\begin{problem*}{8.8}
Two objects are connected by a light string passing over a light,
frictionless pulley as shown in Figure~P8.7.  The object of mass $m_1$
is released from rest at a height $h$ above the table.  Using the
isolated system model, \Part{a} determine the speed of $m_2$ just as
$m_1$ hits the table and \Part{b} find the maximum height above the
table to which $m_2$ rises.
%  ceiling
%   pulley
%  |     |
%  |     1
%  |       } h
%  2       }
%--table-----
\end{problem*}

\begin{solution}
\Part{a}
Conserving energy, we just have to worry about gravitational potential
energy and kinetic energy (there are no springs, friction, external
forces, etc.).  While they are attached by a taught string (i.e. while
$m_1$ is falling), the speeds of the two masses must match.  Labeling
the release point $a$ and $m_1$ just hitting the floor $b$, we have
\begin{align}
  E_a &= E_b \\
  K_{1a} + U_{1a} + K_{2a} + U_{2a} &= K_{1b} + U_{1b} + K_{2b} + U_{2b} \\
  0 + m_1 gh + 0 + 0
    &= \frac{1}{2} m_1 v_b^2 + 0 + \frac{1}{2} m_2 v_b^2  + m_2 gh \\
  (m_1-m_2) gh &= \frac{1}{2} (m_1+m_2) v_b^2 \\
  v_b^2 &= 2gh\frac{m_1-m_2}{m_1+m_2} \\
  v_b &= \pm\sqrt{2gh\frac{m_1-m_2}{m_1+m_2}}
    = \ans{\sqrt{2gh\frac{m_1-m_2}{m_1+m_2}}} \;,
\end{align}
where we dropped the $\pm$ because we only want the magnitude of the
velocity, not its direction.

\Part{b}
After $m_1$ hits the table, the string goes slack as $m_2$ sails up in
a parabola $h(t)$ and peaks at some point $c$.  Conserving energy (now
just for $m_2$, because $m_1$ just sits there on the table being
boring)
\begin{align}
  E_{2b} &= E_{2c} \\
  K_{2b} + U_{2b} &= K_{2c} + U_{2c} \\
  \frac{1}{2} m_2 v_b^2 + m_2 gh &= 0 + m_2 g h_c \\
  h_c &= \frac{v_b^2}{2g} + h
    = \frac{2gh\frac{m_1-m_2}{m_1+m_2}}{2g} + h
    = h\p({\frac{m_1-m_2}{m_1+m_2} + 1}) \\
    &= h\p({\frac{m_1-m_2}{m_1+m_2}+\frac{m_1+m_2}{m_1+m_2}})
    = h\frac{m_1-m_2+m_1+m_2}{m_1+m_2}
    = \ans{\frac{2m_1 h}{m_1+m_2}}
\end{align}
\end{solution}
