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60 plus 60 or 120 into a wall

Aren’t you forgetting that a wall is vitually rigid whereas the front ends of cars

absorb some of the force of impact by collapsing. Thus, relatively speaking,

the crashing of 2 cars is twice as “gradual”

as one car crashing into a rigid wall.

Wolfgang called in and pointed out that click and clack were wrong.

The way I explain it is that if the cars are identical, nothing proceeds beyond the plane where the two vehicles meet. One car pushes just as hard as the other car pushes back. Hence, you end up with two crumpled masses of steel and plastic which meet at a plane. That sounds exactly like what would happen if one car hit a wall.

Not exactly. There is more energy in the two cars hitting each other.

I’m still left scratching my head even after Wolfgang’s explanation. The change in momentum is the same whether you hit an immovable wall, or another car of equal mass. But wouldn’t your deceleration be 2x as fast in the 60+60MPH head-on collision? (and therefore much more, um, injurious?)

Rather than starting a new thread, why not read the hundreds of postings on “60 miles + 60 miles is not force, it’s only the velocity vector”? Please don’t duplicate all that discussion here.

Not exactly. There is more energy in the two cars hitting each other.

And each absorbers 50% of the total energy by deformation or other methods, while the wall in the other case does not “absorb” any of the energy as it does not deform or do anything to absorb the energy (assuming the impossible immoveable wall.)

Your friend Wolfgang as a physicist uses an idealized model to conclude that there is no difference between hitting a wall at 60 mph vs. hitting another car coming head-on also at 60 mph. Adding a bit of realism would tell you the following: If the choice the motorist makes is between the two alternatives, I?d take the second one: hit the other car.

Deceleration rate is the crucial factor here, and that would depend on the kind of car you are driving and the car you are hitting. If the cars are equal in their ability to absorb an impact, then it would not matter. However, if what you are hitting is better able to absorb the shock in the sense that it absorbs more energy in the impact than a wall, perhaps collapsing more slowly than your own, then you would be better off with this outcome (if the word ?better? can be used here.) This assumes that the wall is solid, like 3 feet of concrete or stone.
In advance, unless you can tell what?s coming at you, you would be better off taking the risk of hitting the car. There is a chance that you would be hurt less. What odds you need to make that choice is another, even more complex subject. However, if you see that a truck is coming at you, take the wall. Unless, of course, there?s a ditch you can aim for.

The crash-plane of two identical bodies, at identical speeds, will have absolute rigidity for each body; if not, two bodies would be occupying the same plane at the same time. In the regular world, this isn’t going to happen. It may be better to hit the wall, since the wall may “give” a little.

It seems to me that there is more energy in the one car going twice as fast since kinetic energy is proportional to the velocity squared. So the car going 120 has four times the energy that either car going 60 mph. Hence it has twice the energy of the two 60 mph cars together. You can visuallize that the two slow cars colliding head-on will crunch down to a mutual dead stop but each from 60 mph. The 120 mph car hitting an immovable wall will not only crunch down from a twice the speed to the “dead stop” point but will bounce back (momentum preservation). This has to be a much more violent crash and is not equivalent to the head-on at half the speed.

As with all physics questions, Wolfgang (as must we) assumes that all things are equal and it’s an ideal world. The cars are the same mass, same speed, would crumble in exactly the same way, etc.

Now picture the two cars hitting, and picture the point of impact. It would be like a mirror image. The two cars would crumble in the front. People would fly forward, etc. On each side of the “mirror”, or point of impact.

Now the only difference is the mirror can also be a brick wall. The two situations are identical in an ideal world of hypothetical Physics problems.

However, my family posits that Wolfgang is completely wrong.

Don’t get me wrong. Wolfgang’s physics was spot on. You asked last week if you were “better off” hitting a wall or an oncoming car at the same speed as you. The physics answer Wolfgang clarified on this week’s show is right. There is no difference, momentum-wise, as hitting a wall or a moving car (with the same mass and speed as you) results in basically the same thing - an impact with a stationary object.

However, you didn’t ask if there would be more damage, or if you’d move forward or backward, you asked which would make you “better off.”

We (my family and I) contend that Wolfgang (clearly a physicist, and not a philosopher) is incorrect. You would clearly be “better off” if you hit a wall.

