4 ms·

The damage is the same for (50mph car / 50 mph car) and (50 mph car / 0 mph wall) because of symmetry. Definitely not the same as (100 mph car / 0 mph wall) bec

The damage is the same for (50mph car / 50 mph car) and (50 mph car / 0 mph wall) because of symmetry. Definitely not the same as (100 mph car / 0 mph wall) because that has the car absorbing 4x the energy. To see this, you can imagine the car/car case as having an invisible wall between them. The wall has no net force so it doesn't move and acts the same as the brick wall despite not being either rigid, massive, or fixed to the ground.

s1artibartfast 2y ago
Strange that you are downvoted, Symmetry is a fair way of looking at it.
- ndriscoll 2y ago
  Because it's wrong. Change coordinates and you can see that 50/50 is the same as 100/0, so the system has 4x the kinetic energy of 50/0 before the collision. Then assuming it requires X energy to crumple a car and assuming the wall is strong enough to not yield at all (so no work is done on it), it requires 2X energy to crumple 2 cars, so there's X less energy available to destroy the passenger in the 2 car scenario vs the wall. So the 2 car scenario is slightly better, but probably not meaningfully so (X will be much less than the kinetic energy of a 100 mph car). The correct symmetry principles to invoke here are conservation of energy and invariance of Newtonian physics under Galilean transformations.
  - s1artibartfast 2y ago
    >Change coordinates and you can see that 50/50 is the same as 100/0, so the system has 4x the kinetic energy of 50/0 before the collision. Do you know how crazy that is? the collision doesn't dissipate more or less kinetic energy depending on the reference plane you pick. If you pick a 100/0 reference frame, the combined cars are moving 50 in one direction after. Therefore, they both only expereince a 50mph change. one from 100 to 50, the other from 0 to -50. your frame doesnt change the collision energy. If that was the case, you could pick a 100,000mph reference frame of an astronaut flying by, and when the cars collide there would be an atomic explosion. to put this another way, the energy of the collision comes from the deltaV before and after. The deltaV is the same in EVERY reference frame you can construct.
    - ndriscoll 2y ago
      Yeah you're right. Late night brain fart wasn't thinking that in the 100mph reference frame obviously they're still moving after the collision. Silly me didn't conserve momentum.
      - s1artibartfast 2y ago
        No worries! I think it is a fun example, mostly because it IS counterintuitive and flies against conventional sayings.
    - MostlyStable 2y ago
      Well, at least his comment caused you to make yours, which was the first of this conversation (and the similar one that occurred in the comments on the article) that made the difference in the two situations clear to me. So he may have been wrong, but it was still useful!
      - s1artibartfast 2y ago
        Happy to help you both. It is fun and somewhat counterintuitive given almost everyone gets told the opposite.
  - kazinator 2y ago
    50/50 is the same as 100/0 between two cars. 50/50 is not the same as 100/0 between car and an ideal wall: an immovable barrier of practically infinite inertial mass that suffers no damage. car/car 50/50 is the same as car/ideal-barrier 50.

s1artibartfast 2y ago

Strange that you are downvoted, Symmetry is a fair way of looking at it.

ndriscoll 2y ago
Because it's wrong. Change coordinates and you can see that 50/50 is the same as 100/0, so the system has 4x the kinetic energy of 50/0 before the collision. Then assuming it requires X energy to crumple a car and assuming the wall is strong enough to not yield at all (so no work is done on it), it requires 2X energy to crumple 2 cars, so there's X less energy available to destroy the passenger in the 2 car scenario vs the wall. So the 2 car scenario is slightly better, but probably not meaningfully so (X will be much less than the kinetic energy of a 100 mph car). The correct symmetry principles to invoke here are conservation of energy and invariance of Newtonian physics under Galilean transformations.
- s1artibartfast 2y ago
  >Change coordinates and you can see that 50/50 is the same as 100/0, so the system has 4x the kinetic energy of 50/0 before the collision. Do you know how crazy that is? the collision doesn't dissipate more or less kinetic energy depending on the reference plane you pick. If you pick a 100/0 reference frame, the combined cars are moving 50 in one direction after. Therefore, they both only expereince a 50mph change. one from 100 to 50, the other from 0 to -50. your frame doesnt change the collision energy. If that was the case, you could pick a 100,000mph reference frame of an astronaut flying by, and when the cars collide there would be an atomic explosion. to put this another way, the energy of the collision comes from the deltaV before and after. The deltaV is the same in EVERY reference frame you can construct.
  - ndriscoll 2y ago
    Yeah you're right. Late night brain fart wasn't thinking that in the 100mph reference frame obviously they're still moving after the collision. Silly me didn't conserve momentum.
    - s1artibartfast 2y ago
      No worries! I think it is a fun example, mostly because it IS counterintuitive and flies against conventional sayings.
  - MostlyStable 2y ago
    Well, at least his comment caused you to make yours, which was the first of this conversation (and the similar one that occurred in the comments on the article) that made the difference in the two situations clear to me. So he may have been wrong, but it was still useful!
    - s1artibartfast 2y ago
      Happy to help you both. It is fun and somewhat counterintuitive given almost everyone gets told the opposite.
- kazinator 2y ago
  50/50 is the same as 100/0 between two cars. 50/50 is not the same as 100/0 between car and an ideal wall: an immovable barrier of practically infinite inertial mass that suffers no damage. car/car 50/50 is the same as car/ideal-barrier 50.