4 ms·

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

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
- LegionMammal978 2y ago
  The difference is whether the collision is allowed to make the wall move. A 100-mph car won't be halted by a physical brick wall, but will just bust right through it. Afterward, both the car and wall remnants will be moving at 50 mph (minus however much energy it takes to tear the section of wall from the ground). In that case, the change in kinetic energy will be not much more than in the car/car case. For a truly immovable wall, the ground has to exert a lot of force to keep it in place, which goes into the car, causing the entire discrepancy.
- roarcher 2y ago
  Kinetic energy is relative too. Two cars moving at 100MPH in the same direction (relative to the reference frame) have no kinetic energy relative to each other. You're right that if you match your reference frame to one car, that car becomes "stationary" until the collision, but after the collision it is moving relative to the reference frame. But that's the key--it now has kinetic energy relative to the reference frame, but not to the other car. As far as the pair of cars is concerned, all of the kinetic energy between them has been expended, because they are now moving together. Of course this is assuming a scenario where the cars crumple into a single mass, i.e. they don't "bounce". I think part of the confusion in these comments stems from the problem being poorly defined. Some people are considering crumple zones, others are considering whether a brick wall is movable and still others replace it with another car for simplicity. I do think the driver's ed manual is correct in some Spherical Cow sense, but I don't think its authors ever intended for the problem to stand up to Hacker News scrutiny. It's quite funny to me how innocuous things like this can generate these "unladen swallow"-type discussions on here, even though I'm participating.
  - s1artibartfast 2y ago
    Kinetic energy is relative to the frame of reference, but Changes in kinetic energy is never relative to fixed reference frame. I think people dont realize they are changing their reference frames, and getting insane results like the amount of kinnetic energy release in the collision depend on what frame you pick. two crumple zones don't matter for head on collision, which is another thing that trips people up.
hn_throwaway_99 2y ago
Your highlighted link is the wrong model to use here. The part you highlighted in your link is about elastic collisions in one dimension. Even with "spherical cow" physics, collisions like this are nearly always modeled as perfectly inelastic collisions. You don't model the cars bouncing off each other going back with the same speed that they were originally going - you model then as crashing into each other and stopping, with all of that kinetic energy converted to heat, a perfectly inelastic collision. As per your "stationary" objections, the issue is that this is modeled as a very sturdy wall anchored to the Earth. Yes, the Earth will "recoil" some, but for all intents and purposes it can be modelled as completely stationary because its mass is so much greater than the car.
- roarcher 2y ago
  I don't know what you mean--there's nothing highlighted in that link when I open it. Part of it is about elastic collisions and part is about inelastic. > the issue is that this is modeled as a very sturdy wall anchored to the Earth. That's not what you modeled in the comment you linked. Your math (incorrectly) attempts to describe a car colliding with another car of equal mass. But if you're considering a "very sturdy wall anchored to the earth", then you effectively have a car colliding with the earth. Either one is fine with me, but pick one. It grinds my gears when you authoritatively call me "absolutely incorrect", "correct" me with an incorrect answer of your own, and then try to claim you were modeling something entirely different all along.
  - hn_throwaway_99 2y ago
    Dude, you seriously need to work on your reading comprehension, this conversation is like talking to a dining room table. The link I posted, https://news.ycombinator.com/item?id=40628932 https://news.ycombinator.com/item?id=40628932, clearly showed the kinetic energy math for both a car hitting a stationary wall at speed 2X, and 2 cars hitting each other each with speed X. This is basic high school math of inelastic collisions.
    - roarcher 2y ago

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

LegionMammal978 2y ago
The difference is whether the collision is allowed to make the wall move. A 100-mph car won't be halted by a physical brick wall, but will just bust right through it. Afterward, both the car and wall remnants will be moving at 50 mph (minus however much energy it takes to tear the section of wall from the ground). In that case, the change in kinetic energy will be not much more than in the car/car case. For a truly immovable wall, the ground has to exert a lot of force to keep it in place, which goes into the car, causing the entire discrepancy.
roarcher 2y ago
Kinetic energy is relative too. Two cars moving at 100MPH in the same direction (relative to the reference frame) have no kinetic energy relative to each other. You're right that if you match your reference frame to one car, that car becomes "stationary" until the collision, but after the collision it is moving relative to the reference frame. But that's the key--it now has kinetic energy relative to the reference frame, but not to the other car. As far as the pair of cars is concerned, all of the kinetic energy between them has been expended, because they are now moving together. Of course this is assuming a scenario where the cars crumple into a single mass, i.e. they don't "bounce". I think part of the confusion in these comments stems from the problem being poorly defined. Some people are considering crumple zones, others are considering whether a brick wall is movable and still others replace it with another car for simplicity. I do think the driver's ed manual is correct in some Spherical Cow sense, but I don't think its authors ever intended for the problem to stand up to Hacker News scrutiny. It's quite funny to me how innocuous things like this can generate these "unladen swallow"-type discussions on here, even though I'm participating.
- s1artibartfast 2y ago
  Kinetic energy is relative to the frame of reference, but Changes in kinetic energy is never relative to fixed reference frame. I think people dont realize they are changing their reference frames, and getting insane results like the amount of kinnetic energy release in the collision depend on what frame you pick. two crumple zones don't matter for head on collision, which is another thing that trips people up.

hn_throwaway_99 2y ago

Your highlighted link is the wrong model to use here. The part you highlighted in your link is about elastic collisions in one dimension. Even with "spherical cow" physics, collisions like this are nearly always modeled as perfectly inelastic collisions. You don't model the cars bouncing off each other going back with the same speed that they were originally going - you model then as crashing into each other and stopping, with all of that kinetic energy converted to heat, a perfectly inelastic collision. As per your "stationary" objections, the issue is that this is modeled as a very sturdy wall anchored to the Earth. Yes, the Earth will "recoil" some, but for all intents and purposes it can be modelled as completely stationary because its mass is so much greater than the car.

roarcher 2y ago
I don't know what you mean--there's nothing highlighted in that link when I open it. Part of it is about elastic collisions and part is about inelastic. > the issue is that this is modeled as a very sturdy wall anchored to the Earth. That's not what you modeled in the comment you linked. Your math (incorrectly) attempts to describe a car colliding with another car of equal mass. But if you're considering a "very sturdy wall anchored to the earth", then you effectively have a car colliding with the earth. Either one is fine with me, but pick one. It grinds my gears when you authoritatively call me "absolutely incorrect", "correct" me with an incorrect answer of your own, and then try to claim you were modeling something entirely different all along.
- hn_throwaway_99 2y ago
  Dude, you seriously need to work on your reading comprehension, this conversation is like talking to a dining room table. The link I posted, https://news.ycombinator.com/item?id=40628932 https://news.ycombinator.com/item?id=40628932, clearly showed the kinetic energy math for both a car hitting a stationary wall at speed 2X, and 2 cars hitting each other each with speed X. This is basic high school math of inelastic collisions.
  - roarcher 2y ago