= Solution
Interpret $m$ as the <rest mass> of each initial ball and assume that no energy or <momentum> escapes the collision. Let $\gamma=(1-v^2/c^2)^{-1/2}$ be its <Lorentz factor>. The total initial energy and <momentum> are
$$
E=mc^2(\gamma+1),\qquad P=\gamma mv.
$$
Their conservation determines the final <four-momentum>. For a single lump, $P=\gamma'm'v'$ and $E=\gamma'm'c^2$, hence
$$
\boxed{v'=\frac{Pc^2}{E}=\frac{\gamma v}{\gamma+1}.}
$$
The final <invariant mass> follows from $m'^2c^4=E^2-c^2P^2$:
$$
m'^2=m^2\left[(\gamma+1)^2-\gamma^2\frac{v^2}{c^2}\right]
=m^2(2+2\gamma),\qquad
\boxed{m'=m\sqrt{2(1+\gamma)}.}
$$
This is the <rest mass of a coalescing relativistic collision>. For $v>0$, $m'>2m$: some <kinetic energy> becomes <internal energy> in the final rest frame. Setting $m'=2m$ would discard that energy and contradict conservation. At small speeds $v'\sim v/2$, while $v=0$ gives $m'=2m$.
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