Question

A sphere $P$ of mass m and velocity $\overrightarrow{v_i}$- 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

Answer: (C)

According to law of conservation of linear momentum, we get
$m{\vec{v}}_i+m\times 0=m{\vec{v}}_{Pf}+m{\vec{v}}_{Qf}$
where ${\vec{v}}_{pf}$ and ${\vec{v}}_{Qf}$ are the final velocities of spheres $P$ and $Q$ after collision respectively.
${\vec{v}}_i={\vec{v}}_{Pf}+{\vec{v}}_{Qf}$
$\left({\vec{v}}_i\cdot {\vec{v}}_i\right)=\left({\vec{v}}_{Pf}+{\vec{v}}_{Qf}\right)\cdot \left({\vec{v}}_{Pf}+{\vec{v}}_{Qf}\right)$
$={\vec{v}}_{Pf}\cdot {\vec{v}}_{Pf}+{\vec{v}}_{Qf}\cdot {\vec{v}}_{Qf}+2{\vec{v}}_{Pf}\cdot {\vec{v}}_{Qf}$
or $v_i^2=v_{Pf}^2+v_{Qf}^2+2v_{Pf}{v}_{Qf}\cos \theta$ ...(i)
According to conservation of kinetic energy, we get
$\frac{1}{2}m{v}_i^2=\frac{1}{2}m{v}_{pf}^2+\frac{1}{2}m{v}_{Qf}^2$
$\Rightarrow v_i^2=v_{Pf}^2+v_{Qf}^2$ ...(ii)
Comparing $(i)$ and $(ii)$, we get
$2v_{pf}{v}_{qf}\cos \theta =0$
$\Rightarrow \cos \theta =0$
or $\theta =90^{\circ }$