alexander (jra2623) – homework 19 – Turner – (92510)
1
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printout
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have
13
questions.
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001
10.0 points
What velocity must a car with a mass of
1290 kg have in order to have the same mo
mentum as a 2370 kg pickup truck traveling
at 24 m
/
s to the east?
Correct answer: 44
.
093 m
/
s.
Explanation:
Let :
m
1
= 1290 kg
,
m
2
= 2370 kg
,
and
v
2
= 24 m
/
s to the east
.
vectorp
=
m
1
vectorv
1
=
m
2
vectorv
2
v
1
=
m
2
v
2
m
1
=
(2370 kg) (24 m
/
s)
1290 kg
=
44
.
093 m
/
s
to the east.
002
10.0 points
Note:
Take East as the positive direction.
A(n) 80 kg fisherman jumps from a dock
into a 137 kg rowboat at rest on the West side
of the dock.
If the velocity of the fisherman is 3
.
6 m
/
s
to the West as he leaves the dock, what is the
final velocity of the fisherman and the boat?
Correct answer:
−
1
.
32719 m
/
s.
Explanation:
Let West be negative:
Let :
m
1
= 80 kg kg
,
m
2
= 137 kg kg
,
and
v
i,
1
=
−
3
.
6 m
/
s m
/
s
.
The boat and fisherman have the same final
speed, and
v
i,
2
= 0 m/s, so
m
1
vectorv
i,
1
+
m
2
vectorv
i,
2
= (
m
1
+
m
2
)
vectorv
f
m
1
vectorv
i,
1
= (
m
1
+
m
2
)
vectorv
f
v
f
=
m
1
v
i
m
1
+
m
2
=
(80 kg) (
−
3
.
6 m
/
s)
80 kg + 137 kg
=
−
1
.
32719 m
/
s
,
which is 1
.
32719 m
/
s to the West.
003
10.0 points
A(n) 871 N man stands in the middle of a
frozen pond of radius 8
.
8 m.
He is un
able to get to the other side because of a
lack of friction between his shoes and the
ice. To overcome this difficulty, he throws his
1
.
5 kg physics textbook horizontally toward
the north shore, at a speed of 4
.
2 m
/
s.
The acceleration of gravity is 9
.
81 m
/
s
2
.
How long does it take him to reach the
south shore?
Correct answer: 124
.
02 s.
Explanation:
Let :
W
m
= 871 N
,
r
= 8
.
8 m
,
m
b
= 1
.
5 kg
,
and
v
′
b
= 4
.
2 m
/
s
.
The mass of the man is
m
m
=
W
m
g
.
From conservation of momentum,
m
m
v
m
+
m
b
v
b
=
m
m
v
′
m
+
m
b
v
′
b
0 =
m
m
v
′
m
+
m
b
v
′
b
v
′
m
=
−
m
b
m
m
v
′
b
=
−
g m
b
W
m
v
′
b
=
−
(
9
.
81 m
/
s
2
)
(1
.
5 kg)
871 N
(4
.
2 m
/
s)
=
−
0
.
0709564 m
/
s
.
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alexander (jra2623) – homework 19 – Turner – (92510)
2
The time to travel the 8
.
8 m to shore is
t
=
Δ
x

v
′
m

=
8
.
8 m
0
.
0709564 m
/
s
=
124
.
02 s
.
004
10.0 points
An 67
.
1 kg object moving to the right at
47
.
2 cm
/
s overtakes and collides elastically
with a second 51
.
1 kg object moving in the
same direction at 28
.
1 cm
/
s.
Find the velocity of the second object after
the collision.
Correct answer: 49
.
7854 cm
/
s.
Explanation:
Let :
m
1
= 67
.
1 kg
,
m
2
= 51
.
1 kg
,
v
1
= 47
.
2 cm
/
s
,
and
v
2
= 28
.
1 cm
/
s
.
Momentum conservation gives
m
1
v
1
i
+
m
2
v
2
i
=
m
1
v
1
f
+
m
2
v
2
f
(1)
For headon elastic collisions the centerof
mass motion remains constant
−
(
v
1
f
−
v
2
f
) =
v
1
i
−
v
2
i
,
that is
v
2
f
=
v
1
i
−
v
2
i
+
v
1
f
,
or
v
1
f
=
v
2
i
−
v
1
i
+
v
2
f
.
(2)
Substituting Eq. 2 into Eq. 1, we have
m
1
v
1
i
+
m
2
v
2
i
=
m
1
v
2
i
−
m
1
v
1
i
+(
m
1
+
m
2
)
v
2
f
2
m
1
v
1
i
+ (
m
2
−
m
1
)
v
2
i
= (
m
1
+
m
2
)
v
2
f
v
2
f
=
bracketleftbigg
2
m
1
m
1
+
m
2
bracketrightbigg
v
1
+
bracketleftbigg
m
2
−
m
1
m
1
+
m
2
bracketrightbigg
v
2
=
bracketleftbigg
2 (67
.
1 kg)
(67
.
1 kg) + (51
.
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 Summer '08
 Kaplunovsky
 Energy, Kinetic Energy, Mass, Momentum, Work, kg

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