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 15.4 - Impact - Problems - Page 275: 65



Work Step by Step

We can determine the required time as follows: According to the conservation of linear momentum $m_A(v_A)1+m_B(v_B)1=m_A(v_A)2+m_B(v_B)2$ This simplifies to: $(\frac{1}{32.2})(20)+(\frac{10}{32.2})(0)=\frac{1}{32.2}(v_A)2+(\frac{10}{32.2})(v_B)2$ This simplifies to: $(v_A)2=20-10(v_B)2~~~$[eq(1)] We know that: $e=\frac{(v_B)2-(v_A)2}{20-0}$ This simplifies to: $(v_A)2=(v_B)2-12~~~$[eq(2)] After solving eq(1) and (2), we obtain: $(v_B)2=2.909ft/s\rightarrow$ and $(v_A)2=9.09ft/s\leftarrow$ Now we apply the impulse momentum principle $m_B(v_B)1+\Sigma \int Fdt=m_B(v_B)2$ We plug in the known values to obtain: $(\frac{10}{32.2})(2.909)-\mu_kN\Delta t=(\frac{10}{32.2})(0)$ This simplifies to: $t=0.226s$
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