EXTENDED object (rod) where a perpendicular impact will produce no reactive SHOCK at the pivot.Let us consider a rod of length L and mass m and select a pivot point at a distance p from the center of mass. We strike the rod a short blow a distance bi from the center of mass as shownfrom Newtons' second law\({\rm{m}}\FRAC{{\rm{d}}}{{{\rm{dt}}}}{{\rm{V}}_{{\rm{centre\;of\;mass\;}}}} = {\rm{F}}\) .... (I)or \({{\rm{I}}_{{\rm{COM}}}}\frac{{{\rm{d\omega }}}}{{{\rm{dt}}}} = {\rm{F}} \times {{\rm{b}}_{\rm{i}}}\) ....(II)now for any point P at a distance p from the center of mass, the change of its velocity due to the rotation is GIVEN bydVp = dVcom – pdω\(\frac{{d{V_p}}}{{dt}} = \frac{{d{V_{COM}}}}{{dt}} - \frac{{pdw}}{{dt}}\)from equation (I) and (II)\(\frac{{{\rm{d}}{{\rm{V}}_{\rm{p}}}}}{{{\rm{dt}}}} = {\rm{F}}\left( {\frac{1}{{\rm{m}}} - \frac{{{\rm{p}}{{\rm{b}}_{\rm{i}}}}}{{{{\rm{I}}_{{\rm{COM}}}}}}} \right)\) since p is point of center of mass hence\(\frac{{{\rm{d}}{{\rm{V}}_{\rm{p}}}}}{{{\rm{dt}}}}=0\)\({\rm{F}}\left( {\frac{1}{{\rm{m}}} - \frac{{{\rm{p}}{{\rm{b}}_{\rm{i}}}}}{{{{\rm{I}}_{{\rm{COM}}}}}}} \right)=0\)\({b_i} = \frac{{{I_{COM}}}}{{pm}}\) ....(I)since in question, it is given that rod is pivoted at on of their endpoint hence \(p=\frac{l}{2}\)and \({{\rm{I}}_{{\rm{COM}}}} = \frac{1}{{12}}{\rm{m}}{{\rm{l}}^2}\)hence after putting values of ICOM and p in equation (I)\(b_i=\frac{l}{6}\)center of percussion = p + bi center of percussion = \(\frac{l}{2}+\frac{l}{6}\)center of percussion = \(\frac{2l}{3}\)