I've intuitively believed this to be true for a long time. But the article made me think about it, and the textbook is quite wrong.
It's a test similar to if you're in a train in motion, and you drop a ball in the aisle, does it fall towards the front or the back of the train?
The momentum vectors cancel out, but the relative velocity is identical and the total kinetic energy is the same.
The damage isn't the same because car/car has twice the crumple zone. More of the energy will be dissipated through elastic deformation, so the deceleration will be slightly gentler. (Although almost certainly still fatal.)
That aside - the driver's ed book is basically saying that head-ons are much more dangerous and deadly than most people understand, even at low speeds. All of which is true.
To see this, you can imagine the car/car case as having an invisible wall between them. The wall has no net force so it doesn't move and acts the same as the brick wall despite not being either rigid, massive, or fixed to the ground.
Then assuming it requires X energy to crumple a car and assuming the wall is strong enough to not yield at all (so no work is done on it), it requires 2X energy to crumple 2 cars, so there's X less energy available to destroy the passenger in the 2 car scenario vs the wall. So the 2 car scenario is slightly better, but probably not meaningfully so (X will be much less than the kinetic energy of a 100 mph car).
The correct symmetry principles to invoke here are conservation of energy and invariance of Newtonian physics under Galilean transformations.
Do you know how crazy that is? the collision doesn't dissipate more or less kinetic energy depending on the reference plane you pick.
If you pick a 100/0 reference frame, the combined cars are moving 50 in one direction after. Therefore, they both only expereince a 50mph change. one from 100 to 50, the other from 0 to -50.
your frame doesnt change the collision energy. If that was the case, you could pick a 100,000mph reference frame of an astronaut flying by, and when the cars collide there would be an atomic explosion.
to put this another way, the energy of the collision comes from the deltaV before and after. The deltaV is the same in EVERY reference frame you can construct.
50/50 is not the same as 100/0 between car and an ideal wall: an immovable barrier of practically infinite inertial mass that suffers no damage.
car/car 50/50 is the same as car/ideal-barrier 50.
Is it though? Genuinely wondering if I'm missing something in my ponderings...
I would assume that a collision between two identical cars traveling at the same velocity would each decelerate to zero within the distance of one crumple zone. ie due to the other one also decelerating the same way. Assuming that they don't end up in some sort of intermingled mashup (spherical cows and all that) which is maybe what you're describing?
And that a car at the same speed hitting a practically immovable object would also decelerate the same amount in the same single crumple zone distance.
Anyway as an irrelevant aside, hitting a brick wall isn't the same as hitting an immovable object - a car will mostly go right through a fairly standard unreinforced brick wall. Which unless a brick coming through the windscreen hits you directly, is probably going to be a lot safer than the deceleration of hitting an oncoming car or immovable object.
Imagine a car hitting a car with the wheels magically glued to the road at 100 km/h.
But if the car just busts straight through a wall of equal mass (and a comparable crumple zone), then it will be just like the car/car case energetically: the car and wall both continue at 50 mph following the collision.
And yeah "immovable" isn't literal, just shorthand for practically immovable compared to the car.
Together, they have the energy of two cars at 50 each hitting a wall, and this energy is split between the two cars.
If somehow ALL of the kinetic energy was absorbed only by one car , it still wouldnt be equal to 100mph, do the the square.
Together, two cars traveling at 50mph have the same energy as one car traveling at 70mph.
The only case when two objects colliding at 50mph is like 100mph is if the final reference frame is still traveling at 50mph. a Car going 50pmh hitting a train going 50mph would be like the car hitting a wall at 100mph (net velocity change of 100mph.
The mass of the wall is required to determine which system has more energy.
Or, take it from the perspective of car B or the wall. In both cases, car A is approaching with a relative speed of 100mph.
It goes from sitting still to going backwards at 50pmh, for a net change of 50mph.
The difference of velocity between cars 100mph in every reference frame you can pick. Each of the cars always experience a change of 50mph from the collision, no matter what reference frame you pick.
People are confused because the pick a reference frame for car A, but don't hold it constant for their calculations, as required by physics.
The wall approaches at 100mph and, in a totally elastic collision, it recedes at 100mph, as does car B.
>No, if you’re taking car A as a reference point it behaves as if stationary before and after collision - it’s everything else that moves.
