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(B) Initial momentum of the system: $\vec{P}_i = m(u\hat{i}) + m\left(\frac{u}{2}\hat{i} + \frac{u}{2}\hat{j}\right) = \frac{3}{2}mu\hat{i} + \frac{1}

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

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(B) Initial momentum of the system: $\vec{P}_i = m(u\hat{i}) + m\left(\frac{u}{2}\hat{i} + \frac{u}{2}\hat{j}\right) = \frac{3}{2}mu\hat{i} + \frac{1}{2}mu\hat{j}$.Since the collision is completely inelastic,the particles stick together and move with a common velocity $\vec{v}_f$. By conservation of linear momentum: $\vec{P}_i = (m+m)\vec{v}_f = 2m\vec{v}_f$.$\vec{v}_f = \frac{1}{2m} \left(\frac{3}{2}mu\hat{i} + \frac{1}{2}mu\hat{j}\right) = \frac{3}{4}u\hat{i} + \frac{1}{4}u\hat{j}$.Initial kinetic energy: $K_i = \frac{1}{2}mu^2 + \frac{1}{2}m\left|\frac{u}{2}\hat{i} + \frac{u}{2}\hat{j}\right|^2 = \frac{1}{2}mu^2 + \frac{1}{2}m\left(\frac{u^2}{4} + \frac{u^2}{4}\right) = \frac{1}{2}mu^2 + \frac{1}{4}mu^2 = \frac{3}{4}mu^2$.Final kinetic energy: $K_f = \frac{1}{2}(2m)|\vec{v}_f|^2 = m\left(\left(\frac{3}{4}u\right)^2 + \left(\frac{1}{4}u\right)^2\right) = m\left(\frac{9}{16}u^2 + \frac{1}{16}u^2\right) = \frac{10}{16}mu^2 = \frac{5}{8}mu^2$.Energy lost: $\Delta K = K_i - K_f = \frac{3}{4}mu^2 - \frac{5}{8}mu^2 = \frac{6-5}{8}mu^2 = \frac{1}{8}mu^2$.

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

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