Version 013 – SECOND EXAM – Kleinman – (58225)
1
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001
10.0 points
Two identical balls (labelled A and B) move
on a frictionless horizontal tabletop. Initially,
ball A moves at speed
v
A,
0
= 10 m
/
s while
ball B is at rest (
v
B,
0
= 0).
The two balls collide offcenter, and after
the collision ball A moves at speed
v
A
= 6 m
/
s
in the direction
θ
A
= 53
◦
from its original
velocity vector:
10 m
/
s
A
before
B
after
6 m
/
s
A
0 m/s
53
◦
Which of the following diagrams best repre
sents the motion of ball B after the collision?
1.
0 m/s
B
after
2.
after
6 m
/
s
B
37
◦
3.
8 m
/
s
B
after
4.
after
6 m
/
s
B
53
◦
5.
after
8 m
/
s
B
53
◦
6.
after
4 m
/
s
B
53
◦
7.
4 m
/
s
B
after
8.
6 m
/
s
B
after
9.
after
4 m
/
s
B
37
◦
10.
after
8 m
/
s
B
37
◦
correct
Explanation:
Because we are given both the speed and
the direction of ball A after the collision, the
problem can be solved in terms of momentum
conservation only.
Indeed, there are no (horizontal) external
forces acting on the two balls, so their net
(horizontal) momentum vector is conserved
during the collision:
vector
P
net
=
mvector
v
A,
0
+
mvector
v
B,
0
[before]
=
mvector
v
A
+
mvector
v
B
[after]
and since
vector
v
B,
0
=
vector
0
(ball B is initially at rest),
vector
v
A
+
vector
v
B
=
vector
v
A,
0
.
Pictorially, this means
vector
v
A
vector
v
B
θ
A
θ
B
vector
v
A,
0
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Version 013 – SECOND EXAM – Kleinman – (58225)
2
and in components,
v
A
×
cos
θ
A
+
v
B
×
cos
θ
B
=
v
A,
0
,
v
A
×
sin
θ
A
−
v
B
×
sin
θ
B
= 0
.
Solving for the two components of ball B’s
velocity, we find
v
B,x
≡
v
B
×
cos
θ
B
=
v
A,
0
−
v
A
cos
θ
A
(1)
= (10 m
/
s)
−
(6 m
/
s) cos 53
◦
= 6
.
4 m
/
s
,
v
B,y
≡
v
B
sin
θ
B
=
v
A
sin
θ
A
(2)
= (6 m
/
s) sin 53
◦
= 4
.
8 m
/
s
,
and hence
v
B
=
radicalBig
v
2
B,x
+
v
2
B,y
=
radicalBig
(6
.
4 m
/
s)
2
+ (4
.
8 m
/
s)
2
= 8 m
/
s
,
θ
B
= arctan
v
B,y
v
B,x
= arctan
4
.
8 m
/
s
6
.
4 m
/
s
= 37
◦
.
Solving an Elastic Collision:
If we know that the collision in question
is perfectly elastic — and it is — then we
do not need to be given both the speed
v
A
and the direction
θ
A
of the ball A after the
collision:
Any one of these two parameters
would determine the other.
The reason for
this is conservation of kinetic energy in an
elastic collision:
K
net
=
m v
2
A,
0
2
+
m v
2
A,
0
2
[before]
=
m v
2
A
2
+
m v
2
B
2
[after]
and therefore
v
2
A
+
v
2
B
=
v
2
A,
0
.
(3)
At the same time, eqs. (1) and (2) above imply
v
2
B
=
v
2
B,x
+
v
2
B,y
=
parenleftBig
v
A,
0
−
v
A
cos
θ
A
parenrightBig
2
+
parenleftBig
v
A
sin
θ
A
parenrightBig
2
=
v
2
A,
0
−
2
v
A,
0
×
v
A
×
cos
θ
A
+
v
2
A
(the cosine theorem), and hence
v
2
A
+
v
2
B
−
v
2
A,
0
=
(4)
= 2
v
2
A
−
2
v
A,
0
×
v
A
×
cos
θ
A
.
Comparing this formula to eq. (3), we imme
diately see that
2
v
2
A
−
2
v
A,
0
v
A
cos
θ
A
= 0
and therefore
v
A
=
v
A,
0
×
cos
θ
A
.
(5)
Note that the problem data indeed satisfy this
equation, which confirms that the collision in
question is perfectly elastic.
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 Spring '08
 Turner
 Physics, Friction, Kinetic Energy, Mass, Moment Of Inertia, m/s

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