Tesla head on collision with a Honda

[flar]

If the two vehicles fuse into a single lump, this is a good way to think about it. If both the Tesla and the Honda were doing 40 mph before the collision, the TesHonlada lump would be doing, say, 10mph in the same direction as the Tesla because the Tesla is heavier. Only for an instant, of course, since the lump wouldn't roll very well. Taking that as the frame of reference, the Honda might as well have been doing 50 mph in terms of the damage it took, while the Tesla would feel like it was only doing 30.

... but having written that, I'm not sure it's correct. It works for momentum but not necessarily energy. Have to think some more.

Collisions preserve momentum. Elastic collisions also preserve energy, but inelastic collisions do not. Car collisions are inelastic and the energy that isn't preserved is turned into noise, heat, and deformation of metal.

Here is a site that has a good description: http://www.physicsclassroom.com/mmedia/momentum/cthoi.cfm

Note that the energy given up is not "1/2 m (v1 - v2)^2", it is "1/2 m v1^2 - 1/2 m v2^2" or "1/2 m (v1^2 - v2^2)". Saying "car A was going v1 before the collision and v2 after so it lost the energy equivalent of (v1-v2)" doesn't give you the right amount. The amount will be off by the difference in the two equations above which is 2*v1*v2.

Note also that I realize I was not thinking clearly above when I said that if the Tesla was doing 80 before and 40 after then it would have given up "40mph worth of energy". That amount would be 1/4 of its original energy, but it went from a large (80 squared) amount of energy before the collision down to a much smaller (40 squared) amount after and so it really would have given up 3/4 of its energy to the collision. Still, the point I was trying to make about not giving up all of its 80mph worth of energy in the collision is correct. It's just that you'd have to run the equations to find a before and after velocity that reduced its energy by the proper amount and "80 before and 40 after" aren't those numbers...

 

[flar]

No, it's not. Remember a cars' speed (in this case 40mph) is not an absolute number, but a number relative to the earth. The cars didn't crash into the earth, so why take speed relative to the earth into account at all? If at the same time a Boeing was flying overhead at 500mph ground speed, then relative to the Boeing, one car would have been travelling at 460mph and the other one at 540mph. Saying one car imparted damage as if it was travelling at 460mph is equally as invalid as saying it was travelling at 40mph.

What matters is that the speed of one car relative to the other car at the point of impact is 80mph. Whatever the car's speed is relative to some other arbitrary point in space is irrelevant. So relative to the earth, whether one car was 0 and one at 80, or one at 40 and the other at 40, or one at 100 and the other reversing at 20 - it's all the same thing. (If it helps, take the earth out of the equation and imagine the collision happens in space).

Actually, no. He was correct. You might want to review the Mythbusters video I posted above: http://www.youtube.com/watch?v=r8E5dUnLmh4

40 vs. 40 is the same as 40 into brick wall.

What matters is using the same frame of reference during the entire transaction. If you are taking the measurements of one car "relative to the other" then your frame of reference is involved in the collision and the math gets really complicated. You can sort of rescue that model if you decouple the frame of reference from the car itself and have it simply follow one of the cars and continue on away from the collision after at a constant velocity. But the frame of reference cannot come to a grinding halt along with the cars or you've modified your frame of reference. If you use the earth as the frame of reference for both cars during the entire transaction then your frame of reference did not undergo any transformations and you get good math.

If you use a passing plane then it still works out as RDoc showed earlier. From the perspective of the plane, one car was going 460 before and the other was going 540. After, if they both came to a dead stop at the point of impact, both are now going 500 relative to the plane. Both cars had a change in velocity of 40mph and the math works out. If you start relative to the plane then you have to keep observing relative to the plane.

 

[Jeff Miller]

I think it may be less confusing to focus on acceleration (change in velocity) rather than energy. It's the acceleration which destroys the car and harms the passengers in an accident. Energy is frame dependent as flar has noted so you need to think about how to use it. Acceleration however is the same in all frames, since all observers have to agree in the end on what happened (people killed, cars destroyed).

Let's do a back of the envelope calculation to estimate the acceleration. You are going at speed v. Over a short distance L your speed is reduced to zero. Assume that the speed reduction is linear (it's back of the envelope). The average speed during deceleration is then v/2. The time it takes to stop is L/(v/2). So the acceleration is dv/dt = v/(L/v/2) = v^2/(2L). This proportional to the car's initial energy in the frame where the car is at rest after the accident (our normal everyday rest frame).

