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: 9

Answer

$2.84m/s$

Work Step by Step

We can determine the required velocity as follows: $v_{a,2}^2=\frac{2}{m_a}\cdot[\frac{m_av_{a,1}^2}{2}+m_agh]$ We plug in the known values to obtain: $\implies v_{a,2}^2=\frac{2}{5}\cdot[\frac{5(25)}{2}+5(1.5)(9.81)]$ $\implies v_{a,2}=7.38m/s$ Now according to the law of conservation of momentum $m_av_{a,2}+m_bv_{b,1}=(m_a+m_b) v$ This simplifies to: $v=\frac{m_a v_{a,2}}{m_a+m_b}$ We plug in the known values to obtain: $v=\frac{5(7.38)}{5+8}$ $\implies v=2.84m/s$
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