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. The last block starts moving at `t = n (n - 1) (L)/(2v)`
B. The last block starts moving `t = (n - 1) (L)/(v)`
C. The centre of mass of the system will have a final speed `(v)/(n)`
D. The centre of mass of the system will have a final speed `v`.

Answer

A, C

Explanation

`t_(1) = (L)/(v)` (time for `1st` collision)
`t_(2) = (2L)/(v)` (time for IInd collision )
`t_(3) = (3L)/(v)` (time for 3rd collision)
`t_(n - 1) = (L)/(v) (n - 1)` (time for `(n^(th))` collision)
`underset(1 - 1)overset(h)sum t_(1) = (n(n - 1))/(2) (L)/(v)`