A homogeneous square block of mass m = 1mvkg and side length l = 6 m rests on aтhorizontal surface. The bottom right cornerpivotof the block is attached to the surface bya pivot. A point particle of the same massm = 1 kg moving horizontally with speed v = 64 m/s, as shown in thefigure, hits the block at its upper left corner and sticks to the corner whereit hits. (Moment of inertia of a homogeneous square of mass m and sidelength l around its center of mass is I = ml²/6.)What is the moment of inertia of the particle, when it is at the upper leftcorner of the block it stuck, around the pivot?(a) 72 kg-m²(e) 48 kg·m²(b) 36 kg-m²(c) 96 kg-m²(d) 648 kg-m²

Solution: Given values, Mass of the point particle(m)=1 Kg side length of the square (l)=6 m…

Answered: A homogeneous square block of mass m =…

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