Two identical balls of masses $5 kg$ each are kept on a frictionless surface as shown in the figure below. One of the balls moving at $4 m/s$ makes a head-on collision with the other at rest initially. It is observed that final kinetic energy of the balls is half of the initial kinetic energy. Find the value of the coefficient of restitution.

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Solution

Given, mass of each ball $m_{1} = 5 kg; m_{2} = 5 kg$
speed of ball 1 before collision $u_{1} = 4 m/s$
speed of ball 2 before collision $u_{2} = 0 m/s$
Let $v_{1}$ and $v_{2}$ be the velocities after collision.
Applying linear momentum conservation principle :
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
or $v_{1} + v_{2} = 4$ ......(i)
(since $m_{1} = m_{2}$ )
Also $e = \frac{v_{2} - v_{1}}{u_{1} - u_{2}} == \frac{v_{2} - v_{1}}{4 - 0}$
or $v_{2} - v_{1} = 4 e$ .......(ii)

Given $K_{f} = \frac{1}{2} K_{i}$
$\Rightarrow \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2} = \frac{1}{2} [\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2}]$
$\Rightarrow 2 (v_{1}^{2} + v_{2}^{2}) = (u_{1}^{2} + u_{2}^{2})$
(since $m_{1} = m_{2}$ )
$\Rightarrow (v_{1} + v_{2})^{2} + (v_{1} - v_{2})^{2} = 4^{2} + 0^{2}$
$\Rightarrow 16 + 16 e^{2} = 16$ (using (i) and (ii))
$\Rightarrow e = 0$

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