A set of a identical cubical blocks lies at rest parallel to each other along a line on a smooth horizontal surface. The separation between the near surface of any two adjacent blocks is `L`. The block at one and is given a speed `v` towards the next one at time `t = 0`. All collision are completely inelastic , then the last block starts moving at

Options

A. `(i) ,(iii)`
B. `(i) ,(iv)`
C. `(ii) ,(iii)`
D. `(ii) ,(iv)`

Answer

D

Explanation

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)`