PHY303K-Spring07-Exam2-Moore

# PHY303K-Spring07-Exam2-Moore - midterm 02 – SAENZ LORENZO...

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Unformatted text preview: midterm 02 – SAENZ, LORENZO – Due: Mar 7 2007, 11:00 pm 1 Question 1 part 1 of 1 10 points Energy is required to move a 802 kg mass from the Earth’s surface to an altitude 2 . 55 times the Earth’s radius R E . ( R E =6 . 37 × 10 6 m) What amount of energy is required to ac- complish this move? Hint: G and M E can be eliminated in favor of parameter of the gravitational acceleration at the surface of the earth; i.e. , the accelera- tion of gravity on Earth is 9 . 8 m / s 2 . 1. 3 . 16259 × 10 10 J 2. 3 . 26332 × 10 10 J 3. 3 . 37628 × 10 10 J 4. 3 . 48709 × 10 10 J 5. 3 . 59627 × 10 10 J correct 6. 3 . 7081 × 10 10 J 7. 3 . 82607 × 10 10 J 8. 3 . 95757 × 10 10 J 9. 4 . 0871 × 10 10 J 10. 4 . 22253 × 10 10 J Explanation: Let : R E = 6 . 37 × 10 6 m , hR E = 2 . 55 R E , so R mass = (1 + h ) R E , and m = 802 kg . Using the formulae U = − G mM E R and g = G M E R 2 E , (where M E is the mass of the Earth and g is the gravitational acceleration at the Earth’s surface), we obtain that the change, Δ U , in the potential energy of the mass-Earth system is Δ U = ( U final − U initial ) = GmM E parenleftbigg 1 R initial − 1 R final parenrightbigg = GmM E parenleftbigg 1 R E − 1 R E (1 + h ) parenrightbigg = mg R E parenleftbigg 1 − 1 1 + h parenrightbigg = mg R E h 1 + h = 802 kg × 9 . 8 m / s 2 × 6 . 37 × 10 6 m × 2 . 55 1 + 2 . 55 = 3 . 59627 × 10 10 J . Question 2 part 1 of 2 10 points A 44 . 1 kg girl is standing on a 109 kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat, frictionless sup- porting surface. The girl begins to walk along the plank at a constant speed of 1 . 05 m / s to the right relative to the plank. What is the magnitude of her velocity rela- tive to the ice surface? 1. . 617036 m / s 2. . 636735 m / s 3. . 656462 m / s 4. . 677731 m / s 5. . 698915 m / s 6. . 722262 m / s 7. . 747551 m / s correct 8. . 772165 m / s 9. . 798705 m / s 10. . 824342 m / s Explanation: v g = velocity of girl relative to ice v p = velocity of plank relative to ice v gp = relative velocity of girl. Let : m g = 44 . 1 kg , m p = 109 kg , and v gp = 1 . 05 m / s . By conservation of momentum 0 = m g v g + m p v p , v p = − m g v g m p midterm 02 – SAENZ, LORENZO – Due: Mar 7 2007, 11:00 pm 2 The relative velocity is v gp = v g − v p (1) = v g − parenleftbigg − m g v g m p parenrightbigg = v g m p + m g v g m p = v g ( m p + m g ) m p . Thus v g = m p v gp m g + m p = (109 kg) (1 . 05 m / s) 109 kg + 44 . 1 kg = . 747551 m / s . Question 3 part 2 of 2 10 points What is the velocity of the plank relative to the ice surface? 1. − . 365093 m / s 2. − . 353973 m / s 3. − . 343032 m / s 4. − . 332439 m / s 5. − . 322453 m / s 6. − . 312083 m / s 7. − . 302449 m / s correct 8. − . 293065 m / s 9. − . 283846 m / s 10. − . 27507 m / s Explanation: Using Eq. (1), v p = v g − v gp = 0 . 747551 m / s − 1 . 05 m / s = − . 302449 m / s , directed opposite to the girl’s direction.directed opposite to the girl’s direction....
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## This note was uploaded on 02/09/2010 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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PHY303K-Spring07-Exam2-Moore - midterm 02 – SAENZ LORENZO...

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