Sphere of mass M and radius R is kept on a rough horizontal floor. A small particle of mass m, moving horizontally with velocity, collides with the sphere and sticks to it. Line of motion of particle before collision is at a heighth above the floor. Assume that mase of sphere is very large in comparison to particle so that we may assume that centre of mass of the combined system remains at the centre of the sphere after the collision. Angular velocity of the sphere just after the collision is

Sphere of mass M and radius R is kept on a rough horizontal floor. A s

1.	Sphere of mass M and radius R is kept on a rough horizontal floor. A small particle of mass m, moving horizontally with velocity, collides with the sphere and sticks to it. Line of motion of particle before collision is at a heighth above the floor. Assume that mase of sphere is very large in comparison to particle so that we may assume that centre of mass of the combined system remains at the centre of the sphere after the collision. Angular velocity of the sphere just after the collision is
Answer» `(5"mu")/(2MR)` `(5"mu"(h-R))/(2MR^(2))` `(2"mu"(h-R))/(5MR^(2))` `(2"Mu"(h-R))/(2mR^(2))` Solution :We can apply conservation of angular momentum about CENTRE of sphere just before and just after the collision. `L_(1)=L_(2)` `m u(h-R)=[(2)/(5)MR^(2)+mR^(2)] omega` `omega=(m u(h-R))/((2)/(5) MR^(2)+mR^(2))=(m u(h-R))/(((2)/(5)M+m)R^(2))` SINCE `M gt gt m` so we can neglect m in the denominator in COMPARISON to M. `omega~~(5 m u)(h-R))/(2MR^(2))` Hence option (B) is corrrect.