A thin uniform rod of mass m and length l is hinged at the upper end and free to rotate about its upper end in vertical plane . When it is at rest, it receives an impulse J at its lowest point, normal to its length. Immediately after impact,

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Angular momentum $$=$$ linear momentum x perpendicular distance from the point of rotation  During the course of collision angular momentum of the rod about the axis of rotation remains conserved .                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                       Therefore  $$L=JI$$ Also, $$I=ml23$$        ω$$=LI=JIml2/3=3Jml$$ Kinetic energy $$=L22l=J2l22ml23=3J22m$$        $$vc=$$ω$$l2=3J2m$$

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