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  #51  
Old 07-09-2000, 03:49 PM
sailor sailor is offline
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Mauve Dog, I am not sure I can say i am glad to see this thread kept alive but... I think you are mistaken and it is quite simple to see:

Build a train 20 miles long and place one car at each end and have them head at each other at 35mph with respect to the train.

Case one: train stationary. They collide at 70mph relative speed to each other and going 35 mph each in opposite directions with respect to ground

case two: train moving frward at 35 MPH. The two cars will collide just the same as before but with respect to ground one is stationary and the other is doing 70 MPH (here I swear that, if someone mentions relativity I'll knock his teeth out!)

Case three: Train moving forward at 70 mph: cars collide just the same as before except with respect to ground zer one is backing at 35 mph and the other is doing 105 mph

case four: train backing at 35 mph......

case five: train going at ANY speed! Do you get it now? Only the relative speed between the two cars counts. Remove the train and it works the same way

Yes?
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  #52  
Old 07-09-2000, 04:37 PM
Chronos Chronos is offline
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Hold on, Mauve Dog... if I'm going forward at 100 mph relative to the ground, and the guy in from of me is going 30 mph relative to the ground, he's not moving away from me... He's moving towards me, at 70 mph. The only difference between the two cases is frictional effects, either with the air (assumed to be at rest relative to the ground) or the ground itself, and on the timescales that we're looking at, friction is absolutely negligible (besides, this is physics: Physicists always assume that friction is negligible unless there's absolutely no choice but to include it).

And don't worry, sailor, relativity actually doesn't change the answer one bit, as long as the relative speed is the same.
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  #53  
Old 07-10-2000, 07:36 AM
C K Dexter Haven C K Dexter Haven is offline
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I think my example has been misunderstood.

The OP seemed to be saying that, if you are going 35 mph, a collision with an immovable object (wall) would produce the same impact on you as a collision with another car coming at you at 35 mph. My example was to reframe that: if that were true, then if you were going 1 mph, a collision with an immovable object (wall) would produce the same impact on you as a collision with another car coming at you at 35 mph. That's patently nonsense.
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  #54  
Old 07-10-2000, 08:34 AM
waterj2 waterj2 is offline
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As I pointed out, that is not a similar scenario. In a head-on collision like the one you describe, the initial momentum of the two cars does not equal zero. The total momentum immediately before the crash is the same as the total momentum immediately after the crash. In this example the momentum of your car is -1mv and the momentum of the other car is 35mv, which means a total momentum of 34mv. The two cars will collide, and will stick together, causing each to have a momentum of 34mv, and thus a speed of 17 mph. The change in momentum of your car is therefore equal to 18mv

The reason that hitting an oncoming car travelling at the same speed as you has the same effect as hitting a solid, stationary wall is that in both cases your car undergoes the same change in momentum.

Let's look at this from the frame of reference of your car. Your car is stationary, and another car approaches at 70 mph. This is obviously not the same circumstance as a wall with infinite mass approaching at 70 mph.
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  #55  
Old 07-10-2000, 01:32 PM
Mauve Dog Mauve Dog is offline
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Quote:
Originally posted by Chronos
Hold on, Mauve Dog... if I'm going forward at 100 mph relative to the ground, and the guy in from of me is going 30 mph relative to the ground, he's not moving away from me... He's moving towards me, at 70 mph. The only difference between the two cases is frictional effects, either with the air (assumed to be at rest relative to the ground) or the ground itself, and on the timescales that we're looking at, friction is absolutely negligible (besides, this is physics: Physicists always assume that friction is negligible unless there's absolutely no choice but to include it).

And don't worry, sailor, relativity actually doesn't change the answer one bit, as long as the relative speed is the same.
<sigh>In my current mood, I am content to simply acknowledge that I am wrong and you and sailor are right.

I will however throw this out in parting:

There are three phases for analyzing collisions:
1) What are the objects doing immediatley before the collision?
2) What happens during the collision?
3) What are the subsequent motions of the objects involved immediately after the collision?

When discussing theoretical collisions, only phases 1 and 3 are important. Phase 2 is like a big black box: we know what things were doing before they hit, and we know what things were doing after they hit, but we don't really know what happened inside the box.

In all of the cases mentioned, we know phase 1, because those are 'given'. We also can easily determine phase 3, since we know momentum must be conserved. Assuming no change in mass, we can determine the final velocites of the objects involved. I have never contended, nor do I now contend, that the resulting speeds will be anything other than what everyone here has said they will be.
The differences lie with what happens during phase 2 - during the collision. Without the aid of force gauges and such, we have only the definition of impulse to guide us. That is, impulse = Fdt. In turn, the impulse is equal to the change in momentum.

When the time interval is extended, by whatever means, the force must be lessened. Correct?
OK. Now, consider the following two scenarios:
1) Two non-deformable objects of equal mass; one moving at 100mph, collides with the other moving in the same direction at only 30mph. Because this is an inelastic collision, the two 'stick' together, and continue moving, in the same direction, with the speed of the combined mass being 50mph. I assume there is no disagreement here...?
2) Two deformable objects of equal mass; one moving at 100mph, collides with the other moving in the same direction at only 30mph. Because this is still an inelastic collision, the two stick togeth, and continue moving, in the same direction, with the speed of the combined mass being 50mph. Again, I assume there is no disagreement here...?

So, what's the difference? My simple (and obviously erroneous) contention is that the force of impact involved in the second scenario is less than the force of impact of the first.

It is true that before the collision, the relative speeds were such that the rear object was travelling at 70mph relative to the second. However, starting from the instant they come into contact, this relative speed begins to change; the faster object begins to slow down, the slower object begins to speed up (it is at this point that I meant the slower vehicle would be moving away from the other one - not before the collisions!). And this, I thought, would alter the time interval during which the collison takes place, relative to, say, a car moving at 70mph and hitting a brick wall, a stationary car, or another car moving towards it. Further, the way in which the speed changes will be, I thought, different from a head-on case because one vehicle would be slowing down while the other speeds up (as opposed to both slowing to zero). And further yet, it is the way in which the speed changes that determines the force of impact. So, I hope you can at least see what was going through my clouded mind.

However, it appears that I have misunderstood, and I apologize for beating this dead horse. All collsions where the relative speed difference is 70mph are identical.
I apologize to anyone who may have been misled by my ignorance (although, it appears, that would be no-one).
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  #56  
Old 07-10-2000, 02:19 PM
Mauve Dog Mauve Dog is offline
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Oops...

I erred above...I meant to say that the result after the rear-end collision will be that both vehicles will be moving at 65mph, not 50mph.
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  #57  
Old 07-10-2000, 10:16 PM
sailor sailor is offline
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Does this mean we are *all* in agreement and there's nothing else to discuss? No! it cannot possibly be! There has to be a misunderstanding somewhere!
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  #58  
Old 07-11-2000, 06:16 PM
iampunha iampunha is offline
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We seem to be forgetting a key concept here. Why would anyone want to prove this point? I don't think anyone actually wants to test this theory of being at 0 or 70 mph when a freaking CAR hits them!

