The mass of the wall is required to determine which system has more energy.
I believe the correct approach is to look at it from the perspective of the object that you in car A are striking. Taken from the perspective of the wall, car A strikes it at 100mph. From the perspective of car B, car A also strikes it with the same speed (and kinetic energy).
IF the two cars have equal mass, this will also be the impact energy on car A. And that’s what we’re trying to find for both cases.
The most convenient reference frame for colliding objects is the combined center of mass of those objects. It isn't affected by the collision, and for perfectly inelastic collisions the objects will come to rest with respect to that frame. This makes the math a lot simpler that working in some other frame.
It goes from sitting still to going backwards at 50pmh, for a net change of 50mph.
The difference of velocity between cars 100mph in every reference frame you can pick. Each of the cars always experience a change of 50mph from the collision, no matter what reference frame you pick.
People are confused because the pick a reference frame for car A, but don't hold it constant for their calculations, as required by physics.
The wall approaches at 100mph and, in a totally elastic collision, it recedes at 100mph, as does car B.
>No, if you’re taking car A as a reference point it behaves as if stationary before and after collision - it’s everything else that moves.
If that were true, I could set a reference plan as a bullet in a gun, and when I fired it everyone in the world dies from rapid acceleration as their kinetic energy goes from 0 to 3000mph.
you change 50 if you hit a unmoving wall.
IF you start at 0 or 100mph, you also change 50.
Each car approaches the impact point at 50mph, each car recedes at 50 in the opposite direction after a perfect collision, so each has a dv of 100.
And if my initial reference frame is car A, travelling at 50, then before impact car A appears to be travelling at 0 while car A is approaching at 100. After collision both cars experience a dv of 100, so car A appears to be receding at 100mph while B appears stationary.
For the wall, the car approaches at 100 and recedes at 100 for a delta of 200mph. But in any case the velocity delta doesn’t affect the energy, it just has to be conserved.
As I’ve said elsewhere we are trying to find the collision energy in the moment of impact from the perspective of car A (the question is whether you driving into a wall the same as you driving into another opposite car at half the speed). For that you have to consider a frame of reference relevant to car A, and not a stationary reference that is not following either car.
Together, they have the energy of two cars at 50 each hitting a wall, and this energy is split between the two cars.
If somehow ALL of the kinetic energy was absorbed only by one car , it still wouldnt be equal to 100mph, do the the square.
Together, two cars traveling at 50mph have the same energy as one car traveling at 70mph.
The only case when two objects colliding at 50mph is like 100mph is if the final reference frame is still traveling at 50mph. a Car going 50pmh hitting a train going 50mph would be like the car hitting a wall at 100mph (net velocity change of 100mph.