Problem source

Two small beads, $A$ and $B$, of the same mass, are threaded onto a vertical wire on which they slide without friction, and which is fixed to the ground at $P$. 
They are released simultaneously from rest, $A$ from a height of $8h$ above $P$ and $B$ from a height of $17h$ above $P$. 
When $A$ reaches the ground for the first time, it is moving with speed $ V$. It then rebounds with coefficient of restitution $\frac{1}{2}$ and subsequently collides with $B$ at height $H$ above $P$.
Show that $H= \frac{15}8h$ and find, in terms of $g$ and $h$, the speeds $u_A$ and  $u_B$ of the two beads just before the collision.
 
When $A$ reaches the ground for the second time,  it is again moving  with speed $ V$. Determine the coefficient of restitution between the two beads.

Solution source

\begin{align*}
&& v^2 &= u^2 +2as \\
\Rightarrow && V^2 &= 2 g \cdot (8h)\\
\Rightarrow && V &=4\sqrt{hg}\\
\end{align*}

When the first particle collides with the ground, the second particle is at $9h$ traveling with speed $V$, the first particle is at $0$ traveling (upwards) with speed $\tfrac12 V$.

For a collision we need:

\begin{align*}
&& \underbrace{\frac12 V t- \frac12 g t^2}_{\text{position of A}} &= \underbrace{9h - Vt - 
\frac12 gt^2}_{\text{position of B}} \\
\Rightarrow && \frac32Vt &= 9h \\
\Rightarrow && t &= \frac{6h}{V} \\
\\
&& \underbrace{\frac12 V t- \frac12 g t^2}_{\text{position of A}} &= \frac12 V \frac{6h}{V}  - \frac12 g t^2 \\
&&&= 3h - \frac12 g\frac{36h^2}{16hg} \\
&&&= 3h - \frac{9}{8}h \\
&&&= \frac{15}{8}h
\end{align*}

Just before the collision, $A$ will be moving with velocity (taking upwards as positive)

\begin{align*}
&& u_A &= \frac12 V-gt \\
&&&= 2\sqrt{hg}-g \frac{6h}{V} \\
&&&= 2\sqrt{hg} - g \frac{6h}{4\sqrt{hg}} \\
&&&= 2\sqrt{hg}-\frac32\sqrt{hg} \\
&&&= \frac12 \sqrt{hg}
\end{align*}

Similarly, for $B$.

\begin{align*}
&& u_B &= -V -gt \\
&&&= -4\sqrt{hg} - \frac32\sqrt{hg} \\
&&&= -\frac{11}{2}\sqrt{hg}
\end{align*}


Considering $A$, to figure out $v_A$.

\begin{align*}
&& v^2 &= u^2 + 2as \\
&& V^2 &= v_A^2 + 2g\frac{15}{8}h \\
&& 16hg &= v_A^2 + \frac{15}{4}gh \\
\Rightarrow && v_A^2 &= \frac{49}{4}gh \\
\Rightarrow && v_A &= -\frac{7}{2}\sqrt{gh}
\end{align*}

\begin{center}
    \begin{tikzpicture}[scale=0.7]
        \def\h{2.75};
        \def\arrowheight{1.5};
        \def\arrowwidth{0.4};
        \def\massa{$$};
        \def\massb{$$};
        \def\velocityua{$\frac12\sqrt{hg}$};
        \def\velocityub{$\frac{11}{2}\sqrt{hg}$};
        \def\velocityva{$-\frac72\sqrt{gh}$};
        \def\velocityvb{$v_B$};


        \draw (0,0) circle (1);
        \draw (2.5,0) circle (1);

        \draw (8,0) circle (1);
        \draw (10.5,0) circle (1);
        
        
        \node at (1.25, \h) {before collision};
        \node at (9.25, \h) {after collision};

        \draw[dashed] (5.25,-\arrowheight) -- (5.25, \arrowheight);

            \node at (0,0) {\massa};
            \draw[->, thick, red] (-\arrowwidth, \arrowheight) -- (\arrowwidth, \arrowheight);
            \node[above, red] at (0, \arrowheight) {\velocityua};
            \node at (2.5,0) {\massb};
            \draw[<-, thick, red] (2.5-\arrowwidth, \arrowheight) -- (2.5+\arrowwidth, \arrowheight);  \node[above, red] at (2.5, \arrowheight) {\velocityub};
            \node at (8,0) {\massa};
            \draw[->, thick, red] (8-\arrowwidth, \arrowheight) -- (8+\arrowwidth, \arrowheight);  
            \node[red,above] at (8, \arrowheight) {\velocityva};
            \node at (10.5,0 ) {\massb};
            \draw[->, thick, red] (10.5-\arrowwidth, \arrowheight) -- (10.5+\arrowwidth, \arrowheight);  
            \node[red,above] at (10.5, \arrowheight) {\velocityvb};
    \end{tikzpicture}
\end{center}


To keep things clean, lets use units of $\sqrt{hg}$ so we don't need to focus on that for now:

\begin{align*}
\text{COM}: && \frac12 - \frac{11}{2} &= -\frac{7}{2}+v_B \\
\Rightarrow && v_B& =-\frac{3}{2} \\
\text{NEL}: && e &= \frac{2}{6} = \frac13 
\end{align*}