Question

There are $100$ identical blocks equally spaced on a frictionless track as shown in the figure. Initially all the blocks are at rest. Block $\left(1\right)$ is pushed with velocity $v$ towards block $2$. If each of the collisions is elastic, then the velocity of the final ${100}^{th}$ block is

• A. $\frac{v}{99}$
• B. $\frac{v}{100}$
• C. $v$
• D. zero

Answer 1

The correct option is C $v$


As we know, for elastic collision
$e=\frac{\text{Speed of separation}}{\text{Speed of approach}}=1$
Let $m$ be the mass of each block.
From momentum conservation between blocks $\left(1\right)$ and $\left(2\right)$
( $\because$ no horizontal force is present )
$mv+0=m{v}_1+m{v}_2$
$v=v_1+v_2...\left(i\right)$
Here $v_1$ & $v_2$ are the speeds of $1$ & $2$ block after the collision towards right.
Also $e=\frac{v_2-v_1}{v-0}=1$
$v_2-v_1=v...\left(ii\right)$

From equation $\left(i\right)$ & $\left(ii\right)$,
$v_1=0$ & $v_2=v$

Thus, we can see that after the collision, first block will come to rest, and the second block will start moving with speed $v$ towards right, which is equal to the speed of $1st$ block before the collision. The same process will contiue for the rest of blocks.
Hence for ${99}^{th}$ & ${100}^{th}$ block :
After the ${99}^{th}$ collision between ${99}^{th}$ & ${100}^{th}$block, speed of ${100}^{th}$ block will be $v$ towards right.