The fact that when the two cars come to rest their speed is 0 is irrelevant. Their net speed is going to be 0 once they've stopped. Imagine for a second:

A car hits you at 35 mph. It's going to hurt a helluva lot. You'll probably sustain serious injuries.

You're riding in a car at 35 mph. You hit a car which, relative to you, is traveling 70 mph. So we can say you, relative to that 70 mph car, you're stationary. Ouch.

Either way, I don't see anyone lining up to test if they get hit at 35 mph or 70 mph.
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  #59  
Old 07-11-2000, 08:48 PM
nola andy nola andy is offline
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Physics and cars

Think about it. If two cars going 35 MPH hit head on at an equivalent impact of 70 MPH each, where does the extra energy come from? A rough idea of the effect can be experienced by clapping one hand against the other at a repeatable speed. Practice some and you should be able to clap at a relatively steady rate. Now substitute a good, solid wall with some padding that imitates the hand. Three thicknesses of bath towel seems to be about right.

Now clap the educated hand against the wall. You will note that the impact is virtually identical to clapping against the moving hand.

For those of you who slept through all this in school (I know I did and had to learn it again so I wouldn't look like a fool on the job) the key term is "conservation of momentum".

Sorry to sound so didactic. I've been reading Cecil for a long time.
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  #60  
Old 07-11-2000, 09:07 PM
Punoqllads Punoqllads is offline
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Everyone is wrong

Including me. But I'll try to be as non-wrong in this explanation as I can.

Two cars colliding, one going 35 mph, the other -35 mph, both coming to a complete stop, is not the same as one car going 70 mph colliding with an immoble wall.

The two cars colliding is, however, the same as two cars colliding, one going 70 mph, the other stationary, with the two cars afterwards going 35 mph in the direction the original moving car was going.

For example, consider two gumdrops, each 1 g, moving at 1 km/s and -1 km/s, respectively, sticking together, immoble after colliding.

Kinetic energy is 1/2 m * v^2, so the initial kinetic energy for gumdrop #1 (g1) is 1/2 kJ (J == joule), as is g2's energy. The final energy in the system is 0, since both are unmoving in this inertial frame of reference. Therefore, the amount of energy released, as sound, or heat, or fracturing the space-time continuum, etc. is 1 kJ.

In case #2, g1 is moving at 2 m/s, g2 is stationary, final velocity 1 m/s, both stuck together. Initial energy is 1/2 * 1g * (2 km/s)^2 = 2 kJ. Final energy is 1/2 * 2g * (1km/s)^2 = 1kJ. Energy released: 1 kJ.

In contrast, consider one gumdrop, moving at 2 km/s, hitting an immovable wall, which stops it dead in its tracks. Initial energy is 2 kJ, final energy is 0 kJ, since nothing is moving. Energy released: 2 kJ, twice as much as the others.

I can see how some would feel that the two cases (2 cars, 1 car and 1 wall) were equivalent. The difference in velocities is the same, so intuitively the damage should be the same. However, the damage done is done by the energy removed from the system. As kinetic energy is related to the square of the velocity, however, this intuitive feeling is wrong.

Everyone happy?
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  #61  
Old 07-11-2000, 09:14 PM
Strainger Strainger is offline
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I have too much time on my hands and I'm beating a dead horse.

But nevertheless, I did an analysis of the two scenarios presented in the OP. The thing is, it's in a Word document complete with pictures and equations. Therefore, it is unpostable. If you'd like a copy of the document, send me an email. Feel free to check my math and assumptions if I send it to you.

Short answer: Bigtrout is right and his friend is wrong.

Bigtrout, I would've emailed this to you already, but I see you've chosen to remain private.
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  #62  
Old 07-11-2000, 09:16 PM
OldMan OldMan is offline
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First post on SDMB, after lurking for several months. It was this thread that prompted me to register.

Okay, hands up, how many of you flunked or didn't take physics in high school? How can so many ostensibly bright people create so much confusion around such a simple question? Two cars colliding head-on, each going 35 mph, is exactly equivalent to one car piling into a parked vehicle at 70. Same thing if one car is going 20 and the other's going 50, one's doing 100 and the other's backing away at 30, or any other combination that creates a closing speed of 70, it makes no difference. It's their relative velocity that matters. The confusion in here is over frames of reference, and the most common mistake is according special status to a certain point of view, the observer standing stationary on the ground. All measurements of physical quantities are specific to a frame of reference, and different observers may get different numerical values for measured quantities like speeds, time intervals, magnetic fields, etc., but the relationships among those measured quantities (i.e. the laws of physics) must always be the same for all observers. Knucklehead the First, referred to by Bigtrout in the message that started this, is right. There is no preferred frame of reference, they're all equivalent. In other words, if you're flying along above the road at 35 on your magic carpet and a car blows by underneath you at 70 (measured with respect to the road) and hits a parked car ahead of you head-on, you would see exactly the same thing you'd see if you were standing on the ground and both cars were approaching each other at 35, because from your moving reference frame on the magic carpet, that's exactly what's happening. The parked car is approaching you at 35, the other guy is going away from you at 35, and after the crash the point of collision will move along the road at 35 in the direction you're going, so from your point of view it's stationary. Gotta keep your frames of reference straight folks, or you'll get hopelessly confused.

Yeah, yeah, I hear the purists in the back muttering about inertial versus non-inertial reference frames, but for the scale of this problem, the surface of the earth is close enough to an inertial reference frame as makes no difference. This ain't rocket science; in rocket science, you do have to think about the rotation of the earth and the curvature of the surface. For this problem, we don't. Then there are some relativists over in the corner making rude noises about velocities not adding linearly: quiet down lads, relativistic effects are vanishingly small at this scale.

As for hitting the brick wall, that's a different problem, because the elastic properties of brick walls and cars are a little different. However, hitting a brick wall at 35 is *nothing* like the head-on collision we started out with, it's more like a head-on collision with each car doing 17.5 mph. That should be perfectly obvious with a little thought. How can you believe that if you're going 35 and hit something, it makes no difference what the velocity of the thing you hit is? How about head-on into a big 18-wheeler doing 80? Bigtrout is partly right, the time interval in which you decelerate is critical, but if you do the math, you'll find that the interval for hitting a brick wall at 35 is twice the interval when hitting a car head-on that's also doing 35, which means the impulse is different. Assuming, of course, that your reference frame is the one in which the brick wall is stationary. The speeds and time intervals you measure will vary, depending on your state of motion. As long as the difference in speed between the cars, or between one car and a brick wall, is fixed at 70 mph, you can always find a reference frame in which the speed of any one component of this system is pretty much anything you want it to be, and it can't make any difference to what happens. If it does, you've overturned all of physics since the time of Newton.
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  #63  
Old 07-11-2000, 09:20 PM
Myrr21 Myrr21 is offline
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First off, there is a difference between deformable objects and non-deformable ones:

A non-deformable moving object hits another one of the same mass that is stationary. Object 1 stops, object 2 moves away at whatever v object 1 was travelling at. Think pool balls. Momentum is conserved and KE=1/2 mv^2

A deformable moving at v hits another one of equal mass that is stationary. The two deform, and move as one in the original direction at 1/2 v (conservation of momentum--double the mass, and you move at half the velocity). HOWEVER, the kinetic energy is sgnificantly less: 1/2 2m (1/2v)^2. Note that you are squaring the 1/2, so overall there is half as much KE as in case one. This energy went to deforming the object--breaking molecular bonds, moving molecules arounds, etc. Moreover, the negative acceleration for Object 1 is significantly less and therefore the force on it. So which car do you want to be in (non-deformable vs deformable)? The second, obviously (that is, assuming it doesn't deform enough to impale you on the steering column).

