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Unformatted text preview: practice 07 GADHIA, TEJAS Due: Feb 27 2007, 10:00 pm 1 Question 1 part 1 of 2 10 points Given: G = 6 . 672 10 11 N m 2 / kg 2 A coordinate system (in meters) is con structed on the surface of a pool table, and three objects are placed on the coordinate sys tem as follows: a 1 . 2 kg object at the origin, a 3 . 5 kg object at (0,1 . 3 m), and a 4 . 3 kg object at (3 . 6 m,0). Find the resultant gravitational force ex erted on the object at the origin by the other two objects. Correct answer: 1 . 67927 10 10 N (tolerance 1 %). Explanation: Given : G = 6 . 672 10 11 N m 2 / kg 2 , m 1 = 1 . 2 kg , m 2 = 3 . 5 kg , and r = 1 . 3 m . 1.2 kg 3.5 kg 4.3 kg F x F y 3.6 m 1.3 m The force exerted on the 1 . 2 kg mass by the 3 . 5 kg mass is in the positive y direction and given by F y = Gm 1 m 2 r 2 = ( 6 . 672 10 11 N m 2 / kg 2 ) (1 . 2 kg) (3 . 5 kg) (1 . 3 m) 2 = 1 . 65813 10 10 N . The force exerted on the 1 . 2 kg object by the 4 . 3 kg object is in the positive x direction and is F x = Gm 1 m 2 r 2 = ( 6 . 672 10 11 N m 2 / kg 2 ) (1 . 2 kg) (4 . 3 kg) (3 . 6 m) 2 = 2 . 65644 10 11 N . The resultant F is found from the Pythagorean Theorem: F R 2 = F y 2 + F x 2 = (1 . 65813 10 10 N) 2 + (2 . 65644 10 11 N) 2 = 2 . 81996 10 20 N 2 . so that F R = radicalbig 2 . 81996 10 20 N 2 = 1 . 67927 10 10 N . Question 2 part 2 of 2 10 points At what angle does it act with respect to the positive x axis? Let counterclockwise be positive, within the limits of 180 to 180 . Correct answer: 80 . 8981 (tolerance 1 %). Explanation: tan = F y F x = tan 1 parenleftbigg F y F x parenrightbigg = tan 1 parenleftbigg 1 . 65813 10 10 N 2 . 65644 10 11 N parenrightbigg = 80 . 8981 above the + xaxis. Question 3 part 1 of 1 10 points An object is dropped from rest from a height 4 . 8 10 6 m above the surface of the earth. The acceleration of gravity is 9 . 81 m / s 2 . If there is no air resistance, what is its speed when it strikes the earth? Correct answer: 7 . 32847 km / s (tolerance 1 %). practice 07 GADHIA, TEJAS Due: Feb 27 2007, 10:00 pm 2 Explanation: Let : h = 4 . 8 10 6 m . g = GM E R 2 E and the potential energy at a distance r from the surface of the earth is U ( r ) = GM E m r . Using conservation of energy to relate the initial potential energy of the system to its energy as the object is about to strike the earth ( K i = 0), K f + U f U i = 0 K ( R E ) + U ( R E ) U ( R E + h ) = 0 1 2 mv 2 GM E m R E + GM E m R E + h = 0 . Solve for v : v = radicalBigg 2 parenleftbigg GM E R E GM E R E + h parenrightbigg = radicalBigg 2 g R E parenleftbigg h R E + h parenrightbigg = radicalBigg 2 (9 . 81 m / s 2 ) (6 . 37 10 6 m) 4 . 8 10 6 m + 6 . 37 10 6 m radicalbig 4 . 8 10 6 m 1 km 1000 m = 7 . 32847 km / s ....
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This note was uploaded on 03/09/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner

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