Problem source

 Points $A$ and $B$ are at the same height and a distance $\sqrt{2}r$ apart. Two small, spherical particles of equal mass, $P$ and $Q$, are suspended from $A$ and $B$, respectively, by light inextensible strings of length $r$. Each particle individually may move freely around and inside a circle centred at the point of suspension.
The particles are projected simultaneously from points which are a distance $r$ vertically below their points of suspension, directly towards each other and each with speed $u$. When the particles collide, the coefficient of restitution in the collision is $e$.
\begin{questionparts}
\item Show that, immediately after the collision, the horizontal component of each particle's velocity has magnitude $\frac{1}{2}ev\sqrt{2}$, where $v^2 = u^2 - gr(2 - \sqrt{2})$ and write down the vertical component in terms of $v$.
\item Show that the strings will become taut again at a time $t$ after the collision, where $t$ is a non-zero root of the equation
\[(r - evt)^2 + \left(-r + vt - \frac{1}{2}\sqrt{2}gt^2\right)^2 = 2r^2.\]
\item Show that, in terms of the dimensionless variables
\[z = \frac{vt}{r} \quad \text{and} \quad c = \frac{\sqrt{2}v^2}{rg}\]
this equation becomes
\[\left(\frac{z}{c}\right)^3 - 2\left(\frac{z}{c}\right)^2 + \left(\frac{2}{c} + 1 + e^2\right)\left(\frac{z}{c}\right) - \frac{2}{c}(1 + e) = 0.\]
\item Show that, if this equation has three equal non-zero roots, $e = \frac{1}{3}$ and $v^2 = \frac{9}{2}\sqrt{2}rg$. Explain briefly why, in this case, no energy is lost when the string becomes taut.
\item In the case described in (iv), the particles have speed $U$ when they again reach the points of their motion vertically below their points of suspension. Find $U^2$ in terms of $r$ and $g$.
\end{questionparts}

Solution source

\begin{center}
    \begin{tikzpicture}
        \coordinate (A) at ({-sqrt(2)},0);
        \coordinate (B) at ({sqrt(2)},0);
        \coordinate (P) at ({-sqrt(2)},-2);
        \coordinate (Q) at ({sqrt(2)},-2);
        
        \filldraw (A) circle (0.05);
        \filldraw (B) circle (0.05);
        \draw (P) circle (0.15);
        \draw (Q) circle (0.15);

        \draw (B) -- ($(Q)+(0,0.15)$);
        \draw (A) -- ($(P)+(0,0.15)$);

        \draw[dashed] (A) circle (2);
        \draw[dashed] (B) circle (2);

        
        
    \end{tikzpicture}
\end{center}

\begin{questionparts}
\item Assuming the particles have mass $m$, and speed $v$ just before collision, then \begin{align*}
\text{COE}: && \underbrace{\frac12 m u^2}_{\text{initial kinetic energy}} + \underbrace{0}_{\text{initial GPE}} &= \underbrace{\frac12m v^2}_{\text{kinetic energy just before collision}} + \underbrace{mgr\left(1-\frac1{\sqrt{2}}\right)}_{\text{GPE just before collision}} \\
\Rightarrow && v^2 &= u^2 - gr(2-\sqrt{2})
\end{align*}

Therefore the particles has velocity $\frac{\sqrt{2}}2v \binom{\pm 1}{1}$ before the collision. By symmetry, the impulse between the particles will be horizontal, so the vertical velocities will be unchanged at $\frac{\sqrt{2}}{2}v$. By conservation of momentum (or symmetry) the particles will have equal but opposite velocities after the collision (say $w$) satisfying:

\[ e = \frac{2w}{2\frac{\sqrt{2}}{2}v} \] ie $w = \frac{\sqrt{2}}2 e v$ as required.

\item Once the particles have rebounded, they will be projectiles whilst the strings are slack. If we consider the left-most point $A = (0,0)$ then the particles colide at $\left ( \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right)$ and the position at time $t$ after the collision (before the string goes slack) will be:

\begin{align*}
\mathbf{x}_t &= \frac{\sqrt{2}}{2}r\binom{1}{-1} + \frac{\sqrt{2}}{2} vt \binom{-e}{1} + \frac12 gt^2 \binom{0}{-1}
\end{align*}

The string will go taught when $|\mathbf{x}_t|^2 = r^2$, ie

\begin{align*}
&& r^2 &= \left ( \frac{\sqrt{2}}{2} r - \frac{\sqrt{2}}{2}evt \right)^2 + \left (-\frac{\sqrt{2}}{2} r + \frac{\sqrt{2}}{2}vt -\frac12 gt^2 \right)^2 \\
\Rightarrow && r^2 &= \frac12 \left (r - evt \right)^2 + \frac12 \left (-r+vt - \frac{\sqrt{2}}{2}gt^2 \right)^2 \\
\Rightarrow && 2r^2 &= \left (r - evt \right)^2 +  \left (-r+vt - \frac{\sqrt{2}}{2}gt^2 \right)^2 \\
\end{align*}

as required.

