A walk of shame from the Great Sept to the Red Keep should be sufficient.
It’s a fair cop. Let’s see when this fits in my schedule… (opens calendar app): In two weeks time, OK?
All true. But work isn’t simply force, it’s force multiplied by distance. In the two-car case, which involves twice the energy, the force is not what’s doubled - it’s the distance that’s doubled. Each car is crumpled by a distance d, so the total summed distance of collapse for both cars is 2d. 2d multiplied by 1F = 2df = 2 * 0.5mV2.
Agreed.
You appear to be using the word “force” in some sense other than its meaning in physics. In physics, both the wall and the car have exactly the same amount of force acting on them. Indeed, they must, by Newton’s third law.
What you might be meaning to say is that the effect of that force is different for the car and for the wall. For the car, the force causes a significant acceleration, in accord with F = ma. For the wall, the same force causes a much smaller, negligible, acceleration, because the mass of the wall is much greater.
Not unreasonable: Two cars, each going 60 MPH, and colliding head-on, is roughly the same effect as one car, going 120 MPH, colliding with a parked car. Where it gets tricky is that colliding with a parked car is very different from colliding with an immobile, or nearly-so, object like a concrete wall.
I don’t think so - here you have the K.E. going as the square of velocity, so you have 4 times the energy dissipated by 2 cars crumpling.
Half the KE ends up in the wreck sliding along the pavement, so it works out the same.
I find that hard to believe. A small amount will be dissipated by the whole mess sliding along the road and generating heat through friction until it all comes to a halt, but not half the energy that needs to be dissipated. Most of the energy is still surely going to go into both cars crumpling and disintegrating.
Conservation of momentum dictates otherwise. Assuming a frictionless surface (accurate in the short term), the final wreck has to be traveling at 60 mph. So the KE of the final wreck is 0.5*(2*M)*60^2=3600*M, as compared to the original KE of 0.5*M*120^2=7200M.
Hmm, hard to argue with that. So you have the impact with crumpling and deceleration from 120mph to 60mph, then the combined mess of two cars all screeching along the road together at 60mph until friction brings them to a halt.
Yep. Of course, there will be some debris shooting off in different directions, at various speeds–but that was true in the case of the symmetrical wreck, also. So it still works out the same, excepting that chunks that got launched out at 40 mph would now be traveling at 100 mph or 20 mph relative to Earth, and these will behave differently with respect to the air (and the ground when they hit). But that’s a relatively slow dissipation over time.
I thought that we had established that for the passengers in the car ( which is what matters) the effect of a head-on is not the sum of the speeds of both cars.
Of course, colliding with something relatively energy-absorbent like a parked car will reduce the consequences to passengers of the impact compared to hitting the theoretical immovable object.
Correct, and there’s no contradiction here. It’s a refinement - when two cars collide, the effect for passengers is approximately the same as hitting a rigid wall with half of the the combined closing speed.
For two cars both traveling at 60mph, the effect is (60+60)/2 = 60
For one car traveling at 120mph and one car parked (120+0)/2 = 60
All three of these are equivalent, in terms of damage (to any individual car, or to their occupants):
A car going 60 MPH, and hitting a massive, unyielding wall.
A car going east at 60 MPH, colliding head-on with a car going west at 60 MPH.
A car going 120 MPH, colliding with a parked car.
So you can say that a 60 MPH head-on crash is like a 120 MPH crash, depending on what kind of 120 MPH crash.
Oh, and if it’s surprising that the mass of what you hit matters, consider the opposite extreme: If you’re going at 60 MPH and collide with a housefly, how much damage does that do to your car? Almost none, because the mass of what you hit matters.
No quibble with that, but I disagree that “A car going 120 MPH, colliding with a parked car.” is the equivalent of “A car going 60 MPH, and hitting a massive, unyielding wall”.
It is certainly the case that a parked car is a softer target than a wall, but, extending your analogy of hitting a fly, it’s obvious that the passengers in a 120mph car would suffer more than those in a 20mph car.
This seems like a non sequitur to me. What’s the relevance of a 20mph car here?
You agree that hitting the softer target of a parked car will be less harmful than hitting a wall. So the issue is just how much less, right? I got this wrong at first at post #65, but follow through the subsequent explanation from @Dr.Strangelove that explains why (at least to a first order approximation) the effect of an impact with a parked car is in fact similar to that on a car traveling at half the speed hitting a wall. The explanation is that damage is principally attributable to impact and crushing. But by the time the 120mph car has decelerated after impact from 120mph to 60mph, it has transferred half its momentum to the parked car, so both it and the previously stationary parked car are skidding along the road together at 60mph so no more crushing is happening.
I think Chronos is correct here. IANAP, but for it to be accurate, I believe the transmission of the parked car must also be neutral, and the parking brake not engaged. In other words, the parked car must be able to freely roll.
For perfect equivalence, yes. But even if the parked car is in park with the emergency brake on, it makes almost no practical difference. Friction with the road is going to be negligible compared to the forces between the cars in the collision.
And it’s not so much that the car is softer than the wall: If all cars were made of concrete, and just as hard and rigid as the wall, the three cases I described would still be equivalent. The main thing that makes the parked car different from the wall is that it can move.
Nope, Chronos is dead on balls accurate:
In all three cases, each car’s crumple zone deforms by the same distance, absorbs the same amount of energy, changes each car’s velocity by the same amount, and exerts the same deceleration on its passengers.
This. When hitting an unyielding wall, the front surface of the car collapses onto a flat plane. When hitting an identical car head-on, the fronts of both cars collapse onto a flat plane between them. The plane moves with the combined center of the mass of the two vehicles. For two cars headed toward each other at the same speed, this means the plane of collision is stationary with respect to the earth. For one car headed toward a parked car, the combined center of mass (and plane of collision) is moving toward the parked car at half the speed of the moving car.
The hardness or softness of the car or wall doesn’t change the equivalency of all three collision scenarios.
For completeness we should add that secondary/peripheral effects are going to be worse in the case of the 120mph car hitting a parked car. In all cases you have a similar primary collision and crushing and rapid 60mph deceleration. But in the case where you started at 120mph (or you were sitting in the parked car) after the collision the two wrecks are still careering along the tarmac together at 60mph with an uncertain future. And if you didn’t have your seatbelt on and are thrown out of the 120mph car through the windshield, you’re moving much faster relative to the ground so in subsequent impacts things will not go well for you, especially if you hit a spherical cow.