Question

Two identical billiard balls are in contact on a smooth table. A third identical ball strikes them symmetrically and comes to rest after impact. The coefficient of restitution is

• A. $\frac{2}{3}$
• B. $\frac{1}{3}$
• C. $\frac{1}{6}$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

The correct option is A $\frac{2}{3}$

As we know,
Coefficient of restitution $e=\frac{\text{Velocity of seperation}}{\text{Velocity of approach}}$
The condition of the balls are as shown below



Let the radius of one ball is $r$, joining balls center will make equilateral triangle.
Hence, every angle inside the triangle will be ${60}^{\circ }$

By momentum conservation, assume $u$ be initial velocity of the first ball and $v_1$ and $v_2$ respectively be the final velocity of the two stationary balls after the collision.
Then, by symmetry we can say $v_1=v_2=v$. So, we have
$mu=mv\cos {30}^{\circ }+mv\cos {30}^{\circ }$
$\Rightarrow mu=2mv\cdot \frac{\sqrt{3}}{2}$
$\Rightarrow u=\sqrt{3}v$ ...(1)
Also, we know, from coefficient of restitution, along the line of impact,
$e=\frac{v-0}{u\cos {30}^{\circ }-0}$
$\Rightarrow e=\frac{v}{\frac{u\sqrt{3}}{2}}$
from eq (1)
$\Rightarrow e=\frac{v}{\frac{\sqrt{3}v\times \sqrt{3}}{2}}$
$\Rightarrow e=\frac{2}{3}$
Hence, coefficient of restitution is $e=\frac{2}{3}$.