View Full Version : Head-on Collision Physics

Bigtrout

07-07-2000, 12:59 AM

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. I'm long past my days of high school physics, and can't quote the various laws which lay waste to his folly. Any takers?

tcburnett

07-07-2000, 01:07 AM

I must be a knucklehead as well. Two cars crashing head-on at 35 mph is like one crashing into a wall at 70. See, both vehicles have what's called inertia which may be calculated by the speed and the weight of each vehicle. This inertia is dissipated at the moment of the crash, but not willingly because objects in motion tend to stay in motion.

Believe me. If you get into an accident, hit a parked car instead of one approaching you at the same speed you are traveling.

Bigtrout

07-07-2000, 01:14 AM

Sure, but the parked car moves. The wall doesn't. It's the same as a counteveiling force. Clearly, you would not suffer as great of force if you were traveling 45 mph and your doppelganger auto was moving 35. You would continue to move, albeit significantly slower. Your poor schnook twin, however, would suffer even greater force than had he simply run into a brick wall, as his momentum was even more violently made retrograde. But I'm still looking for the physics of this.

Mousseduck

07-07-2000, 01:21 AM

If you hit a brick wall (assuming it stays in one place and doesn't break), it exerts the same force back to you as you put upon it.

If you hit a car coming toward you, while you exert force on it, it exerts the same force back at you (like the brick wall) and also more force bacause its moving (your car is now like the brick wall).

I know that isn't very well explained, but bascically, I agree with TC Burnett - hit a stationary object or one that is moving in the same direction as you.

Bigtrout

07-07-2000, 01:26 AM

But it exerts no greater force than your car does on it. Cancellation of momentum here. Here's what was said in Scientific American on the subject. It bolsters my theory, but doesn't give the theorom. http://www.sciam.com/askexpert/physics/physics27.html

Even the Director of the Center for Transportation Analysis at Oak Ridge National Laboratory gets this one wrong, it appears.

Bear_Nenno

07-07-2000, 01:29 AM

Bigtrout... sorry man, but your friend is not the knucklehead here.

With your idea of Physics, you say that it is the stopping force that matters. Since one car was going 35 mph and suddenly stopped to 0, that is where all the damage and force comes from. You say that it does not matter if he hit a wall or a car coming at him... he was still going 35 and slowed to 0 almost instantly.

Think for a second about the wall...... do you think that the wall somehow does not feel the car crashing into it because the wall started at 0 and stayed at 0 it was uneffected and felt no force???

Ok so you say a wall feels nothing, it is just a wall. Ok if you were standing against that wall and a car going 35 mph hit you, you would definately feel the momentum of that car. You would feel it right before you got squished. Why would you think that because you started at 0 and then stayed at 0 that you would feel nothing?

Now... lets secure that wall, and you standing against it, to the hood of another car going 35 mph. You are now traveling 35 mph toward a car that is coming 35 mph toward you. Your friend is correct in saying that you would feel like the car coming at you was going 70 mph. Why? It is relative. The other car, relative to YOU, IS going 70 mph. The car you are attached to, relative to you, is not moving at all. So it would be like you standing against a stationary brick wall and getting squished by a car going 70.

And relative to me the observer... I just see some fool getting squished between to cars going about 35 mph each. Poor fella.

Luckie

07-07-2000, 01:31 AM

BigTrout is in fact correct.

if you hit the other car head on, both cars will come to a complete stop ( in theory with perfect symmetry, in practice things aren't perfectly symetric and the cars hit slightly off center, etc., etc...)

the impulse force on the car required to bring it to a complete stop is the same and it doesnn't matter if it was provided by another car in a head on or a brick wall.

if you hit a stationary car (of the same mass) then both cars are moving at half the original speed after the collision, and that would be equivilent to hitting a brickwall at half the speed.

-luckie

a physicist

Bigtrout

07-07-2000, 01:33 AM

Thank you, Luckie. Now can you give a bit more detail in the realm of physics? It would be greatly appreciated.

sailor

07-07-2000, 01:38 AM

bigtrout is right, the other trout is wrong.

tcburnett

07-07-2000, 01:39 AM

I can see this needs more explanation....OK, if we go with your theory BigTrout (and luckie), a bat will interact with a baseball with the same force whether the bat is swung or not. Or for that matter, at no matter what the speed of the baseball. No offense to the physicists here, but if you compare hiting a wall in a car from 35 mph and then hiting an oncoming car (closing speed 70 mph), I guarantee that your car will stop faster by hitting the oncoming car. Look. Hit yourself in the head with a hammer, only moderately hard. Then double the speed of the swing. Is the answer becoming clear yet?

Achernar

07-07-2000, 01:40 AM

Since about half the people in this thread are wrong already (whichever way the answer actually is) I can go ahead and post my take on it, without standing out. I think, Bigtrout, that you are right, but only because you don't know what you're saying. Specifically, there is no such thing as "a[n] immovable brick wall", one which no Force can accelerate or deform. In real life, though, that's ridiculous. If a car smashes into a wall, it's going to give a certain amount. This amount may not be much for the wall, but it's a significant amount for the car. So, if a car going 35MPH crashes into a real life wall, it's going to feel less of an impact than if it crashed into a car going 35MPH in the opposite direction.

Mauve Dog

07-07-2000, 01:42 AM

The force of the collision is determined by what is called the impulse. The impulse, in simple terms, is defined as the change in momentum per change in time. In more complex terms, the average force exerted on an object is equal to the integral of the derivative of the momentum with respect to time:

F(avg) = (integral) dp / dt

What this basically means is this: In a head on collison, the net momentum of both cars will be reduced to zero. How quickly it is reduced determines the magnitude of the force. If both cars have the same mass and are travelling at the same speed (thus, have the same momentum), then the force on each car will be the same as if either had struck a non-moving object. In either case, the time increment is very small, leading to a very large force (and, consequently, the work done on your car and you).

Bear_Nenno

07-07-2000, 01:44 AM

Oh Bigtrout I just read that link....

Here is what is going on. Your original OP mentioned "the resulting collision".

The man in the link is talking about a person inside one of the cars. We he hits something going 35 mph and suddenly stops, he exerts 35 mph worth of force on the windshield... or hopefully the seatbelt. His collision, do to his inertia is 35 mph against his steering wheel. If he hits something that gives then he will slow down more and not break the force of friection holding him in the seat and therefore will not hit his steering wheel. So of course it is best to slow gradually from 35 mph, not instantly.

What does this mean? It means, you and your friend can both be right depending on the specifics of the question. If you are talking about the total collision of two cars going 35 hitting each other being the same as 1 car going 70 and hitting a wall-that is correct. Your friend wins.

But if you are asking "Which is better to the driver?" Then you are right by saying it doesnt matter because he still hits his dashboard at 35mph, whether he hits a wall or another car coming at him....

When the poor guy's car hits the other and stops instantly, the man is no longer connected to his car and he is not part of that total collision anymore, he conitiues to go 35mph himself until he hits something. That is why it does not matter to HIM what he hits. But it still matters to the poor car, or the poor sap CONNECTED to a wall on the hood of the car.

Hope this helps....

Luckie

07-07-2000, 01:49 AM

archnar:

true, most any real object will have SOME give. the term brickwall is frequently used for a abstract that has no give, but if you have a giant block of reinforced concrete ( or hardened steel) that weighs much more than the car in question, then the difference between the reality and the abstraction will be negligible.

BigTrout: I don't know how you can convince said knuckleheaded friend...

-luckie

Bigtrout

07-07-2000, 01:50 AM

Sorry, Bear, I don't buy into the addition of force theory. And I only used the Scientific American site to point out that it also debunked the 1+1 force theory, if you read carefully. I'm not concerned with the nuances of passenger effects here. What's important, as noted in the threads, is how fast you stop. Hit a wall or hit an equal mass moving the same speed, the effect is the same.

Luckie

07-07-2000, 01:53 AM

bear said:

"If you are talking about the total collision of two cars going 35 hitting each other being the same as 1 car going 70 and hitting a wall-that is correct. Your friend wins."

sorry, that is still incorrect.

Luckie

07-07-2000, 01:58 AM

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

Bear_Nenno

07-07-2000, 01:59 AM

He IS NOT debinking the addition of forces! There is something called the conservation of momentum.

If you put a force probe on the bumber of the car, you would get the same data from both collisions.

The writer in the article is talking about the effect on the driver.

Bear_Nenno

07-07-2000, 02:02 AM

Luckie...

