fryer (vdf96) – Assignment 6 Universal Gravitation / Circular Motion – graves – (6)
1
This
printout
should
have
25
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
10.0points
Two spheres have equal densities and are sub
ject only to their mutual gravitational attrac
tion.
Which quantity must have the same mag
nitude for both spheres?
1.
kinetic energy
2.
acceleration
3.
displacement from the center of mass
4.
gravitational force
correct
5.
velocity
Explanation:
Two spheres with the same density have
different masses due to their relative sizes.
Using Newton’s third law,
vector
F
1
=

vector
F
2
.
All of the other quantities (acceleration, ve
locity, kinetic energy, and displacement from
the center of mass) have different magnitudes
because the two spheres have different masses.
002(part1of2)10.0points
A ball is tossed straight up from the surface
of a small, spherical asteroid with no atmo
sphere. The ball rises to a height equal to the
asteroid’s radius and then falls straight down
toward the surface of the asteroid.
What forces, if any, act on the ball while it
is on the way up?
1.
Only a decreasing gravitational force that
acts downward
correct
2.
Both a constant gravitational force that
acts downward and a decreasing force that
acts upward
3.
Only an increasing gravitational force
that acts downward
4.
No forces act on the ball.
5.
Only a constant gravitational force that
acts downward
Explanation:
There is no friction in the system, and the
ball doesn’t have any contact with other ob
jects, so the only force acting on the ball is
the attractive gravitational force, which acts
downward.
From
vector
F
=

G
M m
r
2
ˆ
r
, the force will de
crease as the ball rises.
003(part2of2)10.0points
The acceleration of the ball at the top of its
path is
1.
zero.
2.
equal to the acceleration at the surface of
the asteroid.
3.
at
its
maximum
value
for
the
ball’s
flight.
4.
equal to onefourth the acceleration at the
surface of the asteroid.
correct
5.
equal to onehalf the acceleration at the
surface of the asteroid.
Explanation:
F
=
m a
∝
1
r
2
,
so
a
∝
1
r
2
and
a
′
∝
1
(2
r
)
2
=
1
4
1
r
2
∝
1
4
a .
004
10.0points
In introductory physics laboratories, a typical
Cavendish balance for measuring the gravita
tional constant
G
uses lead spheres of masses
0
.
85 kg and 17
.
2 g whose centers are separated
by 3
.
51 cm.
Calculate the gravitational force between
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fryer (vdf96) – Assignment 6 Universal Gravitation / Circular Motion – graves – (6)
2
these spheres, treating each as a point mass
located at the center of the sphere.
The
value of the universal gravitational constant
is 6
.
67259
×
10
−
11
N
·
m
2
/
kg
2
.
Correct answer: 7
.
91822
×
10
−
10
N.
Explanation:
Let :
m
1
= 0
.
85 kg
,
m
2
= 17
.
2 g = 0
.
0172 kg
,
r
= 3
.
51 cm = 0
.
0351 m
,
and
G
= 6
.
67259
×
10
−
11
N
·
m
2
/
kg
2
.
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 Fall '10
 Graves
 Physics, Circular Motion, Force, Mass, General Relativity, Gravitational constant, fryer

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