Q. 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[ans.w-(15vu/4R)

b.)the impulsive ground reaction during the collision.{ans :3mv/2 * root of 1+u^2}

Dear Raj, 

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)

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Askiitians Expert

Chetan Mandayam Nayakar – IIT Madras

as told by chetan nayakar , the method is correct but the vertical impulse is 3mv/2 not mv/2  .  so multiply his answer by 3 and you will get your answer