233k views
2 votes
. 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​

User Overburn
by
7.2k points

1 Answer

2 votes

Answer:

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.

User Drewjoh
by
7.1k points