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Unformatted text preview: ang (nza67) HW #8 Antoniewicz (57420) 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Three particles are placed in the xy plane. A m 1 = 78 g particle is located at ( x 1 , y 1 ), where x 1 = 4 . 6 m and y 1 = 9 . 1 m and a m 2 = 72 g particle is located at ( x 2 , y 2 ), where x 2 = 9 . 4 m and y 2 = 5 . 5 m. What must be the x coordinate of the m 3 = 38 g particle so that the center of mass of the threeparticle system is at the origin? Correct answer: 8 . 36842 m. Explanation: We know that the x coordinate of the center of mass is defined by x CM = m 1 x 1 + m 2 x 2 + m 3 x 3 m 1 + m 2 + m 3 Now, x CM is zero, and using the input values for m 1 , m 2 , m 3 and x 1 and x 2 , we find and solving for x 3 : x 3 = m 1 x 1 m 2 x 2 m 3 = 8 . 36842 m 002 (part 1 of 2) 5.0 points A square plate is produced by welding to gether four smaller square plates, each of side a . The weight of each of the four plates is shown in the figure. x y 90 N 20 N 80 N 80 N (0 , 0) (2 a, 0) (0 , 2 a ) (2 a, 2 a ) Find the xcoordinate of the center of grav ity (as a multiple of a ). Correct answer: 1 . 09259 a . Explanation: Let : x 1 = x 2 = 1 2 a , x 3 = x 4 = 3 2 a , W 1 = 90 N , W 2 = 20 N , W 3 = 80 N , and W 4 = 80 N . y x W 1 W 2 W 3 W 4 a 2 3 a 2 a 2 3 a 2 The total weight is W = W 1 + W 2 + W 3 + W 4 = 270 N . Applying the definition of center of gravity, x cg = i W i x i W 1 + W 2 + W 3 + W 4 = ( W 1 + W 2 ) a 2 + ( W 3 + W 4 ) 3 a 2 W = (90 N + 20 N) + 3 (80 N + 80 N) 2 (270 N) a = 1 . 09259 a . 003 (part 2 of 2) 5.0 points Find the ycoordinate of the center of gravity (as a multiple of a ). Correct answer: 0 . 87037 a . Explanation: Let : y 1 = y 4 = 1 2 a and y 2 = y 3 = 3 2 a . ang (nza67) HW #8 Antoniewicz (57420) 2 y cg = i W i y i W 1 + W 2 + W 3 + W 4 = ( W 1 + W 4 ) a 2 + ( W 2 + W 3 ) 3 a 2 W = (90 N + 80 N) + 3 (20 N + 80 N) 2 (270 N) a = . 87037 a . 004 (part 1 of 2) 5.0 points A(n) 3 . 8 kg object moving with a speed of 6 . 8 m / s collides with a(n) 3 . 6 kg object mov ing with a velocity of 10 . 4 m / s in a direction 61 . 7788 from the initial direction of motion of the 3 . 8 kg object. 6 . 8 m / s 1 . 4 m / s 3 . 6 kg 3 . 8 kg 61 . 7788 What is the speed of the two objects after the collision if they remain stuck together? Correct answer: 7 . 38242 m / s. Explanation: Let : m 1 = 3 . 8 kg , m 2 = 3 . 6 kg , m f = m 1 + m 2 = 7 . 4 kg , v 1 = 6 . 8 m / s , v 2 = 10 . 4 m / s , p 1 = m 1 v 1 = 25 . 84 kg m / s , p 2 = m 2 v 2 = 37 . 44 kg m / s , 2 p 1 p 2 = 2 m 1 v 1 m 2 v 2 = 1934 . 9 kg 2 m 2 / s 2 , p x = p 1 + p 2 cos = 43 . 5445 kg m / s , p y = p 2 sin = 32 . 9895 kg m / s , = 61 . 7788 , and  = 118 . 221 ....
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This note was uploaded on 04/23/2011 for the course PHYSICS 303K taught by Professor Antoniewicz during the Spring '11 term at University of Texas at Austin.
 Spring '11
 Antoniewicz
 Physics, Work

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