It's also clear from this that in inelastic collisions (cars don't ricochet off each other), running into a cliff at 40 is going to produce the same acceleration as running head on into another car which is doing 40 in the opposite direction.

It's also clear that crashing into a parked car at 80 mph is identical to two cars hitting each other head on going 40 mph. This is the meaning of Galilean invariance.

 

[Yggdrasill]

We should use the earth as our frame of reference because we are the ones doing the observing, and we live on earth.

If you view the situation as car A moving at 80 km/h and car B moving at 0 km/h, the end result isn't that both cars end up at 0 km/h, because 0 km/h would be relative to the earth. Car A ends up at 40 km/h and car B ends up at 40 km/h. (Same speed same direction)

The calculations using this frame of reference is thus:

0,5 * m * 80^2= 3200m

Minus remaining potential energy:

2 * 0,5 * m * 40^2 = 1600m

3200m - 1600m = 1600m

This is incidentally the exact same result as if one uses the earth as the frame of reference.

With car A traveling at 40 km/h and car B traveling at -40 km/h, and they come to a rest at 0 km/h, the calculations for total expended energy are like this:

2 * 0,5 * m * 40^2 = 1600m

The frame of reference one uses does not affect the result, all it affects is the complexity of the calculations.

 

[flar]

It's also clear that crashing into a parked car at 80 mph is identical to two cars hitting each other head on going 40 mph. This is the meaning of Galilean invariance.

Are you assuming that 80 into parked car ending up continuing on at 40 (in a clump)? Or 80 into parked car stopping at the point of impact?

- - - Updated - - -

The calculations using this frame of reference is thus:

0,5 * m * 80^2= 3200m

Minus remaining potential energy:

2 * 0,5 * m * 40^2 = 1600m

Just a terminology nit - if your calculation involves velocity then it is kinetic energy. Potential energy is what a ball has when it is sitting stationary on a shelf. If the shelf disappears then it converts that potential energy into kinetic energy as it accelerates due to gravity...

http://www.diffen.com/difference/Kinetic_Energy_vs_Potential_Energy

 

[Jeff Miller]

Are you assuming that 80 into parked car ending up continuing on at 40 (in a clump)? Or 80 into parked car stopping at the point of impact?

I'm only assuming the cars are equal mass. In that case, center of mass would continue at 40 (conservation of momentum) But you don't need to assume anything about whether the collision is elastic or inelastic. It's just a very general statement that everything must be the same
whether one guy is going 80 and the other is standing still, or both are going 40. 40 and 40 vs 80 and 0 is just a statement of which arbitrary frame you are in. All inertial frames give the same physics.

 

[eledille]

I'm only assuming the cars are equal mass. In that case, center of mass would continue at 40 (conservation of momentum) But you don't need to assume anything about whether the collision is elastic or inelastic. It's just a very general statement that everything must be the same
whether one guy is going 80 and the other is standing still, or both are going 40. 40 and 40 vs 80 and 0 is just a statement of which arbitrary frame you are in. All inertial frames give the same physics.

If a 1000 kg car crashes into a wall at 20 m/s the crash energy is 1/2*1000*20^2 = 200 kJ.
If a 1000 kg car crashes into a wall at 10 m/s the crash energy is 1/2*1000*10^2 = 50 kJ.
If a 1000 kg car travelling at 10 m/s crashes into another 1000 kg car travelling at the same speed but opposite direction, both cars will stop and the crash energy is 2*1/2*1000*10^2 = 100 kJ, with half of that going into crumpling each car. The result is identical to both cars crashing into a wall.

If two cars of identical mass crash, with one car stationary but free to start moving and the other travelling at 20 m/s, then both cars will be travelling at 10 m/s in the same direction after the crash. The moving car has lost 3/4 of its energy, but the stationary car has gained 1/4 of the energy the moving car had before the crash. Only half of the initial the energy goes into the crash, because both cars are moving at 10 m/s after the impact. Crash energy is 100 kJ, evenly split between the cars, and the cars will again look like they had both crashed into a wall at 10 m/s.

Tethering the stationary car produces a different result - it's now a wall with a crumple zone. In this case both cars are stationary after the crash, and the crash energy doubles compared to the case with an untethered stationary car. Both cars should look about twice as bad as a 10 m/s crash into a wall, and about half as bad as a single car at 20 m/s crashed into a wall.

 

[Jeff Miller]

I think we agree about the first 2 paragraphs.

