Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 18 - The Micro/Macro Connection - Exercises and Problems - Page 522: 9

Answer

a) $0\;\rm m/s$ b) $56.6\;\rm m/s$ c) $60\;\rm m/s$

Work Step by Step

a) The best way to solve this problem is to do as the author did in Example 18.2. \begin{array}{|c|c|c|} \hline {\rm Paticle}& v_x\;{\rm (m/s)}& v_y\;{\rm (m/s)}& v_x^2\;{\rm (m/s)^2}&v_y^2\;{\rm (m/s)^2}&v^2\;{\rm (m/s)^2}&v\;{\rm (m/s)}\\ \hline 1 & 20&-30 &400 &900 &1300 & 10\sqrt{13} \\ \hline 2 &40 &70 & 1600& 4900& 6500 & 10\sqrt{65} \\ \hline 3& -80&20 &6400 &400 & 6800 & 20\sqrt{17} \\ \hline 4& 30 &0 & 900 &0 & 900 & 30 \\ \hline 5 & 40 & -40 & 1600 &1600 & 3200 &40\sqrt{2} \\ \hline 6 & -50&-20 & 2500 &400 & 2900 & 10\sqrt{29} \\ \hline {\rm Average}& 0 & 0 & - &- & 3600 & 56.6 \\ \hline \end{array} where $v^2=v_x^2+v_y^2$, and hence, $v=\sqrt{v^2}$. Also $v_{avg,x}=\dfrac{v_{x1}+....+v_{x6}}{6}$. And hence the average velocity $$\vec v_{avg}=\sqrt{v_{avg,x}^2+v_{avg,y}^2}=\sqrt{0+0}=\color{red}{\bf0}\;\rm m/s$$ ______________________________________________________________ b) $v_{avg}$ is calculated in the table above the last down-right cell. $$v_{avg}=\color{red}{\bf 56.6}\;\rm m/s$$ ______________________________________________________________ c) Now we need to find the root-mean-square speed $v_{\rm rms}$ which is given by $$v_{\rm rms}=\sqrt{(v^2)_{avg}}$$ Plugging from the table; $$v_{\rm rms}=\sqrt{(3600)}=\color{red}{\bf 60}\;\rm m/s $$
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