Answer
The magnitude of the velocity is 5.87 m/s, and the direction is $57.7^{\circ}$ north of east.
Work Step by Step
$p_x = m_1~v_x = (85~kg)(7.2~m/s)$
$p_x = 612~kg~m/s$
$p_y = m_2~v_y = (110~kg)(8.8~m/s)$
$p_y = 968~kg~m/s$
We can find the magnitude of the momentum.
$p = \sqrt{(612~kg~m/s)^2+(968~kg~m/s)^2}$
$p = 1145~kg~m/s$
We can find the magnitude of the velocity.
$m~v = p$
$v = \frac{p}{m} = \frac{1145~kg~m/s}{195~kg}$
$v = 5.87~m/s$
We can find the angle $\theta$ north of east.
$tan(\theta) = \frac{968~kg~m/s}{612~kg~m/s}$
$\theta = arctan(\frac{968~kg~m/s}{612~kg~m/s})$
$\theta = 57.7^{\circ}$
The magnitude of the velocity is 5.87 m/s, and the direction is $57.7^{\circ}$ north of east.
