Car physics

[Brazen]

Originally Posted by MAD

I think you'll find though the offset frontal accident is the most common accident and in that test ANCAP use a deformable object. So it may be that they're one step ahead of you.

Ah good point.

[MAD]

Originally Posted by talk2tiny

Thanks Gecko. Tempting to offer proof but mine would not seem any more logical than the other dodgy ones on here. If you use the right terminology you can prove almost anything to those who don't understand the physics whether it's true or not. If only I could get red pen to show on the threads I could spend days marking these posts, reminiscing about teaching first year uni kiddies. It still wouldn't convince those who don't believe me though.


I suspect we already have a few engineers here which is why there's so much astoundingly bad physics posted already.

There's been plenty of examples put up on the "engineers" physics. Why not show us your examples?

[XCXDBABF]

Originally Posted by flappist

And again there is confusion.

The difference in speed of the first car at the beginning of the event to the end of the event in 100km/h NOT 200km/h.

Same for the second car.

The collision point is fixed so it does not matter what it is hitting the differential is the same.
The relative view of the observer or either car is irrelvent.

The only thing that matters is the state of each car before and after the event.

Both doing 100km/h then both stopped.

If all the cars and the wall were on a giant conveyer (and not taking off ) doing 2000km/h when the events happened would the damage be more or less?

If both cars are on the surface of the earth doing several hundered km/h as it rotates and whatever else as we are big banging about the universe would the damage be more or less?.

If the event were to be viewed as one car doing 200km/h running into another that is stopped then at the end of the event the combined mess would have to be doing -100km/h as the second car must be doing 100km/h less than it was at the beginning......

The difference is the forces involved. The wall will only use the level of force required to stop the car from 100km/h. If the car was driven in to the wall at 200km/h, the wall would only apply the force required to stop the car from 200km/h.

To provide an example of that, a person weighing 100kg standing on the ground will not accelerate away from the earth because the force applied by the earth on the person is only enough to stop the person sinking into the earth, but not enough to push them off the earth. Now, if the earth applied this level of force (required for the 100kg person) universally to every one, then a 50kg person would be propelled away from the earth every time they landed/collided (because they have been returned due to gravity). Alternatively, a person weighing 150kg would be on their way to the centre of the earth if the earth only applied a force against them sufficient to hold a 100kg person. Fortunately, the earth only provides the level of force sufficient to allow a person to remain constantly on the earth against each person individually. Therefore, the amount of force the earth provides on each person is equal to their weight. (Your weight is your bodies mass under the acceleration of gravity, so it is a force).

The car hitting the wall has a lesser combined total force than two cars colliding. The combined total force is what acts on the driver, which is why it is important to consider the scenario as a single event, not two distinct effectively non-related events, and relative to the object of impact.

I'm comfortable with my understanding of the events. I'm not doing the modelling to provide the proof.

[Wally]

Originally Posted by talk2tiny

..............
I suspect we already have a few engineers here which is why there's so much astoundingly bad physics posted already.

Nice dig, but probably very accurate. In the spirit of discussion forums however, it's very important to always throw red herrings.

Lewis Epstein must have a continuous smirk on his face.

[auxr]

Bewdy - schools open again. :

[flappist]

Originally Posted by talk2tiny

Thanks Gecko. Tempting to offer proof but mine would not seem any more logical than the other dodgy ones on here. If you use the right terminology you can prove almost anything to those who don't understand the physics whether it's true or not. If only I could get red pen to show on the threads I could spend days marking these posts, reminiscing about teaching first year uni kiddies. It still wouldn't convince those who don't believe me though.


I suspect we already have a few engineers here which is why there's so much astoundingly bad physics posted already.

Well Doctor.

As you are pushing the physics barrow.

Remember these are not particles they are large objects that have mass, length and are designed to collapse absorbing energy by deformation over a period of time and during the event parts of each object are moving at different speeds.

At the beginning of the event the car is moving at 100km/h relative to the end of the event which is 0 km/h.

At the instant of the collision with the wall the front of the car sees something in its way and starts to deform.