If that were true, I could set a reference plan as a bullet in a gun, and when I fired it everyone in the world dies from rapid acceleration as their kinetic energy goes from 0 to 3000mph.
you change 50 if you hit a unmoving wall.
IF you start at 0 or 100mph, you also change 50.
Each car approaches the impact point at 50mph, each car recedes at 50 in the opposite direction after a perfect collision, so each has a dv of 100.
And if my initial reference frame is car A, travelling at 50, then before impact car A appears to be travelling at 0 while car A is approaching at 100. After collision both cars experience a dv of 100, so car A appears to be receding at 100mph while B appears stationary.
For the wall, the car approaches at 100 and recedes at 100 for a delta of 200mph. But in any case the velocity delta doesn’t affect the energy, it just has to be conserved.
As I’ve said elsewhere we are trying to find the collision energy in the moment of impact from the perspective of car A (the question is whether you driving into a wall the same as you driving into another opposite car at half the speed). For that you have to consider a frame of reference relevant to car A, and not a stationary reference that is not following either car.
I believe the correct approach is to look at it from the perspective of the object that you in car A are striking. Taken from the perspective of the wall, car A strikes it at 100mph. From the perspective of car B, car A also strikes it with the same speed (and kinetic energy).
IF the two cars have equal mass, this will also be the impact energy on car A. And that’s what we’re trying to find for both cases.
The most convenient reference frame for colliding objects is the combined center of mass of those objects. It isn't affected by the collision, and for perfectly inelastic collisions the objects will come to rest with respect to that frame. This makes the math a lot simpler that working in some other frame.
Twice the crumple zone divided by two cars is the same as one car and one crumple zone.
The parked car will move away, providing cushion.
When you have two cars in opposite directions, they cant both give way for each other.
The trampoline doesn't deflect, and acts exactly like a concrete wall.
The trampoline cant simultaneously deflect in opposite directions.
Same amount of momentum, two collisions. In one, there is one crumple zone, we will assume the brick wall is completely undamaged, for the sake of simplicity. In the other, there are two crumple zones.
All of the crumpling is dissipated energy. What doesn't get crumpled, goes into the driver's face.
Clearly, the situation with two crumple zones is superior. There's a limit to how much energy can be absorbed by crumpling. With two crumple zones, that limit is 2x, with one, 1x.
Therefore, damage to the driver or drivers is much worse in the brick wall scenario.
For another analogy, picture a go-kart, with a half-meter block of styrofoam, hitting a brick wall. Now, picture the same go-kart, hitting at the same speed, but this time, there's a meter of styrofoam.
That's the one-crumple and two-crumple scenarios.
If you're standing next to the stationary brick wall, it falls straight down relative to the wall.
To me this is a clear demonstration that motion is relative, therefore the two collision scenarios ARE equivalent.
Imagine that the two cars are in space. Are they both moving at 50MPH towards each other? Or is one stationary and one moving at 100MPH, like the wall scenario? Or perhaps moving the same direction, but one at 100MPH and the other at 200MPH, about to catch up to the first one? The answer is that it doesn't matter, because the only difference between these scenarios is the reference frame. The collision is the same in all of them.
But I maintain that whatever the mathematical procedure, it is absolutely impossible for the stationary-ness of one of the objects to affect the outcome of the collision, assuming we're ignoring things like static friction, air resistance, etc. This is a simple matter of reference frames. If you disagree, put the two objects in space as I suggested and tell me which one is "stationary" and how that could possibly affect the outcome of the collision.
And if you think about it further, the difference is not that you're choosing one object as still and one as moving but that in the car vs. car case, the second car takes on some of the energy from the collision, whereas the wall does not. This makes some intuitive sense if you imagine this collision happening. Forget all the crumple zone stuff - a car hitting a stationary car is going to make that car move - and if you ignore friction and assume the cars stick together the two cars will be moving in the same direction as the initially moving car, just slower. This means not all the initial kinetic energy from the moving car was "used up" in the collision.
[1] https://en.wikipedia.org/wiki/Inertial_frame_of_reference
For a truly immovable wall, the ground has to exert a lot of force to keep it in place, which goes into the car, causing the entire discrepancy.
You're right that if you match your reference frame to one car, that car becomes "stationary" until the collision, but after the collision it is moving relative to the reference frame. But that's the key--it now has kinetic energy relative to the reference frame, but not to the other car. As far as the pair of cars is concerned, all of the kinetic energy between them has been expended, because they are now moving together.
