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Physics Old Midterm 4 - Solutions

# Physics Old Midterm 4 - Solutions - oldmidterm 04 FRENNEA...

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oldmidterm 04 – FRENNEA, KYLE – Due: Apr 28 2007, 4:00 am 1 Question 1 part 1 of 1 10 points Given: G = 6 . 6726 × 10 11 N m 2 / kg 2 Three 3 kg masses are located at points in the x-y plane as shown in the figure. 36 cm 55 cm What is the magnitude of the resultant force (caused by the other two masses) on the mass at the origin? Correct answer: 5 . 04111 × 10 9 N (tolerance ± 1 %). Explanation: Let : m = 3 kg x = 36 cm , and y = 55 cm . Basic Concepts: Newton’s Law of Grav- itation: F g = G m 1 m 2 r 2 We calculate the forces one by one, and then add them using the superposition prin- ciple. The force from the mass on the right is pointing in the x direction, and has magni- tude f 1 = G m m x 2 = G m 2 x 2 = (6 . 6726 × 10 11 N m 2 / kg 2 )(3 kg) 2 (0 . 36 m) 2 = 4 . 63375 × 10 9 N . The other force is pointing in the y direction and has magnitude f 2 = G m m y 2 = G m 2 y 2 = (6 . 6726 × 10 11 N m 2 / kg 2 )(3 kg) 2 (0 . 55 m) 2 = 1 . 98524 × 10 9 N . f 2 f 1 F θ Now we simply add the two forces, using vector addition. Since they are at right angles to each other, however, we can use Pythago- ras’ theorem as well: F = radicalBig f 2 1 + f 2 2 = bracketleftBig (4 . 63375 × 10 9 N) 2 + (1 . 98524 × 10 9 N) 2 bracketrightBig 1 2 = 5 . 04111 × 10 9 N . As you see, this force is very small. If the masses in the picture are standing on a table (the view being from above), typically the force of static friction will not be overcome. In agreement with common sense, then, the masses will not move. Note: The angle θ shown in the figure is θ = arctan bracketleftbigg f 2 f 1 bracketrightbigg = arctan bracketleftbigg (1 . 98524 × 10 9 N) (4 . 63375 × 10 9 N) bracketrightbigg = 23 . 1917 . Question 2 part 1 of 1 10 points

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oldmidterm 04 – FRENNEA, KYLE – Due: Apr 28 2007, 4:00 am 2 Planet X has five times the diameter and three times the mass of the earth. What is the ratio of gravitational accelera- tion at the surface of planet X to the gravita- tional acceleration at the surface of the Earth, g x g e ? 1. g x g e = 2 9 2. g x g e = 3 25 correct 3. g x g e = 5 36 4. g x g e = 4 25 5. g x g e = 5 9 6. g x g e = 6 49 7. g x g e = 9 49 8. g x g e = 1 7 9. g x g e = 2 49 10. g x g e = 3 16 Explanation: Let : M X = 3 M e , R X = 5 R e . Because m g e = G M e m R 2 e , where M e is the mass of the earth and R e is the radius. So that g e = G M e R 2 e . Similar equation applies to the planet X g x = G M x R 2 x . The ratio between the surface gravitational accelerations is g x g e = M x R 2 e M e R 2 x = 3 M e R 2 e M e (5 R e ) 2 = 3 25 . Question 3 part 1 of 1 10 points An astronaut weighs 147 N on the Moon’s surface. He is in a circular orbit about the Moon at an altitude of two-ninth the moon’s radius. What gravitational force does the Moon exert on him? Correct answer: 98 . 405 N (tolerance ± 1 %). Explanation: Let : h = 2 9 R moon , R astronaut = parenleftbigg 1 + 2 9 parenrightbigg R moon , W = m g = 147 N . The astronaut’s weight on the Moon is w = m g moon = 147 N , where the gravitational acceleration g moon on the Moon’s surface is g moon = G M moon R 2 moon .
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Physics Old Midterm 4 - Solutions - oldmidterm 04 FRENNEA...

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