Something I was taught in High School Driver Education class.

[Towlie]

It's discussed under Reviews, here.http://www.cartalk.com/ct/review/show.jsp?showid=200821

Thanks. It does appear that those people are in unanimous agreement with alexi_drago, modified, and myself, but I'll be the first to point out that they're certainly no more authoritative that we are.

 

[Unlike a Bull]

But what if this is the other car?

http://travel.webshots.com/photo/2330010320065818567uYDTTk


If it's O.K., I would like to submit this to Mythbusters with a link back to this thread. It seems like something they would enjoy doing, and it would provide us with some data. Wonderful, wonderful data.

 

[sol invictus]

Thanks. It does appear that those people are in unanimous agreement with alexi_drago, modified, and myself, but I'll be the first to point out that they're certainly no more authoritative that we are.

Good question. I agree with you guys - a car crashing into a (very solid) wall at 50mph is equivalent to two identical cars colliding head-on, each moving at 50mph (and also equivalent to a 100mph car colliding with a stationary, otherwise identical car, except for a few minor effects related to wheel motion and friction).

This is actually a pretty familiar situation in physics, because it's common to think of reflecting boundary conditions (the wall) in terms of what's called an "image problem". You replace the boundary with another half-space, and for each object in the original space you have an identical one in the other space. Bouncing off the wall is colliding with your mirror image.

 

[Jeff Corey]

Thanks. It does appear that those people are in unanimous agreement with alexi_drago, modified, and myself, but I'll be the first to point out that they're certainly no more authoritative that we are.

The Tappet Bros. sure weren't in this case. They just added the velocities. On the show, they were corrected by a former prof of theirs at MIT. He explained it in terms of impetus vs velocity, as I recall.

 

[pgwenthold]

Good question. I agree with you guys - a car crashing into a (very solid) wall at 50mph is equivalent to two identical cars colliding head-on, each moving at 50mph (and also equivalent to a 100mph car colliding with a stationary, otherwise identical car, except for a few minor effects related to wheel motion and friction).

that is only true if you assume that the second car in the car/car collision is absolutely stationary, immutable, upon collision, but it isn't. As such, the energy deposition into the car is not the same in the two collisions.

From a physics pov, we should look at the center-of-mass collision energy. For an object running into a target, the COM collision energy (Ecom) is related to the lab energy (Elab) by

Ecom = Elab*Mtarget/(Mobject+Mtarget)

What that means is that there is a great energy in the object if the target mass is larger. Think of it in terms of balls colliding. If you have a bowling ball sitting on the ground, and you roll a ping pong ball at it, what happens after the colliison? Well, the bowling ball is still sitting where it is, while the ping-pong ball bounces away. Because the mass of the target is so much larger than the mass of the object, then effectively all of the initial energy of the system is deposited into the target. In the expression above, Mtarget/(Mobject + Mtarget) = 1, or pretty close to it.

Now, set a ping-pong ball on the ground and roll another ping-pong ball at it. What happens is that both ping-pong balls scatter, as the energy is distributed between both objects. In the expression above, Mtarget/(Mobject+Mtarget) = 1/2, so only half of the energy goes into the object.

I have not done the exercise with cars, but I have done this with ions and molecules. I can take an ion and have it collide with a target. It is a trivial experiment to show that more "damage" occurs to the ion when the target is larger. For the same nominal lab collision energy for the same ion, the extent of dissociation is larger with a larger collision target.

 

[sol invictus]

I have not done the exercise with cars, but I have done this with ions and molecules. I can take an ion and have it collide with a target. It is a trivial experiment to show that more "damage" occurs to the ion when the target is larger. For the same nominal lab collision energy for the same ion, the extent of dissociation is larger with a larger collision target.

I agree with all that, but it doesn't bear on the question. Everyone agrees that a 50mph car colliding with a wall is different (and worse for the car) than a 50mph car colliding with a stationary car (in fact if the stationary car is identical to the moving one and the collision is head-on, that would be equivalent to a car at 25mph colliding with a wall).

The analogous claim is that an ion colliding with a stationary target takes as much "damage" as an ion colliding perfectly head-on with an identical ion moving the other way at the same speed. In practice that second scenario is probably impossible to arrange with ions - it certainly is with particles and with protons at high energies - because they'll scatter rather than collide head-on.

 

[casebro]

E=MCC

Doubling the mass = double the energy.

Doubling the speed = four times the energy. It's called inverse square law.

So, no, two @50mph is not the same as one at 100mph. The 100 mph car would have 4x the energy of a single car, or 2x as the two that hit head on.

Therefor, being in the one @ 100, slamming into a solid object, would make a teenage driver four times as dead as being in a car hitting head on into another car.

 

[Towlie]

E=MCC

Doubling the mass = double the energy.

