University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 8 - Momentum, Impulse, and Collision - Problems - Exercises - Page 267: 8.63

Answer

(a) $\frac{m}{m_0} = e^{-150}$ (b) The fraction of the initial mass that is not fuel is 0.223.

Work Step by Step

(a) $v-v_0 = v_{ex}~ln\frac{m_0}{m}$ $\frac{v-0}{v_{ex}} = ln\frac{m_0}{m}$ $e^{\frac{v}{v_{ex}}} = \frac{m_0}{m}$ $e^{\frac{(1.00\times 10^{-3})~(3.0\times 10^8~m/s)}{2000~m/s}} = \frac{m_0}{m}$ $\frac{m_0}{m} = e^{150}$ $\frac{m}{m_0} = e^{-150}$ We can see that most of the initial mass must be fuel. (b) $v-v_0 = v_{ex}~ln\frac{m_0}{m}$ $\frac{v-0}{v_{ex}} = ln\frac{m_0}{m}$ $e^{\frac{v}{v_{ex}}} = \frac{m_0}{m}$ $e^{\frac{3000~m/s}{2000~m/s}} = \frac{m_0}{m}$ $\frac{m_0}{m} = e^{1.5}$ $\frac{m}{m_0} = e^{-1.5}$ $\frac{m}{m_0} = 0.223$ The fraction of the initial mass that is not fuel is 0.223.
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