= Solution
\b[Invariant and collision.] For a particle of <rest mass> $m$, <special relativity> gives <relativistic energy> and <momentum>
$$
E=\gamma mc^2,\qquad \mathbf p=\gamma m\mathbf v,
\qquad \gamma=\frac{1}{\sqrt{1-v^2/c^2}}.
$$
Therefore the <relativistic energy-momentum relation> is
$$
\boxed{E^2-p^2c^2=\gamma^2m^2c^4(1-v^2/c^2)=m^2c^4.}
$$
It is the squared <Minkowski norm> of the <four-momentum>, multiplied by $c^2$, and hence is preserved by every <Lorentz transformation>. Explicitly a boost with <speed> $U$ along $x$ gives $E'=\Gamma(E-Up_x)$, $p'_x=\Gamma(p_x-UE/c^2)$, with $p'_y=p_y$, $p'_z=p_z$. Expanding $E'^2-c^2p_x'^2$ gives $E^2-c^2p_x^2$ because $\Gamma^2(1-U^2/c^2)=1$. Spatial rotations preserve $p^2$, completing the invariance argument for all <inertial frames>.
For the incoming moving particle, $v=3c/5$ and $\gamma=5/4$. The total laboratory <energy> and <momentum> before the <perfectly inelastic collision> are
$$
E_i=\left(\frac54+2\right)m_0c^2=\frac{13}{4}m_0c^2,
\qquad p_i=\frac54m_0\frac{3c}{5}=\frac34m_0c.
$$
<Four-momentum conservation> identifies these with the final particle's <energy> and <momentum>. Its <rest mass> is therefore
$$
\boxed{m'=\frac{\sqrt{E_i^2-p_i^2c^2}}{c^2}
=\frac{m_0}{4}\sqrt{13^2-3^2}=\sqrt{10}\,m_0.}
$$
The result exceeds the sum $3m_0$ of the incoming <rest masses>; some initial <kinetic energy> has become internal <rest energy>. Since $pc^2/E=v$ for a massive particle,
$$
\boxed{v'=\frac{p_ic^2}{E_i}=\frac3{13}c,\qquad
\gamma'=\frac{13}{4\sqrt{10}}.}
$$
\b[Two-body decay.] Work first in the parent's rest frame, denoted by a star. <Four-momentum conservation> requires equal and opposite daughter <momenta>. Their equal <rest masses> give equal <energies>:
$$
E_* =\frac{m'c^2}{2}=\frac{\sqrt{10}}2m_0c^2,
\qquad
p_*c=\sqrt{E_*^2-\left(\frac{m_0c^2}{2}\right)^2}
=\frac32m_0c^2.
$$
Thus their rest-frame <velocities> are $v_{*\pm}=\pm3c/\sqrt{10}$. Applying the <relativistic velocity-addition formula> with boost $v'=3c/13$ gives the two laboratory <velocities>:
$$
\boxed{\frac{v_\pm}{c}
=\frac{3/13\pm3/\sqrt{10}}{1\pm9/(13\sqrt{10})}.}
$$
Their laboratory <energies> follow directly from the <Lorentz transformation> of <four-momentum>:
$$
\boxed{E_\pm=\gamma'(E_*\pm v'p_*)
=\left(\frac{13}{8}\pm\frac{9}{8\sqrt{10}}\right)m_0c^2.}
$$
The corresponding signed laboratory <momenta> are
$$
p_\pm=\left(\frac38\pm\frac{39}{8\sqrt{10}}\right)m_0c.
$$
Their sums recover $E_i$ and $p_i$, and each satisfies $E_\pm^2-p_\pm^2c^2=(m_0c^2/2)^2$. One daughter moves forwards and the other backwards in the laboratory; both are collinear with the original flight axis. Numerically, $v_+\simeq0.967615c$, $v_-\simeq-0.919138c$, $E_+\simeq1.980756m_0c^2$ and $E_-\simeq1.269244m_0c^2$.
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