Physics 102
Spring 2008
HW 1 Key
Giancoli Chapters 6 and 7
Chapter 6 Questions
12. The spring can leave the table if it is compressed enough. If the spring is compressed an
amount x0 , then the gain in elastic PE is 12 kx02 . As the spring is compressed, its center of
mass is lowered by some amount. If the spring is uniform, then the center of mass is
lowered by x0 2 , and the amount of decrease in gravitational PE is 12 mgx0 . If the gain in
elastic PE is more than the loss in gravitational PE, so that 12 kx02  12 mgx0 or x0  mg k , then
the released spring should rise up off of the table, because there is more than enough elastic
PE to restore the spring to its original position. That extra elastic energy will enable the
spring to “jump” off the table – it can raise its center of mass to a higher point and thus rise
up off the table. Where does that “extra” energy come from? From the work you did in
compressing the spring.
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”.
17.
(a)If there is no friction to dissipate any of the energy, then the gravitational PE that the
child has
at the top of the hill all turns into kinetic energy at the bottom of the hill. The same
kinetic energy will be present regardless of the slope – the final speed is completely
determined by the height. The time it takes to reach the bottom of the hill will be longer
for a smaller slope.
(b)If there is friction, then the longer the path is, the more work that friction will do, and so
the
slower the speed will be at the bottom. So for a steep hill, the sled will have a greater
speed at the bottom than for a shallow hill.
21. The superball cannot rebound to a height greater than its original height when dropped. If it
did, it would violate conservation of energy. When a ball collides with the floor, the KE of
the ball is converted into elastic PE by deforming the ball, much like compressing a spring.
Then as the ball springs back to its original shape, that elastic PE is converted to back to KE.
But that process is “lossy” – not all of the elastic PE gets converted back to KE. Some of
the PE is lost, primarily to friction. The superball rebounds higher than many other balls
because it is less “lossy” in its rebound than many other materials.
24. The climber does the same amount of work whether climbing straight up or via a zigzag
path, ignoring dissipative forces. But if a longer zigzag path is taken, it takes more time to
do the work, and so the power output needed from the climber is less. That will make the
climb easier. It is easier for the human body to generate a small amount of power for long
periods of time rather than to generate a large power for a small period of time.
Chapter 6 Problems:
35. The forces on the sled are gravity and the normal force. The normal
force is perpendicular to the direction of motion, and so does no work.
Thus the sled’s mechanical energy is conserved. Subscript 1 represents
the sled at the bottom of the hill, and subscript 2 represents the sled at the
top of the hill. The ground is the zero location for PE  y  0  . We have
FN
mg 
y1  0 , v2  0 , and y2  1.35 m . Solve for v1, the speed at the bottom.
1
2
mv12  mgy1  12 mv22  mgy2 

1
2
mv12  0  0  mgy2 

v1  2 gy2  2 9.80 m s 2 1.35 m   5.14 m s
Notice that the angle is not used in the calculation.
37. (a)Since there are no dissipative forces present, the mechanical energy of the person –
trampoline –
Earth combination will be conserved. The level of the unstretched trampoline is the
zero level for both the elastic and gravitational PE. Call up the positive direction.
Subscript 1 represents the jumper at the top of the jump, and subscript 2 represents the
jumper upon arriving at the trampoline. There is no elastic PE involved in this part of
the problem. We have v1  5.0 m s , y1  3.0 m , and y2  0 . Solve for v2, the speed upon
arriving at the trampoline.
E1  E2 
1
2
mv12  mgy1  12 mv22  mgy2 
v2   v12  2 gy1  
 5.0 m s 2  2  9.8 m
s2
1
2
mv12  mgy1  12 mv22  0 
  3.0 m   9.154 m s  9.2 m s
The speed is the absolute value of v2 .
(b) Now let subscript 3 represent the jumper at the maximum stretch of the trampoline. We
have v2  9.154 m s , y2  0 , x2  0 , v3  0 , and x3  y3 . There is no elastic energy at
position 2, but there is elastic energy at position 3. Also, the gravitational PE at
position 3 is negative, and so y3  0 . A quadratic relationship results from the
conservation of energy condition.
E2  E3 
1
2
1
2
mv22  mgy2  12 kx22  12 mv32  mgy3  12 kx32 
mv22  0  0  0  mgy3  12 ky32 
y3 

