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?

Answer 1

We consider initial velocity of the ball as u1=-10 m/s with reference to ground, We have taken downward direction as negative. The initial velocity of lift is u2=1m/s. After the collision let the velocities be v1 and v2. Since collision is elastic ,the coefficient of restitution e=1.

Then,  1=(v1-v2)/(u2-u1) or u2-u1=v1-v2. Putting values of u1 and u2 and taking v2=u2 as lift is very heavy, we have ,  v1=2+10=12m/s for observer on ground.

For observer in the lift the lift is stationary and velocity of the ball before and after collision has same value with reversed direction . So, answer for that observer will be 10 m/s.

Answer 2

Answer:

The velocity with which the ball rebounds after the elastic collision will be 12 m/s.

Explanation:

We know that, in an elastic collision of two objects in which mass of one object is very much greater than the other and if they are travelling in opposite direction, their velocity is given by;-

                v = 2u₁ + u₂

Let the velocity of ball moving downwards on the floor be u₁. Let the velocity of lift going upwards be u₂. Let the velocity with which the ball rebounds be v. Then;-

                 v = 2 × 1 + 10

                 v = 2 + 10

            v = 12 m/s

Hope it helps! ;-))