[reidacdc]

I think yes?

Kinetic energy scales as velocity squared, so a car at 100 MPH has 4x the energy of a car at 50 MPH. But in the 50 MPH scenario there are two cars, so the total energy dissipated in the 50 MPH head-on collision is half that of the 100 MPH brick-wall collision. In the brick-wall case, presumably all the energy is available to demolish the one car, but in the head-on case, the energy is spread out demolishing two cars.

Impact force is maybe trickier, it depends on the acceleration, but if the two 50 MPH cars end up with zero momentum, then they have 50 MPH of delta-v each over some collision time. The single car of course has 100 MPH of delta-v. If the collision times are the same (arguably a reasonable approximation if the head-on is a highly symmetric), then the impact force in the head-on case is half that of the brick-wall case.

[HPsquared]

Although the single car at 100 MPH has twice the initial kinetic energy of two cars at 50 MPH, not all of this gets dissipated in the collision. After the collision in the "single car 100 mph hitting an identical stationary car" case, we have both cars moving at 50 MPH in the same direction, NOT two stationary cars. Half the original kinetic energy is still in the form of kinetic energy, in other words. The 100MPH car does not experience a velocity change of 100 MPH, but only 50 MPH - the same as if it had hit an oncoming car at 50MPH.