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Head-on collision math

Consider two scenarios:

Two identical vehicles (same size and mass) travel at the same speed, let's say 50 km/h, in opposite directions, and they collide with each other head-on. One of those vehicles hits a rock wall (which doesn't break nor budge in any significant way) head-on at 50 km/h.

From the point of view of one of the vehicles, which collision is more severe?

Most people would instantly answer that the first collision is more severe because the effective collision speed is 100 km/h, and thus the collision has twice as much force than the second collision, which happens only at 50 km/h.

This answer is wrong, wrong, and utterly wrong. Many people just don't get this one, not even people who should know better. I can't even count how many times I have heard people getting this one wrong.

The most prominent and severe case which I have seen was Jamie Hyneman from the show MythBusters getting this exact problem wrong in their "demolition derby special" episode, where he stated that two trucks travelling at 50 mph each and colliding head-on were subject to a collision force equivalent to hitting a rock wall at 100 mph. Maybe he is not a phycisist, but nevertheless he of all people should know this.

The correct answer is: The two collisions are completely equivalent. From the point of view of one of the vehicles it makes absolutely no difference whether it hits a rock wall at 50 km/h or another identical vehicle which was traveling at the same speed in the opposite direction. The amount of force applied to the vehicle is the same in both situations.

(Ok, in reality there will be some differences because the consistency of a rock wall is very different from a consistency of a vehicle, but this only means that hitting the rock wall will be more severe than hitting the other car, although probably not by a lot.)

I know that no matter how much this is explained, some people just don't get it. They just can't get rid the misconception that the two-vehicle collision must have double the force. There are a few things which might make it easier to accept:

When the vehicle hits the rock wall at 50 km/h, the rock wall causes a force large enough to stop the vehicle right there. In other words, the vehicle hits the rock wall with a momentum equivalent to its speed times its mass. Conversely, by Newton's law, at the moment of the collision the rock wall causes an equal force to the vehicle in the opposite direction, causing it to stop. That is, the rock wall causes a force equivalent to the 50 km/h times the mass of the vehicle.

The misconception in the two-vehicle scenario is basically that this applied force is double that, ie. the equivalent to 100 km/h times the mass of the vehicle.

However, think about where this force is coming from in the two-vehicle scenario: It's coming from the second vehicle. But the second vehicle is also traveling at 50 km/h and has the same mass.

So we have two forces: Vehicle 1 applies the equivalent of 50 km/h times its mass to vehicle 2, and vehicle 2 applies an equal force to vehicle 1. This causes both vehicles to stop right there.

Where would the additional 50 km/h times the mass of the vehicle come from? Vehicle 1 cannot apply that force to itself. It's applying it to vehicle 2. So where is it coming from?

The answer is that it's not coming from anywhere because the force applied to vehicle 1 is not 100 km/h times the mass, but only 50 km/h times the mass. The same as with the rock wall.

Think also about this: If you applied a force equivalent to 100 km/h times the mass in the opposite direction of vehicle 1, that would actually make vehicle 1 change direction and go backwards at 50 km/h after the collision. Conversely it would also make vehicle 2 do the same. That doesn't happen.

Think about it like this: If vehicle 1 couldn't "see" what it hits, how can it tell if it hit a rock wall or vehicle 2? The "point of impact" remains stationary in the two-vehicle case, in the exact same way as in the one-vehicle-and-rock-wall case. From the point of view of vehicle 1, there's no difference.

Now, if vehicle 2 was stationary and vehicle 1 hit it at 50 km/h, that would make a big difference compared to the rock wall. That's because now vehicle 2, having the same mass as vehicle 1 (rather than "infinite" mass, as the rock wall), is applying much less force to vehicle 1.

A rather different approach to the same problem: Consider the two scenarios depicted in the image. In which case is a stronger force applied to the rope, in other words, in which case is the tension of the rope larger? Or is it the same in both cases? This problem is actually equivalent to the vehicle collision scenario.

If your intuitive answer is that in the first case the force is twice that of the second case, then you are again wrong.

Think about it like this: If in the first scenario you grab the rope from the middle (without pulling in any direction), does anything change with respect to any force? If you think about it, you'll see that this doesn't change anything.

Now keep a tight grasp of the rope, without allowing your hand to move anywhere, and cut the rope from the right side.

Now, from the point of view of the weight at the left: Has anything changed? Is the tension of the rope changed in any way?

Now, isn't this situation identical to the second scenario at this moment?

Nothing changed from the point of view of the weight at the left. It has no way of "noticing" that the rope was cut. It still weights the same, and applies the same force to the rope. In both scenarios the tension of the rope is the same: That of 1 kg of weight.

This situation is completely equivalent to the vehicles scenarios, but with pulling forces rather than pushing ones.



Update: In a recent episode of Mythbusters they tested this conundrum in practice because fans complained about Hyneman's comment, and it showed the theory to be completely right: Two identical cars hitting each other head-on at 50mph is equivalent to one of the cars hitting a wall at 50mph (and nowhere even close to hitting the wall at 100mph).

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