Doubling the speed = four times the energy. It's called inverse square law.

Professor Irwin Corey, is that you?

(Professor Corey once explained that heat causes expansion and cold causes contraction, and that's why the days are longer in the Summer.)

 

[sol invictus]

E=MCC

Doubling the mass = double the energy.

Unless one of the cars in an anti-car, that equation is totally irrelevant to this question.

Doubling the speed = four times the energy. It's called inverse square law.

No, it's not called that (hint: it's not inverse, and c doesn't vary).

 

[fuelair]

And, check this out: http://sjbaker.org/wiki/index.php?title=Cars_crashing_head-on_or_into_a_wall

 

[Towlie]

And, check this out: http://sjbaker.org/wiki/index.php?title=Cars_crashing_head-on_or_into_a_wall

Yet another source is in full agreement with my own position.

 

[pgwenthold]

I agree with all that, but it doesn't bear on the question. Everyone agrees that a 50mph car colliding with a wall is different (and worse for the car) than a 50mph car colliding with a stationary car (in fact if the stationary car is identical to the moving one and the collision is head-on, that would be equivalent to a car at 25mph colliding with a wall)..

Huh?

No, my description is exactly in line with the initial scenerio: Two 50 mph cars hitting head-on or a 100 mph hitting a brick wall?

From a relativity perspective, the first situation is exactly the same as having a 100 mph hitting a car that is stationary. Therefore, we can draw the following comparison:

Which is worse? A car going 100 mph running into a another car? Or one running into a brick wall? The answer is, hitting a brick wall is much worse than hitting another car, because the mass is much higher.

Remember the initial claim:

One of the things we were taught is that if you're driving at 50 MPH and you meet another car head-on that's also going 50 MPH, it would be like hitting a brick wall at 100 MPH.

This isn't true. It is most certainly not "like hitting a brick wall at 100 mph" because the center-of-mass collision energy is only half as much.

I haven't wasted time looking at links since I checked out the first one (from the Univ of Illiois) which claimed they were the same based on the assumption (implicit in their scenerio) that one of the cars had an infinite effective mass, basically equal to a brick wall.

To the extent they made that assumption, that is correct. However, it is wrong, because a car is not a brick wall.

 

[Tim Thompson]

Kinetic Energy

Consider this:

1) Two cars of identical mass m collide head on with velocity v for each car, relative to the center of mass. The total kinetic energy of the collision is simply the sum of the kinetic energies of the two cars ...

(1/2)mv² + (1/2)mv² = mv²
2) One car slams into a stationary immovable object ("wall") at double the speed from the case above. Now the kinetic energy of the car is ...

(1/2)m(2v)² = (1/2)m*4v² = 2mv²
Now, unless I am making some elementary mistake, the kinetic energy of the car hitting the wall is double the kinetic energy of the two cars colliding. This implies to me that the two scenarios are not at all equivalent, with the latter being in fact far more damaging.

 

[JoeTheJuggler]

You are very unlikely to hit a car exactly head on, as our instinct is to swerve out of the way of certain death. If you don't hit the car head-on the physics I will attempt to apply is irrelevant.

I think this falls under ceteris parabus. If you're not taking into account your instinct to swerve away from a car, why should you take it into account when talking about one of the two moving vehicles as if it were stationary (which is legitimate in physics).

I disagree. The two cars going 50mph are only going 50mph relative to some third point of reference (the ground). Relative to car A, car B is going 100mph and vice-versa.

Well said.

You could even frame running into a brick wall at 100 mph as the car being stationary and the brick wall rushing toward the car at 100 mph. Of course, you'd also say the pavement was rushing in the same direction as the brick wall underneath the car's wheels at 100 mph, and the air and so on.

 

[JoeTheJuggler]

Which is worse? A car going 100 mph running into a another car? Or one running into a brick wall? The answer is, hitting a brick wall is much worse than hitting another car, because the mass is much higher.

I didn't think the teacher specified a mass for the brick wall. I would have assumed (again trying to keep all things as equal as possible) that its mass was the same as a car.

In fact, maybe the teacher should have just said that two cars colliding head on each going 50 mph is the same as one car going 100 mph and hitting a parked car square on.

ETA: I see everyone's already come to this realization long since. I also see that Towlie said he inferred that the teacher meant the brick wall would be unaffected by the collision. If that's the case (and I don't know why it would be), then it's obvious that if only one object gets all the damage, the damage will be much greater than when both objects get damaged.

 

[Towlie]

Now, unless I am making some elementary mistake, the kinetic energy of the car hitting the wall is double the kinetic energy of the two cars colliding. This implies to me that the two scenarios are not at all equivalent, with the latter being in fact far more damaging.

Tim, what we're really concerned with here is the impact on a single car, so your math appears to show a difference of four times, not just double.

