Ph1a Final Exam Solution
Chefung Chan, Fall 2007
Problem 1: Stay and Sway (10 points)
(a) (3 points)
Notice the diagram uses capital
M
while the question uses
m
. Here
uses
m
for the whole question. By conservation of momentum:
m
1
(0)

m
2
v
ˆ
x
= (
m
1
+
m
2
)
~v
0
~v
0
=
m
1
m
1
+
m
2
v
(

ˆ
x
)
(b) (3 points)
By Hooke’s law,
F
=

kx
m
¨
x
=

kx
¨
x
=

k
m
1
+
m
2
x
Above equation is simple harmonic oscillation in
x
with
ω
2
0
=
k
m
1
+
m
2
. Therefore,
m
2
=
k
ω
2
0

m
1
(c) (4 points)
We know
x
=
A
sin (
ω
0
t
)+
B
cos (
ω
0
t
). We are going to match with
the initial conditions to find
A
and
B
.
x
(
t
= 0) = 0
⇒
B
= 0
˙
x
(
t
= 0) =
v
0
=

m
1
m
1
+
m
2
v
ω
0
A
=

m
1
m
1
+
m
2
v
A
=

m
1
m
1
+
m
2
v
ω
0
Problem 2: Misdirected Meteor (8 points)
(a) (2 points)
First, we want to find the initial speed of the meteor.
Since the
meteor is in a circular orbit that the gravitational force on the meteor from the
planet is equal to the centripetal force.
mv
2
i
r
=
GMm
r
2
v
2
i
=
GM
r
=
GM
R
+ 3
R
v
i
=
r
GM
4
R
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
After the 50% kinetic energy loss,
K
f
=
1
2
K
i
⇒
1
2
mv
2
f
=
1
4
mv
2
i
v
2
f
=
1
2
v
2
i
=
GM
8
R
v
f
=
r
GM
8
R
K
f
=
1
2
mv
2
f
=
GMm
16
R
U
f
=
U
i
=

GMm
4
R
L
f
=
m
(4
R
)
v
f
=
m
√
2
GMR
(b) (2 points)
Total energy
E
=
U
f
+
K
f
=

3
GMm
16
R
<
0.
The new orbit is
either circular or elliptical. Then with
This is the end of the preview.
Sign up
to
access the rest of the document.
 Winter '08
 McKewn
 Momentum, Pallavolo Modena

Click to edit the document details