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#51




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? 
#52




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. 
#53




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. 
#54




As I pointed out, that is not a similar scenario. In a headon 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. 


#55




Quote:
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 nondeformable 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 headon 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 noone). 
#56




Oops...
I erred above...I meant to say that the result after the rearend collision will be that both vehicles will be moving at 65mph, not 50mph.

#57




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!

#58




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. 
#59




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. 


#60




Everyone is wrong
Including me. But I'll try to be as nonwrong 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 spacetime 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? 
#61




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. 
#62




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 headon, 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 headon, 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 noninertial 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 headon collision we started out with, it's more like a headon 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 headon into a big 18wheeler 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 headon 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. 
#63




First off, there is a difference between deformable objects and nondeformable ones:
A nondeformable 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 momentumdouble 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 objectbreaking 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 (nondeformable 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 sameyou want a car that deformselastic vs inelastic collisions, very different species. 
#64




<sigh>



#65




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 onboth 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 realworld 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 realworld wall or the more interesting case of a really cool theorectical wall. 
#66




did you read the entire thread before posting that?

#67




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.

#68




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? 
#69




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.
__________________
MaDa: Making Sense of the Nonsensical... Sensibly. 


#70




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? 
#71




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? 
#72




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 headon 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 betweensince the conditions that result in either extreme are gross approximations. 
#73




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. 
#74




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? 


#75




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 headon, 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 playdough) 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 crumplezones 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 
#76




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. 
#77




Oops. Last sentence should read ...hitting another car head on at 35mph.

#78




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,nonelasticity, 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. 
#79




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.
__________________
In Italy for 30 years under the Borgias they had warfare, terror, murder, and bloodshed, but they produced Michelangelo, Leonardo da Vinci, and the Renaissance. In Switzerland they had brotherly love  they had 500 years of democracy and peace, and what did that produce? The cuckoo clock.  The Third Man 


#80




Yes, sailor, I read the entire thread
Bigtrout
Quote:
Quote:
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 midswing (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?
__________________
This space for hire. 
#81




five, take comfort in the fact that your observation was probably obvious to most readers from the getgo.

#82




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 nonmoving car is attached very firmly to the ground), or the frame of reference is noninertial. 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. 
#83




From the NHTSA page
Here is a quote from the info page on the NHTSA site:
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The link is http://www.nhtsa.dot.gov/ncap/infopage.html#crash 
#84




CKDextHavn
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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 nonelastic 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:
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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. 


#85




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.

#86




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. 
#87




Quote:
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. 
#88




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 
#89




Bigtrout is right



#90




Thanks BearNunno, 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.

#91




But I think that, given the spirit of the question, and the context of the debate,
Bigtrout is Correct 
#92




Bigtrout is correct*
Strainger, you can email me your writeup. I'd hate for someone to start criticizing it, and me not have a copy.
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It is too clear, and so it is hard to see. 
#93




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. 
#94




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... 


#95




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:
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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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"It is not from the benevolence of the butcher, the brewer, or the baker that we expect our dinner, but from their regard to their own interest."  Adam Smith 
#96




Quote:

#97




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. 
#98




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. =) 
#99




Do the math, dammit
Quote:
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 twocar 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:
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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




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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