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 15.2 - Principle of Linear Impulse and Momentum for a System of Particles - Problems - Page 252: 25



Work Step by Step

It is given that $v_1=18Km/h=5m/s$ We know that $S=S_{\circ}+v_{\circ}t -\frac{1}{2}gt^2$ $\implies -10=0+5t-\frac{1}{2}(9.81)t^2$ $\implies t=2.026s$ Now according to the impulse momentum principle $mv_{y1}+\Sigma \int^{t_2}_{t_1} F_ydt=mv_{y_2}$ We plug in the known values to obtain: $(400-4)(5)+2.026(2.026)-(400-40)(9.81)(2.026)=(400-40)v_2$ $\implies v_2=7.208m/s$
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