= Solution
About a specified axis, the <moment of inertia> is $I=\int r_\perp^2\,dm$, giving <rotational kinetic energy> $I\omega^2/2$ for fixed-axis rotation. With square coordinates centred on the plate,
$$
I_x=\frac{M}{l^2}\int_{-l/2}^{l/2}\int_{-l/2}^{l/2}y^2\,dx\,dy
=\boxed{\frac{Ml^2}{12}}.
$$
The <in-plane inertia isotropy of a square plate> follows using the <perpendicular axis theorem>, which gives $I_z=I_x+I_y=Ml^2/6$. For any in-plane central axis at angle $\phi$, the perpendicular in-plane axis has the same inertia by the square's $90^\circ$ rotational symmetry. Their sum is $I_z$, so \b[every such in-plane axis has $I=Ml^2/12$].
The <parallel axis theorem> gives hinge inertia $I_H=Ml^2/3$. During the short <impulse> <gravity> is negligible, and the hinge <impulse> has no moment about the hinge. Therefore
$$
I_H\omega=K(l/2+a),\qquad
\boxed{v=\frac{\omega l}{2}=\frac{3K}{4M}\left(1+\frac{2a}{l}\right).}
$$
The <hinge impulse of an edge-pivoted square plate> follows from <linear momentum>: $J_H=Mv-K$. Taking its positive direction to be that of the applied <impulse>,
$$
\boxed{J_H=K\left(\frac{3a}{2l}-\frac14\right),\qquad
N=|J_H|.}
$$
Generally $K$ alone does not supply the centre-of-mass <momentum> compatible with rotation about the fixed hinge, so the hinge supplies the difference. Its <impulse> vanishes at the <centre of percussion>,
$$
\boxed{a=l/6.}
$$
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