Read your last post. I see what you mean and I will let you answer things for now on :)

tcburnett

07-07-2000, 02:11 AM

I can't provide any further input. But I know who can.

The National Institute for Highway Safety:

http://www.highwaysafety.org/

and

The National Transportation and Safety Board:

http://www.ntsb.gov/

Of course, we can always ask for help from Above.

tcburnett

07-07-2000, 02:21 AM

Now I agree with Lucky. I wonder what question I was trying to answer earlier? Obviously not the one I thought I read.

C K Dexter Haven

07-07-2000, 07:27 AM

May I try to settle this by changing the example? Instead of both cars going at 35 mph, suppose that the first car A is going very slow, say 1 mph, and the other car B is going at the 35 mph.

OK, now ask your question. What's the result for Car A? Is the resulting impact the same as if Car A hit a wall at 1 mph? Certainly not.

This example should make it clear that the speed (OK, mass and acceleration) of the other car increases the magnitude of the impact, si?

sailor

07-07-2000, 09:24 AM

(A) a wall built on the side of a mountain (i,e. solidly connected to the rest of the earth) is, for all purposes an immovable object when compared with a car.

(B) Consider *this* example: a very thin wall, of insignifican mass compared to a car but rigid enough to not deform and mounted on wheels with no friction. In other words, if a car hits it, it just pushes it along without slowing down.

OK. So we place this wall in the middle of a road and two cars of equal mass travelling in opposite directions hit it simultaneously at 35mph. will the wall even move? I think not. the forces are exactly equal from both sides.

So, from my side I have just hit a wall that did not move even by a fraction of an inch. How is this different from hitting a wall with a mountain behind it?

zwaldd

07-07-2000, 11:02 AM

a 35mph head on crash has the effect of a 70mph immovable object crash, since the gap between you closes at 70mph.

douglips

07-07-2000, 11:24 AM

a 35mph head on crash has the effect of a 70mph immovable object crash, since the gap between you closes at 70mph.

No, no, no.

It has been explained nearly to death above, but let me try a different tack.

In a 35 mph head on crash, there is the same energy being dissipated as in a 70mph 'immovable wall' crash.

The difference is, in a 35 mph head on crash, there are two cars to share the energy dissipation! Anytime you can get some shmuck to absorb half the energy of your impact you are getting a break. Look again at Luckie's example of hitting a parked car at 70 mph.

sailor

07-07-2000, 11:36 AM

zwaldd, man, not content with being wrong in another thread you come here to be wrong again? And by you low post count it seems you have been nothing *but* wrong. Wake up man!

A car at 35 mph hitting another car head on at 35 mph will suffer the same damage if it was traveling at 70 mph and hit a stationary CAR of equal mass but a WALL has infinite mass compared to the car. Don't you get it man? Read the earlier posts and THINK about them before you shoot your posts. Sheesh!

Bigtrout

07-07-2000, 11:39 AM

Good point, sailor. The wall stays put. Which is the same as if it were solid. Which means you can discount the force of the car on the opposite side. Which equates to a 35 mph crash into a solid wall.

zwaldd

07-07-2000, 11:47 AM

sailor after reading this and some of your other interactions i believe you have a serious case of 'thread rage'. get help and get educated. also, get one of those swinging ball and string doohickeys and you can see how these forces actually work.

Saltire

07-07-2000, 12:08 PM

It's been stated that the important point here is how fast your car stops. The change in momentum over time (impulse) is the important factor. I agree with that.

But no one followed up by analyzing that fact. If you hit a wall at 35 mph, your car will stop in a very short time, mostly determined by how much the structure crumples. If you hit a wall doing 70, your car stops in a much shorter time, so the impulse is much larger.

It seems to me that hitting another car, with both of you travelling at 35mph would also cut the deceleration time to much shorter than you saw in the first case (the wall at 35). This is because the two cars are approaching each other at a relative speed of 70 mph and they will get from the point of first contact to the point where the front end can't crumple any further in less time. Therefore, you see a larger value for the impluse, and thus feel a larger force.

Can anyone convince us that the deceleration time is identical in all the collisions described? I don't think so.

C K Dexter Haven

07-07-2000, 12:36 PM

I stand by my original example. Suppose the two cars are not going at equal speed. And let's replace the immovable wall with another car, but one that is stationary.

Now, you're in a car moving 1 mph.

Which collision would you rather be in?

- A crash with a car coming straight at you at 35 mph?

Or

- A crash with a stationary car?

You will have to argue long and loud to convince me that those two crashes are the same. If those are different, then why would it matter if your car is moving at two miles and hour? at three? ... or at 35? The crashes are still different, and the speed of the OTHER car is a critical factor that makes them different.

ZenBeam

07-07-2000, 12:37 PM

It seems to me that hitting another car, with both of you travelling at 35mph would also cut the deceleration time to much shorter than you saw in the first case (the wall at 35) This is because the two cars are approaching each other at a relative speed of 70 mph and they will get from the point of first contact to the point where the front end can't crumple any further in less time.

But you have two cars crumpling, so you will have twice the crumple distance. This exactly cancels the factor of two in speed.

Chronos

07-07-2000, 01:48 PM

OK, by now I've mostly lost track of who said what, but I think that Sailor's got it, with his example of the massless rigid wall. As to CK's example, if you specify what's colliding, all that matters is the relative speed: Car A at 1 mph colliding with car B at -35 mph is exactly equivalent to car A at 18 mph and car B at -18 mph. Similarly, car A at 70 colliding with an infinitely massive wall at 0 is equivalent to car A at zero colliding with wall at -70. The important distinction is, that Car A at 70 hitting car B at 0 is not the same as car A at 70 hitting brick wall at 0: In the fist example, the end speed will be 35, while in the second, it'll be 0.

All clear now? What? OK, then, can we pretend that it's all clear?

Spiritus Mundi

07-07-2000, 02:05 PM

CK:

If I travel at 35 and you travel at 1 and we have a head on collision, then our relative velocity is 36 and our "crumple factor" (the amount of force asorbed by the collapse of the frame) is 2cars. This means my relative impact (and yours) is 18 (expressed in speed/crumples). If you travel 15 then we each get 25 s/c of impact. If you travel 35 then we each absorb 35 s/c. This is the same force either of us would absorb if we ran head first into an immovable object at 35.

tcburnett

07-07-2000, 02:27 PM

Originally posted by Chronos

All clear now? What? OK, then, can we pretend that it's all clear?

Yes. Let's pretend. First, let me pretend that I know what I am talking about.

There are two separate questions being answered here. In the original post, if I am reading it correctly, Knucklehead says that if two cars of equal mass, each traveling at 35 mph meet head-on the resultant collision will dissipate inertia roughly equivalent to one car traveling at 70 mph hitting a parked car of the same mass.

Then BigTrout tossed an immovable brick wall into the equation and the entire nature of the problem changed.

The point I (tried to) argue is that two cars traveling at 35 mph each and colliding head-on is not equivalent to one car traveling at 35mph and striking a parked car. I know I SAID wall but, since I need to backpedal out of that, the cars crashed before they built the wall. (yeah, that's it)

I THINK that is what lucky, sailor, Chronos and others said as well:

A car at 35 mph hitting another car head on at 35 mph will suffer the same damage if it was traveling at 70 mph and hit a stationary CAR of equal mass but a WALL has infinite mass compared to the car. - sailor

I apologize if it was me who screwed up the nature of the problem by answering the wrong question.

COGITO EGGO SUM: I think; therefore I am a waffle.

Mauve Dog

07-07-2000, 03:06 PM

OK. It seems that we have a number of scenarios being discussed.

Let's start with the initial example: 2 cars, same mass, same speed, opposite direction, colliding head-on.

For this scenario, the force is determined by the change in momentum per change in time - both cars go from whatever speed they were travelling to zero in a very small amount of time, thus the impulse is very large, thus the force exerted is very large. OK, no-one, I'm sure, disagrees with this.

Now, scenario two: one of those same cars collides with a wall of sufficient mass that it is for all intents and purposes immovable (I say this because in a collision, momentum must be conserved. Thus, if two objects collide, both will experience a change in momentum. However, if the mass is very large, the speed increase will be very small), at the same speed it was travelling before. The change in momentum is exactly the same, and the time increment is probably on the same order. Thus, the impulse is the same, thus the force exerted on the car would be the same. This is the point made in the SciAm article referenced earlier.

