Question

In the diagram shown block $\text{a}$ moves to right and $\text{f}$ and $\text{g}$ move towards left with a speed $v$. All collisions are elastic, and the blocks are identical. Then after sufficient time

• A. $\text{a}$ will move to left with speed $v$
• B. $a$ will be at rest
• C. $\text{f}$ will keep moving toward left
• D. $\text{g}$ will be at rest

Answer 1

The correct option is A $\text{a}$ will move to left with speed $v$

Given that all are identical blocks and all collisions are elastic, so for this scenario we know that velocities will be interchanged between two bodies after collision.

So, after collision between $\text{a}$ and $\text{b},\text{b}$ will move towards right with velocity $v$ and $\text{a}$ will come to rest.

Likewise, $\text{e}$ will move towards left with velocity $v$ and $\text{f}$ will come to rest.

This process continues and at one stage, $\text{c}$ and $\text{d}$ will be moving in opposite direction with equal velocities and after collision their direction will be reversed and the reverse process happens.

So, after sufficient time Block $\text{a}$ and $\text{b}$ will move towards left and block $\text{g}$ will move towards right and all other blocks will come in rest.


Hence, option $\left(\text{A}\right)$ is the correct choice.