A particle of mass m is dropped from a height | vedclass

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(D) Let the downward direction be positive. The position of particle $A$ (dropped) is $y_A = h - \frac{1}{2}gt^2$. The position of particle $B$ (throw

Vedclass · Accepted Answer

$\sqrt{\frac{3}{2}}$

Vedclass · Answer

(D) Let the downward direction be positive. The position of particle $A$ (dropped) is $y_A = h - \frac{1}{2}gt^2$. The position of particle $B$ (thrown up) is $y_B = \sqrt{2gh}t - \frac{1}{2}gt^2$. Collision occurs when $y_A = y_B$: $h - \frac{1}{2}gt^2 = \sqrt{2gh}t - \frac{1}{2}gt^2$,which gives $t = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{h}{2g}}$. The height of collision is $y = h - \frac{1}{2}g(\frac{h}{2g}) = h - \frac{h}{4} = \frac{3h}{4}$. At collision,the velocity of $A$ is $v_A = -gt = -g\sqrt{\frac{h}{2g}} = -\sqrt{\frac{gh}{2}}$. The velocity of $B$ is $v_B = \sqrt{2gh} - gt = \sqrt{2gh} - \sqrt{\frac{gh}{2}} = \sqrt{\frac{gh}{2}}$. Since the collision is completely inelastic,the combined mass $2m$ moves with velocity $v_{cm} = \frac{m v_A + m v_B}{2m} = \frac{-\sqrt{gh/2} + \sqrt{gh/2}}{2} = 0$. The combined mass is at rest at height $H = \frac{3h}{4}$. The time taken to fall from height $H$ is $t' = \sqrt{\frac{2H}{g}} = \sqrt{\frac{2(3h/4)}{g}} = \sqrt{\frac{3h}{2g}} = \sqrt{\frac{3}{2}} \sqrt{\frac{h}{g}}$. Thus,the time in units of $\sqrt{\frac{h}{g}}$ is $\sqrt{\frac{3}{2}}$.

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