Consider a collision between two pool balls. Ball 1 is at rest and ball 2 is moving towards it with a speed of 2 ms-1 . The mass of each ball is 0.3 kg. After the balls collide elastically, ball 2 comes to an immediate stop and ball 1 move off. What is the final velocity of ball 1?
Answers
Answer:
ANSWER
Solving the equation of conservation of momentum give us that the relative velocity of approach is equal to the relative velocity of separation. Hence coefficient of restitution is 1.
Which means collision is elastic.
Explanation:
ANSWER
Let v
0
be the initial velocity of the first ball before collision and v be the velocity of the other ball after collision.
From conservation of linear momentum,
m
1
v
0
=−m
1
3
v
0
+m
2
v.....(1)
Velocityofapproach
Velocityofseparation
=e
⇒
v
0
v+
3
v
0
=e=1 (for elastic collision e=1)
⇒v+
3
v
0
=v
0
⇒v=
3
2v
0
from equation (1)
m
1
v
0
=−m
1
3
v
0
+m
2
3
2v
0
⇒m
1
v
0
+m
1
3
v
0
=m
2
3
2v
0
⇒
3
4
m
1
v
0
=
3
2
m
2
v
0
$
⇒m
2
=
2
3
×
3
4
m
1
=2m
1
m
2
=2×0.1=0.2kg
