Problem source

A smooth billiard ball moving on a smooth horizontal table strikes another identical ball which is at rest. The coefficient of restitution between the balls is $e(<1)$. Show that after the collision the angle between the velocities of the balls is less than $\frac{1}{2}\pi.$
Show also that the maximum angle of deflection of the first ball is
\[
\sin^{-1}\left(\frac{1+e}{3-e}\right).
\]

Solution source

\begin{center}
    \begin{tikzpicture}
        \draw (0,0) circle (1);
        \draw (2,0) circle (1);

        \draw[dashed] (-1.5,0) -- (3.5,0);
        \draw[dashed] (1,-1.5) -- (1,1.5);
        
        \filldraw (0,0) circle (1pt);
        \filldraw (2,0) circle (1pt);

        \draw[-latex, ultra thick, red] (-0.5, -0.5) -- (0,0);
    \end{tikzpicture}
\end{center}

Set up the coordinate frame so that the $x$-direction is the line of centres of the spheres. 

Then if the initial velocities are $\displaystyle \binom{u_x}{u_y}$ and $\displaystyle \binom{0}{0}$. 

Then the final velocities must be: $\displaystyle \binom{v_{x1}}{u_y}$ and $\displaystyle \binom{v_{x2}}{0}$

where $mu_x = mv_{x1}+mv_{x2}$ by conservation of energy and $\frac{v_{x1}-v_{x2}}{u_x} = -e$.

\begin{align*}
&& \begin{cases} v_{x1}+v_{x2} &= u_x \\
v_{x1}-v_{x2} &= -eu_x \\
\end{cases} \\
\Rightarrow && 2v_{x1} &= (1-e)u_x \\
\Rightarrow && v_{x1} &= \frac{(1-e)}{2} u_x \\
&& v_{x2} &= \frac{1+e}{2} u_x 
\end{align*}

Notice that since $0 < e < 1$ we must have $v_{x1} > 0$ and so the ball on the left is still continuing in the positive direction, therefore the angle will be less than $\frac12 \pi$.

The angle the first ball is deflected through is the angle between:  $\displaystyle \binom{u_x}{u_y}$ and  $\displaystyle \binom{\frac{1-e}{2}u_x}{u_y}$.

We can scale the velocities so $u_y = 1$. So we are interested in the angle between  $\displaystyle \binom{x}{1}$ and  $\displaystyle \binom{\frac{1-e}{2}x}{1}$.

To maximise $\theta$ we can maximise $\tan \theta$, so:
\begin{align*}
&& \tan \theta &= \frac{\frac{2}{(1-e)x-\frac{1}{x}}}{1+\frac{2}{(1-e)x^2}} \\
&&&= \frac{2x-(1-e)x}{(1-e)x^2+2} \\
&&&= \frac{(1+e)x}{(1-e)x^2+2} \\
\\
\frac{\d}{\d t}: &&&= \frac{(1+e)((1-e)x^2+2)-2(1+e)(1-e)x^2}{\sim} \\
&&&= \frac{2(1+e)-(1+e)(1-e)x^2}{\sim}\\
\frac{\d}{\d t} = 0: &&0 &= 2(1+e)-(1+e)(1-e)x^2  \\
\Rightarrow && x &= \pm \sqrt{\frac{2}{1-e}} \\
\\
\Rightarrow && \tan \theta &= \frac{\pm(1+e)\sqrt{\frac{2}{1-e}}}{2+2} \\
&&&= \pm \frac{\sqrt{2}(1+e)}{4\sqrt{1-e}} \\
\Rightarrow && \cot^2 \theta &= \frac{8(1-e)}{(1+e)^2} \\
\Rightarrow && \cosec^2 \theta &= \frac{8(1-e)}{(1+e)^2} + 1 \\
&&&= \frac{8-8e+1+2e+e^2}{(1+e)^2} \\
&&&= \frac{9-6e+e^2}{(1+e)^2} \\
&&&= \frac{(3-e)^2}{(1+e)^2} \\
\Rightarrow && \theta &= \sin^{-1} \left ( \frac{1+e}{3-e}\right)
\end{align*}