Physics 203 – Section 4
QUIZ 9
20 April 2015
Name:
1. (5 points) The mass of Mars is 0.107 times the mass of Earth, and its mean radius is
0.532 times the mean radius of Earth. Find the ratio of the acceleration due to gravity
on Earth to that on Mars gE /g M .
Solution: The acceleration on Earth for an object of mass m
gE =
Fgrav,E
1 GME m
GME
=
=
.
m
m R2E
R2E
(1)
As expected this does not depend on the m as expected and will be equal to
9.8 m/s2 . The acceleration on Mars for an object of mass m
gM =
Fgrav,E
GM M
=
.
m
R2M
(2)
Therefore, the ratio
gE
ME R2M
ME (0.532R E )2
0.5322
= 2
= 2
=
= 2.65.
gM
0.107
RE MM
R E 0.107ME
(3)
2. A mass is attached to a spring with spring constant 90.0 N/m. The mass is pulled a
distance of 15.0 cm away from its equilibrium point where it is released and oscillates
with a period of 1.00 s.
(a) (2 points) Find the angular frequency of the oscillation.
Solution: The angular frequency
w = 2⇡ f =
(b) (3 points) What is the mass?
2⇡
2⇡
=
= 6.28 s
T
1.00 s
1
.
(4)
Physics 203 – Section 4
QUIZ 9
20 April 2015
Solution: The period is related to the mass attached on the spring by
T = 2⇡
r
m
.
k
(5)
Solving this equation for m gives
m=
kT 2
(90.0 N/m)(1.00 s)2
=
= 2.28 kg.
4⇡2
4⇡2
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Physics 203 – Section 5
QUIZ 9
20 April 2015
Name:
1. (5 points) The mass of Saturn’s moon Titan is 0.0225 times the mass of Earth, and its
mean radius is 0.404 times the mean radius of Earth. Find the ratio of the acceleration
due to gravity on Earth to that on Titan gE /gT .
Solution: The acceleration on Earth for an object of mass m
gE =
Fgrav,E
1 GME m
GME
=
=
.
m
m R2E
R2E
(1)
As expected this does not depend on the m as expected and will be equal to
9.8 m/s2 . The acceleration on Mars for an object of mass m
gM =
Fgrav,E
GM M
=
.
m
R2M
(2)
Therefore, the ratio
gE
M R2
M (0.404R E )2
0.4042
= 2E T = 2E
=
= 7.25.
gT
0.0225
R E MT
R E 0.0225ME
(3)
2. A mass is attached to a spring with spring constant 40.0 N/m. The mass is pulled a
distance of 20.0 cm away from its equilibrium point where it is released and oscillates
with a an angular frequency of 4.00 s 1 .
(a) (2 points) Find the period of the oscillation.
Solution: The period
T=
1
2⇡
2⇡
=
=
f
w
4.00 s
(b) (3 points) What is the mass?
1
= 1.57 s.
(4)
Physics 203 – Section 5
QUIZ 9
20 April 2015
Solution: The period is related to the mass attached on the spring by
w=
r
k
.
m
(5)
Solving this equation for m gives
m=
k
40.0 N/m
=
= 2.50 kg.
w2
(4.00 s 1 )2
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