Q.
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 collision are completely elastic. Then
A) The last block starts moving at time $t = (n - 1)\frac{L}{v}\ $
B) The last block starts moving at time $t = \frac{(n - 1)L}{2v}$
C) The centre of mass of the system will have a final speed $v$
D) The centre of mass of the system will have a final speed $v/n$
Explanation
(d) Time taken by first block to reach second block $= \frac{L}{v}$
Since collision is 100% elastic, now first block comes to rest and 2nd block starts moving towards the 3rd block with a velocity $v$ and takes time $= \frac{L}{v}$ to reach 3rd block and so on
$\therefore$ Total time $= t + t + ...(n - 1)$ time $= (n - 1)\frac{L}{v}$
Finally only the last $n$th block is in motion velocity $v$, hence final velocity of centre of mass
$v_{CM} = \frac{mv}{nm} = \frac{v}{n}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.