I’m still confused about why the textbook is wrong. Is there something about the energy or momentum scaling differently in the brick wall case vs. the head-on collision case?
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??)
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.
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
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).
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.
He definitely isn’t
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.
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.
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.