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CHAPTER 1: Introduction, Measurement, Estimating
Answers to Questions
4.
The problem is that the precision of the two measurements are quite different. It would be more
appropriate to give the metric distance as 11 km, so that the numbers are given to about the same precision
(nearest mile or nearest km).
7.
The result should be written as 8.32 cm. The factor of 2 used to convert radius to diameter is exact – it has
no uncertainty, and so does not change the number of significant figures.
9.
Since the size of large eggs can vary by 10%, the random large egg used in a recipe has a size with an
uncertainty of about 5% . Thus the amount of the other ingredients can also vary by about 5% and not
adversely affect the recipe.
Solutions to Problems
1.
13.
15.
(a)
14 billion years  1.4  1010 years
(b)
1.4 10 y  3.156 10 s 1 y   4.4 10
(a)
1 106 volts
(b)
2  10 6 meters
(c)
6 103 days
6 kilodays  6 kdays
(d)
18  10 2 bucks
18 hectobucks  18 hbucks
10
7
17
s
1 megavolt  1 Mvolt
2 micrometers  2  m
9
8 nanopieces  8 npieces
(e)
8 10 pieces
(a)
93 million miles  93  106 miles 1610 m 1 mile   1.5  1011 m
(b)
1.5  1011 m  150  109 m  150 gigameters or 1.5  1011 m  0.15  1012 m  0.15 terameters
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CHAPTER 2: Describing Motion: Kinematics in One Dimension
Answers to Questions
3.
When an object moves with constant velocity, its average velocity over any time interval is exactly equal to
its instantaneous velocity at all times
6.
The acceleration of both the motorcycle and the bicycle are the same, since the same change in velocity
occurred during the same time interval. If you do a further calculation, you will find that the distance
traveled by the motorcycle during the acceleration is 17 times the distance traveled by the bicycle.
11. Assume that north is the positive direction. If a car is moving south and gaining speed at an increasing rate,
then the acceleration will be getting larger in magnitude. However, since the acceleration is directed
southwards, the acceleration is negative, and is getting more negative. That is a decreasing acceleration as
the speed increases.
Another example would be an object falling WITH air resistance. As the object falls, it gains speed, the air
resistance increases. As the air resistance increases, the acceleration of the falling object decreases, and it
gains speed less quickly the longer it falls.
17. The elevator moving from the second floor to the fifth floor is NOT an example of constant acceleration.
The elevator accelerates upward each time it starts to move, and it accelerates downward each time it stops.
Ignoring air resistance, a rock falling from a cliff would have a constant acceleration. (If air resistance is
included, then the acceleration will be decreasing as the rock falls.) A dish resting on a table has an
acceleration of 0, so the acceleration is constant.
19. If an object is at the instant of reversing direction (like an object thrown upward, at the top of its path), it
instantaneously has a zero velocity and a non-zero acceleration at the same time. A person at the exact
bottom of a “bungee” cord plunge also has an instantaneous velocity of zero but a non-zero (upward)
acceleration at the same time.
20. An object moving with a constant velocity has a non-zero velocity and a zero acceleration at the same time.
So a car driving at constant speed on a straight, level roadway would meet this condition.
21. The object starts with a constant velocity in the positive direction. At about t = 17 s, when the object is at
the 5 meter position, it begins to gain speed – it has a positive acceleration. At about t = 27 s, when the
object is at about the 12 m position, it begins to slow down – it has a negative acceleration. The object
instantaneously stops at about t = 37 s, reaching its maximum distance from the origin of 20 m. The object
then reverses direction, gaining speed while moving backwards. At about t = 47 s, when the object is again
at about the 12 m position, the object starts to slow down, and appears to stop at t = 50 s, 10 m from the
starting point.
22. Initially, the object moves in the positive direction with a constant acceleration, until about t = 45 s, when it
has a velocity of about 37 m/s in the positive direction. The acceleration then decreases, reaching an
instantaneous acceleration of 0 at about t = 50 s, when the object has its maximum speed of about 38 m/s.
The object then begins to slow down, but continues to move in the positive direction. The object stops
moving at t = 90 s and stays at rest until about t = 108 s. Then the object begins to move to the right again,
at first with a large acceleration, and then a lesser acceleration. At the end of the recorded motion, the
object is still moving to the right and gaining speed.
Solutions to Problems
1.
The average speed is given by:
v  d t  235 km 3.25 h  72.3 km h .
2.
The time of travel can be found by rearranging the average speed equation.
v  d t  t  d v  15 km   25km h   0.60 h  36 min
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3.
The distance of travel (displacement) can be found by rearranging the average speed equation. Also note
that the units of the velocity and the time are not the same, so the speed units will be converted.
d
 1h 
v
 d  v t  110 km h  
  2.0 s   0.061 km  61 m
t
 3600 s 
7.
The time for the first part of the trip is calculated from the initial speed and the first distance.
d
d
130 km
ave speed1  v1  1  t1  1 
 1.37 h  82 min
t1
v1 95 km h
The time for the second part of the trip is therefore
t2  ttot  t1  3.33 h  1.37 h  1.96 h  118 min
The distance for the second part of the trip is calculated from the average speed for that part of the trip and
the time for that part of the trip.
d
ave speed 2  v2  2  d 2  v2 t2   65 km h 1.96 h   127.5 km  1.3  102 km
t2
2
(a) The total distance is then d total  d1  d 2  130 km  127.5 km  257.5 km  2.6  10 km
(b) The average speed is NOT the average of the two speeds. Use the definition of average speed.
d
257.5 km
ave speed  total 
 77 km h
ttotal
3.33 h
9.
The distance traveled is 2.0 miles  8 laps  0.25 mi lap  . The displacement is 0 because the ending point
is the same as the starting point.
d
2.0 mi
 2 mi   1610 m  1 min 