Add 20 mph to each object, the resut is the same--you want a car that deforms--elastic vs inelastic collisions, very different species.
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  #64  
Old 07-11-2000, 10:52 PM
sailor sailor is offline
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<sigh>
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  #65  
Old 07-11-2000, 11:02 PM
Myrr21 Myrr21 is offline
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OK, last post from me for tonight...

Something just occured to me:
I (and likely others) have been working off of a incredibly theorectical wall that is completely immoveable and as such does not absorb momentum. This is a terrible assumption when correlating to real life.
However, in terms of really theoretical models, this lends creedence to the 35 mph theory. Imagine a car that doesn't crumple at all. Two cars hit at 35 mph head on--both bounce backwards at 35 mph. So if you're in car 1, you go from +35 mph to -35 mph pretty damn fast. Overall change...70 mph. Say you hit this perfect wall at 35 mph. You bounce back at -35 mph. Change- 70 mph.

Now, say you have a much more real-world wall, that does absorb momentum (but still does not appear to move). So you hit it at 35 mph, and stop. Net change= 35 mph. You hit it at 70 mph and stop. Net Change = 70 mph.

So I guess the question is whether you want to deal with a real-world wall or the more interesting case of a really cool theorectical wall.
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  #66  
Old 07-12-2000, 12:17 AM
sailor sailor is offline
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did you read the entire thread before posting that?
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  #67  
Old 07-12-2000, 02:26 AM
Revedge Revedge is offline
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Physics

Unless I missed one of the posts. You all are on the right track butt are forgeting the formula. E= 1/2m times v squared. A car mass 1 at speed of one hits a wall and releases enrgy of 1/2. Another car going in the oppisite direction adds its energy to the equation. Total release of evergy of 1. Two times the energy of the brick wall. How ever, if you hit the wall at double the speed (2 squared) you have a release of energy that is four times the initial conditions. So both arguments are right in a way. Two cars hitting head on at 35 mph is not as bad as hitting a brick wall at 70mph. However, it is twice as bad as hitting a brick wall at 35 mph.
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  #68  
Old 07-12-2000, 02:37 AM
Speaker for the Dead Speaker for the Dead is offline
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I don't have the patience to read all of these, so don't flame me if this has been stated

What about the law of relativity? The other car would be going 70 mph relative to you (the car you're in). So wouldn't hitting it be like hitting a wall at 70?
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  #69  
Old 07-12-2000, 03:23 AM
SPOOFE SPOOFE is offline
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I think the smartest thing would be to avoid crashing into anything at all...

ANYway, since I DO occasionally want to contribute to a thread...

I would suggest forgetting the mathematical formulae. Well, not forgetting per se, but just setting them aside for a moment. Mostly, set aside this notion of "35 mph and -35 mph". After all, no speedometer has a speed rating for "-35 mph". So it's not a matter of "cancelling each other out", that only happens with sound waves. Set aside all the physics. Look at this from a different point of view... or, rather, from the point of view of another classic "traveling in opposite directions" question...

Two cars are traveling towards each other, both at 35 mph. Their starting points are 70 miles apart. Now, assuming that neither stop or alter speed in the slightest, they'd meet after one hour of travel. Ergo, these two cars just traveled 70 miles in one hour... or, 70 mph.

Now, you have a car and a wall. The car begins traveling towards the wall at 35 mph. The wall, of course, travels at its top speed of 0 mph. They begin 70 miles apart. Now, assuming that neither stop or alter speed in the slightest, they'd meet after TWO hours of travel. Ergo, these two... er... objects just traveled 70 miles in TWO hours... or, 35 mph.

Derive what conclusions you want from that. And, before Sailor chimes in with his usual retort, yes, I DID read the whole thread.
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  #70  
Old 07-12-2000, 05:35 AM
Bear_Nenno Bear_Nenno is offline
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What happened?????? EVERYONE READ THIS

OK, I thought we all agreed and everyone understood what was going on here. Now all of a sudden people are jumping in with nonsense and clouding the issue with silly formulas that, for the most part, have nothing to do with the OP.
People, Bigtrout is right, his knuckleheaded friend is, well, a knucklehead.
Any confusion should be cleared up by luckie's previous post. He said:
"let me toss this thought in:
bigtrout: point out to your friend that , while the 2 cars going head on at 35mph is NOT equivilent to hitting a brick wall at 70mph, it IS equivilent to hitting a stationary car at 70mph (that is a simple change of refernce frames). clearly hitting a stationary car at 70 is not nearly as bad as hitting a brick wall at 70! in fact , it is like hitting a brick wall at 35 (QED).
if he still doesn't get it after thinking about that, give it up as a hopeless case.
-luckie"

He is a physicist for crying out loud. You all keep yapping about their relative speed being equal to 70mph. OK fine, so you can add up the speed of the cars anyway you want. As luckie said, as long as you use TWO cars. They can both be going 35, on go 70 and one 0, one go 5 and one 20... you get the point. But their mass comes in to play too. The wall, the imovable wall does not have the same mass as a car. The wall, because by definition it does not move, will exert back on any object the same force EQUAL to what is exerted on it. If i push on it with my finger, it exerts that same pressure back to my finger. If a car hits it at 35 mph, it exerts back to that car a force equal to another car going 35 mph toward it. JUST LIKE A HEAD ON CRASH. If a car hits it at 70mph the wall will exert a force equal to a car going 70mph. So a car hitting a wall at 70mph is like two cars of equal mass in a head on collision at 70mph. Are we getting this yet???

Magic walls are not the only thing that do this. If I sit in a chair, I exert on it 220lbs of force. It exerts back toward me 220lbs of force (because neither of us are moving). If someone, say 500lbs, sits in the chair, the chair exerts 500lbs of force back at the person. This is what imovable objects do.

Can we finally be done with this?
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  #71  
Old 07-12-2000, 05:44 AM
Bear_Nenno Bear_Nenno is offline
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Typos

OOOPS, I said:

"As luckie said, as long as you use TWO cars. They can both be going 35, on go 70 and one 0, one go 5 and one 20... you get the point."

That should be "one go 50 and one 20" but I think you smart people could already tell I missed a button.


