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: 26


$7.4m/s$ $5.4m/s$

Work Step by Step

We apply the impulse momentum principle in the x- direction for observer A $mv_{x_1}+\Sigma \int ^{t_2}_{t_1} F_x dt=mv_{x_2}$ We plug in the known values to obtain: $10(5)+6(4)=10v_{x_2}$ $\implies v_{x_2}=7.4m/s$ Now we apply the impulse momentum principle for observer B as follows: $mv_{x_1}+\Sigma \int^{t_2}_{t_1} F_x dt=mv_{x_2}$ $\implies 10(3)+6(4)=10v_{x_2}$ $\implies v_{x_2}=5.4m/s$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.