= Solution
Choose the smaller disc's initial sense as positive. For either uniform <disk>, the axial <moment of inertia> is $I=\tfrac12mr^2$, so $I_1=I_2/2$. The signed initial <angular velocities> are $\Omega_1=2\omega_2$ and $\Omega_2=-\omega_2$, giving total axial <angular momentum> zero. The frictionless axle supplies no axial <torque>, and the pins' negligible <mass> supplies no extra stored <kinetic energy>. Thus
$$
I_1\Omega'_1+I_2\Omega'_2=0,\qquad
\frac12I_1(\Omega'_1)^2+\frac12I_2(\Omega'_2)^2=\frac12I_1(2\omega_2)^2+\frac12I_2\omega_2^2.
$$
These equations allow the original state or its simultaneous reversal. A nontrivial elastic pin impact selects reversal of the relative contact <velocity>, rather than continued interpenetration. Therefore
$$
\boxed{\Omega'_1=-2\omega_2=-\omega_1,\qquad \Omega'_2=+\omega_2.}
$$
The requested magnitudes are unchanged: $|\Omega'_1|=\omega_1$ and $|\Omega'_2|=\omega_2$. The intersecting pin paths make the impact possible; no additional lever arm is needed once axial <angular momentum> and <energy> conservation have been imposed.
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