 mg  m 2 g 2  4  12 k   12 mv22
2  12 k 
1
2

ky32  mgy3  12 mv22  0 

 mg  m 2 g 2  kmv22
k



  65 kg  9.8 m s 2 
 65 kg 2  9.8 m
   6.2 10
 6.2 10 N m 
2
s2
4

N m  65 kg  9.154 m s 
2
4
 0.307 m , 0.286 m
Since y3  0 , y3  0.31 m .
The first term under the quadratic is about 1000 times smaller than the second term,
indicating that the problem could have been approximated by not even including
gravitational PE for the final position. If that approximation would have been made, the
result would have been found by taking the negative result from the following solution.
E2  E3 
1
2
mv22  12 ky32  y3  v2
m
k
  9.2 m s 
65 kg
6.2 104 N m
42. Consider this diagram for the jumper’s fall.
(a)
The mechanical energy of the jumper is
conserved. Use y
for the distance from the 0 of gravitational PE and x
for the amount of bungee cord “stretch” from its
unstretched length. Subscript 1 represents the jumper
at the start of the fall, and subscript 2 represents the
jumper at the lowest point of the fall. The bottom of
the fall is the zero location for gravitational PE
 y  0  , and the location where the bungee cord just
 0.30 m
Start of fall
12 m
Contact with bungee
cord, 0 for elastic PE
19 m
Bottom of fall, 0 for
gravitational PE
starts to be stretched is the zero location for elastic PE  x  0  . We have v1  0 ,
y1  31 m , x1  0 , v2  0 , y2  0 , and x2  19 m . Apply conservation of energy.
E1  E2 
k
2mgy1
2
2
x

1
2
mv12  mgy1  12 kx12  12 mv22  mgy2  12 kx22  mgy1  12 kx22 

2  62 kg  9.8 m s 2
19 m 
2
  31 m   104.4 N m  1.0 10
2
N m
(b) The maximum acceleration occurs at the location of the maximum force,
which
occurs when the bungee cord has its maximum stretch, at the bottom of the
fall. Write Newton’s 2nd law for the force on the jumper, with upward as
positive.
Fcord
mg
Fnet  Fcord  mg  kx2  mg  ma 
a
kx2
m
g 
104.4 N m 19 m 
 9.8 m
 62 kg 
s 2  22.2 m s 2  22 m s 2
53. Since there is a non-conservative force, consider energy conservation with non-conservative
work included. Subscript 1 represents the roller coaster at point 1, and subscript 2 represents
the roller coaster at point 2. Point 2 is taken as the zero location for gravitational PE. We
have v1  1.70 m s , y1  35 m , and y2  0 . Solve for v2 . The work done by the non-
conservative friction force is given by WNC  Ffr d cos180o  0.20mgd , since the force is onefifth of mg, and the force is directed exactly opposite to the direction of motion.
WNC  E1  E2   0.2mgd  12 mv12  mgy1  12 mv22  mgy2 

v2  0.4 gd  v12  2 gy1  0.4 9.80 m s 2
  45.0 m   1.70 m s 
2

 2 9.80 m s 2
  35 m 
 22.64 m s  23 m s
54. Consider the free-body diagram for the skier in the midst of the
motion.
Write Newton’s 2nd law for the direction perpendicular to the
plane, with an acceleration of 0.
 F  FN  mg cos   0  FN  mg cos  
Ffr  k FN  k mg cos 
d
FN
Ffr
 mg

Apply conservation of energy to the skier, including the nonconservative friction force. Subscript 1 represents the skier at the bottom of the slope, and
subscript 2 represents the skier at the point furthest up the slope. The location of the skier at
the bottom of the incline is the zero location for gravitational PE  y  0  . We have
v1  12.0 m s , y1  0 , v2  0 , and y2  d sin  .
WNC  E1  E2  Ffr d cos180o  12 mv12  mgy1  12 mv22  mgy2 
  k mgd cos   12 mv12  0  0  mgd sin  
k 
1
2
v12  gd sin 
gd cos 
12.0 m s 

 tan  
 tan18.0o
2 gd cos 
2  9.80 m s   12.2 m  cos18.0o
v12
2
 0.308
62.
(a)
(b)
 1000 W  3600 s  1 J/s 
6


  3.6  10 J
 1 kW  1 h  1 W 
1 kW  30 d  24 h 
 520 W 1 month    520 W 1 month  


  374 kW h
 1000 W  1 month  1 d 
1 kW h  1 kW h 
 370 kW h
 3.6  106 J 
 1.3  109 J

 1 kW h 
$0.12 
(d)
 374 kW h  
  $44.88  $45
 1 kW h 
Kilowatt-hours is a measure of energy, not power, and so no, the actual rate at which the
energy is used does not figure into the bill. They could use the energy at a constant rate, or
at a widely varying rate, and as long as the total used is 370 kilowatt-hours, the price would
be $45.
(c)
374 kW h  374 kW h 
63.
The energy transfer from the engine must replace the lost kinetic energy. From the two
speeds,
calculate the average rate of loss in kinetic energy while in neutral.
 1m s 
 1m s 
v1  85 km h 
 23.61m s
v2  65 km h 

