Two balls of same mass are dropped from the same height h, on to the floor. The first ball bounces to a height `h//4`, after the collision & the second ball to a height `h/16`. The impulse applied by the first & second ball on the floor are `I_(1)` and `I_(2)` respectively. Then :-

Options

A. `5I_(1) = 6I_(2)`
B. `6I_(1) = 5I_(1)`
C. `I_(1) = 2I_(2)`
D. `2I_(1) = I_(2)`

Answer

A

Explanation

For the `I^(st)` ball : `(h)/(4) = e_(1)^(2)h`
For the `II^(nd)` ball `(h)/(16) = e_(2)^(2)h`
Impulse on first ball
`= I_(1) = mv_(0) (1 + e_(1)) = (3)/(2)mv_(0)`
Impulse on second ball
`I_(2) = mv_(0)(1 + e_(2)) = (5)/(4)mv_(0)`
`rArr (I_(1))/(I_(2)) = (3//2 mv_(0))/(5//4 mv_(0)) = (6)/(5) rArr 5I_(1) = 6I_(2)`