Past exam of the physics 1 paper of the University of Oxford / 2013 / 3

For the relativistic head-on elastic collision with a stationary target, use $c=1$ as stipulated, so masses are measured in energy units. The incoming momentum is $p=\sqrt{E^2-M^2}$ and the invariant squared total four-momentum isThe centre-of-mass frame moves at ${\beta }_{\ast }=p/(E+m)$, with ${\gamma }_{\ast }=(E+m)/\sqrt{s}$. In this frame the incident particle hasA nontrivial head-on elastic collision reverses $p_M^{\ast }$ and preserves $E_M^{\ast }$. Boosting back to the laboratory frame givesThe no-interaction forward branch instead leaves $E^{\prime }=E$; the requested formula describes the actual one-dimensional encounter and recoil.

With $M\gg m$ and $Em\gg M^2$, one also has $E\gg M^2/m\gg M$. The denominator in a is then dominated by $2Em$, and the numerator by $E{M}^2$:This residual total energy is still much larger than the incident particle's rest energy $M$, but is a small fraction of its incident energy $E$. The target acquires most of the available energy in this high-energy, heavy-projectile limit.

For $m\gg M$, retain the ratio $E/m$ until a further limit is specified. Since $E\ge M$, the leading expression from a isFor an infinitely heavy target at fixed incident energy, $E/m\to 0$ and $\boxed{E^{\prime }\to E}$: the projectile reverses with negligible target recoil energy. If the incident particle is ultrarelativistic even relative to the target, $E\gg m\gg M$, the same leading expression instead gives $E^{\prime }\sim m/2$. The mass hierarchy alone does not justify replacing $E/m$ by zero.