Car crash physics/mechanics

NIMAN

I was just thinking about the physics of a car crash whilst reading another thread here, and have a question.

If 2 cars are in a head-on crash (similar car, same weight etc) would the person in the faster car have more chance of coming off with less injuries?

Basically Car A doing 30mph, and Car B coming the opposite direction doing 50mph. Car B obviously has more momentum and power behind it, is Driver A likely to come off worse or does it not matter?

goz83

Collision speed = 80mph for both cars

Anan1

The energy will be equally shared between the two cars, the effect on each car will be the same as if it had hit an immovable object at 40mph.

randy hickey

http://www.sciforums.com/%2520%253c/Collision-physics-explanation-needed-t-104387.html

HTH.

CiniO

NIMAN wrote: »

I was just thinking about the physics of a car crash whilst reading another thread here, and have a question.

If 2 cars are in a head-on crash (similar car, same weight etc) would the person in the faster car have more chance of coming off with less injuries?

Basically Car A doing 30mph, and Car B coming the opposite direction doing 50mph. Car B obviously has more momentum and power behind it, is Driver A likely to come off worse or does it not matter?

If both cars are the same weight and model, then they both will be affected like each on of them crashed at 40MPH into a wall.

goz83

CiniO wrote: »

If both cars are the same weight and model, then they both will be affected like each on of them crashed at 40MPH into a wall.

Am I missing something here? If one car is doing 30 and the other is doing 80, then the combined speed is 80. So would the collision speed not be 80mph?

The above post would suggest that collision speeds are halved without any reason being given. If i hit a car head on, and I am doing 100 mph and the other car is doing 20mph, the inertia/collision force is 120mph, not 60mph. Lets say the other car is stationary....If i hit it at 100mph, what is the collision force/speed that the other car will experience? Not 50 as far as I can tell.

goz83

With head on collisions of objects of the same type, weight, mass etc (two identical cars), then both speeds are added together to get the end collision speed. If both cars were travelling in the same direction and the car in front was struck, then you minus the speed of the front car from the rear car to get the collision speed.

Anan1

goz83 wrote: »

Am I missing something here? If one car is doing 30 and the other is doing 80, then the combined speed is 80. So would the collision speed not be 80mph?

The above post would suggest that collision speeds are halved without any reason being given. If i hit a car head on, and I am doing 100 mph and the other car is doing 20mph, the inertia/collision force is 120mph, not 60mph. Lets say the other car is stationary....If i hit it at 100mph, what is the collision force/speed that the other car will experience? Not 50 as far as I can tell.

The combined speed is indeed 80mph, but it's being shared between two cars. The effect on each car is the same as had that car hit an immovable object at 40mph.

Del2005

Anan1 wrote: »

The energy will be equally shared between the two cars, the effect on each car will be the same as if it had hit an immovable object at 40mph.

goz83 wrote: »

Collision speed = 80mph for both cars

CiniO wrote: »

If both cars are the same weight and model, then they both will be affected like each on of them crashed at 40MPH into a wall.

Wrong, the energies involved is the square of the speed.



The car going at 50 will have a K.E of massx1250 and the car at 30 will have a K.E of massx450, non SI units used so only a rough estimation. The faster travelling car has more kinetic energy so should do more damage to the slower travelling car.

Anan1

Del2005 wrote: »

The faster travelling car has more kinetic energy so should do more damage to the slower travelling car.

Why?

NIMAN

Del2005, thats what my original question was hinting at.

If one car is doing 10mph and the other 60mph, surely the 60mph car will push through the 10mph car and push it back, and go through it with more force than the slower car.

So you saying that it would be better to be in a faster moving car in a head-on?

goz83

The kinetic enery at the point of impact is the same for both cars. The only part I can't wrap my head around is why the impact is halved, just because the impact is with another car....but I guess it's got something to do with the fact that the cars are moveable objects. Still...it doesnt mean that a car hitting a stationary car at 100mph reduces the impact force to that equivalent to 50mph does it? There would have to be 0 friction for that to happen.

BrianD3

Del2005 wrote: »

Wrong, the energies involved is the square of the speed.



The car going at 50 will have a K.E of massx1250 and the car at 30 will have a K.E of massx450, non SI units used so only a rough estimation. The faster travelling car has more kinetic energy so should do more damage to the slower travelling car.

