Homework 9-solutions

Homework 9-solutions - yindeemark(rry82 Homework 9...

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yindeemark (rry82) – Homework 9 – chelikowsky – (59005) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – fnd 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 . = fnal distance oF child From pier . d = fnal distance oF boat From pier . D L X d X We will use the pier as the origin oF the x -coordinate. Initially, the center oF mass oF the boat-child system is x cm = p D + L 2 P M + D m M + m ±inally, the center oF mass oF the boat-child system is x cm = p D + L 2 − X P 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 p D + L 2 P M + D m M + m = p D + L 2 − X P M + ( D + L − X ) m M + m and, solving For X , we have p D + L 2 P M + D m = p D + L 2 − X P M + ( D + L − X ) m 0 = −X M + ( L − X ) m X ( M + m ) = L m X = m m + M L = (57 kg) (57 kg) + (83 kg) × (7 . 2 m) = 2 . 93143 m .
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yindeemark (rry82) – Homework 9 – chelikowsky – (59005) 2 The fnal distance oF the child From the pier is = D + L − X = (8 . 3 m) + (7 . 2 m) (2 . 93143 m) = 12 . 5686 m . The fnal 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 fred 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? 1. d c = 3 R 2. d c = 1 2 R 3. d c = 3 2 R correct 4. d c = 3 2 R 5. d c = 2 3 R 6. d c = 2 R 7. d c = 1 3 R 8. d c = R 9. d c = 2 R Explanation: Basic Concepts: Conservation oF mo- mentum Center oF Mass: vr cm = i m i i i m i Solution: R R/2 3R/2 Because the Fragments have the same zero component oF vertical velocity as the unex- ploded shell, the Fall time oF each Fragment aFter creation is the same as what the Fall time From peak altitude oF the original shell would have been, had it not exploded. Also, this Fall time is equal to the rise time oF the original shell to its peak altitude. Let V fx be the x -component oF velocity oF the Faster Fragment and V shx be the x -component oF the
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Homework 9-solutions - yindeemark(rry82 Homework 9...

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