IIT JEE , AIEEE Forum - Physics , Mechanics

swetalina

SIR,

A BULLET OF MASS 10g AND SPEED 500 m/sIS FIRED INTO A DOOR AND GETS EMBEDDED EXACTLY AT THE CENTRE OF THE DOOR.THE ROAD IS 1m WIDE AND WEIGHS 12kg.IT IS HINGED AT ONE END AND ROTATES ABOUT A VERTICAL AXIS PRACTICALLY WITHOUT FRICTION.FIND THE ANGULAR SPEED OF THE DOOR JUST AFTER THE BULLET EMBEDS INTO IT.

cvramana

Since the torque about the vertical axis is zero, we can conserve angular momentum about the vertical axis.

m v d = I w + m d² w -------------- (1)

Where 'm' is the mass of the bullet, 'd' is the distance of the center of the door from the axis, 'w' is the angular speed after impact, 'v' is the velocity of the bullet before impact, 'I' is the moment of inertia of the door about the axis of rotation.

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