Correct Answer - D
The time after which the `1 st` collision takes place `= (L)/(v)`
Velocity after the `1 st` collision , `mv = ( m + m) v_(1) rArr v_(1) = (v)/(2)`
Time between the `1 st` and the `2^(nd)` collision ` = (L)/(v//2) = (2L)/(v)`
Velocity after the `2^(nd)` collision ,
`2m xx (v)/(2) = ( 2m + m) rArr v_(2) = (v)/(3)`
Time between the `2^(nd)` and the `3^(rd)` collision `= (L)/(v//3) = (3 L)/(v)`
Total time up to the `(n - 1)^(th)` collision is
`(L)/(v) + ( 2L)/(v) + (3L)/(v) + ...+ ((n - 1)L)/(v) = ( n ( n -1)L)/( 2v)`
Velocity of the center of mass after the `(n - 1)^(th)` collision `= (v)/(n)`
