Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 15 - Kinetics of a Particle: Impulse and Momentum - Section 15.3 - Conservation of Linear Momentum for a System of Particles - Fundamental Problems - Page 261: 10


$2.31m/s$ $3.46m/s$

Work Step by Step

According to the law of conservation of energy $\frac{Ks^2}{2}=\frac{m_av_{a,2}^2}{2}+\frac{m_bv_{b,2}^2}{2}$ This simplifies to: $v_{b,2}^2=\frac{Ks^2}{m_b\cdot(\frac{m_b}{m_a}+1)}$ We plug in the known values to obtain: $v_{b,2}^2=\frac{5000(0.04)}{15(\frac{15}{10}+1)}$ $\implies v_{b,2}=2.31m/s$ Now to determine the velocity of block A, we use the law of conservation of momentum $m_av_{a,1}+m_bv_{b,1}=m_av_{a,2}+m_b v_{b,2}$ $\implies 0=m_av_{a,2}+m_bv_{b,2}$ $v_{a,2}=-\frac{m_b v_{b,2}}{m_a}$ We plug in the known values to obtain: $v_{b,2}=-\frac{15(2.31)}{10}=3.46m/s$
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