You could test the theory just the same with two cars going 5 mph.
Just so we’re on the same page:
A and B are different or the same?
Tell that the OP. It’s a their bar bet, not mine. 
But yes, you are right
It may or may not be the same problem. It depends on what exactly is colliding. It matters, for instance, whether the plane is not just immobile, but totally indestructible–perhaps the plane is so massive that it will not move, but still crumples.
For the case of symmetrical cars, the fly will not be double-crushed (as compared to one car hitting a wall). The wall pushes back with the same force as the other car does–as it must, since the point of contact is not accelerating, and any force from the moving car must be exactly countered by the wall.
Alternatively, imagining a 140 mph car vs. a stationary one, the results are the same. But in this interpretation, the greater speed is countered by the fact that the effective crush distance is doubled. So the acceleration is the same and therefore so is the force.
If you collide head on with a similar car coming toward you at a similar speed, you are no worse off than if you hit a wall.
But if you hit a very massive and incompressible truck coming towards you at a similar speed, and you hit it so precisely head on that you end up traveling backwards at roughly the original velocity of the truck, then it’s going to be more like hitting a wall at double the speed.
But the original question in post one was not “what happens to the driver”, but:
At the point of impact, are the forces the same?
This is what I meant when I wrote that it depends on the definition (i.e.: what you mean) by “the forces” and “the point of impact”. As often happens in bar bets, it is not clear enough the morning after.
Emphasis on the “incompressible.” If the truck has “infinite” mass but still has a crush region similar to the car, it would be better to hit that than the wall, as that portion of the kinetic energy would then be shared between the two vehicles.
Right, and that’s not realistic - but maybe it helps in seeing why the realistic case is not the same as hitting a wall at double the speed.
Suppose you mount a pressure/force sensor of some description on the front of a car. It hits the incompressible plane at 70mph and registers pressure P.
Now collide two cars head-on. If you mount a sensor on only one of the cars, it still registers P. If you mount a sensor on both cars, they each register P. But I don’t see how you can argue that the mere existence of a second sensor means the pressure is double.
I guess this does depend on the way the sensor works - I guess you have to assume that the sensor itself and its mounting are no more rigid than the rest of the car?
The exact same force is double because you measured it twice instead of once?
eta Riemann was editing as I replied. This post might not be relevant.
Yes, I just edited my post because the way I originally wrote it is obviously a silly way to look at it, isn’t it.
If I put five sensors on each car, is the force now 10 times as much?
Note that @Snarky_Kong 's reply was in response to a stupid first draft of my prior post which made it sound that way. I edited my post.
I find this easy to visualize :
A. Imagine a strong wall (plane) made of thick steel (a rigid plane ). It gets hit by a car at 70 mph and you can visualize the impact.
B. Now imagine the same steel wall being hit by cars from both sides at 70mph: The damage to each car is the same as in A since the wall is rigid.
Also in case B, there are no net forces on the wall since they cancel out (both the cars collide at the same time). If there are no forces on the wall, you can make it thinner and thinner….ultimately it can be paper thin.
So case B can be visualized as two cars traveling at 70mph each, in opposite directions, colliding with a wall at the same instant. Once you can visualize this, just visualize removing the wall since it has no role to play.
I see now my error was in adding the ‘impenetrable’ part of the wall. All walls, even brick take SOME force.
But you guys are right, wall or opposing-equally sized and sped car it’s the same, shit.
Hopefully he forgets tomorrow.
I think a better reference would be the sensor is hanging from a thread, either just touching the immovable wall or at the point of impact of the two cars. It’s measuring how much it’s being crushed - in other words the gross forces involved. Let’s stipulate that the cars weigh 1000 pounds each and the crash duration is 0.36 second - .0001 hour. In the car vs wall scenario, the car has momentum of 70,000 pound-miles/hour. The wall has to impose a force of 700,000,000 pound-miles/hour-squared against the car to stop it motionless.
In the car vs car scenario, each car has to impose the same force against the opposite car. So the gross amount, the crushing force experienced by the sensor, is 1.4 billion pound-miles/hour-squared. However the net force per car is the same as above - 700,000,000 pound-miles/hour-squared.
If I was adjudicating the bar bet in the OP, I’d say that it’s the gross force that answers the question, and the forces in the two scenarios are not the same.
What’s the force from the wall on the sensor?
700,000,000 pound-miles/hour-squared. If there was no wall, and no friction or any other opposing forces, the car would continue on it’s merry way at the same velocity conserving its momentum. When the car hits the wall, it decelerates. That deceleration comes from an opposing force which is the wall.
Since I’ve brought up conservation of momentum, something I’m trying to remember is the transfer of energy. In the car vs wall scenario, if the car was 100% elastic, it would bounce off the wall and travel in the opposite direction at the same speed. Force would have occurred (1.4 billion pound-miles/hour-squared), but no kinetic energy would have been lost. However, I think that relies on the wall having some force enacting on it to keep it in place or, as Bear Nenno touched upon, the wall having infinite mass. I think it would also have to be 100% inelastic, aka 0% elastic, aka incompressible. In the OP scenario, the car is 0% elastic so all the kinetic energy is lost, I believe resulting in heat. Theoretically, is the lost energy equally distributed between the bodies involved in the collision?
So in one scenario the sensor is hit with for P by car1 and P by car2 and in the other it’s hit by P by car1 and P by the wall?
Car vs wall. In terms of motion, no force is enacted upon the wall by the car. The wall doesn’t move. It’s counterintuitive because it’s the car that has the momentum and the kinetic energy. Nevertheless, it’s the car that is the recipient of the force as it’s the object that decelerates. The sensor records P as the force of the impact.
Car vs car. Both objects decelerate. Each imposes force in the collision onto the opposite car. The sensor records 2P as the force of the impact. This is gross force, which I’ve also described as crushing force. The sensor is external to the cars. If there was an accelerometer within a car, it would record the same amount of negative acceleration whether the car hit the wall or the opposite car. The onboard sensor would be recording net force for the car it was in.