choi (hc5878) – hw7 – Demkov – (59910)
1
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48
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001
(part 1 of 2) 10.0 points
A revolutionary war cannon, with a mass of
2170 kg, fires a 23
.
4 kg ball horizontally. The
cannonball has a speed of 102 m
/
s after it has
left the barrel.
The cannon carriage is on a
flat platform and is free to roll horizontally.
What is the speed of the cannon immedi
ately after it was fired?
Correct answer: 1
.
09991 m
/
s.
Explanation:
Let :
m
= 23
.
4 kg
,
M
= 2170 kg
,
and
v
= 102 m
/
s
.
The cannon’s velocity immediately after it
was fired is found by using conservation of
momentum along the horizontal direction:
M V
+
m v
= 0
⇒ −
V
=
m
M
v
where
M
is the mass of the cannon,
V
is the
velocity of the cannon,
m
is the mass of the
cannon ball and
v
is the velocity of the cannon
ball. Thus, the cannon’s speed is

V

=
m
M

v

=
23
.
4 kg
2170 kg
(102 m
/
s)
=
1
.
09991 m
/
s
.
002
(part 2 of 2) 10.0 points
The same explosive charge is used, so the total
energy of the cannon plus cannonball system
remains the same.
Disregarding
friction,
how
much
faster
would the ball travel if the cannon
were
mounted rigidly and all other parameters re
mained the same?
Correct answer: 0
.
548479 m
/
s.
Explanation:
By knowing the speeds of the cannon and
the cannon ball, we can find out the total
kinetic energy available to the system
K
net
=
1
2
m v
2
+
1
2
M V
2
.
This is the same amount of energy available
as when the cannon is fixed.
Let
v
′
be the
speed of the cannon ball when the cannon is
held fixed. Then,
1
2
m v
′
2
=
1
2
(
m v
2
+
M V
2
)
.
⇒
v
′
=
radicalbigg
v
2
+
M
m
V
2
=
v
radicalbigg
1 +
m
M
= (102 m
/
s)
radicalBigg
1 +
23
.
4 kg
2170 kg
= 102
.
548 m
/
s
.
Thus, the velocity difference is
v
′
−
v
= 102
.
548 m
/
s
−
102 m
/
s
=
0
.
548479 m
/
s
.
003
(part 1 of 3) 10.0 points
Tony (of mass 70 kg) coasts on his bicycle (of
mass 10 kg) at a constant speed of 5 m
/
s,
carrying a 5 kg pack.
Tony throws his pack
forward, in the direction of his motion, at
2 m
/
s relative to the speed of bicycle just
before the throw.
What is the initial momentum of the system
(Tony, the bicycle, and the pack)?
Correct answer: 425 kg
·
m
/
s.
Explanation:
Let :
m
T
= 70 kg
,
m
B
= 10 kg
,
m
P
= 5 kg
,
and
v
i
= 5 m
/
s
.
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choi (hc5878) – hw7 – Demkov – (59910)
2
Since
p
=
m v .
Thus
p
i
= (
m
T
+
m
B
+
m
P
)
v
i
= (70 kg + 10 kg + 5 kg)(5 m
/
s)
=
425 kg
·
m
/
s
.
004
(part 2 of 3) 10.0 points
What is the momentum of the system im
mediately after the pack is thrown?
Correct answer: 425 kg
·
m
/
s.
Explanation:
Since momentum is conserved,
p
f
=
p
i
=
425 kg
·
m
/
s
.
005
(part 3 of 3) 10.0 points
What is the bicycle speed immediately after
the throw?
Correct answer: 4
.
875 kg
·
m
/
s.
Explanation:
Let :
p
f
= 425 kg
·
m
/
s
,
v
p
= 7 m
/
s
,
m
T
= 70 kg
,
m
B
= 10 kg
,
and
m
P
= 5 kg
,
p
f
= (
m
T
+
m
B
)
v
b
+
m
P
v
P
p
f
−
m
P
v
p
= (
m
T
+
m
B
)
v
b
v
b
=
p
f
−
m
P
v
p
m
T
+
m
B
v
b
=
425 kg
·
m
/
s
−
(5 kg)(7 m
/
s)
70 kg + 10 kg
=
4
.
875 kg
·
m
/
s
.
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 Spring '07
 Swinney
 mechanics, Kinetic Energy, Mass, Momentum, kg

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