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Unformatted text preview: pasha (sep635) HW #21 Antoniewicz (56445) 1 This printout should have 29 questions. Multiplechoice 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 #21 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 nonrelativistic, 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 / sh 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
 Turner
 Momentum

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