Answer
(a) The final velocity is $25~m/s$
(b) The change in momentum during the $~4.0~s~$ is $~60~kg~m/s~$ to the east.
Work Step by Step
(a) The impulse exerted on the object is equal to the change in momentum. We can find the east component of the change in momentum:
$\Delta p_x = F~t = (15~N)(4.0~s) = 60~kg~m/s$
We can find the north component of momentum:
$p_y = m~v_y = (3.0~kg)(15~m/s) = 45~kg~m/s$
We can find the magnitude of the momentum after the force is applied:
$p = \sqrt{p_x^2+p_y^2} = \sqrt{(60~kg~m/s)^2+(45~kg~m/s)^2} = 75~kg~m/s$
We can find the final velocity:
$m~v = p$
$v = \frac{p}{m}$
$v = \frac{75~kg~m/s}{3.0~kg}$
$v = 25~m/s$
The final velocity is $25~m/s$
(b) Since the north component of momentum did not change, the only change in momentum is the east component of momentum. The change in momentum during the $4.0~s$ is $60~kg~m/s$ to the east.
