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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 the time, $t = \frac{(n - 1) L}{v}$

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

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

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

answer is C.

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Detailed Solution

So, total time

$= \frac{L}{v} + \frac{2 L}{v} + \frac{3 L}{v} + ... \frac{(n - 1) L}{v}$ $= \frac{n (n - 1) L}{2 v}$

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