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Final-solutions

# Final-solutions - Version 037 Final gentle(56245 This...

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Version 037 – Final – gentle – (56245) 1 This print-out should have 36 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points A homogeneous cylinder of radius 18 cm and mass 50 kg is rolling without slipping along a horizontal floor at 7 m / s. How much work was required to give it this motion? 1. 3.12 2. 2.75625 3. 0.135 4. 2.4 5. 1.8375 6. 0.84375 7. 1.21875 8. 0.21 9. 3.84 10. 3.6 Correct answer: 1 . 8375 kJ. Explanation: Let : r = 18 cm , m = 50 kg , and v = 7 m / s . Because the cylinder rolls without slipping, the translational speed is v = r ω and the total kinetic energy of the cylinder is K total = 1 2 m v 2 + 1 2 I ω 2 = 1 2 m v 2 + 1 2 parenleftbigg 1 2 m r 2 parenrightbigg parenleftBig v r parenrightBig 2 = 3 4 m v 2 = 3 4 (50 kg) (7 m / s) 2 · 1 kJ 1000 J = 1 . 8375 kJ , which is also the work required to give the cylinder this motion. keywords: 002 10.0points At a certain instant of time, a toy car is traveling at a constant speed of v 0 to the right, while a distance D ahead is another toy car starting from rest and traveling to the left with constant acceleration a . The cars are moving along the same line, directly toward each other. At what time do they collide, in terms of D , v 0 and a ? 1. The cars will never collide. 2. v 0 a 3. 1 2 parenleftbigg 2 D a + v 0 a parenrightbigg 4. radicalbigg parenleftBig v 0 a parenrightBig 2 + 2 D a - v 0 a correct 5. radicalbigg 2 D a - parenleftBig v 0 a parenrightBig 2 - v 0 a 6. radicalbigg parenleftBig v 0 a parenrightBig 2 + 2 D a + v 0 a 7. radicalbigg parenleftBig v 0 a parenrightBig 2 - 2 D a - v 0 a 8. 2 D a Explanation: The first toy car starts at a position x = 0 and moves to the right at a constant speed v 0 , so its position is given by x 1 = v 0 t . Mean- while, the second toy car starts at a distance D ahead of the first and accelerates to the left with an acceleration a , so its position is given by x 2 = D - 1 2 a t 2 . The two cars collide when x 1 = x 2 v 0 t = D - 1 2 a t 2 0 = a t 2 + 2 v 0 t - 2 D

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Version 037 – Final – gentle – (56245) 2 t = - 2 v 0 ± radicalBig 4 v 2 0 + 8 D a 2 a = - v 0 ± radicalBig v 2 0 + 2 D a a = - v 0 a + radicalbigg parenleftBig v 0 a parenrightBig 2 + 2 D a t > 0 when the two cars collide, so the negative root must be rejected. 003 10.0points Three masses are arranged in the ( x, y ) plane as shown. 7 kg 1 kg 3 kg y (m) 1 3 5 7 9 x (m) 1 3 5 7 9 What is the magnitude of the resulting force on the 7 kg mass at the origin? The value of the universal gravitational constant is 6 . 6726 × 10 11 N · m 2 / kg 2 . 1. 1.48252e-10 2. 5.58488e-11 3. 3.80511e-11 4. 3.71041e-11 5. 2.04682e-10 6. 1.75476e-11 7. 5.74069e-11 8. 1.32567e-10 9. 2.69041e-11 10. 3.50343e-10 Correct answer: 3 . 50343 × 10 10 N. Explanation: Let: m o = 7 kg , ( x o , y o ) = (0 m , 0 m) , m a = 1 kg , ( x a , y a ) = (10 m , 0 m) , and m b = 3 kg , ( x b , y b ) = (0 m , 2 m) . Applying Newton’s universal gravitational law for m o and m a , F ao = G m o m a ( x a - x o ) 2 + ( y a - y o ) 2 = (6 . 6726 × 10 11 N · m 2 / kg 2 ) × (7 kg) (1 kg) (10 m) 2 + (0 m) 2 = 4 . 67082 × 10 12 N acts along the x axis.
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Final-solutions - Version 037 Final gentle(56245 This...

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