In the case of the wall it is a stationary object.
The front of the car starts to crumple back into the rear.

In the case of the other car it is an object that is deforming at EXACTLY the same rate so the event horizon appears to be stationary (or a wall).
Both fronts start to crumple back into the rear.

The end result is both cars are stationary.

IF one was doing 200km/h and the other was stationary then the result would be the combined object doing 100km/h UNLESS there was an extra force preventing the stationary car from moving.

Now with your maths and physics please show what would happen if a 1kg object travelling at 10m/s were to suddenly increase to 2kg by combining with a stationary 1 kg object with no energy added?

Would it continue at 10m/s, reduce to 5m/s or stop?

In the same way if 2 1 Kg objects were travelling at 5m/s in opposite directions what will be the resultant velocity of the combined object?

[MAD]

Originally Posted by XCXDBABF

The car hitting the wall has a lesser combined total force than two cars colliding. The combined total force is what acts on the driver, which is why it is important to consider the scenario as a single event, not two distinct effectively non-related events, and relative to the object of impact.

There is no dispute that there is double the energy involved in a two car head on collision.
What seems to be the problem is grasping the fact that there is also twice as much bodywork/crumple zones involved in the collision also.

Lets say there is 1m of travel of the car from the point of impact to the point of rest (Eg. 100km/h - 0km/h)

Driver A (single car wall collision) will decelerate from 100-0 in 1m

Drivers B & C (two car head on collision) hit each other at the exact same speed (100km/h each). Since there are two cars involved there is now 2m of total travel from the point of impact to the point of rest, but this distance is shared equally between the two cars. Each driver still decelerates from 100-0 in 1m.

[mik]

there`s no way around it a car that is made basicly of folded sheet metal with a few solid bits attached to it is going to have a bad day if it hits something solid at high speed, my young brother showed me this from another forum its an eye opener.

[talk2tiny]

I've been trying to think of an easy way to illustrate the difference in the forces involved in the collisions without resorting to equations.

Basically the point to grasp is the concept of frames of reference as others have noted. Consider two cars travelling in the same direction. The front car travels at 99km/hr, the rear car at 100km/hr. When the rear car collides with the front car, they will experience a gentle bump (though admittedly probably lose control of their vehicles). This is because, to the driver of the rear car, he was travelling at 1km/hour relative to the object he hit. Not a serious bump.

Now consider that the front car travels at 90km per hour instead. The bump would be slightly larger as the car behind would be travelling at 10km/hour relative to the object that he hit (the front car).

Try 50km per hour (still in the same direction). The rear car now runs into the front car at a speed of 50km/hour relative to the car in front.

Now consider that the front car is moving at 0km/hour (ie a brick wall). The rear car would slam into it at 100km per hour relative to the object he hit.

The same mathematics applies when the velocities go into the negatives. Lets say the rear car is still going 100km per hour, but now the car in front is reversing at 1km per hour. When they collide, the rear car was travelling at 101km/hr relative to the car he hit, experiencing a slightly larger force than when he hit the stationary object.

This could continue if the car in front was reversing at 50 km/hr making the rear car crash travelling at 150km/hr relative to the object he hit.

Now extend to the front car reversing at 100km per hour and slamming into the rear car travelling at 100km/hour. The driver of the rear car would have been travelling at 200km per hour relative to the object he hit.


The only other illustration of frames of reference I can think of without resorting to equations is to consider throwing an orange inside your car. If you throw it at the dashboard whilst driving at 100km per hour, the orange travels slightly faster than 100km per hour (say 101km/hour). When it hits the dash though, the dash is moving at 100km per hour so the force experienced by the orange is equivalent to that of a collision with a stationary object when the orange travels at 1km per hour.

Now consider (don't actually do it) throwing the orange out the window at a stationary street sign. Vastly different forces because the the object the orange hits is moving at 101 km/hr relative to the orange.