I had not considered the case of a tethered car; it seems in most cases of interest cars aren't tethered, but it is true that when I say the
80/0 and 40/40 crashes are the same, I am assuming that the friction between the car and the road is the same. In particular I am assuming the cars are in drive/park and drive/drive and not in neutral.

 

[bonnie]

I would like to take a moment to congratulate everyone for having the single nerdiest car club website on the net. Let the math continue....

+1 ... probably my favorite thing about the site. Feels comfy.

 

[CapitalistOppressor]

Just to be clear. The energy of a single car into a wall is 1/2mv^2. The energy of 2 cars into each other is 1/2*m1*v1^2 + 1/2*m2*v2^2, not 1/2(m1+m2)(v1+v2)^2 so you cannot compare "40mph into 40mph" with "80mph into parked car". The latter has twice the energy of the former because you add the squares, not square the sum.



True, as long as you understand that it goes up with the square of the velocity of each individual car separately and then gets added into a sum, not the square of all of the velocities involved.



A combined impact of 140 could be one car at 140 into a brick wall which would still have a huge amount of energy, or it could be 70 into 70 which would have a bunch more energy than your 60mph car you used as an example, but not nearly as much as 140 squared compared to 60 squared.

The rest of what you said I generally agree with, but I keep seeing this "adding the velocity of the cars" mistake when people talk about collisions and wanted to make sure the math was right.

Also, if you watch Mythbusters they were rather surprised that doubling the speed didn't quadruple the damage to their car, but they chose a velocity where the car was smashed so much that the crumpling wasn't very linear when they doubled the speed. http://www.youtube.com/watch?v=r8E5dUnLmh4

All correct, and I did not intend to imply otherwise. But it's also not really relevant to the point I was attempting to make. Neither you, nor I, nor anyone in this forum has the necessary information to correctly model this accident, let alone identify the cause.

Do I know this is what happened? No, I am not a trained accident investigator, and I lack the logs from the Tesla and analysis of the skid marks and impact speeds that would help properly model the accident. It's an opinion based on my own (potentially flawed) interpretation of the evidence that we have.

But I do know that your hypothesis is almost entirely unsubstantiated and speculative in the extreme. The idea that these cars were drag racing is nothing but pure gossip based on a highly shaky interpretation of an incomplete set of "facts". If the police make an arrest it wont be because they relied on so called evidence generated by this forum. They will look at vehicle logs, witness statements and their own reconstruction of the accident based on actual science instead of amateur musings that are trying to pass themselves off as informed analysis.

That said, I see a lot of math being bandied about further down this thread. I don't feel like engaging, but its utterly false to assume that a 40+40mph head on crash is equivalent to a 40mph crash into a wall. The mass of the Model S is dramatically more than that of the Honda and that ends up being a dominant factor that results in the Honda experiencing far worse deceleration forces and damage.

If this had been a head on crash the Honda would not have simply stopped (as in the case of a wall) but would have been accelerated backwards and experienced more damage, while the Model S would have experienced less damage than it would have by crashing into a wall at that speed. The actual result is more complicated because the crash was offset.

As to how much remaining energy the Model S had, again we do not know. We can probably come up with a good estimate for how far it traveled after the impact based on available photographs (at this point "several hundred feet" is an unsubstantiated guess, and a quick glance at the photo with the labeled impact points indicates it traveled about a hundred feet between the initial impact and where the Model S ended up on the guardrail). But the idea that it was "dragging" a wheel requires pure speculation as to how much actual friction was occurring.

The front left wheel well suffered damage of an unknown nature, and I don't see evidence of drag marks. Just skid marks from the two front tires and at least one of the rear tires which end shortly after the accident and appear to indicate the car was in a bit of a sideways curving skid immediately after the accident. After the car (presumably) straightened out and headed towards the guardrail I don't see any additional skid marks. Regardless our available information is limited and we are forced to speculate.

I still see no evidence which allows us to determine whether the cars in question were drag racing. I certainly do not believe it to be the case that the Model S was doing 80 at the time of impact, but that is just an opinion based on what I see. The original poster that I was responding to was pretty adamant that they must have been racing, and he was right that there is quite a distance between where the lanes merge and the ultimate impact point. But he, along with everyone here, is probably wrong to think we have enough data determine the impact speeds of the two vehicles, let alone what actually caused the crash.

 

[eledille]

I think we agree about the first 2 paragraphs.

