Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
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
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.
Find solution of question on doubt solving app Instasolv
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -