A glass marble dropped from a certain height above the horizontal surface reaches the surface in time t and then continues to bounce up and down. The time in which the marble finally comes to rest is
(where e is the coefficient of restitution) 

Options

A. `e^(n)t`
B. `e^(2)t`
C. `t[(1 + e)/(1 - e)]`
D. `t[(1 - e)/(1 + e)]`

Answer

C

Explanation

Time taken by glass marble to fall through height h is `t = sqrt((2h)/(g))`.
After first rebound, time taken to rise to height `h_1` and coming back
`= 2sqrt((2h_1)/(g)) = 2sqrt((2e^2h)/(g)) = 2e sqrt((2h)/(g))`
Time taken in rising to height, `h_2` and coming back = `2e^2 sqrt((2h)/(g))`
Total time taken by the glass marble in coming to rest
`= sqrt((2h)/(g)) + 2e sqrt((2h)/(g)) + 2e^2 sqrt((2h)/(g)) + .........`
`= sqrt((2h)/(g)) + 2e sqrt((2h)/(g)) [1 + e + e^2 + ........]`
`= sqrt((2h)/(g)) + 2e sqrt((2h)/(g)) xx 1/(1 - e) = sqrt((2h)/(g)) [1 + (2e)/(1 - e)] = t [(1 + e)/(1 - e)]` .