Question

A sphere $P$ of mass $m$ and velocity $v_i$ 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

• A. $0$
• B. ${45}^{\circ }$
• C. ${90}^{\circ }$
• D. ${180}^{\circ }$

Answer 1

The correct option is C ${90}^{\circ }$

According to the 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 $Q$ after collision respectively.

$\overrightarrow{v_i}=\overrightarrow{v_{Pf}}+\overrightarrow{v_{Qf}}$

$(\overrightarrow{v_i}\cdot \overrightarrow{v_i})=(\overrightarrow{v_{Pf}}+\overrightarrow{v_{Qf}})\cdot (\overrightarrow{v_{Pf}}+\overrightarrow{v_{Qf}})$

$=\overrightarrow{v_{Pf}}\cdot \overrightarrow{v_{Qf}}\cdot \overrightarrow{v_{Qf}}+2\overrightarrow{v_{Pf}}\cdot \overrightarrow{v_{Qf}}$

or $v_i^2=v_{Pf}^2+v_{Qf}^2+2v_{Pf}{v}_{Qf}\cos \theta ....\left(i\right)$

According to the conservation of kinetic energy, we get

$\frac{1}{2}m{v}_{i}2=\frac{1}{2}m{v}_{Pf}^2+\frac{1}{2}m{v}_{Qf}^2\Rightarrow v_i^2=v_{Pf}^2+v_{Qf}^2....\left(ii\right)$

Comparing (i) and (ii), we get

$2v_{Pf}{v}_{qf}\cos \theta =0$

$\Rightarrow \cos \theta =0$ or $\theta ={90}^{\circ }$.