Problem source

Two identical smooth spheres $P$ and $Q$ can move on a smooth horizontal table. Initially, $P$ moves with speed $u$ and $Q$ is at rest. Then $P$ collides with $Q$. The direction of travel of $P$ before the collision makes an acute angle $\alpha$ with the line joining the centres of $P$ and $Q$ at the moment of the collision. The coefficient of restitution between $P$ and $Q$ is $e$ where $e < 1$.
As a result of the collision, $P$ has speed $v$ and $Q$ has speed $w$, and $P$ is deflected through an angle $\theta$.
\begin{questionparts}
\item Show that
$$u \sin \alpha = v \sin(\alpha + \theta)$$
and find an expression for $w$ in terms of $v$, $\theta$ and $\alpha$.
\item Show further that
$$\sin \theta = \cos(\theta + \alpha) \sin \alpha + e \sin(\theta + \alpha) \cos \alpha$$
and find an expression for $\tan \theta$ in terms of $\tan \alpha$ and $e$.
Find, in terms of $e$, the maximum value of $\tan \theta$ as $\alpha$ varies.
\end{questionparts}

Solution source

\begin{center}
    \begin{tikzpicture}[scale=1]


        \def\beforeheight{0};
        \def\afterheight{-4};

        \def\textx{-1};
        \def\circleonex{2};
        \def\circler{1};
        \def\circletwox{5};
        \def\endscaling{1.5};

        \def\circleoneangle{25};
        \def\circletwoangle{-30};

        \def\circleoneafterangle{60};

        \def\velocityscale{1.3};

        \def\circleonevelocity{$u$};
        \def\circletwovelocity{$u_B$};

        \def\circletwoafterxvel{$w$};
        \def\circleoneaftervel{$v$};
    
        \node at (\textx,\beforeheight) {Before:};
        \node at (\textx,\afterheight) {After:};

        \draw[dotted] (\textx, {0.5*(\beforeheight+\afterheight)}) -- ({\circletwox+\endscaling*\circler}, {0.5*(\beforeheight+\afterheight)});

        \draw[dashed] ({\circleonex - \endscaling*\circler},\beforeheight) -- ({\circletwox+\endscaling*\circler}, \beforeheight);
        \draw[dashed] ({\circleonex - \endscaling*\circler},\afterheight) -- ({\circletwox+\endscaling*\circler}, \afterheight);

        \coordinate (TRowEnd) at (\textx, \beforeheight);
        \coordinate (TRowFEnd) at (100, \beforeheight);
        \coordinate (BRowEnd) at (\textx, \afterheight);
        \coordinate (BRowFEnd) at (100, \afterheight);
        \coordinate (CToneCentre) at (\circleonex, \beforeheight);
        \coordinate (CTtwoCentre) at (\circletwox, \beforeheight);
        \coordinate (CBoneCentre) at (\circleonex, \afterheight);
        \coordinate (CBtwoCentre) at (\circletwox, \afterheight);
        \coordinate (CToneVel) at ({\velocityscale*\circler*cos(180+\circleoneangle)+\circleonex},{\velocityscale*\circler*sin(180+\circleoneangle)+\beforeheight});
        \coordinate (CTtwoVel) at ({\velocityscale*\circler*cos(\circletwoangle)+\circletwox},{\velocityscale*\circler*sin(\circletwoangle)+\beforeheight});

        \coordinate (CBoneVel) at ({\velocityscale*\circler*cos(\circleoneafterangle)+\circleonex},{\velocityscale*\circler*sin(\circleoneafterangle)+\afterheight});
        \coordinate (CBoneOVel) at ({\velocityscale*\circler*cos(\circleoneangle)+\circleonex},{\velocityscale*\circler*sin(\circleoneangle)+\afterheight});

        \draw[-latex, ultra thick, red] (CToneVel) -- (CToneCentre);
        \pic [draw, angle radius=0.6cm, angle eccentricity=1.3, "$\alpha$"] {angle = TRowEnd--CToneCentre--CToneVel};
        \node[red] at ($(CToneCentre)!1.1!(CToneVel)$) {\circleonevelocity};
        
        % \draw[-latex, ultra thick, red] (CTtwoVel) -- (CTtwoCentre);
        % \node[red] at ($(CTtwoCentre)!1.1!(CTtwoVel)$) {\circleonevelocity};

        \draw (CToneCentre) circle (\circler);
        \draw (CTtwoCentre) circle (\circler);
        \draw (CBoneCentre) circle (\circler);
        \draw (CBtwoCentre) circle (\circler);