Now obviously I realise that hitting a car with crumple zones and only 4x20 square centimetre patches of rubber attaching it to the road will be different to hitting a wall with no crumple zones and securely anchored to the ground but the physical concept asked about in the beginning of the thread is illustrated here. I'm sure there'll still be those I can't convince though.

[MAD]

That is exactly what we have been saying.

There IS twice the energy expelled/absorbed/whatever in the collision, but the forces exerted on the driver(s) will be no different because there are two cars involved and sharing the energy absorption.

[flappist]

Yes you have part of it correct talk2tiny, the bit you are missing is that while the initial impact is twice the absorbsion is also twice.

Comparing head on 100km/h to 200km/h wall.


If the 200km/h car pushed the stationary car up to 100km/h the system balances.

If the 200km/h car comes to rest then there must be the equivalent of a 200km/h car "pushing" in the other direction to balance otherwise it would move.

Remember it is the difference between the initial state and the final state of the objects

In one event two objects change velocity by 100km/h, in the other event one object changes velocity by 200km/h.

Energy is the square of velocity. 100^2 + 100^2 <> 200^2

[XCXDBABF]

Nice example and explanation talk2tiny.

Originally Posted by MAD

Drivers B & C (two car head on collision) hit each other at the exact same speed (100km/h each). Since there are two cars involved there is now 2m of total travel from the point of impact to the point of rest, but this distance is shared equally between the two cars. Each driver still decelerates from 100-0 in 1m.

The driver actually decelerates relative to the other car from 200km/h to 0km/h in 2m in this example.

[talk2tiny]

Originally Posted by MAD

That is exactly what we have been saying.

There IS twice the energy expelled/absorbed/whatever in the collision, but the forces exerted on the driver(s) will be no different because there are two cars involved and sharing the energy absorption.

I'm not sure I can explain why that's not true without equations but I'll try and think of an example. Let me have a think.

[flappist]

Originally Posted by XCXDBABF

Nice example and explanation talk2tiny.



The driver actually decelerates relative to the other car from 200km/h to 0lm/h in 2m in this example.

NO.

The initial speed in 100, the final is 0, the driver only travels 1 metre.

[flappist]

Originally Posted by talk2tiny

I'm not sure I can explain why that's not true without equations but I'll try and think of an example. Let me have a think.

You must use a common frame of reference.

Event A, 2 cars head on at 100. right is + left is -

1) Car on left is ref point.
Car on left is fixed.
Car on right is moving at -200
Crash point is moving at -100

2) Crash is ref point.
Car on left is moving +100
Crash is fixed.
Car on right is moving -100

3) Car on right is ref point.
Car on left is moving at +200
Crash point is moving at +100
Car on right is fixed.

These are all the same thing

Event B, one car into a wall at 200km/h

1) Car is ref point.
Car is fixed
Wall is moving at -200km/h

2) Wall is ref point.
Car is moving at +200km/h
Wall is fixed.

In event A the crash is always at 100+/- from each car
In event B the crash is always at 200+/- from the car.

Remember in both events, all objects finish at rest.

[talk2tiny]

Originally Posted by talk2tiny

Let me have a think.

Nope. I can't think of a way without equations. My equations would be no more more credible than any of the others though so I'll have to give up. Lets just hope no-one has to find out the hard way.

[MAD]

Originally Posted by talk2tiny

My equations would be no more more credible than any of the others though so I'll have to give up.

What makes you say the equations already given in this thread already are not credible? why wouldn't yours be credible? and why give up?
Surely you can, and would like to, prove that your claim is correct.

Originally Posted by talk2tiny

I suspect we already have a few engineers here which is why there's so much astoundingly bad physics posted already.

While I'm at it, why did you feel the need to disrespect any engineers among us like this? Yes, I'm an engineer.
I dont think there was any need for an attack of that nature. Up until that point the discussion had been very civil with clear explanations of each persons claims and their reasons behind it. Yet for some reason you felt the need to reduce yourself to that level.

Originally Posted by talk2tiny

Lets just hope no-one has to find out the hard way.

