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Unformatted text preview: Adams, Russell Homework 8 Due: Nov 3 2006, 11:00 pm Inst: Opyrchal, H 1 This printout should have 9 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A 4 kg steel ball strikes a wall with a speed of 11 . 9 m / s at an angle of 30 . 7 with the normal to the wall. It bounces off with the same speed and angle, as shown in the figure. x y 1 1 . 9 m / s 4 kg 1 1 . 9 m / s 4 kg 30 . 7 30 . 7 If the ball is in contact with the wall for . 254 s, what is the magnitude of the average force exerted on the ball by the wall? Correct answer: 322 . 275 N. Explanation: Let : M = 4 kg , v = 11 . 9 m / s , and = 30 . 7 . 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(4 kg)(11 . 9 m / s) cos30 . 7 . 254 s k ~ F k = 322 . 275 N . Note: The direction of the force is in negative x direction, as indicated by the minus sign. keywords: 002 (part 1 of 1) 10 points A superball with a mass of 25 . 3 g is dropped from a height of 2 . 9 m. It rebounds to a height of 2 . 4 m. The acceleration of gravity is 9 . 8 m / s 2 . What is the change in its linear momentum during the collision with the floor? Correct answer: 0 . 364264 kgm / s. Explanation: Given the height, the speed of the ball right before it hits the floor is v i = p 2 g h i . Like wise, thespeedoftheballrightafter theithits the floor is v f = q 2 g h f . Thus the change in momentum, or the impulse is I = mv f mv i = m hq 2 g h f + p 2 g h i i = (25 . 3 g) [(7 . 53923 m / s) + (6 . 85857 m / s)] = 0 . 364264 kgm / s , where v i = p 2 g h i = q 2(9 . 8 m / s 2 )(2 . 9 m) = 7 . 53923 m / s and v f = q 2 g h f = q 2(9 . 8 m / s 2 )(2 . 4 m) = 6 . 85857 m / s and = 6 . 85857 m / s ....
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 Spring '08
 moro
 Physics, Work

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