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 collisiona are completely inelastic , then the last block starts moving at

Options

A. `((n-1)L)/(v)`
B. `(n (n-1)L)/(2v)`
C. `((n+1)L)/(v)`
D. `(n (n+1)L)/(2v)`

Answer

B

Explanation

Since collision is perfectly inelasticso all the blocks will stick together one by one and move in a form of combined mass.
Time required to cover a displace 'L' by first block `= (L)/(v)`
Now first and second block will stick together and move with `v//2` velocity (by applying conservation of momentum) and combined system will take time
`(L)/(v//2) = (2L)/(v)` to reach up to block third.
Now these three will move with velocity `v//3` and combined system will take time `(L)/(v//3) = (3L)/(v)` to reach upto the block fourth.
So, total time
`= (L)/(v) + (2L)/(v) + (3L)/(v) + .... ((n - 1)L)/(v) = (n(n - 1)L)/(2v)`