I had not considered the case of a tethered car; it seems in most cases of interest cars aren't tethered, but it is true that when I say the
80/0 and 40/40 crashes are the same, I am assuming that the friction between the car and the road is the same. In particular I am assuming the cars are in drive/park and drive/drive and not in neutral.

I wrote:

Tethering the stationary car produces a different result - it's now a wall with a crumple zone.

I could have expressed myself more clearly, let me try again:

Tethering the stationary car produces a different result - it essentially turns into Mother Earth with a crumple zone in front. Earth is the heaviest solid object around and *really* heavy. That has a dramatic impact, momentum is preserved, but as Earth is so insanely heavy, it gains almost zero speed and therefore almost zero kinetic energy. You can safely assume that all the kinetic energy goes into crumpling the cars. But there is twice as much car to crumple, of course, compared to a single car crashing into a wall (a.k.a. Earth).

I agree that this is an unusual scenario, but it might happen if a car crashes into a car parked in front of a wall, for example.

I'm just trying to explain the physics of collisions, I'm not saying anything about what happened in this particular crash. The above calculations assume that the crashes are completely inelastic. They aren't, but they are fairly close.

That said, I see a lot of math being bandied about further down this thread. I don't feel like engaging, but its utterly false to assume that a 40+40mph head on crash is equivalent to a 40mph crash into a wall. The mass of the Model S is dramatically more than that of the Honda and that ends up being a dominant factor that results in the Honda experiencing far worse deceleration forces and damage.

I completely agree, a 40 + 40 mph crash where the masses of the cars differ substantially is going to produce a very different result from what would happen if both of those cars crashed head-on into a wall at 40 mph.

 

[PeterW]

I completely agree, a 40 + 40 mph crash where the masses of the cars differ substantially is going to produce a very different result from what would happen if both of those cars crashed head-on into a wall at 40 mph.

A question more than a statement. Would it be correct that a 60 + 20 mph crash would differ from a 20 + 60 mph crash due to the different masses of the vehicles?

 

[Yggdrasill]

A question more than a statement. Would it be correct that a 60 + 20 mph crash would differ from a 20 + 60 mph crash due to the different masses of the vehicles?

If the masses of the cars aren't the same, definitely. If you assume you have a 1 ton vehicle and a 2 ton vehicle, the difference would be:

2 ton vehicle at 60 mph:

0,5 * 2000 * 26,8^2 = 718 kJ

1 ton vehicle at 60 mph:

0,5 * 1000 * 26,8^2 = 359 kJ

2 ton vehicle at 20 mph:

0,5 * 2000 * 8,9^2 = 79 kJ

1 ton vehicle at 20 mph:

0,5 * 1000 * 8,9^2 = 39 kJ

So for the 2 ton vehicle at 60 mph/ 1 ton vehicle at 20 mph scenario, the kinetic energy released will be 757 kJ, while the 1 ton vehicle at 60 mph/ 2 ton vehicle at 20 mph scenario will release 438 kJ. Basically if the 2 ton vehicle is the car doing 60 mph, the impact will be almost twice as energetic.

 

[yobigd20]

A question more than a statement. Would it be correct that a 60 + 20 mph crash would differ from a 20 + 60 mph crash due to the different masses of the vehicles?

The force exerted on each car will be exactly the same. If a bug hits your windshield, the force the car exerts on the bug is the same that the bug exerts on the car ... this is Newton's third law of "equal and opposite reaction forces.". So it doesn't matter is one car is heavier (more massive) than the other. However, why does 'bug' not fly through your windshield like a bullet? That's because the resulting motion (mph/acceleration after the impact) will be much different based on how much each object absorbs energy (e.g. the "crumple zone").

 

[PeterW]

The force exerted on each car will be exactly the same. If a bug hits your windshield, the force the car exerts on the bug is the same that the bug exerts on the car ... this is Newton's third law of "equal and opposite reaction forces.". So it doesn't matter is one car is heavier (more massive) than the other. However, why does 'bug' not fly through your windshield like a bullet? That's because the resulting motion (mph/acceleration after the impact) will be much different based on how much each object absorbs energy (e.g. the "crumple zone").

Yes I follow what you are saying. There were previous posts on this thread that suggested that part of the reason that the Model S feared better in the crash was because it is a much heavier vehicle. What you are saying is that would be incorrect.

