InterviewSolution
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InterviewSolution
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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//6`. The impulse applied by the first & second ball on the floor are `I_(1)` and `I_(2)` respectively. Then :-A. `5I_(1) = 6I_(2)`B. `6I_(1) = 5I_(1)`C. `I_(1) = 2I_(2)`D. `2I_(1) = I_(2)` |
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Answer» Correct Answer - A 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)` |
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