Answer
The speed of the two cars after the collision is $6.0~m/s$ at an angle of $21.3^{\circ}$ south of east.
Work Step by Step
By conservation of momentum, the final momentum of the system is equal to the initial momentum.
We can find the east component $p_x$ of the momentum:
$p_x = (1700~kg)(14~m/s)~cos~45^{\circ} = 16,829~kg~m/s$
We can find the south component $p_y$ of the momentum:
$p_y = (1300~kg)(18~m/s) - (1700~kg)(14~m/s)~cos~45^{\circ}$
$p_y = 6571~kg~m/s$
We can find the magnitude of the momentum:
$p = \sqrt{p_x^2+p_y^2}$
$p = \sqrt{(16,829~kg~m/s)^2+(6571~kg~m/s)^2}$
$p = 18,066~kg~m/s$
We can find the speed of the two cars after the collision:
$mv = p$
$v = \frac{p}{m}$
$v = \frac{18,066~kg~m/s}{1700~kg+1300~kg}$
$v = 6.0~m/s$
We can find the angle $\theta$ south of east:
$tan~\theta = \frac{6571}{16,829}$
$\theta = tan^{-1}(\frac{6571}{16,829})$
$\theta = 21.3^{\circ}$
The speed of the two cars after the collision is $6.0~m/s$ at an angle of $21.3^{\circ}$ south of east.
