Head-on Collision Physics

[matt]

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!

[matt]

… table TO pound on!

[zut]

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]

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]

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]

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]

*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]

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]

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]

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]

*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]

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]

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]

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]

*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]

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]

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]

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]

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]

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.

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).