5 ms·

It falls straight down relative to the train, regardless of the train's speed. If you're standing next to the stationary brick wall, it falls straight down rel

It falls straight down relative to the train, regardless of the train's speed. If you're standing next to the stationary brick wall, it falls straight down relative to the wall. To me this is a clear demonstration that motion is relative, therefore the two collision scenarios ARE equivalent. Imagine that the two cars are in space. Are they both moving at 50MPH towards each other? Or is one stationary and one moving at 100MPH, like the wall scenario? Or perhaps moving the same direction, but one at 100MPH and the other at 200MPH, about to catch up to the first one? The answer is that it doesn't matter, because the only difference between these scenarios is the reference frame. The collision is the same in all of them.

_dark_matter_ 2y ago
One big difference is the car has a crumple zone to absorb some impact, but the brick wall does not.
- roarcher 2y ago
  In reality yes, but I assume the thought experiment is meant to take place in Spherical Cow Land where all colliding objects can be considered the same. I'm also not sure how relevant the crumple zones were when this was first thought up, as cars haven't always had them.
- AlexandrB 2y ago
  The simple fix to that is have the car hitting another, stationary car at 100mph. It's still a horrendous collision despite the 2x crumple zones.
- s1artibartfast 2y ago
  both are crumpling, but the line between the cars does not change and acts as a wall. Imagine compressing two springs in a line vs one against a wall.
hn_throwaway_99 2y ago
This is absolutely incorrect. I could repeat myself but see https://news.ycombinator.com/item?id=40628932 https://news.ycombinator.com/item?id=40628932.
- deleted 2y ago
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- deleted 2y ago
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- roarcher 2y ago
  I posted a long reply about how your math was wrong, and it is, but the correction I wrote was wrong as well. The correct way to solve this problem, which I do not have time to go through at the moment, can be found here: https://phys.libretexts.org/Bookshelves/University_Physics/Physics_(Boundless)/7%3A_Linear_Momentum_and_Collisions/7.3%3A_Collisions#:~:text=An%20elastic%20collision%20is%20a,which%20kinetic%20energy%20is%20conserved.&text=An%20elastic%20collision%20is%20a,the%20bodies%20after%20the%20collision https://phys.libretexts.org/Bookshelves/University_Physics/P.... But I maintain that whatever the mathematical procedure, it is absolutely impossible for the stationary-ness of one of the objects to affect the outcome of the collision, assuming we're ignoring things like static friction, air resistance, etc. This is a simple matter of reference frames. If you disagree, put the two objects in space as I suggested and tell me which one is "stationary" and how that could possibly affect the outcome of the collision.
  - deleted 2y ago
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  - AlexandrB 2y ago
    This tripped me up too, but one of the StackOverflow answers pointed out something I hadn't considered. If the cars are moving towards each other and you choose one of them as the reference frame. After they collide and "stop" on the road the reference frame keeps moving (because we're using Newtonian physics and "inertial" reference frames[1] here). This means the cars are still "moving" relative to the reference frame and this kinetic energy needs to be subtracted from the energy of the collision since it's residual kinetic energy in the chosen reference frame (and the velocity vectors of the motion before and after collision point in the same direction relative to the reference frame as well). This is in contrast to car vs. brick wall where choosing the brick wall as a reference frame means that the car is not moving relative to the reference frame after the collision and the kinetic energy is 0. And if you think about it further, the difference is not that you're choosing one object as still and one as moving but that in the car vs. car case, the second car takes on some of the energy from the collision, whereas the wall does not. This makes some intuitive sense if you imagine this collision happening. Forget all the crumple zone stuff - a car hitting a stationary car is going to make that car move - and if you ignore friction and assume the cars stick together the two cars will be moving in the same direction as the initially moving car, just slower. This means not all the initial kinetic energy from the moving car was "used up" in the collision. [1] https://en.wikipedia.org/wiki/Inertial_frame_of_reference https://en.wikipedia.org/wiki/Inertial_frame_of_reference

_dark_matter_ 2y ago

One big difference is the car has a crumple zone to absorb some impact, but the brick wall does not.

roarcher 2y ago
In reality yes, but I assume the thought experiment is meant to take place in Spherical Cow Land where all colliding objects can be considered the same. I'm also not sure how relevant the crumple zones were when this was first thought up, as cars haven't always had them.
AlexandrB 2y ago
The simple fix to that is have the car hitting another, stationary car at 100mph. It's still a horrendous collision despite the 2x crumple zones.
s1artibartfast 2y ago
both are crumpling, but the line between the cars does not change and acts as a wall. Imagine compressing two springs in a line vs one against a wall.

hn_throwaway_99 2y ago

This is absolutely incorrect. I could repeat myself but see https://news.ycombinator.com/item?id=40628932 https://news.ycombinator.com/item?id=40628932.

deleted 2y ago
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deleted 2y ago
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roarcher 2y ago
I posted a long reply about how your math was wrong, and it is, but the correction I wrote was wrong as well. The correct way to solve this problem, which I do not have time to go through at the moment, can be found here: https://phys.libretexts.org/Bookshelves/University_Physics/Physics_(Boundless)/7%3A_Linear_Momentum_and_Collisions/7.3%3A_Collisions#:~:text=An%20elastic%20collision%20is%20a,which%20kinetic%20energy%20is%20conserved.&text=An%20elastic%20collision%20is%20a,the%20bodies%20after%20the%20collision https://phys.libretexts.org/Bookshelves/University_Physics/P.... But I maintain that whatever the mathematical procedure, it is absolutely impossible for the stationary-ness of one of the objects to affect the outcome of the collision, assuming we're ignoring things like static friction, air resistance, etc. This is a simple matter of reference frames. If you disagree, put the two objects in space as I suggested and tell me which one is "stationary" and how that could possibly affect the outcome of the collision.
- deleted 2y ago
  [deleted]
- AlexandrB 2y ago
  This tripped me up too, but one of the StackOverflow answers pointed out something I hadn't considered. If the cars are moving towards each other and you choose one of them as the reference frame. After they collide and "stop" on the road the reference frame keeps moving (because we're using Newtonian physics and "inertial" reference frames[1] here). This means the cars are still "moving" relative to the reference frame and this kinetic energy needs to be subtracted from the energy of the collision since it's residual kinetic energy in the chosen reference frame (and the velocity vectors of the motion before and after collision point in the same direction relative to the reference frame as well). This is in contrast to car vs. brick wall where choosing the brick wall as a reference frame means that the car is not moving relative to the reference frame after the collision and the kinetic energy is 0. And if you think about it further, the difference is not that you're choosing one object as still and one as moving but that in the car vs. car case, the second car takes on some of the energy from the collision, whereas the wall does not. This makes some intuitive sense if you imagine this collision happening. Forget all the crumple zone stuff - a car hitting a stationary car is going to make that car move - and if you ignore friction and assume the cars stick together the two cars will be moving in the same direction as the initially moving car, just slower. This means not all the initial kinetic energy from the moving car was "used up" in the collision. [1] https://en.wikipedia.org/wiki/Inertial_frame_of_reference https://en.wikipedia.org/wiki/Inertial_frame_of_reference