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

{\bf Question}

Two particles, one of mass $m$ and one of mass $\alpha m \,
(\alpha >0),$ both have speed $U.$  They suffer a head-on
collision.  What are their speeds after the collision of their
coefficient of restitution is $e?$

\vspace{.25in}

{\bf Answer}

$$$$
\begin{center}
\setlength{\unitlength}{.5in}
\begin{picture}(7,2)

\put(1,2.2){\makebox(0,0){Before}}

\put(5,2.2){\makebox(0,0){After}}

\put(0,1.5){\line(1,0){0.5}}

\put(0,1.5){\vector(1,0){0.25}}

\put(0.25,1.7){\makebox(0,0){$u$}}

\put(0.25,0.75){\circle{0.5}}

\put(0.25,0.25){\makebox(0,0){$m$}} %first ball%

\put(1.75,1.5){\line(1,0){0.5}}

\put(2.25,1.5){\vector(-1,0){0.25}}

\put(2,1.7){\makebox(0,0){$u$}}

\put(2.,0.75){\circle{0.75}}

\put(2.,0.25){\makebox(0,0){$\alpha m$}} %second ball%


\put(4,1.5){\line(1,0){0.5}}

\put(4,1.5){\vector(1,0){0.25}}

\put(4.25,1.7){\makebox(0,0){$V$}}

\put(4.25,0.75){\circle{0.5}}

\put(4.25,0.25){\makebox(0,0){$m$}} %third ball%

\put(5.75,1.5){\line(1,0){0.5}}

\put(5.75,1.5){\vector(1,0){0.25}}

\put(6,1.7){\makebox(0,0){$W$}}

\put(6.,0.75){\circle{0.75}}

\put(6.,0.25){\makebox(0,0){$\alpha m$}} %fourth ball%

\end{picture}
\end{center}

\bigskip

$\begin{array}{rcl} \ds{\rm Conservation\ of\ mometum} \\  mU -
\alpha m U & = & m V + \alpha m W \\  U (1 - \alpha) & = & V +
\alpha W \hspace{.1in}(1) \\ {\rm Newton's\ law\ of\ restitution\
} \\  -e 2 U & = & V - W \hspace{.1in}(2) \\ (1) - (2) \\ u(1 -
\alpha + 2e) & = & \alpha W + W \\  W & = & \ds\frac{u(1 - \alpha
+ 2e)}{1+\alpha} \\ {\rm sub\ into\ (1)\ and\ rearrange} \\ V & =
& u(1-\alpha) - \ds\frac{\alpha u (1 - \alpha + 2e)}{1+ \alpha} \\
& = & \ds\frac{u(1-\alpha)(1+\alpha) - u\alpha (1 - \alpha +2e)}{1
+ \alpha} \\  & = & \ds\frac{u(1 - \alpha(1 + 2e))}{1+ \alpha} \\
\end{array}$

Thus the answers are:

$$ W = \ds\frac{u(1 - \alpha + 2e)}{1+\alpha};\ \  V =
\ds\frac{u(1 - \alpha(1 + 2e))}{1+ \alpha}$$



\end{document}