I also said:
"So a car hitting a wall at 70mph is like two cars of equal mass in a head on collision at 70mph. "

Understand that it should read " A car hitting a wall at 70mph is like two cars of equal mass in a head on collision at 70mph EACH!" TWO cars heading at each other, both going 70mph and crashing is the same as ONE car going 70mph into a brick wall. Because the wall supplies a force equal to the car hitting it. How else would the car stop?
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  #72  
Old 07-12-2000, 08:40 AM
Myrr21 Myrr21 is offline
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OK, I like a good arguement as much as anybody, but I'm afraid that I'm going to have to sorta split the difference on this one.
From what I've gone through of the physics of it, basically it depends on the assumptions you make. It seems that two cars hitting head-on at 35 mph can range from one hitting a wall at 35 mph to one hitting a wall at 70 mph. Most likely, in reality, it's somewhere in between--since the conditions that result in either extreme are gross approximations.
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  #73  
Old 07-12-2000, 11:21 AM
Myrr21 Myrr21 is offline
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OK, just to be sure, I checked w/ a professor here; he is also of the opinion that it is most likely somewhere in between is most accurate, but leaning towards the 35 mph side.
Bigtrout: maybe you should just get a couple cars, a brick wall, and a few crash test dummies. That'd make for a good demonstration.
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  #74  
Old 07-12-2000, 12:45 PM
HalloweenMan HalloweenMan is offline
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I will put this arguement to rest

The Original Post said that you have to ignore things like crumple zone and what not. So...

If you have a car traveling at 35mph and it hits another car traveling at 35mph they will both stop dead. (In an ideal situation). If a car traveling at 35mph hits a brick wall then it will stop dead. Therefore no matter whether you hit the oncoming car, or the brick wall, the impact will be he same, the stopping time will be the same. Therefore, there is no difference.

Where is the confusion?
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  #75  
Old 07-12-2000, 12:52 PM
MetallicAsh MetallicAsh is offline
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Ouch.
I just read this thread and it gave me a headache...and I studied physics.

Let me see if I can boil things down.

The key concepts are:

1) Conservation of Momentum
2) Elastic vs. Inelastic Collisions
3) The Principle of Relativity
4) Cars vs. Walls (ambiguity in the OP)

Taking the side of the underdog, if we interpret the OP's statement:

I have a knucklehead friend I am unable to convince of a fairly simple physics problem. He contends if two cars of equal mass (discounting crumple zones and the like) both traveling 35 mph meet head-on, the resulting collision is multiplied to a 70 mph crash.

in a manner charitable to the knucklehead, we can presume he (the knucklehead) meant that for two cars in a head on collision, it is the same for each as though they had been stationary and hit by another car moving 70mph (or as if they had been moving 70mph and hit a stationary car).

I presume this is what the knucklehead meant, otherwise, the progression would go as follows:
Two cars colliding at 35mph multiply to a 70mph collision.
What is a 70mph collision like?
Well: two cars colliding at 70mph multiply to a 140mph collision.
...
...
Two cars colliding at 5600mph multiply to a 11200mph collision...
...and so on...

You can start at 1mph and end up at C (the speed of light) or beyond.

Its a rare knucklehead who would suggest all collisions are equivalent to impact with infinite velocity.

Using the Principle of Relativity (also useful for deriving the Theory of Relativity, but let's save that for another day), it becomes clear that the knucklehead is more head than knuckle. If you have Car A and Car B, one moving East at 35mph relative to you and one moving West at 35mph relative to you; without an external reference point, it is indistinguishable to Car A whether:

1) he is moving at 70mph and Car B is stationary;
2) he is stationary and car B is approaching at 70mph;
3) they are each moving towards each other at 35mph;
4) he is moving at velocity X and Car B is moving at velocity Y, where the difference between X and Y (how fast they approach each other) is 70mph.

If you've got a problem with that, take it up with Einstein (actually, since we're talking principle of relativity, rather than Theory of Relativity, take it up with Mach, but that's not important right now).

(All of this also applies to Car B.)

Now, the trick is to remember to stay in the same reference frame after the collision.
If you consider case:
1) both cars appear to be moving at 35mph in the direction of Car A's original motion after the collision;
2) both cars appear to be moving at 35mph in the direction of Car B's original motion after the collision;
3) both cars appear stationary after the collision;
4) left as an exercise to the reader. (That's physics lingo for: I'm too lazy to write out the equations.)

Conservation of momentum (the sum of both velocities (speed + direction) has to sum to the same number before and after...) is the only thing we need to worry about here, because we are dealing with an inelastic collision. In an elastic collision (billiard balls) things bounce off each other and both momentum and energy are conserved. In an inelastic collision (car crashes, football tackles, billiard balls made of play-dough) objects fuse and only momentum is conserved (note that in each of the four reference frames, there is the same total momentum before and after...but you cannot compare momentum across different inertial reference frames).

It was actually Bigtrout who brought up the brick wall.

His friend, if we stuck to talking about cars, would be correct, provided the second car in the 70mph collision wasn't some magical car that began stationary, and remained stationary after the collision without being acted upon by an outside force.

By changing the problem from one of cars to one of a car and a wall, Bigtrout basically did just that: the wall is a "magic car".

If you collide with a wall at 35mph you feel the same force as if you hit a (neglecting crumple-zones and the like as we have been in previous posts) car which is held perfectly still both before and after the collision. The problem is, the wall needs to bring outside forces to bear to stop the moving car (being fixed to the ground, a mountain, whatever); the same thing would happen if you had a car that was stationary both before and after but in the latter case, it is more obvious that an outside force has to hold the car still.
Note that there is no reference frame where car B is stationary both before and after the crash!!!
That's why we have trouble comparing the two.

Now, since the wall brings you to a complete stop, neglecting differing physical properties of walls and cars, you as a driver in car A moving 35mph experience the same phenomenon if you:
1) Drive into a brick wall at 35mph and come to a complete instantaneous stop; (techically no such thing, but just assume the same stopping time for all the collisons)
2) Drive into an oncoming car at 35mph, bringing you both to a complete instantaneous stop;
3) Drive into a stationary car at 70mph, and both you and resultant wreckage (both cars) continue on at 35mph;
4) Drive into a stationary "magic car" at 35mph which does not move when you hit it, bringing you to a complete stop.

So two cars driving into each other at 35mph is the same as hitting an initially stationary car at 70mph (what your friend probably meant) but not hitting a wall at 70mph (the words you put in his mouth).

I think when you put it that way, Bigtrout, your knucklehead friend doesn't sound like so much of a knucklehead.

Tell him he owes me a beer.
I say you owe him a new moniker.

- MetallicAsh
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  #76  
Old 07-12-2000, 12:55 PM
Revedge Revedge is offline
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As I Said.

