PHY303Kquiz2soln (2005) - Johnson Matthew Quiz 2 Due Mar 9...

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Johnson, Matthew – Quiz 2 – Due: Mar 9 2005, 10:00 pm – Inst: Kleinman 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A(n) 84 kg boat that is 4 . 7 m in length is initially 5 m from the pier. A 46 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. How far is the child from the pier when she reaches the far end of the boat? Correct answer: 8 . 03692 m. Explanation: Let D = distance of the boat from the pier , L = length of the boat , M = mass of the boat , m = mass of the child , and X = change in position of the boat . D L l χ Initially, the center of mass of the boat-child system is x cm = D + L 2 M + D m M + m Finally, the center of mass of the boat-child system is x cm = D + L 2 - X 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 D + L 2 M + D m M + m = D + L 2 - X M + ( D + L - X ) m M + m and, solving for X , we have D + L 2 M + D m = D + L 2 - X M + ( D + L - X ) m 0 = -X M + ( L - X ) m X ( M + m ) = L m X = m m + M L = (46 kg) (46 kg) + (84 kg) × (4 . 7 m) = 1 . 66308 m . The final distance of the child from the pier is = D + L - X = (5 m) + (4 . 7 m) - (1 . 66308 m) = 8 . 03692 m . 002 (part 1 of 1) 10 points An ore car of mass 36000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 24 m lower vertically, is a spring with constant 550000 N / m. The acceleration of gravity is 9 . 8 m / s 2 . Ignore friction. How much is the spring compressed in stop- ping the ore car? Correct answer: 5 . 54886 m. Explanation: Energy is conserved, so the change of po- tential energy from when the car is at rest to when it just hits the spring is m g h = 1 2 m v 2 .
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Johnson, Matthew – Quiz 2 – Due: Mar 9 2005, 10:00 pm – Inst: Kleinman 2 The kinetic energy is then converted to po- tential energy in the spring as the cart comes to rest. When the spring is fully compressed by an amount d , all of the kinetic energy has been converted to potential energy so 1 2 m v 2 = 1 2 k d 2 . Thus, 1 2 k d 2 = m g h , and solving for d we have d = r 2 m g h k = s 2 (36000 kg) (9 . 8 m / s 2 ) (24 m) (550000 N / m) = 5 . 54886 m . 003 (part 1 of 1) 10 points A projectile of mass 0 . 553 kg is shot from a cannon, at height 6 . 7 m, as shown in the figure, with an initial velocity v i having a horizontal component of 5 . 7 m / s. The projectile rises to a maximum height of Δ y above the end of the cannon’s barrel and strikes the ground a horizontal distance Δ x past the end of the cannon’s barrel. The acceleration of gravity is 9 . 8 m / s 2 .
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