I think momentum and change in velocity rather than KE is what the OP is asking about. The KE does indicate the potential for damage but it will be dissipated by both cars as both have the exact same crumple zones.

A car colliding with an identical stationary car at 80 km/h results in the same damage as two identical cars colliding with each other doing 40 km/h

If KE before the collision was the full story then if a moving car hits a stationary truck you'd be better off in the car. This clearly isn't the case.

Going back to the OPs query i think for the occupants in identical vehciles a 50 km/h + 30 km/h collision is the same as an 80 km/h + 0 km/h collision or a 40 km/h + 40 km/h collison.

The human body doesn't distinguish between being decelerated from 80 km/h to 0 km/h in 1 second by a seatbelt or being accelerated from 0 km/h to 80 km/h in 1 second by a seatbelt. Therefore it doesn't matter whether you are in the car doing 80 km/h or the car doing 0 km/h.

I may be wrong of course and some of the posts at the below link would suggest so but others would agree with me.
http://www.abc.net.au/science/k2/stn/q&a/notes/060518-6.htm

AvAv

If you collide, the forces are as if the cars hit immovable objects. The car cannot force the object to move, so the object forces the car to stop.

Unfortunately for the driver, if you are sitting in a car going at a high speed, your body is also moving at the high speed. When the car is stopped, the stopping force is only acting on the car, not yourself and you will continue to travel with high speed until forced to stop by seatbelt etc.

If you are travelling at a higher speed it takes more force to stop you.

If you travel and collide at 60 km/h, you will need a big enough force to slow you down from 60 to 0.

If you travel at 10, you need a force to slow you down from 10 to 0.

Which speed is better is relative to whether you're talking about your car or yourself.

Mikros

NIMAN wrote: »

I was just thinking about the physics of a car crash whilst reading another thread here, and have a question.

If 2 cars are in a head-on crash (similar car, same weight etc) would the person in the faster car have more chance of coming off with less injuries?

Basically Car A doing 30mph, and Car B coming the opposite direction doing 50mph. Car B obviously has more momentum and power behind it, is Driver A likely to come off worse or does it not matter?

The momentum of each car is given by p = mass x velocity. Assuming both cars weigh the same the car travelling at 50 mph will have a higher momentum than a car travelling at 30 mph.

In the event of a head on collision the total momentum is conserved. Because the momentum of the faster car is greater, the final velocity of the combined system (2 cars together) will be in the direction of the faster car after the collision. In other words the slower car will be pushed backwards.

That means the driver of the faster car experiences less acceleration (same direction of velocity) than the driver of the slower car does (direction of velocity is reversed). The person in the slower car is likely to come off worse, all things being equal.

In the real world there are a lot of different factors at play - the elasticity of the collision will be determined by the types of vehicles - crumble zones, airbags etc. The actual physics at play is a tad more complicated that 30mph+50mph = 80mph collision for both drivers. But in general the heavier / faster vehicle comes out better in a straight head on collision - i.e. the one with the greatest momentum.

Anan1

goz83 wrote: »

The only part I can't wrap my head around is why the impact is halved, just because the impact is with another car....

The force of the impact isn't halved, but there is twice as much crumple zone to absorb it.

Mikros wrote: »

In the event of a head on collision the total momentum is conserved. Because the momentum of the faster car is greater, the final velocity of the combined system (2 cars together) will be in the direction of the faster car after the collision. In other words the slower car will be pushed backwards.

This is true.

Mikros wrote: »

That means the driver of the faster car experiences less acceleration (same direction of velocity) than the driver of the slower car does (direction of velocity is reversed). The person in the slower car is likely to come off worse, all things being equal.

No, even though one ends up going backwards, both are still subject to the same deceleration.

07Lapierre

Send an email to "mythbusters"!!. They'll sort it!

tacofries

Del2005 wrote: »

Wrong, the energies involved is the square of the speed.



The car going at 50 will have a K.E of massx1250 and the car at 30 will have a K.E of massx450, non SI units used so only a rough estimation. The faster travelling car has more kinetic energy so should do more damage to the slower travelling car.

wrong.

both cars exert the same force on each other no matter what the difference in speed is. This can be seen by newtons third law: if a body 'a' exerts a force on a body 'b' ,then body 'b' exerts an equal but opposite force on body 'a'.

thus each car experiences the exact same amount of force in the crash.

