Classical car collision I have a very confusing discussion with a friend of mine.
2 cars ($car_a$ and $car_b$) of the same mass $m$ are on a collision course. Both cars travel at $50_\frac{km}{h}$ towards each other.
They collide. Ignoring any shreds and collateral damage, what is the speed of collision that the driver of $car_a$ felt?
What I mean is, if $car_a$ were to be driven into an infinite mass wall, what would the velocity be to replicate the damage caused by the initial collision?
 A: I think you are asking how much damage would be done to the driver in the two cases you described.  If that is your question, then the single car that is driven at a speed of $50\frac{km}{hr}$ into an infinite mass wall would experience the same damage as two identical cars being driven exactly head on at a speed relative to the ground of $50\frac{km}{hr}$ into each other.
You can easily understand this if you imagine an infinitely thin wall between the two colliding cars at exactly the plane where they collide.  Assuming no shards or other pieces breaking off and going through this imaginary wall you can see that this is exactly the same as having an infinitely massive wall in place of either car since everything is exactly symmetric about this infinitely thin imaginary wall.
To be absolutely correct, this answer would actually require that the cars be left-right symmetric so that the centers of mass exactly line up perpendicular to the plane of the collision.  If they were asymmetric it would be similar to symmetric cars hitting slightly off center - so there would be some torque around the point on the collision plane where the line determined by centers of mass intersect the plane.  Clear?
EDIT: A situation which is equivalent to 2 cars hitting head on at 50kmph is the following: one car sitting stationary (with brakes off) while the other car hits it head on at 100kmph. This assumes a perfectly inelastic collision so that the two cars will then proceed (joined together) in the original direction of the 100kmph car but they will both be going at 50kmph. In both of these cases the change in speed of each car is 50kmph in a short time so the damages will be equivalent (either 50->0, 100->50 or 0->-50). However, if a car that is traveling at 100kmph hits an infinite mass wall the change in speed in a short time will be 100kmph so it is not equivalent to the two cases.
A: There is no defined quantity called "speed of collision" and you cannot really measure "damage" in that sense either. I've heard this before when people at the drivers licence center talk about collisions and it is a bit strange.
If your question is with what force a car will be subject to when it comes to a halt it is the same as always: $F = ma$, where $a$ here will be the deceleration the car is subject to in the collision. So if you collide with an imaginary inelastic wall, as you imply, that brings you to an absolute stop in no time the force will be infinite. Of course this is impossible in reality since there is no such wall and the car itself will also compress and therefore slow the deceleration.
If you're interested in an inelastic crash it could give you the speed before and after the crash if you look at the momentum, $p=mv$. The total momentum must be the same before and after the crash, called the conservation of linear momentum, so $m_av_a + m_b+v_b = 0$ where $v_a = -v_b$ and $m_a = m_b$ as stated in the question. Here you can see that depending on the initial movement of the cars it is impossible to come to a halt if they do not move in opposite directions. You could simulate you problem by making the wall very heavy and making it move the opposite way with a small velocity. This is of course not what happens in reality either though but maybe closer to your question.
A: Dispense with most of the math and think of this question in more practical terms and in an easier to understand way: You have 3 theoretically "perfectly identical", "perfectly symmetrical" cars "A", "B" and "C".  If car "A" traveling at "exactly 50kp/h" hits a theoretically "immovable wall" "head on", (that is to say, exactly perpendicular to the plane of their masses) it will undergo, in effect, an "instantaneous deceleration" from 50kp/h to 0kp/h.  If the other two theoretically "perfectly identical", "perfectly symmetrical" cars "B" and "C", moving in "exactly opposite directions" crash "exactly head on" (that is to say, coming from exactly opposite points and moving perpendicular to their respective planes of mass and in direct line with the perfect centers of their respective symmetries) at "exactly 50kp/h each", then upon impact they will each transfer 100% of their kinetic energy into the other vehicle and will each undergo, in effect, an "instantaneous deceleration" from 50kp/h to 0kp/h.   BY DEFINITION, all 3 vehicles will undergo, in effect, IDENTICAL, instantaneous rates of deceleration!  As for the "occupants" of any or all of these cars, they would feel the "effects" of a 50kp/h crash into an immovable object.  Remember, of course, this is all happening in the "perfect world" of theoretical physics and the ACTUAL consequences, due to the unpredictability of "real life" physics and its, at times, uncontrolled/uncontrollable variables, might vary SIGNIFICANTLY from those represented above!                
