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Past exam of the physics 1 paper of the University of Oxford

2018

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.

Solution

Past exam of the physics 1 paper of the University of Oxford / 2018 / 9 / i / Solution

Past exam of the physics 1 paper of the University of Oxford / 2018 / 9 / i / Solution

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