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