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SOLUTION MANUAL PHYSICS FOR SCIENTISTS AND ENGINEERS 9TH EDITION SERWAY

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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
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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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