Homework_12 - Nguyen, Thanh – Homework 12 – Due: Oct 16...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Nguyen, Thanh – Homework 12 – Due: Oct 16 2007, 7:00 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 . 7 kg particle has a velocity of v x = 1 . 25 m / s and v y = 6 . 96 m / s. Find the magnitude of its total momen- tum. Correct answer: 12 . 0213 kg · m / s. Explanation: Let : m = 1 . 7 kg , v x = 1 . 25 m / s , and v y = 6 . 96 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 . 7 kg) q (1 . 25 m / s) 2 + (6 . 96 m / s) 2 = 12 . 0213 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 . 0213 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 . 31 × 10 7 m / s, what would be the recoil speed of the thorium nucleus? Correct answer: 395320 m / s. Explanation: Let : v α = 2 . 31 × 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 . 31 × 10 7 m / s)(6 . 64 × 10- 27 kg) 3 . 88 × 10- 25 kg = 395320 m / s ....
View Full Document

This note was uploaded on 10/24/2011 for the course PHYS 1710 taught by Professor Weathers during the Winter '08 term at North Texas.

Page1 / 5

Homework_12 - Nguyen, Thanh – Homework 12 – Due: Oct 16...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online