How is momentum conserved

Or is it not? I am working this problem involving two cars. One is initially sitting at a red light, and the other car (driving behind the first) ends up hitting the car sitting at the light. It then says that both cars travel a certain distance, and that friction acts on the cars as they are moving those distances.

How is momentum conserved under these circumstances? I just don't get it. Please cure my ignorance.

Note the meaning of "conservation".
Change in momentum of one car equals the change in momentum of the other car due to collision.
It is not that the momentums are equal before and after collision.
=======================================…
Car A hits car B.
In Physics we say car A exerts a force on car B.
Car B was initially moving with a speed u1.
Since car A exerts a force on carB, by Newton’s second law the force is equal to the rate of change of momentum of car B.
If the car B moves with the speed v1 because of the force, then

Force F1 = m1 (v1-u1)/t = [m1v1 – m1u1] /t, where t is the time taken for the change in speed and m1 is the mass of carB.

Thus there is a change in the momentum of the car B.
---------------------------------------…
Now consider CarA
Car A hits carB. It is the same as saying car B hits car A.
In Physics we say car A exerts a force F2 on Car B.
Car A was initially moving with a speed u2.
Since car B exerts a force on car A, by Newton’s second law the force is equal to the rate of change of momentum of car A.
If the car A moves with the speed v2 because of the force, then

Force F2= m2 (v2-u2)/t = [m2v2 – m2u2] /t, where t is the time taken for the change in speed and m2 is the mass of car A.

Thus there is a change in the momentum of the car A.
---------------------------------------…
By Newton’s third law F1 = F2 in magnitude and are oppositely directed.
.
[m1v1 – m1u1] /t = = [m2v2 – m2u2] /t,