 

[Yggdrasill]

Yes I follow what you are saying. There were previous posts on this thread that suggested that part of the reason that the Model S feared better in the crash was because it is a much heavier vehicle. What you are saying is that would be incorrect.

The reason why it fares batter is not just that it is heavier. If the extra weight was manifested in a block of lead behind the drivers seat, the driver of the heavier car would die first.

All vehicles are designed to survive hitting a wall at X* mph with no deaths. This means that the crumple zone of a 2 ton car is twice as strong as the crumple zone of a 1 ton car. (As it needs to be twice as strong to decelerate twice the mass in an approximately equal amount of time.) This means that in a collision, the smaller car, with the weaker crumple zone, will lose.

The crumple zone of the heavier car might crumple by 50%, while the crumple zone of the smaller car might crumple by 150%. (This depends on the specific vehicles in the specific situation, it can't be solved on a napkin.)

* Not taking into account different model years. It might be X for model year A and Y for model year B.

 

[yobigd20]

Yes I follow what you are saying. There were previous posts on this thread that suggested that part of the reason that the Model S feared better in the crash was because it is a much heavier vehicle. What you are saying is that would be incorrect.

Sort of. You can't defy physics. I'm simple saying that the initial force exerted on each vehicle is THE SAME. However, newton's second law is what counts. Newton's second law states that the net force on an object is equal to the rate of change of its linear momentum over times it's mass.

A critical component of a vehicle's rate of change of momentum is the design of the crumple zone. A lighter vehicle's crumple zone is much smaller, and the change in motion will be faster. therefore a lighter vehicle's deceleration will be greater in general and this is what you feel - this is the remaining force exerted upon it's occupants.

The Model S feared better in the crash because, yes the car is heavier, but Tesla's engineers are also genius' and the crumple zone in the front of the Model S is incredibly brilliant's designed. While vehicle mass is a very important factor in the equations, it is not the 'mass of the Model S' that saves you, but instead the design of the crumple zone. In fact, the mass of the your may work against you, not for you. The heavier your vehicle, the greater the force of impact. But this also depends on what your hitting and the resulting mph after the impact. In other words, consider heavy car A and light car B in head on collision. Heavy car A's change in velocity is from 60mph before impact to 15mph after impact. A's change in velocity is -45mph. Lighter car B goes from 60mph to -15mph (bounces backwards), thus B's change in velocity is 75mph. But what's even more critical than mass or the change in velocity is the time it takes to make this change. Let's put this into a very easy perspective. If it takes the heavier car .1 seconds to go from 60mph to 15mph but the lighter vehicle took 10 seconds to go from 60mph to -15mph, then the heavier car's occupants are not going to fair so well while the lighter car's occupants are doing just fime. This is where crumple zones come into play. It's a car's crumple zone that saves your life.

To understand further, you need to know how crumple zones work. This write up is pretty decent (4 pages) http://auto.howstuffworks.com/car-driving-safety/safety-regulatory-devices/crumple-zone1.htm and I'll try to summarize below

Force of impact = mass * acceleration

Acceleration (deceleration) = delta velocity (change in velocity) / time

So basically, if you go from 60-0 in a very short about of time, say .8 seconds, then resulting force of impact is 60/.8=75.

The point of the crumple zone is to reduce the initial force of the crash and redistribute this force before it reaches the occupants.

The best way to reduce the initial force of the crash is by extending the time of the crash. So crumple zones are designed to 'crumple'. Definition: 'Crush so that it becomes creased and wrinkled.' The act of crushing to crease/wrinkle takes time. The longer the better. Webster def of crumple zone: 'a section of an automobile body designed to absorb the force of an impact in order to protect the passengers.'

If the crumple zone extends the rate of deceleration from .8 seconds to 2 seconds, then 60/2 = 30. Force of impact is MUCH smaller. That is a *HUGE* life-saving result.

Now the remaining part of the equation is what to do with the remaining force. So it's not the weight of your vehicle that saves you (as obvious from the equations, it works against you, not for you), it's how the car is designed to redistribute the remaining force. You don't want all the remaining force to be exerted on the occupants. Crumple zones create a buffer around the perimeter of the car and help redistribute the remaining force of impact. It's gotta go somewhere, and you want it away from the occupants. It can be transferred to the object your hitting, or spent if your car is spinning around in circles after the crash, or it can be absorbed by your car itself (and hopefully not by you).

"Crumple zones spend as much force as possible so that other parts of the car as well as the occupants don't suffer the effects."

