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Unformatted text preview: pasha (sep635) HW #2-1 Antoniewicz (56445) 1 This print-out should have 29 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A ball moves in the direction of the arrow labeled c in the following diagram. The ball is struck by a stick that briefly exerts a force on the ball in the direction of the arrow labeled e . Which arrow best describes the direction of ~ p , the change in the balls momentum? a b c d e f g h 1. g 2. a 3. d 4. f 5. b 6. e correct 7. c 8. h Explanation: Recall the definition of impulse: Impulse = ~ F net t = ~ p. Therefore, whatever direction the net force points in, that will be the direction of the change in the balls momentum. We are told that the force is in the direction of e , so that is the correct answer. 002 (part 1 of 3) 4.0 points A hockey puck is sliding along the ice with nearly constant momentum ~ p i = h 10 , , 8 i kg m / s when it is suddenly struck by a hockey stick with a force ~ F = h , , 1800 i N that lasts for only 4 milliseconds (0 . 004 s). What is the final momentum ~ p f of the puck? Begin by finding just the x component, p f,x . Answer in kg m / s. Correct answer: 10 kg m / s. Explanation: This is a straightforward application of the momentum principle. Let the system be the puck, and assume the only significant interac- tion on the system is with the hockey stick. ~ p = ~ F net , sys t ~ p f- ~ p i = ~ F net , sys t ~ p f = ~ p i + ~ F net , sys t ~ p f = h 10 , , 8 i kg m / s + h , , 1800 i N (0 . 004 s) ~ p f h 10 , , 15 . 2 i kg m / s . This gives us the answers to all three parts. The x component of the final momentum is p f,x = 10 kg m / s . 003 (part 2 of 3) 3.0 points Find p f,y . Answer in kg m / s. Correct answer: 0 kg m / s. Explanation: See the explanation to part 1. 004 (part 3 of 3) 3.0 points Find p f,z . Answer in kg m / s. Correct answer: 15 . 2 kg m / s. pasha (sep635) HW #2-1 Antoniewicz (56445) 2 Explanation: See the explanation to part 1. 005 (part 1 of 9) 2.0 points You were driving a car with velocity ~ v i = h 27 , , 12 i m / s . You quickly turned and braked, and your ve- locity became ~ v f = h 10 , , 18 i m / s . The mass of the car was 1000 kg. What are the components of the (vector) change in momentum ~ p during this maneuver? Start with the x component, p x , and give your answer in kg m / s. Pay attention to signs. Correct answer:- 17000 kg m / s. Explanation: Lets go ahead and do the vector algebra for all three components. Were given the initial and final speeds, which are non-relativistic, and the mass of the car. Let the system consist of the car. We have ~ p sys = ~ p f- ~ p i = m car ~ v f- m car ~ v i = m car ( ~ v f- ~ v i ) = (1000 kg) ( h 10 , , 18 i m / s-h 27 , , 12 i m / s) = h- 17000 , , 6000 i kg m / s 006 (part 2 of 9) 1.0 points Find p y . Answer in kg m / s.s....
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This note was uploaded on 11/17/2011 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas at Austin.
- Fall '08