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quiz1031-solutions

quiz1031-solutions - Version 045/AACDB quiz1031...

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Version 045/AACDB – quiz1031 – Demkov – (59910) 1 This print-out should have 7 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A cylindrical pulley with a mass oF 5 . 1 kg, ra- dius oF 0 . 883 m and moment oF inertia 1 2 M r 2 is used to lower a bucket with a mass oF 2 kg into a well. The bucket starts From rest and Falls For 3 . 6 s. r M m How Far does it drop? The acceleration oF gravity is 9 . 8 m / s 2 . 1. 24.5 2. 25.6517 3. 41.2571 4. 19.6951 5. 13.1116 6. 27.9138 7. 30.2565 8. 17.8025 9. 13.4742 10. 36.0624 Correct answer: 27 . 9138 m. Explanation: Let : M = 5 . 1 kg , r = 0 . 883 m , m = 2 kg , and g = 9 . 8 m / s 2 . Let T be the tension in the cord and α the angular acceleration oF the wheel. Newton’s equation For the mass m is mg - T = ma T = m ( g - a ) and For the disk T r = I α ( - ) rh = p 1 2 M r 2 P ± a r ² 2 - 2 = M a 2 = ( M + 2 m ) a a = 2 M + 2 m ±rom kinematics, since v 0 = 0 , y = v 0 t + 1 2 a t 2 = 1 2 a t 2 = 1 2 p 2 M + 2 m P t 2 = mg t 2 M + 2 m = (2 kg) (9 . 8 m / s 2 ) (3 . 6 s) 2 5 . 1 kg + 2 (2 kg) = 27 . 9138 m . 002 10.0 points A(n) 7 kg object moving at 9 m / s in the posi- tive x direction has a one-dimensional elastic collision with an object initially at rest. AFter the collision the object oF unknown mass has a velocity oF 8 m / s in the positive x direction. What is the unknown mass? 1. 8.75 2. 10.6667 3. 5.0 4. 2.1 5. 3.2 6. 1.4 7. 3.6 8. 6.3 9. 2.13333 10. 4.0 Correct answer: 8 . 75 kg. Explanation: Given: v 2 i = 0. ±rom conservation oF momentum, we have m 1 v 1 i = m 1 v 1 f + m 2 v 2 f m 1 ( v 1 i - v 1 f ) = m 2 v 2 f . (1)
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Version 045/AACDB – quiz1031 – Demkov – (59910) 2 Because of elastic collision, energy is con- served. Therefore 1 2 m 1 v 2 1 i = 1 2 m 1 v 2 1 f + 1 2 m 2 v 2 2 f m 1 ( v 2 1 i - v 2 1 f ) = m 2 v 2 2 f m 1 ( v 1 i - v 1 f ) ( v 1 i + v 1 f ) = m 2 v 2 2 f (2) Dividing Eq. (2) by Eq. (1) we have v 1 i + v 1 f = v 2 f v 1 f = v 2 f - v 1 i (3) Substituting Eq. (3) in Eq. (1) we have m 1 v 1 i = m 1 ( v 2 f - v 1 i ) + m 2 v 2 f m 2 v 2 f = m 1 v 1 i - m 2 ( v 2 f - v 1 i
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