= Solution
Put $\beta=v/c$ and $\gamma=(1-\beta^2)^{-1/2}$, the common <Lorentz factor>. The initial total <relativistic energy> and <momentum> are
$$
E=\gamma mc^2+\gamma\frac m2c^2=\frac32\gamma mc^2,\qquad
P_x=\gamma mv-\gamma\frac m2v=\frac12\gamma mv.
$$
Assume that coalescence emits no additional radiation. By <four-momentum conservation>, the final <rest mass> $m_f$ is the <rest mass of a coalescing relativistic collision>:
$$
m_f^2c^4=E^2-c^2P_x^2,
\qquad
\boxed{m_f=\frac{\gamma m}{2}\sqrt{9-\beta^2}
=\frac m2\sqrt{\frac{9-\beta^2}{1-\beta^2}}.}
$$
The final <velocity> is $V=c^2P_x/E=v/3$, the <centre-of-momentum frame> <velocity>.
For $v\ll c$,
$$
\boxed{m_f=\frac32m+\frac23m\frac{v^2}{c^2}+O(v^4/c^4).}
$$
The leading term is the classical sum of <masses>. The first correction is the internal <energy> retained after sticking, divided by $c^2$: the initial classical <kinetic energy> is $3mv^2/4$, while the common-motion <kinetic energy> is $\tfrac12(3m/2)(v/3)^2=mv^2/12$. Their difference is $2mv^2/3$, consistent with <mass-energy equivalence>.
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