40. REASONING Considering the boat and the
stone as a single system, there is no net
external horizontal force acting on the stoneboat system, and thus the horizontal
component of the system’s linear momentum
is
conserved:
m1vf1x  m2vf2x  m1v01x  m2v02x (Equation
7.9a). We have taken east as the positive
direction.
vf1
v01x
12°
vf1x
15°
v01
Initial and final velocities of the stone
SOLUTION We will use the following symbols in solving the problem:
m1 = mass of the stone = 0.072 kg
vf1 = final speed of the stone = 11 m/s
v01 = initial speed of the stone = 13 m/s
m2 = mass of the boat
vf2 = final speed of the boat = 2.1 m/s
v02 = initial speed of the boat = 0 m/s
Because the boat is initially at rest, v02x = 0 m/s, and Equation 7.9a reduces to
m1vf1x  m2vf2x  m1v01x . Solving for the mass m2 of the boat, we obtain
m2vf2x  m1v01x  m1vf1x
or
m2 
m1  v01x  vf1x 
vf2x
As noted above, the boat’s final velocity is horizontal, so vf2x = vf2. The horizontal
components of the stone’s initial and final velocities are, respectively, v01x = v01 cos 15° and
vf1x = vf1 cos 12° (see the drawing). Thus, the mass of the boat must be
m2 
m1  v01x  vf1x 
vf2

 0.072 kg  13 m/s  cos15  11 m/s  cos12
2.1 m/s

  0.062 kg
41. SSM REASONING The two skaters constitute the system. Since the net external force
acting on the system is zero, the total linear momentum of the system is conserved. In the x
direction (the east/west direction), conservation of linear momentum gives Pf x  P0 x , or
( m1  m2 ) v f cos  m1v 01
Note that since the skaters hold onto each other, they move away
with a common velocity vf. In the y direction, Pf y  P0 y , or
( m1  m2 ) v f sin  m2 v 02
These equations can be solved simultaneously to obtain both the
angle  and the velocity vf.
SOLUTION
a. Division of the equations above gives
  tan 1
Fm v I  tan L(70.0 kg)(7.00 m / s) O 73.0
G
Hm v J
K M
N(50.0 kg)(3.00 m / s) P
Q
2
1
02
1 01
b. Solution of the first of the momentum equations gives
vf 
m1 v 01
( m1  m2 ) cos

(50.0 kg)(3.00 m / s)
 4.28 m / s
(50.0 kg  70.0 kg)(cos 73.0  )
42. REASONING
+355 m/s
a. The conservation of linear
Block 1
Block 2
momentum can be applied to this
three-object system (the bullet and
(a) Before collisions
the two blocks), even though the
second collision occurs later than
the first one. Since there is no
vblock 2
+0.550 m/s
friction between the blocks and the
horizontal surface, air resistance is
negligible, and the weight of each
mblock 2 = 1530 g
mblock 1 = 1150 g
block is balanced by a normal
mbullet = 4.00 g
force, the net external force acting
on this system is zero, and the
(b) After collisions
conservation of linear momentum
applies. This principle will allow us to determine the velocity of the second block after the
bullet imbeds itself.
b. The total kinetic energy of the three-body system is not conserved. Both collisions are
inelastic, and the collision with block 2 is completely inelastic since the bullet comes to rest
within the block. As with any inelastic collision, the total kinetic energy after the collisions
is less than that before the collisions.
SOLUTION
a. The conservation of linear momentum states that the total momentum of the system after
the collisions [see part (b) of the drawing] is equal to that before the collisions [part (a) of
the drawing]:
mblock 1vblock 1 +  mblock 2 + mbullet  vblock 2 
Total momentum after collisions
mbullet vbullet
Total momentum
before collisions
Solving for the velocity vblock 2 of block 2 after the collisions gives
vblock 2 
mbullet vbullet  mblock 1vblock 1
mblock 2 + mbullet
4.00  103 kg   355 m/s 


 1.150 kg  0.550 m/s 
1.530 kg + 4.00  103 kg
 0.513 m/s
b. The ratio of the total kinetic energy (KE) after the collision to that before the collision is
KEafter

KE before

1m
v2
2 block 1 block 1

1
2
2
 mblock 2 + mbullet  vblock
2
1m
v2
2 bullet bullet
1
2
1.530 kg + 4.00  103 kg  0.513 m/s2  1.49  103
1 4.00  10 3 kg 355 m/s 2


2
1.150 kg  0.550 m/s 2

1
2
EXAM Sample Questions:
3. A stunt person jumps from the roof of a tall building, but no injury occurs because the person
lands on a large, air-filled bag. Which one of the following best describes why no injury occurs?
A) The bag provides the necessary force to stop the person.
B) The bag reduces the impulse to the person.
C) The bag increases the amount of time the force acts on the person and reduces the change in
momentum.
D) The bag decreases the amount of time during which the momentum is changing and reduces
the average force on the person.
E) The bag increases the amount of time during which the momentum is changing and reduces
the average force on the person.
Ans: E
Difficulty: Hard
SectionDef: Section 7-1
7. A 0.2-kg steel ball is dropped straight down onto a hard, horizontal floor and bounces straight
up. The ball's speed just before and just after impact with the floor is 10 m/s. Determine the
magnitude of the impulse delivered to the floor by the steel ball.
A) zero N  s
B) 2 N  s
C) 4 N  s
D) 20 N  s
E) 200 N  s
Ans: C
Difficulty: Medium
SectionDef: Section 7-1
14. The head of a hammer (m = 1.5 kg) moving at 4.5 m/s strikes a nail and bounces back with
the same speed after an elastic collision lasting 0.075 s. What is the magnitude of the average
force the hammer exerts on the nail?
A) 6.8 N
B) 60 N
C) 90 N
D) 180 N
E) 240 N
Ans: D
Difficulty: Hard
SectionDef: Section 7-1
44. A 35-kg girl is standing near and to the left of a 43-kg boy on the frictionless surface of a
frozen pond. The boy throws a 0.75-kg ice ball to the girl with a horizontal speed of 6.2 m/s.
What are the velocities of the boy and the girl immediately after the girl catches the ice ball?
girl
boy
A) 0.81 m/s, left
0.67 m/s, right
B) 0.17 m/s, left
0.14 m/s, left
C) 0.18 m/s, right 0.13 m/s, left
D) 0.42 m/s, left
0.49 m/s, right
E) 0.13 m/s, left
0.11 m/s, right
Ans: E
Difficulty: Hard
SectionDef: Section 7-3 and 7-4