        \draw[-latex, ultra thick, red] (CBoneCentre) -- (CBoneVel);
        \node[red] at ($(CBoneCentre)!1.1!(CBoneVel)$) {\circleoneaftervel};
        \draw[dashed] ($(CBoneCentre)!-1!(CBoneOVel)$) -- (CBoneOVel);
        \pic [draw, angle radius=0.6cm, angle eccentricity=1.3, "$\alpha$"] {angle = BRowFEnd--CBoneCentre--CBoneOVel};
        \pic [draw, angle radius=0.7cm, angle eccentricity=1.3, "$\theta$"] {angle = CBoneOVel--CBoneCentre--CBoneVel};

        % \draw[-latex, ultra thick, red]  (CBoneCentre) -- ++ ({\velocityscale*\circler}, 0) node[above] {$v_{Ax}$};
        % \draw[-latex, ultra thick, red]  (CBoneCentre) -- ++ (0,{\velocityscale*\circler}) node[above] {$v_{Ay}$};
        \draw[-latex, ultra thick, red]  (CBtwoCentre) -- ++ ({\velocityscale*\circler}, 0) node[above] {\circletwoafterxvel};
        % \draw[-latex, ultra thick, red]  (CBtwoCentre) -- ++ (0,{\velocityscale*\circler})node[above] {$v_{By}$};

        
    \end{tikzpicture}
\end{center}

\begin{questionparts}
\item Since the impulse is along the line of centres, the velocities are as show in the diagram. Additionally, vertical velocity is unchanged, so:

$v \sin (\theta + \alpha) = u \sin \alpha$


\begin{align*}
\text{COM}(\rightarrow): && u \cos\alpha &= v \cos(\alpha + \theta) + w \\
\Rightarrow && w &= u \cos \alpha - v \cos (\alpha + \theta)
\end{align*}

\item 
Since the approach speed (horizontally) is $u \cos \alpha$ the speed of separation is $e u \cos \alpha$, in particular $w - v \cos(\theta + \alpha) = e u \cos \alpha$ or $w = v \cos (\theta + \alpha) + e u \cos \alpha$.

\begin{align*}
&& w &= w \\
&& v \cos (\theta + \alpha) + e u \cos \alpha &= u \cos \alpha - v \cos (\alpha + \theta) \\
\Rightarrow && \frac{u \sin \alpha}{\sin (\alpha + \theta)} \cos (\theta + \alpha) + e u \cos \alpha &= u \cos \alpha - \frac{u \sin \alpha}{\sin (\alpha + \theta)} \cos (\alpha + \theta) \\
\Rightarrow && \sin \alpha \cos(\theta + \alpha) + e \sin (\alpha+\theta)\cos \alpha &= \sin(\alpha+\theta) \cos \alpha - \cos(\alpha+\theta)\sin \alpha \\
&&&= \sin ((\alpha+\theta)-\alpha) \\
&&&= \sin \theta
\end{align*}
as required.

\begin{align*}
&& \sin \theta &= \cos(\theta+ \alpha)\sin \alpha + e \sin (\theta + \alpha) \cos \alpha \\
&&&= \cos \theta \cos \alpha \sin \alpha - \sin \theta \sin^2 \alpha  + e \sin \theta \cos ^2 \alpha + e \cos \theta \sin \alpha \cos \alpha \\
\Rightarrow && \tan \theta \sec^2 \alpha &= \tan \alpha - \tan \theta \tan^2 \alpha + e \tan \theta + e \tan \alpha  \\
\Rightarrow && \tan \theta (1 + \tan^2 \alpha+\tan^2 \alpha-e) &= \tan \alpha + e \tan \alpha \\
\Rightarrow && \tan \theta &= \frac{(1+e)\tan \alpha}{1-e + 2\tan^2 \alpha}
\end{align*}


We seek to maximise $y = \frac{x}{c+2x^2}$, 
\begin{align*}
&& \frac{\d y}{\d x} &= \frac{c+2x^2-4x^2}{(c+2x^2)^2} \\
&&&= \frac{c-2x^2}{(c+2x^2)^2}
\end{align*}

Therefore the maximum will occur at $x = \sqrt{c/2}$, ie $\tan \alpha = \sqrt{(1-e)/2}$ and theta will be $\displaystyle \frac{(1+e)\sqrt{(1-e)/2}}{2(1-e)} =\frac{1}{2\sqrt{2}} \frac{1+e}{\sqrt{1-e}}$
\end{questionparts}