Homework 12 - Gozick, Brandon – Homework 12 – Due: Oct...

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Unformatted text preview: Gozick, Brandon – Homework 12 – Due: Oct 20 2006, 8:00 pm – Inst: D Weathers 1 This print-out should have 10 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 1 . 44 kg particle has a velocity of v x = 1 . 23 m / s and v y =- 4 . 91 m / s. Find the magnitude of its total momen- tum. Correct answer: 7 . 28888 kg · m / s. Explanation: Let : m = 1 . 44 kg , v x = 1 . 23 m / s , and v y =- 4 . 91 m / s . Given the velocity vector, the magnitude of the velocity vector is v = q v 2 x + v 2 y The momentum is defined by p = m v = m q v 2 x + v 2 y = (1 . 44 kg) q (1 . 23 m / s) 2 + (- 4 . 91 m / s) 2 = 7 . 28888 kg · m / s . Alternatively, the momentum vector is de- fined by p x = mv x and p y = mv y , so p = q p 2 x + p 2 y = q ( m v x ) 2 + ( m v y ) 2 = m q v 2 x + v 2 y = m v = 7 . 28888 kg · m / s . keywords: 002 (part 1 of 1) 10 points A uranium nucleus 238 U may stay in one piece 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 2 . 1 × 10 7 m / s, what would be the recoil speed of the thorium nucleus? Correct answer: 359381 m / s. Explanation: Let : v α = 2 . 1 × 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, theUraniumnucleushadzeromomentum (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 = (2 . 1 × 10 7 m / s)(6 . 64 × 10- 27 kg) 3 . 88 × 10- 25 kg = 359381 m / s ....
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This note was uploaded on 09/12/2008 for the course PHYS 1710 taught by Professor Weathers during the Fall '08 term at North Texas.

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Homework 12 - Gozick, Brandon – Homework 12 – Due: Oct...

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