Scenario three: One of those same cars collides with a wall at twice the speed. Obviously, the force of impact in this case would be twice as great, since the change in momentum is twice as great, etc., etc.

Scenario four: One of those cars moving at twice the speed collides with the other car, which is stationary. In this case, it is true that the force of impact will be the same. However, this is a different collsion, since neither car will be at rest after the collision - momentum must be conserved. Thus, if one car, travelling 70mph, collides with an identical but stationary one, the result will be both cars continuing in the direction of the orignal motion, but at half the speed. This would result in the same change in momentum, probably in about the same amount of time, thus the force of impact would be the same as the head-on case. This would also be true from the point of view of either vehicle.

Scenario five: A car travelling at 1mph collides head-on with a car travelling 35mph. This is very different indeed. Remember, momentum must be conserved. Thus, the faster car will continue travelling in its original direction after the collision, but at half of the difference between the original speeds, or, 17mph. It should be obvious that the change in momentum (going from 35mph to 17mph) is significantly less than the original head-on case. The slower car would experience a change in velocity of 18mph (it was travelling 1mph in one direction, and would end uo travelling 17mph in the other, along with the other car), thus it would experience a slightly greater change in momentum, thus a greater force of impact than the faster car experienced. In any case, all of the forces involved are less than they would be for either car as compared to the equal-speed head-on. And, of course, the force of collision for the 1mph car would be 18 times larger than a collision with our wall!

So, the end result is, for a head-on collision between two cars of equal mass and speed, the forces will be the same (but not the subsequent motions) as a car moving twice as fast colliding with a stationary one. This will also be true from the point of view of either vehicle, in both cases.

douglips

07-07-2000, 03:26 PM

So, CKDext, did the above explanations cut it? You are right in your examples, but you are taking the example beyond its applicability when you start to talk about fixed objects.

The main point is that a collision with 2 cars of equal mass and equal crumple zones is going to be just as bad for either driver as a collision of one car with a fixed (i.e. infinite mass, non-deformable) object at half the combined speed.

So, in your example of car A being driven at 1 MPH and car B being driven variously at 0 or 35 MPH, in the first example it is equivalent to hitting the proverbial brick wall at 0.5 MPH, and in the second example it is equivalent to hitting a brick wall at 18 MPH.

The main point is that not all stationary targets are equal - a parked car will absorb a lot of energy by crumpling and by moving, so the combined collided cars will have some non-zero final velocity which reduces the amount of acceleration the driver of the moving car must undergo. A fixed object will not absorb any (in the ideal case) energy, nor will it move to reduce the acceleration.

The crumple zones increase the time of acceleration, and the final motion decreases the magnitude of the acceleration, so the impulse (what kills you) is reduced greatly by colliding with a car instead of a wall. It is in the ideal case reduced by half as described in my 3rd paragraph here.

P.S. I made an incorrect and somewhat confusing statement earlier:

In a 35 mph head on crash, there is the same energy being dissipated as in a 70mph 'immovable wall' crash. This is not true. This is the same amount of energy being dissipated as in a 70mph 'parked car' crash, but twice as much energy is dissipated in the 'immovable wall' crash at 70 because kinetic energy goes as the square of velocity but only directly as mass. Also, since only one car absorbs damage it is 4 times as bad to hit a fixed object at 70 as it is to hit a car head on at 35 each.

What my previous post should have said is that a 35 mph head-on crash has twice as much energy to dissipate as a 35 mph fixed object crash, but since there are two cars to dissipate the energy in the same amount of time you end up with the same amount of energy absorbed by each car in the same amount of time. The more energy you have to absorb (and the more quickly, a la impulse), the more mangled your car is and the more dead you are.

sailor

07-07-2000, 05:42 PM

Zwaldd said:

sailor after reading this and some of your other interactions i believe you have a serious case of 'thread rage'. get help and get educated. also, get one of those swinging ball and string doohickeys and you can see how these forces actually work.

I do not have any thread rage. I just wish you would read the earlier postings but that is OK. Some other posters who were mistaken initially have come to realize it after reading the thread. I wish you would do the same but I do not need to buy any doohickeys, I am ready to make you the same offer I made in another thread: How much do you want to bet?

Bigtrout

07-07-2000, 05:56 PM

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.

Take two objects (granite boulders, if you wish) equal in all respects, which collide head-on while both are moving at 35 mph. (This takes the crumple zone business out of the equation.) Their forces should cancel each other. This, therefore, should be the same as if one of the objects were to hit an essentially immovable object at 35 mph. It stops, discounting some inherent rebound which can be ignored for purposes of this exercise.

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.

I hope I haven't caused anyone to needlessly wreck automobiles in the pursuit of future barroom sagacity.

sailor

07-07-2000, 06:23 PM

Bigtrout, I think we have pretty much all agreed that in your OP you were right and your knucklehead friend is mistaken. If I am not mistaken tcburnett and others who were not with you at first are all in agreement with you and the only one holding out is Zwald (I believe, if there's anyone else, please raise your hand).

Since Zwaldd is quite certain I have given him the opportunity to make some fast cash and make me look like a fool at the same time by winning a bet with me. I just hope he does not want to put on this more than I can possibly cover. ;)

waterj2

07-08-2000, 01:46 AM

Wow, this is interesting. So much ignorance and so many red herrings. OK, first off, inertia doesn't mean what many of you people think it does. There are a couple of physics equations and concepts that are going to be relevant here.

First is momentum, which is mass times velocity. p=mv

Next is kinetic energy, which is one half mass times velocity squared. k=½mv2

There is also impulse, which is the integral of the force over the time elapsed, as well as the change in momentum.

The basic principle in collisions is that the total momentum is always preserved. It is important to note that momentum is a vector.

The thing that separates a car crash from a baseball hitting a bat, billiard balls, or those little metal balls hanging by string is that energy is not conserved. Hit two billiard balls into each other at equal speeds and they bounce off each other, unlike cars. Note that energy is not a vector.

OK, let's look at the two cars approaching. Since they have the same mass, each has the same momentum in opposite directions. The total momentum is the sum of these two, which is zero. After the collision, the total momentum must also be zero. This is easy, since both cars stop completely. Each car has an impulse equal to its change in momentum, which is given by mv.

Since each car stops, the kinetic energy of each car goes from ½mv2 to 0. Since energy is never actually lost, all the energy is converted to other forms, such as heat and sound.

Now for a car crashing into a wall at the same speed, we will assume that the wall has infinite mass, and zero velocity. The result of the car stopping completely still happens (and to be fair, the wall moves back at an infinitessimal velocity), thus the car loses the same amount of momentum, and the impulse is the same. To figure out the force, you would have to know the comparative amounts of time that each collision takes. This is tricky, but it seems that since we aren't assuming any crumple zones or anything, that it would be about the same as for two cars hitting each other.

Of course, much more energy is dissipated in the collision between two cars. Thus there is a bit more noise and heat, and some other damage, which is certainly not entirely insignificant, but doesn't really affect the driver. However, for the case of a car hitting a wall at twice the original speed, there is actually twice the energy dissipated as in the two cars hitting each other. Not to mention the fact that there is twice as much impulse on each car.

So in summation, if you are travelling 35 mph and hit a car travelling at 35 mph in a head on collision, it is not the same as hitting a brick wall at 35 mph, but it is similar in many ways. It is nothing like hitting a brick wall at 70 mph.

Zwaald, since you presumably have one of those dohickeys, I suggest that you try this experiment. Hopefully you can use just two of the balls with it. Take two balls and raise them to equal heights on opposite sides. Simultaneously drop them. The ball in your left hand is your car. Observe how far back this ball bounces. Now drop this same ball from the same height, but this time hold the other ball perfectly motionless. Observe the results. If you were good with holding it motionless, the first ball should bounce back just the same as in the first experiment.

CKDexHavn, the reason that the cars in your experiment do not have similar results is that the total momentum is not zero. Your car has a great deal of momentum imparted to it in this example, meaning a high impulse, and a high force.

sailor

07-08-2000, 09:30 AM

Waterj2, how do you define "doohickey"?

Your post seems right in every point and answers more questions than the OP.

Oh my goodness! A little red light just went off inside my brain and I think I was wrong all along! I need to analyze further but I think I was wrong. Boy, is my face going to be red!

Let me get some coffee and wake up and I'll be right back. (and to think I have to pick my own arguments apart!)

BRB

sailor

07-08-2000, 10:03 AM

OK; here I go.

