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

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The last block starts moving at $t = \frac{(n - 1) L}{v}$

The last block starts moving at $t = \frac{n (n - 1) L}{2 v}$

The centre of mass of the system will have a final speed v

The centre of mass of the system will have a final speed $\frac{v}{n}$

answer is B, D.

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Detailed Solution

$t_{1} = \frac{l}{v}, t_{2} = \frac{l}{v / 2} = \frac{2 l}{v}, t_{3} = \frac{3 l}{v}, ........, t_{n - 1} = \frac{(n - 1)}{v} l$

$S o, T = \frac{l}{v} + \frac{2 l}{v} + \frac{3 l}{v} + ..... + \frac{(n - 1)}{v} l = \frac{l}{v} [1 + 2 + 3 + .... + (n - 1)] = \frac{n (n - 1) l}{2 v}$

$v_{c m} = \frac{m v + 0 + 0 + ....... + 0}{n m} = \frac{v}{n}$

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