Two particles each of mass $m$ and travelling with velocities $u_{1}$ and $u_{2}$ collide perfectly inelastically. The loss of energy will be :

\frac{1}{4} m (u_{1} - u_{2})^{2}

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\frac{1}{2} m (u_{1} - u_{2})^{2}

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2 m (u_{1} - u_{2})^{2}

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m (u_{1} - u_{2})^{2}

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Solution

The correct option is A $\frac{1}{4} m (u_{1} - u_{2})^{2}$
Loss in K.E during inelastic collision
If a body of mass $m_{1}$ moving with velocity $u_{1}$ collides inelastically with a stationary mass $m_{2}$ , then the loss in kinetic energy is given by
$K E = \frac{1}{2} \frac{m_{1} m_{2}}{m_{1} + m_{2}} (u_{1} - u_{2})^{2} (1 - e^{2})$
In case of perfectly inelastic collision $e = 0$
then $K E = \frac{1}{2} \frac{m_{1} m_{2}}{m_{1} + m_{2}} (u_{1} - u_{2})^{2}$

For same mass replace $m_{1}$ and $m_{2}$ with $m$ .
that gives us loss of energy $= \frac{1}{4} m {(u_{1} - u_{2})}^{2}$

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