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(D) Let the velocity of the ball of mass $m$ just before the collision be $u$. By conservation of energy, $\frac{1}{2}mu^2 = mgh$, so $u = \sqrt{2gh}$

Vedclass · Accepted Answer

$\frac{h}{9}$

Vedclass · Answer

(D) Let the velocity of the ball of mass $m$ just before the collision be $u$. By conservation of energy, $\frac{1}{2}mu^2 = mgh$, so $u = \sqrt{2gh}$.In an elastic collision, both kinetic energy and linear momentum are conserved.Let $v_1$ be the velocity of the ball (in the opposite direction) and $v_2$ be the velocity of the block after the collision.Conservation of momentum: $mu = 2mv_2 - mv_1 \Rightarrow u = 2v_2 - v_1 \quad \dots (i)$Conservation of kinetic energy: $\frac{1}{2}mu^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}(2m)v_2^2 \Rightarrow u^2 = v_1^2 + 2v_2^2 \quad \dots (ii)$From $(i)$, $2v_2 = u + v_1$. Substituting this into $(ii)$:$u^2 = v_1^2 + 2\left(\frac{u+v_1}{2}\right)^2 = v_1^2 + \frac{(u+v_1)^2}{2}$$2u^2 = 2v_1^2 + u^2 + v_1^2 + 2uv_1$$u^2 - 2uv_1 - 3v_1^2 = 0$$(u - 3v_1)(u + v_1) = 0$Since $v_1$ is the speed in the opposite direction, $v_1 = \frac{u}{3}$.The new height $h'$ reached by the ball is given by $mgh' = \frac{1}{2}mv_1^2$.$h' = \frac{v_1^2}{2g} = \frac{(u/3)^2}{2g} = \frac{u^2}{18g} = \frac{2gh}{18g} = \frac{h}{9}$.

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