Question

Two balls of identical masses undergo perfectly elastic oblique collsion. One of the balls (Ball 1) is initially at rest while the other (Ball 2) is moving with velocity $v$. After the collisions, what angle will the final velocities of the balls make with each other?

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

Answer 1

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

Applying conservation of linear momentum,


In horizontal direction: $mu+0=m{v}_1\cos {\theta }_1+m{v}_2\cos {\theta }_2...\left(1\right)$
In vertical direction:
$0=m{v}_1\sin {\theta }_1-m{v}_2\sin {\theta }_2...\left(2\right)$

If the collision is perfectly elastic, kinetic energy is also conserved.
$\frac{1}{2}m{u}^2=\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2$
$\Rightarrow u^2=v_1^2+v_2^2...\left(3\right)$
Squaring and adding (1) and (2), we get
$u^2=v_1^2+v_2^2+2v_1{v}_2(\cos {\theta }_1\cos {\theta }_2-\sin {\theta }_1\sin {\theta }_2)$
or $u^2=v_1^2+v_2^2+2v_1{v}_2\cos ({\theta }_1+{\theta }_2)...\left(4\right)$

From equation (3) and (4), we get
$2v_1{v}_2\cos ({\theta }_1+{\theta }_2)=0$
$\Rightarrow \cos ({\theta }_1+{\theta }_2)=0$ or ${\theta }_1+{\theta }_2=\frac{\pi }{2}$ or ${90}^{\circ }$
i.e the masses move at right angle to each other after the collision.