A ball collides elastically with another ball of the same mass. The collision is oblique and initially one of the ball was at rest. After the collision, the two balls move with same speed. What will be the angle between the velocity of the balls after the collision ?

Applying the law of conservation of momentum along X and Y axes, we have

$$\begin{gathered} \mathrm{mu}=\mathrm{mvcos}{\mathrm{\theta }}_1+\mathrm{mvcos}{\mathrm{\theta }}_2 \\ \text{ and }\mathrm{mvsin}{\mathrm{\theta }}_1=\mathrm{mvsin}{\mathrm{\theta }}_2 \\ \therefore {\mathrm{\theta }}_1={\mathrm{\theta }}_2=\mathrm{\theta }\text{ (say) } \\ \text{ So, }\mathrm{u}=\mathrm{vcos}\mathrm{\theta }+\mathrm{vcos}\mathrm{\theta }\text{ or }\cos \mathrm{\theta }=\frac{\mathrm{u}}{2\mathrm{v}} \\ \mathrm{Applying} \mathrm{the} \mathrm{law} \mathrm{of} \mathrm{conservatioa} \mathrm{of} \mathrm{energy}, \mathrm{we} \mathrm{get} \\ \frac{1}{2}{\mathrm{mu}}^2=\frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}{\mathrm{mv}}^2=\frac{1}{2}\mathrm{m}\left(2{\mathrm{v}}^2\right) \\ \mathrm{or} \mathrm{u}=\sqrt{2}\mathrm{v} \\ \mathrm{From} \mathrm{eqs} \left(3\right) \mathrm{ad} \left(4\right), \mathrm{we} \mathrm{have} \\ \cos \mathrm{\theta }=\frac{\sqrt{2}\mathrm{V}}{2\mathrm{V}}=\frac{1}{\sqrt{2}}\text{ or }\mathrm{\theta }={45}^{\circ } \\ \therefore {\mathrm{\theta }}_1+{\mathrm{\theta }}_2={90}^{\circ } \end{gathered}$$