Problem source

Two particles $A$ and $B$ of masses $m$ and $km$,
respectively, are at rest on a smooth horizontal surface. The direction of the line passing through $A$ and $B$ is perpendicular to a vertical wall which is on the other side of $B$ from $A$. The particle $A$ is now set in motion towards $B$ with speed $u$. The coefficient of restitution between $A$ and $B$ is $e_1$ and between $B$ and the wall is $e_2$. Show that there will be a second collision between $A$ and $B$ provided 
$$
k< \frac {1+e_2(1+e_1)} {e_1}\;.
$$
Show that, if $e_1=\frac13$, $e_2=\frac12$ and $k<5$, then the kinetic energy of $A$ and $B$ immediately after $B$ rebounds from the wall is greater than $mu^2/27$.

Solution source

First collision:

\begin{center}
    \begin{tikzpicture}[scale=0.7]
        \def\h{2.75};
        \def\arrowheight{1.5};
        \def\arrowwidth{0.4};
        \def\massa{$m$};
        \def\massb{$km$};
        \def\velocityua{$u$};
        \def\velocityub{$0$};
        \def\velocityva{$v_A$};
        \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}

Since the $e = e_1$, the speed of approach is $u$ the speed of separation will be $e_1u$ and so $v_B = v_A + e_1u$.

\begin{align*}
\text{COM}: && mu &= mv_A + km(v_A + e_1u) \\
\Rightarrow && v_A(1+k) &= u(1-ke_1) \\
\Rightarrow && v_A &= \frac{1-ke_1}{1+k} u \\
&& v_B &= \frac{1-ke_1}{1+k} u  + e_1 u \\
&&&= \frac{1-ke_1 + e_1+ke_1}{1+k}u \\
&&&= \frac{1+e_1}{1+k}u
\end{align*}

Once the ball rebounds from the wall it will have velocity (still taking towards the wall as +ve) of $-\frac{1+e_1}{1+k}e_2u$. There will be another collision if it is travelling faster than $A$, ie if:

\begin{align*}
-\frac{1+e_1}{1+k}e_2u &< \frac{1-ke_1}{1+k} u \\
\Leftrightarrow && 0 &< (1-ke_1) + (1+e_1)e_2 \\
\Leftrightarrow && ke_1 &< 1 +e_2 (1+e_1) \\
\Leftrightarrow && k &< \frac{1 +e_2 (1+e_1)}{e_1} \\
\end{align*}

If $e_1 = \frac13, e_2 = \frac12$, then $v_A = \frac{1-\frac13k}{1+k}u = \frac{3-k}{3(1+k)}u$  and $v_B = \frac{4}{3(1+k)}u$.

Therefore

\begin{align*}
&& \text{total k.e.} &= \underbrace{\frac12 m v_A^2}_{\text{k.e. of }A} + \underbrace{\frac12 (km) (e_2 v_B)^2}_{\text{k.e. of }B} \\
&&&= \frac12 m \frac{(3-k)^2}{9(1+k)^2}u^2 + \frac12 km \frac14 \frac{16}{9(1+k)^2}u^2 \\
&&&= \frac12mu^2 \frac{1}{9(1+k)^2}\left ( (3-k)^2+4k \right) \\
&&&= \frac12mu^2 \frac{1}{9(1+k)^2}\left ( 9-2k+k^2 \right) \\
&&&= \frac{mu^2}{18} \frac{9-2k+k^2}{1+2k+k^2}
\end{align*}

We wish to minimize this as a function of $k$.

\begin{align*}
\frac{\d}{\d k} \left ( \frac{9-2k+k^2}{1+2k+k^2}\right) &= \frac{(1+k)^2(2k-2)-2(1+k)(k^2-2k+9)}{(1+k)^4} \\
&= \frac{2(k^2-1) - 2(k^2-2k+9)}{(1+k)^3} \\
&= \frac{2(2k-10)}{(1+k)^3} 
\end{align*}

Therefore the minimum will be when $k = 5$ can't be a maximum by considering $k \to 0$.

This value is $\frac{2}{3}$ and therefore $\frac{mu^2}{18} \frac{2}{3} = \frac{mu^2}{27}$ is the smallest energy (which isn't quite achievable since $k < 5$.