A car of mass m drives at speed v towards a concrete wall. Its kinetic energy is ½ mv2, and all of that energy is dissipated when it hits the wall.
Now consider an identical car driving at speed 2v towards the same wall. Its kinetic energy is ½ m(2v)2 = ½ m4v2 = 2 mv2 — four times as much as the original car, since kinetic energy goes with the square of the velocity. I hope that up to this point, all this is uncontroversial.
Now consider two more identical cars, each driving at the original speed v, but this time directly towards each other, so that their collision speed is 2v. In this case, the total kinetic energy that is dissipated in the collision is (obviously) the sum of the energy the two cars have: that is 2 × ½ mv2 = mv2.
So in one crash at speed 2 v (the head-on collision), there is twice as much energy than in the other (the double-speed car into wall).
How can that be?