homework 13 – RAHMAN, TARIQUE – Due: Mar 21 2008, 3:00 am
1
Question 1, chap 10, sect 1.
part 1 of 2
10 points
When the velocity of an object is doubled,
by what factor is its momentum changed?
1.
4
2.
3
3.
2
correct
4.
1
Explanation:
Momentum is proportional to speed.
Question 2, chap 10, sect 1.
part 2 of 2
10 points
When the velocity of an object is doubled,
by what factor is its kinetic energy changed?
1.
2
2.
3
3.
1
4.
4
correct
Explanation:
Kinetic Energy is proportional to speed
squared.
Question 3, chap 10, sect 1.
part 1 of 1
10 points
Bill (mass
m
) plants both feet solidly on
the ground and then jumps straight up with
velocity
→
v
.
The earth (mass
M
) then has velocity
1.
V
Earth
= +
→
v
man
.
2.
V
Earth
= +
parenleftBig
m
M
parenrightBig
→
v
man
.
3.
V
Earth
=
−
parenleftbigg
M
m
parenrightbigg
→
v
man
.
4.
V
Earth
= +
radicalbigg
m
M
→
v
man
.
5.
V
Earth
=
−
parenleftBig
m
M
parenrightBig
→
v
man
.
correct
6.
V
Earth
=
−
radicalbigg
m
M
→
v
man
.
7.
V
Earth
=
−
→
v
man
.
8.
V
Earth
= +
parenleftbigg
M
m
parenrightbigg
→
v
man
.
Explanation:
The momentum is conserved. We have
m
→
v
man
+
M
→
V
Earth
= 0
So
→
V
Earth
=
−
parenleftBig
m
M
parenrightBig
→
v
man
.
Question 4, chap 10, sect 1.
part 1 of 1
10 points
An
open
train
car
moves
with
speed
14
.
7 m
/
s on a flat frictionless railroad track,
with no engine pulling the car.
It begins to
rain. The rain falls straight down and begins
to fill the train car.
The speed of the car
1.
stays the same.
2.
decreases.
correct
3.
increases.
Explanation:
Using Newton’s second law, we have
d P
dt
= 0
,
since no external forces act on the train in the
horizontal direction. With no rain, the train
will move at a constant velocity; however,
when it starts to rain, and the rain starts to
fill the car, the mass of the train changes.
Thus,
m
d v
dt
=
−
v
d m
dt
.
Since
d m
dt
is positive;
i.e.
, the mass of the
train is increasing with accumulating rain,
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homework 13 – RAHMAN, TARIQUE – Due: Mar 21 2008, 3:00 am
2
d v
dt
should be negative;
i.e.
, the speed of the
train should decrease.
Question 5, chap 10, sect 1.
part 1 of 2
10 points
A revolutionary war cannon, with a mass of
1980 kg, fires a 18
.
6 kg ball horizontally. The
cannonball has a speed of 103 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: 0
.
967576 m
/
s (tolerance
±
1
%).
Explanation:
Let :
m
= 18
.
6 kg
,
M
= 1980 kg
,
and
v
= 103 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

=
18
.
6 kg
1980 kg
(103 m
/
s)
=
0
.
967576 m
/
s
.
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