Problem source

Three particles, $A$, $B$ and $C$, of masses $m$, $km$ and $3m$ respectively, are initially at rest lying in a straight line on a smooth horizontal surface. Then $A$ is projected towards $B$ at speed $u$. After the collision, $B$ collides with  $C$. The coefficient of restitution between  $A$ and $B$ is $\frac12$ and the coefficient of restitution between $B$ and $C$ is $\frac14$.
\begin{questionparts}
\item Find the range of values of $k$ for which $A$ and $B$ collide for a second time.
\item Given that $k=1$ and that $B$ and $C$ are initially a distance $d$ apart, show that the time that elapses between the two collisions of $A$ and $B$ is $\dfrac{60d}{13u}\,$.
\end{questionparts}

Solution source

\begin{questionparts}
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}

The speed of approach is $u$ and $e = \frac12$, so $v_B = v_A + \frac12 u$.

\begin{align*}
\text{COM}: && mu &= mv_A + km(v_A + \frac12 u) \\
\Rightarrow && v_A(k+1) &= \left (1-\frac{k}{2} \right)u \\
\Rightarrow && v_A &= \frac{2-k}{2(k+1)}u \\
&& v_B &= \frac{2-k}{2(k+1)}u + \frac12 u \\
&&&= \frac{3}{2(k+1)}u
\end{align*}

Second collision

\begin{center}
    \begin{tikzpicture}[scale=0.7]
        \def\h{2.75};
        \def\arrowheight{1.5};
        \def\arrowwidth{0.4};
        \def\massa{$km$};
        \def\massb{$3m$};
        \def\velocityua{$v_B$};
        \def\velocityub{$0$};
        \def\velocityva{$w_B$};
        \def\velocityvb{$w_C$};


        \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 is $v_B$, and $e = \frac14$ so $\displaystyle w_C = w_B + \frac{3}{8(k+1)}u$, and

\begin{align*}
\text{COM}: && \frac{3k}{2(k+1)}mu &= kmw_B + 3m\left (w_B + \frac{3}{8(k+1)}u \right) \\
\Rightarrow && w_B \left (k+3 \right) &= \left ( \frac{3k}{2(k+1)} - \frac{9}{8(k+1)} \right)u \\
\Rightarrow && w_B &= \frac{1}{k+3}  \left ( \frac{12k-9}{8(k+1)} \right)u \\
&&&= \frac{3(4k-3)}{8(k+1)(k+3)}u
\end{align*}

There will be a further collision between $A$ and $B$ if $v_A > w_B$, ie

\begin{align*}
&& \frac{2-k}{2(k+1)}u &> \frac{3(4k-3)}{8(k+1)(k+3)}u \\
\Leftrightarrow && 4(k+3)(2-k) &> 3(4k-3) \\
\Leftrightarrow && 4(-k^2-k+6) &> 12k-9 \\
\Leftrightarrow && 0&>4k^2+16k-33 \\
\Leftrightarrow && 0&> (3-2k)(11+2k) \\
\Leftrightarrow && k &< \frac32
\end{align*}

\item After the first collision, it takes $B$, $\frac{d}{v_B} = \frac{d}{u} \frac{2(k+1)}{3} = \frac{4d}{3u}$ to collide with $C$. During which time $B$ and $A$ have been moving apart with speed $\frac12u$ and so are a distance $\frac{2d}{3}$ apart.

After the second collision, $w_B = \frac{3(4\cdot 1 - 3)}{8(1+1)(1+3)}u = \frac{3}{64}u$ and $v_A = \frac{1}{4}u$ so they are moving together at speed $\frac{16-3}{64}u  = \frac{13}{64}u$.

It will take them $\frac{2d}{3} \div \frac{13}{64}u = \frac{128d}{3 \times 13u}$ to do this for a total time of $\frac{128d}{3 \times 13u} + \frac{4d}{3u} = \frac{(128+52)d}{3 \times 13 u} = \frac{60d}{13u}$

\end{questionparts}