PHY303K-Spring07-Exam2-Moore

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 Earths surface to an altitude 2 . 55 times the Earths 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 Earths surface), we obtain that the change, U , in the potential energy of the mass-Earth system is 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 girls direction.directed opposite to the girls direction....
View Full Document

Page1 / 17

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online