Law of Conservation of Momentum = M1V1 + M2V2 = M1V1' + M2V2'

Since M1 = M2 and since mass of each object will remain the same on both sides of the equasion (assuming we're not talking near-speed of light speeds), we get:

V1 +V2 = V1' + V2'

Now, since V2 is say, twice V1 we get:
V1 + 2(V2) = V1' +2(V2)'

Since the two cars are completely inelastic and stick together, V1' and V2' are the same, so we can combine to get:

V1 + 2(V2) = 3V'

Therefore:
.```` V1+2(V2)
V' = -------------
.``````` 3

If we insert real numbers (V=50)we get:

`````50 + 2(-50)
V' = --------------
````````` 3

So:
-50/3 = -16.6666666666 kph is the momentum after collision. Since both cars come to a complete stop, the car travelling slower (V1) will absorb 16.666666kpm X (mass) more energy than the faster car.