Version 037 – Final – gentle – (56245)
1
This
printout
should
have
36
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
10.0points
A homogeneous cylinder of radius 18 cm and
mass 50 kg is rolling without slipping along a
horizontal floor at 7 m
/
s.
How much work was required to give it this
motion?
1. 3.12
2. 2.75625
3. 0.135
4. 2.4
5. 1.8375
6. 0.84375
7. 1.21875
8. 0.21
9. 3.84
10. 3.6
Correct answer: 1
.
8375 kJ.
Explanation:
Let :
r
= 18 cm
,
m
= 50 kg
,
and
v
= 7 m
/
s
.
Because the cylinder rolls without slipping,
the translational speed is
v
=
r ω
and the total kinetic energy of the cylinder is
K
total
=
1
2
m v
2
+
1
2
I ω
2
=
1
2
m v
2
+
1
2
parenleftbigg
1
2
m r
2
parenrightbigg
parenleftBig
v
r
parenrightBig
2
=
3
4
m v
2
=
3
4
(50 kg) (7 m
/
s)
2
·
1 kJ
1000 J
=
1
.
8375 kJ
,
which is also the work required to give the
cylinder this motion.
keywords:
002
10.0points
At a certain instant of time, a toy car is
traveling at a constant speed of
v
0
to the
right, while a distance
D
ahead is another toy
car starting from rest and traveling to the left
with constant acceleration
a
.
The cars are
moving along the same line, directly toward
each other.
At what time do they collide, in terms of
D
,
v
0
and
a
?
1.
The cars will never collide.
2.
v
0
a
3.
1
2
parenleftbigg
2
D
a
+
v
0
a
parenrightbigg
4.
radicalbigg
parenleftBig
v
0
a
parenrightBig
2
+
2
D
a

v
0
a
correct
5.
radicalbigg
2
D
a

parenleftBig
v
0
a
parenrightBig
2

v
0
a
6.
radicalbigg
parenleftBig
v
0
a
parenrightBig
2
+
2
D
a
+
v
0
a
7.
radicalbigg
parenleftBig
v
0
a
parenrightBig
2

2
D
a

v
0
a
8.
2
D
a
Explanation:
The first toy car starts at a position
x
= 0
and moves to the right at a constant speed
v
0
,
so its position is given by
x
1
=
v
0
t
.
Mean
while, the second toy car starts at a distance
D
ahead of the first and accelerates to the left
with an acceleration
a
, so its position is given
by
x
2
=
D

1
2
a t
2
.
The two cars collide when
x
1
=
x
2
v
0
t
=
D

1
2
a t
2
0 =
a t
2
+ 2
v
0
t

2
D
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Version 037 – Final – gentle – (56245)
2
t
=

2
v
0
±
radicalBig
4
v
2
0
+ 8
D a
2
a
=

v
0
±
radicalBig
v
2
0
+ 2
D a
a
=

v
0
a
+
radicalbigg
parenleftBig
v
0
a
parenrightBig
2
+
2
D
a
t >
0 when the two cars collide, so the
negative root must be rejected.
003
10.0points
Three masses are arranged in the (
x, y
)
plane as shown.
7 kg
1 kg
3 kg
y
(m)
1
3
5
7
9
x
(m)
1
3
5
7
9
What is the magnitude of the resulting
force on the 7 kg mass at the origin?
The
value of the universal gravitational constant
is 6
.
6726
×
10
−
11
N
·
m
2
/
kg
2
.
1. 1.48252e10
2. 5.58488e11
3. 3.80511e11
4. 3.71041e11
5. 2.04682e10
6. 1.75476e11
7. 5.74069e11
8. 1.32567e10
9. 2.69041e11
10. 3.50343e10
Correct answer: 3
.
50343
×
10
−
10
N.
Explanation:
Let:
m
o
= 7 kg
,
(
x
o
, y
o
) = (0 m
,
0 m)
,
m
a
= 1 kg
,
(
x
a
, y
a
) = (10 m
,
0 m)
,
and
m
b
= 3 kg
,
(
x
b
, y
b
) = (0 m
,
2 m)
.
Applying
Newton’s
universal
gravitational
law for
m
o
and
m
a
,
F
ao
=
G
m
o
m
a
(
x
a

x
o
)
2
+ (
y
a

y
o
)
2
= (6
.
6726
×
10
−
11
N
·
m
2
/
kg
2
)
×
(7 kg) (1 kg)
(10 m)
2
+ (0 m)
2
= 4
.
67082
×
10
−
12
N
acts along the
x
axis.
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 Fall '07
 Swinney
 mechanics, Acceleration, Force, Mass, Correct Answer, Standard gravity, θ

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