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 - Problems - Page 265: 53


$ s_B=71.4mm$

Work Step by Step

According to conservation of linear momentum $m_A(v_A)_{\circ}+m_B(v_B)_{\circ}=m_Av_A+m_Bv_B$ We plug in the known values to obtain: $5(0)+30(0)=5(v_A)+30v_B$ This simplifies to: $v_A=6v_B$ But we know that: $v_{B/A}=v_B-(-v_A)$ $\implies v_{B/A}=v_B+6v_B=7v_B$ We integrate with respect to time to obtain: $\implies s_{B/A}=7s_B$ As $s_{B/A}=0.5m$ $\implies s_B=\frac{s_{B/A}}{7}$ $\implies s_B=\frac{0.5}{7}$ $\implies s_B=0.071m=71.4mm$
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