(a) Average speed =

  4.3 m s

t 12.5 min  12.5 min   1 mi  60 s 
(b) Average velocity = v  x t  0 m s
21.
By definition, the acceleration is a 
v  v0

25 m s  13 m s
t
6.0 s
The distance of travel can be found from Eq. 2-11b.

x  x0  v0t  12 at 2  13 m s  6.0 s   12 2.0 m s 2
  6.0 s 
2
 2.0 m s 2 .
 114 m
33. Choose downward to be the positive direction, and take y0  0 at the top of the cliff. The initial velocity
is v0  0 , and the acceleration is a  9.80 m s 2 . The displacement is found from equation (2-11b), with x
replaced by y.
y  y0  v0t  12 at 2  y  0  0 
1
2
 9.80 m s   3.25 s 
2
2
 y  51.8 m
37. Choose upward to be the positive direction, and take y0  0 to be the height from which the ball was
thrown. The acceleration is a  9.80 m s 2 . The displacement upon catching the ball is 0, assuming it
was caught at the same height from which it was thrown. The starting speed can be found from Eq. 2-11b,
with x replaced by y.
y  y0  v0t  12 at 2  0 
y  y0  12 at 2


 3.0 s   14.7 m s  15 m s
t
The height can be calculated from Eq. 2-11c, with a final velocity of v  0 at the top of the path.
v0 
  12 at   12 9.80 m s 2
v  v  2 a  y  y0   y  y0 
2
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2
0
v 2  v02
2a
 0
0  14.7 m s 

