Two particles of equal mass m have respective initial velocities $u \hat{i}$ and $u (\frac{\hat{i} + \hat{j}}{2})$ . They collide completely inelastically. The energy lost in the process is :

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$\frac{1}{3} m u^{2}$

$\frac{1}{8} m u^{2}$

$\frac{3}{4} m u^{2}$

$\sqrt{\frac{2}{3}} m u^{2}$

answer is B.

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Detailed Solution

$\begin{array}{l} A p p l y i n g M o m e n t u m C o n s e r v a t i o n, \\ m_{1} {\vec{u}}_{1} + m_{2} {\vec{u}}_{2} = (m_{1} + m_{2}) {\vec{V}}_{c} \\ \Rightarrow m (u \hat{i}) + m u (\frac{\hat{i} + \hat{j}}{2}) = 2 m {\vec{V}}_{c} \\ \Rightarrow {\vec{V}}_{c} = \frac{3}{4} u \hat{i} + \frac{1}{4} u \hat{j} \\ N o w, m a g n i t u d e o f t h e s p e e d s a r e, \\ u_{1} = u \\ u_{2} = \frac{u}{\sqrt{2}} \\ V_{c} = \frac{\sqrt{10}}{4} u \\ K . E l o s s = K_{i} - K_{f} \\ \Rightarrow K . E l o s s = (\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2}) - \frac{1}{2} (m_{1} + m_{2}) V_{c}^{2} \\ = (\frac{1}{2} m u^{2} + \frac{1}{2} m \frac{u^{2}}{2}) - \frac{1}{2} (2 m) \frac{10}{16} u^{2} \\ = \frac{1}{8} m u^{2} \end{array}$