Collision speed is 80mph.

Mikros

Anan1 wrote: »

The force of the impact isn't halved, but there is twice as much crumple zone to absorb it. :)This is true.

No, even though one ends up going backwards, both are still subject to the same deceleration.

No not quiet. Taking the direction of travel of car1 as positive. The figures below are illustrative but hopefully you will get the idea:

Car1 travelling ---> +30 mph
Car 2 travelling <---- -50 mph

Car1car2system after collisions <

---

-20mph

Change in velocity of car1: -50 mph
Change in velocity of car2: -30 mph

There is a greater deceleration on car1 and the resulting forces on the driver would be higher.

Mikros

tacofries wrote: »

wrong.

both cars exert the same force on each other no matter what the difference in speed is. This can be seen by newtons third law: if a body 'a' exerts a force on a body 'b' ,then body 'b' exerts an equal but opposite force on body 'a'.

thus each car experiences the exact same amount of force in the crash.

Collision speed is 80mph.

You don't understand Newtons laws and their application to collisions.

The force each car experiences is determined by the acceleration. Acceleration is change in velocity over time. When two cars collide there is a change in velocity in a very short period of time. The change in velocity is based on the conservation of momentum - the momentum before the collision is the same as the momentum after.

Momentum is mass x velocity. The faster and heavier car will lose less velocity and therefore will experience less acceleration / force.

Think about if a train hit a car head-on both travelling at 80 km/h. The train will not experience the same force as the car because the change in velocity of the train will be almost 0. This is because the mass of the train is many many times that of a car. That is the conversation of momentum.

The "speed of the collision is 80mph" means nothing in physics.

Anan1

Mikros wrote: »

Car1 travelling ---> +30 mph
Car 2 travelling <---- -50 mph

Car1car2system after collisions <

---

-20mph

Change in velocity of car1: -50 mph
Change in velocity of car2: -30 mph

There is a greater deceleration on car1 and the resulting forces on the driver would be higher.

Where are you getting those figures from? If one car is doing +30mph, and the other is doing -50 then the wreckage will be doing -10, ie a reduction in velocity for both cars of 40.

Mikros

Anan1 wrote: »

Where are you getting those figures from? If one car is doing +30mph, and the other is doing -50 then the wreckage will be doing -10, ie a reduction in velocity for both cars of 40.

Your right of course. For some reason I was thinking one car was heavier than the other.

Something about reading the question comes to mind...

CiniO

Del2005 wrote: »

Wrong, the energies involved is the square of the speed.



The car going at 50 will have a K.E of massx1250 and the car at 30 will have a K.E of massx450, non SI units used so only a rough estimation. The faster travelling car has more kinetic energy so should do more damage to the slower travelling car.

In your calculation you assumed, that during the accident car going 50 would stop to 0 and car going 30 would stop to 0.
That's wrong, as this would break the rule of conservation of linear momentum.

In fact what is going to happen, car going 50 will slow down to 10, and car going 30, will be accelerated backwards to 10 (so both cars will travel 10 in the same direction).
Obviously this speed will be lost very quick due to friction between cars and the ground (wheels might not operate anymore, brakes might get engaged, etc).
But generally assumption that cars will stop still at the place of accident is only valid when they hit each other in exactly straight line at the same speed, and they will be both the same kind of cars with the same weight.
If speed or weight differs, then after the accidents car will still move.

So according to rule of conservation of momentum (assuming that those 2 cars are isloate system without any other forces working on them) we have somethign like that (assuming cars weight the same).

p=m*v (momentum = mass * velocity)


p(total) = p1(car 1 momentum) + p2 (car 2 momentum)
assume m=1 (tonne)
p1 = 1*50 = 50
p2 = 1*-30 = -30 (minus because other car is going in opposite direction)

p(total) = p1 + p2 = 20


So the total momentum of both cars is 20, and it must remain the same after the accident (due to rule of conservation of momentum)

After the accident:
p(total after) = 20
P(total after) = (m1+m2) * v

v= P(total after) / (m1+m2)

v = 20 / 2 = 10

So the speed after the accident will be 10MPH.

And as you said with kinetic energry - both cars will suffer change of speed of 40mph, so their kinetic energy loss will be equal, therefore damage should be equal.

tacofries

Mikros wrote: »

The train will not experience the same force as the car because the change in velocity of the train will be almost 0.