2 9.8 m s 2
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2

 11 m
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CHAPTER 3: Kinematics in Two Dimensions; Vectors
Answers to Questions
1. Their velocities are NOT equal, because the two velocities have different directions.
6.
If the two vectors are in the same direction, the magnitude of their sum will be a maximum, and will be 7.5
km. If the two vectors are in the opposite direction, the magnitude of their sum will be a minimum, and
will be 0.5 km. If the two vectors are oriented in any other configuration, the magnitude of their sum will
be between 0.5 km and 7.5 km.
18. The arrow should be aimed above the target, because gravity will deflect the arrow downward from a
horizontal flight path. The angle of aim (above the horizontal) should increase as the distance from the
target increases, because gravity will have more time to act in deflecting the arrow from a straight-line path.
If we assume that the arrow when shot is at the same height as the target, then the range formula is
applicable: R  v02 sin 2 0 g    12 sin 1  Rg v02  . As the range and hence the argument of the
inverse sine function increases, the angle increases.
19. The horizontal component of the velocity stays constant in projectile motion, assuming that air resistance is
negligible. Thus the horizontal component of velocity 1.0 seconds after launch will be the same as the
horizontal component of velocity 2.0 seconds after launch. In both cases the horizontal velocity will be
given by vx  v0 cos    30 m s   cos 30o   26 m s .
Solutions to Problems
1.
The resultant vector displacement of the car is given by
DR  Dwest  Dsouth- . The westward displacement is
Dwest

DR
Dsouth-
west
west
215  85 cos 45  275.1 km and the south displacement is
o
85sin 45o  60.1 km . The resultant displacement has a magnitude of
275.12  60.12  281.6 km
 282 km . The direction is   tan 1 60.1 275.1  12.3o  12o south of west .
17. Choose downward to be the positive y direction. The origin will be at the point where the tiger leaps from
the rock. In the horizontal direction, vx 0  3.5 m s and ax  0 . In the vertical direction, v y 0  0 ,
ay  9.80 m s 2 , y0  0 , and the final location y  6.5 m . The time for the tiger to reach the ground is
found from applying Eq. 2-11b to the vertical motion.
y  y0  v y 0t  12 a y t 2

6.5m  0  0 
1
2
 9.8 m s  t
2
2

t
2  6.5m 
9.8 m s 2
The horizontal displacement is calculated from the constant horizontal velocity.
 1.15 sec
x  vxt   3.5m s 1.15 sec   4.0 m
21. Choose downward to be the positive y direction. The origin will be at the point where the ball is thrown
from the roof of the building. In the vertical direction, v y 0  0 , ay  9.80 m s 2 , y0  0 , and the
displacement is 45.0 m. The time of flight is found from applying Eq. 2-11b to the vertical motion.
y  y0  v y 0t  12 a y t 2
45.0 m 
 9.80 m s  t
1
2
2
2
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
t
2  45.0 m 
 3.03 sec
9.80 m s 2
The horizontal speed (which is the initial speed) is found from the horizontal motion at constant velocity:
x  vx t  vx  x t  24.0 m 3.03 s  7.92 m s .
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37. Call the direction of the flow of the river the x direction, and the direction of Huck walking relative to the
raft the y direction.
v Huck
v Huck  v Huck  v raft rel.   0, 0.60  m s  1.70, 0  m s
rel. bank
rel. bank
rel. raft
v Huck
bank
 1.70, 0.60  m s
Magnitude: vHuck
rel. raft

v raft
 1.702  0.602  1.80 m s
rel. bank
 current 
rel. bank
Direction:   tan 1
38.
0.60
1.70
 19o relative to river
We have vcar rel.  25 m s . Use the diagram, illustrating vsnow rel.  vsnow rel.  v car rel. , to calculate the
ground
ground
other speeds.
cos 30 
ground
vsnow rel.
ground
vsnow rel.
vcar rel.
o
car
 vsnow rel.  25 m s cos 30  29 m s
o
car
vsnow rel.
car
ground
30o
car
vsnow rel.
tan 30 o 
ground
vcar rel.
vcar rel.
 vsnow rel.   25 m s  tan 30 o  14 m s
ground
ground
ground
39. Call the direction of the flow of the river the x direction, and the direction the boat is headed the y direction.
(a) vboat rel. 
2
2
vwater
 vboat
 1.20 2  2.30 2  2.59 m s
rel.
rel.
shore
shore
  tan 1
1.20
2.30
water
 27.6o ,   90o    62.4o relative to shore
(b) The position of the boat after 3.00 seconds is given by
d  vboat rel.t  1.20, 2.30  m s   3.00 sec 
shore
  3.60 m downstream, 6.90 m across the river 
As a magnitude and direction, it would be 7.8 m away from the starting point, at an angle of 62.4o
relative to the shore.
v water rel.
shore
v boat rel.
shore

