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Ch2-HW1-solutions

# Ch2-HW1-solutions - schopper(as55864 Ch2-HW1...

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schopper (as55864) – Ch2-HW1 – Antoniewicz – (56465) 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 10.0points 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. d 2. b 3. h 4. e correct 5. c 6. a 7. g 8. f 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(part1of3)4.0points A hockey puck is sliding along the ice with nearly constant momentum vectorp i = ( 10 , 0 , 8 ) kg · m / s when it is suddenly struck by a hockey stick with a force vector F = ( 0 , 0 , 1400 ) N that lasts for only 8 milliseconds (0 . 008 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 , 0 , 8 ) kg · m / s + ( 0 , 0 , 1400 ) N × (0 . 008 s) vectorp f ≈ ( 10 , 0 , 19 . 2 ) 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(part2of3)3.0points Find p f,y . Answer in kg · m / s. Correct answer: 0 kg · m / s. Explanation: See the explanation to part 1. 004(part3of3)3.0points Find p f,z . Answer in kg · m / s. Correct answer: 19 . 2 kg · m / s.

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schopper (as55864) – Ch2-HW1 – Antoniewicz – (56465) 2 Explanation: See the explanation to part 1. 005(part1of9)2.0points You were driving a car with velocity vectorv i = ( 27 , 0 , 13 ) m / s . You quickly turned and braked, and your ve- locity became vectorv f = ( 9 , 0 , 19 ) 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: 18000 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) ( ( 9 , 0 , 19 ) m / s −( 27 , 0 , 13 ) m / s) = (− 18000 , 0 , 6000 ) kg · m / s 006(part2of9)1.0points Find Δ p y . Answer in kg · m / s.
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Ch2-HW1-solutions - schopper(as55864 Ch2-HW1...

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