Physics - Center of Mass Question with Solution | TestHub

For the system of the block and the wedge, horizontal momentum is conserved. Let $u$ be the horizontal speed of the block as it leaves the wedge and $V$ be the speed of the wedge. So $0=mu-MV$. By energy conservation, $mgH=\frac{1}{2}m{u}^2+\frac{1}{2}M{V}^2$. Substituting $V=mu/M$, we get $mgH=\frac{1}{2}m{u}^2(1+m/M)$. Thus, $u^2=\frac{2gH}{1+m/M}=\frac{2(10)(2.4)}{1+1/3}=\frac{48}{4/3}=36$. So $u=6\text{ m/s}$. This block then collides inelastically with a stationary block of mass $2m$. By momentum conservation, $mu=(m+2m)V_{final}=3m{V}_{final}$. Therefore, $V_{final}=u/3=6/3=2\text{ m/s}$. The correct answer is 2.