NEET-XII-Physics
pp year:2018
- #16A moving block having mass m, collides with
another stationary block having mass 4m. The
lighter block comes to rest after collision.
When the initial velocity of the lighter block
is v, then the value of coefficient of
resistitution (e) will be :-
(1) 0.5
(2) 0.25
(3) 0.8
(4) 0.4digAnsr: 2Ans : (2)
Sol. By conservation of linear momentum
mv = 4mv' &implies;ï€ =
v
v '
4
coefficient of restitution (e) =
Velocity of separation
Velocity of approach
ï€
=
ï€
v
0
4
v 0
= =
1
0.25
4