^^^I agree.
In the end, with either of our claims, the best option is still the tree.
If your explanation is correct the tree impact will be less, if the other claim is correct, the impact with the tree will be no worse than that of the car so why hurt more people than necessary?

[Piotr]

Originally Posted by talk2tiny

The only other illustration of frames of reference I can think of without resorting to equations is to consider throwing an orange inside your car. If you throw it at the dashboard whilst driving at 100km per hour, the orange travels slightly faster than 100km per hour (say 101km/hour). When it hits the dash though, the dash is moving at 100km per hour so the force experienced by the orange is equivalent to that of a collision with a stationary object when the orange travels at 1km per hour.

Now consider (don't actually do it) throwing the orange out the window at a stationary street sign. Vastly different forces because the the object the orange hits is moving at 101 km/hr relative to the orange.

The orange does not travel at 101km/h.

Same argument as the classic train traveling at the speed of light. If you work forward you will NOT be breaking the speed of light

[barbarian]

You better want to be a VW Golf if you want to survive 100kmh solid object crash

http://www.youtube.com/watch?v=_B8M_FfNw3I

[XCXDBABF]

Originally Posted by MAD

If your explanation is correct the tree impact will be less, if the other claim is correct, the impact with the tree will be no worse than that of the car so why hurt more people than necessary?

Not quite. The consideration here has been full frontal impact, not a partial frontal impact. If you drive into a wall or another vehicle exactly the same as yours directly head on, then the forces are working on your vehicle across a much greater area than what would occur if you crashed into a tree or power pole.

For example, if some one was to throw a javelin at a tree, the javelin would penetrate into the tree because the forces involved in the impact work only on a very specific area of the tree. Now, if you get the same javelin, and instead of throwing it pointy end first, threw it side on instead, such that the entire length, rather than just the point, of the javelin impacted against the tree, then the javelin would not penetrate the surface of the tree.

I wouldn't like to take a guess as to which would be worse, hitting a power pole at 100km/h or hitting a another vehicle head on, that is also traveling at 100km/h in the opposite direction, at 100km/h. Given the seriousness of the injuries to anyone involved, I would have to agree with the paramedic (sorry, I can't remember who you are at the moment) who posted earlier, the two car collision will have more people involved, therefore it's worse.

==================

Flappist - I'll have one final attempt to explain the issue with the point of reference.

1. If you are standing still and I throw a ball at you at 20km/h, then when it impacts with you, you will feel the forces relative to the initial velocity of the ball and physical characteristics of the ball. To the observer watching this experiment, he sees me throw the ball at you at 20km/h, and the ball decelerate to 0km/h when it hits you.

2. Now, if instead of me throwing the ball at you, you run towards me at 20km/h. Just as you are about to get to me, I drop the ball in front of you, such that it impacts with you in the same spot as the ball I threw at you. As you are traveling at 20km/h and the ball is traveling at 0km/h at the time of impact, you will feel the same forces on the impact point as when I threw the ball at you at 20km/h and you were moving at 0km/h. This time however, the observer sees you running at the ball making impact, not the ball traveling to you making impact.

But, here's the thing with the point of reference. If you are the point of reference, so you only consider what you see through your eyes, and not what your body is doing, then even though you are running at the ball, to you, you see the ball coming towards you at 20km/h, even though the casual observer actually sees you running at the ball at 20km/h (you both have different points of reference - your own eyes).

3. So now if I combine both 1 and 2 and throw the ball at you at 20km/h and you run towards me at 20km/h, then the observer will see both objects (you and the ball) traveling at 20km/h towards each other.

When the ball hits you, you will feel the combined total affect of you traveling towards the ball at 20km/h and the ball traveling towards you at 20km/h.

From your point of view (through your eyes) you see the ball traveling towards you at 40km/h, and when it impacts with you, you will feel the forces relative to if the ball was traveling at 40km/h and you were standing still.

4. What I haven't told you till now is the ball I am throwing at you weighs exactly the same as you, and at the point of impact, both you and the ball a completely rigid.

Now this is what both the observer and you see.

