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
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.
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…
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.
sailor writes:
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Your intuition is correct. See my earlier post.
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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.
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.
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.
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.
>> 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.
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.
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?
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.
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.
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.
<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:
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:
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).
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.
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!
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.
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.
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?