"Sample Work Shikhar Mittal May 22, 2017 Question A ball of mass m and radius r is rolled(pure rolling) towards wall with a speed v and an angular o velocity ! . The coe\u0000cient of restitution and the coe\u0000cient of friction between the wall and the o ball are e and \u0000 , respectively. The coe\u0000cient of restitution and the coe\u0000cient of friction between the ground and the ball are 0 and \u0000 , respectively. Find the value of e and \u0000 so that the ball may come to rest completely. See Fig.1 Figure 1 1Sample Work Shikhar Mittal May 22, 2017 Solution From the de\u0000nition of coe\u0000cient of restitution, the horizontal component of the velocity of the ball is ev , after the collision. The horizontal component of change in linear momentum gives us the o impulse of the normal force as:- Z N dt =m\u0000 ev \u0000 (\u0000 mv ) =mv (1 +e) (1) 1 o o o Let the velocity imparted to the ball by the friction impulse in the upwards direction be v, then, Z F dt =mv\u0000 0 (2) 1 * F =\u0000N 1 1 Z Z ) F dt =\u0000 N dt (3) 1 1 Hence from (??), (??) and (??), we get the vertical component of the velocity:- v =\u0000 (1 +e)v o Friction impulse also changes the angular velocity of the ball, hence let the \u0000nal angular velocity be ! in the anti-clockwise direction Z 2 2 2 2 \u0000 F rdt = mr !\u0000 mr ! 1 o 5 5 v o Substituting ! for o r \u0000 \u0000 5\u0000 (1 +e) ! =! 1\u0000 (4) o 2 After moving like a projectile it hits the ground. For the ball to come to rest completely the \u0000nal velocity should be 0, i.e., vertical component, horizontal component and the angular velocity all three should be 0. Clearly the vertical component will become 0 as e = 0 for the ground. For the 1horizontal component:- Z N dt = 0\u0000 (\u0000 m\u0000 (1 +e)v ) =m\u0000 (1 +e)v 2 o o Z \u0000 F dt = 0\u0000 m\u0000 ev 2 o F =\u0000N 2 2 Simplifying the above equation we get:- 2 \u0000 (1+e) =e (5) Similarly, for the angular velocity to become 0:- Z 2 2 \u0000 F rdt = 0\u0000 mr ! 2 5 Using the above equation and (??) we get:- 2 \u0000 (1+\u0000 )(1+e) = (6) 5 Solve (??) and (??) to get 2 4 \u0000 = and e = 7 45 2"