Doing the math. Hitting another car head on at 35mph is twice as bad as hitting a brick wall at 35mph. It is roughly equivalent to hitting the brick wall at 50mph. If you hit the brick wall at 70mph it is twice as bad as hitting the wall at 50mph, or hitting another car at 30mph.
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  #77  
Old 07-12-2000, 12:57 PM
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Oops. Last sentence should read ...hitting another car head on at 35mph.
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  #78  
Old 07-12-2000, 02:29 PM
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Vectors vs. Systems

IANAPB (I am not a physicist, but)

It seems to me the confusion stems from mixing vectors with 'systems'. Vectors are additive and in the 'perfect' situation of identical mass,velocity,non-elasticity, no friction etc... a vector with energy equal to a verctor
'car' going 35mph meeting another in exactly the opposite way reduces to 0 at once. A vector hitting 'a mass' that is stationary would be devided by two and move in the same direction. A vector hitting an 'immovable' object would also drop to 0 but the energy has to go somewhere so that one gets weird because it's not a fair question.

Here's how I did it... You know that little desk toy with the 5 balls on string? Take 2 on one side, 2 on the other and drop them at about the same time. 'Pretty much' they will hit each other and stop. Now, take two balls and lift them up about twice the height as before and take one ball out of the equation on the other side (so there are only 2 stationary) and drop the two balls your holding. The four balls will move (actually, because they have some elasticity they'll bounce around a bit) but won't cancel out the energy gets divided more or less in half and shared.

However, the 'energy' in the overall system is the same for the situation except the 'brick wall' theory which as I said, is weird.

If I had a Net Cam I could show you right now...

The reason this has gone on so long is that the question was really asked about vectors but you're all trying to describe the 'system' of the car along with all the variables which goes to show how hard the application of physics can be. We all 'know' that cars don't stop instantaneously, have passengers, crumple zones etc and imperfect transfer of inertia so it becomes a mess of a problem... but, in short, Knucklehead is wrong and the other guys is right I think. Show him the balls on a string trick. Paint little cars on them and he'll get it.

Do the experiment with tennis balls on a string, don't try the car thing at home.
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  #79  
Old 07-12-2000, 03:05 PM
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Okay, everyone is wrong.

Let's change the example to minecars just for fun.

If your job is to push a minecar along a track and you push it at 5 mph until you hit another, stationary minecar which has its brake on and therefore doesn't move and your foreman comes along and sees you standing there, he'll inform you that you've done no <B>work</B>.

Try all the fancy equations you want, you're still fired.

Since it is necessary that work be done in order to change the velocities of the various cars, brick walls, etc., and we have shown that no work is getting done around here, I think I have reasonably proven that, no matter what the speed of impact, the cars have no effect whatsoever on each other and, in fact, do not change speed at all, but rather go on to live healthy, prosperous lives.

Thank you.
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  #80  
Old 07-12-2000, 03:05 PM
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Yes, sailor, I read the entire thread

Bigtrout
Quote:
We've gotten off the beaten path of my original query, it seems (although it's been interesting.) Perhaps if we get rid of cars it will simplify things.
No kidding. I like your granite boulders idea; that simplifies things nicely.

Quote:
That another completely equal counterforce is part of the equation in the two boulder scenario makes no difference. The effect is the same. It does not result in a collision with forces equal to 70 miles per hour. That's my spin on this.
And that spin is wrong, Bigtrout. I can't let this thread die until I've made a personal one-on-one attempt to make you understand this. Forget all the others' posts about impulse, momentum, speed of stopping, deforming objects etc., because those complications simply distract us from your failure to understand this basic Newtonian principle: it does so result in a collision with forces equal to 70 miles an hour!

Both boulders are rolling toward each other at 35 miles an hour. This will always equal an additive closing speed of 70 miles an hour, and your friend, who is no knucklehead at all, is entirely correct to say so. Your posts make clear that you simply don't understand this.

An earlier post mentioned baseballs and bats, and I'm surprised no one stressed that more. If the world worked the way you say, a batter would not have to swing his bat! A thrown ball that hit a bat held rigid would be the same as if it hit a bat mid-swing (make the ball a tiny boulder and the rigid bat a wall for consistency).

Have you ever seen a home run hit off a bat held in bunting position? Has your own experience (assuming you've played Little League or softball in your life) not shown you that the faster you swing your bat, the faster and farther the ball will fly away from it?

Do you see what I'm saying? Please?
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  #81  
Old 07-12-2000, 03:14 PM
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five, take comfort in the fact that your observation was probably obvious to most readers from the get-go.
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  #82  
Old 07-12-2000, 03:47 PM
Punoqllads Punoqllads is offline
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Read a book

<sigh>. One more try...

It amazes me how many people post to this thread saying, "If you do the math..." without actually doing the math. To everyone out there: read a book; specifically, a book on physics. If you can't find a book, find a physicist, and I don't mean some frosh straight out of introductory mechanics.

Honestly, I originally believed that two cars moving at 35 mph in opposite directions colliding, and one car moving at 70 mph hitting a wall and stopping were the same. I read a book ("Physics for Scientists & Engineers", third edition, Raymond A. Serway). I did the math (see previous post). I was wrong, they weren't the same, and I posted the results. Few listened, apparently.

Someone said that 'no speedometer has a speed rating for "-35 mph."' Duh. Read a book. Velocity is a vector, therefore direction is significant. A car "moving at -35 mph" is moving at 35 mph in the direction opposite to the admittedly arbitrary frame of reference. For any two cars moving directly towards each other, each going 35 mph, to do the math, you pick one to be going in the positive direction, and the other will by definition be moving at -35 mph.

The Newtonian theory of relativity (as distinguished from Einstein's special and general theories of relativity) is that, in any two inertial frames of reference, the laws of physics are the same. Notice I said inertial. Accelerating frames of reference have different rules, hence the appearence of the centrifugal force for observers in a rotating frame of reference. To any other observer in an inertial frame, there is no centrifugal force acting on the principles; rather, there is centripetal acceleration acting on them.

People who set up the gedankin experiment need to be very careful to pick two inertial frames of reference when comparing the (-35, 35) collision to the (0, 70) collision. I'll attempt to demonstrate.

In the first experiment, one car is moving at -35 mph, the other at 35 mph. They collide, and stop moving. Boom. For the second experiment, we want to see one car moving at 0 mph, and the other at 70 mph towards it. Essentially, we're following the car going -35 mph in the first experiment. So, we're going -35 mph relative to the first experiment.

The apparent 70 mph car approaches the 0 mph car and they collide, and stop. But, wait! They stop relative to the first experiment's frame of reference. We're going -35 mph relative to the first frame of reference. The cars stop in the first frame of reference, but to us, moving at -35 mph relative to that frame, the cars appear to be moving at 35 mph.

For the cars to stop moving in our frame of reference, either we're describing two separate collisions, i.e., not (necessarily) equal, or our frame of reference must accelerate by 35 mph to be moving at 0 mph relative to the first frame of reference. Therefore, it's not an inertial frame of reference, therefore, again we are describing two different collisions.

Proof via math:
35 mph = 15.65 m/s

Cars masses are unstated, but assumed to be equal. Call them each half a ton. So:
1000 lbs = 453.6 kg (at 1 g, i.e., standard Earth gravity).

In the SI system (meters, kilograms, seconds; the standard scientific measurement system, sometimes called MKS), energy is in units of joules (J), where 1 J = 1 kg * m^2 / s^2.

