 Solution

Let $u>0$ be the projectile's final speed, taking its deflection above the $x$-axis. Momentum conservation givesSubstituting these into kinetic energy conservation, $V^2=u^2+2|{\mathbf{v}}_2{|}^2$, gives $3u^2-2Vu\cos 30^{\circ }-V^2=0$. Its positive root is $u=V(\sqrt{3}+\sqrt{15})/6$. HenceThese vectors satisfy ${\mathbf{v}}_1+2{\mathbf{v}}_2=(V,0)$ and $|{\mathbf{v}}_1{|}^2+2|{\mathbf{v}}_2{|}^2=V^2$. For the reflected deflection, reverse both $y$ components.