newtonian mechanics - Is two cars colliding at 50mph the same as one car colliding into a wall at 100 mph?

I was watching a youtube video the other day where an economist said that he challenged his physics professor on this question back when he was in school. His professor said each scenario is the same, while he said that they are different, and he said he supplied a proof showing otherwise.

He didn't say whether or not the cars are the same mass, but I assumed they were. To state it more clearly, in the first instance each car is traveling at 50mph in the opposite direction and they collide with each other. In the second scenario, a car travels at 100 mph and crashes into a brick wall. Which one is "worse"?

When I first heard it, I thought, "of course they're the same!" But then I took a step back and thought about it again. It seems like in the first scenario the total energy of the system is the KE of the two cars, or $\frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$. In the second scenario, it's the KE of the car plus wall, which is $\frac{1}{2}m(2v)^2 + 0 = 2mv^2$. So the car crashing into the wall has to absorb (and dissipate via heat) twice as much energy, so crashing into the wall is in fact worse.

Is this correct?

To clarify, I'm not concerned with the difference between a wall and a car, and I don't think that's what the question is getting at. Imagine instead that in the second scenario, a car is crashing at 100mph into the same car sitting there at 0mph (with it's brakes on of course). First scenario is the same, two of the same cars going 50mph in opposite directions collide. Are those two situations identical?

PS: This scenario is also covered in an episode of mythbusters.

Answer

I don't think any of the other answers have made the following point clear enough, so I am going to give it a try. Both scenarios are very similar before the collision, but they differ greatly afterwards...

From a stationary reference, you see the cars driving towards each other at 50mph, but of course if you choose a reference frame moving with the first car, then the second will be headed toward it at 100 mph. How is this different from the wall scenario?

Well, from a stationary reference frame, after the crash both cars remain at rest, so the kinetic energy dissipated is $2\times \frac{1}{2}mv^2$.

From the reference frame moving with the first car, the kinetic energy before the crash is $\frac{1}{2}m(2v)^2=4\times\frac{1}{2}mv^2$, but after the crash the cars do not remain at rest, but keep moving in the direction of the second car at half the speed. So of course the kinetic energy after the crash is $2\times\frac{1}{2}mv^2$, and the total kinetic energy lost in the crash is the same as when considering a stationary reference frame.

In the car against a wall, you do have the full dissipation of a kinetic energy of $4\times\frac{1}{2}mv^2$.