Crumple zones are also the reason why nascar racers can survive 150+mph crashes. These cars are designed with ridiculous crumple zones with parts flying off everywhere and twisted wreckage but yet the driver and climb right out of it. Most of the remaining force (kinetic energy) is wasted while the car is being destroyed and you see these crashes take a lot of time to complete.

Car crash (model) without crumple zone:

Car crash (model) with crumple zone:

Elon speaks about safety and Model S crumple zone. he talks about how long the crumple zone is and how it saves your life, watch from 1:40

For a detailed look at the crumple zone, Tesla's CTO talks about the unique features they have done for the crumple zone because there is no conventional gasoline engine taking up space:

So in short, the initial force exerted on each vehicle is the same, but it's the crumple zone's design that affects how much of that force is absorbed by each vehicle and how long it takes to complete the change in velocity. "in general" while mass affects the remaining force, the longer the time it takes for a vehicle to crumple is what's going to save your life. yes yes mass still matters though, especially when there is a big difference between them. if a mack truck hits a motorcycle, the initial force exerted upon each object is the same, but the net force on the mack truck is basically nothing because it's change in velocity wont be very big if much at all(i'm making up numbers here) force = m * (v/t) = 10000 * (0/.1) = 0. unfortunately the same can't be said for the motorcycle. force = 200 * (50/.1) = 100000. mack truck wins. motorcycle not so much.

 

[yme]

If the masses of the cars aren't the same, definitely. If you assume you have a 1 ton vehicle and a 2 ton vehicle, the difference would be:

2 ton vehicle at 60 mph:

0,5 * 2000 * 26,8^2 = 718 kJ

1 ton vehicle at 60 mph:

0,5 * 1000 * 26,8^2 = 359 kJ

2 ton vehicle at 20 mph:

0,5 * 2000 * 8,9^2 = 79 kJ

1 ton vehicle at 20 mph:

0,5 * 1000 * 8,9^2 = 39 kJ

So for the 2 ton vehicle at 60 mph/ 1 ton vehicle at 20 mph scenario, the kinetic energy released will be 757 kJ, while the 1 ton vehicle at 60 mph/ 2 ton vehicle at 20 mph scenario will release 438 kJ. Basically if the 2 ton vehicle is the car doing 60 mph, the impact will be almost twice as energetic.

You're just computing the total kinetic energy of the two cars before the collision. I don't see the relevance of that number to the severity of the collision.

Consider two different collisions. Both involve two 1-ton cars. In one, a 51-mph car collides head-on with a 50-mph car going in the opposite direction. In the other, a 51-mph car rear-ends a 50-mph car going in the same direction. Obviously, the first collision is much worse. But the total kinetic energy before the collision in the same in both cases.

- - - Updated - - -

[...] when I say the
80/0 and 40/40 crashes are the same, I am assuming that the friction between the car and the road is the same. In particular I am assuming the cars are in drive/park and drive/drive and not in neutral.

Are you sure?

I think that 80/0 and 40/40 are equivalent only if we suppose that there is no friction between the cars and the road. Otherwise, the 80/0 cars will not end up moving at 40 after they collide, but rather at some lower speed. Or, in other words, any friction between the 0 car and the road serves to tether it somewhat.

 

[Yggdrasill]

You're just computing the total kinetic energy of the two cars before the collision. I don't see the relevance of that number to the severity of the collision.

Consider two different collisions. Both involve two 1-ton cars. In one, a 51-mph car collides head-on with a 50-mph car going in the opposite direction. In the other, a 51-mph car rear-ends a 50-mph car going in the same direction. Obviously, the first collision is much worse. But the total kinetic energy before the collision in the same in both cases.

I was talking about head-on collisions.

Of course the time it takes for the energy to be expended is very important, but this time will be virtually the same for any head-on collision with two cars of specific types and a given difference in speed.

 

[Jeff Miller]

Are you sure?

I think that 80/0 and 40/40 are equivalent only if we suppose that there is no friction between the cars and the road. Otherwise, the 80/0 cars will not end up moving at 40 after they collide, but rather at some lower speed. Or, in other words, any friction between the 0 car and the road serves to tether it somewhat.

No, I'm not sure. When you start involving friction, it's hard to be sure without doing experiments. But my sense that the 80/0 crash can be accurately modeled as occurring in two phases: the initial impact with very rapid accelerations where most of the damage is done, and then a second part, with much less rapid accelerations, where the center of mass comes to rest because of friction with the road.