The observer sees me throw a ball at you at 20km/h. The observer sees you run towards the ball at 20km/h. When you and the ball collide, the observer sees you decelerate from 20km/h to 0km/h, and the ball decelerate from 20km/h to 0km/h per hour. The observer therefore jumps to the conclusion that the forces you experienced would only be the same as if you'd run into a brick wall at 20km/h, and therefore you are only so injured.

From your perspective however, the events are very different. You see the ball weighing the same as you approaching at 40km/h. When the ball impacts with you, you see the ball decelerate from 40km/h to 0km/h per hour. You feel the forces acting on you relative to if the ball's actual velocity before the collision was 40km/h. Therefore the injuries you suffer are worse than if you had just run into a wall at 20km/h.

I hope this helps. If not, we will just have to agree to disagree.

[scoupedy]

Originally Posted by MAD

That is exactly what we have been saying.

There IS twice the energy expelled/absorbed/whatever in the collision, but the forces exerted on the driver(s) will be no different because there are two cars involved and sharing the energy absorption.

Correct

twice the energy twice the cars to absorb it, think of the wall as a mirror -it sends all of the energy back into the car

We did this exact problem at uni and discussed it for two hours in a tutorial as far as phsics is concerned its the same

[MAD]

Originally Posted by XCXDBABF

Not quite. The consideration here has been full frontal impact, not a partial frontal impact. If you drive into a wall or another vehicle exactly the same as yours directly head on, then the forces are working on your vehicle across a much greater area than what would occur if you crashed into a tree or power pole.

True, I was just considering the tree as a REALLY big one...
The smaller contact area of a pole or tree would be far more devastating to the vehicle.

[flappist]

Originally Posted by XCXDBABF

Not quite. The consideration here has been full frontal impact, not a partial frontal impact. If you drive into a wall or another vehicle exactly the same as yours

Flappist - I'll have one final attempt to explain the issue with the point of reference.

1. If you are standing still and I throw a ball at you at 20km/h, then when it impacts with you, you will feel the forces relative to the initial velocity of the ball and physical characteristics of the ball. To the observer watching this experiment, he sees me throw the ball at you at 20km/h, and the ball decelerate to 0km/h when it hits you.

2. Now, if instead of me throwing the ball at you, you run towards me at 20km/h. Just as you are about to get to me, I drop the ball in front of you, such that it impacts with you in the same spot as the ball I threw at you. As you are traveling at 20km/h and the ball is traveling at 0km/h at the time of impact, you will feel the same forces on the impact point as when I threw the ball at you at 20km/h and you were moving at 0km/h. This time however, the observer sees you running at the ball making impact, not the ball traveling to you making impact.

But, here's the thing with the point of reference. If you are the point of reference, so you only consider what you see through your eyes, and not what your body is doing, then even though you are running at the ball, to you, you see the ball coming towards you at 20km/h, even though the casual observer actually sees you running at the ball at 20km/h (you both have different points of reference - your own eyes).

3. So now if I combine both 1 and 2 and throw the ball at you at 20km/h and you run towards me at 20km/h, then the observer will see both objects (you and the ball) traveling at 20km/h towards each other.

When the ball hits you, you will feel the combined total affect of you traveling towards the ball at 20km/h and the ball traveling towards you at 20km/h.

From your point of view (through your eyes) you see the ball traveling towards you at 40km/h, and when it impacts with you, you will feel the forces relative to if the ball was traveling at 40km/h and you were standing still.

4. What I haven't told you till now is the ball I am throwing at you weighs exactly the same as you, and at the point of impact, both you and the ball a completely rigid.

Now this is what both the observer and you see.

The observer sees me throw a ball at you at 20km/h. The observer sees you run towards the ball at 20km/h. When you and the ball collide, the observer sees you decelerate from 20km/h to 0km/h, and the ball decelerate from 20km/h to 0km/h per hour. The observer therefore jumps to the conclusion that the forces you experienced would only be the same as if you'd run into a brick wall at 20km/h, and therefore you are only so injured.