Mostly because it’s better than hitting another person’s car, perhaps injuring that person, and certainly damaging their car. How bad would that make you feel? What kind of insurance mess would you have to deal with? Will you be sued?

Hitting a wall, the most harm you do is to yourself and your car. (And perhaps an inanimate wall.) Hitting a wall, the wall owner may get angry, but no third party gets hurt because you collided with them.

So unless there are people lined off in front of the wall, you are far “better off” hitting the wall.

Sean Huxter and Family.

If Wolfgang is right high-energy particle physicists are fooling themselves by using oppositely directed beams of elementary particles. It would be enough to direct one beam at a fixed target.

Here?s a thought experiment. Instead of running a car at 60 mph into a wall, sit in a parked car and have a second car of equal mass hit it head on at 60 mph. Compare that with the two cars hitting head on, each at 60 mph.

What Wolfgang overlooks is that it is the relative velocity that is involved, not the velocity with respect to the stationary earth. The relative velocity is 120 mph. Wolfgang comes close to a correct undestanding when he admits that the relative masses also come into consideration. As an example, at 60 mph, hitting a wall, the energy dissipated may “only” crumple the front end; with head-on collision the front end will be crumpled and the engine driven back into the passenger compartment.

Mv goes to zero, perhaps, in either case. But in one v=60 mph, in the other it is 1120. There are different inertial frames in the two cases.

I am a retired computer engineer. Years ago when I was an ?electronic engineer? my boss used to say that physics was a special case of electronics. That is why I dare contradict a Harvard physics professor.

No. think of any single part, like maybe a side mirror. That mirror might come off and flow towards the like (but mirror image) mirror coming from the other car. They would strike each other each traveling at 60 mph, each stopping totally as it contacted the other mirror. As far as the mirror knows it is hitting a brick wall since it is stopping form 60 to zero the same as it would hitting a brick wall.

In real life you don’t have exact mirror image cars, so some parts will go past the contact plane so that will slightly reduce the total damage since some parts will loose some energy by wind resistance or sliding on grass etc.

You are not considering the same system of particles as Wolfgang. See the post by Tyler (the caller) in the other thread on this subject. It is all about how the question is phrased. To his detriment, Wolfgang does not restate the question he is addressing properly and has generated more confusion.

What needs to be understood in this problem is that energy is not conserved in this system. I am not saying energy is destroyed, but rather released in the form of heat. It is complicated to predict how much heat will be generated, but we can assume momentum will be conserved, which is how Wolfgang approaches the problem (correctly, for the type of analysis done here).

As for your thought experiment, it is important to consider whether the parked car is parked and immobile, or simply stopped and in neutral. If it is parked, in order to be consistent, we need to assume the parked car won’t crumple or skid, and the problem is identical to the wall problem. If in neutral, momentum would be conserved, so depending on the type of collision (think of two billiard balls hitting), the end result could be the moving car is stopped and the neutral car is going 60 mph backward, both cars are moving together at 30 mph, or something in between.

When two cars collide head-on as in the question, they evenly share the energy of the collision, as if each were either side of an immovable wall.

Sorry, my friend. This is a high school level problem, and the correct answer is 60 into wall is equivalent to 60 into an identical car doing 60 coming at you.

The problem with your particle physics analogy is there is no such thing a a fixed, rigid target for particles at the energy levels being studied. What material would you suggest the fixed target be constructed of so that it would not be partially penetrated by the high-energy particle? Kyrptonite perhaps?

As an engineer rather than a physicist, I can’t help but point out that in real life, I’d rather hit the wall (and not just to save the occupants in the other car). If you did the actual experiment with identical cars, etc etc, the two cars hitting each other head on each have half the total energy and absorb half the total energy. With one car hitting a wall, you have the same energy (per car), but a real wall will always absorb some of it. So the car hitting the wall will have less energy to absorb, and will thus always be better off.

Depending on the wall, it may not be much better off. (Bales of cotton are better than reinforced concrete.)

Well, this is a physics question, and the assumption is the wall is completely rigid. In real life, forgetting the fact that you are involving others in the crash, I’ll take the head-on, because it won’t be exactly head-on so the collision will be reduced by some small amount. Of course it is silly to talk about which one of the assumptions breaks down the most in the real world, this was a theoretical question.