Something I was taught in High School Driver Education class.

[sol invictus]

I think that when complexities aren't stated one is expected to use the most simple interpretation which means (to most) a fixed, inelastic, largely undamaged wall. Also, the purpose of the instructor in the OP was to dramatize the hazards of driving speed and intended the audience to think of the "brick wall" as an immovable object that was going to inflict most all the damage onto the car.

But if so, the instructor was wrong in his/her comparison.

I asked a very specific question, but you did not give me a specific answer. I described the same event using three reference frames. Are those three descriptions accurate or not given the stipulations (perfectly elastic, no friction, no air resistance)?

I'm not Modified, but they looked fine to me.

About this "several feet in front" thing, I think alexi just meant that if at the start the origin of your reference frame is hovering over the center of the roof of car A with zero relative velocity to it, car A's fender is a few feet in front and below you. When car B collides with car A, if the collision is both perfectly elastic and instantaneous car A will go flying backwards (assuming car B was coming from in front of you), and car B will stop with its fender several feet in front and below you and the center of its roof a bit further away.

If as you say car B is moving the same speed as me after the collision (remember, I wasn't involved in the collision), then by definition the coefficient of restitution was 1. If the collision is going to take time, which it will in the real world, then the relative velocity will be lower as well. You can't have it both ways.

Coefficient of restitution and collision time are not the same thing, and one can vary without the other. If you're willing to consider ideal situations like CoR=1, I don't see why you can't allow them to take as little or as much time as you like. Taking more time just means some of the kinetic energy is stored inside the objects (or in some other form) temporarily.

I disagree with that. Hitting a stationary car at 100mph even if both cars have the same weight will result in the stationary car being pushed backwards, and the hitting car slowing down quickly while pushing it, but certainly not ending up at the collision point.

Not at the same point, true. They'll be moving.

However, when the two cars of the same weight collide head-on both going at 50 mph, assuming sufficient alignment of the impact, both wrecked cars will end up pretty much at the collision point. So the experienced deccelaration is much worse in the latter scenario.

How did you jump to that conclusion? It's wrong, and it doesn't follow from what you said. I think you need to brush up on Galilean relativity a little...

First case: 100mph + 0mph -> 50mph+50mph
Second case: 50mph + 50mph -> 0mph + 0mph
Both cases: acceleration is 50mph over the collision time.

That's assuming perfectly inelastic collisions, but the general answer won't change the fact that the acceleration must be the same. The only difference in the two scenarios comes from friction with ground and air, which isn't relevant for how much damage the cars incur (at least assuming the stationary car doesn't have its brake on).

 

[rsaavedra]

That's assuming perfectly inelastic collisions

Yes, precisely, which isn't the case here.

 

[sol invictus]

Yes, precisely, which isn't the case here.

As I already told you, that makes no difference. For example, if the collision is completely elastic, the situation is as follows:

First case: 100mph + 0mph -> 0mph+100mph
Second case: 50mph + -50mph -> -50mph + 50mph (same speeds, opposite velocities)
Both cases: the magnitude of the acceleration of each car is 100mph over the collision time.

The same thing will work for an arbitrary level of inelasticity. That follows from the Lorentz (or Galilean) invariance of the laws of physics.

 

[rsaavedra]

Both cases: acceleration is 50mph over the collision time.

That's assuming perfectly inelastic collisions

Both cases: the magnitude of the acceleration of each car is 100mph over the collision time.

The same thing will work for an arbitrary level of inelasticity.

Notice that in the first scenario (full inelasticity) the accelaration will be 50, while in the later (full elasticity) it will be 100. You can't say it makes no difference then. And moreover, a head-on collision of two cars in the real world is neither fully elastic nor fully inelastic.

But I see your point Sol, the deccelaration for one of the cars when two of them hit head-on at 50 mph each would be in general, closer to the fully inelastic scenario (some crash tests could be required to support that though) than to the fully elastic, so each car will experience a decceleration of approximately 50, or in any case, closer to 50 than to 100.

The most similarly damaging decceleration would be then having one of those cars hit directly an unmovable titanium wall when the car was travelling just at that same 50 mph (with no stationary car in between to cushion any further that collision with the wall.)

 

[alexi_drago]

Exactly. It's the same as me riding on top of the car (ignoring friction, etc). This third car is now moving parallel to the second car (I stipulated a perfectly elastic collision). As far as the third car is concerned, the second car is now stationary and the first car is moving away. If you introduce their ground speeds into this, you are introducing another frame of reference, so don't do that.

What I was trying to show was that if you use A as your frame of reference before the collision and still use A as your frame of reference after the collision your frame of reference has changed so comparing before and after becomes more complicated. Choose one inertial frame of reference and stick to it, you could choose one of the cars but both their velocities are going to change in the collision and personally I find using the ground more convenient as it's never going to change relative to the reference frame (unless the moon falls).

 

[marting]

But if so, the instructor was wrong in his/her comparison.

Of course he was wrong but keep in mind his purpose: to impress inexperienced drivers with the associated risks. An accurate physics lesson it wasn't designed to be. This brings up an interesting question. When is it appropriate for those with specialized knowledge to stretch facts for a greater good?