Regardless of the measurement system, kinetic energy is 1/2 * mass * velocity^2. So, in the first experiment, car #1's, kinetic energy is 1/2 * 453.6 kg * 244.9 m^2/s^2, or 55543.3 kg * m^2/s^2, which conveniently is equal to 55543.3 J. The #2 car's kinetic energy is equal to the first car's kinetic energy, as (15.65)^2 is equal to (-15.65)^2. Therefore, in the initial system, there is a total energy of approximately 111.1 kJ, to 4 significant figures.

In the second experiment, car #1's kinetic energy is 1/2 * 1000 lbs * (70 mph)^2 = 1/2 * 453.6 kg * 979.7 m^2 / s^2 = 222.2 kJ, while car #2, since it's at rest, has a kinetic energy of 0 kJ, so the total energy is 222.2 kJ.

Now, for the original comparison, where, at the end, both cars are not moving, the final kinetic energy in each system is 0 kJ. Therefore, the amount of energy dissipated (via heat, light, sound, damage to the cars and/or occupants, etc.) in the first experiment is 111.1 kJ, while in the second, is 222.2 kJ. Notice that these numbers are not the same.

Notice also that in the first experiment, the initial momentum is 453.6 kg * 15.65 m/s + 453.6 kg * -15.65 m/s = 0 kg * m/s (sorry, there's no SI base unit for momentum). Since everything is at rest at the end, the ending momentum is also 0. However, in the second experiment, the initial momentum is 453.6 kg * 31.3 m/s = 1420 kg * m/s. Note that, if everything stops at the end, the final momentum will be 0 kg * m/s.

Notice that 1420 is not equal to 0.

By the standard laws of physics, momentum in any closed system is conserved. Therefore, either there is an outside body acting on the principles (e.g., the non-moving car is attached very firmly to the ground), or the frame of reference is non-inertial. In either case, the two experiments are not equivalent.

I hope by now I have shown why the two different experiments are just that -- different. If you believe you have found a flaw in my reasoning, by all means, post why. But do the math first, and show your work in your post. Just saying "it's obvious" without doing the math is equivalent to those math books saying, "it is easy to see that this equation can be derived from the earlier one" when it's not easy, because it can't.

So, please, if you have anything more to contribute, either perform the experiments in real life a statistically significant amount of times and show your experimental data. Or do the math and show your work. The most important facet of a real scientist is their willingness to accept when they're wrong. You can't be taken seriously any other way.
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  #83  
Old 07-12-2000, 05:24 PM
Telemark Telemark is online now
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From the NHTSA page

Here is a quote from the info page on the NHTSA site:

Quote:
All of the vehicles are crashed into a fixed barrier at a speed of 35 mph. The impact is the same as if two identical vehicles, each going 35 mph, collided head-on.
I think the people who do the crash tests know what they are testing. It also says elsewhere on the site that it is the equivilant of a car traveling 70 MPH colliding with an identical parked car.

The link is http://www.nhtsa.dot.gov/ncap/infopage.html#crash
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  #84  
Old 07-12-2000, 05:57 PM
The Ryan The Ryan is offline
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CKDextHavn
Quote:
The OP seemed to be saying that, if you are going 35 mph, a collision with an immovable object (wall) would produce the same impact on you as a collision with another car coming at you at 35 mph. My example was to reframe that: if that were true, then if you were going 1 mph, a collision with an immovable object (wall) would produce the same impact on you as a collision with another car coming at you at 35 mph.
Five
Quote:
An earlier post mentioned baseballs and bats, and I'm surprised no one stressed that more. If the world worked the way you say, a batter would not have to swing his bat! A thrown ball that hit a bat held rigid would be the same as if it hit a bat mid-swing (make the ball a tiny boulder and the rigid bat a wall for consistency).
You both are pulling conclusions out of thin air. Nowhere has anyone said that hitting a wall at 1 mph is the same as hitting a car at 35 mph, or that hitting a wall that's at rest is the same as hitting a wall that's coming towards you at 35 mph. Neither of these statements are logical consequences of anything that anyone has said, either.

CKDextHavn:
A car going 1 mph would exert 1/35 the force on a wall as a car going 35 mph, and therefore would have 1/35 the force returned to it. Hitting a wall at 1 mph is the same as two cars hitting each other, each going 1 mph. What part of this do you not understand?

Five:
The question of what a baseball does is completely different from what a car does. Baseballs bounce much more than cars. If you were using a non-elastic ball (NEB) imnstead of a baseball, then: a NEB with a velocity v hitting a stationary bat (assuming that the batter were able to hold the ball still, and have it not bounce back from the impact at all) would be the same as a NEB hitting a NEB with a velocity of -v. A NEB going v hitting another NEB going -v would clearly not be the same as a NEB going v hitting a bat going -v; a bat is much more massive than a ball, and therefore would exert more force than a ball.


MetallicAsh:
Quote:
It was actually Bigtrout who brought up the brick wall.
according to the OP (emphasis mine):
Quote:
I have a knucklehead friend I am unable to convince of a fairly simple physics problem. He contends if two cars of equal mass (discounting crumple zones and the like) both traveling 35 mph meet head-on, the resulting collision is multiplied to a 70 mph crash. I told him it's the speed of stopping that's important, that this scenario is no different than had one car traveling 35 mph run into a immovable brick wall --- the car goes from 35 to zero near-instantaneously. He disagrees, in the face of logic, and I can suffer his foolishness no longer.
It is quite clear to me that "knucklehead" is saying that two cars at 35= one car at 70 + wall at zero.

Revedge
[quote]Unless I missed one of the posts. You all are on the right track butt are forgeting the formula. E= 1/2m times v squared. A car mass 1 at speed of one hits a wall and releases enrgy of 1/2. Another car going in the oppisite direction adds its energy to the equation. Total release of evergy of 1. Two times the energy of the brick wall. How ever, if you hit the wall at double the speed (2 squared) you have a release of energy that is four times the initial conditions. So both arguments are right in a way. Two cars hitting head on at 35 mph is not as bad as hitting a brick wall at 70mph. However, it is twice as bad as hitting a brick wall at 35 mph.[/quoteh]
No, you're the on that's forgetting something: if you hit another car, there is twice as much energy, but also twice as many cars. So the energy per car is the same.
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  #85  
Old 07-12-2000, 07:19 PM
Strainger Strainger is offline
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Guys, I have equations, pictures, etc. in a Word document proving that Bigtrout is correct. Only two people (including Bigtrout) have asked me to email this to them. I put enough time and effort into it to where if more of you don't email me a request for this document, I'm going to start sending out to y'all at random.
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  #86  
Old 07-12-2000, 08:02 PM
Wee Man Wee Man is offline
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I don't follow the math well enough, so I may be wrong about this, but it seems to me that Oldman is right in terms of point of reference.

Car A @ 35 MPH hits wall @ 0. Car a sustains 35 MPH worth of impact/damage/inertia/whatever.

Car B @ 35 MPH hits wall @ 0. Car a sustains 35 MPH worth of impact/damage/inertia/whatever.

