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

Answer:

24 m/s

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,

$v^2=0+2\times 10 \times 20 \times h$

$v=\sqrt{400}=20$ 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, $\mu=1$

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

$\Rightarrow 1=\frac{V_2-2}{22} \\\\\Rightarrow V_2-2=22 \\\\\Rightarrow V_2=22+2$

$\Rightarrow V_2=24$ m/s

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