hw18 - Overton, Mays – Homework 18 – Due: Mar 8 2005,...

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Unformatted text preview: Overton, Mays – Homework 18 – Due: Mar 8 2005, 4:00 am – Inst: Turner 1 This print-out should have 13 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 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 1 . 84 × 10 7 m / s, what would be the recoil speed of the thorium nucleus? Correct answer: 314887 m / s. Explanation: Let : v α = 1 . 84 × 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 . 84 × 10 7 m / s)(6 . 64 × 10- 27 kg) 3 . 88 × 10- 25 kg = 314887 m / s . 002 (part 1 of 1) 10 points A 4 kg steel ball strikes a wall with a speed of 6 . 99 m / s at an angle of 43 . 2 ◦ with the surface. It bounces off with the same speed and angle, as shown in the figure. x y 6 . 9 9 m / s 4 kg 6 . 9 9 m / s 4 kg 43 . 2 ◦ 43 . 2 ◦ If the ball is in contact with the wall for . 307 s, what is the magnitude of the average force exerted on the ball by the wall? Correct answer: 132 . 781 N. Explanation: Let : M = 4 kg , v = 6 . 99 m / s , and θ = 43 . 2 ◦ . The y component of the momentum is un- changed. The x component of the momentum is changed by Δ P x =- 2 M v cos θ . Therefore, using impulse formula, F = Δ P Δ t =- 2 M v cos θ Δ t =- 2(4 kg)(6 . 99 m / s) sin43 . 2 ◦ . 307 s k ~ F k = 132 . 781 N ....
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This note was uploaded on 03/22/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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hw18 - Overton, Mays – Homework 18 – Due: Mar 8 2005,...

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