Car A @ 35 MPH hits Car B @ 35 MPH. Each car sustains 35 MPH worth of impact/damage/inertia/whatever, but the total amount of impact/damage/inertia/whatever = 70 MPH worth (A+B).

If you are measuring from the point of view or either A or B, then the impacts are the same. If you are an outside observer, the total combined impact to both cars = a 70 MPH crash into a brick wall.

I think that both Bigtrout and Knucklehead are both correct, it just depends on who is measuring.
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  #87  
Old 07-12-2000, 08:34 PM
The Ryan The Ryan is offline
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Quote:
Originally posted by Wee Man
I don't follow the math well enough, so I may be wrong about this, but it seems to me that Oldman is right in terms of point of reference.

Car A @ 35 MPH hits wall @ 0. Car a sustains 35 MPH worth of impact/damage/inertia/whatever.

Car B @ 35 MPH hits wall @ 0. Car a sustains 35 MPH worth of impact/damage/inertia/whatever.

Car A @ 35 MPH hits Car B @ 35 MPH. Each car sustains 35 MPH worth of impact/damage/inertia/whatever, but the total amount of impact/damage/inertia/whatever = 70 MPH worth (A+B).

If you are measuring from the point of view or either A or B, then the impacts are the same. If you are an outside observer, the total combined impact to both cars = a 70 MPH crash into a brick wall.

I think that both Bigtrout and Knucklehead are both correct, it just depends on who is measuring.
I think that your post illustrates another problem that people have with respect to this problem: the idea that mph is a valid unit for measuring the severity of a crash. Not all crashes at x mph are the same. 70 mph into a stopped car is very different from 70 mph into a brick wall. You can't just say "oh, they're both at 70 mph, therefore they'll cause the same amount of damage." Well, you can, but you shouldn't.
I think that another problem is the idea that you can just switch around frame of references. Once you choose a frame of refernce, you're committed to that frame of reference for all the calculations. If you work out the problem with the ground being at rest, you see that the car goes from 35 to 0 mph, for a change of 35 mph. If you look at it from the point of view of the other car, the original car does indeed start out at 70 mph but (and this is what many people seem to be missing) it ends up at 35 mph. Remember, if you consider the car to be traveling at 70 mph, then the road must be traveling at 35 mph. So if the car ends up traveling at the same velocity as the road, it must be traveling 35 mph as well. So it goes from 70 mph to 35 mph, and the net change is still 35 mph not 70 mph. In the case of the wall, however, the car goes from 70 mph to 0 mph, and change of 70 mph, not 35 mph as in the original case.
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  #88  
Old 07-12-2000, 08:35 PM
Bear_Nenno Bear_Nenno is offline
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Darnit Wee Man, you were almost right until you said this:

"If you are measuring from the point of view or either A or B, then the impacts are the same. If you are an outside observer, the total combined impact to both cars = a 70 MPH crash into a brick wall. "

WRONG WRONG WRONG WRONG.

Cmon people!!! The total combined impact, Wee Man, is equal to a 70mph crash into a stationary vehicle, NOT A BRICK WALL!!!! It doesnt matter if you are watching that crash from the ground, from another car, from a blimp traveling backwards at 35mph, from a train or from a car in the crash.

You guys are putting your emphasis on the spead the cars are going. The object they crash into matters!!!! So it is like a 70mph crash into a stationary car because 70 + 0 equals 70. But when you say it is like crashing into a wall at 70, then you are wrong because the wall will hit that car as hard as another car going 70mph. 70 + 70 is 140mph. Hitting a wall at 70 is like hitting a parked car at 140mph.



::: OK, everyone who thinks Bigtrout is right, Say, "Bigtrout is right". Everyone who thinks his friend is right, say. "Bigtrout is a knucklehead":::


BIGTROUT IS RIGHT
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  #89  
Old 07-12-2000, 08:56 PM
Wee Man Wee Man is offline
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Bigtrout is right
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  #90  
Old 07-12-2000, 09:01 PM
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Thanks Bear-Nunno, I stand corrected about the brick wall. All I meant was that from each car's perspective, 35 MPH worth of damage is done, so the cumulative result is a 70 MPH crash. In that respect, Knucklehead could be said to be correct.
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  #91  
Old 07-12-2000, 09:05 PM
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But I think that, given the spirit of the question, and the context of the debate,


Bigtrout is Correct
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  #92  
Old 07-12-2000, 09:30 PM
ZenBeam ZenBeam is offline
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Bigtrout is correct*

Strainger, you can e-mail me your write-up. I'd hate for someone to start criticizing it, and me not have a copy.
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  #93  
Old 07-12-2000, 09:59 PM
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I'll send a copy to you tomorrow, ZenBeam. I have it saved on my drive at work.

What kills me is that Telemark has provided a very concise anwer from a very reputable source, and he's getting blown off left and right.
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  #94  
Old 07-12-2000, 10:20 PM
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Read the test data

Strainger, thanks. I spent 30 minutes searching the web sites of the people who run the crash tests for the right quote, I'm glad someone noticed it.

Look folks, read the rationale and test results of the actual crash tests. The people who conduct the tests know exactly what they are testing. That's why there is all the handwaving about don't compare tests for cars of different weight classes! Hitting a wall simulates hitting an identical car head on at the exact same speed.

Now if you're talking about the 40% offset frontal tests into a deformable barrier, the physics gets really interesting...
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Old 07-12-2000, 10:48 PM
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Strainger, if you email it to me, I can convert it to html and post it on the web for you, if you'd like. No offense, but I'm not really terribly interested in reading it, because I already know that Bigtrout is correct.

There is some difference between a crash at 35 mph into a wall and a crash at 35 mph into an oncoming car travelling at 35 mph. In the crash between a car and a wall, the wall does not dissipate any energy, and will absorb some of the energy dissipated in the crash, due to being heated up. It does not deform at all, and therefore does not absorb a substantial protion of the energy. The difference is negligible.

OK, now to correct some of the opposing side's assertions (I have a feeling that this will be the next Monty Hall problem):

Quote:
Unless I missed one of the posts. You all are on the right track butt are forgeting the formula. E= 1/2m times v squared. A car mass 1 at speed of one hits a wall and releases enrgy of 1/2. Another car going in the oppisite direction adds its energy to the equation. Total release of evergy of 1. Two times the energy of the brick wall. How ever, if you hit the wall at double the speed (2 squared) you have a release of energy that is four times the initial conditions. So both arguments are right in a way. Two cars hitting head on at 35 mph is not as bad as hitting a brick wall at 70mph. However, it is twice as bad as hitting a brick wall at 35 mph.
You are looking at the energy that affects the entire system here, when the problem deals with energy that affects your car. Each car absorbs 0.5*m*(35 mph)2 worth of energy. This is the same as the amount of energy absorbed by the wall and the car in the case of the car hitting the wall at 35 mph. Since the wall does not deform, and the car does, the energy is almost entirely absorbed by the car. Thus, it is the same as the energy absorbed by your car in the case of oncoming cars hitting.

