Physics 102
Spring 2008
Homework 2 Solutions
Giancoli Chapter 7
Questions:
13. (a)
The downward component of the momentum is unchanged. The horizontal component of
momentum changes from rightward to leftward. Thus the change in momentum is to the left in
the picture.
(b) Since the force on the wall is opposite that on the ball, the force on the wall is to the right.
14. (a) The momentum of the ball is not conserved during any part of the process, because there is an
external force acting on the ball at all times — the force of gravity. And there is an upward
force on the ball during the collision. So considering the ball as the system, there are always
external forces on it, and so its momentum is not conserved.
(b) With this definition of the system, all of the forces are internal, and so the momentum of the
Earth-ball system is conserved during the entire process.
(c) The answer here is the same as for part (b).
15. In order to maintain balance, your CM must be located directly above your feet. If you have a heavy
load in your arms, your CM will be out in front of your body and not above your feet. So you lean
backwards to get your CM directly above your feet. Otherwise, you would fall over forwards.
Problems:
22. Let A represent the 0.440-kg ball, and B represent the 0.220-kg ball. We have vA  3.30 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  mBvB  mA vA  mBvB
vA 

mA vA  mA vA  mB vA  vA  
mA  mB v  0.220 kg 3.30 m s 


mA  mB  A 0.660 kg
1.10 m s east 
vB  vA  vA  3.30 m s  1.10 m s  4.40 m s east 
27. Let the original direction of the cars be the positive direction. We have vA  4.50 m s and
vB  3.70 m s
(a)
Use Eq. 7-7 to obtain a relationship between the velocities.
vA  vB   vA  vB  
vB  vA  vB  vA  0.80 m s  vA
Substitute this relationship into the momentum conservation equation for the collision.
mA vA  mBvB  mA vA  mBvB
vA 

mA vA  mB vB  0.80 m s  450 kg 4.50 m s  550 kg 2.90 m s 

 3.62 m s
mA  mB
1000 kg
vB  0.80 m s  vA  4.42 m s
(b)
mA vA  mBvB  mA vA  mB 0.80 m s  vA  
Calculate p  p  p for each car.
pA  mA vA  mA vA  450 kg 3.62 m s  4.50 m s   3.96  10 2 kg  m s
 4.0  10 2 kg  m s
pB  mBvB  mBvB  550 kg 4.42 m s  3.70 m s   3.96  10 2 kg  m s
 4.0  10 2 kg  m s
The two changes are equal and opposite because momentum was conserved.
34. Use conservation of momentum in one dimension, since the particles will separate and travel in
opposite directions. Call the direction of the heavier particle’s motion the positive direction. Let A
represent the heavier particle, and B represent the lighter particle. We have mA  1.5 mB , and
vA  vB  0.
pinitial  pfinal 
0  mA vA  mBvB

vA  
mBvB
  23 vB
mA
The negative sign indicates direction.
Since there was no mechanical energy before the explosion, the kinetic energy of the particles after
the explosion must equal the energy added.
Eadded  KEA  KEB  12 mA vA 2  12 mB vB 2 
KEB  53 Eadded 
3
5
7500 J   4500 J
1
2
1.5mB 23 vB 
2
 12 mB vB 2 
5
3
 m v 
1
2
2
B B
(b)
px : mA vA  mA vA cos A  mBvB cos B
 sin  A  mB vB sin  B
p y : 0  mA v A
Solve the x equation for cos  B and the y equation for sin  B and then
find the angle from the tangent function.
 sin  A
mA v A
 sin  A
sin  B
mB vB
vA
tan B 




mA vA  vA cos A  vA  vA
 cos A 
cos B
mB vB
 B  tan 1
vA sin  A
1.10 m s sin 30 .0º
 tan 1
 33 .0º
v A  vA cos A
1.80 m s  1.10 m s cos 30 .0º
With the value of the angle, solve the y equation for the velocity.
vB 
mA vA sin  A 0.400 kg 1.10 m s  sin 30.0º

 0.808 m s
mB sin  B
0.500 kg sin 33.0º
47. Choose the carbon atom as the origin of coordinates.
xCM 

KEB
KEA  Eadded  KEB  7500 J  4500 J  3000 J
Thus KEA  3.0  10 3 J KEB  4.5  10 3 J
42. (a)
5
3

10
mC xC  mO xO 12 u 0   16 u  1.13  10 m

 6.5  10 11 m from the C atom.
mC  mO
12 u  16 u
Giancoli Chapter 16
Questions:
3.
Water is a polar molecule – it has a positive region and a negative region. Thus it is easily attracted
to some other charged object, like an ion or electron in the air.
4.
The positively charged rod slightly polarizes the molecules in the paper. The negative charges in the
paper are slightly attracted to the part of the paper closest to the rod, while the positive charges in the
paper are slightly repelled from the part of the paper closest to the rod. Since the opposite charges
are now closer together and the like charges are now farther apart, there is a net attraction between
the rod and the paper.
 

 
8.
The force of gravity pulling down on the leaves, tending to return them to the vertical position.