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.
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.
Imagine a car hitting a car with the wheels magically glued to the road at 100 km/h.
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.
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 deflect.
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.
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.
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.
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.
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.