## University Physics with Modern Physics (14th Edition)

The magnitude of the velocity is 5.87 m/s, and the direction is $57.7^{\circ}$ north of east.
$p_x = m_1~v_x = (85~kg)(7.2~m/s)$ $p_x = 612~kg~m/s$ $p_y = m_2~v_y = (110~kg)(8.8~m/s)$ $p_y = 968~kg~m/s$ 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.