Problem source

Three particles, $A$, $B$ and $C$, each of mass $m$, lie on a smooth horizontal table. Particles $A$ and $C$ are attached to the two ends of a light inextensible string of length $2a$ and  particle $B$ is attached to the midpoint of the string. Initially, $A$, $B$ and $C$ are at rest at points $(0,a)$, $(0,0)$ and $(0,-a)$, respectively. 
An  impulse is delivered to $B$, imparting to it a speed $u$ in the positive $x$ direction. The string remains taut throughout the subsequent motion.
\begin{center}
\begin{tikzpicture}[x=1.0cm,y=1.0cm]
    % Clip to match the \begin{pspicture*} boundaries
    \clip (-2.18,-3.26) rectangle (6.26,3.18);
    % Axes
    \draw (-2,0) -- (6,0);
    \draw (0,-3) -- (0,3);
    % Thick lines
    \draw[line width=2pt] (2,1.5) -- (3,0);
    \draw[line width=2pt] (3,0) -- (2,-1.5);
    % Angle arc \theta
    % The start angle is exactly 180 - atan(1.5/1) ≈ 123.69 degrees, ending at 180 degrees.
    \draw (3,0) ++({180-atan(1.5)}:0.7) arc ({180-atan(1.5)}:180:0.7);
    % Labels
    \node[anchor=north west, inner sep=1pt] at (-0.25,2.96) {$y$};
    \node[anchor=north west, inner sep=1pt] at (5.76,-0.1) {$x$};
    \node[anchor=north west, inner sep=1pt] at (2.5,0.38) {$\theta$};
    \node[anchor=north west, inner sep=1pt] at (2.1,1.94) {$A$};
    \node[anchor=north west, inner sep=1pt] at (3.2,0.4) {$B$};
    \node[anchor=north west, inner sep=1pt] at (2.16,-1.46) {$C$};
    % Points
    % dotsize=8pt corresponds to a radius of 4pt (or you can adjust to standard 2.5pt)
    \fill (2,1.5) circle (3pt);
    \fill (3,0) circle (3pt);
    \fill (2,-1.5) circle (3pt);
\end{tikzpicture}
\end{center}
\begin{questionparts}
\item At time $t$, the angle between the $x$-axis and the string joining $A$ and $B$ is $\theta$, as shown in the diagram, and $B$ is at $(x,0)$. Write down the coordinates of $A$ in terms of $x,a$ and $\theta$. Given that the velocity of $B$ is $(v,0)$, show that the velocity of $A$ is $(\dot x + a\sin\theta \,\dot \theta\,,\, a\cos\theta\, \dot\theta)$, where the dot denotes differentiation with respect to time. 
\item Show that,  before particles $A$ and $C$ first collide,
\[
3\dot x + 2a \dot\theta \sin\theta =u \text{      and       } \dot \theta^2  = \frac{u^2}{a^2(3-2\sin^2\theta)}
\,.
\]
\item When $A$ and $C$ collide, the collision is elastic (no energy is lost). At what value of $\theta$ does the second collision between particles $A$ and $C$ occur? (You should justify your answer.)
\item When $v=0$, what are the possible values of $\theta$? Is $v =0$ whenever $\theta$ takes these values?
\end{questionparts}

Solution source

\begin{questionparts}
\item $A$ has coordinates $(x-a\cos \theta, a\sin \theta)$. Differentiating with respect to $t$ the velocity of $A$ is $(\dot{x}+a\sin \theta \cdot \dot{\theta}, a \cos \theta \cdot \dot{\theta})$

\item By considervation of momentum $\rightarrow$ we must have $mu = m(\dot{x}+a\dot{\theta}\sin \theta) + m\dot{x} + m(\dot{x}+a\dot{\theta}\sin \theta) = m(3\dot{x} + 2a \dot{\theta} \sin \theta)$ and the first equation follows.

By conservation of energy, we must have \begin{align*}
&& \frac12 m u^2 &= \frac12 m \dot{x}^2 + \frac12m((\dot{x}+a\dot{\theta}\sin \theta)^2 + a^2 \dot{\theta}^2 \cos^2\theta ) + \frac12m((\dot{x}+a\dot{\theta}\sin \theta)^2 + a^2 \dot{\theta}^2 \cos^2\theta ) \\
&&&= \frac32m\dot{x}^2 + 2m a\dot{x}\dot{\theta}\sin \theta + ma^2\dot{\theta}^2(\sin^2\theta+\cos^2\theta) \\
\Rightarrow && u^2 &= \dot{x}(3\dot{x} + 4a \dot{\theta} \sin \theta) + 2a^2\dot{\theta}^2 \\
&&&= \left ( \frac{u-2a\dot{\theta}\sin \theta}{3}\right)\left ( 3\left ( \frac{u-2a\dot{\theta}\sin \theta}{3}\right)+ 4a \dot{x}\dot{\theta} \sin \theta \right) + 2a^2\dot{\theta}^2 \\
\Rightarrow && 3u^2 &= (u - 2a\dot{\theta} \sin \theta)^2 + 4a(u - 2 a \dot{\theta} \sin \theta) \dot{\theta}\sin \theta + 6a^2 \dot{\theta}^2 \\
&&&= u^2  + 4a^2\dot{\theta}^2 \sin^2 \theta - 8a^2\dot{\theta}^2\sin^2\theta + 6a^2 \dot{\theta}^2 \\
\Rightarrow && \dot{\theta}^2 &= \frac{u^2}{a^2(3-2\sin^2\theta)}
\end{align*}

\item Since $\dot{\theta}^2 > 0$ $\theta$ is strictly increasing or decreasing, therefore the first collision will be when $\theta = 0$, the second when $\theta = \pi$

\item If $v = 0$, from our first equation we have $2a \dot{\theta} \sin \theta = u \Rightarrow \dot{\theta}^2  = \frac{u^2}{4a^2 \sin^2 \theta} = \frac{u^2}{a^2(3-2\sin^2\theta)}$ so $4\sin^2 \theta = 3 - 2\sin^2 \theta \Rightarrow \sin^2 \theta = \frac{1}{2}$ therefore the angles are all the multiples of $\frac{\pi}{4}$. 
\end{questionparts}