 Solution

Nonrelativistic momentum conservation gives $\mathbf{u}={\mathbf{v}}_1+{\mathbf{v}}_2$, while an elastic collision conserves kinetic energy, $u^2=v_1^2+v_2^2$. Squaring the momentum equation yields ${\mathbf{v}}_1\cdot {\mathbf{v}}_2=0$. The two nonzero outgoing velocities are perpendicular. In the degenerate case where one outgoing particle is at rest, its direction is undefined.

For relativistic collisions conserve total four-momentum, with $E=\gamma m{c}^2$, $\mathbf{p}=\gamma m\mathbf{v}$ and $E^2-p^2{c}^2=m^2{c}^4$. Rest energies must be included; for photons $E=pc$.

For the symmetric relativistic equal-mass scattering angle, the incident momentum is $mc\sqrt{\alpha (\alpha +2)}$. Energy conservation gives each final particle $E_f=(1+\alpha /2)m{c}^2$, hence $p_f=(mc/2)\sqrt{\alpha (\alpha +4)}$. Transverse momenta cancel and $2p_f\cos \theta =p_{\mathrm{in}}$, givingThe nonrelativistic limit has $\theta \to 45^{\circ }$, so the outgoing angle is $90^{\circ }$ as above.

For the symmetric two-photon annihilation angle in the specified electron-positron annihilation, each photon has energy $E_{\gamma }=(\alpha +2)m_e{c}^2/2$. Momentum conservation gives $2(E_{\gamma }/c)\cos \varphi =m_{e}c\sqrt{\alpha (\alpha +2)}$, soAs $\alpha \to 0^{+}$ the two photons become back-to-back, transverse to the limiting incident direction; as $\alpha \to \infty$ they are forward collimated. Equal photon energies follow from the specified equal angles, and are not a general property of annihilation in a moving laboratory frame.