A bullet of mass 10, g and speed 500, m/s is | vedclass

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(B) The angular momentum of the bullet about the hinge is given by $L = mvr$,where $r$ is the distance from the hinge to the point of impact. Since th

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$0.625\, rad/s$

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(B) The angular momentum of the bullet about the hinge is given by $L = mvr$,where $r$ is the distance from the hinge to the point of impact. Since the bullet hits the center of the door,$r = 0.5\, m$.$L = (10 	imes 10^{-3}\, kg) 	imes (500\, m/s) 	imes (0.5\, m) = 2.5\, kg \cdot m^2/s$.The moment of inertia of the door about the hinge is $I = \frac{1}{3} M l^2 = \frac{1}{3} 	imes 12\, kg 	imes (1.0\, m)^2 = 4\, kg \cdot m^2$.By the principle of conservation of angular momentum,the angular momentum of the system just before and after the impact is conserved: $L = I\omega$.$\omega = \frac{L}{I} = \frac{2.5}{4} = 0.625\, rad/s$.

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