Question

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

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is D

${90}^{\circ }$

Applying law of conservation of momentum along direction of initial velocity we get

$mu=mv(cos{\theta }_1+cos{\theta }_2)$

Applying law of conservation of momentum perpendicular to direction of initial velocity

$0=mvsin{\theta }_1-mvsin{\theta }_2\Rightarrow {\theta }_1={\theta }_{2}mu=2mvcos\theta [{\theta }_1={\theta }_2=\theta ]cos\theta =\frac{u}{2v}\left(ii\right)$

According to law of conservation of KE,

$\frac{1}{2}m{u}^2=\frac{1}{2}m{v}^2+\frac{1}{2}m{v}^2{u}^2=2v^{2}oru=\sqrt{2}v\left(iii\right)$ From Eqs. (ii) and (iii), we have $cos\theta =\frac{\sqrt{2}v}{2v}$

$orcos\theta =\frac{1}{\sqrt{2}}or\theta ={45}^{\circ }{\theta }_1+{\theta }_2={90}^{\circ }$