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Exam2-solutions - Version 011 Exam2 distler(56295 This...

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Version 011 – Exam2 – distler – (56295) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Two ice skaters approach each other at right angles. Skater A has a mass of 78 . 8 kg and travels in the + x direction at 2 . 82 m / s. Skater B has a mass of 43 . 5 kg and is moving in the + y direction at 2 . 24 m / s. They collide and cling together. Find the final speed of the couple. 1. 0.75961 2. 1.98398 3. 1.43353 4. 1.05818 5. 1.48904 6. 1.03438 7. 1.71108 8. 1.35638 9. 1.60274 10. 1.31843 Correct answer: 1 . 98398 m / s. Explanation: From conservation of momentum Δ p = 0 m A v A ˆ ı + m B v B ˆ = ( m A + m B ) v f Therefore v f = radicalbig ( m A v A ) 2 + ( m B v B ) 2 m A + m B = radicalbig (222 . 216 kg m / s) 2 + (97 . 44 kg m / s) 2 78 . 8 kg + 43 . 5 kg = 1 . 98398 m / s 002 10.0points A wheel rotating with a constant angular ac- celeration turns through 21 revolutions during a 2 s time interval. Its angular velocity at the end of this interval is 15 rad / s. What is the angular acceleration of the wheel? Note that the initial angular veloc- ity is not zero. 1. 0.893806 2. -6.42257 3. -18.6549 4. -50.9734 5. -8.27876 6. -1.90874 7. -2.72639 8. -0.16182 9. -2.54513 10. -17.8623 Correct answer: 50 . 9734 rad / s 2 . Explanation: Let : N = 21 , t = 2 s , and ω = 15 rad / s . From kinematics α t = ω f ω 0 ω 0 = ω f α t and Δ θ = N 2 ( π ) , so Δ θ = ω 0 t + 1 2 α t 2 2 N π = ( ω f α t ) t + 1 2 α t 2 = ω f t 1 2 α t 2 α = 2 ω f t 2 N π t 2 = 2 (15 rad / s) (2 s) 2 (21) π (2 s) 2 = 50 . 9734 rad / s 2 . keywords: 003 10.0points A bowling ball has a mass of 2 . 5 kg, a moment of inertia of 0 . 0529 kg · m 2 , and a radius of 0 . 23 m. It rolls along the lane without slipping at a linear speed of 2 . 9 m / s. What is the total kinetic energy of the rolling ball?
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Version 011 – Exam2 – distler – (56295) 2 1. 13.1404 2. 27.5562 3. 73.9508 4. 53.6648 5. 34.3 6. 14.875 7. 68.9087 8. 8.6751 9. 14.7175 10. 25.7488 Correct answer: 14 . 7175 J. Explanation: Let : v = 2 . 9 m / s , r = 0 . 23 m , m = 2 . 5 kg , and I = 0 . 0529 kg · m 2 . E = E rot + E kin = 1 2 I ω 2 + 1 2 m v 2 = 1 2 I v 2 r 2 + 1 2 m v 2 = 1 2 (0 . 0529 kg · m 2 ) (2 . 9 m / s) 2 (0 . 23 m) 2 + 1 2 (2 . 5 kg)(2 . 9 m / s) 2 = 14 . 7175 J . 004 10.0points When it orbited the Moon, the Apollo 11 spacecraft’s mass was 12700 kg, and its mean distance from the Moon’s center was 1 . 7028 × 10 6 m. Assume its orbit was circular and the Moon to be a uniform sphere of mass 7 . 36 × 10 22 kg.
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