I should be able to find the answer to this question, but my google-fu is failing me.
I teach a drivers safety class, and one of the topics we cover is head on collisions. During my training 6 years ago my instructor told me that if two vehicles traveling at 60 mph hit head on it’s equivalent to hitting a wall at 120 miles an hour. That the forces are additive.
During today’s class I had a person challenge me on that and say that if two vehicles travelling 60 mph hit head on it’s like hitting a wall at 60 mph.
In a class-room physics problem, if the cars have the same mass, then in a head-on collision the will both decelerate equally and come to a stop exactly in the same way that a car would decelerate if it hit a completely immobile and non-crumpling wall. Yo could imagine placing a thin sheet of paper at the crash point which (in a huge leap of faith) would not move at all as the two cars crumpled exactly the same way.
In other words in a two car crash there is twice the kinetic energy of motion being converted into smashy energy (technical term), but it will be shared equally by the two cars. In the wall collision against a perfect wall all the (half as much) energy goes into the car.
So it’s more like a car into a wall at 60 mph, and the wall is probably even a little less than like 60 mph head on of two cars if the wall is smashed some.
Wall at 60 with identical vehicles, like everyone else said. If the other vehicle has greater mass, or doesn’t crumple as easily, then the “brick wall equivalent” speed may be >60, or even >120.
Then it’s more like 60/sqrt(2) ~ 40 mph, since the stationary car crumples too, so less energy goes in to your car. It’s the same idea as the barrels of water/sand placed in front of a bridge abutment on the highway, or features in guardrails that are intentionally designed to crumple on impact.
Two cars at 60 mph hitting head on is the same as one car hitting a wall at 60 mph, OR one car traveling at 120 mph and crashing into a stationary car. The crumple zones and such would change the math slightly, but assuming perfect elastic collisions, this is how it works and helps explain where the confusion comes from.
The only way it can feel like > 120 is if the truck accelerates during or after the crash.
Consider both scenarios, but in the first one you are in the wall’s reference frame, and in the second, you are in the truck’s reference frame (an the truck is, for this experiment, as solid as a wall, but on wheels).
Once the vehicles hit, they will de-accelerate due to friction with the road, and eventually stop. But the truck will have dissipated some energy due to friction, too, whereas the wall will not. As soon as the crash happens, the truck’s speed will decrease. The wall, OTOH, had zero speed before and after the crash. (If we get super nit picky, the wall will move slightly, but this is nothing compared to the deceleration of the truck due to road friction.)
I have to think about my assumptions here. So do you agree that (a) a head-on collision between the car and an infinitely heavy and perfectly rigid truck, at 60 mph each, and (b) a collision between the car and a perfectly rigid and stationary wall, at 120 mph, in the frames of the truck and wall respectively, are identical? But my thought was that at the end of the truck case, the crashed vehicles were still moving with respect to the road, and the truck could still deliver energy to the car while they skidded to a stop.
When two cars are traveling at each other, both going 60 mph, then the relative speed of one car to the other car is 120 mph. According to the physicists in the linked thread, its the relative speed that matters.
This is true. Are you agreeing with me, or disagreeing, and how? Note that the total kinetic energy in a system will vary with the frame chosen.
The “identical cars” case has a clean answer because after an inelastic collision with (a) a stationary brick wall, and (b) an object of equal mass with opposite velocity, the velocity of the stuck-together objects is in both cases zero. More assumptions are needed in other cases.
That’s true for the initial collision, but after a perfect inelastic collision between vehicles of equal mass at 60 and 60, the two vehicles are stopped. After 120 and zero, the two vehicles are still moving. For real, those vehicles will eventually stop, due to friction with the road and stuff, and that’s where octopus’s extra energy gets dissipated. That probably makes the collision worse, but I’m not sure how to model that cleanly.
Are you expecting the tires to supply a force during the collision that’s significant compared to the collision forces? If not, I don’t see how the velocity of anything with respect to the road can be very relevant there. Velocity with respect to the road obviously does matter afterward, since it determines whether you’re stopped after the crash or still moving.
Okay, forget the road. Two spacecraft traveling at each other going 60 mph each, versus Ship A traveling at 120 mph relative to Ship B. You don’t think that is equal?
The road can only interact with the cars through the tires. The acceleration that a car sees during a crash can be much larger than the maximum acceleration that the tires can supply without losing traction. So I don’t think the road can contribute much to the total force during that initial collision, so I don’t see how the velocity of anything with respect to the road matters much then.
Basically, I think the kinetic energy dissipated in the crash falls in to two categories: (a) the energy lost in the initial collision, and (b) the kinetic energy remaining after the collision, which gets dissipated while the stuck-together vehicles (assuming an inelastic collision) skid/roll/bounce/etc. to a stop. In the particular case of identical vehicles head-on, (b) is zero, so it doesn’t have to be considered. If you maintain the relative velocity of the vehicles, but change their velocity with respect to the road, then (a) stays the same but (b) changes.
I’m thinking that a good model for “crash deadliness” would consider both, but assign less weight to (b), since that energy probably gets dissipated slower, decreasing the harm. This gets beyond simple physics, though, and it’s outside my area.
Seems measured velocities would be equal to me. But in a collision where the assumption is they have stopped and all energy goes into the deformation of the body of the vehicle after the collision that seems relative to the road for cars. For spaceships? If the assumption is they crash and aren’t moving with respect to each other after I’d assume that’s the coordinate system you’d use for that case. Where V2 is = 0.
Perhaps I’m not thinking of this the right way. It just seems intuitive to me that if the shape of the vehicles is dependent on the energy absorbed then it should be possible to determine how much kinetic energy was transformed into deformation energy.