Problem source

A particle rests at a point $A$ on a horizontal table and is joined to a point $O$ on the table by a taut inextensible string of length $c$. The particle is projected vertically upwards at a speed $64\surd(6gc)$. It next strikes the table at a point $B$ and rebounds. The coefficient of restitution for any impact between the particle and the table is ${1\over 2}$. After rebounding at $B$, the particle will rebound alternately at $A$ and $B$ until the string becomes slack. Show that when the string becomes slack the particle is at height $c/2$ above the table.
Determine whether the first rebound \textit{between} $A$ and $B$ is nearer to $A$ or to $B$.

Solution source

\begin{align*}
\text{N2}(radially): && T + mg \cos \theta &= m\frac{v^2}{r} \\ 
\Rightarrow && v^2-gc \cos \theta &\geq 0 \\
\text{COE}: && \frac12 m u^2 &= \frac12mv^2 + mgc\cos \theta \\
\Rightarrow && u^2 &= v^2 + 2gc\cos \theta \\
&& u^2 &\geq gc \cos \theta+2gc\cos \theta \\
\Rightarrow && u^2 &\geq 3gc\cos \theta
\end{align*}

Therefore it will complete bounces with the string taught if it leaves the table with $u^2 \geq 3gc$.

After $6$ bounces it will leave the table with speed $\sqrt{6gc} > \sqrt{3gc}$ and after $7$ bounces it will leave the table with speed $\sqrt{\frac{3}{2} gc} < \sqrt{3 gc}$. 

When it leaves the table with speed $\sqrt{\tfrac32 gc}$, the string will go slack when

\begin{align*}
&& \tfrac32 gc &= 3gc \cos \theta \\
\Rightarrow && \cos \theta &= \frac{1}{2}
\end{align*}

ie at a height $c\cos \theta = \frac12c$ above the table.

Once the string goes slack, the particle travels under circular motion,

\begin{align*}
&& u^2 &= \frac12 gc \\
\Rightarrow && u_\rightarrow &= \sqrt{\tfrac12 gc} \cos \theta \\
&& u_{\uparrow} &= \sqrt{\tfrac12 gc} \sin \theta \\
\Rightarrow && s &= ut - \tfrac12 gt^2 \\
\Rightarrow && -\frac{c}{2} &= \sqrt{\tfrac12 gc} \frac{\sqrt{3}}{2} t - \tfrac12 g t^2 \\
\Rightarrow && t &= \sqrt{\frac{c}{g}} \left ( \frac{\sqrt{6}+\sqrt{22}}{4} \right) \\
\Rightarrow && s_{\rightarrow} &= \tfrac12 \sqrt{\tfrac12 gc} \sqrt{\frac{c}{g}} \left ( \frac{\sqrt{6}+\sqrt{22}}{4} \right) - \frac{\sqrt{3}}{2}c \\
&&&=  \left ( \frac{\sqrt{11}-3\sqrt{3}}{8} \right)c \\
\end{align*}

We need to establish whether this position is positive or negative (ie if we cross the centre line. Clearly $\sqrt{11} < 3\sqrt{3}$ so we haven't crossed the centre line and we land closer to where we took off. Since it's the 7th take off, this is closer to $B$.