Of course this is assuming a scenario where the cars crumple into a single mass, i.e. they don't "bounce". I think part of the confusion in these comments stems from the problem being poorly defined. Some people are considering crumple zones, others are considering whether a brick wall is movable and still others replace it with another car for simplicity.
I do think the driver's ed manual is correct in some Spherical Cow sense, but I don't think its authors ever intended for the problem to stand up to Hacker News scrutiny. It's quite funny to me how innocuous things like this can generate these "unladen swallow"-type discussions on here, even though I'm participating.
I think people dont realize they are changing their reference frames, and getting insane results like the amount of kinnetic energy release in the collision depend on what frame you pick.
two crumple zones don't matter for head on collision, which is another thing that trips people up.
As per your "stationary" objections, the issue is that this is modeled as a very sturdy wall anchored to the Earth. Yes, the Earth will "recoil" some, but for all intents and purposes it can be modelled as completely stationary because its mass is so much greater than the car.
> the issue is that this is modeled as a very sturdy wall anchored to the Earth.
That's not what you modeled in the comment you linked. Your math (incorrectly) attempts to describe a car colliding with another car of equal mass. But if you're considering a "very sturdy wall anchored to the earth", then you effectively have a car colliding with the earth.
Either one is fine with me, but pick one. It grinds my gears when you authoritatively call me "absolutely incorrect", "correct" me with an incorrect answer of your own, and then try to claim you were modeling something entirely different all along.
The link I posted, https://news.ycombinator.com/item?id=40628932, clearly showed the kinetic energy math for both a car hitting a stationary wall at speed 2X, and 2 cars hitting each other each with speed X. This is basic high school math of inelastic collisions.
Imagine compressing two springs in a line vs one against a wall.
If the wall is immovable and perfectly fixed to the ground, then the car/wall collision will indeed have a greater effect than the car/car collision.
But if the wall's connection to the ground is negligibly weak (e.g., an actual brick wall), then after the collision, both the car wreckage and destroyed wall will continue forwards at 50 mph, until some other force slows them down. In this case, the effect (change in kinetic energy) is the same as the car/car collision, not counting the impact of crumple zones.
In the former case, the kinetic energy of the car relative to the ground is relevant only because the ground is exerting a massive force to keep the wall in place.
In the Newtonian inertial frame of the "stationary" Earth, it accelerates downwards at 9.81m/s² until it hits the floor and comes to a stop and is no longer accelerating ("gravity", which is a thing here, and the normal force now summing to zero).
This is fairly academic for sporting goods, as if working with these two frames results in substantially different outcomes for you or your ball, your day is probably going badly, or at least fairly energetically, and baseball isn't your priority right now.
But yes, the textbook is wrong. Unless the brick wall (and only the wall) is 100% inelastic so it "catches" the 100 MPH car, slowing it to 0. Quite unrealistic.
If the kinetic energy for a car at 50 miles per hour is 1x, at 100 miles per hour, it’s 4x. After the collision, the total kinetic energy is 2x, and the other 2x was absorbed by damaging both cars. That’s different from hitting a wall where all the kinetic energy goes into damaging the car.
The damage from hitting the parked car matches the head-on collision, which starts with 2x kinetic energy and afterwards, it’s zero. (If they are stopped afterwards.)
If the cars were rigid, they would bounce off and go in opposite directions, like balls on a pool table. That would be very bad for anyone inside. Crumple zones absorb energy for good reasons.
I expect that guard rails absorb more energy than a brick wall?
Once people are done with that they might be amused by try this one:
• You are on a stationary bus, holding a helium balloon on a string. The balloon is floating directly above you. The bus starts accelerating forward. Does the balloon:
1. Move toward the front of the bus relative to you,
2. Continue to float directly above you, or
3. Move toward the back of the bus relative to you.
Alternatively, imagine one car is parked (in neutral, with its brake off) and the other car hits it at 100. The center of mass of the two cars is moving at 50 both before and after the collision (conservation of momentum); after it, the cars will be moving at that average speed. The impact will again be equivalent to the original scenario.
The original claim probably results from a conflation of these two scenarios.
