Download Physics 107 HOMEWORK ASSIGNMENT #7

Physics 107 HOMEWORK ASSIGNMENT #7
Cutnell & Johnson, 7th edition
Chapter 7: Problems 10, 16, 19, 34, 39
*10 A student (m=63 kg) falls freely from rest and strikes the ground. During the collision with
the ground, he comes to rest in a time of 0.010 s. The average force exerted on him by the ground
is + 18 000 N, where the upward direction is taken to be the positive direction. From what height
did the student fall? Assume that the only force acting on him during the collision is that due to
the ground.
16 A 55-kg swimmer is standing on a stationary 210-kg floating raft. The swimmer then runs
off the raft horizontally with a velocity of +4.6 m/s relative to the shore. Find the recoil velocity
that the raft would have if there were no friction and resistance due to the water.
*19 To view an interactive solution to a problem that is very similar to
this one, go to and select Interactive Solution 7.19 A fireworks rocket
is moving at a speed of 45.0 m/s. The rocket suddenly breaks into two
pieces of equal mass, which fly off with velocities v1 and v2, as shown
in the drawing. What is the magnitude of (a) v1, and (b) v2 ?
*34 A mine car (mass= 440 kg) rolls at a speed of 0.50 m/s on a
horizontal track, as the drawing shows. A 150-kg chunk of coal has a
speed of 0.80 m/s when it leaves the chute. Determine the velocity of
the car/coal system after the coal has come to rest in the car.
Problem
19
**39 Starting with an initial speed of 5.00 m/s at a height of 0.300 m, a
1.50-kg ball swings downward and strikes a 4.60-kg ball that is at rest, as the drawing shows. (a)
Using the principle of conservation of mechanical energy, find the speed of the 1.50-kg ball just
before impact. (b) Assuming that the collision is elastic, find the velocities (magnitude and
direction) of both balls just after the collision. (c) How high does each ball swing after the
collision, ignoring air resistance?
Problem
34
Problem
39
10. REASONING This solution can be divided into two parts. First, the student falls freely
from rest and attains a certain velocity (which is unknown) just before hitting the ground.
This impact velocity depends on the height H from which the student falls, and we will use
an equation of kinematics from Chapter 3 to relate the height to the impact velocity. Second,
the student then collides with the ground and comes to rest. We will employ the impulsemomentum theorem to determine the impact velocity from a knowledge of the average
impact force and the time of impact, both of which are known.
SOLUTION Since the student falls freely from rest, we know that v0y = 0 m/s, and ay =
2
−9.8 m/s . The acceleration is negative, because the downward direction is taken to be the
negative direction. The displacement y of the student is y = −H, where H is the height; the
minus sign indicates that the student falls downward. These variables are related by
Equation 3.6b from the equations of kinematics as
2
v 2y = v02y + 2a y y = ( 0 m/s ) + 2a y ( − H )
where vy is the velocity of the student just before hitting the ground. Solving this expression
for the height yields
−v 2y
H=
(1)
2a y
At this point we do not know the impact velocity vy. However, it can be determined by
examining the collision of the student with the ground. Just before the collision, the velocity
of the student is vy. Since the student comes to rest, the final velocity is vf = 0 m/s. The
average force exerted on the student by the ground is F = +18 000 N, and the time of
collision is ∆t = 0.010 s. The impulse-momentum theorem, Equation 7.4, relates these
variables:
F ∆t = mvf − mv0 = m ( 0 m/s ) − mv0 = − mv0
(2)
Now v0 is the impact velocity of the student, which was labeled as vy in Equation (1);
therefore, v0 = vy. Thus, Equation (2) becomes F ∆t = − mv y . Solving this equation for vy
and substituting the result into Equation (1) yields
2
 − F ∆t 
−
2
2
2

− F 2 ( ∆t )
− ( +18 000 N ) ( 0.010 s )
 m 
H=
=
=
= 0.42 m
2
2a y
2a y m 2
2 −9.80 m/s 2 ( 63 kg )
(
)
16. REASONING The sum of the external forces acting on the swimmer/raft system is zero,
because the weight of the swimmer and raft is balanced by a corresponding normal force and
friction is negligible. The swimmer and raft constitute an isolated system, so the principle of
conservation of linear momentum applies. We will use this principle to find the recoil
velocity of the raft.
