Problem source

Three identical particles lie, not touching one another, in a straight line on a smooth horizontal surface. One particle is projected with speed $u$ directly towards the other two which are at rest. The coefficient of restitution in all collisions is $e$, where $0 < e < 1\,$.
\begin{questionparts}
\item Show that, after the second collision, the speeds of the particles are $\frac12u(1-e)$, $\frac14u (1-e^2)$ and $\frac14u(1+e)^2$.  Deduce that there will be a third collision whatever the value of $e$.
\item Show that there will be a fourth collision if and only if  $e$ is less than a particular value which you should determine. 
\end{questionparts}

Solution source

\begin{questionparts}
\item 
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{$m$};
        \def\velocityua{$u$};
        \def\velocityub{$0$};
        \def\velocityva{$v_1$};
        \def\velocityvb{$v_2$};


        \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}

By NEL, $v_2 = v_1 + eu$, so

\begin{align*}
\text{COM}: && mu &= mv_1 + m(v_1 + eu) \\
\Rightarrow && 2mv_1 &= mu(1-e) \\
\Rightarrow && v_1 &= \frac12 u(1-e) \\
&& v_2 &=  \frac12 u(1-e) + eu \\
&&&= \frac12 u(1+e)
\end{align*}

The second collision is identical to the first except replacing $u$ with $\frac12u(1+e)$, therefore after that collision:

\begin{align*}
&& \text{first particle} &= \frac12 u(1-e) \\
&& \text{second particle} &= \frac12 \left (\frac12 u(1+e) \right)(1-e) \\
&&&= \frac14 u(1-e^2) \\
&& \text{third particle} &= \frac12 \left (\frac12 u(1+e) \right)(1+e) \\
&&&= \frac14 u(1+e)^2
\end{align*}

After all these collisions, all particles are moving in the same direction (since they all have positive velocity), but the first particle is now travelling faster than the second particle (as $\frac12(1-e) < 1$). Therefore they will collide again.

\item The third 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{$m$};
        \def\velocityua{$\frac12 u(1-e)$};
        \def\velocityub{$\frac14u(1-e^2)$};
        \def\velocityva{$w_1$};
        \def\velocityvb{$w_2$};


        \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 will be $\frac12u(1-e) - \frac14u(1-e^2) = \frac14u(1-e)(2 - (1+e)) = \frac14 u(1-e)^2$, therefore by NEL, $w_2 = w_1 + \frac14ue(1-e)^2$

\begin{align*}
\text{COM}: && m\frac12u(1-e)  + m \frac14u(1-e^2) &= mw_1 + m\left (w_1 + \frac14ue(1-e)^2 \right) \\
\Rightarrow && \frac14u(1-e)(2+(1+e)) &= 2w_1 + \frac14ue(1-e)^2 \\
\Rightarrow && 2w_1 &= \frac14u(1-e)(3+e)-\frac14ue(1-e)^2 \\
&&&= \frac14u(1-e)(3+e-e(1-e)) \\
&&&= \frac14u(1-e)(3+e^2) \\
\Rightarrow && w_1 &= \frac18 u(1-e)(3+e^2) \\
&& w_2 &=  \frac18 u(1-e)(3+e^2) + \frac14ue(1-e)^2 \\
&&&= \frac18u(1-e)(3+e^2+2e(1-e)) \\
&&&= \frac18u(1-e)(3+2e-e^2) \\
&&&= \frac18u(1-e)(1+e)(3-e) \\
\end{align*}

A fourth collision is possible, iff 
\begin{align*}
&& \frac18u(1-e)(1+e)(3-e)&> \frac14 u(1+e)^2 \\
\Leftrightarrow && (1-e)(3-e)&> 2 (1+e) \\
\Leftrightarrow &&3-4e-e^2&> 2+2e \\
\Leftrightarrow &&1-5e-e^2&>0 \\
\Leftrightarrow && e &< 3-\sqrt{2}






\end{align*}

\end{questionparts}