A set of n identical cubical blocks lies at r | vedclass

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(B) Since the collisions are perfectly inelastic,all the blocks will stick together one by one and move as a combined mass.Let the mass of each block

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Since the collisions are perfectly inelastic,all the blocks will stick together one by one and move as a combined mass.Let the mass of each block be $m$.Time required for the first block to cover distance $L$ is $t_1 = \frac{L}{v}$.After the first collision,the first and second blocks stick together. By conservation of momentum,$mv = (2m)v_1$,so $v_1 = \frac{v}{2}$.The time taken by this combined system to cover the next distance $L$ is $t_2 = \frac{L}{v/2} = \frac{2L}{v}$.After the second collision,the three blocks move with velocity $v_2 = \frac{mv}{3m} = \frac{v}{3}$.The time taken to cover the next distance $L$ is $t_3 = \frac{L}{v/3} = \frac{3L}{v}$.Continuing this process,the time taken to cover the distance between the $(n-1)$-th and $n$-th block is $t_{n-1} = \frac{(n-1)L}{v}$.The total time taken for the last block to start moving is the sum of these intervals:$T = \frac{L}{v} + \frac{2L}{v} + \frac{3L}{v} + ... + \frac{(n-1)L}{v} = \frac{L}{v} (1 + 2 + 3 + ... + (n-1)) = \frac{L}{v} \cdot \frac{(n-1)n}{2} = \frac{n(n-1)L}{2v}$.Thus,option $(b)$ is correct.

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