Question

. A rubber ball dropped from a height of 20m

(g=10m/s2) falls on the roof of an elevator
going up at 2 m/s. If the collision of the ball
with the elevator is elastic, the ball
rebounds with a velocity of​

Answer 1

24 m/s

Step-by-step explanation:

Given that the height of the ball from the elevator, h=20 m

The acceleration due to gravity, g=10 m/
`s^2`

The initial velocity of the ball is zero, so, by using the equation of the motion
`v^2=u^2+2as` the velocity of the ball, v, when reached the elevator,

`v^2=0+2* 10 * 20 * h`

`v=√(400)=20` m/s

The velocity of the ball just before the collision is 20 m/s in the downward direction.

The velocity of the lift just before the collision is 2 m/s in the upward direction.

As the ball and the lift moves in the opposite direction, so the speed of approach
`= 20+2=22` m/s.

Assuming that there is negligible change in the speed of the lift, so the velocity of the lift after the collision is 2 m/s

Let the velocity of the ball after the collision is
`V_2` m/s in the direction of the velocity of the lift.

So, the speed of the separation
`=V_2 -2`

As the collision is elastic, so the coefficient of restitution,
`\mu=1`

`\mu = \frac {\text{Speed of seperation}}{\text{speed of approach}}`

`\Rightarrow 1=(V_2-2)/(22) \\\\\Rightarrow V_2-2=22 \\\\\Rightarrow V_2=22+2 \\\\`

`\Rightarrow V_2=24` m/s

Hence, the ball rebounds with a velocity of 24 m/s.