Explain the special cases of elastic collision in one dimension.

Case $1:$ If the two masses are equal $\left(m_{1}=m_{2}\right)$

$v_{1 f}=0, v_{2 f}=v_{1 i}$

The first mass comes to rest and pushes off the second mass with its initial speed on collision.

$\therefore$ Velocity of first mass after collision,

$v_{1 f} =\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right) v_{1 i}$

$=\left(\frac{0}{2 m_{1}}\right) v_{1 i} ~\\ =0 \text { and velocity of second mass }$

$v_{2 f} =\left(\frac{2 m_{1} v_{1 i}}{m_{1}+m_{2}}\right) $

$=\frac{2 m v_{1 i}}{m+m} \quad\left[\because m_{1}=m_{2}=m \text { supposed }\right]$

$v_{2 f} =v_{1 i}$

$\therefore$ Velocity of second mass (ball) is same as velocity of first mass (ball) after collision.

Case $2:$ If $m_{2} \gg m_{1}$ then second ball is more heavier then first one.

$\therefore$ The velocity of first ball after collision,

$v_{1 f}=\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right) v_{1 i}$

Neglecting the $m_{1}$ compared to $m_{2}$,

$v_{1 f}=\frac{\left(0-m_{2}\right)}{\left(0+m_{2}\right)} v_{1 i}$

$\therefore v_{1 f}=-v_{1 i}$

$\therefore$ First ball having same velocity after collision and

$v_{2 f} =\frac{2 m_{1} v_{1 i}}{m_{1}+m_{2}}, \text { here } m_{1}=0$

$=\frac{2 \times 0 \times v_{1 i}}{0+m_{2}}=0$

Hence, there is no effect in velocity of heavier ball and so it remain static at its position.