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 263: 44



Work Step by Step

The required velocity can be determined as follows: According to the law of conservation of energy $\frac{1}{2}(m_t+m_g+m_b) v_1^2+(m_t+m_g+m_b) g_h=\frac{1}{2}(m_t+m_g+m_b) v_2^2+0$ This simplifies to: $v_2=7.672m/s$ Now we use the principle of impulse and momentum in the x-direction $mv_{x_1}+\Sigma \int F_xdt=mv_{x_2}$ $\implies (m_t+m_g+m_b) v_2=(m_t+m_g)v_t+m_bv_b$ We plug in the known values to obtain: $(10+40+45)(7.672)=(10+40)v_t+45v_b$ but $v_b=v_t-2$ $\implies (10+40+45)(7.672)=(10+40)v_t+45(v_t-2)$ This simplifies to: $v_t=8.62m/s$
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