A small body starts from rest at the top of a smooth…

A small body starts from rest at the top of a smooth hemisphere of radius $R$ standing on a horizontal floor. After leaving the hemisphere it hits the floor and bounces elastically (the speed does not change). (a) At what height does it leave the hemisphere? (b) How high does it rise after the bounce? (c) What is its speed at the highest point of that flight?

Explanation of Method

Sliding off a hemisphere and bouncing.

Step 1: (a) Leaving the hemisphere

At angle $\alpha$ from the top, the normal force vanishes when the radial part of gravity alone provides the centripetal force; energy gives the speed: $\frac{mv_B^2}{R} = mg\cos\alpha$.

\[ \tfrac12 v_B^2 = gR\,(1 - \cos\alpha). \]

\[\begin{gathered} \cos\alpha = \tfrac23, \quad v_B^2 = \tfrac23 gR \\ h = R\cos\alpha = \tfrac23 R \end{gathered}\]

Step 2: (b) Height after the bounce

At B the velocity is tangent to the sphere, at angle $\alpha$ below the horizontal, so its horizontal part is $v_x = v_B\cos\alpha$. This never changes (the bounce only reverses the vertical part). At the top of the bounce only $v_x$ is left, so by energy conservation (total energy $mgR$):

\[ gR = gh_{\max} + \tfrac12 v_x^2, \]

\[ v_x^2 = \tfrac23 gR \cdot \tfrac49 = \tfrac{8}{27}gR. \]

\[h_{\max} = R - \tfrac{4}{27}R = \tfrac{23}{27}R \approx 0.85\,R\]

Step 3: (c) Speed at the highest point

\[v_{\text{top}} = v_B\cos\alpha = \sqrt{\tfrac{8}{27}gR} \approx 0.544\sqrt{gR}\]

$\text{leaves: } h = \tfrac23 R$ $R$ $\tfrac{23}{27}R$ $\alpha$

Drawn to scale. The body leaves the hemisphere at $\cos\alpha = 2/3$, bounces elastically off the floor and rises to $23R/27$: below the starting height, because it keeps its horizontal velocity at the top.

Result

\[\boxed{\,\begin{gathered} \cos\alpha = \tfrac23, \quad v_B^2 = \tfrac23 gR \\ h = R\cos\alpha = \tfrac23 R \end{gathered}\,}\]

\[\boxed{\,h_{\max} = R - \tfrac{4}{27}R = \tfrac{23}{27}R \approx 0.85\,R\,}\]

\[\boxed{\,v_{\text{top}} = v_B\cos\alpha = \sqrt{\tfrac{8}{27}gR} \approx 0.544\sqrt{gR}\,}\]

Remark

About the original solution. Part (a) is correct. In (b) and (c) it used $\sin^2\alpha = 1/9$ and $\cos\alpha = \sqrt3/2$, neither of which follows from $\cos\alpha = 2/3$ (in fact $\sin^2\alpha = 5/9$). Its height $\tfrac{23}{22}R$ would even be above the starting point, which energy conservation forbids. The correct results are $h_{\max} = \tfrac{23}{27}R$ and $v_{\text{top}} = \sqrt{8gR/27}$.