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 15.4 - Impact - Problems - Page 274: 60


$v_{A_2}=1.53m/s \leftarrow$ $v_{B_2}=1.27m/s \rightarrow$

Work Step by Step

The required velocities can be determined as follows: According to the conservation of linear momentum $m_Av_{A_1}+m_Bv_{B_1}=m_Av_{A_2}+m_Bv_{B_2}$ We plug in the known values to obtain: $(2)(5)+4(-2)=2v_{A_2}+4v_{B_2}$ This simplifies to: $v_{A_2}+2v_{B_2}=1~~~$[eq(1)] We know that: $e=\frac{v_{B_2}-v_{A_2}}{v_{A_1}-v_{B_1}}$ $\implies 0.4=\frac{v_{B_2}-v_{A_2}}{5-(-2)}$ This simplifies to: $0.8=v_{B_2}-v_{A_2}~~~$[eq(2)] After solving eq(1) and eq(2), we obtain: $v_{A_2}=1.53m/s \leftarrow$ and $v_{B_2}=1.27m/s \rightarrow$
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