Question

A sphere P of mass m and velocity ${\overrightarrow{V}}_1$ indergoes an oblique and perfectly elastic collision with an identical sphete Q initially at rest. The angle $\theta$ between the velocities of the spheres after the collision shall be :-

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

Answer 1

The correct option is C ${90}^o$

According to law of conservation of linear momentum, we get

$m\overrightarrow{v_i}+m\times 0=mv\overrightarrow{P_f}+mv\overrightarrow{Q_f}$

where $v\overrightarrow{P_f}$ and $v\overrightarrow{Q_f}$ are the final velocities of spheres $P$ and $Q$ after collision respectively.

$\overrightarrow{v_i}=v\overrightarrow{P_f}+v\overrightarrow{Q_f}$

$(\overrightarrow{v_i}.\overrightarrow{v_i})=(v\overrightarrow{P_f}+v\overrightarrow{Q_f}).(v\overrightarrow{P_f}+v\overrightarrow{Q_f)}$

$=v\overrightarrow{P_f}.v\overrightarrow{Q_f}.v\overrightarrow{Q_f}+2v\overrightarrow{P_f}.v\overrightarrow{Q_f}$

or

$v_i^2=v^2{P}_f+v^2{Q}_f+2v{P}_{f}v{Q}_{f}cos\theta$ ...........(1)

According to conservation of kinetic energy, we get

$\frac{1}{2}m{v}_i^2=\frac{1}{2}m{v}^2{P}_f+\frac{1}{2}m{v}^2{Q}_f⟹v_i^2=v^2{P}_f+v^2{Q}_f$ .............(2)

Comparing (1) and (2) we get,

$2v{P}_{f}v{Q}_f=0⟹cos\theta =0$ or $\theta ={90}^0$