\begin{problem*}{9.50}
Ted and Mary are playing a game of catch in frame $S'$, which is
moving at $0.600\U{c}$ with respect to frame $S$, while Jim, at rest
in frame $S$, watches the action (Fig.~P9.50).  Ted throws the ball to
Mary at $0.800c$ (according to Ted), and their seperation (measured in
$S'$) is $1.80\E{12}\U{m}$.  \Part{a} According to Mary, how fast is
the ball moving?  \Part{b} According to Mary, how long does the ball
take to reach her?  \Part{c} According to Jim, how far apart are Ted
and Mary, and how fast is the ball moving?  \Part{d} According to Jim,
how long does it take the ball to reach Mary?
\end{problem*} % problem 9.50

\begin{solution}
\Part{a}
Mary is in the same frame as Ted, so she also feels the ball is moving
at $\ans{-0.800c}$.

\Part{b}
Mary thinks the ball takes
\begin{equation}
  \Delta t' = \frac{\Delta x'}{v'}
    = \frac{1.80\E{12}\U{m}}{0.800\cdot3.00\E{8}\U{m/s}}
    = \ans{7500\U{s}} = \ans{2.08\U{hours}}
\end{equation}

\Part{c}
Jim sees the $S'$ frame as length contracted, so he sees the proper
length $\Delta x'$ between Mary and Ted (proper length is length
measured in the frame where the two ends are stationary, here the $S'$
frame) as
\begin{align}
  \gamma &= \frac{1}{\sqrt{1-v^2/c^2}}
    = \frac{1}{\sqrt{1-0.6^2}} = 1.25 \\
  \Delta x &= \frac{\Delta x'}{\gamma} = \frac{1.8\E{12}\U{m}}{1.25}
    = \ans{1.44\E{12}\U{m}} \;.
\end{align}
You can use the Lorentz velocity transformations to find the ball
speed in Jim's frame.
\begin{equation}
  u_x = \frac{u_x'+v}{1+\frac{vu_x'}{c^2}}
   = \frac{-0.8c+.6c}{1-0.8\cdot0.6}
   = \ans{-0.385c}
\end{equation}


\Part{d}
You can do this several ways.  One way is to use the total distance
between Mary and Ted in the $S$ frame
\begin{align}
  (v - u_x)\Delta t &= \Delta x \\
   \Delta t &= \frac{\Delta x}{v - u_x}
    = \frac{1.44\E{12}\U{m}}{0.600c + 0.385c}
    = \ans{4870\U{s}} = \ans{1.35\U{hours}}
\end{align}

Another way is to convert from the $S'$ frame to the ball frame to get
proper time
\begin{equation}
  \Delta t_0 = \frac{\Delta t'}{\gamma_b'} 
    = 2.08\U{hours}\cdot\sqrt{1-0.8^2}
    = 1.25\U{hours}
\end{equation}
Then you can time dilate from the ball frame into the $S$ frame
\begin{equation}
  \Delta t = \gamma_b \Delta t_0 = \frac{1.25\U{hours}}{\sqrt{1-0.385^2}}
    = \ans{1.35\U{hours}}
\end{equation}
\end{solution}
