Question

A sphere P of mass m and velocity $\overrightarrow{v}$ undergoes an oblique and perfectly elastic collision with an identical sphere Q initially at rest. The angle $\theta$ between the velocities of the spheres after the collision shall be

Answer 1

According to law of conservation of linear momentum, we get
$m\overrightarrow{v_i}+m\times 0=m{\overrightarrow{v}}_{pf}+m{\overrightarrow{v}}_{Qf}$
where ${\overrightarrow{v}}_{pf}$ and ${\overrightarrow{v}}_{Qf}$ are the final velocities of spheres P and Q respectively.
$\overrightarrow{v_i}=\overrightarrow{v_{pf}}+\overrightarrow{v_{Qf}}$

$(\overrightarrow{v_i}.\overrightarrow{v_i})=({\overrightarrow{v}}_{pf}+{\overrightarrow{v}}_{Qf}).({\overrightarrow{v}}_{pf}+{\overrightarrow{v}}_{Qf})$
$={\overrightarrow{v}}_{pf}.{\overrightarrow{v}}_{pf}+{\overrightarrow{v}}_{Qf}.{\overrightarrow{v}}_{Qf}+2{\overrightarrow{v}}_{pf}.{\overrightarrow{v}}_{Qf}$
or $v_i^2=v_{pf}^2+v_{Qf}^2+2v_{pf}{v}_{Qf}\text{cos}\theta \left(1\right)$
According to conservation of kientic energy, we get
$\frac{1}{2}m{v}_i^2=\frac{1}{2}m{v}_{pf}^2+\frac{1}{2}m{v}_{Qf}^2$
$v_i^2=v_{pf}^2+v_{Qf}^2$
Comparing Eqs. (1) and (2), we get $\text{cos}\theta =0\Rightarrow \theta ={90}^{\circ }$