v boat rel.
water

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CHAPTER 4: Dynamics: Newton’s Laws of Motion
Answers to Questions
1.
The child tends to remain at rest (Newton’s 1st Law), unless a force acts on her. The force is applied to the
wagon, not the child, and so the wagon accelerates out from under the child, making it look like the child
falls backwards relative to the wagon. If the child is standing in the wagon, the force of friction between
the child and the bottom of the wagon will produce an acceleration of the feet, pulling the feet out from
under the child, also making the child fall backwards.
2.
(a) Mary sees the box stay stationary with respect to the ground. There is no horizontal force on
the box since the truck bed is smooth, and so the box cannot accelerate. Thus Mary would describe
the motion of the box in terms of Newton’s 1st law – there is no force on the box, so it does not
accelerate.
(b) Chris, from his non-inertial reference frame, would say something about the box being
“thrown” backwards in the truck, and perhaps use Newton’s 2nd law to describe the effects of that
force. But the source of that force would be impossible to specify.
3.
If the acceleration of an object is zero, then by Newton’s second law, the net force must be zero. There can
be forces acting on the object as long as the vector sum of the forces is zero.
4.
If only once force acts on the object, then the net force cannot be zero. Thus the object cannot have zero
acceleration, by Newton’s second law. The object can have zero velocity for an instant. For example, an
object thrown straight up under the influence of gravity has a velocity of zero at the top of its path, but has
a non-zero net force and non-zero acceleration throughout the entire flight.
13. Let us find the acceleration of the Earth, assuming the mass of the freely falling object is m = 1 kg. If the
mass of the Earth is M, then the acceleration of the Earth would be found using Newton’s 3rd law and
Newton’s 2nd law.
FEarth  Fobject  MaEarth  mg  aEarth  g m M
Since the Earth has a mass that is on the order of 1025 kg, then the acceleration of the Earth is on the order
of 1025 g , or about 1024 m s 2 . This tiny acceleration is undetectable.
Solutions to Problems
1.
Use Newton’s second law to calculate the force.
 F  ma   60.0 kg  1.25 m s  
2
9.
75.0 N
The problem asks for the average force on the glove, which in a direct calculation would require knowledge
about the mass of the glove and the acceleration of the glove. But no information about the glove is given.
By Newton’s 3rd law, the force exerted by the ball on the glove is equal and opposite to the force exerted by
the glove on the ball. So calculate the average force on the ball, and then take the opposite of that result to
find the average force on the glove. The average force on the ball is its mass times its average acceleration.
Use Eq. 2-11c to find the acceleration of the ball, with v  0 , v0  35.0 m s , and x  x0  0.110 m . The
initial direction of the ball is the positive direction.
aavg 
v 2  v02
2  x  x0 

0   35.0 m s 
2  0.110 m 

2
 5568 m s 2

Favg  maavg   0.140 kg  5568 m s 2  7.80  102 N
Thus the average force on the glove was 780 N, in the direction of the initial velocity of the ball.
23. Consider the point in the rope directly below Arlene. That point can
analyzed as having three forces on it – Arlene’s weight, the tension
the rope towards the right point of connection, and the tension in the
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10
o
FT
FT
mg
10
o
be
in
rope
Page 6 of 11
towards the left point of connection. Assuming the rope is massless, those two tensions will be of the same
magnitude. Since the point is not accelerating the sum of the forces must be zero. In particular, consider
the sum of the vertical forces on that point, with UP as the positive direction.
 F  FT sin10.0o  FT sin10.0o  mg  0 
FT 
mg
2sin10.0
o

