Physics - Center of Mass Question with Solution | TestHub

$(\mathrm{P})$ Up to $t_1$: momentum conservation gives $mv=3mv\prime$, so $v\prime =\frac{v}{3}$. The block rises on the wedge; work by gravity $=-\Delta PE=-\frac{m{v}^2}{3}$ (P to 4).

$(\mathrm{Q})$ Work-energy on block A: $W_N+W_g=\Delta KE=\frac{m{v}^2}{18}-\frac{m{v}^2}{2}=-\frac{4m{v}^2}{9}$, so $W_N=-\frac{4m{v}^2}{9}+\frac{m{v}^2}{3}=-\frac{m{v}^2}{9}$ (Q to 1).

$(\mathrm{R})$ From $t_1$ to $t_2$ the interaction is elastic: the block rebounds at $-\frac{v}{3}$ and the wedge moves at $\frac{2v}{3}$; work by the block's normal on the wedge $=\Delta K{E}_{wedge}=\frac{4m{v}^2}{9}-\frac{m{v}^2}{9}=\frac{m{v}^2}{3}$ (R to 2).

$(\mathrm{S})$ At $t_2$ the block is back on the horizontal plane, so gravity's net work is zero (S to 3). Answer (D).