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Phys 2101
Homework 6 Solution Spring `11
The numerical answers to the problems are calculated with the values used in the problems found
in the printed textbook; the values for your particular WebAssign assignment will be different.
1.P952. We think of this as having two parts: the first is the collision itself
–
where the bullet
passes through the block so quickly that the block has not had time to move through any distance
yet
–
and then the subsequent
―leap‖
of the block into the air (up to height
h
measured from its
initial position).
The first part involves momentum conservation (with +
y
upward):
0.01kg 1000m s
5.0kg
0.01kg
400m s
v
which yields
v
12
. m s
. The second part involves either the freefall equations from Ch. 2
(since we are ignoring air friction) or simple energy conservation from Ch. 8. Choosing the latter
approach, we have
2
2
1
1.2m s
9.8m s
2
h
which gives the result
h
= 0.073 m.
2. We think of this as having two parts: the first is the collision itself, where the blocks
―join‖
so
quickly that the 1.0kg block has not had time to move through any distance yet, and then the
subsequent motion of the 3.0 kg system as it compresses the spring to the maximum amount
x
m
.
The first part involves momentum conservation (with +
x
rightward):
m
1
v
1
= (
m
1
+
m
2
)
v
( .
( .
2 0
30
kg)(4.0 m s)
kg)
v
which yields
v
2 7
.
.
m s
The second part involves mechanical energy conservation:
1
2
1
2
( . kg) (2.7 m s)
(200 N m)
2
m
2
x
which gives the result
x
m
= 0.33 m.
3. (a) Let
m
1
be the mass of one sphere,
v
1
i
be its velocity before the collision, and
v
1
f
be its
velocity after the collision. Let
m
2
be the mass of the other sphere,
v
2
i
be its velocity before the
collision, and
v
2
f
be its velocity after the collision. Then, according to Eq. 975,
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m
m
m
m
v
m
m
m
v
f
i
i
1
1
2
1
2
1
2
1
2
2
2
.
Suppose sphere 1 is originally traveling in the positive direction and is at rest after the collision.
Sphere 2 is originally traveling in the negative direction. Replace
v
1
i
with
v
,
v
2
i
with
–
v
, and
v
1
f
with zero to obtain 0 =
m
1
–
3
m
2
. Thus,
21
/3
(300 g)/3 100 g
mm
.
(b) We use the velocities before the collision to compute the velocity of the center of mass:
1 1
2 2
com
12
300 g
2.00 m s
100 g
1.00 m/s.
300 g 100 g
ii
m v
m v
v
4. . First, we find the speed
v
of the ball of mass
m
1
right before the collision (just as it reaches its
lowest point of swing). Mechanical energy conservation (with
h
= 0.700 m) leads to
2
11
1
2
3.7 m s.
2
m gh
m v
v
gh
(a) We now treat the elastic collision using Eq. 967:
1
0.5 kg
2.5 kg
(3.7 m/s)
2.47 m/s
0.5 kg
2.5 kg
f
vv
which means the final speed of the ball is
2 47
.
.
m s
(b) Finally, we use Eq. 968 to find the final speed of the block:
1
2
2
2(0.5 kg)
(3.7 m/s)
1.23 m/s.
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 Spring '07
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