 

[Careyp74]

to keep it simple, if you were traveling 50 MPH inside a car, whose speed was reduced to zero almost instantaneously, it doesn't matter which one you hit, the car going 50, or the wall, both would look similar in outcome, and you would rapidly decelerate from 50 MPH to zero. If you were traveling 100 MPH in the car, your bodily reaction when hitting the wall will be twice as harsh, since you are deccelerating from 100 MPH to zero.

 

[Phage0070]

What changes is the total mass in the system and the elasticity of what the two vehicles are impacting.

No, this does not matter. The elasticity of what the vehicles are impacting is identical if they are equally matched. If we examine one car, it spreads the force of the impact out over the crumple zone of the car. Basically it has a few feet of "give" over which to dissipate the energy of the crash. For the elasticity of the object it hit to matter it would have to "give" inward, basically moving the wall in the direction of the car's travel.

Now lets picture two cars crashing together. For the car they are impacting to act like a more giving surface than the brick wall it would need to "give" inward; this is impossible unless you are expecting them to somehow phase into each other.

 

[JoeTheJuggler]

It is absolutely true that the average "brick wall" might not put up all that much resistance to a 50 MPH car. I think that when complexities aren't stated one is expected to use the most simple interpretation which means (to most) a fixed, inelastic, largely undamaged wall.

I'm with UncaYimmy on this. It makes more sense to me to assume the teacher was making a point about relative velocity and think of a wall whose mass is equal to the other car in a head-on collision.

I don't think it's less complex to assume a magical wall.

The point the teacher was trying to make is that in a head-on collision, both velocities matter.

I'm not sure where people got this idea that hitting a brick wall at 100 mph will leave the wall undamaged.

 

[Ron Swanson]

Of course he was wrong but keep in mind his purpose: to impress inexperienced drivers with the associated risks. An accurate physics lesson it wasn't designed to be. This brings up an interesting question. When is it appropriate for those with specialized knowledge to stretch facts for a greater good?

I'd say it could be appropriate only when an example with actual facts would not do the job just as well.

 

[Uncayimmy]

What I was trying to show was that if you use A as your frame of reference before the collision and still use A as your frame of reference after the collision your frame of reference has changed so comparing before and after becomes more complicated.

That sentence is self-contradictory. If I stick with car A I am by definition not changing reference frames. What you don't realize is that you keep jumping back into the reference frame of the ground because that's what you prefer. Car A always remains stationary in the reference frame of car A because the only thing I am looking at is car B. I never need to look at the ground.

Choose one inertial frame of reference and stick to it, you could choose one of the cars but both their velocities are going to change in the collision and personally I find using the ground more convenient as it's never going to change relative to the reference frame (unless the moon falls).

You are misunderstanding the situation. I am sticking with one frame. You, on the other hand, are switching frames and not realizing it. Car A will always remain stationary in the reference frame of car A before and after the collision. Car B will approach car A, then recede from car A.

What's confusing you is that at after the collision you're looking at things from the perspective of the ground. Why? We didn't need to look at the ground before the collision.

From the reference frame of car A, I only need fire my laser at car B. I never need to fire it at the ground because I don't care how fast I am going relative to the ground. The description would be same if the cars were in the vast emptiness of space with no other reference points except each other.

The only thing I care about is how fast car B is approaching me. I will measure car B as approaching me up until the point where it hits me. Then I will measure it moving away from me. I never have to take any measurements using the ground.

As for which reference frame is simpler, that depends. From a standpoint of effort to measure the collision, it's easier to use car A than the ground. Remember, with car A I only need to fire my laser at car B. From the reference point of the ground, I need two lasers. I have to monitor car A and car B the whole time. I end up with two sets of measurements. This method of describing the collision is just as valid as doing it from car A.

Since we humans live and drive on the ground, we actually care where the cars end up after the collision. So, yeh, measuring from the ground is probably a good idea. It's more practical, but it is also more complicated.


What I think you are doing with the reference frame of car A is asking yourself what happens to car A relative to the ground after the collision. That's jumping reference frames. From the reference frame of car A, we don't need the ground. Like I said, car B will approach at 100mph and in a perfectly elastic collision recede at 100mph.

The above description will be the same as long as their relative velocities are 100mph. From the reference frame of the ground, it won't matter if it's 50-50 or 1-99. Hell, it wouldn't matter if car B was going 125mph and car A ran up its ass at 225mph. From the reference frame of car A it will always look the same.

 

[Uncayimmy]

No, this does not matter. The elasticity of what the vehicles are impacting is identical if they are equally matched. If we examine one car, it spreads the force of the impact out over the crumple zone of the car. Basically it has a few feet of "give" over which to dissipate the energy of the crash. For the elasticity of the object it hit to matter it would have to "give" inward, basically moving the wall in the direction of the car's travel.

Now lets picture two cars crashing together. For the car they are impacting to act like a more giving surface than the brick wall it would need to "give" inward; this is impossible unless you are expecting them to somehow phase into each other.