 50.0 kg   9.80
2sin10.0
m s2
o
  1.4110 N
3
27. Choose the y direction to be the “forward” direction for the motion of the snowcats, and the x direction to
be to the right on the diagram in the textbook. Since the housing unit moves in the forward direction on a
straight line, there is no acceleration in the x direction, and so the net force in the x direction must be 0.
Write Newton’s 2nd law for the x direction.
 Fx  FAx  FBx  0   FA sin 50o  FB sin 30o  0 
FB 
FA sin 50o
o

 4500 N  sin 50o
o
 6.9  103 N
sin 30
sin 30
Since the x components add to 0, the magnitude of the vector sum of the two forces will just be the
sum of their y components.
F
y
39.
 FAy  FBy  FA cos 50o  FB cos 30o   4500 N  cos 50o   6900 N  cos 30o  8.9 103 N
A free-body diagram for the accelerating car is shown. The car does not
accelerate vertically, and so FN  mg . The static frictional force is the
FN
Ffr
accelerating force, and so Ffr  ma . If we assume the maximum acceleration,
we need the maximum force, and so the static frictional force would be its
maximum value of s FN . Thus we have
then
mg
Ffr  ma   s FN  ma   s mg  ma 


a   s g  0.80 9.8 m s 2  7.8 m s 2
y
x
FN
Ffr


mg
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CHAPTER 6: Work and Energy
Answers to Questions
3.
The normal force can do work on an object if the normal force has a component in the direction of
displacement of an object. If someone were to jump up in the air, then the floor pushing upward on the
person (the normal force) would do positive work and increase the person’s kinetic energy. Likewise when
they hit the floor coming back down, the force of the floor pushing upwards (the normal force) would do
negative work and decrease the person’s kinetic energy.
6.
While it is true that no work is being done on the wall by you, there is work being done inside your arm
muscles. Exerting a force via a muscle causes small continual motions in your muscles, which is work, and
which causes you to tire. An example of this is holding a heavy load at arm’s length. While at first you
may hold the load steady, after a time your arm will begin to shake, which indicates the motion of muscles
in your arm.
7.
(a) In this case, the same force is applied to both springs. Spring 1 will stretch less, and so more
work is done on spring 2.
(b) In this case, both springs are stretched the same distance. It takes more force to stretch spring 1,
and so more work is done on spring 1.
10. Since each balloon has the same initial kinetic energy, and each balloon undergoes the same overall change
in gravitational PE, each balloon will have the same kinetic energy at the ground, and so each one has the
same speed at impact.
14. When water at the top of a waterfall falls to the pool below, initially the water’s gravitational PE is turned
into kinetic energy. That kinetic energy then can do work on the pool water when it hits it, and so some of
the pool water is given energy, which makes it splash upwards and outwards and creates outgoing water
waves, which carry energy. Some of the energy will become heat, due to viscous friction between the
falling water and the pool water. Some of the energy will become kinetic energy of air molecules, making
sound waves that give the waterfall its “roar”.
19. If we assume that all of the arrow’s kinetic energy is converted into work done against friction, then the
following relationship exists:
W  KE  KE f  KEi  Ffr d cos180o  12 mv 2f  12 mv02   Ffr d   12 mv02 
d
mv02
2 Ffr
Thus the distance is proportional to the square of the initial velocity. So if the initial velocity is doubled,
the distance will be multiplied by a factor of 4. Thus the faster arrow penetrates 4 times further than the
slower arrow.
Solutions to Problems
1.
The force and the displacement are both downwards, so the angle between them is 0o.

