= Solution
In the initial <centre-of-mass frame>, equal fragments acquire opposite velocities $\pm w$ by <conservation of momentum>. The released energy is
$$
\Delta K=2\left[\frac12\frac m2w^2\right]=\frac12mw^2.
$$
The <equal-mass fragmentation energy budget> needs $w=500\,\mathrm{m\,s^{-1}}$ merely to match the other ship, costing $1.25\,\mathrm{GJ}$. To close an existing gap requires a little more than this. Using the full reserve gives
$$
\boxed{w_{\max}=\sqrt{\frac{2(2\times10^9)}{10^4}}=632\,\mathrm{m\,s^{-1}},\qquad
v_{\rm front}=2.632\,\mathrm{km\,s^{-1}}.}
$$
\b[The reserve is sufficient to overtake.] The rear half moves at $1.368\,\mathrm{km\,s^{-1}}$, preserving momentum. The kinetic-energy increase is the same in any inertial frame because total momentum is unchanged. Subsequent capture or braking would require additional interaction details; no target mass is needed for the interception test.
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