\(M v+m \times 0=M v_{1}+m v_{2}\)

\(\Rightarrow M\left(v-v_{1}\right)=m v_{2} \ldots\) \((i)\)

Again, from the conservation of kinetic energy (as collision is of elastic nature)

\(\frac{1}{2} M v^{2}+\frac{1}{2} m \times 0=\frac{1}{2} M v_{1}^{2}+\frac{1}{2} m v_{2}^{2}\)

\(\Rightarrow M\left(v^{2}-v_{1}^{2}\right)=m v_{2}^{2} \ldots(\) ii \()\)

On solving equations \((i)\) and \((ii),\) we get

\(\frac{M\left(v-v_{1}\right)}{M\left(v+v_{1}\right)\left(v-v_{1}\right)}=\frac{m v_{2}}{m v_{2}^{2}}\)

\(v_{2}=v+v_{1} \ldots\) \((iii)\)

Now, solving equations \((i)\) and \((iii)\), we get

\(v_{1}=\frac{(M-m) v}{(M+m)}\) and \(v_{2}=\frac{2 M v}{(M+m)}\)

As \(M \gg m\)

\(So , v_{1}=v \Rightarrow v_{2}=v+v=2 v\)