Problem source

Two small discs of masses $m$ and $\mu m$  
lie on a smooth horizontal surface.  
The disc of mass $\mu m$ is at rest,  
and the disc of mass $m$ is projected towards it with velocity $\mathbf{u}$.  
After the collision, the disc of mass $\mu m$ moves in the direction  
given by unit vector $\mathbf{n}$. The collision is perfectly elastic. 
 
\begin{questionparts} 
\item
Show that  the speed of the disc of mass $\mu m$ after the collision  
is \ \ 
$ 
\dfrac {2\mathbf{u} \cdot \mathbf{n}}{1+\mu}. 
$ 
 
\item
 Given that  the two discs have equal kinetic energy  
after the collision, 
find an expression for the cosine of the angle between 
 $\bf n$ and $\bf u$ and show that  
$3-\sqrt8\le \mu \le 3+\sqrt8$.  
\end{questionparts}

Solution source

\begin{questionparts} 
\item
In the direction of $\mathbf{n}$,

Conservation of momentum gives: $m \mathbf{u} \cdot \mathbf{n} = m v_m + \mu m v_{\mu m}$
Newton's experimental law gives: $\frac{\mathbf{u} \cdot \mathbf{n}}{v_{\mu m} - v_m} = 1$

Therefore 

\begin{align*}
 && \mathbf{u} \cdot \mathbf{n} &= v_m + \mu v_{\mu m} \\
&& \mathbf{u} \cdot \mathbf{n} &= v_{\mu m} - v_m \\
\Rightarrow && 2 \mathbf{u} \cdot \mathbf{n} &= (1 + \mu)v_{\mu m} \\
\Rightarrow && v_{\mu m} &= \frac{2 \mathbf{u} \cdot \mathbf{n}}{1 + \mu} \\
\end{align*}

\item Kinetic energy after the collision for the second mass is: $\frac12 m \mu \frac{4 (\mathbf{u} \cdot \mathbf{n})^2 }{(1 + \mu)^2}$

For the first mass the final speed (in the direction $\mathbf{n}$ is: $\mathbf{u} \cdot \mathbf{n}- \frac{2 \mathbf{u} \cdot \mathbf{n}}{1 + \mu} = \frac{(\mu - 1) \mathbf{u} \cdot \mathbf{n}}{1 + \mu}$

It's velocity perpendicular to $\mathbf{n}$ is unchanged, which is $\mathbf{u} - (\mathbf{u} \cdot \mathbf{n}) \mathbf{n}$, so it's speed squared is $\mathbf{u}\cdot \mathbf{u} - (\mathbf{u} \cdot \mathbf{n})^2$

Therefore the total kinetic energy is:

$\frac12 m  \frac{(\mu - 1)^2 (\mathbf{u} \cdot \mathbf{n})^2}{(1 + \mu)} + \frac12 m (\mathbf{u}\cdot \mathbf{u} - (\mathbf{u} \cdot \mathbf{n})^2)$

Therefore since the kinetic energies are equal we have:

\begin{align*}
&& \frac12 m  \frac{(\mu - 1)^2 (\mathbf{u} \cdot \mathbf{n})^2}{(1 + \mu)} + \frac12 m (\mathbf{u}\cdot \mathbf{u} - (\mathbf{u} \cdot \mathbf{n})^2) &= \frac12 m \mu \frac{4 (\mathbf{u} \cdot \mathbf{n})^2 }{(1 + \mu)^2} \\
\Rightarrow && \frac{(\mu - 1)^2 (\mathbf{u} \cdot \mathbf{n})^2}{(1 + \mu)} + (\mathbf{u}\cdot \mathbf{u} - (\mathbf{u} \cdot \mathbf{n})^2) &=  \mu \frac{4 (\mathbf{u} \cdot \mathbf{n})^2 }{(1 + \mu)^2} \\
\Rightarrow && (\mathbf{u} \cdot \mathbf{n})^2 \l 1 + \frac{4\mu}{(1+ \mu)^2} - \frac{(1-\mu)^2}{(1+ \mu)^2} \r &= (\mathbf{u} \cdot \mathbf{u}) \\
\Rightarrow && (\mathbf{u} \cdot \mathbf{n})^2 \l \frac{(1 + \mu)^2 + 4\mu - (1-\mu)^2}{(1+ \mu)^2} \r &= (\mathbf{u} \cdot \mathbf{u}) \\
\Rightarrow && (\mathbf{u} \cdot \mathbf{n})^2 \l \frac{8\mu}{(1+ \mu)^2} \r &= (\mathbf{u} \cdot \mathbf{u}) \\
\Rightarrow && \cos^2 \theta &=\frac{(1 + \mu)^2}{8\mu} \\
\end{align*}

We need 

\begin{align*}
&& \frac{(1 + \mu)^2}{8\mu} & \leq 1 \\
\Rightarrow && 1 +2 \mu + \mu^2 \leq 8 \mu \\
\Rightarrow && 1 - 6 \mu + \mu^2 \leq 0
\end{align*}

This quadratic has roots at $3 \pm \sqrt{8}$ and therefore our quadratic inequality is satisfied if:

$\boxed{3 - \sqrt{8} \leq \mu \leq 3 + \sqrt{8}}$

\end{questionparts}