0 kg = 4.71 m/s

meaning she moves at 4.71 m/s east .

(b) original chemical potential energy in girl’s body = total final

kinetic energy

Uchemical = 1

2

mboy vboy

2 +

1

2

mgirl

vgirl

2

= 1

2

(65.0 kg)(2.90 m/s)

2

+

1

2

(40.0 kg)(4.71 m/s)

2

= 717 J

(c) Yes. System momentum is conserved with the value zero.

(d) The forces on the two siblings are internal forces, which cannot

change the momentum of the system— the system is isolated .

(e) Even though there is motion afterward, the final momenta are

of equal magnitude in opposite directions so the final momentum

of the system is still zero.

Chapter 9 449

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*P9.9 We assume that the velocity of the blood is constant over the 0.160 s.

Then the patient’s body and pallet will have a constant velocity of

6 × 10−5

m

0.160 s = 3.75 × 10−4

m/s in the opposite direction. Momentum

conservation gives

p1i +

p2i =

p1 f +

p2 f :

0 = mblood (0.500 m s) + (54.0 kg) −3.75 × 10−4 ( m/s)

mblood = 0.040 5 kg = 40.5 g

P9.10 I have mass 72.0 kg and can jump to raise my center of gravity 25.0 cm.

I leave the ground with speed given by

vf

2 − vi

2 = 2a xf − x ( i): 0 − vi

2 = 2 −9.80 m/s2 ( )(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 × 1024 ( kg) −v ( e ) + (85.0 kg)(2.20 m/s)

ve 10−23 m/s

P9.11 (a) For the system of two blocks Δp = 0, or i f p = p . Therefore,

0 = mvm + (3m)(2.00 m/s)

Solving gives vm = −6.00 m/s (motion toward the left).

(b) 1

2

kx2 = 1

2

mvM

2 +

1

2

(3m)v3M

2

= 1

2

(0.350 kg)(−6.00 m/s)

2 +

3

2

(0.350 kg)(2.00 m/s)

2

= 8.40 J

(c) The original energy is in the spring.

(d) A force had to be exerted over a displacement to compress the

spring, transferring energy into it by work.

The cord exerts force, but over no displacement.

(e) System momentum is conserved with the value zer

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