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: 11



Work Step by Step

The maximum compression of the spring can be determined as: According to the law of conservation of momentum $m_av_{a,1}+m_bv_{b,1}=(m_a+m_b) v$ $\implies v=\frac{m_av_{a,1}+m_bv_{b,1}}{m_a+m_b}$ $\implies v=\frac{0+10(15)}{15+10}=6m/s$ Now according to the law of conservation of energy $\frac{Ks^2}{2}=\frac{(m_a+m_b) v^2}{2}$ $\implies s^2=\frac{(m_a+m_b) v^2}{K}$ We plug in the known values to obtain: $s^2=\frac{(15+10)36}{10000}$ $s^2=0.09m^2$ $\implies s=0.3m$
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