\item Suppose $z = \frac{vt}{r}$, $c = \frac{\sqrt{2}v^2}{rg}$, then 

\begin{align*}
&& 2r^2 &= \left (r - evt \right)^2 +  \left (-r+vt - \frac{\sqrt{2}}{2}gt^2 \right)^2 \\
\Leftrightarrow && 2 &= \left (1 - e\frac{vt}{r} \right)^2 + \left (-1 + \frac{vt}{r}- \frac{\sqrt{2}}{2} \frac{gt^2}{r} \right)^2 \\
\Leftrightarrow && 2 &=  \left (1 - ez \right)^2 + \left (-1 +z- \frac{v^2t^2}{r^2} \frac{gr}{\sqrt{2}v^2}\right)^2 \\
\Leftrightarrow && 2 &=  \left (1 - ez \right)^2 + \left (-1 +z- \frac{z^2}{c} \right)^2 \\
\Leftrightarrow && 2 &= 1-2ez + e^2z^2 + 1 + z^2 +\frac{z^4}{c^2} - 2z-2\frac{z^3}{c}+2\frac{z^2}{c} \\
\Leftrightarrow && 0 &= z(-2e-2) + z^2(e^2+1 + \frac{2}{c}) + z^3(-\frac{2}{c}) + z^4 \frac{1}{c^2} \\
\underbrace{\Leftrightarrow}_{z \neq 0} && 0 &=  \left ( \frac{z}{c} \right)^3 - 2\left ( \frac{z}{c} \right)^2 + \left ( \frac{z}{c} \right) (1 + e^2 + \frac{2}{c} ) - \frac{2}{c}(1+e)
\end{align*}
as required, (where on the last step we divide by $z/c$). 

\item If a cubic has $3$ equal, non-zero roots then it must have the form $(z-a)^3 = z^3 -3az^2 + 3a^2 z -a^3 = 0$, so $3a = 2$, and so the expansion must be $\left ( \frac{z}{c} \right)^3 - 2\left ( \frac{z}{c} \right)^2 + \frac{4}{3}\left ( \frac{z}{c} \right) - \frac{8}{27} = 0$

\begin{align*}
&& \frac{2}{c}(1+e) &= \frac{8}{27} \\
\Rightarrow && \frac{2}{c} &= \frac{8}{27} \frac{1}{1+e} \\ 
&& 1 + e^2 + \frac{2}{c} &= \frac43 \\
\Rightarrow &&  e^2 + \frac{8}{27(1+e)} &= \frac{1}{3} \\
\Rightarrow &&  27(1+e)e^2+8 &= 9(1+e) \\
\Rightarrow && 27e^3 + 27e^2-9e-1 &= 0 \\
\Rightarrow && (3e-1)(9e^2+12e+1) &= 0
\end{align*}

The only (positive) root is $e = \frac13$, therefore $e = \frac13$.

We must also have
\begin{align*}
&& \frac{2}{c} \frac43 &= \frac{8}{27} \\
\Rightarrow && c &= 9 \\
\Rightarrow && \frac{\sqrt{2}v^2}{rg} &= 9 \\
\Rightarrow && v^2 &= \frac{9\sqrt{2}rg}{2}
\end{align*}

as required.

If we consider the path of the particle acting as a projectile, iff the path is tangent to the circle then there will be exactly one solution for $z/c$ and (importantly) it will be a repeated root. Therefore the particle rejoins the circle at a tangent and the tension is acting perpendicularly to the direction of motion (ie no energy loss).

\item Since the only energy lost is lost in the collision, we can apply conservation of energy again:

\begin{align*}
\text{COE:} && \frac12 m U^2  &= \frac12 m \frac12v^2(1+e^2) + mgr\left (1 - \frac1{\sqrt{2}} \right) \\
\Rightarrow && U^2 &= \frac12 \frac{9 \sqrt{2}}{2}gr(1+\frac19) + gr(2 - \sqrt{2}) \\
&&&= \left (\frac{5\sqrt{2}}{2}+2 - \sqrt{2} \right)gr \\
&&&= \left (\frac{4+3\sqrt{2}}{2} \right)gr
\end{align*}
\end{questionparts}