## College Physics (4th Edition)

The speed of the two cars after the collision is $6.0~m/s$ at an angle of $21.3^{\circ}$ south of east.
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.