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Practice Homework 9 Solutions

Practice Homework 9 Solutions - practice 09 – ALIBHAI...

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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 left-hand 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 right-hand 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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Practice Homework 9 Solutions - practice 09 – ALIBHAI...

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