 Solution

Boost along $+x$ at $0.6c$ into the centre-of-momentum frame. The composite is stationary. The incoming particle has $p_x^{\prime }=1.25(2.4-0.6\times 3)mc=0.75mc$, while the target has $p_x^{\prime }=-0.75mc$. After decay the two photons have energies $M{c}^2/2=1.6m{c}^2$ and opposite momenta.

Equal laboratory photon energies require the rest-frame photons to have zero momentum component along the boost: $E_{\mathrm{lab}}=\gamma (E^{\prime }+V{p}_x^{\prime })$ must be equal for two opposite photons. Their directions are therefore $+y$ and $-y$ in the centre-of-momentum diagram, perpendicular to the incoming collision axis. Boosting those transverse photons gives exactly the laboratory momenta in part (a). A stationary composite has no direction-of-travel arrow in this frame.