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(B) The velocity of each mass just before impact is $v_0 = \sqrt{2gh}$.After the collision,the velocities of the two masses are $v_1 = e_1 v_0$ and $v

Vedclass · Accepted Answer

$(e_1 + e_2)^2 \frac{h}{4}$

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(B) The velocity of each mass just before impact is $v_0 = \sqrt{2gh}$.After the collision,the velocities of the two masses are $v_1 = e_1 v_0$ and $v_2 = e_2 v_0$ in the upward direction.The velocity of the centre of mass $(v_{cm})$ immediately after the collision is given by the average of the individual velocities:$v_{cm} = \frac{v_1 + v_2}{2} = \frac{e_1 \sqrt{2gh} + e_2 \sqrt{2gh}}{2} = \frac{(e_1 + e_2) \sqrt{2gh}}{2}$.Using the kinematic equation $v^2 = u^2 + 2as$,where $v=0$ at maximum height $h'$,$u = v_{cm}$,and $a = -g$:$0 = \left( \frac{(e_1 + e_2) \sqrt{2gh}}{2} \right)^2 - 2gh'$$2gh' = \frac{(e_1 + e_2)^2 (2gh)}{4}$$h' = \frac{(e_1 + e_2)^2 h}{4}$.

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