This
printout
should
have
21
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
10.0 points
An automobile has 57 percent of its weight on
the front wheels. The front and back wheels
are separated by 1
.
5 m.
Where is the center of gravity located with
respect to the front wheels?
Correct answer: 0
.
645 m.
Explanation:
Let :
W
1
= 0
.
57
W
,
W
2
= 0
.
43
W
,
x
1
= 0 m
,
and
x
2
= 1
.
5 m
.
x
cg
=
∑
i
W
i
x
i
W
=
W
1
x
1
+
W
2
x
2
W
=
(0
.
57
W
)
x
1
+ (0
.
43
W
)
x
2
W
= (0
.
57) (0 m) + (0
.
43) (1
.
5 m)
=
0
.
645 m
.
002
10.0 points
Three particles are placed in the
xy
plane.
A
m
1
= 23 g particle is located at (
x
1
,
y
1
),
where
x
1
= 1
.
5 m and
y
1
= 4
.
5 m and a
m
2
= 43 g particle is located at (
x
2
,
y
2
),
where
x
2
=

1
.
3 m and
y
2
=

4
.
8 m.
What must be the
x
coordinate of the
m
3
=
22 g particle so that the center of mass of the
threeparticle system is at the origin?
Correct answer: 0
.
972727 m.
Explanation:
We know that the x coordinate of the center
of mass is defined by
x
CM
=
m
1
x
1
+
m
2
x
2
+
m
3
x
3
m
1
+
m
2
+
m
3
Now,
x
CM
is zero, and using the input values
for
m
1
, m
2
, m
3
and
x
1
and
x
2
, we find and
solving for
x
3
:
x
3
=

m
1
x
1

m
2
x
2
m
3
= 0
.
972727 m
003
10.0 points
A(n) 68 kg boat that is 8
.
8 m in length is
initially 7
.
9 m from the pier.
A 43 kg child
stands at the end of the boat closest to the
pier. The child then notices a turtle on a rock
at the far end of the boat and proceeds to
walk to the far end of the boat to observe the
turtle.
Assume:
There is no friction between boat
and water.
7
.
9 m
8
.
8 m
How far is the child from the pier when she
reaches the far end of the boat?
Correct answer: 13
.
291 m.
Explanation:
Let
D
= distance of the boat from the pier
,
= 7
.
9 m
,
L
= length of the boat
,
= 8
.
8 m
,
M
= mass of the boat
,
= 68 kg
,
and
m
= mass of the child
,
and
= 43 kg
,
X
= change in position of the boat
.
ℓ
= final distance of child from pier
.
d
= final distance of boat from pier
.
D
L
oldmidterm3 02 – Turner – (58185)
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
hinojosa (jlh3938) – oldmidterm3 02 – Turner – (58185)
2
X
d
ℓ
X
We will use the pier as the origin of the
x
coordinate.
Initially, the center of mass of the boatchild
system is
x
cm
=
parenleftbigg
D
+
L
2
parenrightbigg
M
+
D m
M
+
m
Finally, the center of mass of the boatchild
system is
x
cm
=
parenleftbigg
D
+
L
2
 X
parenrightbigg
M
+ (
D
+
L
 X
)
m
M
+
m
,
where
X
is the change in position of the center
of mass of the boat. Since the center of mass
of the system does not move, we can equate
the above two expressions for
x
cm
parenleftbigg
D
+
L
2
parenrightbigg
M
+
D m
M
+
m
=
parenleftbigg
D
+
L
2
 X
parenrightbigg
M
+ (
D
+
L
 X
)
m
M
+
m
and, solving for
X
, we have
parenleftbigg
D
+
L
2
parenrightbigg
M
+
D m
=
parenleftbigg
D
+
L
2
 X
parenrightbigg
M
+ (
D
+
L
 X
)
m
0 =
X
M
+ (
L
 X
)
m
X
(
M
+
m
) =
L m
X
=
m
m
+
M
L
=
(43 kg)
(43 kg) + (68 kg)
×
(8
.
8 m)
= 3
.
40901 m
.
The final distance
ℓ
of the child from the
pier is
ℓ
=
D
+
L
 X
= (7
.
9 m) + (8
.
8 m)

(3
.
40901 m)
=
13
.
291 m
.
The final distance
d
of the near end of the
boat to the pier is
d
=
D
 X
= (7
.
9 m)

(3
.
40901 m)
= 4
.
49099 m
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Turner
 Angular Momentum, Mass, Moment Of Inertia, Correct Answer, Hinojosa

Click to edit the document details