\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\begin{document}
\parindent=0pt

{\bf Question}

A set of $n$ wagons, each of mass $m,$ standing in a railway
siding is set in motion by a locomotive of mass $rm$ starting with
speed $V.$  Initially the couplings between the wagons are slack,
as is the coupling between the first wagon and the locomotive, and
each wagon moves a distance $s$ before it jerks the succeeding
wagon in motion.  The coefficient of restitution of the coupling
is $e.$  Find the time which ensues between the movement of the
first and last wagons, neglecting the effect of friction and
assuming that each coupling does not remain taut until the last
wagon moves.

\vspace{.25in}

{\bf Answer}

$$$$
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\put(1.75,1){\makebox(0,0){rm}}

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\put(2.9,0.25){\framebox(0.5,0.5)[c]{$m$}}

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\put(4.7,0.25){\framebox(0.5,0.5)[c]{$m$}}

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Consider the locomotive and the first wagon.

\bigskip

Before the coupling is taut:
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\put(1,.5){\vector(-1,0){0.5}}

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\put(1.6,.1){\circle{.2}}

\put(2,.1){\circle{.2}}

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\put(3.4,.45){\makebox(0,0){$m$}}

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\put(4.6,.2){\makebox(0,0)[l]{0}}

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After the coupling is jerked:
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\put(1.3,.2){\line(0,1){0.5}}

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\put(2.3,.2){\line(0,1){0.5}}

\put(1.4,.45){\makebox(0,0)[l]{LOCO}}

\put(3.2,.1){\circle{.2}}

\put(3.6,.1){\circle{.2}}

\put(2.9,.2){\line(1,0){1}}

\put(2.9,.2){\line(0,1){0.5}}

\put(2.9,.7){\line(1,0){1}}

\put(3.9,.2){\line(0,1){0.5}}

\put(3.4,.45){\makebox(0,0){$m$}}

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\put(4.6,.3){\makebox(0,0)[l]{$w$}}

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Using the law of conservation of momentum

$\ds \begin{array}{rcl} rmV + 0 & = & rmu + m w \\ rV & = & ru + w
\hspace{.5in}(1) \\ {\rm Restitution\ law:\ \ \ \ \ \ } -eV & = &
u - w \hspace{.58in}(2) \\ {\rm Solving\ (1)\ and\ (2)\ gives:\ \
\ \ \ \ \ \ \ \ \ \ } w & = & rV\ds \frac{1 + e}{1 + r}
\hspace{.4in}(3)
\end{array}$

Note that equation (2) implies that $w>u,$ so the coupling is
slack after the wagon is jerked forward.

\newpage

Consider the $i$ and $i+1^{th}$ wagon.

\bigskip

Before:
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\put(2,.1){\circle{.2}}

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\put(1.3,.2){\line(0,1){0.5}}

\put(1.3,.7){\line(1,0){1}}

\put(2.3,.2){\line(0,1){0.5}}

\put(1.7,.45){\makebox(0,0)[l]{$m$}}

\put(1.7,1){\makebox(0,0)[l]{$i$}}

\put(3.2,.1){\circle{.2}}

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\put(2.9,.2){\line(0,1){0.5}}

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\put(3.9,.2){\line(0,1){0.5}}

\put(3.4,.45){\makebox(0,0){$m$}}

\put(3.2,1){\makebox(0,0)[l]{$i+1$}}

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After:

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\put(1,.5){\vector(-1,0){0.5}}

\put(0.4,0.3){\makebox(0,0)[l]{$u_{i+1}$}}

\put(1.6,.1){\circle{.2}}

\put(2,.1){\circle{.2}}

\put(1.3,0.2){\line(1,0){1}}

\put(1.3,0.2){\line(0,1){0.5}}

\put(1.3,0.7){\line(1,0){1}}

\put(2.3,0.2){\line(0,1){0.5}}

\put(1.7,0.45){\makebox(0,0)[l]{$m$}}

\put(3.,.1){\circle{.2}}

\put(3.4,.1){\circle{.2}}

\put(2.7,0.2){\line(1,0){1}}

\put(2.7,0.2){\line(0,1){0.5}}

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\put(3,1){\makebox(0,0)[l]{$i+1$}}

\put(3.2,0.45){\makebox(0,0){$m$}}

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\put(4.9,.5){\vector(-1,0){0.5}}

\put(4.4,0.2){\makebox(0,0)[l]{$v_{i+1}$}}

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\bigskip

The same procedure gives (3) with $r = 1, v = v_i$

Therefore $ v_{i+1} = v_i \frac{1 + e}{2}$ (Wagons successively
move more slowly as $ \frac{1 + e}{2} < 1$)

If T$_i$ is the time taken to elapse between wagon $i$ and wagon
$i+1$ starting to move then the total time for the last wagon to
move is

$T = T_1 + T_2 + ... + T_{n-1}$

$T = T_1 + \alpha T_1 + \alpha ^2T_1 + ... + \alpha ^{n-2}T_1$

Where $\alpha = \frac{2}{1+e}$ because the time is the distance
moved by wagon(s) divided by the speed $v_i$.

Thus $T_1 = \frac{s}{w} = \frac{s(1+r)}{V(1+e)r}$

$\ds T = \frac{s}{Vr}\left(\frac{1+r}{1+e}\right) \left(\frac{1 -
\alpha^{n-1}}{1-\alpha}\right) = \frac{s}{Vr} \cdot
\frac{1+r}{1+e} \cdot \frac{(1+e)^{n-1} -
2^{n-1}}{(e-1)(1+e)^{n-1}}$



\end{document}