WG  mgd cos    265 kg  9.80 m s 2
5.
  2.80 m  cos 0
 7.27 103 J
Since the acceleration of the box is constant, use Eq. 2-11b to find the distance moved. Assume that the
box starts from rest.
2
x  x  x0  v0t  12 at 2  0  12 2.0 m s 2  7 s   49 m


Then the work done in moving the crate is

W  F x cos 0o  max   5 kg  2.0 m s 2
15.
o
  49 m  
4.9  102 J
Find the velocity from the kinetic energy, using Eq. 6-3.
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2  KE 
KE  12 mv 2  v 
m


2 6.21 1021 J
5.31 10
26

484 m s
19. The force exerted by the bow on the arrow is in the same direction as the displacement of the arrow. Thus
W  Fd cos 0o  Fd  110 N  0.78 m   85.8 J . But that work changes the KE of the arrow, by the
work-energy theorem. Thus
Fd  W  KE2  KE1  12 mv22  12 mv12  v2 
2Fd
m
 v12 
2  85.8 J 
0.088 kg
 0  44 m s
26. The elastic PE is given by PEelastic  12 kx 2 where x is the distance of stretching or compressing of the spring
from its natural length.
x
2 PEelastic
k

2  25 J 
440 N m
 0.34 m
33. The only forces acting on Jane are gravity and the vine tension. The tension
pulls in a centripetal direction, and so can do no work – the tension force is
perpendicular at all times to her motion. So Jane’s mechanical energy is
conserved. Subscript 1 represents Jane at the point where she grabs the vine, and
subscript 2 represents Jane at the highest point of her swing. The ground is the
location for PE  y  0  . We have v1  5.3m s , y1  0 , and v2  0 (top of
swing). Solve for y2, the height of her swing.
1
mv12  mgy1  12 mv22  mgy2  12 mv12  0  0  mgy2 
2
y2 
v12
2g

 5.3 m s 

zero
v2 , y2
v1 , y1
2
2 9.8 m s 2

 1.4 m
No, the length of the vine does not enter into the calculation, unless the vine is less than 0.7 m long. If that
were the case, she could not rise 1.4 m high. Instead she would wrap the vine around the tree branch.
51. (a) Calculate the energy of the ball at the two maximum heights, and subtract to find the amount of
energy “lost”. The energy at the two heights is all gravitational PE, since the ball has no KE at those
maximum heights.
Elost  Einitial  Efinal  mgyinitial  mgyfinal
Elost

mgyinitial  mgyfinal
yinitial  yfinal

2.0 m  1.5 m
 0.25  25%
Einitial
mgyinitial
yinitial
2.0 m
(b) Due to energy conservation, the KE of the ball just as it leaves the ground is equal to its final PE.
2
PEfinal  KEground  mgyfinal  12 mvground




vground  2gyfinal  2 9.8 m s 2 1.5 m   5.4 m s
(c) The energy “lost” was changed primarily into heat energy – the temperature of the ball and the ground
would have increased slightly after the bounce. Some of the energy may have been changed into
acoustic energy (sound waves). Some may have been lost due to non-elastic deformation of the ball or
ground.
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CHAPTER 7: Linear Momentum
Answers to Questions
1.
For momentum to be conserved, the system under analysis must be “closed” – not have any forces on it
from outside the system. A coasting car has air friction and road friction on it, for example, which are
“outside” forces and thus reduce the momentum of the car. If the ground and the air were considered part
of the system, and their velocities analyzed, then the momentum of the entire system would be conserved,
but not necessarily the momentum of any single component, like the car.
4.
If the rich man would have faced away from the shore and thrown the bag of coins directly away from the
shore, he would have acquired a velocity towards the shore by conservation of momentum. Since the ice is
frictionless, he would slide all the way to the shore.
5.
When a rocket expels gas in a given direction, it puts a force on that gas. The momentum of the gas-rocket
system stays constant, and so if the gas is pushed to the left, the rocket will be pushed to the right due to
Newton’s 3rd law. So the rocket must carry some kind of material to be ejected (usually exhaust from some
kind of engine) to change direction.
6.
The air bag greatly increases the amount of time over which the stopping force acts on the driver. If a hard
object like a steering wheel or windshield is what stops the driver, then a large force is exerted over a very
short time. If a soft object like an air bag stops the driver, then a much smaller force is exerted over a much
longer time. For instance, if the air bag is able to increase the time of stopping by a factor of 10, then the
average force on the person will be decreased by a factor of 10. This greatly reduces the possibility of
serious injury or death.
7.
“Crumple zones” are similar to air bags in that they increase the time of interaction during a collision, and
therefore lower the average force required for the change in momentum that the car undergoes in the
collision.
12. The turbine blades should be designed so that the water rebounds. If the water rebounds, that means that a
larger momentum change for the water has occurred than if it just came to a stop. And if there is a larger
momentum change for the water, there will also be a larger momentum change for the blades, making them
spin faster.
Solutions to Problems
p  mv   0.028 kg  8.4 m s   0.24 kg m s
1.
2.
From Newton’s second law, p  Ft . For a constant mass object, p  mv . Equate the two
expressions for p .
Ft  mv  v 
Ft
.
m
If the skier moves to the right, then the speed will decrease, because the friction force is to the left.
F t
 25 N  20 s 
v  

