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
- A6 v 32 a
- B6 v 33 a
- C6 v 40 a
- 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