## Engineering Mechanics: Statics & Dynamics (14th Edition)

$7.4m/s$ $5.4m/s$
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: $mv_{x_1}+\Sigma \int^{t_2}_{t_1} F_x dt=mv_{x_2}$ $\implies 10(3)+6(4)=10v_{x_2}$ $\implies v_{x_2}=5.4m/s$