Answer
(a) $p_x = -35,000~kg~m/s$
$p_y = 34,500~kg~m/s$
(b) The magnitude of the net momentum is $49100~kg~m/s$.
The direction of the net momentum is $44.6^{\circ}$ north of west.
Work Step by Step
(a) $p_x = mv_x$
$p_x = (2500~kg)(-14.0~m/s)$
$p_x = -35000~kg~m/s$
$p_y = mv_y$
$p_y = (1500~kg)(23.0~m/s)$
$p_y = 34500~kg~m/s$
(b) We can find the magnitude of the net momentum.
$p = \sqrt{(-35000~kg~m/s)^2+(34500~kg~m/s)^2}$
$p = 49100~kg~m/s$
We can find the angle $\theta$ north of west.
$\tan(\theta) = \frac{34500~kg~m/s}{35000~kg~m/s}$
$\theta = \arctan(\frac{34500~kg~m/s}{35000~kg~m/s})$
$\theta = 44.6^{\circ}$
The magnitude of the net momentum is $49100~kg~m/s$.
The direction of the net momentum is $44.6^{\circ}$ north of west.
