Question

A sphere has an elastic oblique collision with another identical sphere which is initially at rest. The angle between their velocities after the collision is

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

Answer 1

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

Hint: “Elastic oblique collision”

Formula used:

$P_i=P_f$

$K{E}_i=K{E}_f$

Solution:

Let the equal masses be m and their initial velocities be $u$ and zero respectively.

Since momentum is a vector, according to conservation of momentum we have
$mu=m{v}_1\cos {\theta }_1+m{v}_2\cos {\theta }_2$
i.e., $u=v_1\cos {\theta }_1+v_2\cos {\theta }_2\cdot \cdot \left(i\right)$
and $0=v_1\sin {\theta }_1-v_2\sin {\theta }_2\cdots \left(ii\right)$
Being an elastic collision, kinetic energy is also conserved.
$\therefore \frac{1}{2}m{u}^2=\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2$
$\therefore u^2=v_1^2+v_2^2\cdots \left(iii\right)$
Squaring and adding$\left(i\right)$ and $\left(ii\right)$, we get
$u^2=v_1+v_2^2+2v_1{v}_2(\cos {\theta }_1\cos {\theta }_2-\sin {\theta }_1\sin {\theta }_2)$
$u^2=v_1^2+v_2^2+2v_1{v}_2\cos ({\theta }_1+{\theta }_2)$
Using $\left(iii\right)$ in the above equation, we have
$2v_1{v}_2\cos ({\theta }_1+{\theta }_2)=0$
$\therefore \cos ({\theta }_1+{\theta }_2)=0$
${\theta }_1+{\theta }_2=\pi /2$
i.e., the masses move at right angles after the collision.
Final Answer: (d)