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# HW 18 - Morby Grant Homework 18 Due Mar 6 2006 noon Inst...

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Morby, Grant – Homework 18 – Due: Mar 6 2006, noon – Inst: Drummond 1 This print-out should have 14 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 An object of mass m is moving with speed v 0 to the right on a horizontal frictionless surface, as shown, when it explodes into two pieces. Subsequently, one piece of mass 16 19 m moves with a speed v 16 / 19 = v 0 2 to the left. m v 0 before 16 19 m v 0 2 3 19 m v 3 / 19 after The speed v f = k ~v 3 / 19 k of the other piece the object is 1. None of these is correct. 2. v f = 5 v 0 . 3. v f = 9 v 0 . correct 4. v f = 10 v 0 . 5. v f = 8 v 0 . 6. v f = 4 v 0 . 7. v f = 2 v 0 . 8. v f = 6 v 0 . 9. v f = 7 v 0 . 10. v f = 3 v 0 . Explanation: The horizontal component of the momen- tum is conserved, so 0 + m v 0 = 16 19 m v 16 / 19 + 3 19 m v 3 / 19 0 + m v 0 = 16 19 m - v 0 2 · + 3 19 m v 3 / 19 m v 0 = - 16 38 m v 0 + 3 19 m v 3 / 19 3 19 v 3 / 19 = 38 38 + 16 38 v 0 3 19 v 3 / 19 = 54 38 v 0 v 3 / 19 = 54 38 19 3 v 0 k ~v 3 / 19 k = 9 v 0 . 002 (part 1 of 1) 10 points Bill (mass m ) plants both feet solidly on the ground and then jumps straight up with ve- locity v . The earth (mass M ) then has velocity 1. V Earth = - m M · v man . correct 2. V Earth = - v man . 3. V Earth = + M m v man . 4. V Earth = - r m M v man . 5. V Earth = + r m M v man . 6. V Earth = + v man . 7. V Earth = + m M · v man . 8. V Earth = - M m v man . Explanation: The momentum is conserved. We have m v man + M V Earth = 0 So V Earth = - m M · v man . 003 (part 1 of 1) 10 points A uranium nucleus 238 U may stay in one piece

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Morby, Grant – Homework 18 – Due: Mar 6 2006, noon – Inst: Drummond 2 for billions of years, but sooner or later it de- cays into an α particle of mass 6 . 64 × 10 - 27 kg and 234 Th nucleus of mass 3 . 88 × 10 - 25 kg, and the decay process itself is extremely fast (it takes about 10 - 20 s). Suppose the uranium nucleus was at rest just before the decay. If the α particle is emitted at a speed of 1 . 11 × 10 7 m / s, what would be the recoil speed of the thorium nucleus? Correct answer: 189959 m / s. Explanation: Let : v α = 1 . 11 × 10 7 m / s , M α = 6 . 64 × 10 - 27 kg , and M Th = 3 . 88 × 10 - 25 kg . Use momentum conservation: Before the de- cay, the Uranium nucleus had zero momentum (it was at rest), and hence the net momentum vector of the decay products should total to zero: ~ P tot = M α ~v α + M Th ~v Th = 0 . This means that the Thorium nucleus recoils in the direction exactly opposite to that of the α particle with speed k ~v Th k = k ~v α k M α M Th = (1 . 11 × 10 7 m / s) (6 . 64 × 10 - 27 kg) 3 . 88 × 10 - 25 kg = 189959 m / s .
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