[Curious Cat (OP)]

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[alancalverd]

Because mv and ½mv² are quantities called momentum and energy.

[alancalverd]

But why does speed really kill?

Because the transfer of large amounts of momentum and energy to the human body tends to disrupt its vital functions.

If I have a head-on with another car doing 50 each, I would expect to walk away, unscathed, in my big car,

Prepare to be disappointed. Briefly.

[Halc]

But why does speed really kill?

It doesn't kill at all. It's unequally distributed acceleration which tends to disassemble an object. Absent the acceleration, one has no problem.

So I might die by being accelerated by a bridge abutment moving at 60 km/hr, but the guys in the ISS are moving at 26000 km/hr (approx) and are not hurt at all by it.

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,

That's the same accident. Maybe the airbags didn't go off in the parked car because it isn't armed when the engine is off. I'm assuming the same two cars colliding. You're equally protected in both situations.

A car doing 100 has twice the KE of 2 cars doing 50 each, combined.

But KE is utterly abstract, and is frame dependent. In a different frame, both cars might have massive KE. It isn't a function of KE.
I had an accident on the highway 10 years ago with both vehicles moving at well over 100 km/hr and it resulted in a minor dent in my car and no visible damage to the other larger one. The fairly large KE we both had (him more than me) had nothing to do with it. It's the acceleration imparted to my vehicle that caused the damage, and there was very little of that.

The abutment on the other hand is a problem. Hitting that at a delta-V of 100km/hr will likely cause serious injury because the acceleration is far higher than when hitting the parked car.

My mother, in her 80's, hit (broadsided) a much larger nearly stationary vehicle at at least that speed and came away with massive bruises but nothing broken except what little remained of the car. Try that with a car made in the 70's and see what happens.

Take care. Drive careful, hey?

[Colin2B]

...Yes of course I've heard of "It's not the speed that/which kills, U. It's the sudden deceleration/stopping".

Come on, guys. Get Ur head/s out of Ur physics books. Get real.

It’s nothing to do with physics books, just common sense.
Two people of same weight jump from same floor of a burning building.
They both have same terminal velocity, momentum and KE.
One hits the deck, the other into a fireman’s blanket (or whatever they are called).
Who survives?
Change of momentum f=ma. A human body can only withstand a certain amount of g force.

Obviously, you need enough speed to get the deceleration, which is where I suspect the confusion comes from, add in crumple zones etc.

Dunces, with cars, know this, instinctively!

I wonder why they are dunces? 

[evan_au]

Why does double the velocity double the momentum and quadruple the KE?

It comes down to Integration.

When you see E = ½mv², the association of the  ½ and the v² suggests that there is integration going on here

I'll leave it to others to describe what is being integrated (and why).

[Just thinking]

Why does double the velocity double the momentum and quadruple the KE?

This is a fantastic question and very difficult to give an example of why it is so. I will try an example if we have two circles and one is twice the diameter of the other one we get 4x the surface area. Now with a projectile, if we double the velocity the impact force is increased by a factor of 4x the diameter of the projectile has not changed but the speed and the kinetic energy has the impact force is measured by the force applied over the diameter of the front the mass of the projectile that is made up of particles that are all doubling the velocity and each particle must contend with the particles behind them each particle is loaded up by the mass of particles behind them the first particle is x2 second particle is also x2 this is true for all the particles and the net result is 2x2 a factor of x4. If we only double the weight but not the speed each particle remains the same and will be a factor of x1. 1x1 = 1 double the mass 1+1 = 2. This is how my funny head worked it out I hope it makes sense It could be wrong.

[Just thinking]

JT, please tell 'em:
Would U rather crash into a stationary car at 100
or have a head-on with another car at 50, each?
The relative velocity is the same but the damage is twice in the first case, coz the (total) KE is twice.

Yes, I see what you're saying my first response was regarding only one object in motion. The second is the sharing of impact this is a different story. I do believe that the initial impact is the same but due to the cars having their own momentum this will reduce the overall damage to the front as the rear will take some of the load if both are in motion if we take the example of a car hitting a wall at 100 and the wall is still we hit at 100 if we are travelling at 50 and the wall is travelling at 50 that will be the same as hitting a still wall the wall has no rear collapse but cars do.

[Just thinking]

KE and stopping distance go up exponentially, with speed"!

KE and stopping distance go up exponentially, with speed"!
If U double the speed, U quadruple the KE and the stopping distance.

Yes this I have already agreed with it is true. It is in agreeance with the factor of four.

