Question

A ball of mass m moving at a speed v makes a head-on collision with an identical ball at rest. The kinetic energy of the balls after the collision is three fourths of the original. Find the coefficient of restitution.  

Answer 1

Given:
The mass of the both balls is m.
Initial speed of first ball = v  
Initial speed of second ball = 0

Let the final of balls be v₁ and v₂ respectively.

$$e=\frac{\text{ velocity of separation}}{\text{ velocity of approach}}$$

$$\Rightarrow e=\frac{v_1-v_2}{v}$$

$$\Rightarrow v_1-v_2=ev...\left(1\right)$$

On applying the law of conservation of linear momentum, we get:

$$m(v_1+v_2)=mv$$

$$\Rightarrow v_1+v_2=v...(2)$$

$$\text{ According to the given condition, }$$

$$\text{ Final K . E .}=\frac{3}{4}\text{ Initial K . E }.$$

$$\Rightarrow \frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2=\frac{3}{4}\times \frac{1}{2}m{v}^2$$

$$\Rightarrow v_1^2+v_2^2=\frac{3}{4}{v}^2$$

$$\Rightarrow \frac{(v_1+v_2{)}^2+(v_1-v_2{)}^2}{2}=\frac{3}{4}{v}^2$$

$$\Rightarrow \frac{(1+e^2)v^2}{2}=\frac{3}{4}{v}^2\left[\text{ using the equations }\left(1\right)\text{ and }\left(2\right)\right]$$

$$\Rightarrow 1+e^2=\left(\frac{3}{2}\right)$$

$$\Rightarrow e^2=\frac{1}{2}$$

$$\Rightarrow e=\frac{1}{\sqrt{2}}$$
Hence, the coefficient of restitution is found to be$$\frac{1}{\sqrt{2}}$$