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AP EAMCET2006PhysicsRotational Motion

A uniform rod of length 8 a and mass 6 m lies on a smooth horizontal surface. Two point masses m and 2 m moving in the same plane with speed 2 v and v respectively strike the rod perpendicularly at distances a and 2 a from the mid point of the rod in the opposite directions and stick to the rod. The angular velocity of the system immediately after the collision is :

Options

  1. A6 v 32 a
  2. B6 v 33 a
  3. C6 v 40 a
  4. D6 v 41 a

Correct answer

D. 6 v 41 a

Step-by-step solution

Applying conservation of angular momentum about point O , m(a)(2 v)+2 m(2 a)(v)=I or = 6 m a v I ...(i) Now, I= 6 m(8 a)^2 12 +m (a^2 )+2 m(2 a)^2=32 m a^2+m a^2+8 m a^2=41 ma ^2 Hence, from Eq. (i) -= 6 m a v 41 m a^2 = 6 v 41 a

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