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 262: 37



Work Step by Step

We apply the law of conservation of momentum to determine the required velocity $m_t(v_t)_1+m_c(v_c)_1=(m_t+m_c) v$ We plug in the known values to obtain: $(2500)(\frac{30\times 1000}{3600})+1500(0)=(2500+1500)v$ This simplifies to: $v=5.208m/s$ The kinetic energy of the system before coupling is given as $T_1=\frac{1}{2}m_t(v_t)^2_1+\frac{1}{2}m_c(v_c)^2_1$ $\implies T_1=\frac{1}{2}(2500)(\frac{30\times 1000}{3600})^2+0=86805.55J$ and the kinetic energy after coupling is given as $T_2=\frac{1}{2}(m_t+m_c)(v)^2$ $\implies T_2=\frac{1}{2}(2500+1500)(5.208)^2=54246.528J$ Now the difference in kinetic energy is $\Delta T=T_1-T_2$ $\implies \Delta T=86805.55-54246.528=32559.022J$
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