kuruvila (lk5992) – HW 8 – opyrchal – (11113)
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) 70
.
8 kg astronaut becomes separated
from the shuttle, while on a space walk. She
finds herself 31
.
8 m away from the shuttle
and moving with zero speed relative to the
shuttle. She has a(n) 0
.
556 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: 5
.
62411 min.
Explanation:
Let :
M
= 70
.
8 kg
,
m
= 0
.
556 kg
,
d
= 31
.
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
.
556 kg) (12 m
/
s)
70
.
8 kg
= 0
.
0942373 m
/
s
.
And the time it takes for her to reach the
shuttle
t
=
d
V
=
31
.
8 m
0
.
0942373 m
/
s
·
1 min
60 s
=
5
.
62411 min
.
002
10.0 points
A child bounces a 49 g superball on the side
walk. The velocity change of the superball is
from 28 m
/
s downward to 17 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: 1764 N.
Explanation:
Let :
m
= 49 g = 0
.
049 kg
,
v
u
= 17 m
/
s
,
v
d
= 28 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
=
(49 g) (17 m
/
s + 28 m
/
s)
0
.
00125 s
=
1764 N
.
003 (part 1 of 2) 10.0 points
A(n) 20 g object moving to the right at
24 cm
/
s overtakes and collides elastically with
a 33 g object moving in the same direction at
11 cm
/
s.
Find the velocity of the slower object after
the collision.
Correct answer: 7
.
81132 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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kuruvila (lk5992) – HW 8 – opyrchal – (11113)
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
(20 g) + (33 g)
×
bracketleftBig
(20 g)(24 cm
/
s)
+ (33 g)(11 cm
/
s)

(33 g)
{
(24 cm
/
s)

(11 cm
/
s)
}
bracketrightBig
= 7
.
81132 cm
/
s
.
004 (part 2 of 2) 10.0 points
Find the velocity of faster object after the
collision.
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 Fall '08
 moro
 Momentum, vy, 6.3 cm, 4.99 m, 0.0942373 m/s, 1.22848 m/s

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