Now I agree with Lucky. I wonder what question I was trying to answer earlier? Obviously not the one I thought I read.
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?
(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?
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.
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!
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.
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.
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.
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.
But you have two cars crumpling, so you will have twice the crumple distance. This exactly cancels the factor of two in speed.
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?
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.
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:
- 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.
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.
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:
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.
Zwaldd said:
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?
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.
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. 
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=½mv[sup]2[/sup]
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 ½mv[sup]2[/sup] 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.