Given
Car 1
m1 = 1300 kg
v1 = 20 m/s
m2 = 900 kg
v2 = -15 m/s
(Negative sign shows that direction of car 2 is opposite to car 1)
Procedure
As per the conservation of linear momentum, "The total momentum of the system before the collision must be equal to the total momentum after the collision". And this applies to the perfectly inelastic collision as well. Then the expression is,
![\begin{gathered} m_1v_1+m_2v_2=(m_1+m_2)v \\ v=(m_1v_1+m_2v_2)/(m_1+m_2) \\ v=\frac{1300\operatorname{kg}\cdot20m/s-900\operatorname{kg}\cdot15m/s}{1300\operatorname{kg}+900\operatorname{kg}} \\ v=5.681m/s \end{gathered}]()
Thus, we can conclude that the speed and direction of the cars after the impact is 5.68 m/s towards the first car.