Premise: In one case my car, of mass M, is moving at speed V and hits another car coming head-on at the same speed. The alternative case is that it hits a stationary wall of infinite mass. (And we may add just for illustration a third case where it hits a stationary car of equal mass). We are only concerned with the effects to my car (not to the other cars or wall).

Case one: my car, of mass M = 1, traveling at speed V = 1, has a kinetic energy E = 0.5 * M * V^2 = 0.5 ... When my car hits another car head on it comes to a complete stop instantly and all this energy (0.5) is instantly dissipated in crumpling my car (and maybe my face). We are not concerned with the energy (same amount of 0.5) that does the same thing to the other car.

Case two: My car has exactly the same amount of kinetic energy (0.5) which is instantly dissipated in the same way.

ERGO: for the purposes of what happens to my car cases 1 and 2 are equal and the OP was right and I can rest easy that I was also right. Wow, I was confused there for a moment. I really need to get that coffee *before* I do anything else.

OK, bonus questions: What happens if my car at speed V=2 hits the immovable wall and what happens if my car at speed V=2 hits a parked car.

Case three: my car at speed V=2 has a kinetic energy of E=2 (four times what it had before). When it hits a wall the energy dissipated in crumpling my car will be four times as much and the damage much worse that in cases one and two.

case four: my car at speed V=2 hits a stationary car of equal mass M. This case is a bit more complicated but it is obvious the effects are smaller than hitting a wall at the same speed of 2.

To resolve this case we have to make certain assumptions regarding the crumpling and the conservation of movement. How much kinetic energy is transferred to the other vehicle to impart it motion and how much is dissipated in my crumpling? This depends on the construction of the cars. Elastic cars (bumper cars) would suffer no crumpling and conserve motion and energy. In the real world you have to define this before you can resolve the problem. But this question was not asked anyway. I added it for illustration.

So the fact remains unshaken: bigtrout was right all along and his friend Chuckie Knucklehead was, and continues to be, wrong.

sailor

07-08-2000, 10:41 AM

actually.. in case four above, to preserve consistency, I am constricted to assume the following: My car at V=2 hits a stationary car of equal mass. Right after impact they are both traveling together at the same speed of V=1. My intuition tells me hitting a stationary car at 70 MPH is similar to going at 35 MPH and hitting another car head on which is go8ing at the same speed.

Heck, of course, this should be obvious! As long as the masses remain the same and the combined speed difference remains the same, then it is the same case. In other words, I can be going 100 MPH and rear end a car going 30 MPH. But it has to be the same car all the time. If you make it a wall the problem changes.

Well, there's case 4 resolved as well.

I am still wondering why waterj2's post, which essentialy corroborates everything we've said, made me doubt and think he may be wrong, and therefore me too. I have to remember to get my coffee first...

Danalan

07-08-2000, 11:46 PM

I think we have to do an experiment to prove these various theorys. We'll actually need several experiments, cars hitting each other both at 35, cars hitting each other at 35 and 1, 35 and stationary, etc. Also cars hitting brick walls.

Okay, everybody post what type of car you drive so we can select 'identical' cars. Then we'll simply pick a place and time, and settle this once and for all.

Somebody be sure and bring force guages, video cameras, and band-aids.

Mauve Dog

07-09-2000, 01:09 AM

sailor writes:

actually.. in case four above, to preserve consistency, I am constricted to assume the following: My car at V=2 hits a stationary car of equal mass. Right after impact they are both traveling together at the same speed of V=1. My intuition tells me hitting a stationary car at 70 MPH is similar to going at 35 MPH and hitting another car head on which is go8ing at the same speed.

Your intuition is correct. See my earlier post.

Heck, of course, this should be obvious! As long as the masses remain the same and the combined speed difference remains the same, then it is the same case. In other words, I can be going 100 MPH and rear end a car going 30 MPH. But it has to be the same car all the time.

Not quite. A rear-end collision with both cars moving in the same direction will be slightly different. While it is true that the change in momentum will be the same as with a 70mph car hitting a stationary one, the time it takes for this change will be slightly greater because both cars are moving in the same direction. Thus, the force of impact will be reduced.

sailor

07-09-2000, 01:44 AM

Could you expand on that. As long as the relative speed is the same I cannot see how it would matter. As long as you do not have exterior references (which you do not need) what counts is the relative speed between the two. These scenarios would do similar damage to my car:

I am going 35mph and I hit a car head on who is doing 35 mph

I am going 70mph and hit a car that is stationary

I am going 105 mph and hit a car that is backing at 35mph

in every single case the relative speed is 70mph and the consequences for me will be the same. In *my* frame of reference I have hit a car going at 70mph.

Mauve Dog

07-09-2000, 02:19 AM

As mentioned, it is not merely the speed difference that affects the force of impact. The speed difference will result in the same change in momentum, which will result in the same impulse. However, the impulse is the average force x time differential. Thus, the magnitude of the force will be determined by how long it takes the momentum to change. When both cars are going in the same direction, the lead car is moving away from the ramming car; the car being rammed does not remain fixed in place (as would essentially be the case with a head-on collision) while the ramming car crumples. It should, therefore, take slightly longer for this change in momentum to take place, which would result in a smaller force. I'm not saying the difference would be drastic. But there would be a diference.

The point is, if you extend the amount of time over which the collision occurs, the forces involved will decrease.

Thus, the magnitude of the force will be determined by how long it takes the momentum to change.

I agree with that. But shouldn't the duration of the collision depend only on the relative velocity of the two cars? So I think sailor's assertion is correct, that the damage only depends on the relative speed of the cars.

Of course, cars respond differently depending on whether it's hit from the front or back, so there may be a small difference. My WAG is that a car has a longer crumple zone in front so a head-on collision results in a smaller acceleration for the occupants.

sailor

07-09-2000, 08:42 AM

>> cars respond differently depending on whether it's hit from the front or back

we are trying to make abstraction of these details but in anticipation for exactly this objection please notice I said the car is "backing at 35mph" so you still hit the front end.

Mauve Dog

07-09-2000, 03:35 PM

Originally posted by scr4 :

I agree with that. But shouldn't the duration of the collision depend only on the relative velocity of the two cars? So I think sailor's assertion is correct, that the damage only depends on the relative speed of the cars.

I think not. If the object you are colliding with is moving away from you, the collision should take place over a longer time interval.

Consider the OP: two cars collide head-on. Both are effectively colliding with a fixed point. The same is true if one of the cars collides with a wall. Thus, those situations are comparable; the time interval of the collision is the same.

Basically, I think the confusion here arises because at some point in the discussion, crumpling was introduced. If we are talking straight theory, and assuming that for the sake of simplification, all object involved in these collisions are non-deformable, then yes, the only thing that matters is the speed difference before and after, and with same-direction collisions, the only thing that matters is the difference in speed, and the resulting forces would be the same as a half-speed head on colision.

However, once crumpling is introduced, we are no longer discussing the ideal case. The objects now are deformable, and that deformation occurs over a finite time interval - it cannot be 'instantaneous'.

When colliding with a fixed point, the deformation takes a given time. This is true whether that fixed object is another car travelling in the opposite direction, a wall, tree, whatever.

However, if the object being collided with is movable, then things change. Because the second object is accelerating in the direction of motion of the original object, the relative speeds are changing in a different manner; the faster object is slowing down, the slower object is speeding up.

As such, when considering such a non-ideal case (where deformation of the objects is considered), any of the scenarios where a second vehicle is not fixed, and not moving at the same speed, the actual time interval of the collison should be greater than the ideal time of collison, thus reducing the overall force.

The problem is, the real world does not submit easily to reduction to simple mathemitcal terms. That is why auto manufacturers do crash tests; otherwise, they could simply crunch the numbers, so to speak, to determine the effects of a collision.

I guess my point here is that in theory, yes, the situations presented (e.g., a faster-moving vehicle striking another vehicle which is moving in reverse, in the same direction) are as explained. In reality, there will be differences.

sailor

07-09-2000, 03:49 PM

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?

Chronos

07-09-2000, 04:37 PM

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.

C K Dexter Haven

07-10-2000, 07:36 AM

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.

waterj2

07-10-2000, 08:34 AM

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.

Mauve Dog

07-10-2000, 01:32 PM

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).

Mauve Dog

07-10-2000, 02:19 PM

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.

sailor

07-10-2000, 10:16 PM

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!

iampunha

07-11-2000, 06:16 PM

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.

nola andy

07-11-2000, 08:48 PM

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.

