= Solution
At the threshold found in <9/c/i>, both products are stationary in the <centre-of-momentum frame>, so they have the same <velocity> in the <laboratory frame>. Take the energetic incident <photon> along $+x$. The total <energy> is $E+\varepsilon$ and total <momentum> is $(E-\varepsilon)/c$, so the <centre-of-momentum frame> moves at
$$
\boxed{\mathbf V=c\frac{E-\varepsilon}{E+\varepsilon}\hat{\mathbf x}.}
$$
For the numerical threshold,
$$
\frac Vc=1-\frac{2\varepsilon}{E+\varepsilon}
\simeq1-7.66\times10^{-18},\qquad
\gamma_{\rm CM}=\frac{E+\varepsilon}{2m_ec^2}\simeq2.56\times10^8.
$$
This is extremely close to, but strictly below, the <speed of light>. Each product has <energy> $(E+\varepsilon)/2$ in the <laboratory frame>. Above threshold the same ratio of total <momentum> to total <energy> determines the motion of the pair's <centre-of-momentum frame>, but the individual <electron> and <Positron> need not have a common <velocity>.
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