Something I was taught in High School Driver Education class.

[Uncayimmy]

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

Good. We agree.

Now, let's back up. The question was whether two cars colliding at a closing speed of 100mph was the same as one car colliding with a brick wall at 100mph. If we stipulate that the brick wall is hugely massive and solid, then we all agree that is not true. It's more like hitting such a brick wall at 50mph.

What you said was to simply imagine the brick wall there and then not there, same as with the paper being there and not there. The problem with that as a "proof" is that it presumes the answer. Yes, it's correct, but it doesn't get us any closer to the answer. It requires substituting something you already know will result in the answer being the same. That's why I suggested the giant marshmallow and box of pillows. You can't substitute just anything.

In order to resolve the problem we need to work out or at least approximate the math. Relative velocity and mass (and elasticity) together determine the severity of the accident. If we keep the relative velocity the same (100mph) and change one of the masses so it's huge, of course the severity is going to be different. It will also be different if we make one of the masses really small (paper). To me that's self-evident just based on our observations of the world around us.

We can also approach the problem from the other side. Suppose we think of a car striking a massive brick wall at 50mph. We put a car in place of the brick wall. Will the collision be the same? No, because the car is much smaller. If we want it to be the same, we'll need to speed up our car(s).

To stick with the "proof" that you're offering, you would have to tell us to imagine the car speeding up to 100mph. Not 60mph or 250mph, but 100mph. Again, that wouldn't be a "proof" but the answer. That's why I took exception to what you said.

If I misunderstood you, I apologize.

 

[Uncayimmy]

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.

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.

Can you clarify what you mean by identical? And while you're at it can you check my understanding below?

If the collision is nearly perfectly elastic, the two cars will bounce off each other and start going in opposite directions (from the reference frame of the ground). Their relative velocity to each other will be very close to multiplying their old relative velocity by -1. In other words they will recede from each other at close to 100mph.

If the collision is nearly perfectly inelastic, the two cars will bounce off each other, but in this case their relative velocity when receding from each other might only be 1mph.

The difference in the two collisions is that in the latter, something like 99% of the kinetic energy will be "lost" and converted into something else (crumpled metal, heat).

Am I describing that correctly?

 

[69dodge]

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?

You can arbitrarily decide to use the car as your reference frame and measure all velocities relative to it, but you can't arbitrarily decide that this reference frame is inertial. If it isn't, then it just isn't.

Imagine two cars, some distance apart and facing each other, one parked and the other moving quickly. The driver of the moving car slams hard on the brakes, and comes to a stop just in time to avoid a crash.

The two drivers see the same thing, but they don't feel the same thing. Each sees the other car approach, first quickly and then more and more slowly until it stops. But one feels nothing, while the other is thrown forward against his seat belt. This is because the reference frame in which the parked car is stationary is an inertial reference frame, while the reference frame in which the other car is stationary is not an inertial reference frame.

Newton's laws of motion apply only in inertial reference frames. You can't just pick any old reference frame and expect that they will necessarily apply there.

 

[alexi_drago]

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.


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.

I don't think you can look at it that way, that would mean that no matter what forces are applied to car A it will never move implying that it has an infinite mass and that the ground would be accelerated as a result of a collision between two cars.

 

[rsaavedra]

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.

I was wrong, let's move on. I can do that. Let's see if you can Sol.

What do you mean the two scenarios are identical? They just aren't.

Identical might be the final magnitude of the acceleration of each car over the collision time in the first (100 - 0) and second case (50 - 50) within each scenario. But the scenarios themselves are certainly not identical. By definition they aren't: one is called elastic and the other inelastic for a reason. And experiencing a deceleration of 50 mph as in the first sceneraio is very different from experiencing a deceleration of 100 mph as in the second. Where's the identity?

 

[Towlie]

That's pretty bad. You quote a post that says "equivalent" three times, and then respond with "What do you mean the two scenarios are identical?"

 

[sol invictus]

Can you clarify what you mean by identical?

For simplicity, imagine the cars are in otherwise empty space. Situation 1: the cars are moving towards each other, each at 50mph. Situation 2: same two cars, but one is at rest and the other is moving towards it at 100mph.

Those two situations are absolutely identical in every way - there is no experiment you could do to distinguish them. The same is not true on earth, since you could measure the velocity relative to the ground or air - but the ground is not very important for the collision itself, so it doesn't play much of a role in this discussion.

And while you're at it can you check my understanding below?

It's correct.

What do you mean the two scenarios are identical? They just aren't.