Punoqllads

07-11-2000, 09:07 PM

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?

Strainger

07-11-2000, 09:14 PM

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.

OldMan

07-11-2000, 09:16 PM

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.

Myrr21

07-11-2000, 09:20 PM

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.

sailor

07-11-2000, 10:52 PM

<sigh>

Myrr21

07-11-2000, 11:02 PM

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.

sailor

07-12-2000, 12:17 AM

did you read the entire thread before posting that?

Revedge

07-12-2000, 02:26 AM

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.

Speaker for the Dead

07-12-2000, 02:37 AM

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?

SPOOFE

07-12-2000, 03:23 AM

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.

Bear_Nenno

07-12-2000, 05:35 AM

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?

Bear_Nenno

07-12-2000, 05:44 AM

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?

Myrr21

07-12-2000, 08:40 AM

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.

Myrr21

07-12-2000, 11:21 AM

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.

HalloweenMan

07-12-2000, 12:45 PM

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?

MetallicAsh

07-12-2000, 12:52 PM

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

Revedge

07-12-2000, 12:55 PM

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.

Revedge

07-12-2000, 12:57 PM

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

UnderDog

07-12-2000, 02:29 PM

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.

Scupper

07-12-2000, 03:05 PM

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.

Fiver

07-12-2000, 03:05 PM

Bigtrout

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.

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?

zwaldd

07-12-2000, 03:14 PM

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

Punoqllads

07-12-2000, 03:47 PM

<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.

Telemark

07-12-2000, 05:24 PM

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

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

The Ryan

07-12-2000, 05:57 PM

CKDextHavn

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

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:

It was actually Bigtrout who brought up the brick wall.

according to the OP (emphasis mine):

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.

Strainger

07-12-2000, 07:19 PM

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.

Wee Man

07-12-2000, 08:02 PM

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.

The Ryan

07-12-2000, 08:34 PM

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.

Bear_Nenno

07-12-2000, 08:35 PM

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

Wee Man

07-12-2000, 08:56 PM

Bigtrout is right

Wee Man

07-12-2000, 09:01 PM

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.

Wee Man

07-12-2000, 09:05 PM

But I think that, given the spirit of the question, and the context of the debate,

Bigtrout is Correct

ZenBeam

07-12-2000, 09:30 PM

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.

Strainger

07-12-2000, 09:59 PM

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.

Telemark

07-12-2000, 10:20 PM

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...

waterj2

07-12-2000, 10:48 PM

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):

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.

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.

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.

The Ryan

07-13-2000, 12:19 AM

(I have a feeling that this will be the next Monty Hall problem)

Except that this actually has a right answer.

casdave

07-13-2000, 12:45 AM

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.

Chocobo

07-13-2000, 02:01 AM

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. =)

Punoqllads

07-13-2000, 02:04 AM

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.

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.

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.

Myrr21

07-13-2000, 08:48 AM

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.

Is this STILL going on?

Firstly, Bigtrout is right.

Secondly, Bigtrout's friend is NOT a knucklehead. This problem is a little counter-intuitive. I got it wrong until I thought about it a little.

CKdextHavn said:

"May I try to settle this by changing the example? Instead of both cars going at 35 mph, suppose that the first car A is going very slow, say 1 mph, and the other car B is going at the 35 mph.

OK, now ask your question. What's the result for Car A? Is the resulting impact the same as if Car A hit a wall at 1 mph? Certainly not.

This example should make it clear that the speed (OK, mass and acceleration) of the other car increases the magnitude of the impact, si?"

This threw me a little, since I KNEW Bigtrout was right but I couldn't fault the logic. But I've figured it out now.

Head on crash, two identical cars, 35 mph. The kinetic energy partitions evenly between the cars, from the symmetry of the collision. So each car absorbs one "35 mph car's worth" of K.E.

One car, 35 mph, perfectly rigid, massive brick wall. Half the kinetic energy available, but NONE of it goes into the wall, it ALL goes into the car. So the collision energy is the same from the car's point of view. This is the point which is counterintuitive and is messing people up.

Head-on crash, Car A at 1 mph, car B at 35 mph. It is clear that car A will take a much bigger hit than if it went into the wall at 1 mph. BUT this is because some of the K.E. from car B is transferred into car A due to the assymetry of the collision. In the original example, no K.E from car B ends up in car A (or they exchange equal amounts, if you're picky).

This discussion shouldn't be continued without beer and a table pound on!

Man. I can't believe this thread made it into Threadspotting. A question that has a simple answer was sorta complicated in the discussion.

Idealized Case 1: Symmetric, head-on collision of identical cars, each going 35 mph.

Idealized Case 2: The collision of one car at 35 mph with a non-deformable wall.

The results are identical, and can be shown by any of the numerous arguments above (my personal favorite is sailor's, way back near the beginning).

Bigtrout is right, and any counter-argument has a logical flaw.

Punoqllads

07-13-2000, 03:24 PM

Okay, the horse is quite dead, I believe, but I looked back at the original post, and found something quite curious. We've beem arguing back and forth about whether Bigtrout or his knucklehead friend are right. But, looking at what was said in the first post, it seems that Bigtrount and his friend are right.

Originally posted by Bigtrout

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. I'm long past my days of high school physics, and can't quote the various laws which lay waste to his folly. Any takers?

Two cars, each going 35 mph towards each other, colliding inelastically, are equivalent to one car, moving at 70 mph, colliding inelastically with a parked car. So Bigtrout's friend is correct. But, the collision energy absorbed by one car in either case is the same amount of energy absorbed by one car crashing into an immobile, undeformable wall.

It seems that Bigtrout thought his friend, in talking about the 70 mph - 0 mph collision, was talking about a car travelling at 70 mph hitting an immobile, undeformable wall, which clearly causes much more damage to the car than the any of the three scenerios above. Four times as much damage, in fact.

But reading litterally what Bigtrout said, I believe his friend was talking about a 70 mph car crashing into a parked car to be equivalent to the two of them driving at 35 mph towards each other, crashing; which, as has been shown previously, is correct.

Perhaps Bigtrout can enlighten us, by telling us which scenario his friend meant. Bigtrout, could you ask him?

waterj2

07-13-2000, 06:29 PM

Please, for the love of God, people, stop posting irrelevant and repetitive claptrap! Read through the thread. If you think that hitting a wall at 70 mph is equivalent to hitting a car travelling at 35 mph while also travelling 35 mph in the opposite direction yourself, explain why you feel that the NHTSA and the one physicist in this thread are wrong. The problem is clearly explained from every possible angle in this thread, and you are not likely to win anyone over to your side.

Chronos

07-13-2000, 09:40 PM

Minor quibble, waterj2: There've been several real physicists posting to this thread. Of course, since they all agree, this just makes your point stronger. As to the OP, bigtrout is correct, and his friend may or may not be correct, depending on what he meant by a 70 mph collision.

The Ryan

07-14-2000, 12:35 AM

Originally posted by Chocobo

Let's forget the complex equations for a minute,

Let's not. If the equation don't square with your intuition, then it's probably your intuition that's to blame.

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. =)

Yes, that's what would happen to two cars but it's not what would happen with one car and a wall.

Danalan

07-14-2000, 10:10 PM

I still don't think we're never going to end this discussion without performing an experiment. I've got a 2-door Ford Escort GT and a 4-door Mercury Sable I can contribute to the experiment. Can anybody else match those cars? Anybody know of a good, solid, brick wall at the end of a road?

I think we might also need a fire extinguisher. . .

Bigtrout

07-18-2000, 09:21 AM

Knucklehead's scenario is two cars moving 35 mph crashing head on. He believes that is the equivalent of a car moving 70 mph crashing into the proverbial inelastic, immovable wall. No parked cars in his scenario. The two colliding cars are of equal mass and speed.

Arjuna34

07-18-2000, 10:27 AM

I think we've got the car vs. wall situation covered, but what about two immovable walls colliding, each moving at 35 MPH into each other? How does that compare to a 70MPH immovable wall hitting a parked wall?

:)

Arjuna34

UnderDog

07-18-2000, 05:16 PM

Originally posted by Arjuna34

I think we've got the car vs. wall situation covered, but what about two immovable walls colliding, each moving at 35 MPH into each other? How does that compare to a 70MPH immovable wall hitting a parked wall?