[Just thinking]

KE and stopping distance go up exponentially, with speed"!

Yes double the speed the energy goes up x4 double the speed stopping power goes up x4.

[Bored chemist]

I think that a head-on at 50, each, is equivalent to crashing into a solid wall at 50,
coz yes the other car is comin' at U but a wall will not crumple.

Well spotted.
If anyone is wondering if that's true, imagine putting a piece of paper between the two identical cars as they crash.
In principle, the paper gets squeezed and torn up down + sideways, but it never moves forwards or backwards (from the point of view of the vehicles). Each bit of it that is pushed one way by one car is pushed the other way by the other car.
Since the paper doesn't move, it acts like an infinitely stiff wall.

The really important difference is that the KE in a head on crash is shared between the two cars.
Since the wall (in the idealised limit) does not move, no work is done on it so all the energy is dissipated in one car.
So hitting a wall at 50 is pretty much the same experience as hitting another car also traveling at 50.

So, if you have the same energy in the crash, but only one car, you are in more trouble.
To get the same energy with 1 car hitting a wall, it would need to be travelling at 1.414 * 50.


The amount of damage done to the passenger  is more nearly proportional to the energy than to the speed or the momentum. That's not a perfect model.
It's complicated.
If I fall off my push bike at 20 KPH I'm not likely to be severely harmed.
That's also true if you see me fall off while you are going past me in a car at 70KPH.
How much energy I have is frame dependent.
But if we consider my speed WRT the road, that gets rid of that problem (to a great extent).

There's 75 Kg of me.
20 KPH is 5.6 m/s
So 1/2 m v^2 tells me I would be carrying about 1.2KJ of kinetic energy.
But I dissipate millions of KJ in a day, so that much energy isn't the problem.

Part of the problem is how fast I need to shift that energy if I fall off (say a tenth of a second) implying a power of about 12KW or roughly 100 times how much power I normally dissipate.

Part of the problem is the force involved- That's harder to calculate because, essentially, I bounce a bit.

It's not a simple "it's the speed" or "it's the momentum" or whatever; it's a combination.

[Just thinking]

Thank god for collapsible steering columns it helps to keep the skin on ones face. And seat belts.

[Just thinking]

And don't forget the air bags, JT.
They say they are lubricated with talc, to make 'em come out easier,
and which afterwards looks like smoke, so U think U're on fire, for a sec!
Or is there really smoke from the explosive cartridge?

Thankfully I can't say as I have never been in a car accident and had an airbag go off but I have seen the results of a steering wheel on a persons face and it's like pealing one side of a orange.

[Just thinking]

We seem to have drifted from the original question of why double the speed =4x the energy.

[alancalverd]

That was answered in #2 above.

[hamdani yusuf]

Now do it to the parked car situation. One car parked (no KE). One going 100, so KE is 5000. After the hit, both cars are moving at 50, so KE is 2500 total. Total energy dissipated is 2500, shared between the cars so 1250 each.  Same change in KE dissipated equally by both cars, so same damage as the slower speed head-on.

Your analysis assumes that the parked car doesn't have brakes.
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.
In real life cases, they're mostly between those two extremes. So the OP's intuition is justifiable.

[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.

[Bored chemist]

We seem to have drifted from the original question of why double the speed =4x the energy.

Good point.
Imagine that you wanted to "harvest" the energy to do something useful like pull water out of a river for irrigation or something.
Consider hooking the car to a bucket on a rope over a pulley so that the car is slowed down by the tension in the rope.
If we allow the rope to be a bit elastic we can get round the sudden jerk when the rope goes tight.

So, to make the maths easy just imagine that the car is being slowed down by a constant force.
To make the maths easy (albeit an impractical example) consider the case when the car of mass 1000Kg is slowed down from about 30m/s to a halt by that force.

Then consider the case where it's slowed from 15m's to a halt (and then the general case where t is slowed down from a speed of  V  m/s.
Since energy is a conserved property (in this case; none is wasted as heat,  for example), all the car's KE is converted into work done pulling on the rope.
So the work done tells you how much energy the car has.

You can also consider the question the other way- how far would you have to pull a rope at a constant tension to get the car up to a given speed.

[evan_au]

the KE and stopping distance go up exponentially, with speed!
If U double the speed, U quadruple the KE and the stopping distance

That is a square law = v², not an exponential law = e^v.
- Exponentials grow much more quickly than square laws
- Exponentials also start of with a non-zero value when v=0


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[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.