Question

A set of $n$-identical cubical blocks lies at rest along a line on a smooth horizontal surface. The separation between 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

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

Answer: (B), (C)

The last block starts moving at $t=\frac{(n+1)L}{v}$
The centre of mass of the system will have a final speed $v/n$

Time taken by the first block to reach second $=\frac{L}{V}$

Collision is elastic, therefore the velocity of first block becomes zero while that of second block becomes $V$, in the same direction. It will hit the third block afetr an additional time of $\frac{L}{V}$ seconds. Therefore the third starts moving at $(\frac{L}{V}+\frac{L}{V})=\frac{2L}{V}$ secondsfrom start. In a similar way the fourth block starts moving at $\frac{3L}{V}$ second and the $n^{th}$ block starts moving at $(n-1)\frac{L}{V}$ seconds.

In the end only last block will remain moving with a speed $V$, all others will stop. Therefore the velocity of centre of mass of the system will be

$V_{cm}=\frac{mV}{m+m+.......nterms}=\frac{V}{n}$

So the correct answer is option (B) and (C)