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Overton, Mays – Oldquiz 3 – Due: Mar 8 2005, noon – Inst: Turner
1
This
printout
should
have
20
questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
The due time is Central
time.
001
(part 1 oF 2) 10 points
A 48
.
5 kg girl is standing on a 228 kg plank.
The plank, originally at rest, is Free to slide on
a Frozen lake, which is a ±at, Frictionless sup
porting surFace. The girl begins to walk along
the plank at a constant speed oF 1
.
93 m
/
s to
the right relative to the plank.
What is her velocity relative to the ice sur
Face?
Correct answer: 1
.
59146 m
/
s.
Explanation:
v
g
= velocity oF girl relative to ice
v
p
= velocity oF plank relative to ice
v
gp
= relative velocity oF girl.
Let :
m
g
= 48
.
5 kg
,
m
p
= 228 kg
,
and
v
gp
= 1
.
93 m
/
s
.
By conservation oF momentum
0 =
m
g
v
g
+
m
p
v
p
,
v
p
=

m
g
v
g
m
p
The relative velocity is
v
gp
=
v
g

v
p
(1)
=
v
g

µ

m
g
v
g
m
p
¶
=
v
g
m
p
+
m
g
v
g
m
p
=
v
g
(
m
p
+
m
g
)
m
p
.
Thus
v
g
=
m
p
v
gp
m
g
+
m
p
=
(228 kg) (1
.
93 m
/
s)
228 kg + 48
.
5 kg
=
1
.
59146 m
/
s
.
002
(part 2 oF 2) 10 points
What is the velocity oF the plank relative to
the ice surFace?
Correct answer:

0
.
338535 m
/
s.
Explanation:
Using Eq. (1),
v
p
=
v
g

v
gp
= 1
.
59146 m
/
s

1
.
93 m
/
s
=

0
.
338535 m
/
s
,
directed opposite to the girl’s direction.
003
(part 1 oF 1) 10 points
Which oF the Following quantities are con
served in an elastic collision but are
not
con
served in an inelastic collision?
1.
kinetic energy
correct
2.
angular momentum
3.
linear and angular momentum
4.
kinetic energy and linear momentum
5.
total mass
6.
None (the same quantities are conserved
For both types oF collisions).
7.
kinetic energy and angular momentum
8.
linear momentum
Explanation:
In both collisions, the linear and angular
momenta are conserved, as well as mass. An
elastic collision is defned as a collision where
kinetic energy is conserved, which is di²erent
From an inelastic collision.
004
(part 1 oF 2) 10 points
A(n) 589 kg mass is sliding on a horizontal
Frictionless surFace with a speed oF 11 m
/
s
when it collides with a 87 kg mass initially at
rest, as shown in the fgure. The masses stick
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View Full DocumentOverton, Mays – Oldquiz 3 – Due: Mar 8 2005, noon – Inst: Turner
2
together and slide up a frictionless track at
60
◦
from horizontal.
The acceleration of gravity is 9
.
8 m
/
s
2
.
589 kg
87 kg
9
.
8m
/
s
2
11 m
/
s
60
◦
What is the speed of the two blocks imme
diately after the collision?
Correct answer: 9
.
58432 m
/
s.
Explanation:
Let :
v
12
= speed of blocks after collision
m
1
= 589 kg
,
m
2
= 87 kg
,
v
1
= 11 m
/
s
,
and
g
= 9
.
8 m
/
s
2
.
The initial momentum of block 1 is
m
1
v
1
.
When
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 Spring '08
 Turner
 Physics

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