Two particles of equal mass mathrm{m} have re | vedclass

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Question

From momentum conservation

$\operatorname{mu\hat{i}}+\operatorname{mu}\left(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{2}\right)=(\mathrm{m}+\mathrm{

Vedclass · Accepted Answer

$\frac{1}{8} \mathrm{mu}^{2}$

Vedclass · Answer

From momentum conservation

$\operatorname{mu\hat{i}}+\operatorname{mu}\left(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{2}\right)=(\mathrm{m}+\mathrm{m}) \overline{\mathrm{v}}$

$\Rightarrow \overline{\mathrm{v}}=\frac{3}{4} \mathrm{u} \hat{\mathrm{i}}+\frac{\mathrm{u}}{4} \hat{\mathrm{j}}$

$\Rightarrow|\mathrm{v}|=\frac{\mathrm{u}}{4} \sqrt{10}$

Final kinetic energy $=\frac{1}{2} 2 \mathrm{m}\left(\frac{\mathrm{u}}{4} \sqrt{10}\right)^{2}=\frac{5}{8} \mathrm{mu}^{2}$

Initial kinetic energy

$=\frac{1}{2} \mathrm{mu}^{2}+\frac{1}{2} \mathrm{m}\left(\frac{\mathrm{u}}{\sqrt{2}}\right)^{2}=\frac{6}{8} \mathrm{mu}^{2}$

Loss in $\mathrm{K.E.}=\mathrm{k}_{\mathrm{i}}-\mathrm{k}_{\mathrm{f}}=\frac{1}{8} \mathrm{mu}^{2}$

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

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