Problem source

Two parallel vertical barriers are fixed a distance $d$ apart on horizontal ice. A small ice hockey puck moves on the ice backwards and forwards between the barriers, in the direction perpendicular to the barriers, colliding with each in turn. The coefficient of friction between the puck and the ice is $\mu$ and the coefficient of restitution between the puck and each of the barriers is $r$. 
The puck starts at one of the barriers,  moving with speed $v$ towards the other barrier. Show that
\[
v_{i+1}^2 - r^2 v_i^2 = - 2 r^2 \mu gd\,
\]
where $v_i$ is the speed of the puck just after its $i$th collision.
The puck comes to rest against one of the barriers after traversing the gap between them $n$ times. In the case $r\ne1$, express $n$ in terms of $r$ and $k$, where $k= \dfrac{v^2}{2\mu g d}\,$. If $r=\e^{-1}$ (where $\e$ is the base of natural logarithms) show that
\[
n = \tfrac12 \ln\big(1+k(\e^2-1)\big)\,.
\]
Give an expression for $n$ in the case $r=1$.

Solution source

\begin{align*}
\text{W.E.P.}: && \text{change in energy} &= \text{work done on particle} \\
\Rightarrow && \underbrace{\frac12mv^2}_{\text{speed before hitting barrier}} - \underbrace{\frac12mu^2}_{\text{speed leaving first barrier}} &= \underbrace{\left( -\mu mg \right)}_{F} \cdot \underbrace{d}_{d} \\
\Rightarrow && v^2 &= v_i^2-2\mu gd
\end{align*}

Newton's experimental law tells us that the speed leaving the barrier will be $r$ times the speed approaching, ie

\begin{align*}
&& v_{i+1} &= rv \\
\Rightarrow && v_{i+1}^2 &= r^2 v^2 \\
&&&= r^2v_i^2 - 2r^2\mu gd \\
\Rightarrow && v_{i+1}^2 - r^2v_i^2 &= - 2r^2\mu gd
\end{align*}

It must be the case that after $n+1$ collisions the speed is zero, ie $v_{n+1}^2 = 0$.

Not that we can consider $w_i = \frac{v_i^2}{2\mu gd}$ and we have the recurrence:

\begin{align*}
&& w_{i+1} &=r^2w_i  -r^2 \\
\end{align*}

Looking at this we have a linear recurrence with a constant term, so let's try $w_i = C$, then

\begin{align*}
&& C &= r^2 C - r^2 \\
\Rightarrow && C &= \frac{-r^2}{1-r^2} \\
\end{align*}

So $w_i = Ar^{2i} - \frac{r^2}{1-r^2}$.

$w_0 = k \Rightarrow A = k+\frac{r^2}{1-r^2}$

Therefore $w_n = \left (k+\frac{r^2}{1-r^2} \right)r^{2n} - \frac{r^2}{1-r^2}$

Suppose $w_n = 0$ then,

\begin{align*}
&& 0 &= \left (k+\frac{r^2}{1-r^2} \right)r^{2n} - \frac{r^2}{1-r^2} \\
\Rightarrow && r^{2n} &= \frac{r^2}{1-r^2} \frac{1}{k+\frac{r^2}{1-r^2}} \\
&&&= \frac{r^2}{k(1-r^2)+r^2} \\
\Rightarrow && 2n \ln r &= 2\ln r - \ln[k(1-r^2)+r^2] \\
\Rightarrow && n &= 1 - \frac1{2\ln r}  \ln[k(1-r^2)+r^2)]
\end{align*}

If $r = e^{-1}$ then $\ln r = -1$

\begin{align*}
&& n &= 1 + \frac12 \ln [k(1-e^{-2}) + e^{-2}] \\
&&&= 1 + \frac12 \ln [e^{-2}(k(e^2-1)+1)] \\
&&&= 1 + \frac12 \ln e^{-2} + \frac12  \ln [1+k(e^2-1)] \\
&&&= \frac12  \ln [1+k(e^2-1)]
\end{align*}

If $r = 1$ the recurrence becomes:

$w_{i+1} = w_i - 1$, so $w_i = k-n$, so we have $k$ collisions.