:)

Arjuna34

Geez! Don't even suggest such a thing... The last time that happened it created the Big Bang and things have pretty much gone down hill from there!

jbird3000

07-24-2000, 12:10 AM

Looking at the car hitting the other car, both at 35mph, you'll see that both drivers of the cars will continue to move (decelerating) for a fraction of a second after their cars hit (as the cars crumple), while the front bumpers of their cars will halt virtually instantaneously. Assuming both cars crumple the same, neither will give and the car will come to a dead stop.

In comparison, the "unmovable" wall will obviously not give, and the car and driver will decelerate the same as above.

So, since F=ma, the mass of the car will not change in the 2 instances, the deceleration will not change, Therefore, the force will not change.

astro

07-24-2000, 02:03 AM

The physics of dead horses.

The following item," Jackson began, "which is altogether in apposite to this case, was called to my attention by a colleague of mine. The code of tribal wisdom says that when you discover you are riding a dead horse, the best strategy is to dismount.

"In law firms," the judge continued, "we often try other strategies with dead horses, including the following: buying a stronger whip; changing riders; saying things like `this is the way we have always ridden this horse'; appointing a committee to study the horse; arranging to visit other firms to see how they ride dead horses; increasing the standards to ride dead horses; declaring that the horse is better, faster and cheaper dead; and finally, harnessing several dead horses together for increased speed."

In conclusion, Jackson turned to Department of Justice attorney David Boies and added, "That said, the witness is yours."

ticker

07-24-2000, 09:11 AM

Look at it this way:

If a car hits a truly solid brick wall (that mythical immovable object twinned with the irresistable force) then the wall will exert an equal and opposite force on the car.

An identical car travelling towards me at an identical speed in the opposite direction is an equal and opposite force. Therefore the OP is correct.

Punoqllads

07-25-2000, 06:08 PM

Originally posted by jbird3000

Looking at the car hitting the other car, both at 35mph, you'll see that both drivers of the cars will continue to move (decelerating) for a fraction of a second after their cars hit (as the cars crumple), while the front bumpers of their cars will halt virtually instantaneously. Assuming both cars crumple the same, neither will give and the car will come to a dead stop.

In comparison, the "unmovable" wall will obviously not give, and the car and driver will decelerate the same as above.

So, since F=ma, the mass of the car will not change in the 2 instances, the deceleration will not change, Therefore, the force will not change.

Dear jbird3000,

You give a clear, concice explanation that is completely wrong. It's about as good as the 1950 New York Times article that proves how rockets could not fly through outer space "because they would have nothing to push against." I suggest that before you post again, you should read a physics book, and I'm talking about something without illustrations by Richard Scar[r?]y.

Once you've got it, look up the term "kinetic energy." Note that it is based upon the square of the velocity. Note that 2 * 35^2 is less than 70^2; in fact, half. You might, as others have done, believe that if an observer moving along with one of the cars in the two-car collision would see the same kind of velocity change as a stationary observer viewing the single car crashing. But there is a difference: after the collision, the moving observer would still be moving at 35 miles per hour, while the stationary observer viewing the 70 mph crash would not see any motion, by the premises above. Thus, the two collisions would either be different, or one of the observers must not be in an inertial reference frame.

If you're not sure how velocity is different than speed, or exactly what an inertial frame of reference, or even what a frame of reference is, again, please pick up a physics book. If there's something it doesn't explain well, talk to a physics teacher/professor. The life you save might be your own.

ZenBeam

07-25-2000, 09:06 PM

Punoqllads, I believe that in jbird3000's scenario where the car crashes into the wall, he means that it is travelling 35 MPH, not 70 MPH. In that case, his description is both concise and correct. I doubt he needs to consult a physics book.

Bear_Nenno

07-26-2000, 03:18 AM

Jesus Christ!! You guys are still debating this? I thought we settled this a month ago??!!!!!??

I think everything that could possibly be said, has been said already. You guys keep repeating the same crap!!! Before you post, read the other three freakin pages. If what you want to say has been said, do not bother!! Let this thing die already. Some people just cannot be convinced!

Fiver

07-26-2000, 09:34 AM

Bear_Nenno:

Let this thing die already. Some people just cannot be convinced!

At the very, very least, I would hope that the vigor of this debate, and the sharp disagreements among intelligent people, convinces Bigtrout his friend is not a "knucklehead."

I wish Cecil would do a column on this matter and settle it once and for all.

tcburnett

07-26-2000, 03:32 PM

The problem needs to be simplified.

Suppose two identical satellites, traveling in opposite directions at 34mph each (the reduced speed of light) in the same heliocentric orbit crash head-on? By 'head on', we mean the angle of impact for both objects is exactly zero, so any crushing effects would be exactly the same for both.

How would this compare to one satellite having a relative speed of zero and the other bashing into it at 1mph; at 34mph; at 68mph?

Name withheld by request.

margoretep

09-02-2001, 03:18 PM

this is actually the subject of a bet i had with someone. (i am in the 35+35=35 camp.) as an ex-journalist, my first reaction was to look for an authority rather than try to convince my opponent with pop-physics.

below are indications from three pretty good authorities (the Swedish National Road and Transport Research Institute, the U.S. DOT, and the Insurance Institute for Highway Safety).

i pray this will silence the 35+35=70 camp.

1)

Hello Peter

In principle you are right. Imagine putting a sheet of paper between the cars. The pressure on each point of that paper will be the same from both sides and the paper will remain flat like the rigid wall. In practice the cars are not that symmetrical so the stopping distance will be a little longer than when hitting the rigid wall.

Regards

Thomas Turbell

Research Director at the Swedish National Road and Transport Research Institute

2)

U.S. Department of Transportation

http://www.dot.gov/affairs/1997/nht2397.htm

(third paragraph): "Vehicles are crashed so that the entire front goes into a fixed barrier at 35 mph. This crash, which is equivalent to a head-on collision between two identical vehicles, each moving at 35 mph, or with a 70 mph closing speed."

3)

Insurance Institute for Highway Safety

http://www.hldi.org/teachers_guide.pdf

(page 10): "6. Which would be more damaging to your car: having a head-on collision with an identical car traveling at an identical speed or driving head on into the Vehicle Research Center's 320,000 pound deformable crash barrier? Answer: Both crashes produce the same result. Either way the car rapidly decelerates to a stop. In a head-on crash of identical cars traveling at equal speeds, the result is equal impact forces and impact times (according to Newton's Third Law of Motion, and therefore equal changes in momenta).

robby

09-02-2001, 06:02 PM

I've just taught a year of physics (using Serway's Physics for Scientists and Engineers, 5th ed., and Bigtrout is correct.

What I can add to the discussion is an excellent conceptual explanation supplied in the book used by our "low-track" physics course: Hewett's Conceptual Physics, 8th ed. In Hewett's "Next Time Questions" ancillary material, he poses the question:

"Which would be more damaging; driving into a massive concrete wall, or driving at the same speed into a head-on collision with an identical car traveling toward you at the same speed?

Answer: Both cases are equivalent, because either way, your car rapidly decelerates to a dead stop. The dead stop is easy to see when hitting the wall, and a little thought will show the same is true when hitting the car. If the oncoming car were traveling slower, with less momentum, you'd keep going after collision with more "give," and less damage (to you!). But if the oncoming car had more momentum than you, it would keep going and you'd snap into a sudden reverse with greater damage [again, to you]. Identical cars at equal speeds means equal momenta--zero before, zero after collision."

SuperTer

09-02-2001, 08:03 PM

Okay, this is my first post ever and I am about to immortalize myself as the member who knows the least about physics. However, here is my take, if anyone cares to see me give the horse another whack:

I am surprised that the best "real-world" demonstration that you all have come up with is the hand-clapping. Does anyone play pool? I do, and pretty well. Pool is one of the best ways to learn a little about physics. Take the cue ball and roll it into a wall (let us presume it is the magic immovable wall). Watch what the ball does after impact. Now take the cue ball and hit it into a stationary ball (the object ball in poolspeak). Observe how both balls react to the impact. Now take the cue ball in one hand, the object ball in the other, and observe again the reaction of the balls after the collision.

I know we don't know the speed of the cue ball, but at least let's assume that in demo three that both balls are travelling at the same speed. I have no mathmatical revelation to make from this. In fact, I studied Literature, so I am way out of my league. But it seems to me that there is only x amount of energy involved in each impact, and that the impact itself cannot generate energy, but yet the initial energy from all objects involved has to be accounted for. What was that I learned in high school about how energy never disappears, just transfers? In example two, the cue ball not only moves the object ball, but itself continues to move if a normal stroke is used. Only a type of spin put on the ball can cause it to stop on impact and remain stationary. In this circumstance, the object ball moves away from the collision at a higher speed than if the cue ball were allowed to continue on its original path.

