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Unformatted text preview: yindeemark (rry82) Homework 9 chelikowsky (59005) 1 This printout should have 15 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points An 83 kg boat that is 7 . 2 m in length is initially 8 . 3 m from the pier. A 57 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. 8 . 3 m 7 . 2 m How far is the child from the pier when she reaches the far end of the boat? Assume there is no friction between boat and water. Correct answer: 12 . 5686 m. Explanation: Let D = distance of the boat from the pier , = 8 . 3 m , L = length of the boat , = 7 . 2 m , M = mass of the boat , = 83 kg , and m = mass of the child , and = 57 kg , X = change in position of the boat . = final distance of child from pier . d = final distance of boat from pier . D L 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 = (57 kg) (57 kg) + (83 kg) (7 . 2 m) = 2 . 93143 m . yindeemark (rry82) Homework 9 chelikowsky (59005) 2 The final distance of the child from the pier is = D + L X = (8 . 3 m) + (7 . 2 m) (2 . 93143 m) = 12 . 5686 m . The final distance d of the near end of the boat to the pier is d = D X = (8 . 3 m) (2 . 93143 m) = 5 . 36857 m . 002 10.0 points On level ground, a shell is fired from a cannon to hit an enemy bunker situated a distance R away from the cannon. At the highest point of the trajectory, the shell accidentally explodes into two equal parts. The explosion occurs in such a way that the two pieces have no velocity in the vertical direction immediately after the explosion. ( i.e. the internal forces from the explosion act only in the horizontal direction.) One part of the shell is found to hit the ground at R 2 . At what distance d c from the cannon would the second fragment fall?...
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