ETA: So what student drivers should be told is that hitting another car head-on is like hitting a brick wall at the same speed. For this to be exactly true, the momenta (mass * velocity) of the two vehicles have to be equal and opposite, but to communicate the general idea, I don't think we have to go into that.
Assume two perfectly elastic cars. When they collide at 50mph, each car will bounce backwards at 50mph (due to conservation of momentum), representing a change in velocity of 100mph — identical to the brick wall case.
I feel that introducing deformation or other energy dissipation to the equation kind of takes it out of the “high school physics” realm, right? What else am I missing?
Edit: ah I see, the car will bounce off the brick wall at 100mph as well, resulting in a 200mph change in velocity. I guess you could explain it then that the effect of the impact is felt entirely in one car in the brick wall case, and it’s spread out over two cars in the head-on case?
This is the incorrect part. They would both go to zero velocity/momentum.
Momentum is a vector quantity, so has a direction and magnitude. Two identical cars with the same speed going opposite directions would have the same magnitude of momentum, but opposite sign. After colliding, their sum would be zero.
If you watch billiards you would see kinda the same thing going on.
Edit: completely messed this up. Other comments are more correct
In a perfectly elastic collision, both kinetic energy and momentum are conserved. In a perfectly inelastic collision, kinetic energy is not conserved (because it is converted to heat), but momentum is always conserved.
So lets say you have 2 objects of the same mass traveling toward each other at the same speed. In a perfectly elastic collision, the balls objects will "bounce" off each other, going back in the opposite directions. In that case momentum is conserved (as you note, it's a vectored metric, so before and after the the total momentum of the system is 0), but so is kinetic energy, because you still have 2 masses traveling at the same speeds (think about if you have a Newton's cradle and pull both end balls up and drop them at the same time - they'll both bounce back).
In a perfectly inelastic collision, both masses will essentially crush and come to a complete stop where they collide. Again, momentum is conserved (it's still 0 before and after the collision) but kinetic energy is not conserved because it's all converted to heat of the 2 objects.
Thanks for the correction.
(Also, how does a “throwaway” account get that much karma??)
I don't think so. In my high school physics class I learned about both fully elastic and fully inelastic collisions. The math basically works out the same in the fully inelastic collision case, which I did here, https://news.ycombinator.com/item?id=40628932
Perfectly elastic objects bounce apart with mirrored velocities. No energy is lost.
Perfectly inelastic objects just stop. All of the energy is dissipated through noise, heat, and deformation.
Momentum is conserved in both.
For drivers ed, cars are almost perfectly inelastic.
With two cars hitting head on, the energy to be dissipated is twice as much, but it is equally distributed to each car, and so the energy dissipated in each car is the same as if each independently hit a brick wall.
In other words, conservation of energy.
That might be correct but it comes off as a smart-ass answer unless the course covered deformation, which I suspect it did not given it was a high school course.
Assuming the cars are of equal mass. If not, the heavier one will drive the other car backwards faster, and the energy will not be divided equally between them. But the sum will remain the same.
I think it's simpler just to directly look at the equation for kinetic energy: KE = 1/2mv^2. In other words, kinetic energy is directly proportional to the mass, and proportional to the square of the velocity. Also, let's assume that in both cases the car comes to a complete stop at the wall, or both cars come to a complete stop where they collide (a perfectly inelastic collision), so that all of that kinetic energy is essentially released as heat (i.e. the cars crush).
In the case of 2 cars hitting each other, if the cars both have equal mass X and are traveling at velocity Y, then the total kinetic energy dissipated in the collision is 2 * 1/2XY^2, but then of course for each car it's just 1/2XY^2.
In the case of a car hitting a wall but traveling twice as fast, the kinetic energy dissipated in the collision is 1/2X(2Y)^2, or 2XY^2 - so that's essentially 4 times what the car experienced in the head on collision (of course in the real world the wall would dissipate some of the heat, but in this "spherical cow" example I think it's OK to assume the vast majority of the crushing occurs in the car).
It's this assumption that causes the entire discrepancy. For a wall to be truly immovable, the ground must exert a great force into the wall to keep it in place, which goes into the car.
But if the wall isn't anchored very securely (or backed by the ground), then the car will bust through it and continue moving forward. In the limiting case where the wall isn't attached to the ground at all, both the car and wall move at 50 mph following the collision. The dissipated kinetic energy comes out the same as in the car/car case.
