3 ms·

From the reference frame of car A, it is as if it is stationary and car B - or the wall - is travelling towards it at 100mph. The mass of the wall is required

From the reference frame of car A, it is as if it is stationary and car B - or the wall - is travelling towards it at 100mph. The mass of the wall is required to determine which system has more energy. Or, take it from the perspective of car B or the wall. In both cases, car A is approaching with a relative speed of 100mph.

tbrownaw 2y ago
The wall is indestructible and anchored to the earth, so it has effectively infinite mass.
- teamonkey 2y ago
  An infinite mass is impossible. But in any case it just limits the usefulness of that frame. 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.
  - tbrownaw 2y ago
    > I believe the correct approach is to look at it from the perspective of the object that you in car A are striking. 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.
s1artibartfast 2y ago
In the reference frame you propose, car A is traveling 50mph after the collision. 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.
- teamonkey 2y ago
  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. The wall approaches at 100mph and, in a totally elastic collision, it recedes at 100mph, as does car B.
  - s1artibartfast 2y ago
    you cant change reference frames mid problem to follow an object when calculating kinetic energy. The frame is the original velocity of car A. >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.
    - teamonkey 2y ago
      You can compare frames of reference before and after collision and check the relative speeds of the entities involved.
      - s1artibartfast 2y ago
        Yes, and if you do that, The change in speed is always the same 50mph different. you change 50 if you hit a unmoving wall. IF you start at 0 or 100mph, you also change 50.
        teamonkey 2y ago
        I don’t get the 50mph change; the change in velocity for each car is going to be 100mph, regardless of the frame of reference, no? 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.

tbrownaw 2y ago

The wall is indestructible and anchored to the earth, so it has effectively infinite mass.

teamonkey 2y ago
An infinite mass is impossible. But in any case it just limits the usefulness of that frame. 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.
- tbrownaw 2y ago
  > I believe the correct approach is to look at it from the perspective of the object that you in car A are striking. 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.

s1artibartfast 2y ago

In the reference frame you propose, car A is traveling 50mph after the collision. 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.

teamonkey 2y ago
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. The wall approaches at 100mph and, in a totally elastic collision, it recedes at 100mph, as does car B.
- s1artibartfast 2y ago
  you cant change reference frames mid problem to follow an object when calculating kinetic energy. The frame is the original velocity of car A. >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.
  - teamonkey 2y ago
    You can compare frames of reference before and after collision and check the relative speeds of the entities involved.
    - s1artibartfast 2y ago
      Yes, and if you do that, The change in speed is always the same 50mph different. you change 50 if you hit a unmoving wall. IF you start at 0 or 100mph, you also change 50.
      - teamonkey 2y ago
        I don’t get the 50mph change; the change in velocity for each car is going to be 100mph, regardless of the frame of reference, no? 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.