Exam 1 - Version 005 – FIRST EXAM – Kleinman –...

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Unformatted text preview: Version 005 – FIRST EXAM – Kleinman – (58225) 1 This print-out should have 15 questions. Multiple-choice 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 x-coordinates of two objects moving along the x-axis 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.

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Exam 1 - Version 005 – FIRST EXAM – Kleinman –...

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