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A POINT MASS M MOVING WITH VELOCITY U TOWARDS THE CENTRE OF THE DISC COLLIDES WITH A DISCOF MASS M AND RADIUS R RESTING ON A ROUGH HORIZONTAL SURFACE. ITS COLLISION IS PERFECTLY INELASTIC . FIND ANGULAR VELOCITY OF SYSTEM AFTER PURE ROLLING STARTS. A. 2U/7R B. 7U/2R C. 5U/2R D. 2U/5R
The answer is A. 2U/7R. Considering the conservation of momentum, we get the velocity of the combined mass to be U/2. Consider the angular velocity of the system just after the collision about the point of contact of the disc. 2M * U/2 * R = M U R Angular momentum will be conserved about this point because the frictional force will pass through the point of contact, and will thus not provide any torque. The moment of Inertia of the combined mass will be (1/2MR^2 + MR^2). Thus, applying conservation of angular momentum about the point of contact, M U R = (1/2 MR^2 + MR^2)* w + 2M V R When pure rolling starts, w= V R So, M U R = 3/2 MR^2 w +2Mw. After simplification, we get the required result.
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