Identical might be the final magnitude of the acceleration of each car over the collision time in the first (100 - 0) and second case (50 - 50) within each scenario. But the scenarios themselves are certainly not identical. By definition they aren't: one is called elastic and the other inelastic for a reason. And experiencing a deceleration of 50 mph as in the first sceneraio is very different from experiencing a deceleration of 100 mph as in the second. Where's the identity?

You're either trolling or deeply confused. See what I wrote above - so long as the same two cars are involved, there is no difference between the two situations. Whether the collision is elastic or inelastic has to do with the materials and construction of the cars, not whether they're initially at 50-50 or 100-0.

 

[alexi_drago]

Sol is not saying an inelastic collision is the same as an elastic collision.

He is saying that a 100-0 collision will be the same as a 50-50 collision of the same type whether that's elastic or inelastic.

 

[rsaavedra]

You're either trolling or deeply confused. See what I wrote above - so long as the same two cars are involved, there is no difference between the two situations. Whether the collision is elastic or inelastic has to do with the materials and construction of the cars, not whether they're initially at 50-50 or 100-0.

Not trolling, but I don't think I was deeply confused either. Final deceleration in inelastic scenario is 50, and in the elastic scenario is 100. So they are not identical -as far as that goes. Yet that difference is irrelevant to what you were trying to say, well no problem with that.

 

[Uncayimmy]

I don't think you can look at it that way, that would mean that no matter what forces are applied to car A it will never move implying that it has an infinite mass and that the ground would be accelerated as a result of a collision between two cars.

Once again you have brought the ground into it. Why not the moon? A jet flying overhead? There is no preferred reference frame. We can describe the relative velocities from any number of different reference frames, and all of those descriptions will be accurate. We don't need the Earth to describe this collision.

As I go through this, I am going to ask you if you agree or not. This not me being condescending. Consider me a student, which I am when it comes to this stuff. I want to find out where one of us is wrong or where there is a misunderstanding or semantics issue. I figure it's best to do this after each step.

If we have two equal masses (A and B) moving at a relative velocity of X and colliding in a perfectly elastic collision, the individual velocities of those two masses will change equal amounts so that they end up with a relative velocity of -X.

Do we agree on that?

That description accurately describes any number of scenarios. From the reference frame of the ground, that describes A and B each going 50mph and colliding - they each reverse direction and go 50mph the other way. That describes A being stationary and B hitting it at 100mph - B stops and A shoots off in the direction B was going at 100mph. It also describes A going 250mph and hitting B, which is going 150mph. A slows down by 100mph to 150mph and B speeds up by 100mph to 250mph.

Do we agree on that?

Suppose A fires a laser at B and the readout says X. What happens when B fires a laser at A? It will also read X. In our case X is a 100mph closing speed. Now pick any other point in the universe (the ground, for example). Fire your laser at A. You get a readout of Y. Now fire your laser at B. You will always measure a speed that when combined with Y results in a closing speed of X.

Note: For this paragraph positive velocities mean they are going east to west and negative velocities mean west to east. If A is going 200mph, then B must be going 100mph in the same direction. If A is going 50mph, then B must be going -50mph. If A is stationary (0mph) then B is going -100mph.

Do we agree on that?

As described earlier, from the reference frame of A, we measure B at X or 100mph. Do we know if B is moving towards us or if we are moving towards B? No. We only know the relative velocity is 100mph. Using the ground as the reference frame, it could be any one of the pairs I described above as well as an infinite number of other combinations.

Do we agree on that?

Still in the reference frame of A, B collides with us. We fire our laser again. What speed does it read? -100mph, right? It's the opposite of what we measured before the collision.

Do we agree on that?

That description of the event is the same as what I described in the post you're disagreeing with. I wrote, "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 can also say that I measured myself as moving towards B and then away from B. Truth is, the only thing I did was measure two different relative velocities, one before the collision and one afterwards.

I stand by that description.

I think what you are saying is that if I say that B is moving towards a stationary A before the collision, then I must say that B then becomes stationary and A moves away from B. If I say B is stationary before the collision, then I must say that A stops and B zooms away.

Is that correct?

I say that I can describe it in any of the four ways above and still be correct. There's nothing that compels me to say that any one of those descriptions is more accurate than another based solely on what I measure from the reference frame of A, and the only thing I am measuring is relative velocities.

As far as I can tell, the only way I would be "allowed" to pick a description would be if I actually knew the relative masses of A and B. If I know their masses are equal, then I agree with your contention that if B hits a stationary A, then A will accelerate at X and B will accelerate at -X.

From the reference frame of the ground, I will actually measure how much acceleration (positive and negative) A and B each experience. In all of the descriptions above except from the reference frames of A or B, I can deduce that A and B have equal masses.

