Two particles of equal mass m are moving with velocities $v \hat{i}$ and $v (\frac{\hat{i} + \hat{j}}{2})$ respectively. If the particles collide completely inelastically, then the energy lost in the process is $\frac{{mv}^{2}}{n}$ ,then the value of ‘ n ’ is _____

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answer is 8.

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Detailed Solution

Conserving momentum $mv \hat{i} + m (\frac{v}{2} \hat{i} + \frac{v}{2} \hat{j}) = 2 m (v_{1} \hat{i} + v_{2} \hat{j})$

On solving $V_{1} = \frac{3 v}{4}$ and $V_{2} = \frac{v}{4}$ Change in K.E

$[\frac{1}{2} {mv}^{2} + \frac{1}{2} m {(\frac{v}{2} \sqrt{2})}^{2}] - [\frac{1}{2} (2 m) (\frac{9 v^{2}}{16} + \frac{v^{2}}{16})]$

$= \frac{3 {mv}^{2}}{4} - \frac{5 {mv}^{2}}{8} = \frac{{mv}^{2}}{8}$

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