If we consider the force experienced by a passenger in any car, then you also have to take into account the crumple zone (larger in modern cars). Here we imagine the cars have perfectly inelastic collisions coming to a complete rest as they crumple, first calculating the KE in each scenario:
100 mph (44 m/s) into a brick wall -> KE = 1/2 m (44)^2 = m(968)
2 50 mph (22 m/s) into each other -> KE = 2 * 1/2 m (22)^2 = m(484)
Thus the car in the brick wall scenario has to absorb twice the energy of the overall collision in the two-car scenario, in which each car only has to absorb 1/4 of the brick wall total KE (assuming the brick wall absorbs nothing, maybe a giant granite boulder is a better model). Thus in modern cars it's not just the KE but the crumple zone that reduces the force felt by the two-car passengers.
(Incidentally I used Chat GPT4 to help sort this out and it tried to claim that (x/2)^2 == (x^2)/2 so use with care - it did catch the error but only when I said double check that calculation in a step-by-step manner. User beware!)
Imagine the two cars hitting each other in a (unrealistically) symmetric, mirror image way. Where they hit is a plane that neither car crosses. That is your immovable wall, it’s just both cars experience it.
It's the same as saying that if you drive 2 cars into the same wall at 50mph the force felt by each driver is not the same as driving a single car into a wall at 100mph.
That is true obviously.
However if there isn't a perfectly immovable "piece of paper" in between the 2 cars, then you can't just ignore the force acting on each car, from the other car.
F = ma
Also, Newton's 3rd law states that forces generate equal and opposite forces.
So if you punch a wall, you are exerting a force on the wall, and the wall is exerting an equal and opposing force on your hand.
When car A hits car B going 50mph, the force is the mass of the car times the acceleration required to stop the car (ie. deceleration, acceleration in the opposite direction).
However that force is matched by a force in the other direction PLUS the force of each car attempting to decelerate the other car.
You have 2 forces for each car: the force felt by car A as car B tries to accelerate it in the direction car B was travelling, and the force felt by car B from car A that opposes that force, and vice versa.
So if we assume the cars are driving in an Elon Musk hyperloop tunnel under Vegas in a straight line then yeah, the forces felt by the drivers in each vehicle is the same for a car driving 100mph into a "brick wall" (or more accurately a big pile of bricks the same size and shape as a car with equivalent mass).
This is an odd assumption.
If someone asked me what kind of wall cars are most likely to run into at 100 mph, I'd have assumed something like a freeway barrier wall that are fully anchored to the ground or the reinforced concrete walls they use for crash tests - something with a mass substantially greater than a vehicle.
Some people in this thread think that you can take two cars and double the kinetic energy acting on both of them. If this were true, you could create infinite energy by repeatedly colliding cars.
Perhaps it isn't surprising that people who think this is true also disagree with his economics.
https://news.ycombinator.com/item?id=40631089
Although I still think Chicago school/libertarian economics is "confidently wrong".
Of course I don't expect anyone to refute an entire school of thought in an HN post.
He definitely isn’t
Kinetic energy scales as velocity squared, so a car at 100 MPH has 4x the energy of a car at 50 MPH. But in the 50 MPH scenario there are two cars, so the total energy dissipated in the 50 MPH head-on collision is half that of the 100 MPH brick-wall collision. In the brick-wall case, presumably all the energy is available to demolish the one car, but in the head-on case, the energy is spread out demolishing two cars.
Impact force is maybe trickier, it depends on the acceleration, but if the two 50 MPH cars end up with zero momentum, then they have 50 MPH of delta-v each over some collision time. The single car of course has 100 MPH of delta-v. If the collision times are the same (arguably a reasonable approximation if the head-on is a highly symmetric), then the impact force in the head-on case is half that of the brick-wall case.
Maybe half the multiple choice questions were phrased to trick you into giving a wrong answer that broke newtons 3rd law.
I have no idea if that is true in reality, but to me that seems like the most likely explanation. I agree, without this the statement does not make sense.
As a side note on the article, I'm not sure why David Friedman associates being open to being questioned by their children with libertarian economics (with circumstantial evidence at best)? I, myself and many of my friends have been educated in a similar way and our parents were definitely not libertarian economists. In fact I would argue that this philosophy is routed in anti-authoritarian education, which originates in communities more associated with anarchistic/socialist political/economical leanings.