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: 64


$v_A=0.16m/s\leftarrow$ $v_B=8.24m/s\rightarrow$

Work Step by Step

We can determine the required velocity 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: $3(8)+(2)(-4)=3v_{A_2}+2v_{B_2}$ This simplifies to: $3v_A+2v_B=16~~~$[eq(1)] We know that: $e=\frac{v_{B_2}-v_{A_2}}{v_{A_1}-v_{B_1}}$ $\implies 0.7=\frac{v_{B_2}-v_{A_2}}{8-(-4)}$ $\implies 8.4=v_{B_2}-v_{A_2}~~~$[eq(2)] After solving eq(1) and eq(2), we obtain: $v_A=0.16m/s\leftarrow$ and $v_B=8.24m/s\rightarrow$
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