[hamdani yusuf]

Both cars change velocity by the exact same amount in the same time, regardless of frame, because acceleration is absolute, not relative.

Someone who are free falling with their aeroplane don't accelerate according to the aeroplane. But they accelerate according to someone on earth.

[Bored chemist]

On the other extreme, we can assume that the car is perfectly braked, or bolted to the ground. In this case, the whole KE is dissipated by the moving car.

No; the other car still crumples and that dissipates energy.

I was describing two possible extreme cases. Ideally, the parking car is much stronger than cybertruck. Its crumple would be negligible.
That's why I said that real life cases are likely between those extremes.

You just re-defined a car as being a wall.

[Bored chemist]

but it just doesn't sound right to say SQUARELY.

Fine; The stopping distance rises quadratically with the speed.

[alancalverd]

We disposed of the actual question some while back. The discussion is mostly about the corollary - why do people die in crashes?

Two reasons: gross internal disruption due to the sudden application of force by a non-penetrating object, or disruption of one or more critical organs by a penetrating object. The numbers are very variable, depending on the structure of the vehicle(s), but statistically (and there is no shortage of statistics) the probability of fatality correlates with the square of the closing speed.

[Bored chemist]

In many places you can restrict the range of vehicle speed to "a bit more than  the speed limit" and , over that range the quadratic might fit fairly well.

Quite what one would do with that data is another question.
I guess you might get a PhD out of it.

[hamdani yusuf]

What's the purpose of that last column: e^v-1?
Are they just reminding us not to confuse it with e^(v-1),
which is the form I have seen/used?

To make the output value 0 when v=0.

[Just thinking]

If we use 3 cars that are identical and crash one into a very hard and unforgiving wall head on at 100kph the damage will be the same as the two cars having a head on collision at a closing speed of 200 kph. If we consider solid narrow objects as point of a collision the damage will be far more server.

[hamdani yusuf]

On the other extreme, we can assume that the car is perfectly braked, or bolted to the ground. In this case, the whole KE is dissipated by the moving car.

No; the other car still crumples and that dissipates energy.

I was describing two possible extreme cases. Ideally, the parking car is much stronger than cybertruck. Its crumple would be negligible.
That's why I said that real life cases are likely between those extremes.

You just re-defined a car as being a wall.

Ideal parking car would behave like a wall when being hit. Fortunately, real life cars are not ideal. Just like the other extreme, where the ideal parking car has no friction with the road.

[hamdani yusuf]

acceleration is absolute, not relative.

Someone who are free falling with their aeroplane don't accelerate according to the aeroplane. But they accelerate according to someone on earth.

In Newtonian mechanics, which is what is being discussed in this thread, the aeroplane accelerates the same amount relative to any inertial frame, and thus the acceleration isn't frame dependent. The aeroplane doesn't define an inertial frame, but rather an accelerated reference frame.

Someone on earth don't define an inertial frame either.

[Petrochemicals]

Because momentum is related to duration where as energy is just a total regardless of variables.

In the same time for double the velocity the energy needed to counter it is quadrupled due the body having twice the speed and covering twice the distance. To counter the faster body with the same rate of deceleration , the faster body will have covered 4 times the  distance, so it takes 4 times the opposing force or energy.

If momentum is the effect a body will have on another in a set distance, the faster body has half  the time to exert the mass, but conversely for time, it has double the distance to exert its momentum through, or in other word there is double the opportunity for mass to act due to the doubled distance, the mass dwells for double the distance.

 Doubling the mass for the same energy increaces the momentum.

[hamdani yusuf]

Because momentum is related to duration where as energy is just a total regardless of variables.

How should we interpret this statement? Is total volume becomes energy? Or total intensity?
Energy stored in a battery is often expressed in kWh. It means that energy is also related to duration.

[hamdani yusuf]

In general, objects get deformed when their parts receive non-homogeneous force. A water baloon gets deformed when laid down on the floor or hung to the ceiling because the floor or the string only put force to some parts of the baloon.
In a free falling balloon, every part of it experiences the same gravitational force, hence it doesn't get deformed.

[Just thinking]

If U couldn't avoid a head-on, collision, would U rather have a full
head-on or a single side one,

I think an impact to the side will wash off some of the energy but that could be worse for a person on that side as there could be more penitration in that aria. A T bone is the worst impact as there is very little structior to protect one.

[Petrochemicals]

Because momentum is related to duration where as energy is just a total regardless of variables.

How should we interpret this statement? Is total volume becomes energy? Or total intensity?
Energy stored in a battery is often expressed in kWh. It means that energy is also related to duration.

Joules is also expressed as time dependant. But it is a total. Momentum is a quality.

[Bored chemist]

Because momentum is related to duration where as energy is just a total regardless of variables.

How should we interpret this statement?

I wouldn't try to interpret it.
It's nonsense.

[yor_on]

I not entierly sure Mr cat but it seems the culprit is in this " Why is it that if a car doing 100K crashes into a stationary car, it's a gonner, and yet if I have a head-on with another car doing 50 each, I would expect to walk away, unscathed, in my big car "

It depends on how one read it possibly but are you assuming that there is a difference between those two scenarios, in the kinetic energy experienced by the driver? As if one could assume one of those cars, regardless of which one you choose, as representing something unmovable, as a rock? In which case the ´kinetic energy' experienced then will be 'halved', represented by just one side, no matter which side you choose?

[yor_on]

Heh, missed the second page. It seems that you do? Assume that the the momentum of something isn't there if it is of a speed, equal to what it collides with?

Is that correct?
=

Both cars use earth as their relative frame of no motion. and both cars are moving at a same speed relative earth. You put up a rock in front of any of those cars and the kinetic energy will exist, being the same for both occasions, assuming the cars to be identical clones :) as well as the speed etc.

So that kinetic energy must exist as a momentum for both cars and will transform into kinetic energy at a collision. So it should be the same as if you collide with that rock at a hundred, or crash head-on, doing fifty, into that other car, it too doing fifty although in the opposite direction.

But it is weird, momentum and kinetic energy. You could define something else than earth as your 'stationary point', some patch of space that earth swish by in its 'relative motion' through the universe. At least in relativistic terms. Then vectors and speeds relative that motion will come in and play a role for how that collision will be interpreted etc. But it won't change the kinetic energy involved in the 'system' colliding. So suddenly it's not relative anymore, or is it?

[Bored chemist]

So that kinetic energy must exist as a momentum

That can't make sense; it has the wrong units.

So it should be the same as if you collide with that rock at a hundred, or crash head-on, doing fifty, into that other car, it too doing fifty although in the opposite direction.

No.
See above.
Hitting an oncoming car which is doing 50 is the same "experience" as hitting a wall at 50.

[yor_on]

How can that be BC?
What about the momentum?

[yor_on]

And how do would I get to a kinetic energy without a momentum involved BC?

Hmm, okay, depends on definitions. https://www.sarthaks.com/454279/a-can-kinetic-energy-of-a-system-be-changed-without-changing-its-momentum