oldhomewk 22 – PAPAGEORGE, MATT – Due: Apr 1 2008, 4:00 am
1
Question 1, chap 11, sect 2.
part 1 of 1
10 points
Objects having the same mass travel toward
each other on a flat surface, each with a speed
of 1.0 meter per second relative to the surface.
The objects collide headon and are reported
to rebound after the collision, each with a
speed of 2.0 meters per second relative to the
surface.
Which of the following assessments of this
report is most accurate?
1.
If the surface were inclined, the report
could be true.
2.
If potential energy were released to the
objects during the collision, the report could
be true.
correct
3.
Momentum was not conserved; therefore
the report is false.
4.
If the objects had different masses, the
report could be true.
5.
If there were no friction between the ob
jects and the surface, the report could be
true.
Explanation:
P
i
=
m
1
v
1

m
2
v
2
= 0
P
f
=
m
2
v
′
2

m
1
v
′
1
= 0
The total momentum is conserved here.
However,
v
′
1
=
v
′
2
= 2
v
1
= 2
v
2
, so energy
must be added to the system to make the
process physically realistic.
Thus if potential energy were released to
the objects during the collision, the report
could be true.
Question 2, chap 11, sect 4.
part 1 of 1
10 points
A(n) 7 kg object moving with a speed of
7
.
7 m
/
s collides with a(n) 19 kg object moving
with a velocity of 8 m
/
s in a direction 24
◦
from the initial direction of motion of the 7 kg
object.
7
.
7 m
/
s
8 m
/
s
19 kg
7 kg
24
◦
What is the speed of the two objects after
the collision if they remain stuck together?
Correct answer: 7
.
7858
m
/
s (tolerance
±
1
%).
Explanation:
Let :
m
1
= 7 kg
,
m
2
= 19 kg
,
m
f
=
m
1
+
m
2
= 26 kg
,
v
1
= 7
.
7 m
/
s
,
v
2
= 8 m
/
s
,
p
1
=
m
1
v
1
= 53
.
9 kg m
/
s
,
p
2
=
m
2
v
2
= 152 kg m
/
s
,
p
1
p
2
=
m
1
v
1
m
2
v
2
= 16385
.
6 kg
2
m
2
/
s
2
,
p
x
=
p
1
+
p
2
cos
θ
= 192
.
759 kg m
/
s
,
p
y
=
p
2
sin
θ
= 61
.
824 kg m
/
s
,
θ
= 24
◦
,
and
π

θ
= 156
◦
.
v
1
v
2
m
2
m
1
θ
v
f
φ
m
f
The final momentum is
p
f
= (
m
1
+
m
2
)
v
f
.
(1)
Momentum is conserved
vectorp
1
+
vectorp
2
=
vectorp
f
.
(2)
Using the law of cosines, we have
p
2
1
+
p
2
2

2
p
1
p
2
cos(
π

θ
) =
p
2
f
.
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oldhomewk 22 – PAPAGEORGE, MATT – Due: Apr 1 2008, 4:00 am
2
Solving for
v
f
, we have
v
f
=
radicalBig
p
2
1
+
p
2
2

2
p
1
p
2
cos(
π

θ
)
m
1
+
m
2
=
bracketleftbigg
1
(7 kg) + (19 kg)
bracketrightbigg
×
bracketleftBig
(2905
.
21 kg
2
m
2
/
s
2
)
2
+ (23104 kg
2
m
2
/
s
2
)
2

(16385
.
6 kg
2
m
2
/
s
2
) cos(156
◦
)
bracketrightBig
1
/
2
=
7
.
7858 m
/
s
.
Since the direction of the total momentum
before and after the collision does not change,
the initial and final momentum direction is
φ
= arctan
parenleftbigg
p
y
p
x
parenrightbigg
= arctan
parenleftbigg
61
.
824 kg m
/
s
192
.
759 kg m
/
s
parenrightbigg
= 17
.
7827
◦
.
Alternate Solution:
Since
p
2
f
=
p
2
x
+
p
2
y
,
we have
v
f
=
radicalBig
p
2
x
+
p
2
y
m
1
+
m
2
=
bracketleftbigg
1
(7 kg) + (19 kg)
bracketrightbigg
×
bracketleftBig
(37156 kg
2
m
2
/
s
2
)
+ (3822
.
2 kg
2
m
2
/
s
2
)
bracketrightBig
1
/
2
=
7
.
7858 m
/
s
.
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 Spring '09
 KLEINMAN
 Physics, Energy, Kinetic Energy, Mass, Momentum, m/s

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