Question

A ball falls from a height such that it strikes the floor of lift at 10 m/s. If lift is moving in the upward direction with a velocity 1 m/s, then velocity with which the ball rebounds after elastic collision will be
A. 11 m/s
B. 10 m/s
C. 12 m/s
D. 9 m/s

Answer 1

Hint: Two particles are moving in the same direction in a straight line with velocities $u_1$ and $u_2$ respectively. Before the collision their velocity of approach will be $(u_1-u_2)$
After collision velocity of the first particle and second particle are $v_1$ and $v_2$ respectively. Therefore, velocity of separation will be $(v_2-v_1)$.
Coefficient of restitution is the ratio of relative velocity of separation after collision to relative velocity of approach before collision.
Coefficient of restitution, $e={\displaystyle \frac{v_2-v_1}{u_1-u_2}}$
For, elastic collision, $e=1$.

Complete step by step answer:
Let us consider ball as particle 1 and lift as particle 2.
See the diagram before collision:

Taking upward direction as positive,
Velocity of ball before collision, $u_1=-10$m/s
Velocity of lift before collision, $u_2=1$m/s
Now, we should consider the situation after the collision.
After collision lift will continue it’s motion 1m/s upwards and the ball will rebound with a velocity, say $v_2$ in upward direction.
So, Velocity of lift after collision, $v_2=1$m/s
Velocity of ball after collision be $v_1$m/s (let)
As it is an elastic collision, then$e=1$.
We know, $e={\displaystyle \frac{v_2-v_1}{u_1-u_2}}$
Now, putting the values,
$1={\displaystyle \frac{1-v_1}{-10-1}}$
$⟹1-v_1=-11$
$\therefore v_1=11+1=12$m/s
So, Ball rebounds with 12m/s velocity upwards after elastic collision.

So, the correct answer is “Option C”.

Note:
We can take $e=1$, when the collision is elastic. We can’t take $e=1$for inelastic or partially elastic collision. For, perfectly inelastic collision, $e=0$and for partially elastic collision $0<e<1$.