Solution
= Solution
For $T\gg Mc^2$, the result in <9> gives
$$
\sin^2\theta=\frac{2Mc^2}{T+4Mc^2}\simeq\frac{2Mc^2}{T}.
$$
Thus the angle between the outgoing trajectories is
$$
\boxed{2\theta\simeq2\sqrt{\frac{2Mc^2}{T}}\ \mathrm{rad}\longrightarrow0.}
$$
The two particles are strongly concentrated in the forward direction. This relativistic small-angle limit differs from the right-angle separation of equal-mass particles in the low-speed limit.