Okay, I have exhausted my physics capabilities. Someone who knows what they are talking about should pick this up from here. Now I wonder how many of you are going to seek out a pool hall?

astro

09-02-2001, 08:21 PM

Man this thread has got it goin' on to rise from the dead after over a year in the ground! Even Jesus only managed three days.

robby

09-03-2001, 11:51 AM

Originally posted by SuperTer

Okay, this is my first post ever and I am about to immortalize myself as the member who knows the least about physics. However, here is my take, if anyone cares to see me give the horse another whack:

I am surprised that the best "real-world" demonstration that you all have come up with is the hand-clapping. Does anyone play pool? I do, and pretty well. Pool is one of the best ways to learn a little about physics...

Pool balls closely simulate an elastic collision, in which kinetic energy is conserved. Cars colliding head on are much closer to an inelastic collision (especially if they are stuck together after the collision). In an inelastic collison, kinetic energy is NOT conserved.

For what it's worth, I don't think hand-clapping is the best analogy either. A better analogy is air-hocky pucks wrapped in a band of velcro such that the pucks stick together on collision. If two pucks, initially traveling at the same speed in the opposite direction, collide and stick together, they come to a dead stop. Now put velcro on a wall. If a puck collides with a wall and sticks to the wall, it (obviously) also comes to a dead stop. In each case, looking at one of the pucks, it undergoes the same change in momentum, and thus, the same "damage."

AlainHarvey

02-02-2018, 03:12 PM

Introduction to the Physics Principles of Collisions

The physics of collisions are governed by the laws of momentum.

One of these important laws can be expressed this way:

F x t = m x ∆v

The equation is known as the impulse-momentum change equation.

In a collision, an object experiences a force for a specific amount of time that results in a change in momentum. The result of the force acting for the given amount of time is that the object's mass either speeds up or slows down (or changes direction). The impulse experienced by the object equals the change in momentum of the object. In equation form, F x t = m x Δ v.

In a collision, objects experience an impulse; the impulse causes and is equal to the change in momentum. Consider a football halfback running down the football field and encountering a collision with a defensive back. The collision would change the halfback's speed and thus his momentum.

If there are only two objects involved in the collision, then the momentum change of the individual objects are equal in magnitude and opposite in direction.

In a collision, the impulse experienced by an object is always equal to the momentum change.

One of the most powerful laws in physics is the law of momentum conservation. The law of momentum conservation can be stated as follows.

For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. That is, the momentum lost by object 1 is equal to the momentum gained by object 2.

The above statement tells us that the total momentum of a collection of objects (a system) is conserved - that is, the total amount of momentum is a constant or unchanging value.

Consider a collision between two objects - object 1 and object 2. For such a collision, the forces acting between the two objects are equal in magnitude and opposite in direction (Newton's third law). This statement can be expressed in equation form as follows.

F1 = - F2 (the forces are equal in magnitude and opposite in direction)

The forces act between the two objects for a given amount of time. In some cases, the time is long; in other cases the time is short. Regardless of how long the time is, it can be said that the time that the force acts upon object 1 is equal to the time that the force acts upon object 2. This is merely logical. Forces result from interactions (or contact) between two objects. If object 1 contacts object 2 for 0.050 seconds, then object 2 must be contacting object 1 for the same amount of time (0.050 seconds). As an equation, this can be stated as:

t1 = t2

Since the forces between the two objects are equal in magnitude and opposite in direction, and since the times for which these forces act are equal in magnitude, it follows that the impulses experienced by the two objects are also equal in magnitude and opposite in direction. As an equation, this can be stated as

F1 x t1 =-F2 x t2 (the impulses are equal in magnitude and opposite in direction)

But the impulse experienced by an object is equal to the change in momentum of that object (the impulse-momentum change theorem). Thus, since each object experiences equal and opposite impulses, it follows logically that they must also experience equal and opposite momentum changes. As an equation, this can be stated as

m1 Δv1 = - m2 x Δv2 (the momentum changes are equal in magnitude and opposite in direction)

The above equation is one statement of the law of momentum conservation.

In a collision, the momentum change of object 1 is equal to and opposite of the momentum change of object 2. That is, the momentum lost by object 1 is equal to the momentum gained by object 2. In most collisions between two objects, one object slows down and loses momentum while the other object speeds up and gains momentum. If object 1 loses 75 units of momentum, then object 2 gains 75 units of momentum. Yet, the total momentum of the two objects (object 1 plus object 2) is the same before the collision as it is after the collision. The total momentum of the system (the collection of two objects) is conserved.

A useful analogy for understanding momentum conservation involves a money transaction between two people. Let's refer to the two people as Jack and Jill. Suppose that we were to check the pockets of Jack and Jill before and after the money transaction in order to determine the amount of money that each possesses. Prior to the transaction, Jack possesses $100 and Jill possesses $100. The total amount of money of the two people before the transaction is $200. During the transaction, Jack pays Jill $50 for the given item being bought. There is a transfer of $50 from Jack's pocket to Jill's pocket. Jack has lost $50 and Jill has gained $50. The money lost by Jack is equal to the money gained by Jill. After the transaction, Jack now has $50 in his pocket and Jill has $150 in her pocket. Yet, the total amount of money of the two people after the transaction is $200. The total amount of money (Jack's money plus Jill's money) before the transaction is equal to the total amount of money after the transaction. It could be said that the total amount of money of the system (the collection of two people) is conserved. It is the same before as it is after the transaction.

For any collision occurring in an isolated system, momentum is conserved. The total amount of momentum of the collection of objects in the system is the same before the collision as after the collision. Momentum is conserved for any interaction between two objects occurring in an isolated system. This conservation of momentum can be observed by a total system momentum analysis or by a momentum change analysis.

Scenario Presented by Questioner - Head-on Collision Between Two Vehicles of Identical Masses and Velocities

My understanding of the original question is essentially that if a car traveling at for example, 50 mph collides with an immovable concrete wall, the car will suffer a certain amount of damage. If, however, the same car were to collide with another identical car moving in the opposite direction also at 50 mph, then the resulting head-on collision between the two vehicles would be equivalent to a single vehicle traveling at 50 mph colliding with an immovable cement wall creating what would be equivalent to a collision at 100 mph.

In terms of the physics involved, collisions between objects such as cars are governed by the laws of momentum and energy. When a collision occurs in an isolated system, the total momentum of the system of objects is conserved. Provided that there are no net external forces acting upon the objects, the momentum of all objects before the collision equals the momentum of all objects after the collision. If there are only two objects involved in the collision as here, then the momentum change of the individual objects are equal in magnitude and opposite in direction.

Certain collisions are referred to as elastic collisions. Elastic collisions are collisions in which both momentum and kinetic energy are conserved. The total system kinetic energy before the collision equals the total system kinetic energy after the collision. If total kinetic energy is not conserved, then the collision is referred to as an inelastic collision. The situation described here -- a head-on collision between cars, is an inelastic collision.

A force acting for a given amount of time will change an object's momentum. Put another way, an unbalanced force always accelerates an object - either speeding it up or slowing it down. If the force acts opposite the object's motion, it slows the object down. If a force acts in the same direction as the object's motion, then the force speeds the object up. Either way, a force will change the velocity of an object. And if the velocity of the object is changed, then the momentum of the object is changed.

These concepts are merely an outgrowth of Newton's second law. Newton's second law (Fnet = m x a) stated that the acceleration of an object is directly proportional to the net force acting upon the object and inversely proportional to the mass of the object. When combined with the definition of acceleration (a = change in velocity / time), the following equalities result.

F = m x a

or

F = m x ∆v / t

If both sides of the above equation are multiplied by the quantity t, a new equation results.

F x t = m x ∆v

This equation represents one of two primary principles to be used in the analysis of collisions in physics.

To truly understand the equation, it is important to understand its meaning in words. In words, it could be said that the force times the time equals the mass times the change in velocity. In physics, the quantity Force x time is known as impulse. And since the quantity m•v is the momentum, the quantity mxΔv must be the change in momentum. The equation really says that the:

Impulse = Change in momentum

Let's put some numbers to the described scenario to better understand the physics of a head-on inelastic collision.