Sorry, the question my post was aimed at was one about force(not the original question by the poster).

I do still stand by the car and the train experiencing the same force when they crash due to the fact that when the train hits the car, the car also hits the train. It doesn't matter if the change in velocity will be almost 0 for the train since force is equal to mass by acceleration,, and so even if the de-acceleration of the train is very small the mass of it is large enough to make the force very large. Obviously the change in the cars velocity will be very high, but only in proportion to the mass of the train.. Thus both experience the same force!

CiniO

Mikros wrote: »

You don't understand Newtons laws and their application to collisions.

The force each car experiences is determined by the acceleration.

You forgot about one detail - force is determined by mass and acceleration (no solely acceleration)

Acceleration is change in velocity over time. When two cars collide there is a change in velocity in a very short period of time. The change in velocity is based on the conservation of momentum - the momentum before the collision is the same as the momentum after.

All true here.

Momentum is mass x velocity. The faster and heavier car will lose less velocity and therefore will experience less acceleration / force.

Heavier car - true - will lose less velocity.
But if they weight the same, they will both experience the same velocity change.

Think about if a train hit a car head-on both travelling at 80 km/h. The train will not experience the same force as the car because the change in velocity of the train will be almost 0.

The train and car will experience the same force.
But acceleration is force * mass (a=Fm).
So the same force will give huge acceleration to car, as his mass is small, while the same force will change train's speed only by a negligible bit.

MrDerp

07Lapierre wrote: »

Send an email to "mythbusters"!!. They'll sort it!

They already did
http://ca.autoblog.com/2010/11/21/road-myths-physics-of-a-head-on-collision/

The nuts and bolts of the results are:
1. If two cars collide head on, both doing 50km/h, the result is the same as hitting a wall at 50km/h
2. It is not the same as a 100km/h car to wall collision, that was the myth busted

Forget the maths and equations for a minute:
- If two objects hit at the same speed, it's the same for each as if they hit a wall at that speed
- Car 1 effectively hits a wall at 50km/h; Car 2 effectively hits a wall at 50km/h

Why?
- Newton's 3rd law is often simplified to every action has an equal an opposite reaction
- Roughly speaking, the energy taken into the collision is exerted back.
- It seems intuitive that the relative velocity is 100km/h and so double the impact, but there is 4 actions (2 actions and 2 reactions) dissipating the energy in 4 ways:
1. Car A exerts an action, actionA, onto Car B
2. actionA has an equal and opposite reaction, reactionA, onto Car A
3. Car B exerts an action, actionB, onto Car A
4. actionB has an equal and opposite reaction, reactionB onto Car B
- actionA = reactionA
- actionB = reactionB

Chucking the physics and maths aside, think of the reaction as the impact. If you hit a wall (immovable object), the reaction is equal to the action, and the reaction crushes your car.

A car travelling at the same speed is a brick wall. Since both cars pop with the same energy (assuming equal weight here), they each have the same reaction as if they hit the wall at 50km/h.

Now, where it gets interesting is if the cars are at different speeds. Lets say Car A is at 50km/h and Car B is at 100km/h.
- action A = reaction A
- action B = reaction B
- BUT the speed is not equal

Effective car to wall impact speed? 100km/h.
- Car B hits a 'wall' at 100km/h [action B = reaction B]
- Car A has a 'wall' coming at it at 100km/h. It hits at 50km/h, but the wall hits back at 100km/h.

You see:
- The action brought by car A in hitting car B has an equal and opposite reaction on car A, but car B is hitting car A too, which in turn has an equal and opposite reaction from car A to car B.
- The action/reaction pair on A must be the same as the action/reaction pair on B


The 'effective speed' (car to wall equivalent) of the crash is the speed of the faster car.

In short, the momentum of the car does not change just because there's something coming towards it.

BrianD3

MrDerp wrote: »

Now, where it gets interesting is if the cars are at different speeds. Lets say Car A is at 50km/h and Car B is at 100km/h.
- action A = reaction A
- action B = reaction B
- BUT the speed is not equal

Effective car to wall impact speed? 100km/h.
- Car B hits a 'wall' at 100km/h [action B = reaction B]
- Car A has a 'wall' coming at it at 100km/h. It hits at 50km/h, but the wall hits back at 100km/h.

