= Solution
For a particle of <rest mass> $m_r$, $E=\gamma m_rc^2$ and $p=\gamma m_rv$, with $\gamma=(1-v^2/c^2)^{-1/2}$. The low-speed expansion is $E=m_rc^2+\tfrac12m_rv^2+O(m_rv^4/c^2)$, separating invariant <rest energy> from the nonrelativistic <kinetic energy> term.
Conserve the total laboratory <energy> and <momentum> at coalescence: $E_{\rm tot}=4mc^2$ and $p_{\rm tot}=2.4mc$. Since $V=c^2p_{\rm tot}/E_{\rm tot}$, \b[the combined particle moves at $0.6c$]. Its <rest mass> is $M=\sqrt{4^2-2.4^2}\,m=3.2m$. The incident particle is not of <mass> $m$: its stated <energy> and <momentum> imply $m_{\rm incident}=\sqrt{3^2-2.4^2}\,m=1.8m$. Thus
$$
\boxed{M=3.2m>m_{\rm incident}+m_{\rm target}=2.8m.}
$$
The extra $0.4mc^2$ is internal <rest energy> retained by the composite after a <perfectly inelastic collision>; total <energy> and <four-momentum> are conserved, while the sum of individual <rest masses> need not be.
The composite's <Lorentz factor> is $1.25$. For a rest-frame lifetime $\tau$, the corresponding laboratory time is $1.25\tau$ and its travel <distance> is
$$
\boxed{\ell=V\gamma\tau=0.75c\tau.}
$$
If $\tau$ denotes a mean radioactive lifetime, this is the mean flight <distance>, not a fixed <distance> for every decay. The separate frame diagrams below use the common positive collision axis.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1-two-photon-frames.png]
{title=Original momentum-direction diagrams for the laboratory and centre-of-momentum frames, with equal-energy photons and the composite at rest in its own frame}
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