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.
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.