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second exam - Version 013 SECOND EXAM Kleinman(58225 This...

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Version 013 – SECOND EXAM – Kleinman – (58225) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Two identical balls (labelled A and B) move on a frictionless horizontal tabletop. Initially, ball A moves at speed v A, 0 = 10 m / s while ball B is at rest ( v B, 0 = 0). The two balls collide off-center, and after the collision ball A moves at speed v A = 6 m / s in the direction θ A = 53 from its original velocity vector: 10 m / s A before B after 6 m / s A 0 m/s 53 Which of the following diagrams best repre- sents the motion of ball B after the collision? 1. 0 m/s B after 2. after 6 m / s B 37 3. 8 m / s B after 4. after 6 m / s B 53 5. after 8 m / s B 53 6. after 4 m / s B 53 7. 4 m / s B after 8. 6 m / s B after 9. after 4 m / s B 37 10. after 8 m / s B 37 correct Explanation: Because we are given both the speed and the direction of ball A after the collision, the problem can be solved in terms of momentum conservation only. Indeed, there are no (horizontal) external forces acting on the two balls, so their net (horizontal) momentum vector is conserved during the collision: vector P net = mvector v A, 0 + mvector v B, 0 [before] = mvector v A + mvector v B [after] and since vector v B, 0 = vector 0 (ball B is initially at rest), vector v A + vector v B = vector v A, 0 . Pictorially, this means vector v A vector v B θ A θ B vector v A, 0
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Version 013 – SECOND EXAM – Kleinman – (58225) 2 and in components, v A × cos θ A + v B × cos θ B = v A, 0 , v A × sin θ A v B × sin θ B = 0 . Solving for the two components of ball B’s velocity, we find v B,x v B × cos θ B = v A, 0 v A cos θ A (1) = (10 m / s) (6 m / s) cos 53 = 6 . 4 m / s , v B,y v B sin θ B = v A sin θ A (2) = (6 m / s) sin 53 = 4 . 8 m / s , and hence v B = radicalBig v 2 B,x + v 2 B,y = radicalBig (6 . 4 m / s) 2 + (4 . 8 m / s) 2 = 8 m / s , θ B = arctan v B,y v B,x = arctan 4 . 8 m / s 6 . 4 m / s = 37 . Solving an Elastic Collision: If we know that the collision in question is perfectly elastic — and it is — then we do not need to be given both the speed v A and the direction θ A of the ball A after the collision: Any one of these two parameters would determine the other. The reason for this is conservation of kinetic energy in an elastic collision: K net = m v 2 A, 0 2 + m v 2 A, 0 2 [before] = m v 2 A 2 + m v 2 B 2 [after] and therefore v 2 A + v 2 B = v 2 A, 0 . (3) At the same time, eqs. (1) and (2) above imply v 2 B = v 2 B,x + v 2 B,y = parenleftBig v A, 0 v A cos θ A parenrightBig 2 + parenleftBig v A sin θ A parenrightBig 2 = v 2 A, 0 2 v A, 0 × v A × cos θ A + v 2 A (the cosine theorem), and hence v 2 A + v 2 B v 2 A, 0 = (4) = 2 v 2 A 2 v A, 0 × v A × cos θ A . Comparing this formula to eq. (3), we imme- diately see that 2 v 2 A 2 v A, 0 v A cos θ A = 0 and therefore v A = v A, 0 × cos θ A . (5) Note that the problem data indeed satisfy this equation, which confirms that the collision in question is perfectly elastic.
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