= Solution
Take $+x$ along the upper ship's initial <velocity>, and take $+y$ from the lower ship to the upper ship at closest approach. The upper and lower ships then have initial <velocities> $v_1\widehat{\mathbf x}$ and $-v_2\widehat{\mathbf x}$. Their equal <masses> put each at distance $l/2$ from the <centre of mass>, so
$$
\boxed{I_{\rm CM}=2m(l/2)^2=\frac12ml^2.}
$$
By <momentum conservation>, the <centre of mass> keeps moving with constant <velocity>
$$
\boxed{\mathbf V_{\rm CM}=\frac{v_1-v_2}{2}\widehat{\mathbf x}.}
$$
In that frame the ships have opposite <velocities> of magnitude $u=(v_1+v_2)/2$. Their initial <angular momentum about the centre of mass> is nonzero even though their trajectories were straight:
$$
\mathbf L=-ml\frac{v_1+v_2}{2}\widehat{\mathbf z},\qquad L=\frac{ml(v_1+v_2)}2.
$$
The tether exerts central internal <forces>, so <conservation of angular momentum> gives
$$
\boxed{\omega=\frac L{I_{\rm CM}}=\frac{v_1+v_2}{l}.}
$$
Each ship follows uniform <circular motion> of radius $l/2$ around the translating <centre of mass>, clockwise viewed from $+z$. More explicitly, put the initial centre at the origin and define
$$
\mathbf n(t)=\sin(\omega t)\widehat{\mathbf x}+\cos(\omega t)\widehat{\mathbf y},\qquad\mathbf r_{1,2}(t)=\mathbf V_{\rm CM}t\pm\frac l2\mathbf n(t).
$$
The <kinetic-energy decomposition about the centre of mass> verifies <conservation of mechanical energy> directly:
$$
\begin{aligned}
E_{\rm after}&=\frac12(2m)V_{\rm CM}^2+\frac12I_{\rm CM}\omega^2\\
&=\frac m4(v_1-v_2)^2+\frac m4(v_1+v_2)^2\\
&=\boxed{\frac12mv_1^2+\frac12mv_2^2=E_{\rm before}.}
\end{aligned}
$$
At closest approach the relative radial <velocity> is already zero, so connecting the tether requires no dissipative radial impulse. Subsequently its <tension> is perpendicular to each ship's <velocity> relative to the <centre of mass> and does no net <work> on the pair at fixed $l$.
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