You see:
- The action brought by car A in hitting car B has an equal and opposite reaction on car A, but car B is hitting car A too, which in turn has an equal and opposite reaction from car A to car B.
- The action/reaction pair on A must be the same as the action/reaction pair on B


The 'effective speed' (car to wall equivalent) of the crash is the speed of the faster car.

What if car A is doing 0 km/h and car B is doing 80 km/h - in terms of damage and injury the "car to wall" equivalent is not going to be the speed of the faster car. An 80 km/h impact into a parked car is a lot less severe (for the occupants of both the parked car and the moving car) than an 80 km/h impact into a wall.

Ronnie Beck

BrianD3 wrote: »

What if car A is doing 0 km/h and car B is doing 80 km/h - in terms of damage and injury the "car to wall" equivalent is not going to be the speed of the faster car. An 80 km/h impact into a parked car is a lot less severe (for the occupants of both the parked car and the moving car) than an 80 km/h impact into a wall.

Depends on the wall...

CiniO

Ronnie Beck wrote: »

Depend on the wall...

True.
Some wall lose even with small cars.

https://www.youtube.com/watch?v=2TbNR7L57FQ

MrDerp

BrianD3 wrote: »

What if car A is doing 0 km/h and car B is doing 80 km/h - in terms of damage and injury the "car to wall" equivalent is not going to be the speed of the faster car. An 80 km/h impact into a parked car is a lot less severe (for the occupants of both the parked car and the moving car) than an 80 km/h impact into a wall.

Neither here nor there. I said nothing about damage, I was merely talking about the equivalent speed of impact. Two cars will crumple and take some impact, and a car on wheels is more moveable than a wall. Whatever the obstacle, the impact on the moving car and the other object is the same.

As, car chases on tv show us, a car hitting a bunch of boxes will scatter the boxes, a car hitting a plate glass window on it will go through the glass, and a car hitting a clothes rail will drag it down the street, with hilarious consequences. This has everything to do with what's hit and doesn't change newtons 3rd law.

A trampoline will bounce you better than a sheet, but it doesn't mean the impact is less, it's the elasticity of what your hitting that changes the results. I'd rather hit a modern parked empty car than a wall or tree for sure, just as I'd rather hit that wall in my 2007 car than a 1993 kadett

Anan1

Ronnie Beck wrote: »

Depends on the wall...

For the purposes of this discussion the wall is immovable.

coylemj

In a head-on collision, the car with the greater kinetic energy (same as momentum) will push the other car backwards. Kinetic energy as already stated is 1/2 * mass * velocity squared. People here are talking about acceleration which has nothing to do with it, all that matters is the actual speed at the moment of impact and the weight (mass) of the vehicle.

Only if both cars have the same KE can you compare it to a single car hitting an immovable object, otherwise one car will push the other backwards and there will be significant differences in deceleration.

Say two cars hit head-on, both travelling at 50 kph but one is bigger than the other. Let's say that a fraction of a second after the impact, the two cars are travelling in the original direction of the bigger car at 20 kph. The bigger car has experienced an almost instantaneous deceleration of 30 kph but the smaller car has gone from +50 to -20 i.e. a deceleration of 70 kph. Leaving out the issue of how much damage has been caused to each car (which depends on how they were built), the occupants of the smaller car in this case will be much more likely to suffer internal organ damage due to the rapid deceleration.

Seatbelts and airbags will protect you from a direct impact with metal and other hard parts of the car, what does people in in a lot of head-on crashes is the damage to internal organs caused by the sudden stop i.e. the rapid deceleration, that's what killed Princess Diana.

Ronnie Beck

Anan1 wrote: »

For the purposes of this discussion the wall is immovable.

'wall', not wall

Ronnie Beck

coylemj wrote: »

all that matters is the actual speed at the moment of impact and the weight (mass) of the vehicle.

Angles, lots of angles involved. Different materials too.

Is this not a discussion about a perfect head on collision of identical cars, neither car swerving or braking. In this case, in theory, the occupants of both cars would experience the exact same decelleration and g-forces on internal organs. m1v1+m2v2before =m2v1+m1v2 after. Simple.

In reality I'd rather be in the slower car. A better chance of avoiding the gobsh;te driving towards you, meanwhile he fly's off the road and hits a tree.

