Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 7 - Linear Momentum - Problems - Page 194: 44

Answer

a) See solution below b) $\arctan(1.07)=-46.9^o$ $v_B=1.23\frac{m}{s}$

Work Step by Step

a) $m_Av_A+m_Bv_B=m_Av_A'+m_Bv_B'$ For the x direction $(0.120kg)(2.8\frac{m}{s})+(0.140kg)(0\frac{m}{s})=(0.120kg)(2.8\frac{m}{s})\cos(30^o)+(0.140kg)v_{Bx}$ For the y direction $0=(0.120kg)(\sin(30.0^o))+(0.140kg)(\sin(\theta))$ b) For the x direction $(0.336kg\frac{m}{s})+0=(0.218kg\frac{m}{s})+(0.140kg)v_{Bx}$ $v_B\cos(\theta)=0.841\frac{m}{s}$ For the y direction $-0.9=v_B(\sin(\theta))$ Combining the two equations, we get the following result $\arctan(1.07)=-46.9^o$ $v_B=1.23\frac{m}{s}$
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