= Solution
For an <elastic eccentric impact of a thin disc>, take upward <velocity> as positive and let $J$ be the short upward <impulse>. The disc's <moment of inertia> about a horizontal diameter is $I=Ma^2/4$. With impact offset $d=a/2$,
$$
v_C=-u+J/M,\qquad \omega=Jd/I.
$$
The contact point has upward <velocity> $v_C+d\omega$. Unit restitution against the fixed post requires that <velocity> to be $+u$, so
$$
J\left(\frac1M+\frac{d^2}I\right)=2u.
$$
Since $I=Md^2$, \b[$J=Mu$ and the centre's instantaneous <speed> is zero]. The disc rotates with $\omega=2u/a$; its <rotational kinetic energy> $I\omega^2/2=Mu^2/2$ equals the entire initial <energy>. <Gravity>'s <impulse> during the impact is neglected. Using the polar-axis <moment of inertia> would describe the wrong rotation axis.
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