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All the forces in the circular groove will be perpendicular to the direction of motion. Therefore the particles will continue moving with constant speed at all times (aside from collisions). We can consider the collisions to occur as if along a tangent, (since they will be travelling perfectly perpendicular at the collisions).

\begin{center}
    \begin{tikzpicture}[scale=0.7]
        \def\h{2.75};
        \def\arrowheight{1.5};
        \def\arrowwidth{0.4};
        \def\massa{$M$};
        \def\massb{$m$};
        \def\velocityua{$u$};
        \def\velocityub{$u$};
        \def\velocityva{$v_M$};
        \def\velocityvb{$v_m$};

\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}

The speed of approach at the first collision will be $2u$. Therefore $v_m = v_M + 2eu$

\begin{align*}
\text{COM}: && Mu + m (-u) &= Mv_M + m(v_M + 2eu) \\
\Rightarrow && u(M-m - 2em) &= (M+m)v_M \\
\Rightarrow && v_M &= \left ( \frac{M-m-2em}{M+m} \right) u \\
&& v_m &= \left ( \frac{M-m-2em}{M+m} \right) u + 2eu \\
&&&= \left ( \frac{M-m+2eM}{M+m} \right) u
\end{align*}

Both particles will reverse direction if $v_M < 0$ , ie $M-m-2em < 0 \Rightarrow 2em > M-m$

Since at each collision the velocity of the particles reverses, they must still be travelling in opposite directions, and so by conservation of momentum $mv - MV = u(M-m)$. After each collision, the speed of approach (ie $V+v$) reduces by a factor of $e$, therefore $V+v = 2ue^{2n}$

\begin{align*}
&& mv - M V &= u (M-m) \\
&& v + V &= 2u e^{2n} \\
\Rightarrow && (m+M)v &= u(M-m) + M2ue^{2n} \\
\Rightarrow && v &= \frac{u(M-m) + 2ue^{2n}M}{M+m} \\
\Rightarrow && (m+M)V &= 2ume^{2n} - u(M-m) \\
\Rightarrow && V &= \frac{2um e^{2n} - u(M-m)}{M+m}
\end{align*}

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