solution-hw13

solution-hw13 - lai (yl8859) HW13 Ross (15200) 1 This...

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Unformatted text preview: lai (yl8859) HW13 Ross (15200) 1 This print-out should have 6 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A 9 kg steel ball strikes a wall with a speed of 11 m / s at an angle of 42 . 5 with the normal to the wall. It bounces off with the same speed and angle, as shown in the figure. x y 1 1 m / s 9 kg 1 1 m / s 9 kg 42 . 5 42 . 5 If the ball is in contact with the wall for . 198 s, what is the magnitude of the average force exerted on the ball by the wall? Correct answer: 737 . 277 N. Explanation: Let : M = 9 kg , v = 11 m / s , and = 42 . 5 . The y component of the momentum is un- changed. The x component of the momentum is changed by P x = 2 M v cos . Therefore, using impulse formula, F = P t = 2 M v cos t = 2 (9 kg) (11 m / s) cos 42 . 5 . 198 s bardbl vector F bardbl = 737 . 277 N ....
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This note was uploaded on 02/24/2011 for the course PHYS 152 taught by Professor Button during the Winter '08 term at IUPUI.

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solution-hw13 - lai (yl8859) HW13 Ross (15200) 1 This...

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