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Physics Question on work, energy and power

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

A

34mu2\frac{3}{4} mu^{2}

B

23mu2\sqrt{\frac{2}{3}} mu^{2}

C

13mu2\frac{1}{3} mu^{2}

D

18mu2\frac{1}{8} mu^{2}

Answer

18mu2\frac{1}{8} mu^{2}

Explanation

Solution

From momentum conservation
mui^+mu(i^+j^2)=(m+m)vˉmu\hat{i}+mu\left(\frac{\hat{i}+\hat{j}}{2}\right)=\left(m+m\right)\bar{v}
⇒vˉ=34ui^+u4j^\Rightarrow \bar{v}=\frac{3}{4}u\hat{i}+\frac{u}{4}\hat{j}
⇒∣v∣=u410\Rightarrow \left|v\right|=\frac{u}{4}\sqrt{10}
Final kinetic energy =122m(u410)2=58mu2=\frac{1}{2}2m\left(\frac{u}{4}\sqrt{10}\right)^{2}=\frac{5}{8}mu^{2} Initial kinetic energy
=12mu2+12m(u2)2=68mu2=\frac{1}{2}mu^{2}+\frac{1}{2}m\left(\frac{u}{\sqrt{2}}\right)^{2}=\frac{6}{8}mu^{2}
Loss in K.E.=ki−kf=18mu2K.E.=k_{i}-k_{f}=\frac{1}{8}mu^{2}