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 at `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

`mv = nv'm rArr v' =`
time for first collisen is `t_(1) = (L)/(V)` (`2nd` block)
`2`nd collisions `t_(2) = (2l)/(V) = 2t_(1)` (`3`rd block)
so `t = t_(1) + 2l_(1) + 3t_(1) + at_(1)"….."(n-1)t_(1)`.
`t = t_(1)[1 + 2 + 3]"........."(n-1)]`
`= ((n-1)(n-1+1))/(2) = (n(n-1))/(2)`
so `t = (L)/(2V)n(n - 1)`