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Unformatted text preview: Patel, Kinal Homework 12 Due: Oct 16 2007, 7:00 pm Inst: D Weathers 1 This printout should have 10 questions. Multiplechoice 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 . 96 kg particle has a velocity of v x = 6 . 02 m / s and v y = 1 . 15 m / s. Find the magnitude of its total momen tum. Correct answer: 12 . 0126 kg m / s. Explanation: Let : m = 1 . 96 kg , v x = 6 . 02 m / s , and v y = 1 . 15 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 . 96 kg) q (6 . 02 m / s) 2 + (1 . 15 m / s) 2 = 12 . 0126 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 = 12 . 0126 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 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 particle is emitted at a speed of 2 . 2 10 7 m / s, what would be the recoil speed of the thorium nucleus? Correct answer: 376495 m / s. Explanation: Let : v = 2 . 2 10 7 m / s , M = 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 ~v + M Th ~v Th = 0 . This means that the Thorium nucleus recoils in the direction exactly opposite to that of the particle with speed k ~v Th k = k ~v k M M Th = (2 . 2 10 7 m / s)(6 . 64 10 27 kg) 3 . 88 10 25 kg = 376495 m / s ....
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