Let's take the positive x-direction towards east and the positive y-direction towards south. The momentum must be conserved on both directions, after the collision. On the x-direction, initially we have only momentum from car 1, while on the y-direction initially only car 2 contributes to the total momentum of the system. After the collision, the two cars will move together with a total mass (m1+m2) and with final velocity vf, which can be decomposed on both directions. All of this translates into the equations:
[tex]m_1 v_1 = (m_1+m_2)v_{fx}[/tex] (1)
[tex]m_2 v_2 = (m_1 + m_2)v_{fy}[/tex] (2)
where [tex]m_1 = 1500 kg[/tex], [tex]v_1 = 90 km/h[/tex], [tex]m_2 = 3000 kg[/tex] and [tex]v_2 = 60 km/h[/tex]. [tex]v_{fx}[/tex] and [tex]v_{fy}[/tex] are the components of the final velocity on both axes x and y.
By dividing equation (2) by (1), we get:
[tex] \frac{v_{fy}}{v_{fx}}= \frac{m_2v_2}{m_1v_1}= \frac{(3000 kg)(60 km/h)}{(1500 kg)(90km/h)}=1.33 [/tex]
And the tangent of this ratio gives exactly the angle of the velocity vf in the south-east direction, with respect to the positive x-axis, so it gives us the direction of the final velocity:
[tex]\alpha = \arctan (1.33)=53.1 ^{\circ}[/tex]
