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Unformatted text preview: madrid (tmm2353) – 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 Δ vectorp , the change in the ball’s momentum? a b c d e f g h 1. c 2. g 3. e correct 4. f 5. a 6. d 7. b 8. h Explanation: Recall the definition of impulse: Impulse = vector F net Δ t = Δ vectorp. Therefore, whatever direction the net force points in, that will be the direction of the change in the ball’s 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 vectorp i = ( 10 , , 4 ) kg · m / s when it is suddenly struck by a hockey stick with a force vector F = ( , , 2000 ) N that lasts for only 7 milliseconds (0 . 007 s). What is the final momentum vectorp 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. Δ vectorp = vector F net , sys Δ t vectorp f − vectorp i = vector F net , sys Δ t vectorp f = vectorp i + vector F net , sys Δ t vectorp f = ( 10 , , 4 ) kg · m / s + ( , , 2000 ) N × (0 . 007 s) vectorp f ≈ ( 10 , , 18 ) 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: 18 kg · m / s. madrid (tmm2353) – 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 vectorv i = ( 30 , , 14 ) m / s . You quickly turned and braked, and your ve- locity became vectorv f = ( 11 , , 18 ) m / s . The mass of the car was 1000 kg. What are the components of the (vector) change in momentum Δ vectorp during this maneuver? Start with the x component, Δ p x , and give your answer in kg · m / s. Pay attention to signs. Correct answer: − 19000 kg · m / s. Explanation: Let’s go ahead and do the vector algebra for all three components. We’re 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 Δ vectorp sys = vectorp f − vectorp i = m car vectorv f − m car vectorv i = m car ( vectorv f − vectorv i ) = (1000 kg) ( ( 11 , , 18 ) m / s −(...
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This note was uploaded on 11/15/2011 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.
- Fall '08