Yeah, this was a complicated physics problem that has some quick "common-sense" tricks in critical-thinking, like the bat+ball=$1.10 problem.

You want to calculate total momentum and energies of each car individually. Assume that the collision is perfectly inelastic one where ALL of the energies of each car is dissipated (they don't bounce). That puts ALL of the momentum and kinetic energy into crushing each other and causing maximum damage. What ends up causing the damage is basically the weight of the entire car crushing itself into the front-bumper. Since each car is equal mass and equal velocity, the contact-point where their bumpers touch should stick and not move (there's no backwards movement). The only movement will be gradual crushing to the back of each car. In which case, it looks like this:

20mph --> || <-- 20mph

as if both cars ran into a brick wall simultaneously. The full damage to each car will be similar to each other and it won't exceed the damage of running into a brick wall. One way to look at it is to calculate total momentum and kinetic energy. You'd then divide by two to split the total collision energy and the most energy apiece, is its own momentum in a perfectly inelastic collision, like hitting a brick wall. Another way to imagine it is another car, even moving at you will be "softer" than a brick wall, it'll crush and crumple on impact.

On the impact itself, it's the deceleration rates that determines the G-forces that the car and occupants feel. If you decelerate from 20mph within 2-feet (nose of car crumples 2-feet), that's a certain G-force. if you have this 2-feet of crush in hitting a wall or another car with a dead impact, it's the same G-forces. The key here is that they both have equal mass and velocity, therefore kinetic energy.


Now the situtation you're thinking of is hitting an irresistable force. Let's say you hit something that's moving at 20mph with much higher mass than you, such that it doesn't even slow down, but pushes BACK at you with the same speed unimpeded, then yes, it will be like hitting a brickwall at 40mph. Such as the case of hitting a locomotive or giant big-rig at 20mph head-on:

20mph --> >> <--20mph = 40mph --> ||

In this case, you do get an ADDITIONAL 20mph of momentum because not only did you decelerate from 20mph to 0mph, but you're also getting accelerated to -20mph in the opposite direction. So the total deceleration is 40mph and yes, you'd end up with double the G-forces and impact. That's why in crashes with two differently sized cars, like an SUV vs. a Neon, the passengers of the smaller cars end up dead at a much higher rate.