# Frontal car collision

## Recommended Posts

Hi.

I have this nagging issue with people's physics in all kinds of movies and shows and even on the street in which they claim that if you have a frontal collision between two cars, the speeds add up.

Basically, there's car A, going 100m/s that way --> and car B, going 100m/s that way <--. Frontal ideal collision. I hear people say that it's worse than hitting a wall (ideal, concrete, immovable wall), because the speeds add up, essentially people actually smash at 200m/s.

I think they're nuts.

My problem with this is: Car a is in a crash. Essentially, all its energy is converted into molten crash juice (deformation, mostly), equal to speed difference (100 to 0, 100), divided by stopping time (let's assume 10 ms), basically this means that one gets a negative deceleration of 100m/s in 0.01s, which is -what- some 1000G? Math is unimportant.

However, the same thing happens to car B, assuming identical cars. It also goes from 100 to 0 in the same space and time, basically, there is no difference between an impact of 100m/s to a wall or to another car coming at you at the same speed the wrong way, the acceleration is the same (assuming that's what kills you).

Now, if car A throws a rubber ball out the window, the ball goes from 100m/s to -100m/s, effectively adding the speeds and thus the impact (divided by mass ratio).

Am I wrong? Is some energy transferred to the wall I'm not adding up?

##### Share on other sites

I think you're right, but that things get messier when the cars are different masses, as then they don't come to a complete stop upon impact. If one is smaller, it'll go through a much larger change in velocity and end up going backwards.

##### Share on other sites

If it's a conservation of energy, then it should be difference squared.

I assume energy is divided directly by masses, so the difference is speed should be relatively simple?

##### Share on other sites

Hi.

I have this nagging issue with people's physics in all kinds of movies and shows and even on the street in which they claim that if you have a frontal collision between two cars, the speeds add up.

That's the problem with learning your physics on the streetcorner.

##### Share on other sites

I have this nagging issue with people's physics in all kinds of movies and shows and even on the street in which they claim that if you have a frontal collision between two cars, the speeds add up.

...

I think they're nuts.

Conservation of energy does not apply to changing reference frames. Suppose you have a 1 kg mass, and in your reference frame its stationary. In a different reference frame its moving at 1000 m/s. In one frame it has zero kinetic energy, in the other a million Joules of kinetic energy. Same thing with conservation of momentum.

What you want to consider in a crash in the transfer of energy and momentum, and the speed at which the transfer takes place (power and acceleration). If you travel at 100 mph and crash a stationary car, it is much like the traveling at 50 mph and crashing another car at 50 mph in the other direction -- unless you consider the rest of the Earth. You're probably more likely to flip in the first scenario, or crash a stationary object.

Cars are designed to crumple on impact. The crumple absorbs kinetic energy, and increases the time over which energy and momentum are transferred (for a lower power collision with less acceleration). You really don't want to crash a concrete wall at any reasonable speed.

##### Share on other sites

Conservation of energy does not apply to changing reference frames.

Why would I change reference frames? Ground seems like a good idea. Also, how would that change anything? If my reference would be a satellite in orbit, one car would go 10100 and the other 9900. Resulting acceleration during impact would still be the same?

Also, how does crumple zones change anything (in the context of the original question)? Both cars crumple the same, so the net effect on the passengers regarding peak acceleration is the same in both cases?

@swansont: Me or them?

##### Share on other sites

Cars are designed to crumple on impact. The crumple absorbs kinetic energy, and increases the time over which energy and momentum are transferred (for a lower power collision with less acceleration). You really don't want to crash a concrete wall at any reasonable speed.

But the crumple absorbs an equal amount of energy in the two scenarios; if a car hits its mirror, the point of impact does not move in the CoM frame, and each car absorbs its own KE in the crumple. The system has twice the KE, but two places to put it.

Put another way, if you could not see or hear anything outside of the car, could you distinguish a completely inelastic collision with an identical car vs. a wall?

In reality, it's the collision not being completely inelastic that makes hitting the wall worse, since you tend to recoil.

Merged post follows:

Consecutive posts merged

@swansont: Me or them?

Anyone who thinks a head-on collision at v and -v is the same as hitting a wall at 2v

##### Share on other sites

Why would I change reference frames? Ground seems like a good idea. Also, how would that change anything? If my reference would be a satellite in orbit, one car would go 10100 and the other 9900. Resulting acceleration during impact would still be the same?

The reference frame of the person in the car is a good option.

Also, how does crumple zones change anything (in the context of the original question)? Both cars crumple the same, so the net effect on the passengers regarding peak acceleration is the same in both cases?

Well concrete walls don't have crumple zones (that's why I compare to crashing a stationary car, not a concrete wall with huge mass and no crumple zone).

##### Share on other sites

Well concrete walls don't have crumple zones (that's why I compare to crashing a stationary car, not a concrete wall with huge mass and no crumple zone).

That's true but not consistent with the conditions of the OP.

##### Share on other sites

So, to summarize:

A head on collision between two cars going 100m/s is basically the same as hitting a stationary car going 200m/s, or being hit by a car going 200m/s while stationary.

This is not the same as hitting a concrete wall going 200m/s, because the wall is a lot more massive than your car and gives a lot less. The masses of the colliding bodies matters. Just like it's not as bad for a mosquito to hit your windshield at 200m/s, because, since the mass of your car is so much greater than the mosquito, it's the mosquito that does almost all of the accelerating. Splat for the mosquito, no noticeable acceleration for you. (Or splat for you, but the wall doesn't move.)

