That would only be the case if Friedman, even after getting his PhD in physics, still continued to disagree with that teacher and that textbook.
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Each car crumples against the dotted line, not passing through it at all.
(In our idealized example with a perfectly mirror-like collision in which both cars crumple identically, it behaves like an ideal brick wall: the worst kind.)
Hitting an ideal brick wall at 50 is far less severe than doing so at 100.
If the collision is perfectly inelastic, then in the two car case, the energy dissipated is 2mv² where v is 50 mph. In in the one car case where it is going double (2v) it the energy is m(2v)² which is 4mv². We assume the brick wall has no kinetic energy to contribute. We also assume the brick wall is not damaged so the 4mv² is absorbed by one car. Whereas in the two car case 2mv² energy is absorbed by two cars, so mv² each.
Thus, from the point of view of one car, the 100 mph brick wall collision is 4 times more energetic than the 50 mph head-on collision. From that alone you can intuit that head-on is same as brick wall, since the damage scales 4 times with a doubling in velocity.
(Now yes, hitting an identical parked car at 100 mph would me more similar to the head-on collision. A parked car is not a brick wall. It will move in the direction of the collision in addition to crumpling. Books for drivers should probably not talk about either brick walls or parked cars, but just convey that kinetic energy follows a square law, quadrupling when speed doubles.)
Since in a head on, they are both changing speed then the deceleration must be half the combined relative deceleration.
The way I was imagining it was as if the car collides with another car going 50mph in a head on collision and the other car doesn't slow down due to the collision.
So it appears that it is I who am confidently wrong about physics (although I still hold that Milton Friedman is confidently wrong about economics!).
Two approximately round space rocks of exactly equal size and density have a center of mass that is halfway between them. If one is moving 40 mph (according to the observer's frame of reference) and the other is catching up to it in exactly the same direction at 50 mph, then what it means is that the center of mass between the two rocks is moving at 45 mph. If the rocks collide in such a way that they clump together (inelastic collision) and no pieces fly off, the combined double rock will thereafter be moving at 45 mph.
Vehicle collisions are more complicated because of cars skidding and rolling. In the milliseconds after the main impact though, those effects don't matter that much, I think. E.g. if you hit a parked car that is in neutral with no parking brake, versus one that has its transmission and brakes locked down, and this happens at 100 mph, I suspect those factors hardly make a difference.
Ummmm ... what? I mean, sure if you assume that the dynamics of a brick wall and a car are different, then they won't be identical but if you just take a parked car and a brick wall to both be "immovable objects at rest" which for all intents and purposes is true for the driver, then I don't see how this statement could possibly be true.
If you have 1m of hood to work with, you're decelerating 100km/h to 0km/h over 1m in the wall case.
In the hood-to-hood case, depending on reference frame, you're either decelerating 50km/h to 0km/h over 1m or 100km/h to 0km/h over 2m (both hoods).
Either way it should be trivially obvious that running into a wall is going to be a much more violent experience. Likely the wall crumpling up the front of your car won't be enough to bring you to a stop before you personally impact something.
Then you also have the force exerted by their car attempting to accelerate you backwards.
For each car, there are 2 forces: the force that you exert due to your trying to accelerate them in your direction of travel and the opposing force.
What you're saying is the same as saying that if you fall from the roof of an elevator at rest, the effect is the same as if you fall from the roof of an elevator that travels upwards at the same time as you fall. That's obviously not true, but the physics you need to prove it are F = ma and Newton's 3rd law.
For Car A:
- Force of Car B accelerating Car A opposite to direction of travel
- Force of Car B opposing force of Car A trying to accelerate Car B
And vice versa for Car B.
EDIT: Like if you hit a wall, the wall feels the force of your car. The force that crushes the car is the force opposing that, which is the wall acting on the car.
EDIT2 (because I can't reply): the force acting on the driver will be the sum of the force of Car B acting on Car A opposing the force of Car A acting on Car B plus the force of Car B acting on Car A. There are 2 forces in each direction of travel that are exerted by the cars. The force felt by the driver is the sum of these 2 forces, exerted by the dashboard on the driver's face. If the car hits a brick wall, the only force opposing the direction of travel of the car is the force of the wall acting on the car, which opposes the force of the car acting on the wall. As such the force felt by the driver of the dashboard acting on their face is halved.
EDIT3 (still because I can’t reply): the dynamics of the materials is not the issue. Yes bricks behave differently than cars, but that’s not the basis of the author’s argument.
> EDIT2 (because I can't reply): the force acting on the driver will be the sum of the force of Car B acting on Car A opposing the force of Car A acting on Car B plus the force of Car B acting on Car A. There are 2 forces in each direction of travel that are exerted by the cars. The force felt by the driver is the sum of these 2 forces, exerted by the dashboard on the driver's face. If the car hits a brick wall, the only force opposing the direction of travel of the car is the force of the wall acting on the car, which opposes the force of the car acting on the wall. As such the force felt by the driver of the dashboard acting on their face is halved.
If this was how it worked, crumple zones would do nothing. Next you're going to model the airbag as a rigid body too? Ouch.
Correct me if I'm wrong, but I think you've added non existent extra forces.
In the car/car situation, there are still only two forces - each the reaction of the other depending on which cars point of view you're taking. There aren't two separate forces to sum.
Each car experiences a decelerating force - the reaction of which is the decelerating force for the other car and the reaction of that is the original deceleration force of the original car.
If the wall is also moving towards the car then the force felt by the car is the sum of the reaction force it would have felt if the wall were stationary plus the force due to the wall trying to accelerate the car in the opposite direction.
In the author’s example is says that 2 cars crashing into each other is the equivalent of two cars crashing into a wall, and it’s false.
A parked car will move when struck. The combination of the two cars continues to have kinetic energy, which is lost by friction.
https://news.ycombinator.com/item?id=40629971
It is entirely unassailable; all you can argue with is the realism of my assumptions. (Real head-on collisions won't be precise mirror images.)
When two identical objects are moving toward each other with a relative speed v, the collision is like hitting an ideal barrier at speed v/2.
Okay yep I get it now. The 4mv2 did it for me mathematically but I still couldn't get it intuitively until I imagined a car going 50mph rear ending a car going 40mph. If they wind up going the same speed as a result of the collision then the deceleration felt by the car in the back would only be 10mph if the car in front didn't change speed because it was like, an ocean liner. If the cars are the same size and the collision is perfect and all that then the car in front would gain 5mph and the car behind would lose 5mph.
Since in a head on, they are both changing speed then the deceleration must be half the combined relative deceleration.
The way I was imagining it was as if the car collides with another car going 50mph in a head on collision and the other car doesn't slow down due to the collision.
So it appears that it is I who am confidently wrong about physics (although I still hold that Milton Friedman is confidently wrong about economics!).
As a driver I'd rather run into another car than a brick wall. No amount of simplified textbook physics is going to convince me otherwise.
On a related school anecdote: A test problem once asked us to calculate the length/angle a ladder would have to be extended from a firefighting vehicle to reach a window, given height of the window and distance of the firefighting vehicle to the house. I had about twenty minutes left and was a bit of a cheeky student, so I explained how I could not answer that question since I did not know the height at which the ladder originates from the vehicle. That got me full points since I had a great teacher.
That said, you have to be a bit of a jackass to intentionally miss the point in either case.
Smarts kids probably often have resentment when they first discover their teachers/elders whom they hold up to a naturally realistic high regard are less inquisitive or intellectually rigorous than they are (or hope to be). That sort of thing has a way of sticking in your brain.