Explain the special cases of elastic collision in one dimension.

(A) Case $1$: If the two masses are equal $(m_{1} = m_{2} = m)$:
$v_{1f} = \left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right) v_{1i} + \left(\frac{2m_{2}}{m_{1}+m_{2}}\right) v_{2i}$. Assuming $v_{2i} = 0$,we get $v_{1f} = 0$ and $v_{2f} = v_{1i}$. The masses exchange their velocities.
Case $2$: If $m_{2} \gg m_{1}$ (a light body strikes a very heavy stationary body):
$v_{1f} = \left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right) v_{1i} \approx \left(\frac{-m_{2}}{m_{2}}\right) v_{1i} = -v_{1i}$. The light body rebounds with the same speed.
$v_{2f} = \left(\frac{2m_{1}}{m_{1}+m_{2}}\right) v_{1i} \approx 0$. The heavy body remains practically at rest.
Case $3$: If $m_{1} \gg m_{2}$ (a heavy body strikes a light stationary body):
$v_{1f} = \left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right) v_{1i} \approx \left(\frac{m_{1}}{m_{1}}\right) v_{1i} = v_{1i}$. The heavy body continues to move with almost the same velocity.
$v_{2f} = \left(\frac{2m_{1}}{m_{1}+m_{2}}\right) v_{1i} \approx \left(\frac{2m_{1}}{m_{1}}\right) v_{1i} = 2v_{1i}$. The light body moves with twice the velocity of the heavy body.