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For a body constrained to rotate about a fixed pivot, a perpendicular impulse $K$ delivered at distance $b$ from the pivot produces $\omega =Kb/I_O$. If the centre of mass is distance $h$ from the pivot, its momentum is $M\omega h$. The pivot impulse is $J_O=M\omega h-K$, and vanishes when $b=I_O/(Mh)$. This impact location is the centre of percussion. The result assumes the stated planar rigid-body geometry and a short impulse; it is not a claim that all forces at the pivot vanish afterward.

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,The in-plane inertia isotropy of a square plate follows using the perpendicular axis theorem, which gives $I_z=I_x+I_y=M{l}^2/6$. For any in-plane central axis at angle $\varphi$, the perpendicular in-plane axis has the same inertia by the square's $90^{\circ }$ rotational symmetry. Their sum is $I_z$, so every such in-plane axis has $I=M{l}^2/12$.

The parallel axis theorem gives hinge inertia $I_H=M{l}^2/3$. During the short impulse gravity is negligible, and the hinge impulse has no moment about the hinge. ThereforeThe 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,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,