schroeder (cs39963) – Resource 7: Momentum, Work and Energy – balasubramanya – (1401310) 1
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print-out
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have
40
questions.
Multiple-choice questions may continue on
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before answering.
001 (part 1 of 3) 10.0 points
A 10 metric ton train moves toward the
south at 40 m
/
s.
At what speed must it travel to have five
times its original momentum?
A metric ton
is 1000 kg.
Correct answer: 200 m
/
s.
Explanation:
Let :
v
= 40 m
/
s
and
p
final
= 5
p
initial
.
Since momentum is proportional to speed,
its speed must be increased to
v
f
= 5(40 m
/
s) =
200 m
/
s
.
002 (part 2 of 3) 10.0 points
At what speed must it travel to have a
momentum of 4
×
10
5
kg
·
m
/
s?
Correct answer: 40 m
/
s.
Explanation:
Let :
m
= 10 metric ton
and
p
= 4
×
10
5
kg
·
m
/
s
.
Since a metric ton is 1000 kg,
p
=
m v
v
=
p
m
=
4
×
10
5
kg
·
m
/
s
10 metric ton
·
1 metric ton
1000 kg
=
40 m
/
s
.
003 (part 3 of 3) 10.0 points
If there were a speed limit for this train as
it traveled through a city, but not a weight
limit, what mass must be added to the train
to slow it down to 30 m
/
s while at the same
time keeping the momentum the same as in
the second part?
Correct answer: 3333
.
33 kg.
Explanation:
Let :
p
= 4
×
10
5
kg
·
m
/
s
,
v
= 30 m
/
s
,
and
m
= 10000 kg
.
p
=
m v
m
=
p
v
=
4
×
10
5
kg
·
m
/
s
30 m
/
s
= 13333
.
3 kg
is the required mass, so we must add
m
= 13333
.
3 kg
−
10000 kg =
3333
.
33 kg
.
004 (part 1 of 4) 10.0 points
Calculate the magnitude of the linear momen-
tum for each of the following cases
a) a proton with mass 1
.
67
×
10
−
27
kg mov-
ing with a velocity of 6
×
10
6
m
/
s.
Correct answer: 1
.
002
×
10
−
20
kg
·
m
/
s.
Explanation:
Let :
m
= 1
.
67
×
10
−
27
kg
and
v
= 6
×
10
6
m
/
s
.
Momentum is
vectorp
=
mvectorv ,
so
p
=
(
1
.
67
×
10
−
27
kg
) (
6
×
10
6
m
/
s
)
=
1
.
002
×
10
−
20
kg
·
m
/
s
.
005 (part 2 of 4) 10.0 points
b) a 1
.
6 g bullet moving with a speed of
312 m
/
s to the right.
Correct answer: 0
.
4992 kg
·
m
/
s.

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schroeder (cs39963) – Resource 7: Momentum, Work and Energy – balasubramanya – (1401310) 2
Explanation:
Let :
m
= 1
.
6 g
and
v
= 312 m
/
s
.
p
= (1
.
6 g) (312 m
/
s)
·
1 kg
1000 g
=
0
.
4992 kg
·
m
/
s
.
006 (part 3 of 4) 10.0 points
c) a 8 kg sprinter running with a velocity of
13
.
3 m
/
s.
Correct answer: 106
.
4 kg
·
m
/
s.
Explanation:
Let :
m
= 8 kg
and
v
= 13
.
3 m
/
s
.
p
= (8 kg) (13
.
3 m
/
s)
=
106
.
4 kg
·
m
/
s
.
007 (part 4 of 4) 10.0 points
d) Earth (
m
= 5
.
98
×
10
24
kg) moving with
an orbital speed equal to 29400 m
/
s.
Correct answer: 1
.
75812
×
10
29
kg
·
m
/
s.
Explanation:
Let :
m
= 5
.
98
×
10
24
kg
and
v
= 29400 m
/
s
.
p
=
(
5
.
98
×
10
24
kg
)
(29400 m
/
s)
=
1
.
75812
×
10
29
kg
·
m
/
s
.
008
10.0 points
Two blocks of masses
m
and
M
[
M
= 3
.
35
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
.
23 m
/
s.
m
M
m
M
Before
After
(a)
(b)
v
What is the speed of the block of mass
m
?
Correct answer: 7
.
4705 m
/
s.
Explanation:
From conservation of momentum Δ
p
= 0,
in our case we obtain
0 =
M v
M
−
m v
m
.

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