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Unformatted text preview: This printout should have 21 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points An automobile has 57 percent of its weight on the front wheels. The front and back wheels are separated by 1 . 5 m. Where is the center of gravity located with respect to the front wheels? Correct answer: 0 . 645 m. Explanation: Let : W 1 = 0 . 57 W , W 2 = 0 . 43 W , x 1 = 0 m , and x 2 = 1 . 5 m . x cg = i W i x i W = W 1 x 1 + W 2 x 2 W = (0 . 57 W ) x 1 + (0 . 43 W ) x 2 W = (0 . 57) (0 m) + (0 . 43) (1 . 5 m) = . 645 m . 002 10.0 points Three particles are placed in the xy plane. A m 1 = 23 g particle is located at ( x 1 , y 1 ), where x 1 = 1 . 5 m and y 1 = 4 . 5 m and a m 2 = 43 g particle is located at ( x 2 , y 2 ), where x 2 = 1 . 3 m and y 2 = 4 . 8 m. What must be the x coordinate of the m 3 = 22 g particle so that the center of mass of the threeparticle system is at the origin? Correct answer: 0 . 972727 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 = 0 . 972727 m 003 10.0 points A(n) 68 kg boat that is 8 . 8 m in length is initially 7 . 9 m from the pier. A 43 kg child stands at the end of the boat closest to the pier. The child then notices a turtle on a rock at the far end of the boat and proceeds to walk to the far end of the boat to observe the turtle. Assume: There is no friction between boat and water. 7 . 9 m 8 . 8 m How far is the child from the pier when she reaches the far end of the boat? Correct answer: 13 . 291 m. Explanation: Let D = distance of the boat from the pier , = 7 . 9 m , L = length of the boat , = 8 . 8 m , M = mass of the boat , = 68 kg , and m = mass of the child , and = 43 kg , X = change in position of the boat . = final distance of child from pier . d = final distance of boat from pier . D L oldmidterm3 02 Turner (58185) 1 hinojosa (jlh3938) oldmidterm3 02 Turner (58185) 2 X d X We will use the pier as the origin of the xcoordinate. Initially, the center of mass of the boatchild system is x cm = parenleftbigg D + L 2 parenrightbigg M + D m M + m Finally, the center of mass of the boatchild system is x cm = parenleftbigg D + L 2 X parenrightbigg M + ( D + L X ) m M + m , where X is the change in position of the center of mass of the boat. Since the center of mass of the system does not move, we can equate the above two expressions for x cm parenleftbigg D + L 2 parenrightbigg M + D m M + m = parenleftbigg D + L 2 X parenrightbigg M + ( D + L X ) m M + m and, solving for X , we have parenleftbigg D + L 2 parenrightbigg M + D m = parenleftbigg D + L 2 X parenrightbigg M + ( D + L X ) m 0 =X M + ( L X ) m X ( M + m ) = Lm X = m m + M L = (43 kg) (43 kg) + (68 kg) (8 . 8 m) = 3...
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This note was uploaded on 05/02/2009 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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