Example 1: Inelastic Collision Between Two Identical 1000 kg Cars

Mass of car 1: 1000 kg

Velocity of car: 20 m/s

Momentum of car: 20000 kg*m/s

Mass of car 2: 1000 kg

Velocity of truck: -20 m/s

Momentum of truck: -20000kg*m/s

Note: the velocity and momentum of car 2 are negative because both variables are vector quantities.

In the above example of a collision between two identical car,s the total system momentum is conserved. Before the collision, the momentum of car 1 is +20000 kg*m/s and the momentum of car 2 is -20000 kg*m/s; the total system momentum is 0 kg*m/s. After the collision, the momentum of car 1 is -10000 kg*m/s and the momentum of car 2 is +10000 kg*m/s; the total system momentum is 0 kg*m/s. The total system momentum is conserved. The momentum change of car 1 (-10000 kg*m/s) is equal in magnitude and opposite in direction to the momentum change of car 2 (+10000 kg*m/s).

An analysis of the kinetic energy of the two objects reveals that the total system kinetic energy before the collision is 400000 Joules (200000 J for car 1 plus 200000 J for car 2). After the collision, the total system kinetic energy is 100000 Joules (50000 J for car 1 and 50000 J for car 2). The total kinetic energy before the collision is not equal to the total kinetic energy after the collision. A large portion of the kinetic energy is converted to other forms of energy such as sound energy and thermal energy. A collision in which total system kinetic energy is not conserved is known as an inelastic collision.

In this scenario, what each car independently experiences in the two car collision is exactly the same as what each car would experience in a head-on with an immovable object such as a cement wall or a mountainside.

So, if you now think that two cars possessing similar masses and velocities colliding with one another is equivalent to a single car colliding with fixed wall or the side of a mountain at a comparable velocity, then you are in error. In the original scenario, there are two wrecked cars, not one wrecked car, and the kinetic energy utilized to wreck each of those cars in a collision requires twice the energy required to wreck only one car. (Kinetic energy = 1/2 mass X velocity^2).

Its true that two identical cars in a head-on collision traveling at identical velocities would each experience an equivalent amount of damage to each colliding with a cement wall at the same velocity. This might lull a person into a false sense of security concerning head-on collisions. Let's compare a collision between a truck and a car in the same scenario to dispel that feeling.

Example 2: Inelastic Collision Between a 1000 kg Car and a 3000 kg Truck

The before- and after-collision velocities and momentum are listed below

Mass of car: 1000 kg

Velocity of car: 20 m/s

Momentum of car: 20000 kg*m/s

Mass of truck: 3000 kg

Velocity of truck: -20 m/s

Momentum of truck: -60000kg*m/s

Note: the velocity and momentum of the truck are negative because both variables are vector quantities.

In the above example of a collision between a truck and a car, the total system momentum is conserved. Before the collision, the momentum of the car is +20000 kg*m/s and the momentum of the truck is -60000 kg*m/s; the total system momentum is -40000 kg*m/s. After the collision, the momentum of the car is -10000 kg*m/s and the momentum of the truck is -30 000 kg*m/s; the total system momentum is -40000 kg*m/s. The total system momentum is conserved. The momentum change of the car (-30000 kg*m/s) is equal in magnitude and opposite in direction to the momentum change of the truck (+30000 kg*m/s).

An analysis of the kinetic energy of the two objects reveals that the total system kinetic energy before the collision is 800000 Joules (200000 J for the car plus 600000 J for the truck). After the collision, the total system kinetic energy is 200000 Joules (50000 J for the car and 150000 J for the truck). The total kinetic energy before the collision is not equal to the total kinetic energy after the collision. A large portion of the kinetic energy is converted to other forms of energy such as sound energy and thermal energy. A collision in which total system kinetic energy is not conserved is known as an inelastic collision.

The original question examined whether two identical cars possessing equivalent masses colliding head-on at identical velocities is equivalent to each car individually experiencing a collision in which their individual velocities are combined, thus doubling the velocity at which the collision occurs for each car. This does not occur. Rather, each car experiences the collision at the velocity at which the car is traveling at that instant of impact. The velocities of the cars do not combine. Each car would receive an equivalent amount of damage.

The outcome would be very different if a smaller compact car of less mass collides head-on with a fully loaded semi-truck possessing

a much greater mass traveling at the same velocity as the car but in the opposite direction. The two vehicles would share the kinetic energy involved in the collision evenly but due to their very different masses, the compact car would experience much greater damage, analogous to an insect with relatively low mass colliding with the windshield of a moving semi-truck of much greater mass.

Likewise, according to the equation for impulse (impulse = force X time), the greater the time over which a collision occurs, the correspondingly less peak force there will be upon impact. An example would be running into a haystack versus a fixed brick wall. The collision with the haystack occurs over a longer period of time, correspondingly reducing the force of the impact. In comparison, a collision with a fixed brick wall occurs over a brief time period with a correspondingly larger peak force resulting in s substantial increase in damage.

Any other interpretation is clever, but misleading, wordsmithing.

Relevant formulas for inelastic collisons

Before Collison

Momentum = mass car x velocity car

Kinetic energy = 1/2 mass car x velocity car ^2

Note: This equation reveals that the kinetic energy of an object is directly proportional to the square of its speed. That means that for a twofold increase in speed, the kinetic energy will increase by a factor of four. For a threefold increase in speed, the kinetic energy will increase by a factor of nine. And for a fourfold increase in speed, the kinetic energy will increase by a factor of sixteen. The kinetic energy is dependent upon the square of the speed.

After Collision

momentum = (mass car 1 + mass car 2) x velocity car 2

Kinetic energy = 1/2 (mass car 1 + mass car 2) x velocity^2 of car 2

Ratio of kinetic energies before and after collision:

Kinetic energy final / Kinetic energy initial = mass car 1 / mass car 1 + mass car 2

Fraction of Kinetic Energy Lost in Collision

Kinetic energy initial - kinetic energy final / kinetic energy initial = mass car 2 / mass car 1 + mass car 2

From conservation of momentum:

mass car 1 x velocity car 1 = (mass car 1 + mass car 2) x velocity car 2 yields velocity car 2 = mass car 1 / mass car 1 + mass car 2 x velocity car 1

Experiment

This link is to an animation that portrays the inelastic collision between two 1000-kg cars. The before- and after-collision velocities and momentum are shown in the animation's data tables. Try it!

http://www.physicsclassroom.com/mmedia/momentum/2di.cfm

Hampshire

02-02-2018, 03:31 PM

Holy crap! It took you 16 years to write that?

watchwolf49

02-02-2018, 07:24 PM

Collision management (https://www.youtube.com/watch?v=Mw0riLqQBj0) --- {YouTube 3'27"}

TimeWinder

02-02-2018, 09:35 PM

Holy crap! It took you 16 years to write that?

He's working on planes and treadmills now.

watchwolf49

02-03-2018, 08:07 AM

He's working on planes and treadmills now.

I hope I'm still alive to read that ...

Czarcasm

02-03-2018, 11:36 AM

Mythbusters covered this...and were shocked as everyone else when it turned out the force didn't double.

sich_hinaufwinden

02-03-2018, 07:09 PM

Mythbusters covered this...and were shocked as everyone else when it turned out the force didn't double.

Right, IIRC what they actually found was that the force turned out to go down ever so slightly. I believe they calcuated it as 0.99... repeating.

I remember they did mention this unexpected finding could not be attributed to any limitations on the accuracy of their measurement devices or anything like that.

If memory serves, they weren't able to offer any reasonable explanation for the deviation from theory.

DSYoungEsq

02-04-2018, 10:15 AM

Heh, I read the whole thing, and it seems to me, as is so often the case, that the problem arose out of the difficulty of communicating in seemingly "plain English" simple concepts of Physics. If two people discussing a scenario do not actually agree about what they are saying, it's usually impossible to reach an understanding on the truth of the matter. And that proliferated through the thread, compounded by those who chimed in with an imprecise understanding of Physics. :(

watchwolf49

02-04-2018, 10:36 AM

Of course the force is doubled ... of course there's twice as much work performed ... but in the second case, this work is divided between the two cars ...

The catch in the OP is we are only looking at one car ... and just looking at the one car, we cannot tell whether it hit a brick wall or on-coming traffic ... the amount of work performed on the one car is the same ... assuming we're not using AMC Pacers ...

watchwolf49

02-04-2018, 10:45 AM

Correction ... "but in this second case, the work is divided between the two cars" ... emphasizing the word switch ...

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