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.
 A: Actually, assuming that the oncoming car is the same mass as yours, colliding with an oncoming car at 50 MPH is equal to colliding with an ideal immovable wall at 50 MPH.  Consider this:
I'm going to set up one of two experiments.  I'm either going to ram car A into car B, both of them moving 50 MPH in opposite directions, or I'm going to ram car A into a solid wall at 50 MPH.  However, I'm going to put up a shroud so that you can only see car A, you will be unable to see either car B or the wall, whichever one my coin-flip tells me to use.
Because you can now only see at car A and its contents, how would you tell which experiment I'd decided to do?
A: I think that it makes sense to move away from the  specific walls and cars and consider simply an inelastic collision on of two masses, $m_1$ and $m_2$. Otherwise we get stuck in the details.   
When two bodies collide, the devastating effect in collision depends only on their relative velocity $v_1-v_2$. Kinetic energy, which has the destructive effect is equal to  
$$\frac{1}{2}\frac{m_1m_2}{m_1+m_2}(v_1-v_2)^2$$  
The rest of the kinetic energy is associated with the movement of the center of mass of the system. This energy in the collision does not change, and has no effect of destruction.  
In given case, if faced two identical cars moving toward each other with one and the same velocity $v$ the energy of destruction is  
$$\frac{1}{2}\frac{mm}{m+m}(v+v)^2=mv^2$$   
Now, consider the case where a car collides with a massive barrier at speed $2v$.In this case $m_1=m$, $v_1=2v$, $m_2=\infty$, $v_2=0$   
The energy of destruction:  
$$\frac{1}{2}m(2v)^2=2mv^2$$  
I.e. the latter case is much more dangerous.
A: Certainly they are not exactly the same - a wall is not the same thing as a car, and a crash is a very complicated physical event. Even if simple calculations involving momentum and energy or descriptions involving reference frames suggest that aspects of a car-car and car-wall collision are the same, the real collisions will be fairly different.
In this case, though, simple considerations do reveal that the car-car crash at 50 mph is almost certainly safer than crashing 100 mph into a wall. Your energy calculation is a fine way to see this.
Another is to consider the car-car collision from a frame co-moving with the second car. In this frame, you're going 100 mph and crash into a stationary car. So the question is like asking whether it is worse to crash into a stationary wall or a stationary car when going 100 mph (apart from the fact that the movement relative to the road is a little different). Of course crashing into the car is less dangerous than crashing into the wall, confirming your earlier result.
I have often heard the same problem rephrased so that you consider crashing into a wall at 100 mph or crashing into a car when you're both going 100 mph. It may be that this was the original problem the physics professor mentioned, and it got distorted somewhere in the game of telephone it played since then. 
In that scenario, some people say they are equally bad because the energy dissipated per car is the same. Personally, I would probably go for the wall because at least some of the car's energy should go into the wall, but here the details become important (e.g. what if I fly through the window and then hit the wall?), and the energy alone is not a strong enough difference to say what which is worse. I imagine that either crash is very likely to be fatal at that speed.
Addressing your new question, two cars crashing head-on each at 50 mph is essentially the same as one car going 100 mph and crashing into a stationary car, by the relativity principle. However, relativity is broken by the existence of the road, so to the extent that the cars interact with the road during the collision there may be some differences.
A: 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 $4\times\frac{1}{2}mv^2$.
A: The most straightforward way to see how different the two scenarios are is to:
(1) consider two cars crashing into each other from a reference frame in which one of the cars is stationary and the other has a speed of 100mph
(2) consider the one car crashing into the wall with a speed of 100mph.
Assuming the wall is substantial enough that its mass and physical strength far exceeds that of the stationary car in (1), it's clear that the two scenarios significantly differ.
In (2), the car crashes into a stationary and effectively immovable, indestructible object at 100mph while in (1), the moving car crashes, at 100mph, into a stationary but otherwise identical object that importantly, can both move and deform.
A: Damage should be the same if two cars colliding at $50$mph and if a car travelling $(50*\sqrt{2})$ mph crashes into a wall.
The energy of destruction is an internal energy so:

First case

Equation of energy conservation
$\frac{mv^2}{2}+\frac{mv^2}{2}= T$
where $T$ -  internal energy, $m$ - car mass
so energy of destruction in first case:
$T = mv^2$

Second case

Equation of energy conservation
$\frac{m(v\sqrt{2})^2}{2}=\frac{(m+M)u^2}{2}+ T$
where $T$ -  internal energy, $m$ - car mass, $M$ - wall mass
Equation of momentum conservation
$mv\sqrt{2}=(m+M)u$
so internal energy
$T= \frac{m(v\sqrt{2})^2}{2}\Big(1-\frac{m}{m+M}\Big)$
$M>>m$ so energy of destruction in second case:
$T \approx {mv^2}$
A: I will assume that:


*

*There is car-car collision in both the cases.

*The car is a wooden lifeless cart with no engine.


Now, the Classical difference between the two collisions is-
The wheels of the car moving at 100mph is twice as energetic as the wheels of both the cars combined in the 50mph case. So, in the 100mph case (2nd case), the accident will be worse.
Now, there will also be special relativistic effect, although very tiny. The combined energy of the cars in the second case (100mph) is larger than the first one.
So in both the cases, the second case will be worse, for the reasons stated above.