SOLUTION As the swimmer runs off the raft, the total linear momentum of the
swimmer/raft system is conserved:
ms vs + mr vr =
0
Total
momentum
Total momentum
after swimmer
runs off raft
before swimmer
starts running
where ms and vs are the mass and final velocity of the swimmer, and mr and vr are the mass
and final velocity of the raft. Solving for vr gives
vr = −
ms vs
mr
=−
( 55 kg )( +4.6 m/s )
210 kg
= −1.2 m/s
19. REASONING During the breakup¸ the linear momentum of the system is conserved, since
the force causing the breakup is an internal force. We will assume that the +x axis is along
the original line of motion (before the breakup), and the +y axis is perpendicular to this line
and points upward. We will apply the conservation of linear momentum twice, once for the
momentum components along the x axis and again for the momentum components along the
y axis.
SOLUTION The mass of each piece of the rocket after breakup is m, and so the mass of the
rocket before breakup is 2m. Applying the conservation of momentum theorem along the
original line of motion (the x axis) gives
mv1 cos 30.0° + mv2 cos 60.0° = 2mv0
or
v1 cos 30.0° + v2 cos 60.0° = 2v0
(1)
P0, x
Pf ,x
Applying the conservation of momentum along the y axis gives
mv1 sin 30.0° − mv2 sin 60.0° = 0
Pf , y
P0, y
or
v2 =
v1 sin 30.0°
sin 60.0°
(2)
a. To find the speed v1 of the first piece, we substitute the value for v2 from Equation (2)
into Equation (1). The result is
 v sin 30.0° 
v1 cos 30.0° +  1
 cos 60.0° = 2v0
 sin 60.0° 
Solving for v1 and setting v0 = 45.0 m/s yields
v1 =
2v0
 sin 30.0° 
cos 30.0° + 
 cos 60.0°
 sin 60.0° 
=
2 ( 45.0 m/s )
 sin 30.0° 
cos 30.0° + 
 cos 60.0°
 sin 60.0° 
= 77.9 m/s
b. The speed v2 of the second piece can be found by substituting v1 = 77.9 m/s into
Equation (2):
v sin 30.0° ( 77.9 m/s ) sin 30.0°
v2 = 1
=
= 45.0 m/s
sin 60.0°
sin 60.0°
34. 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 )vf = m1v1 cos 25.0° + m2v2
vf =
m1v1 cos 25.0° + m2v2
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 .
39. SSM REASONING The two balls constitute the system. The tension in the wire is the
only nonconservative force that acts on the ball. The tension does no work since it is
perpendicular to the displacement of the ball. Since Wnc = 0 J, the principle of conservation
of mechanical energy holds and can be used to find the speed of the 1.50-kg ball just before
the collision. Momentum is conserved during the collision, so the principle of conservation
of momentum can be used to find the velocities of both balls just after the collision. Once
the collision has occurred, energy conservation can be used to determine how high each ball
rises.
SOLUTION
a. Applying the principle of energy conservation to the 1.50-kg ball, we have
1
mv f2 + mgh f = 21 mv 02 + mgh0
2
Ef
E0
If we measure the heights from the lowest point in the swing, hf = 0 m, and the expression
above simplifies to
1
2
mv f2 = 21 mv 02 + mgh0
Solving for vf , we have
v f = v 02 + 2 gh0 = ( 5.00 m / s) 2 + 2(9.80 m / s 2 )(0.300 m) = 5.56 m / s
b. If we assume that the collision is elastic, then the velocities of both balls just after the
collision can be obtained from Equations 7.8a and 7.8b:
v f1 =
F m −m Iv
GH m + m JK
1
1
2
01
and
v f2 =
2
F 2m I v
GH m + m JK
1
1
01
2
Since v01 corresponds to the speed of the 1.50-kg ball just before the collision, it is equal to
the quantity vf calculated in part (a). With the given values of m1 = 1.50 kg and
m2 = 4 .60 kg , and the value of v 01 = 5.56 m / s obtained in part (a), Equations 7.8a and
7.8b yield the following values:
v f1 = –2.83 m / s
and
v f2 = +2.73 m / s
The minus sign in vf1 indicates that the first ball reverses its direction as a result of the
collision.
c. If we apply the conservation of mechanical energy to either ball after the collision we
have
1
mv f2 + mgh f = 21 mv 02 + mgh0
2
Ef
E0
where v0 is the speed of the ball just after the collision, and hf is the final height to which the
ball rises. For either ball, h0 = 0 m, and when either ball has reached its maximum height,
vf = 0 m/s. Therefore, the expression of energy conservation reduces to
gh f = 21 v 02
or
hf =
v 02
2g
Thus, the heights to which each ball rises after the collision are
1.50 - kg ball
4.60 - kg ball
v 02
(2.83 m / s) 2
hf =
=
= 0.409 m
2 g 2 (9.80 m / s 2 )
v 02
(2.73 m / s) 2
hf =
=
= 0.380 m
2 g 2 (9.80 m / s 2 )