Am I right?

 

[Towlie]

I didn't think the teacher specified a mass for the brick wall. I would have assumed (again trying to keep all things as equal as possible) that its mass was the same as a car.

Whose teacher, yours or mine? I said in my initial post that a "brick wall" is a wall that's unaffected by the collision. It's sort of like a pillar that holds up a bridge - it doesn't suffer a scratch.

In fact, maybe the teacher should have just said that two cars colliding head on each going 50 mph is the same as one car going 100 mph and hitting a parked car square on.

That's exactly what I said in my post #38.

 

[Modified]

that is only true if you assume that the second car in the car/car collision is absolutely stationary, immutable, upon collision, but it isn't. As such, the energy deposition into the car is not the same in the two collisions.

Two cars head on at 50 mph. Assume they crumple and stick together rather than bouncing off. Final speed of each is zero.

One car at 100 mph hitting stationary car. Assume they crumple and stick together rather than bouncing off. Speed of both cars shortly after collision is 50 mph, then they will skid or roll to a halt.

For the period of interest, the change in speed of both cars is 50 mph in both cases. The skidding or rolling to a halt after collision in the second case has negligible effect. Whether the stationary car is in neutral or park will have a very small effect at 100 mph - I would guess negligible - at 2 mph it would make a difference.

If the second car is immovable, as you suggest, then the change in speed of the first car will be 100 mph, assuming they stick together. The stopping distance will be twice as far, so in the best case the deceleration will be the same as in the head-on crash, but will last twice as long. I suspect that for a real collision, the peak deceleration will be higher if hitting an immovable car.

 

[marting]

This is actually a pretty familiar situation in physics, because it's common to think of reflecting boundary conditions (the wall) in terms of what's called an "image problem". You replace the boundary with another half-space, and for each object in the original space you have an identical one in the other space. Bouncing off the wall is colliding with your mirror image.

Indeed, because of that same training the answer jumps out. I'm a bit surprised the Car Talk guys didn't get it given their MIT background.

 

[Uncayimmy]

Modified - I agree that "hitting a car head on where both are going 50 mph is a lot like hitting a stationary car head on at 100 mph." That analogy would be much more useful in a Driver Education class and would be perfectly honest as well.

When I heard my teacher say essentially the same thing, the above was the lesson I got out of it. In my mind I know that "brick wall" is vague. How big is it? Is it attached to a building? Is it just sitting on the ground or sunk into it? So I figured the teacher meant for us to imagine a brick wall built to be a reasonable but dramatic equivalent to a car. I figured the teacher's goal was relative velocity, not mass, since we're probably never going to find ourselves driving into brick walls.

Perhaps there's a basic misunderstanding here. Of course the head-on collision of two cars traveling at 50 MPH each will involve twice the energy of one car hitting a brick wall at 50 MPH, and therefore the collision will cause twice the damage, but that energy and damage will be split between the two cars, and what we're discussing here is the effect on one car, not both.

As I explain myself, imagine we're two friends discussing this over a couple of beers rather than two knuckleheads arguing over the Internet. The reason I went for the answer I did was because to me it's pretty obvious that if you exchange a car for a massive solid object, things will be very different. It would also be very different if you exchange a car for a chain link fence.

I figured that changing the mass wouldn't confuse people. Everybody knows that you can drive your car through a chain link fence and the fence will be the big loser. If you drive into a telephone pole at the same speed, the car is going to be the big loser. By contrast I have seen lots of people get confused by relative velocities. That's why I went the direction I did.

No, not really. I thought so at first but then I had to edit this. It would actually be more like hitting a brick wall at 1/2 MPH. In both cases you would slow down by 1/2 MPH.

I agree and disagree. I also freely admit that I may be another victim of "a little knowledge is a dangerous thing" syndrome.

If car A is going 50 and strikes car B going 49 (assuming the cars to be perfectly elastic -they bounce off each other), then car slows to 49 and car speeds up to 50. Thus car A has slowed by 1mph.

Suppose we replace car B with a wall 10,000X as massive and make it stationary (and perfectly elastic). If we want car A to slow down by 1mph, then it needs to hit the wall at 0.5mph. It will bounce off the wall and start going backwards (-0.5mph). That's the 1mph slow down.

I checked my numbers using the simulation on this page:
http://qbx6.ltu.edu/s_schneider/physlets/main/momenta4.shtml

Trial 1:
Mass 1: 10
Mass 2: 10
Velocity 1: 2
Velocity 2: 1
Restitution coefficient: 1.0

After running the simulation, V1=1 and V2=2.

Trial 2:
Mass 1: 10
Mass 2: 100,000
Velocity 1: 0.5
Velocity 2: 0
Restitution coefficient: 1.0

After running the simulation, V1=-0.5 and V2=0.