A ball is let fall from a height `h_(0)`. There are `n` collisions with the earth. If the velocity of rebound after `n` collision is `v_(n)` and the ball rises to a height `h_(n)` then coefficient of restitution `e` is given by

Options

A. ` e^(n) = sqrt((h_(n))/(h_(0)))`
B. ` e^(n) = sqrt((h_(0))/(h_(n)))`
C. `n e = sqrt((h_(n))/(h_(0)))`
D. `sqrt(n e) = sqrt((h_(n))/(h_(0)))`

Answer

A

Explanation

In this problem, the velocity of the earth before and after the collision may be assume zero. Hence, collision will be
`e^(n) = (v_(1))/(v_(2)) xx (v_(2))/(v_(1)) xx (v_(3))/(v_(2)) xx …. (v_(n))/(v_(n) - 1)`
where `v_(n)` is the velocity after nth rebounding and `v_(0)` is the velocity with which the ball strickes the earth first time
Hence. `e^(n) = (v_(n))/(v_(0)) = sqrt((2gh_(n))/(2 gh_(0)))`
where `h_(n)` is the height to which the ball rises time after `n^(th)` rebounding: Hence `e^(n) = (v_(n))/(v_(0)) = sqrt((h_(n))/h_(0))`