= Solution
For fixed-target <electron>-<Positron> formation, set $m_1=m_2=m_e$ in <10/a> and solve for the incident total <energy>:
$$
\boxed{E_{e^+}=\frac{m_\rho^2c^4-2m_e^2c^4}{2m_ec^2}=\frac{(768\,\mathrm{MeV})^2}{2(0.511\,\mathrm{MeV})}-0.511\,\mathrm{MeV}\simeq5.7713\times10^5\,\mathrm{MeV}=577.1\,\mathrm{GeV}.}
$$
The incident <kinetic energy> is smaller by another $m_ec^2$, a negligible relative change here. Most of the fixed-target <energy> becomes motion of the final <rho meson> in the laboratory.
With equal and opposite incoming <momenta>, the <centre-of-momentum frame> is the laboratory. Each incoming particle has the same total <energy>, and there is no final translational <kinetic energy> at formation:
$$
\boxed{E_{e^+}=E_{e^-}=\frac{m_\rho c^2}{2}=384\,\mathrm{MeV}.}
$$
This large contrast illustrates the invariant-energy advantage of colliding beams. The given $768\,\mathrm{MeV}/c^2$ resonance <mass> is used rather than replacing it with a modern tabulated value.
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