From your perspective however, the events are very different. You see the ball weighing the same as you approaching at 40km/h. When the ball impacts with you, you see the ball decelerate from 40km/h to 0km/h per hour. You feel the forces acting on you relative to if the ball's actual velocity before the collision was 40km/h. Therefore the injuries you suffer are worse than if you had just run into a wall at 20km/h.

I hope this helps. If not, we will just have to agree to disagree.

FFS

You really dont get it at all.

This would be true if you were still doing 20km/h AFTER the impact.

YOU ARE NOT.

YOU HAVE STOPPED

AT THE INSTANT OF THE IMPACT YOU ARE STOPPED AND ALL YOUR ENERGY HAS BEEN ABSORBED.

YOU DO NOT CONTINUE ON AFTER THE IMPACT.....

AFTER THE IMPACT THE BALL IS GOING 20km/h SLOWER so that is all the enery it has lost that must be absorbed by me.

Because the ball and I are unequal it will not balance and I would not stop.

If it were 2 balls then each would absorb half the total force.

You are getting confused by thinking of unequal objects. The mass and velocity must be the same or after the impact they will not be stopped, they will contine moving.

[flappist]

Actually I have thought of a better way to explain your mistake.

In your experiment you are on a skateboard (or whatever) rolling at 20 km/h.
A ball is thrown at you at 20 km/h and you catch it.

This seems to hurt more than if you were still and a ball was thrown at you at 20km/h.

WHY?

After you catching the ball are you stopped?

Or are you still rolling in the same direction very slightly slower than 20km/h and the ball in now also moving in the opposite direction at slightly slower than 20km/h which is slightly slower than 40km/h less than it was before....

You have not only had to decellerate the ball to 0 but then accellerate the ball to -19.xxxxx.

The forces MUST AWAYS balance.

[XCXDBABF]

Originally Posted by flappist

FFS

You really dont get it at all.

This would be true if you were still doing 20km/h AFTER the impact.

YOU ARE NOT.

YOU HAVE STOPPED

AT THE INSTANT OF THE IMPACT YOU ARE STOPPED AND ALL YOUR ENERGY HAS BEEN ABSORBED.

YOU DO NOT CONTINUE ON AFTER THE IMPACT.....

AFTER THE IMPACT THE BALL IS GOING 20km/h SLOWER so that is all the enery it has lost that must be absorbed by me.

Because the ball and I are unequal it will not balance and I would not stop.

If it were 2 balls then each would absorb half the total force.

You are getting confused by thinking of unequal objects. The mass and velocity must be the same or after the impact they will not be stopped, they will contine moving.

Have another read of the first line of what I wrote in point 4.

You and I can argue this till we're both blue in the face and I still don't think we'll agree...

(There are some convenient over-sights in what I've written if you apply that line then go and re-read the previous points. Where I was trying to get you to though was point 4.).

[flappist]

Originally Posted by XCXDBABF

Have another read of the first line of what I wrote in point 4.

You and I can argue this till we're both blue in the face and I still don't think we'll agree...

And again you miss the point completely.

The colission is not instant it takes time.

At the initial contact there is no energy transfer.

I try to push the ball.
The ball trys to push me.

As I push the ball I slows down.
As the ball pushes me it slows down.

My energy is expended
Its energy is expended.

The ball stops and I stop.
The ball has half the damage and I have half the damage.

At no time does the ball extend past the the impact point.
It is not pushed back nor does it push forward.

The force of the ball is the same as the force of me, otherwise the impact would move.

Therefore the ball acts like a wall fixed in position and the only force I expend is my own force.
The ball expends its own force in stopping.

Rather that argue until you are blue in the face, why not ask others (ones who are prepared to show proofs or equasions not ones that run and hide) who actually work things like this out for a living.

I have.

[XCXDBABF]

Originally Posted by flappist

And again you miss the point completely.

The colission is not instant it takes time.

At the initial contact there is no energy transfer.

I try to push the ball.
The ball trys to push me.

As I push the ball I slows down.
As the ball pushes me it slows down.

My energy is expended
Its energy is expended.