  18.06 m s
 3.6 km h 
 3.6 km h 
KE  12 mv22  12 mv12 
P
W
t

1.330  105 J
6.0 s
1
2
1150 kg  18.06 m s    23.61m s    1.330  105 J
2

 2.216  104 W , or 2.216  10 4 W
2
1 hp
 29.71 hp
 746
W
So 2.2  10 W or 3.0 101 hp is needed from the engine.
4
Chapter 7 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.
2.
Consider this problem as a very light object hitting and sticking to a very heavy object. The
large object – small object combination (Earth + jumper) would have some momentum after
the collision, but due to the very large mass of the Earth, the velocity of the combination is
so small that it is not measurable. Thus the jumper lands on the Earth, and nothing more
happens.
3.
When you release an inflated but untied balloon at rest, the gas inside the balloon (at high
pressure) rushes out the open end of the balloon. That escaping gas and the balloon form a
closed system, and so the momentum of the system is conserved. The balloon and
remaining gas acquires a momentum equal and opposite to the momentum of the escaping
gas, and so move in the opposite direction to the escaping gas.
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.
8.
From Eq. 7-7 for a 1-D elastic collision, vA  vB  vB  vA . Let “A” represent the bat, and let
“B” represent the ball. The positive direction will be the (assumed horizontal) direction that
the bat is moving when the ball is hit. We assume the batter can swing the bat with equal
strength in either case, so that vA is the same in both pitching situations. Because the bat is
so much heavier than the ball, we assume that vA  vA – the speed of the bat doesn’t change
significantly during the collision.
Then the velocity of the baseball after being hit is vB  vA  vA  vB  2vA  vB . If vB  0 , the
ball tossed up into the air by the batter, then vB  2vA – the ball moves away with twice the
speed of the bat. But if vB  0 , the pitched ball situation, we see that the magnitude of
vB  2vA , and so the ball moves away with greater speed. If, for example, the pitching speed
of the ball was about twice the speed at which the batter could swing the bat, then we would
have vB  4vA . Thus the ball has greater speed after being struck, and thus it is easier to hit a
home run. This is similar to the “gravitational slingshot” effect discussed in problem 85.
11. Consider two objects, each with the same magnitude of momentum, moving in opposite
directions. They have a total momentum of 0. If they collide and have a totally inelastic
collision, in which they stick together, then their final common speed must be 0 so that
momentum is conserved. But since they are not moving after the collision, they have no
kinetic energy, and so all of their kinetic energy has been lost.
Problems from Chapter 7:
4.
The throwing of the package is a momentum-conserving action, if the water resistance is
ignored. Let “A” represent the boat and child together, and let “B” represent the package.
Choose the direction that the package is thrown as the positive direction. Apply
conservation of momentum, with the initial velocity of both objects being 0.
 mA  mB  v  mAvA  mB vB 
 6.40 kg 10.0 m s 

 0.901m s
 26.0 kg  45.0 kg 
pinitial  pfinal 
vA  
8.
mB vB
mA
The boat and child move in the opposite direction as the thrown package.
Consider the motion in one dimension, with the positive direction being the direction of
motion of the first car. Let “A” represent the first car, and “B” represent the second car.
Momentum will be conserved in the collision. Note that vB  0 .
pinitial  pfinal  mAv A  mB vB   mA  mB  v 
mB 
mA  v A  v   9300 kg 15.0 m s  6.0 m s 

 1.4  104 kg
v
6.0 m s
17. The impulse given the ball is the change in the ball’s momentum. From the symmetry of the
problem, the vertical momentum of the ball does not change, and so there is no vertical
impulse. Call the direction AWAY from the wall the positive direction for momentum
perpendicular to the wall.
p  mv  mv
final



 m v sin 45o  v sin 45o  2mv sin 45o
initial

 2 6.0  102 km  25 m s  sin 45o  2.1kg m s , to the left
18. (a)The average force on the car is the impulse (change in momentum) divided by the time of
interaction. The positive direction is the direction of the car’s initial velocity.

 1m s  
 0  50 km h  3.6 km h  
p mv

   1.389  105 N  1.4  105 N
F

 1500 kg  


t
t
0.15 s




(b) The deceleration is found from Newton’s 2nd law.
F  ma  a 
F
m
 1.389 10 N   93m s

5
1500 kg
2