 7.7 m s
m
65 kg
The skier loses 7.7 m s of speed.
3.
Choose the direction from the batter to the pitcher to be the positive direction. Calculate the average force
from the change in momentum of the ball.
p  F t  mv 
F m
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v
t
 52.0 m s  39.0 m
3.00  103 s

  0.145 kg  
s
3
  4.40 10 N, towards the pitcher

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15.
(a)
The impulse is the change in momentum. The direction of travel of the struck ball is the
positive direction.
p  mv   4.5  102 kg   45 m s  0   2.0 kg m s
(b)
The average force is the impulse divided by the interaction time.
p 2.0 kg m s
F

 5.8  10 2 N
t
3.5  10 3 s
20. (a) The impulse given the ball is the area under the F vs. t graph. Approximate the area as a
triangle of “height” 250 N, and “width” 0.01 sec.
p  12  250 N  0.01 s   1.25N s
(b)
The velocity can be found from the change in momentum. Call the positive direction the
direction of the ball’s travel after being served.
p
1.25N s
p  mv  m  v f  vi   v f  vi 
 0
 21m s
m
6.0  10-2 kg
23. Let A represent the 0.450-kg puck, and let B represent the 0.900-kg puck. The initial direction of puck A is
the positive direction. We have vA  3.00 m s and vB  0 . Use Eq. 7-7 to obtain a relationship between
the velocities.
vA  vB    vA  vB   vB  vA  vA
Substitute this relationship into the momentum conservation equation for the collision.
mA vA  mB vB  mA vA  mB vB  mA vA  mA vA  mB  vA  vA  
vA 
0.450 kg
 mA  mB 
vA 
 3.00 m s   1.00 m
1.350 kg
 mA  mB 
s  1.00 m s  west 
vB  vA  vA  3.00 m s  1.00 m s  2.00 m s  east 
35 Use conservation of momentum in one dimension. Call the direction of the sports car’s velocity the
positive x direction. Let A represent the sports car, and B represent the SUV. We have vB  0 and
vA  vB . Solve for vA .
pinitial  pfinal  mA vA  0   mA  mB  vA  vA 
mA  mB
vA
mA
The kinetic energy that the cars have immediately after the collision is lost due to negative work done by
friction. The work done by friction can also be calculated using the definition of work. We assume the
cars are on a level surface, so that the normal force is equal to the weight. The distance the cars slide
forward is x . Equate the two expressions for the work done by friction, solve for vA , and use that to find
vA .
Wfr   KEfinal  KEinitial after
collision
 0  12  mA  mB  vA2
Wfr  Ffr x cos180o    k  mA  mB  g x
 12  mA  mB  vA2    k  mA  mB  g x  vA 
vA 
mA  mB
mA
vA 
mA  mB
mA
2  k g x 
2  k g x
920 kg  2300 kg
920 kg

2  0.80  9.8 m s 2
  2.8 m 
 23.191m s  23 m s
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