{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Homework 12

# Homework 12 - Gozick Brandon Homework 12 Due 8:00 pm Inst D...

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

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, 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 = (2 . 1 × 10 7 m / s) (6 . 64 × 10 - 27 kg) 3 . 88 × 10 - 25 kg = 359381 m / s . keywords: 003 (part 1 of 1) 10 points A gadget of mass 19 . 55 kg floats in space without motion. Because of some internal malfunction, the gadget violently breaks up into 3 fragments flying away from each other.

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}