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Problem Text

Particles $P_1$, $P_2$, $\ldots$ are at rest on the $x$-axis, and the  $x$-coordinate of $P_n$ is $n$. 
The  mass of $P_n$ is $\lambda^nm$.
Particle $P$, of mass $m$, is projected  from the origin at speed $u$ towards $P_1$. A series of collisions takes place, and the coefficient of restitution at each collision is $e$, where $0 < e <1$. The speed of  $P_n$ immediately after its first collision is $u_n$ and the speed of $P_n$ immediately after its second collision is  $v_n$. No external forces act on the particles.
\begin{questionparts}
\item Show that $u_1=\dfrac{1+e}{1+\lambda}\, u$ and find  expressions for  $u_n$ and $v_n$  in terms of $e$, $\lambda$, $u$ and $n$.
\item Show that, if $e > \lambda$, then each particle (except $P$)  is involved in exactly two collisions.
\item Describe what happens if $e=\lambda$ and show that, in this case,  the fraction of the initial kinetic energy lost approaches $e$ as the number of  collisions increases.
\item Describe what happens if $\lambda e=1$. What fraction of the initial kinetic energy is \mbox{eventually} lost in this case?
\end{questionparts}

Solution (Optional)

\begin{questionparts}
\item \begin{center}
    \begin{tikzpicture}[scale=0.7]
        \def\h{2.75};
        \def\arrowheight{1.5};
        \def\arrowwidth{0.4};
        \def\massa{$m$};
        \def\massb{$\lambda m$};
        \def\velocityua{$u$};
        \def\velocityub{$0$};
        \def\velocityva{$v$};
        \def\velocityvb{$u_1$};

\draw (0,0) circle (1);
        \draw (2.5,0) circle (1);

\draw (8,0) circle (1);
        \draw (10.5,0) circle (1);
        
        
        \node at (1.25, \h) {before collision};
        \node at (9.25, \h) {after collision};

\draw[dashed] (5.25,-\arrowheight) -- (5.25, \arrowheight);

\node at (0,0) {\massa};
            \draw[->, thick, red] (-\arrowwidth, \arrowheight) -- (\arrowwidth, \arrowheight);
            \node[above, red] at (0, \arrowheight) {\velocityua};
            \node at (2.5,0) {\massb};
            \draw[->, thick, red] (2.5-\arrowwidth, \arrowheight) -- (2.5+\arrowwidth, \arrowheight);  \node[above, red] at (2.5, \arrowheight) {\velocityub};
            \node at (8,0) {\massa};
            \draw[->, thick, red] (8-\arrowwidth, \arrowheight) -- (8+\arrowwidth, \arrowheight);  
            \node[red,above] at (8, \arrowheight) {\velocityva};
            \node at (10.5,0 ) {\massb};
            \draw[->, thick, red] (10.5-\arrowwidth, \arrowheight) -- (10.5+\arrowwidth, \arrowheight);  
            \node[red,above] at (10.5, \arrowheight) {\velocityvb};
    \end{tikzpicture}
\end{center}

\begin{align*}
\text{COM}: && mu &= mv + \lambda m u_1 \\
\Rightarrow && u &= v + \lambda u_1 \tag{1} \\
\text{NEL}: && e &= \frac{u_1-v}{u} \\
\Rightarrow && eu &= u_1 - v \tag{2} \\
(1)+(2) && (1+e)u &= (1+\lambda) u_1 \\
\Rightarrow && u_1 &= \frac{1+e}{1+\lambda}u \\
&& v &= u_1 - eu \\
&&&= \frac{1+e - (1+\lambda)e}{1+\lambda} u \\
&&&= \frac{1-\lambda e}{1+\lambda}u
\end{align*}

Note that subsequent (first (and second)) are the same as these, therefore:

\begin{align*}
u_n &= \left ( \frac{1+e}{1+\lambda} \right)^n u \\
v_n &= \frac{1-\lambda e}{1+\lambda } u_n \\
&=  \frac{1-\lambda e}{1+\lambda } \left ( \frac{1+e}{1+\lambda} \right)^n u
\end{align*}

\item If $e > \lambda$ then $(1-\lambda e) > 1-e^2 > 0$ and 
\begin{align*}
\frac{v_{n+1}}{v_n} &= \frac{1+e}{1+\lambda} > 1
\end{align*}

So the particles are moving away from each other - hence no more collisions.

\item If $e = \lambda$ then $u_n = u$ and $v_n = (1-\lambda)u$ so all the particles end up moving at the same speed.

\begin{align*}
\text{initial k.e.} &= \frac12 m u^2 \\
\text{final k.e.} &= \frac12 m((1-e)u)^2 + \sum_{n = 1}^{\infty} \frac12 \lambda^n m ((1-e)u)^2 \\
&= \frac12mu^2(1-e)^2 \left ( \sum_{n=0}^{\infty} e^n \right) \tag{$e = \lambda$} \\
&= \frac12 mu^2(1-e)^2 \frac{1}{1-e} \\
&= \frac12m u^2 (1-e) \\
\text{change in k.e.} &=  \frac12 m u^2 - \frac12m u^2 (1-e) \\
&= e\frac12m u^2 
\end{align*}

Ie the total energy lost approaches a fraction of $e$.

\item If $\lambda e = 1$, after the second collision the particle will be stationary. ie

\begin{align*}
\text{initial k.e.} &= \frac12 m u^2 \\
\text{k.e. after }n\text{ collisions} &= \frac12 \lambda^n m \left (\left ( \frac{1+e}{1+\lambda} \right)^n u \right)^2\\
&= \frac12 \lambda^n m \left ( \frac{1+\frac1{\lambda}}{1+\lambda} \right)^{2n} u&2\\
&= \frac12 \lambda^n m \left ( \frac{1+\frac1{\lambda}}{1+\lambda} \right)^{2n} u\\
&= \frac12 \lambda^n m \left ( \frac{1}{\lambda} \right)^{2n} u\\
&= \frac12 m \lambda^{-n} u\\
&\to 0
\end{align*}

Eventually we lose all the kinetic energy.
\end{questionparts}

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