Ronnie Beck

From OP

Car A 30kph. Car B 50kph.

mass 1500kg or m

x = velocity of faster car after collision

30m -50m = 0m + xm
-20m=(0+x)m
-20=x

both car experienced a change of velocity of 30kph.

Car A 30 to 0
Car B 50 to 20 or -50 to -20

Ronnie Beck

Ronnie Beck wrote: »

From OP

Car A 30kph. Car B 50kph.

mass 1500kg or m

x = velocity of slower car after collision

30m -50m = xm + 0m
-20m=(0+x)m
-20=x

both car experienced a change of velocity of 50kph.

Car A 30 to -20
Car B 50 to 0 or -50 to 0

FYP

or 50 to 10 and -30 to 10. How elastic or inelastic is the collision in this situation??? I suppose they would just blend together if there was no angles involved.

coylemj

Ronnie Beck wrote: »

Car A 30 to 0
Car B 50 to 20 or -50 to -20

How can one car go to zero and the other car go to -20? They are going to be embedded in one another a fraction of a second after the impact at which point there will be deceleration on both sides. The dissipated energy will be absorbed by crushing metal and the destruction of various other parts.

In the case you're quoting, if one car goes from 50 to 20, that implies that it continues in the same direction at 20 kph which means the other car goes from +30 to -20 i.e. it will experience a more or less instantaneous deceleration of 50 kph.

If both cars are the same mass, the faster car will push the slower car backwards purely because it has more kinetic energy, there is no way that they will both come to rest at the point of impact unless they both have identical kinetic energy in which case the effect of the collision will be the same as either of them hitting an immovable object.

Ronnie Beck

I was thinking out loud. I just realised at the end that the two cars would merge into one mass.

BrianD3

MrDerp wrote: »

Neither here nor there. I said nothing about damage, I was merely talking about the equivalent speed of impact

The OP asked about the risk of injury. While you seem to be correct about Newton's third law, in the context of this thread, making the statement:

The 'effective speed' (car to wall equivalent) of the crash is the speed of the faster car.

Is very likely to be misinterpreted as relating to injury potential.

BrianD3

Posters are now talking about cars of different masses and the KE. This confuses matters.

Same masses, different speeds:
Car A is travelling at 30
Car A weighs 10

Car B is travelling at -50
Car B weighs 10

((30*10)+(-50*10))20 = -10. That's the speed of the wreckage after impact. Both cars slow down by 40. Car A is pushed backwards but the human body doesn't distinguish between going from -50 to -10 or going from 30 to -10.


Different masses, same speeds:
Car C is travelling At 50
Car C weighs 20

Car D is travelling at -50
Car D weighs 10

((50*20)+(-50*10))/30 = 16.66. That's the speed of the wreckage after impact. Car C slows down by 33.33 whereas car D slows by 66.66 and is pushed backwards. The human body certainly does distinguish between slowing down by 33.33 and slowing down by 66.66.

The above shows why heavier cars are generally safer than light cars but why speeding up before an impact to "increase momentum and KE of your car and plough through the other car" is not an effective road safety strategy

MrDerp

BrianD3 wrote: »

The above shows why heavier cars are generally safer than light cars but why speeding up before an impact to "increase momentum and KE of your car and plough through the other car" is not an effective road safety strategy

Also, consider that (especially in older cars) if a frontal collision is unavoidable, you might be better off hitting square on than having a partial overlap collision

http://abcnews.go.com/blogs/headlines/2012/08/partial-collisions-prove-more-dangerous-in-new-crash-test/

BrianD3

MrDerp wrote: »

Also, consider that (especially in older cars) if a frontal collision is unavoidable, you might be better off hitting square on than having a partial overlap collision

http://abcnews.go.com/blogs/headlines/2012/08/partial-collisions-prove-more-dangerous-in-new-crash-test/

Yes, the small overlap is a very interesting test. There have been a huge number of interesting crash tests and stuides done in recent years. Full overlap, 40% overlap, 25% overlap, rigid barriers, deformable barriers, oblique angles, big car vs small car, old car vs new car - and those are just the frontal tests

In terms of the deceleration felt by the occupants, Mercedes S class vs Smart Fortwo offset test is a good one. The Smart is flung backwards violently and rolls over. It looks like a terrible collision for the Smart occupants but IIRC the injury risk is not that high and crucially, the very strong Smart passenger compartment remains relatively intact.

