homework 09 – ALIBHAI, ZAHID – Due: Mar 28 2007, 4:00 am
1
Question 1
part 1 of 1
10 points
Two identical balls are on a frictionless hor
izontal tabletop. Ball
X
initially moves at 10
meters per second, as shown in figure on the
lefthand side. It then collides elastically with
ball
Y
, which is initially at rest. After the col
lision, ball
X
moves at 6 meters per second
along a path at 53
◦
to its original direction,
as shown in on the righthand side.
10 m/s
X
before
Y
after
6 m/s
X
0 m/s
53
◦
Which of the following diagrams best repre
sents the motion of ball
Y
after the collision?
1.
8 m/s
Y
after
2.
after
4 m/s
Y
53
◦
3.
6 m/s
Y
after
4.
after
8 m/s
Y
53
◦
5.
0 m/s
Y
after
6.
after
6 m/s
Y
37
◦
7.
after
8 m/s
Y
37
◦
correct
8.
after
4 m/s
Y
37
◦
9.
4 m/s
Y
after
10.
after
6 m/s
Y
53
◦
Explanation:
From conservation of momentum,
m
1
vectorv
1
i
+
m
2
vectorv
2
i
=
m
1
vectorv
1
f
+
m
2
vectorv
2
f
mvectorv
=
mvectorv
1
+
mvectorv
2
vectorv
=
vectorv
1
+
vectorv
2
.
This relation leads to the vector diagram be
low, where it is used that when two equal
mass have an elastic collision, and one of the
masses was originally at rest, they move per
pendicular to each other after the collision.
vectorv
2
θ
vectorv
1
φ
vectorv
Now 90
◦
=
φ
+
θ
, and with
φ
= 53
◦
θ
= 90
◦
−
53
◦
= 37
◦
Then

vectorv

cos
θ
=

vectorv
2


vectorv
2

= (10 m
/
s) cos 37
◦
= 8 m
/
s
Question 2
part 1 of 1
10 points
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homework 09 – ALIBHAI, ZAHID – Due: Mar 28 2007, 4:00 am
2
Two objects of mass 5 g and 4 g
,
respec
tively, move parallel to the
x
axis, as shown.
The 5 g object overtakes and collides with
the 4 g object. Immediately after the collision,
the
y
component of the velocity of the 5 g
object is 2 m
/
s upward.
v
i
5 g
v
i
4 g
What is the
y
component of the velocity
vectorv
of the 4 g object immediately after the
collision?
1.
vectorv
y
= 6 m
/
s downward
2.
vectorv
y
= 12
.
8 m
/
s downward
3.
vectorv
y
= 2
.
5 m
/
s downward
correct
4.
vectorv
y
= 7
.
2 m
/
s downward
5.
vectorv
y
= 15 m
/
s downward
6.
vectorv
y
= 7 m/s upward
7.
vectorv
y
= 10
.
6667 m
/
s downward
8.
vectorv
y
= 5 m/s upward
Explanation:
Let :
m
1
= 5 g
,
m
2
= 4 g
,
and
v
1
y
= 2 m
/
s
.
Momentum
vectorp
is conserved for collision
m
1
vectorv
1
=
m
2
vectorv
2
.
In particular, the
y
component of the mo
mentum must be the same before or after the
collision. Since the two objects move horizon
tally along the
x
axis, the
y
component of the
momentum before the collision is zero.
Thus, to make the
y
component of the
momentum after the collision zero, the
y

component of the velocity is
0 =
m
1
v
1
y
+
m
2
v
2
y
vectorv
2
y
=
−
m
1
m
2
v
1
y
=
−
(5 g)
(4 g)
(2 m
/
s)
=
−
2
.
5 m
/
s
,
and the direction is downward.
Question 3
part 1 of 1
10 points
Two ice skaters approach each other at right
angles. Skater A has a mass of 71
.
5 kg and
travels in the +
x
direction at 1
.
96 m
/
s. Skater
B has a mass of 55
.
1 kg and is moving in the
+
y
direction at 1
.
09 m
/
s.
They collide and
cling together.
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 Spring '08
 Turner
 Friction, Kinetic Energy, Work, Moment Of Inertia, Correct Answer, kg, Iring

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