Question

A rotating ball hits a rough horizontal plane with a vertical velocity v and angular velocity w.Given that the coefficient of friction is u and the vertical component of the velocity after the collision is v/2.Find



a)the angular velocity after collision

Answer 1

Answer:

Let J1 be the horizontal impulse and J2 be the vertical impulse.

J1=uJ2, J2=mv/2,

let angular velocity after collision=w2,

angular impulse=RJ1=uRJ2=umvR/2=(2mR2/5)(w-w2),5uv=4R(w-w2), w2=w-(5uv/4R), J=√(J12+J22)=(mv/2)√(1+u2)

Explanation: **vertical impulse is 3mv/2 so multiply the ans by 3...

thus , ans is 3(mv/2)√(1+u2)