Physics - Center of Mass Question with Solution | TestHub

Ball falls h:$v=\sqrt{2gh}$. Inelastic hit on plane 1 keeps the along-plane component$v\cos 60^{\circ }=\frac{v}{2}$. Slide down plane 1 (drop h):$w^2=\frac{v^2}{4}+2gh=\frac{gh}{2}+2gh=\frac{5gh}{2}$. Inelastic hit on plane 2 keeps$\frac{w}{2}:h^{\prime }=\frac{(w/2{)}^2}{2g}=\frac{w^2}{8g}=\frac{5h}{16}$ - (A) true. Start and end are both at rest, so total impulse$=\Delta p=0$ - (C) true. Distance$=h+\frac{h}{\sin 30^{\circ }}+\frac{h^{\prime }}{\sin 30^{\circ }}=h+2h+\frac{5h}{8}=\frac{29h}{8}-(\mathrm{D})$ true. Answer: (A), (C), (D)