Version 086/ABBBC – Test #2 – Antoniewicz – (57420)
1
This printout should have 17 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
10.0 points
A satellite circles planet Roton every 8
.
1 h in
an orbit having a radius oF 1
.
9
×
10
7
m.
IF the radius oF Roton is 8
.
17
×
10
6
m, what
is the magnitude oF the FreeFall acceleration
on the surFace oF Roton?
1. 5.39001
2. 26.9821
3. 47.4022
4. 4.77091
5. 8.9794
6. 50.0457
7. 23.3898
8. 444.318
9. 1.82421
10. 6.34291
Correct answer: 4
.
77091 m
/
s
2
.
Explanation:
Basic Concepts:
Newton’s law oF gravi
tation
F
g
=
G
m
1
m
2
r
2
.
Kepler’s third law
T
2
=
p
4
π
2
G M
P
r
3
.
The FreeFall acceleration
a
on the surFace oF
the planet is the acceleration which a body in
Free Fall will Feel due to gravity
F
g
=
G
M m
R
2
=
m a ,
where
M
is the mass oF planet Roton. This
acceleration
a
is
a
=
G
M
R
2
,
(1)
the number which is
g
on Earth. Here, how
ever, the mass
M
is unknown, so we try to
fnd this From the inFormation given about the
satellite. Use Kepler’s third law For the period
oF the orbit
T
2
=
p
4
π
2
G M
P
r
3
.
(2)
By multiplying both sides with
R
2
and com
paring to equation (1), we can identiFy our
a
in the right hand side
T
2
R
2
=
p
4
π
2
a
P
r
3
.
IF we solve For
a
, we obtain
a
=
p
4
π
2
T
2
R
2
P
r
3
= 4
.
77091 m
/
s
2
which is our answer. Although identiFying
a
in this way is a “quick” way oF solving the
problem, we could just as well have calculated
the planet mass
M
explicitly From equation
(2) and inserted into equation (1).
002
10.0 points
A bucket Full oF water is rotated in a verti
cal circle oF radius 1
.
118 m (the approximate
length oF a person’s arm).
What must be the minimum speed oF the
pail at the top oF the circle iF no water is to
spill out?
1. 3.13675
2. 3.31005
3. 2.49656
4. 3.71989
5. 3.84044
6. 2.52389
7. 2.34683
8. 3.5265
9. 3.34831
10. 3.39626
Correct answer: 3
.
31005 m
/
s.
Explanation:
IF we analyze the Forces on the water at the
top oF a vertical circle (using the notation
F
ji
For a two body Force on
ˆ
i
From
ˆ
j
), we see that
N
pail,water
+
W
earth,water
=
m a
centripetal
=
m
v
2
r
.
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2
To minimize the velocity, we minimize the lhs
of the equation. Since we can’t change the
weight of the water, we use the lowest Normal
force that we can, 0. Then,
m g
=
m
v
2
r
and so,
v
=
√
g r
=
r
(9
.
8 m
/
s
2
) (1
.
118 m)
=
3
.
31005 m
/
s
.
003
10.0 points
The horizontal surface on which the objects
slide is frictionless.
The acceleration of gravity is 9
.
8 m
/
s
2
.
1 kg
8 kg
6 kg
F
ℓ
F
r
If
F
ℓ
= 23 N and
F
r
= 7 N, what is the
magnitude of the force exerted on the block
with mass 8 kg by the block with mass 6 kg?
1. 12.0769
2. 8.84615
3. 7.94118
4. 12.5714
5. 10.4286
6. 11.5625
7. 7.8
8. 10.0
9. 9.57143
10. 13.4
Correct answer: 13
.
4 N.
Explanation:
m
1
m
2
m
3
F
ℓ
F
r
Given :
V
F
ℓ
= +23 N ˆ
ı ,
V
F
r
=
−
7 N ˆ
ı ,
m
1
= 1 kg
,
m
2
= 8 kg
,
m
3
= 6 kg
,
and
g
= 9
.
8 m
/
s
2
.
Note:
F
is acting on the combined mass
of the three blocks, resulting in a common
acceleration after accounting for friction.
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 Spring '08
 Turner
 Physics, Force, Friction, Correct Answer, roton, µs tan, Version 086/ABBBC

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