1. The problem statement, all variables and given/known data The question is given in the following pictures. http://imageshack.us/photo/my-images/851/iitjee2011.png/ http://imageshack.us/photo/my-images/844/iitjee20112.png/ This question had one or more than one correct answer(s). This was asked in IIT-JEE 2011 (which is regarded by some as the toughest engineering entrance exam in the world) The problem is that,after much debate,it was deemed as WRONG (I don't know why).Although the institute's website displayed either A or BOTH A and C as the correct answer(s),grace marks were given to all the students who attempted it. All I want to know is the MISTAKE in this problem. 2. Relevant equations I am using - "M,r,v and ω1 and ω2" as the mass,radius,velocity and angular velocity of the ring before and after the collision respectively. "m,Vx,Vy,x and h" as the mass of the ball,velocity of the ball before collision(x-direction),velocity of the ball after collision (y-direction),perpendicular distance between center of ring and the direction of ball's velocity after the collision and the height above the ground at which the ball hits the ring,respectively. Given : M = 2 kg , r = 0.5 m , v = 1 m/sec ; m = 0.1 kg , Vx = 20 m/sec , Vy = 10 m/sec , h = 0.75 m , x = sq.root of (1/4 - 1/16) = 0.433 m . Also I is the moment of inertia of the ring (Mr^2) = 0.5 kg m^2 3. The attempt at a solution Taking the left direction as negative,upward as positive,anticlockwise rotation as positive and applying the law of conservation of angular momentum, → Iω1 + m(Vx)(h-r) - Mvr = m(Vy) + Iω2 → ω1=v/r ( pure rolling ) = 2 rad/sec → 0.5 = 0.433 + Iω2 → ω2 = 0.134 rad/sec (is in the anticlockwise direction because it is positive) .....1 Applying law of conservation of linear momentum, → mV - Mv = 2-2 = 0 So linear velocity of ring after collision will be zero ..... 2 From results ....1 and ....2 we can say that the ring will not move but will only rotate about it center of mass.So option A is correct(according to me) I think option C is also correct becasue friction will be towards left as it has the ring will have a tendency to slip towards right. What is confusing me is that during pure rolling(as mentioned in option A) friction does not act so after the collision it shall first slip and then purely rotate about its com . I want to if I am wrong anywhere and also what is WRONG with this problem ? Thanks.