Collision is elastic, so both linear momentum and kinetic energy are conserved.

We have following situation,

According to figure, $MV=M{V}^{\text{'}}+mv\dots \left(\mathrm{i}\right)$ (linear momentum

conservation)

$\frac{1}{2}M{V}^2=\frac{1}{2}M{V}^2+\frac{1}{2}m{v}^2$

(kinetic energy conservation)

From Eqs. (i) and (ii), we get

$M\left(V-V^{\text{'}}\right)=mv$

and $M\left(V^2-V^2\right)=m{v}^2$

Dividing Eq. (iv) by Eq. (iii), we have

$\frac{M\left(V^2-V^2\right)}{M\left(V-V^{\text{'}}\right)}=\frac{m{v}^2}{mv}$

or $V+V^{\text{'}}=v$