= Solution
For <collinear inelastic capture followed by two photons>, each particle has $\mathbf p=\gamma m\mathbf u$ and $E=\gamma mc^2$, with $\gamma=(1-u^2/c^2)^{-1/2}$. An isolated collision conserves the vector sum of <momentum> and total <energy>, including <rest energy>, internal excitation and radiation. It need not conserve the sum of individual <rest masses>. At low <speed>,
$$
E=mc^2+\tfrac12mu^2+O(mu^4/c^2),
$$
recovering the classical <kinetic energy> after subtracting $mc^2$.
Here the incoming <Lorentz factors> are $5/3$ and $5/4$, so
$$
E_1=E_2=5mc^2,\qquad p_{1x}=4mc,\quad p_{2x}=-3mc.
$$
The composite therefore has
$$
\boxed{v=\frac{pc^2}{E}=\frac c{10}\text{ in the }+x\text{ direction},\qquad
M=\frac{\sqrt{E^2-p^2c^2}}{c^2}=\sqrt{99}\,m=3\sqrt{11}\,m.}
$$
Its <rest mass> exceeds $7m$ because relative <kinetic energy> has become internal <rest energy>.
For the subsequent collinear <photon> pair, $E_++E_-=10mc^2$ and $(E_+-E_-)/c=mc$. Thus $E_+=11mc^2/2$, $E_-=9mc^2/2$, and
$$
\boxed{\nu_+/\nu_-=11/9.}
$$
The positive-$x$ <photon> is the higher-frequency one.
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