\begin{problem*}{8.51}
A small block of mass $m_1 = 0.500\U{kg}$ is released from rest at the
top of a curve-shaped, frictionless wedge of mass $m_2 = 3.00\U{kg}$,
which sits on a frictionless, horizontal surface as sown in
Fig. P8.51a.  When the block leaves the wedge, its velocity is
measured to be $v_1 = 4.00\U{m/s}$ to the right (\ihat) as shown in
Fig. P8.51b.
\Part{a} What is the velocity $\vect{v}_2$ of the wedge after the block reaches the horizontal surface?
\Part{b} What is the height $h$ of the wedge?
\end{problem*} % problem 8.51

\begin{solution}
\Part{a}
Conserving momentum
\begin{align}
 P_i = 0 &= P_f = m_1 v_1 + m_2 v_2 \\
 v_2 &= -v_1 \frac{m_1}{m_2} = -4.00\U{m/s} \frac{0.500\U{kg}}{3.00\U{kg}}
      = \ans{-0.667\U{m/s}}
\end{align}
Where the $-$ sign denotes motion in the $-\ihat$ direction (to the
left in Fig.~P8.51b).

\Part{b}
Let $y = 0$ be the level of the table for the purpose of calculating
gravitational potential energy.  Conserving energy (since none is lost
to friction or other internal energies)
\begin{align}
 E_i = m_1 g h &= E_f = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \\
  m_1 g h &= \frac{1}{2}\left(m_1 v_1^2 + m_2 \frac{m_1^2 v_2^2}{m_2^2}\right) \\
          &= \frac{m_1 v_1^2}{2}\left(1 + \frac{m_1}{m_2}\right) \\
  h &= \frac{v_1^2}{2g}\left(1 + \frac{m_1}{m_2}\right)
     = \frac{(0.667\U{m/s})^2}{2\cdot9.80\U{m/s}^2}\left(1 + \frac{0.500\U{kg}}{3.00\U{kg}}\right)
     = \ans{0.952\U{m}}
\end{align}
\end{solution}
