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.
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.
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.
<sigh>
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.
did you read the entire thread before posting that?
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.
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?
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.
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?
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?
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.
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.
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?
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:
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:
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:
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:
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.
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.
Oops. Last sentence should read …hitting another car head on at 35mph.
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.
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.
Bigtrout
No kidding. I like your granite boulders idea; that simplifies things nicely.
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?