\begin{problem*}{8.24}
An $m_1 = 90\U{kg}$ fullback running east (\ihat) with a speed of $v_1
= 5.00\U{m/s}$ is tackled by an $m_2 = 95\U{kg}$ opponent running
north (\jhat) with a speed of $v_2 = 3.00\U{m/s}$.  Noting that the
collision is perfectly inelastic,

 \Part{a} calculate the speed $v_f$ and direction $\theta$ of the
 players just after the tackle and
 \Part{b} determine the mechanical energy lost as a result of the
 collision.  Account for the missing energy.
\end{problem*} % problem 8.24

\begin{solution}
\Part{a}
Conserving momentum in the \ihat\ and \jhat\ directions
\begin{align}
 P_{ix} = m_1 v_1 &= P_{fx} = (m_1 + m_2) v_{fx} \\
  v_{fx} &= v_1 \frac{m_1}{m_1 + m_2} = 2.43\U{m/s} \\
 P_{iy} = m_2 v_2 &= P_{fy} = (m_1 + m_2) v_{fy} \\
  v_{fy} &= v_2 \frac{m_2}{m_1 + m_2} = 1.54\U{m/s} \\
 v_f &= \sqrt{v_{fx}^2 + v_{fy}^2} = \ans{2.88\U{m/s}} \\
 \theta &= \arctan\left(\frac{v_{fy}}{v_{fx}}\right) = \ans{32.3\dg}
\end{align}

\Part{b}
\begin{align}
 \Delta K &= K_f - K_i = \left( \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \right) - \frac{1}{2} (m_1 + m_2) v_f^2 \\
          &= \frac{1}{2} \left[ 90.0\U{kg} (5.00\U{m.s})^2 + 95.0\U{kg} (3.00\U{kg})^2 - (90.0\U{kg} + 95.0\U{kg}) (2.88\U{m/s})^2 \right]\\
          &= \ans{-786\U{J}}
\end{align}
All of which has been lost as mechanical energy, and is now thermal
energy (warmer football players), noise (a loud crunch), etc.
\end{solution}