But from the reference frame of A or B, how can I deduce the relative masses just by measuring the relative velocity of one to another?

If I say A remains stationary before and after, I am by definition saying that A is massive and B tiny by comparison. I don't actually know that based on just my velocity measurements. That's just one possibility. If I say that A is stationary before and that B is stationary afterwards, then I am by definition saying that they have equal masses.

You presumed to know the masses, so you declared the first description above to be invalid. Thing is, I was only describing a set of velocity measurements without any consideration of mass.

Are we on the same page?

If so, then we can get back your claim that I am changing reference frames. I contend that I am not. If we're in agreement up to this point, then the only way for you to "know" I am changing reference frames is if you know the relative masses in advance. Since we cannot know the relative masses from the reference frame of just A or just B, how do you justify declaring that I am changing reference frames?

I'll tell you how. You change reference frames and jump to a point on the ground. You then calculate the relative masses, find them equal, and then jump back into my reference frame to tell me I must have jumped reference frames. Unless, of course, it turns out that you measure A as being incredibly massive compared to B. In that case you come back to tell me that I didn't jump reference frames.

 

[alexi_drago]

i'm posting from phone atm so just a short post.the inertial frame car A is in before the collision is not the same as the inertial frame car A is in after the collision. There have been forces applied to car A which has a definite mass and yet you say it's always stationary which cannot be the case.take the case of 2 cars each with a mass of 1000kg travelling towards each other at 10m/s each (closing speed of 20m/s and colliding and not bouncing off each other.I think we both agree that the damage will be the same whether it's 10-10 or 0-20 so the kinetic energy lost and used up in deformation of the cars should be equal in both cases.Calculate the change in kinetic energy using car A as your reference.

 

[alexi_drago]

sorry, that should've said:After the collision car A is no longer stationary in the inertial reference frame in which it was stationary before the collision.

 

[sol invictus]

Not trolling, but I don't think I was deeply confused either. Final deceleration in inelastic scenario is 50, and in the elastic scenario is 100. So they are not identical -as far as that goes.

Trolling, then. I'll remind you where we started:

Obviously, hitting a car head on where both are going 50 mph is a lot like hitting a stationary car head on at 100 mph,

I disagree with that.

Oops!

Elastic versus inelastic wasn't even mentioned until I gave some specific examples. At some point in there you realized you were wrong, and proceeded to try to pretend the conversation was about something else to avoid admitting it. I've seen you try that several times before, just as transparently.

Yet that difference is irrelevant to what you were trying to say, well no problem with that.

I explained that difference to you.

 

[sol invictus]

Once again you have brought the ground into it.

It may be that part of the communication problem here is semantic. There are two possible meanings of the word "frame" or "reference frame":

One is when it's short for "inertial reference frame". In that case you cannot properly speak of "A's frame". To be precise, you should say "the frame in which A is at rest at before the collision" or "the frame in which A is at rest after the collision", and it's important to realize that those two frames are not the same.

The other meaning is more loose - "from the point of view of", or "as measured by someone attached to or inside of". I think that's what you mean, but it may not be what Alexi means. It's important to realize that that is not an inertial frame, and therefore (for example) Newton's laws of motion in standard form do not apply in it (they must be supplemented with an additional term while the car is accelerating), and that after the collision is over you have ended up in a different inertial frame.

 

[rsaavedra]

Trolling, then. I'll remind you where we started:

I did say I was wrong. Who is doing the trolling now?

I've seen you try that several times before, just as transparently.

Ýou've seen me try that several times before? That statement would require some backing up, and you seem to want to just keep arguing even if it's off topic. Any old grudges, or are you thinking I'm someone else?

 

[Uncayimmy]

It may be that part of the communication problem here is semantic. There are two possible meanings of the word "frame" or "reference frame":

One is when it's short for "inertial reference frame". In that case you cannot properly speak of "A's frame". To be precise, you should say "the frame in which A is at rest at before the collision" or "the frame in which A is at rest after the collision", and it's important to realize that those two frames are not the same.

The other meaning is more loose - "from the point of view of", or "as measured by someone attached to or inside of". I think that's what you mean, but it may not be what Alexi means. It's important to realize that that is not an inertial frame, and therefore (for example) Newton's laws of motion in standard form do not apply in it (they must be supplemented with an additional term while the car is accelerating), and that after the collision is over you have ended up in a different inertial frame.

That makes sense to a degree, but I'm still struggling with the strict definition. Suppose that I only know that there are objects A and B. I know nothing about their masses, but I know their relative velocity is X (moving closer together). If I use A's inertial frame of reference, I must say that B is moving towards A at X. I can't say that A is moving towards B because A would not be inertial in that case. Is that correct?

