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 264: 48


$v_A=3.29m/s$ ,$v_B=2.19m/s$

Work Step by Step

We can determine the required speed as follows: We know that: $\Sigma mv_1=\Sigma mv_2$ $\implies 0=m_Av_A+m_Bv_B$ $\implies 40v_A=60v_B$ $\implies v_A=1.5v_B~~~$eq(1) The conservation of total energy before the blocks begin to rise is given as $T_1+V_1=T_2+V_2$ $\implies \frac{1}{2}m_Av_{A_{\circ}}^2+\frac{1}{2}m_Bv_{B_{\circ}}^2+\frac{1}{2}ks^2=\frac{1}{2}m_Av_A^2+\frac{1}{2}m_Bv_B^2+0$ $\implies ks^2=m_Av_A^2+m_Bv_B^2$ We plug in the known values to obtain: $(180)(2)^2=40v_A^2+60v_B^2$ From eq(1), $v_A=1.5v_B$ $\implies (180)(2)^2=40(1.5v_B)^2+60v_B^2$ This simplifies to: $v_B=2.19m/s$ We plug in this value in eq(1) to obtain: $v_A=1.5(2.19)=3.29m/s$
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