Obviously at higher speeds the Smart is at a bigger disadvantage. Increase the speed of both cars by 5 mph and the S class occupants may still walk away while the Smart occupants may be dead.

The Smart test shows the rigidity of modern cars. However they are very reliant on the modern restraint systems working correctly. I don't speak German but this article seems to say that in a full head on crash with an immovable object at 50 km/h, a soft Renault 19 without an airbag is safer than a rigid Renault Megane II with non working airbag
http://www.welt.de/motor/article1244415/Warum-ein-altes-Auto-sicherer-ist-als-ein-neues.html

Yet if the Megane hits the R19 in a full head on or especially a partial overlap test, the old car will be far weaker and will likely fold like tinfoil, crushing the occupants. Not much good having a crumple zone and reduced deceleration if your head, legs and chest are in the crumple zone!

BrianD3

While I'm at it - crash compatibility. We've already seen from conservation of momentum that occupants of light cars are more vulnerable than occupants of heavy cars in a two car collision.

As well as that, the light car will probably have a weaker crumple zone. A light car is designed to crash into a wall at 40 mph and keep the passenger compartment intact and keep deceleration for its occupants at a tolerable level. A heavy car is designed to do the same but due to increased KE of a heavy car, it needs a stronger and/or bigger crumple zone.

If you give the small car the crumple zone from the big car it will either have a very long bonnet or a very rigid crumple sone. The former is undesirable due to packaging and if the the latter is done, in a crash with an immovable object, the crumple zone is too strong meaning occupants are subject to high declerations.

If you give the big car the crumple zone off the small car it will be too weak and the passenger compartment may be crushed in a crash with an immovable object.

So, each car has to have a different crumple zone in case they hit an immovable object. But what happens when they hit each other. The heavy car will likely be more aggressive as well as having more momentum. This is reiterated by the quote below (bold emphasis mine) from "Vehicle Structural Crashworthiness with
respect to Compatibility in Collisions
Frei P, Kaeser R, Muser MH, Niederer PF, Walz FH"

This is why EuroNCAP frontal crash test results can't be compared if the cars are significantly different weights and only claim to approximate a crash with an identical car.

However the EuroNCAP side impact result with the moving sled can be compared among weight classes. Also the pole test can be compared as it's unlikely that you will crash sideways into a pole moving towards you, only a stationary pole.

COLLISION MECHANICS
GENERAL - A small, light car will always experience
a higher deceleration level in a real car versus car
collision compared to its heavier opponent, as the deceleration
for each of the colliding cars can be calculated
with a simplified model from the crush load at the interface
of the two cars. Both cars experience the same
crush load vs. time curve with this simplified model.
deceleration = crush load / car mass
Furthermore, this type of frontal collision will
subject the lighter car to a higher total change of velocity.
The velocity change of both cars can be calculated with
the law of conservation of momentum. In real collisions,
the observed Dv lies 5 - 10 % above the theoretical value
due to residual elasticity of the deformed structures.
Dv1 = m2 * ( v1+ v2) / (m1+ m2)
Dv2 = m1 * ( v1+ v2) / (m1+ m2)
As this paper shows, the main reason for the
higher injury risk for the occupants of currently circulating
small cars in frontal collisions is not intrinsic in these two
physical constraints. Instead, it is given by the lack of
compatibility in frontal collisions among cars of different
size and weight.
The restraint system of any car will only be able
to deploy its effectiveness if the passenger compartment
does not collapse in a collision. Intrusion of structural
components into the cabin and deformation of the passenger
compartment must be kept small. An extremely
important parameter for the development of the restraint
system is the mean deceleration level of the car cabin
with its restraint system interface points and surfaces
impacted by the body of the occupant during the
collision.
Current crash test standards still encourage to
design the car front stiffness for moderate cabin deceleration
pulses, allowing the use of conventional restraint
system components. This approach creates car front
stiffnesses that are more or less proportional to the car
mass. A low mass vehicle, designed in such a way, will
be disadvantaged twofold in case of a collision with a
heavier vehicle. Not only will it experience a higher Dv
due to the mass ratio, but its frontal deformation space,
which is relatively soft (according to the reasons mentioned
above), will be crushed before significant deformation
of the heavier car even starts.
In conclusion, the frontal structural stiffness of a
low mass vehicle must be at least equal or slightly higher
than the stiffness of its heavier counterpart [7] [8].