##### Share on other sites

Theoretically, if two cars with the same momentum collided head-on where a line was drawn across the road, each car would stay on its own side of the line, proving that a head-on crash unfolds the same as, for example, colliding with an immovable object.

Realistically, no two cars have the same momentum, thus one ends up suffering a greater impact than the other. For example, one dead person is "worse" than two injured people. And this is in addition to there typically being twice as many injured/killed people and broken/ruined cars involved.

Also, from a human perspective (literally), instead of a victim seeing, say, a bridge abutment approaching at a realistic relative speed of 60 mph, they see another car approaching at an unrealistic relative speed of 120 mph, which makes it seem worse.

##### Share on other sites

So, to summarize:

A head on collision between two cars going 100m/s is basically the same as hitting a stationary car going 200m/s, or being hit by a car going 200m/s while stationary.

The latter two are equivalent but they are not the same as the first. Going 200 m/s give you four times the energy as going 100 m/s, but the 100 m/s case has two cars, so the total energy is twice as large when one cars is going 200 m/s. Momentum dictates the final speed of the cars will be cut in half when there is a stationary target involved and the collision is completely inelastic, which means that there is significantly more kinetic energy that must be dissipated when one of the cars is going 200 m/s.

edit: math error

Edited by swansont
##### Share on other sites

The latter two are equivalent but they are not the same as the first. Going 200 m/s give you four times the energy as going 100 m/s, but the 100 m/s case has two cars, so the total energy is twice as large when one cars is going 200 m/s. Momentum dictates the final speed of the cars will be cut in half when there is a stationary target involved and the collision is completely inelastic, which means that there is significantly more kinetic energy that must be dissipated when one of the cars is going 200 m/s.

There's twice as much energy relative to the ground, but isn't it the car's frame of reference that matters more?

With the 200m/s car going east towards a stationary car, after the collision they're both going 100m/s east, right? So the first car accelerates 100m/s west, and the second car accelerates 100m/s east. Which is the same as what happens when they're both going 100m/s to start with and come to a stop. The only difference is that in the first case they still have kinetic energy relative to the ground, but that shouldn't matter unless they then hit something else before rolling and air resistance bring them to a stop. Right?

##### Share on other sites

Ok, lets put some numbers to this. Lets say we have two cars weighing 1000 kg, traveling at 100 m/s difference (either 50 m/s in opposite directions or one at 100 m/s and the other stationary). Assume we have a 1 m perfect crumple zone on the front of each car, and that the collision is completely inelastic. By perfect crumple zone I mean it gives a constant deceleration.

Case 1: two cars at 50 m/s each in opposite directions:

Both cars have energy of 1000 kg * (50 m/s)^2 = 2.5 MJ each, total 5 MJ. Both cars have a momentum of 1000 kg * 50 m/s = 50,000 kg m/s, in opposite directions, total of zero.

After the collision, both cars are at a standstill, with zero kinetic energy. Energy dissipated = 2 X 2.5 MJ - 0 MJ= 5 MJ. Momentum transferred: 50,000 kg m/s from each to the other. Average acceleration = 50 m/s in 1 m, each.

Case 2: one car at 100 m/s colliding with a stationary car:

One car has 1000 kg * (100 m/s)^2 = 10 MJ, the other car 0 MJ, total 10 MJ. One car has momentum of 1000 kg * 100 m/s = 100,000 kg m/s, the other has zero, total 100,000 kg m/s.

After the collision, conservation of momentum dictates both cars are traveling at 50 m/s since the mass has doubled but the momentum is the same. After the collision, kinetic energy is 2,000 kg * (50 m/s)^2 = 5 MJ. Energy dissipated: 10 MJ - 5 MJ = 5 MJ, just like before. Momentum transferred: 50,000 kg m/s from each to the other. Average acceleration: one decelerates 50 m/s while the other accelerates 50 m/s, over a larger distance but still 1 m crumple zone for each.

No difference the moment of a collision, but after the collision in the second case both cars are traveling 50 m/s (or 112 mph), one of them backward, and are very likely to flip or collide with a stationary object.

##### Share on other sites

There's twice as much energy relative to the ground, but isn't it the car's frame of reference that matters more?

With the 200m/s car going east towards a stationary car, after the collision they're both going 100m/s east, right? So the first car accelerates 100m/s west, and the second car accelerates 100m/s east. Which is the same as what happens when they're both going 100m/s to start with and come to a stop. The only difference is that in the first case they still have kinetic energy relative to the ground, but that shouldn't matter unless they then hit something else before rolling and air resistance bring them to a stop. Right?

Gah, I dropped a factor of 2 in the calculation I did when composing my post. You are quite right.

And I should have caught it, because they are indeed all the same collision, just viewed from different reference frames (each car, and the CoM frame). They have to give the same answer.

##### Share on other sites

• 1 month later...

Well, matter has since been tested by the Mythbusters (Season 8 ep 07 - Mythssion Control, aired May 5). Small scale test and full crash, nicely done. Turns out physics still works, the crash was equivalent to single speed, not double.

Sorry for popping the question and disappearing, the forum was supposed to notify by mail? Huh. I must've messed it up.

## Create an account

Register a new account