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(C) $1$. At maximum height,the velocity of the first particle is $v_x = u \cos 60^{\circ} = \frac{u}{2}$ and $v_y = 0$.$2$. The second particle has ve

Vedclass · Accepted Answer

$\frac{3 \sqrt{3}}{8} \frac{u^{2}}{g}$

Vedclass · Answer

(C) $1$. At maximum height,the velocity of the first particle is $v_x = u \cos 60^{\circ} = \frac{u}{2}$ and $v_y = 0$.$2$. The second particle has velocity $u \hat{i}$.$3$. By conservation of linear momentum in the horizontal direction:$m \left( \frac{u}{2} \right) + m(u) = (m + m) v^{\prime}$$\frac{3mu}{2} = 2mv^{\prime} \implies v^{\prime} = \frac{3u}{4}$.$4$. The maximum height $H$ is given by $H = \frac{u^2 \sin^2 60^{\circ}}{2g} = \frac{u^2 (3/4)}{2g} = \frac{3u^2}{8g}$.$5$. The time taken to fall from height $H$ to the ground is $t = \sqrt{\frac{2H}{g}} = \sqrt{\frac{2}{g} \cdot \frac{3u^2}{8g}} = \sqrt{\frac{3u^2}{4g^2}} = \frac{u \sqrt{3}}{2g}$.$6$. The horizontal distance covered by the combined mass after collision is $d = v^{\prime} \cdot t = \left( \frac{3u}{4} \right) \left( \frac{u \sqrt{3}}{2g} \right) = \frac{3 \sqrt{3} u^2}{8g}$.

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