The ball stops and I stop.
The ball has half the damage and I have half the damage.

At no time does the ball extend past the the impact point.
It is not pushed back nor does it push forward.

The force of the ball is the same as the force of me, otherwise the impact would move.

Therefore the ball acts like a wall fixed in position and the only force I expend is my own force.
The ball expends its own force in stopping.

Rather that argue until you are blue in the face, why not ask others (ones who are prepared to show proofs or equasions not ones that run and hide) who actually work things like this out for a living.

I have.

Agreed! The collision is not instant, it takes time.

Car drives into wall at 100km/h and stops. Forces required? Over what period of time? Both car and wall experience the same force.

Two men decide to go bungee jumping. The first selects a elastic rope (bungee cord) as his method of connection to the mounting point, the other selects a steel wire rope. Both men reach terminal velocity during their fall and reach a stop at the end of their fall. Neither connection method to the mounting point fails and neither man hits the ground, yet while one man has an exhilarating experience, the other dies. Why?

Forces required? Over what period of time?

Car drives into wall at 100km/h and stops.
Car traveling at 100km/h drives into another car exactly the same traveling in the opposite direction at 100km/h and both stop.

There is no doubt that the wall and the car exert the same force over the same period of time on each other.
The two cars involved in the two car collision exert the same force over the same period of time on each other.

BUT - Do the car hitting the wall and the car hitting the other car apply the same force over the same period of time during their independent collisions?

This is why you need to consider the relative velocity.



I know I said it before, but now I am finished.

[flappist]

Originally Posted by XCXDBABF

Agreed! The collision is not instant, it takes time.

Car drives into wall at 100km/h and stops. Forces required? Over what period of time? Both car and wall experience the same force.

Two men decide to go bungee jumping. The first selects a elastic rope (bungee cord) as his method of connection to the mounting point, the other selects a steel wire rope. Both men reach terminal velocity during their fall and reach a stop at the end of their fall. Neither connection method to the mounting point fails and neither man hits the ground, yet while one man has an exhilarating experience, the other dies. Why?

Forces required? Over what period of time?

Car drives into wall at 100km/h and stops.
Car traveling at 100km/h drives into another car exactly the same traveling in the opposite direction at 100km/h and both stop.

There is no doubt that the wall and the car exert the same force over the same period of time on each other.
The two cars involved in the two car collision exert the same force over the same period of time on each other.

BUT - Do the car hitting the wall and the car hitting the other car apply the same force over the same period of time during their independent collisions?

This is why you need to consider the relative velocity.



I know I said it before, but now I am finished.

Yes they do.

And you miss the point of the relative velocity because in the 2 ball at 20 situation the impact is travelling at a different velocity to the one ball at 40 and the wall.

The wall is that situation is traveling at 20km/h relative to the impact in the first case and as the wall does not change its state it must continue at 20km/h relative to the first case.

THAT IS WHY THEY ARE DIFFERENT

In one case 2 objects change state.
In the other case only one object changes state.

[Wally]

NHTSA test:

NHTSA's frontal test is a good indication of how well a vehicle's safety belts and air bags protect the occupants in specific types of impacts. The frontal test runs vehicles into a rigid barrier at 35 mph. That simulates a head-on collision between two vehicles of similar weight, each traveling at 35 mph. Instrumented crash dummies in the two front seats record the crash forces they sustain and scores are assigned for the driver and front passenger.

[flappist]

OK another attempt to get the concept across....

It has been determined that FG GT falcons will stop from 100km/h in 40m

Two GT owners demonstrate how clever they are by driving at each other at 100km/h and then at 80m apart they brake coming to a halt with their noses 1mm apart.

Another FG GT owner has spoken to some people who may or may not understand dynamics and they all decide to demonstrate their understanding by driving at a brick wall at 200km/h braking at 80m because they believe that this is exactly the same as 2 GTs doing 100km/h.....relative speed and all that stuff.

GUESS WHAT HAPPENS..........

And after posting this metaphor I cannot believe I am going to say this but......

Thank you Wally