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:

\[m( v_1 + v_2 ) = mv \]

\[ \Rightarrow v_1 + v_2 = v . . . (2)\]

\[\text{ According to the given condition, }\]

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

\[ \Rightarrow \frac{1}{2}m v_1^2 + \frac{1}{2}m v_2^2 = \frac{3}{4} \times \frac{1}{2}m v^2 \]

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

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

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

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

\[ \Rightarrow e^2 = \frac{1}{2}\]

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