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: 46



Work Step by Step

According to the law of conservation of momentum $\Sigma mv_1=\Sigma mv_2$ $\implies 0=m_A v_A+m_Bv_B$ $\implies v_A=v_B=v$ Now, according to the law of conservation of energy $\frac{1}{2}m_Av_{A_{\circ}}^2+\frac{1}{2}m_Bv_{B_{\circ}^2}+\frac{1}{2}KS^2=\frac{1}{2}mv_A^2+\frac{1}{2}m_Bv_B^2+0$ $\implies KS^2=m_Av^2+m_Bv^2$ $\implies (60)(0.3)^2=5v^2+5v^2$ This simplifies to: $v=0.735m/s$ Now, we can determine the required angle as $\frac{1}{2}mv^2+0=0+mgh(1-cos\theta)$ We plug in the known values to obtain: $\frac{1}{2}(5)(0.735)^2=5(9.81)(2)(1-cos\theta)$ $\implies \theta=\phi=9.52^{\circ}$
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