sindt (as53996) – HW 8 – opyrchal – (11101)
1
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printout
should
have
11
questions.
Multiplechoice questions may continue on
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before answering.
001
10.0 points
A(n) 59
.
3 kg astronaut becomes separated
from the shuttle, while on a space walk. She
finds herself 49
.
8 m away from the shuttle
and moving with zero speed relative to the
shuttle. She has a(n) 0
.
848 kg camera in her
hand and decides to get back to the shuttle
by throwing the camera at a speed of 12 m
/
s
in the direction away from the shuttle.
How long will it take for her to reach the
shuttle? Answer in minutes.
Correct answer: 4
.
83678 min.
Explanation:
Let :
M
= 59
.
3 kg
,
m
= 0
.
848 kg
,
d
= 49
.
8 m
,
and
v
= 12 m
/
s
.
Because of conservation of linear momentum,
we have
0 =
M V
+
m
(

v
)
m v
=
M V
where
V
is the velocity of the astronaut and
it has a direction toward the shuttle.
V
=
m v
M
=
(0
.
848 kg) (12 m
/
s)
59
.
3 kg
= 0
.
171602 m
/
s
.
And the time it takes for her to reach the
shuttle
t
=
d
V
=
49
.
8 m
0
.
171602 m
/
s
·
1 min
60 s
=
4
.
83678 min
.
002
10.0 points
A child bounces a 51 g superball on the side
walk. The velocity change of the superball is
from 26 m
/
s downward to 15 m
/
s upward.
If the contact time with the sidewalk is
1
800
s, what is the magnitude of the average force
exerted on the superball by the sidewalk?
Correct answer: 1672
.
8 N.
Explanation:
Let :
m
= 51 g = 0
.
051 kg
,
v
u
= 15 m
/
s
,
v
d
= 26 m
/
s
,
and
Δ
t
= 0
.
00125 s
Choose the upward direction as positive.
The impulse is
I
=
F
Δ
t
= Δ
P
=
m v
u

m
(

v
d
)
=
m
(
v
u
+
v
d
)
F
=
m
(
v
u
+
v
d
)
Δ
t
=
(51 g) (15 m
/
s + 26 m
/
s)
0
.
00125 s
=
1672
.
8 N
.
003 (part 1 of 2) 10.0 points
A(n) 15 g object moving to the right at
24 cm
/
s overtakes and collides elastically with
a 26 g object moving in the same direction at
13 cm
/
s.
Find the velocity of the slower object after
the collision.
Correct answer: 10
.
0488 cm
/
s.
Explanation:
Basic Concepts:
Momentum conserva
tion gives
m
1
v
1
i
+
m
2
v
2
i
=
m
1
v
1
f
+
m
2
v
2
f
.
Solution:
For headon elastic collisions, we
know that
v
1
i

v
2
i
=

(
v
1
f

v
2
f
)
.
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sindt (as53996) – HW 8 – opyrchal – (11101)
2
For the relative velocities, we have
v
1

v
2
=
v
2
f

v
1
f
v
2
f
=
v
1

v
2
+
v
1
f
.
Momentum is conserved, so
m
1
v
1
+
m
2
v
2
=
m
1
v
1
f
+
m
2
v
2
f
m
1
v
1
+
m
2
v
2
=
m
1
v
1
f
+
m
2
(
v
1

v
2
+
v
1
f
)
m
1
v
1
+
m
2
v
2

m
2
(
v
1

v
2
) =
v
1
f
(
m
1
+
m
2
)
v
1
f
=
m
1
v
1
+
m
2
v
2

m
2
{
v
1

v
2
}
m
1
+
m
2
=
1
(15 g) + (26 g)
×
bracketleftBig
(15 g)(24 cm
/
s)
+ (26 g)(13 cm
/
s)

(26 g)
{
(24 cm
/
s)

(13 cm
/
s)
}
bracketrightBig
= 10
.
0488 cm
/
s
.
004 (part 2 of 2) 10.0 points
Find the velocity of faster object after the
collision.
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 Spring '09
 WANG
 Physics, Momentum, vy, 4.11 m, 0.171602 m/s, 6.46 g, 13.6216 m/s

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