karna (pk4534) – HW 9 – coker – (58245)
1
This
printout
should
have
20
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
I could find very few problems in the ”bank”
that were relevant to Ch. 10 in Ohanian and
Markert.
001
(part 1 of 2) 10.0 points
An object is moving so that its kinetic energy
is 95
.
4 J and the magnitude of its momentum
is 47
.
8 kg
·
m
/
s.
Determine the speed of the object.
Correct answer: 3
.
99163 m
/
s.
Explanation:
Let :
K
= 95
.
4 J
and
p
= 47
.
8 kg
·
m
/
s
.
The kinetic energy is given by
K
=
1
2
m v
2
,
and the magnitude of the momentum is
p
=
m v.
Dividing one equation by the other yields
K
p
=
v
2
.
v
=
2
K
p
=
2 (95
.
4 J)
47
.
8 kg
·
m
/
s
=
3
.
99163 m
/
s
.
002
(part 2 of 2) 10.0 points
Determine the mass of the object.
Correct answer: 11
.
9751 kg.
Explanation:
m
=
p
v
=
47
.
8 kg
·
m
/
s
3
.
99163 m
/
s
=
11
.
9751 kg
.
003
10.0 points
A balloon of mass
M
is floating motionless in
the air. A person of mass less than
M
is on
a rope ladder hanging from the balloon. The
person begins to climb the ladder at a uniform
speed
v
relative to the ground.
How does the balloon move relative to the
ground?
1.
Down with a speed less than
v
correct
2.
Up with speed
v
3.
Down with speed
v
4.
Up with a speed less than
v
5.
The balloon does not move.
Explanation:
Let the mass of the person be
m
.
Total momentum is conserved (because the
exterior forces on the system are balanced),
especially the component in the vertical di
rection.
When the person begins to move, we have
m v
+
M v
M
= 0
,
=
⇒
v
M
=
−
m
M
v < v
=
⇒

v
M

=
m
M
v < v,
since
m < M
=
⇒
m
M
<
1
.
Thus the balloon moves in the opposite
direction.
004
10.0 points
Two blocks of masses
m
and
M
[
M
= 2
.
45
m
]
are placed on a horizontal, frictionless surface.
A light spring is attached to one of them, and
the blocks are pushed together with the spring
between them. A cord holding them together
is burned, after which the block of mass
M
moves to the right with a speed of 2
.
01 m
/
s.
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karna (pk4534) – HW 9 – coker – (58245)
2
m
M
m
M
Before
After
(a)
(b)
v
What is the speed of the block of mass
m
?
Correct answer: 4
.
9245 m
/
s.
Explanation:
From conservation of momentum Δ
p
= 0,
in our case we obtain
0 =
M v
M
−
m v
m
.
Therefore
v
m
=
M
m
v
M
= (2
.
45)
v
M
= (2
.
45) (2
.
01 m
/
s)
= 4
.
9245 m
/
s
.
005
10.0 points
A baseball player uses a pitching machine to
help him improve his batting average.
He
places the 77 kg machine on a frozen pond.
The machine fires a 0
.
153 kg baseball horizon
tally at a speed of 36
.
8 m
/
s.
What is the magnitude of the recoil velocity
of the machine?
Correct answer: 0
.
0731221 m
/
s.
Explanation:
Let :
m
1
= 0
.
153 kg
,
m
2
= 77 kg
,
and
v
1
f
= 36
.
8 m
/
s
.
We take the system to consist of the base
ball and the pitching machine.
Because of
the force of gravity and the normal force, the
system is not really isolated. However, both
of these forces are directed perpendicularly to
the motion of the system. Therefore, momen
tum is constant in the
x
direction because
there are no external forces in this direction
(assuming the surface is frictionless). The to
tal momentum of the system before firing is
zero.
Therefore, the total momentum after
firing must be zero, that is,
m
1
v
1
f
+
m
2
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 Spring '08
 Turner
 Physics, Mass, Momentum, kg, Karna

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