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Unformatted text preview: Version 005 – FIRST EXAM – Kleinman – (58225) 1 This printout should have 15 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A light, inextensible cord passes over a light, frictionless pulley with a radius of 12 cm. It has a(n) 24 kg mass on the left and a(n) 3 . 1 kg mass on the right, both hanging freely. Initially their center of masses are a vertical distance 4 . 2 m apart. The acceleration of gravity is 9 . 8 m / s 2 . 4 . 2 m 12 cm ω 24 kg 3 . 1 kg At what rate are the two masses accelerat ing when they pass each other? 1. 5.03665 2. 6.73012 3. 3.59568 4. 2.18253 5. 4.82687 6. 7.55793 7. 3.87442 8. 5.33235 9. 2.72968 10. 6.53333 Correct answer: 7 . 55793 m / s 2 . Explanation: Let : R = 12 cm , m 1 = 3 . 1 kg , m 2 = 24 kg , h = 4 . 2 m , and v = ω R . Consider the free body diagrams 24 kg 3 . 1 kg T T m 2 g m 1 g a a Since the larger mass will move down and the smaller mass up, we can take motion downward as positive for m 2 and motion up ward as positive for m 1 . Apply Newton’s second law to m 1 and m 2 respectively and then combine the results: For mass 1: summationdisplay F 1 : T − m 1 g = m 1 a (1) For mass 2: summationdisplay F 2 : m 2 g − T = m 2 a (2) We can add Eqs. (1) and (2) above and obtain: m 2 g − m 1 g = m 1 a + m 2 a a = m 2 − m 1 m 1 + m 2 g = 24 kg − 3 . 1 kg 24 kg + 3 . 1 kg (9 . 8 m / s 2 ) = 7 . 55793 m / s 2 . T = m 1 ( g + a ) = (3 . 1 kg) (9 . 8 m / s 2 + 7 . 55793 m / s 2 ) = 53 . 8096 N . 002 10.0 points Hint: x 1 and x 2 never have the same value. The xcoordinates of two objects moving along the xaxis are given below as a function of time t . Version 005 – FIRST EXAM – Kleinman – (58225) 2 x 1 = (4 m / s) t and x 2 = − (207 m) + (28 m / s) t − (1 m / s 2 ) t 2 . Calculate the magnitude of the distance of closest approach of the two objects. 1. 42 m 2. 23 m 3. 30 m 4. 46 m 5. 50 m 6. 33 m 7. 35 m 8. 60 m 9. 63 m correct 10. 56 m Explanation: Given : v 1 = 4 m / s , x = 207 m , v 2 = 28 m / s , and a 2 = 1 m / s 2 . x 1 = v 1 t and x 2 = − x + v 2 t − a 2 t 2 . Δ x = x 2 − x 1 = − x + [ v 2 − v 1 ] t − a 2 t 2 . The distance of closest approach will be when the derivative d Δ x dt = v 2 − v 1 − 2 a 2 t = 0 . Therefore, t = v 2 − v 1 2 a 2 = (28 m / s) − (4 m / s) 2 (1 m / s 2 ) = 12 s . At that time, we have x 2 = − x + v 2 t − a 2 t 2 = − (207 m) + (28 m / s) (12 s) − (1 m / s 2 ) (12 s) 2 = − 15 m x 1 = a t = (4 m / s) (12 s) = 48 m Δ x = x 2 − x 1 = ( − 15 m) − (48 m) = − 63 m  Δ x  = 63 m . A plot of the functions x 1 (upper curve) and x 2 (lower curve) are shown below. One can see that the distance of closest approach occurs at 12 s and the maximum of x 2 occurs at 14 s ....
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This note was uploaded on 04/08/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics, Light

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