Suppose further that B reverses direction at some point before reaching A (there is no collision). I must then say that B is now moving at some negative speed. Let's just say it's -X. I can't say that A is moving away from B now because the A would no longer be inertial. Right?

In the second case, suppose B continues towards A and this time collides with A. Regardless of what their relative masses are, A will have a change of momentum. It could be enormous or incredibly tiny, like the space shuttle running into a huge asteroid or a speck of cosmic dust. At this point I cannot say that A is inertial because I know that cannot be. Is that correct?

From the "point of view" of A, I measure the relative velocity of B to be -X. Only in this limited scenario there's no way for me to accurately describe what happened because I need to account for A's change in momentum, however big or small it might be. In order to do that I need an inertial reference frame that remains inertial before and after. Since that cannot be A's point of view, I need to use some other reference frame such as the a frame where A was stationary before the collision.

Have I got it?

 

[alexi_drago]

That makes sense to a degree, but I'm still struggling with the strict definition. Suppose that I only know that there are objects A and B. I know nothing about their masses, but I know their relative velocity is X (moving closer together). If I use A's inertial frame of reference, I must say that B is moving towards A at X. I can't say that A is moving towards B because A would not be inertial in that case. Is that correct?

OK, at this point you're using A as your inertial reference so A is stationary and B is moving towards it.

Suppose further that B reverses direction at some point before reaching A (there is no collision). I must then say that B is now moving at some negative speed. Let's just say it's -X. I can't say that A is moving away from B now because the A would no longer be inertial. Right?

Yes, A hasn't been affected and will still be stationary in the inertial frame from before, B is now moving away from A.

In the second case, suppose B continues towards A and this time collides with A. Regardless of what their relative masses are, A will have a change of momentum. It could be enormous or incredibly tiny, like the space shuttle running into a huge asteroid or a speck of cosmic dust. At this point I cannot say that A is inertial because I know that cannot be. Is that correct?

Yes.

While velocity is relative, acceleration is not and can be measured. If you have two cars travelling side by side then one pulls away from the other is one car braking or is the other accelerating?

If the drivers head of the car pulling in front is being pushed back into the headrest then he is accelerating or if the drivers head of the car behind is being pushed forward then he is braking (use an accelerometer in each car for more precision).

Initially you know that both cars were stationary in the same inertial frame of reference but afterwards only one was and it's possible to determine which one.

If the speed of the space shuttle is altered by something, hitting something or firing it's rockets it's change in velocity can be measured without using any external reference and it is no longer stationary in the inertial frame of reference that it was stationary in initially, it will now be moving at whatever the measured change was.

From the "point of view" of A, I measure the relative velocity of B to be -X. Only in this limited scenario there's no way for me to accurately describe what happened because I need to account for A's change in momentum, however big or small it might be. In order to do that I need an inertial reference frame that remains inertial before and after. Since that cannot be A's point of view, I need to use some other reference frame such as the a frame where A was stationary before the collision.

Yep, the inertial frame of reference can be anything you want to start off with but should remain constant even if an object you based it on is affected in some way.

Have I got it?

I think so.

 

[Uncayimmy]

If the speed of the space shuttle is altered by something, hitting something or firing it's rockets it's change in velocity can be measured without using any external reference and it is no longer stationary in the inertial frame of reference that it was stationary in initially, it will now be moving at whatever the measured change was.

How?

 

[alexi_drago]

How?

When you're in a car that accelerates or brakes you feel like you're being pushed forward or back, you're body is actually wanting to carry on at the speed it was travelling at and the back of the seat or the seatbelt is applying a force to your body to speed it up or slow it down with the car. Measuring that force and dividing by your mass will give you the acceleration at that moment in time. Graphing that over time you can work out your velocity relative to some previous point in time.

 

[Uncayimmy]

When you're in a car that accelerates or brakes you feel like you're being pushed forward or back, you're body is actually wanting to carry on at the speed it was travelling at and the back of the seat or the seatbelt is applying a force to your body to speed it up or slow it down with the car. Measuring that force and dividing by your mass will give you the acceleration at that moment in time. Graphing that over time you can work out your velocity relative to some previous point in time.

That makes sense, but I'm still struggling a bit. How do you calculate your mass while you're in the space shuttle? In other words suppose I am in my space vehicle with no external references. I don't know my mass or even my weight on Earth. Suddenly I feel a force. What devices could I use to figure out my mass and subsequently my change in velocity?

I already understand that the equivalence principle states that "within sufficiently small regions of space-time, it is impossible to distinguish between a uniform acceleration and a uniform gravitational field." I'm cool with that.