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Unformatted text preview: practice 09 – ALIBHAI, ZAHID – Due: Mar 25 2007, 4:00 am 1 Question 1 part 1 of 1 10 points Assume an elastic collision (ignoring fric tion and rotational motion). A queue ball initially moving at 4 . 2 m / s strikes a stationary eight ball of the same size and mass. After the collision, the queue ball’s final speed is 1 . 8 m / s . 4 . 2 m / s 1 . 8 m / s θ φ Before After Find the queue ball’s angle θ with respect to its original line of motion. Correct answer: 64 . 6231 ◦ (tolerance ± 1 %). Explanation: Let : v q i = 4 . 2 m / s and v q f = 1 . 8 m / s . 4 . 2 m / s 1 . 8 m / s 3 . 7 9 m / s 64 ◦ 26 ◦ 90 ◦ Before After Solution: θ = arccos bracketleftbigg v q f v q i bracketrightbigg = arccos bracketleftbigg (1 . 8 m / s) (4 . 2 m / s) bracketrightbigg = 64 . 6231 ◦ . Question 2 part 1 of 1 10 points Two identical balls are on a frictionless hor izontal tabletop. Ball X initially moves at 10 meters per second, as shown in figure on the lefthand side. It then collides elastically with ball Y , which is initially at rest. After the col lision, ball X moves at 6 meters per second along a path at 53 ◦ to its original direction, as shown in on the righthand side. 10 m/s X before Y after 6 m/s X 0 m/s b b 53 ◦ Which of the following diagrams best repre sents the motion of ball Y after the collision? 1. 8 m/s Y after b 2. after 4 m/s Y 53 ◦ b 3. 6 m/s Y after b 4. after 8 m/s Y 53 ◦ b 5. 0 m/s Y b after 6. after 6 m/s Y 37 ◦ b 7. after 8 m/s Y 37 ◦ b correct practice 09 – ALIBHAI, ZAHID – Due: Mar 25 2007, 4:00 am 2 8. after 4 m/s Y 37 ◦ b 9. 4 m/s Y after b 10. after 6 m/s Y 53 ◦ b Explanation: From conservation of momentum, m 1 vectorv 1 i + m 2 vectorv 2 i = m 1 vectorv 1 f + m 2 vectorv 2 f mvectorv = mvectorv 1 + mvectorv 2 vectorv = vectorv 1 + vectorv 2 . This relation leads to the vector diagram be low, where it is used that when two equal mass have an elastic collision, and one of the masses was originally at rest, they move per pendicular to each other after the collision. vectorv 2 θ b vectorv 1 φ b vectorv Now 90 ◦ = φ + θ , and with φ = 53 ◦ θ = 90 ◦ − 53 ◦ = 37 ◦ Then  vectorv  cos θ =  vectorv 2   vectorv 2  = (10 m / s) cos 37 ◦ = 8 m / s Question 3 part 1 of 3 10 points A hockey puck of mass 0.6 kg with an ini tial speed of 9 . 14 m / s collides elastically with another equally massive puck initially at rest. After the collision, the two hockey pucks are observed to have the same speed, each recoil ing symmetrically at equal magnitude but op positely directed angles from the initial puck’s direction. v v v Φ Before After What is the angle Φ between the two recoil ing pucks?...
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 Spring '08
 Turner
 Friction, Kinetic Energy, Mass, Momentum, Work, Moment Of Inertia, Correct Answer, kg, hockey

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