Quote:
And that spin is wrong, Bigtrout. I can't let this thread die until I've made a personal one-on-one attempt to make you understand this. Forget all the others' posts about impulse, momentum, speed of stopping, deforming objects etc., because those complications simply distract us from your failure to understand this basic Newtonian principle: it does so result in a collision with forces equal to 70 miles an hour
Read luckie's post. Somehow you seem unable to grasp the difference between hitting a car and hitting a wall. Watch a football (American style) game. Watch someone who is not moving much get tackled. Now imagine the tackler trying the same thing with the White Cliffs of Dover. They wouldn't go down quite so easily, would they? Maybe you should try studying the physics behind the problem, rather than dismissing Newtonian mechanics out of hand.

Quote:
An earlier post mentioned baseballs and bats, and I'm surprised no one stressed that more. If the world worked the way you say, a batter would not have to swing his bat! A thrown ball that hit a bat held rigid would be the same as if it hit a bat mid-swing (make the ball a tiny boulder and the rigid bat a wall for consistency).
NO!!! In the case of hitting an approaching car, the analagous situation would be the baseball hitting another baseball, travelling at the same speed. The swinging bat is like a wall that is approaching your car at 35 mph. This is not the same as a car approaching you at 35 mph. What would you rather hit in an equal-speed, head-on collision, a car identical to your own, or a train?

And I will again stress the point, hopefully for the last time, HITTING A CAR IS NOT THE SAME AS HITTING A WALL!!!!

I was trying to find something from the NTSB as well, but Telemark seems to have had more luck. Hopefully, with the professionals and the laws of physics supporting our claims, we can convince the knuckleheads of their wrongness. I kinda doubt it, though.
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  #96  
Old 07-13-2000, 12:19 AM
The Ryan The Ryan is offline
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Quote:
(I have a feeling that this will be the next Monty Hall problem)
Except that this actually has a right answer.
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  #97  
Old 07-13-2000, 12:45 AM
casdave casdave is online now
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Hitting a stationary car at 70mph is not the same as hitting a wall at 35mph because the stationary car is fixed to the ground by the grip of its tyres.

The tyres will transfer ,eventually, all of the impact energy to the ground but over a longer period than the wall would.

In a friction free world then maybe the first statement would be true.

So the impact would be worse at 70mph but not double the impact with wall at 35mph.

This is also a crucial differance to consider when trying to compare with a head on with two 35mph cars.
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  #98  
Old 07-13-2000, 02:01 AM
Chocobo Chocobo is offline
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First time poster to SDMB...I couldn't resist this one. I'm going to have to agree with the people that say two cars hitting each other, both traveling at 35 mph is the equivalent to one car hitting a wall at 70 mph. (Granted, it's not exactly the same, but lighten up; it's close enough)

Let's forget the complex equations for a minute, and instead, try a little science experiment. All you have to do is clap your hands together twice. The first time, hold one hand still and clap. The second time, clap both hands, and have them move approx. the same speed as each other, but each hand moving one half the speed as your hand on the first clap.

Now, if the force you felt on your hands was (or very close to) the same, then you have your answer. The same thing that just happened to your hands would happen to two cars, just on a much larger scale. If it didn't feel the same, I suggest you try it again. =)
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  #99  
Old 07-13-2000, 02:04 AM
Punoqllads Punoqllads is offline
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Location: Silly Cone Valley, CA
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Do the math, dammit

Quote:
Originally posted by casdave
Hitting a stationary car at 70mph is not the same as hitting a wall at 35mph because the stationary car is fixed to the ground by the grip of its tyres.
You are completely wrong, which you would have realized had you bothered to do the math necessary to determine what the forces and energies involved in the collisions were.

In both of these examples, we are assuming a perfectly inelastic collision, which basically means that when it occurs, all objects involved stick together, without bouncing apart.

We can make any assumption about the weight of the cars, as long as we make both cars in the two-car collision weigh the same. Let's assume the cars' weight is half a ton apiece, 1000 lbs.

35 mph is approximately 15.65 m/s, and 70 mph is 31.3 m/s. 1000 lbs, at standard Earth gravity, is 453.6 kg.

In the first example, one car is moving at 70 mph towards the other car. Its momentum is therefore 31.3 m/s * 453.6 kg, ~14200 (missed a 0 in my previous post) kg * m/s. The momentum after the collision will therefore be the same. Dividing 14200 kg * m/s by 907.2 kg (the mass of both cars together, since this is an inelastic collision) we get 15.65 m/s.

The initial kinetic energy of the system (again, 1/2 * mass * velocity^2) is 1/2 * 453.6 kg * (31.3 m/s)^2 + 1/2 * 453.6 kg * (0 m/s)^2, which is about 222.2 kJ. The final kinetic energy is 1/2 * 907.2 kg * (15.65 m/s)^2, about 111.1 kJ. The amount of energy dissipated is, then 111.1 kJ. A reasonable assumption is that both cars, being identical, deform just as much as each other, so each car is deformed by about 55.55 kJ.

In the second example, one car moving at 35 mph towards a static wall, has kinetic energy of 1/2 * 453.6 * (15.65 m/s)^2, which is about 55.55 kJ. Upon hitting the wall, it stops, (Mostly true. The wall is attached to the ground. Divide 14200 kg * m/s by the mass of the Earth, and you get a negligible velocity.) resulting in a final kinetic energy of 0 kJ, meaning that 55.55 kJ of energy is dissipated on the car, since we've said the wall doesn't deform.

Hey, look at that, the amount of energy dissipated on the car is 55.55 kJ for both. What a lovely coincidence. That means they're equivalent.

Quote:
casdave continued
The tyres will transfer ,eventually, all of the impact energy to the ground but over a longer period than the wall would.
This has nothing to do with the energy involved in the collision. An object inelastically colliding with a initially-stationary identical object at a given velocity has the same amount of force applied to it as the object at half that velocity inelastically colliding with an immovable, undeformable wall.

Quote:
casdave continued
In a friction free world then maybe the first statement would be true.

So the impact would be worse at 70mph but not double the impact with wall at 35mph.

This is also a crucial differance to consider when trying to compare with a head on with two 35mph cars.

The crucial thing is to double-check your sodding answer to make sure you don't look like a fool in front of the whole world. The crucial thing is to give references or to provide the math necessary to show your arrogant statements have some basis in reality. Put up or shut up.

Look, the math says that a car moving at 35 mph hitting a fixed barrier takes as much damage as if it had hit an oncoming identical car moving at it at 35 mph. The NTSB, an agency that performs these kinds of experiments, says the same thing. What sort of evidence do you need to believe these things?

P.S.: switch to the unopened door.
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  #100  
Old 07-13-2000, 08:48 AM
Myrr21 Myrr21 is offline
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Bigtrout is Right

Whith the slight sidenote that if these are racing cars, the wall is this nice soft foam rubber and tires, and will absorb a lot of energy...but then again, any racecar driver who loses control at 35 mph...

Besides, all this physics could be avoided by not hitting the bloddy things; saving "knucklehead" the pain of having it proven.
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