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(C) $1$. The velocity of the ball just before impact is $u = \sqrt{2gh}$.$2$. Let the vertical velocity of the ball after impact be $v_b$ and the vert

Vedclass · Accepted Answer

$0.3 \sqrt{2gh}$

Vedclass · Answer

(C) $1$. The velocity of the ball just before impact is $u = \sqrt{2gh}$.$2$. Let the vertical velocity of the ball after impact be $v_b$ and the vertical velocity of the block be $0$ (since it is constrained to move horizontally).$3$. By the definition of the coefficient of restitution $e$ for the vertical direction: $e = \frac{v_b - 0}{u - 0} \implies v_b = e \cdot u = 0.5 \sqrt{2gh}$.$4$. The impulse $J$ exerted by the block on the ball is equal to the change in momentum of the ball: $J = m(v_b - (-u)) = m(v_b + u) = m(0.5\sqrt{2gh} + \sqrt{2gh}) = 1.5m\sqrt{2gh}$.$5$. By Newton's third law,the ball exerts an equal and opposite impulse $J$ on the block in the downward direction.$6$. The normal force $N$ between the block and the surface is $N = mg + J/\Delta t$,where $\Delta t$ is the duration of impact. The frictional impulse on the block is $f \Delta t = \mu N \Delta t = \mu (mg \Delta t + J) = \mu mg \Delta t + \mu J$.$7$. As $\Delta t 	o 0$,the term $\mu mg \Delta t$ becomes negligible. Thus,the change in momentum of the block is $\Delta p_{block} = \mu J = 0.2 	imes 1.5m\sqrt{2gh} = 0.3m\sqrt{2gh}$.$8$. Since $\Delta p_{block} = m(v - v')$,the decrease in velocity is $v - v' = \frac{0.3m\sqrt{2gh}}{m} = 0.3\sqrt{2gh}$.

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