alexander (jra2623) – oldhomework 19 – Turner – (92510)
1
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10
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001
10.0 points
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
= 4
v
2.
v
′
3
m
=
3
2
v
3.
v
′
3
m
=
∞
4.
v
′
3
m
=
2
3
v
5.
v
′
3
m
=
1
2
v
6.
v
′
3
m
= 0
correct
7.
v
′
3
m
=
v
8.
v
′
3
m
= 2
v
9.
v
′
3
m
=
1
3
v
10.
v
′
3
m
= 3
v
Explanation:
Let :
v
1
=
v ,
v
2
=

v ,
m
1
=
m
and
m
2
= 3
m .
Basic Concepts:
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
.
Alternative Solution:
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
10.0 points
Two ice skaters approach each other at right
angles.
Skater A has a mass of 76
.
8 kg and
travels in the +
x
direction at 2
.
03 m
/
s. Skater
B has a mass of 61
.
1 kg and is moving in the
+
y
direction at 2
.
38 m
/
s.
They collide and
cling together.
Find the final speed of the couple.
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alexander (jra2623) – oldhomework 19 – Turner – (92510)
2
Correct answer: 1
.
54602 m
/
s.
Explanation:
From conservation of momentum Δ
p
= 0
m
A
v
A
ˆ
ı
+
m
B
v
B
ˆ
= (
m
A
+
m
B
)
v
f
Therefore
v
f
=
radicalbig
(
m
A
v
A
)
2
+ (
m
B
v
B
)
2
m
A
+
m
B
=
radicalbig
(155
.
904 kg m
/
s)
2
+ (145
.
418 kg m
/
s)
2
76
.
8 kg + 61
.
1 kg
= 1
.
54602 m
/
s
003
10.0 points
A(n) 79 g particle moving with an initial speed
of 49 m
/
s in the positive
x
direction strikes
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 Summer '08
 Kaplunovsky
 Kinetic Energy, Mass, Momentum, Work, Correct Answer, Alexander

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