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 265: 57



Work Step by Step

We can determine the required distance as follows: We know that: $m_bv_{bx}=(m_b+m_B)v_x$ $\implies (0.01)(300cos30)=(10+0)v$ This simplifies to: $v=0.2595m/s$ Now, according to the conservation of energy $T_1+V_1=T_2+V_2$ $\implies 0+\frac{1}{2}(m_b+m_B)v^2=0+(m_b+m_B)gh$ We plug in the known values to obtain: $0+\frac{1}{2}(10+0.01)(0.225)^2=(0.01+10)(9.81)h$ This simplifies to: $h=3.43mm$ Now $d=\frac{h}{sin30}=\frac{3.43}{sin30}=6.87mm$
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