Problem source

Two particles  of masses $m$ and $M$, with $M>m$, lie in a smooth circular groove on a horizontal plane. The coefficient of restitution between the particles is $e$. The particles are initially projected round the groove with the same speed $u$ but in opposite directions. Find the speeds of the particles after they collide for the first time and show that they will both change direction if $2em> M-m$. 
After a further $2n$ collisions, the speed of the particle of mass $m$ is $v$ and the speed of the particle of mass $M$ is $V$. Given that at each collision both particles change their directions of motion, explain why  
\[
mv-MV  = u(M-m),
\]
and find $v$ and $V$ in terms of $m$, $M$, $e$, $u$ and $n$.

Solution source

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*}