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: 8

Answer

$1.6m/s$

Work Step by Step

We can determine the required velocity as follows: We apply the law of conservation of momentum in the x-direction: $m_c\cdot v_{c, 1x}+m_p\cdot v_{p,1x}=(m_c+m_p)v_x$ This simplifies to: $v_x=\frac{m_p\cdot v_{p,1x}}{m_c+m_p}$.eq(1) As $\frac{v_{p,1}}{v_{p,1x}}=\frac{5}{4}$ $\implies v_{p,1x}=\frac{4v_{p,1}}{5}=8m/s$ We plug in the known values in eq(1) to obtain: $v_x=\frac{5.8}{20+5}=1.6m/s$
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