A uniform rod of length l and mass m is kept on a horizontal smooth plane.It is free to rotate and move.A particle of same mass m is moving on the plane with velocity v strikes the rod at a point B making an angle 37° with the rod.BC= .The collision is elastic.After collision the angular velocity of rood will be .The impulse of the impact force is .Find x+y+p+q.
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The ball has v' component of its velocity perpendicular to the length of rod immediately after the collision.u is the velocity of the COM of the rod and ω is the angular velocity of the rod just after collision.The ball strikes the rod with speed vcos53° in the perpendicular direction and its component along the length of the rod after the collision is unchanged.Velocity of separation=Velocity of approach. 5 3 ∗ v = ω 4 l +u+v'.......i.Conserving linear momentum of rod+particle system in the direction perpendicular to the rod 5 3 *mv=mu-mv'.......ii.Conservig angular momentum about point D will yield u= ω 3 l .Now pls don't ask me to solve this equations I will 200% err(I have the correct answer tho :P ).U will get x=72,y=55,p=24,q=55.....as a part u may also try to find the distance the centre of mass will travel in which it makes half rotation....if I am not wrong it will be π 3 l .