le (txl65) – Linear Momentum – mohanty – (81440)
1
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001
10.0points
Two particles of masses
m
and 3
m
are
moving toward each other along the
x
axis
with the same speed
v
. They undergo a head
on elastic collision and rebound along the
x

axis.
m
v
3
m
v
Determine the final speed of the heavier
object.
1.
v
′
3
m
= 0
correct
2.
v
′
3
m
=
v
3.
v
′
3
m
= 2
v
4.
v
′
3
m
=
2
3
v
5.
v
′
3
m
=
3
2
v
6.
v
′
3
m
=
1
2
v
7.
v
′
3
m
=
∞
8.
v
′
3
m
= 3
v
9.
v
′
3
m
= 4
v
10.
v
′
3
m
=
1
3
v
Explanation:
Let :
v
1
=
v ,
v
2
=
−
v ,
m
1
=
m
and
m
2
= 3
m .
BasicConcepts:
Conservation of momentum
Solution:
Denote the smaller mass initially moving to
the right as
m
1
and the larger mass initially
moving to the left as
m
2
.
Since no external
forces act on the two masses, even during the
collision, the total momentum is conserved.
Also, since the masses collide elastically, en
ergy is conserved.
Since this is a head on,
elastic collision we can use
v
′
2
= 2
v
cm
−
v
2
,
where
v
cm
=
m
1
v
1
+
m
2
v
2
m
1
+
m
2
,
so
v
′
2
=
2
m
1
v
1
+ 2
m
2
v
2
m
1
+
m
2
−
v
2
(
m
1
+
m
2
)
m
1
+
m
2
=
2
m v
−
6
m v
+
m v
+ 3
m v
m
+ 3
m
=
0
.
AlternativeSolution:
From the relative
velocities formula,
v
1
−
v
2
=
v
′
2
−
v
′
1
.
(1)
From conservation of momentum,
m
1
v
1
+
m
2
v
2
=
m
1
v
′
1
+
m
2
v
′
2
.
(2)
Combining Eqs. 1 and 2 gives
m
1
v
1
+
m
2
v
2
=
m
1
(
v
′
2
+
v
2
−
v
1
) +
m
2
v
′
2
,
v
′
2
=
m
1
v
1
+
m
2
v
2
−
m
1
(
v
2
−
v
1
)
m
1
+
m
2
=
m v
+ 3
m
(
−
v
)
−
m
(
−
v
−
v
)
m
+ 3
m
=
m
(
v
−
3
v
+
v
+
v
)
4
m
=
0
.
002
10.0points
A 6 kg steel ball strikes a wall with a speed
of 9
.
26 m
/
s at an angle of 35
.
6
◦
with the
normal to the wall.
It bounces off with the
same speed and angle, as shown in the figure.
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le (txl65) – Linear Momentum – mohanty – (81440)
2
x
y
9
.
26 m
/
s
6 kg
9
.
26 m
/
s
6 kg
35
.
6
◦
35
.
6
◦
If the ball is in contact with the wall for
0
.
251 s, what is the magnitude of the average
force exerted on the ball by the wall?
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 Spring '11
 sdsd
 Momentum, m/s

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