CHAPTER 7 IMPULSE AND MOMENTUM
CONCEPTUAL QUESTIONS
2.
REASONING AND SOLUTION Since linear momentum is a vector quantity, the
total linear momentum of any system is the resultant of the linear momenta of the
constituents. The people who are standing around have zero momentum. Those
who move randomly carry momentum randomly in all directions. Since there is such
a large number of people, there is, on average, just as much linear momentum in any
one direction as in any other. On average, the resultant of this random distribution is
zero. Therefore, the approximate linear momentum of the Times Square system is
zero.
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4.
REASONING AND SOLUTION
a. If a single object has kinetic energy, it must have a velocity; therefore, it must
have linear momentum as well.
b. In a system of two or more objects, the individual objects could have linear
momenta that cancel each other. In this case, the linear momentum of the system
would be zero. The kinetic energies of the objects, however, are scalar quantities
that are always positive; thus, the total kinetic energy of the system of objects would
necessarily be nonzero. Therefore, it is possible for a system of two or more objects
to have a total kinetic energy that is not zero but a total momentum that is zero.
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15. REASONING AND SOLUTION For a system comprised of the three balls, there
is no net external force. The forces that occur when the three balls collide are
internal forces. Therefore, the total linear momentum of the system is conserved.
Note, however, that the momentum of each ball is not conserved. The momentum of
any given ball will change as it interacts with the other balls; the momentum of each
ball will change in such a way as to conserve the momentum of the system. It is the
momentum of the system of balls, not the momentum of an individual ball, that is
conserved.
20. REASONING AND SOLUTION A sunbather is lying on a floating, stationary raft.
She then gets up and walks to one end of the raft. The sunbather and the raft are
considered as an isolated system.
a. As the sunbather walks to one end of the raft, she exerts a force on the raft;
however, the force is internal to the isolated system. Since there are no external
forces acting on the system, the linear momentum of the system cannot change.
Since the linear momentum of the system is initially zero, it must remain zero.
Therefore, the velocity of the center of mass of the system must be zero.
b. The sunbather has linear momentum as she walks to one end of the raft. Since the
linear momentum of the isolated system must remain zero, the raft must acquire a
linear momentum that is equal in magnitude and opposite in direction to that of the
sunbather. From the definition of linear momentum, p = mv, we know that the
direction of the linear momentum of an object is the same as the direction of the
velocity of the object. Thus, the raft acquires a velocity that is opposite to the
direction of motion of the sunbather.
PROBLEMS
4.
REASONING AND SOLUTION The magnitude of each arrow's momentum is
p  mv  (0.100 kg)(30.0 m/s )  3.00 kg m/s
The vector momenta of the arrows are perpendicular, so the magnitude of the total
momentum is given by the Pythagorean theorem
PT 
p12  p 22 
b3.00 kg  m / sg b3.00 kg  m / sg 
2
2
4 .24 kg  m / s
The angle of the total momentum vector as measured from due east is
F p I
H1 K
1
  tan 1 G
Gp 2 J
J tan e1.00 j  45.0 south of east
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5.
REASONING AND SOLUTION The impulse can be found from Equation
7.4, the impulse-momentum theorem:
SSM
:
F t  mv f  mv 0
Impulse
:
Final
momentum
:
Initial
momentum
 m ( v f  v 0 )  (0.35 kg) (–21 m / s) – (+4.0 m / s)  –8.7 kg  m / s
The minus sign indicates that the direction of the impulse is the same as that of the
final velocity.
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18.
REASONING Let m be Al’s mass, which means that Jo’s mass is 168 kg – m.
Since friction is negligible and since the downward-acting weight of each person is
balanced by the upward-acting normal force from the ice, the net external force
acting on the two-person system is zero. Therefore, the system is isolated, and the
conservation of linear momentum applies. The initial total momentum must be equal
to the final total momentum.
SOLUTION
Applying the principle of conservation of linear momentum and
assuming that the direction in which Al moves is the positive direction, we find
b gb
gb g b
gb
gb
g
m 0 m / s  168 kg  m 0 m / s  m 0.90 m / s  168 kg  m 1.2 m / s

 
Initial total momentum
Final total momentum
Solving this equation for m and suppressing the units in the interests of clarity, we
find
b gb gb g b g
b168gb1.2 g 96 kg
m
0  m 0.90  168 1.2  m 1.2
0.90  1.2
20. REASONING AND SOLUTION First, find the initial speed of Adolf by using the
conservation of mechanical energy, mA ghA  21 mA v A2 .
vA  2gh A  2(9.80 m / s 2 )( 0.65 m )  3.6 m / s
c
h
Now use the conservation of momentum mE v E   mA v A to find the velocity of
Ed, so that
m
120 kg
vE   A vA  
( 3.6 m / s )   5.5 m / s
mE
78 kg
The conservation of mechanical energy allows us to find the height reached by Ed,
2
v E2
5.5 m / s
hE 

 1.5 m
2 g 2 9 .80 m / s 2
b
c
g
h
26. REASONING Since friction is negligible and since the downward-acting weight of
each person is balanced by the upward-acting normal force from the sidewalk, the
net external force acting on the two-person system is zero. Therefore, the system is
isolated, and the conservation of linear momentum applies. The initial total
momentum must be equal to the final total momentum.
SOLUTION Applying the principle of conservation of linear momentum and
assuming that the direction in which Kevin is moving is the positive direction, we
find
87 kg g
v
b
22 kg g
87 kg  22 kg g
b
b0 m /sg b
b2.4 m / sg



Kevin
Final total momentum
Initial total momentum
v Kevin 
b87 kg  22 kg gb2.4 m / sg
87 kg
3.0 m / s
32. REASONING The net external force acting on the two-puck system is zero (the
weight of each ball is balanced by an upward normal force, and we are ignoring
friction due to the layer of air on the hockey table). Therefore, the two pucks
constitute an isolated system, and the principle of conservation of linear momentum
applies.
SOLUTION Conservation of linear momentum requires that the total momentum is
the same before and after the collision. Since linear momentum is a vector, the x
and y components must be conserved separately. Using the drawing in the text,
momentum conservation in the x direction yields
mAv0A  mAvfA (cos 65) + mBvfB (cos 37)
(1)
while momentum conservation in the y direction yields
0  mAvfA (sin 65) – mBvfB (sin 37)
(2)
Solving equation (2) for vfB, we find that
vfB 
mA vfA (sin 65)
mB(sin 37)
(3)
Substituting equation (3) into Equation (1) leads to
m v (sin 65) 
mA v0A  mAv fA (cos 65) +  A fA
(cos 37)
sin 37
a. Solving for vfA gives
vfA 
v0A
5.5 m/s

 3.4 m/s
sin 65 
sin 65 


cos 65 +
cos 65 +
tan 37 
tan 37
b. From equation (3), we find that
vfB 
(0.025 kg) (3.4 m/s) (sin 65 )
 2.6 m/s
(0.050 kg) (sin 37)
36. REASONING AND SOLUTION
Momentum is conserved in the horizontal
direction during the "collision". Let the coal be object 1 and the car be object 2.
Then
(m1  m2 )v f  m1v 1 cos25.0  m2 v2
vf 
m1 v1 cos 25.0  m2 v 2
m1  m2

(150 kg)(0.80 m/s)cos 25.0  (440 kg)(0.50 m/s)
 0.56 m/s
150 kg + 440 kg
The direction of the final velocity is to the right .