Explanation

(b) Total initial momentum=Total final momentum
$ie\ \ \ \ \ \ m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}$
$\therefore\ \ \ M \times v + m \times 0 = Mv_{1} + mv_{2}$
$or\ \ Mv = Mv_{1} + Mv_{2}$
Or $M\left( v - v_{1} \right) = mv_{2}\ \ \ \ \ \ \ldots\ldots(i)$
Again kinetic energy is also conserved.
$\ \ \ \ \ \ \ \frac{1}{2}m_{1}u_{1}^{2} + \frac{1}{2}m_{2}u_{2}^{2} = \frac{1}{2}m_{1}v_{1}^{2} + \frac{1}{2}m_{2}v_{2}^{2}$
$\therefore\ \ \ Mv^{2} + m \times 0 = Mv_{1}^{2} + mv_{2}^{2}$
$or\ \ Mv^{2} = Mv_{1}^{2} + mv_{2}^{2}$
Or $M\left( v^{2} - v_{1}^{2} \right) = mv_{2}^{2}\ \ \ldots\ldots..(ii)$
Dividing Eq.(ii)by Eq.(i), we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{M\left( v^{2} - v_{1}^{2} \right)}{M\left( v - v_{1} \right)} = \frac{mv_{2}^{2}}{mv_{2}}$
Or $\ v + v_{1} = v_{2}$
As M $> > m,\ so\ v_{1} = v\ $
$\therefore{\ \ \ \ v}_{2} = v + v = 2v$