You're making all sorts of assumptions that you're not conveying. That's why I asked you to substitute a giant marshmallow. Let's run the accident four times:

1) Cars hit head on with nothing in between.
2) Put a piece of paper in between.
3) Put 10 meter cube of bricks in between.
4) Put 100 meter cube cardboard box filled with pillows in between.

Will the vehicles look essentially the same in all four scenarios?

 

[Uncayimmy]

About this "several feet in front" thing, I think alexi just meant that if at the start the origin of your reference frame is hovering over the center of the roof of car A with zero relative velocity to it, car A's fender is a few feet in front and below you. When car B collides with car A, if the collision is both perfectly elastic and instantaneous car A will go flying backwards (assuming car B was coming from in front of you), and car B will stop with its fender several feet in front and below you and the center of its roof a bit further away.

Ah, I see. That makes sense.

Coefficient of restitution and collision time are not the same thing, and one can vary without the other. If you're willing to consider ideal situations like CoR=1, I don't see why you can't allow them to take as little or as much time as you like. Taking more time just means some of the kinetic energy is stored inside the objects (or in some other form) temporarily.

Is it called collision time? I ask in the interests of doing further reading.

 

[Phage0070]

Will the vehicles look essentially the same in all four scenarios?

No, the first three would be nearly identical and the 4th would be much softer for both cars.

 

[marting]

I'm with UncaYimmy on this. It makes more sense to me to assume the teacher was making a point about relative velocity and think of a wall whose mass is equal to the other car in a head-on collision.

I don't think it's less complex to assume a magical wall.

The point the teacher was trying to make is that in a head-on collision, both velocities matter.

I'm not sure where people got this idea that hitting a brick wall at 100 mph will leave the wall undamaged.

I think if the teacher was doing that he would have used a car at 100 MPH colliding with a stopped car. The "brick wall" is more proverbial than real. There just aren't many "brick walls" in the path of cars compared to other cars, trees, utility poles, people, deer, etc. I think it is altogether obvious that the teacher intended the student to visualize this as a car impacting a fixed structure where the car crunches and the wall remains relatively fixed in place.

However, the wall being completely undamaged is just a simplification we make. It's not strictly necessary, but makes the problem easier to understand. One can then extrapolate to actual structures of increasing complexity.

 

[JoeTheJuggler]

I'm not sure what's come up here but remember that if you use a moving car as your frame of reference, and the car changes speed (as would occur during a collision) your frame of reference does not change with the car.

I don't follow this. If my inertial reference frame is the car, then the car is stationary and everything else is moving relative to it. The car does not change speed during a collision--everything else does.

Or am I missing something?

 

[JoeTheJuggler]

I think it is altogether obvious that the teacher intended the student to visualize this as a car impacting a fixed structure where the car crunches and the wall remains relatively fixed in place.

Then we'll just have to disagree. I think it's altogether obvious that the teacher was making the point that relative velocity is what's important. The fact that I'm "only" going 50 mph isn't so important if my relative speed wrt the other car is 100 mph.

Actually, I've always heard these things done with much lower speeds to make that point. 25 mph hour sounds very slow, but if you have a head on collision with another car coming straight at you at 25 mph, the relative speed between you is 50 mph. This is one of the reason why cross-over accidents (where there is no median barrier) are so often fatal.

 

[GlennB]

I'm not sure where people got this idea that hitting a brick wall at 100 mph will leave the wall undamaged.

In these analogies, "a brick wall" is typically used to mean something immovable and undamageable. Theoretical. You don't need to factor it in to your equations.

eta: what marting said

 

[Ron Swanson]

In these analogies, "a brick wall" is typically used to mean something immovable and undamageable ...

Similar to the phrase "Built like Brick OutHouse" I guess

 

[sol invictus]

Notice that in the first scenario (full inelasticity) the accelaration will be 50, while in the later (full elasticity) it will be 100. You can't say it makes no difference then.

You claimed two cars approaching head on at 50mph is not equivalent to one car approaching a stationary car at 100mph. When I showed you they are equivalent in the inelastic case, you said that was because I had assumed the collision was inelastic. When I showed you they are equivalent in the elastic case too, you tried to make it sound as if we were discussing something else. Just admit you were wrong and move on - it's easier.

And moreover, a head-on collision of two cars in the real world is neither fully elastic nor fully inelastic.

Sounds like you agree you were wrong when the collision is inelastic, and you agree you were wrong when it's elastic, but now you claim you were right when it's in between?

Well, you're wrong again. Those two scenaria are exactly identical regardless of the level of elasticity, or anything else apart from differences due to friction with air, ground, or wheel bearings.

But I see your point Sol, the deccelaration for one of the cars when two of them hit head-on at 50 mph each would be in general, closer to the fully inelastic scenario (some crash tests could be required to support that though) than to the fully elastic, so each car will experience a decceleration of approximately 50, or in any case, closer to 50 than to 100.

No, that's not my point. My point is simply that those two scenaria are identical, no matter what the other characteristics of the collision.