To echo robby and Chessic Sense, the collisions are identical. What differs is post-collision. In case #1, post collision is two wrecked cars at rest. In case #2, you have two wrecked cars going 50km/h.
If the cars are on neat, weightless, fricitonless sleds, and are collided under controlled conditions, and are brought to a clean damage-free stop, you will see little difference. In the real world, two wrecked cars going 50km/h down a street will eventually stop, but will likely hit other things and become more damaged along the way.
From a “let’s assume the cow is a sphere” standpoint, the cases are the same. From a “let’s pry these poor bastards out of the car” standpoint, the first case will likely be somewhat less damaging than the second. However, even though the second case has twice the energy as the first, most of that extra energy will be dissipated by the cars scraping to a stop, and not via additional impact damage.
This is why the car going 100 into a brick wall is way worse than the 50-50 head on, all the extra energy is impact energy.
No. It does if the car hits an immovable object (the proverbial brick wall), but not if it hits a stationary car of equal weight, which will be propelled backwards at close to 50 km/h, disregarding inelastic effects.
So – a car going 100 hitting a stationary car is the same as two cars going 50 hitting head on, and also the same as a car going 50 hitting a brick wall.
Your first statement is where you err. The word “all” does not belong in that sentence. If we slam a 100 km/hr car into a stopped one, they don’t end with zero energy. They still have energy at the end. That is what your “team” is failing to consider.
Second, it’s not important how much energy is in the collision. What’s important is what energy change there is. Here’s a thought experiment for you:
There’s a parked car and a car moving at .05 mph that wrecks into it. How much energy is in the collision? Colloquially, how “violent” would you describe it?
There are two cars in a NASCAR race. One is moving at 200 mph and another behind it at 200.05 mph. The faster, rear car catches up to the front one and bumps it perfectly squarely from behind. How much energy is there in that scenario? Forgetting for a moment how in the real world this might cause the front driver to lose control or spin out after impact , how violent is this collision?
Reviewing your answers, are you sure you want to stick with the assertion that initial energy is what matters?
OK, I actually switched from my iPad to my work computer (which I don’t normally like to do), because extensive typing on the iPad is a pain in the neck. However, it is my lunch break, so here goes…
OK, first off, realize that the starting energy in both cases is kinetic energy, and kinetic energy is not conserved in an inelastic collision (like this one).
Now, total energy is always conserved, but realize that in the first case, all of the kinetic energy is dissipated in the collision, because the vehicles end up at rest immediately after the collision (because momentum is conserved).
However, in the second case, some of the initial kinetic energy is dissipated in the collision, but realize that the two mangled vehicles then have some residual kinetic energy (because they slide away from the collision at a speed of 50 km/hr, again due to conservation of momentum). This residual kinetic energy is then dissipated by friction as the two cars slide to a stop after the collision. Note that this slide to a stop doesn’t actually hurt the occupants because it will take place over a much longer time interval.
As I have stated several times now, the detriment to the people in the vehicle would be due to the peak forces experienced by the occupants.
Whoa, whoa, whoa. Hold up. First, how do you know that both cars come to rest in the first collision. Cars bounce, y’know. Second, how could the tangled mass in the second collision be going at 50 km/hr and also be damaged? They’d lose energy in the collision itself, not just from scraping.
I’m not sure where you’re getting that momentum would be conserved. Unless the cars are in pristine, pre-crash condition, they’re not coming away from this at 50 km/hr.
And energy is also and always conserved - no exceptions. Does not matter if the collision is elastic or inelastic. Energy is always conserved. The term elastic only says which side of a collision gets more energy. Energy is always conserved. What comes out is always what goes in.
The 100 km/hr car puts in twice the energy. What results in both cars is twice the energy. Obviously conservation of energy cannot be denied.
So where do you post your definition of detrimental in each post? You don’t. Detrimental. The energy imparted into both cars. Even if one car is still moving, the energy remains. Your ‘whoas’ come from this fact, You don’t define ‘detrimental’ before making conclusions. And you are not the only one doing what is common in junk science.
I was making the simplying assumption that in both cases, the collisions are perfectly inelastic. (There is no question that the collisions are inelastic.) However, considering an inelastic collision that is not perfectly inelastic adds a needless complication to the problem.
What the effect of making the collisions not perfectly inelastic is to make the collisions worse, because while momentum is still conserved, you end up with a larger change in momentum for each vehicle (because momentum, unlike kinetic energy, is a vector quantity).
Easy, you’d have a big crash, followed by a mangled heap of two cars sliding away at 50 km/hr (and rapidly slowing to a stop).
If the cars stick together, which is not a bad assumption, they must be moving away from the point of collison at 50 km/hr. In this universe, we follow the law of conservation of momentum.
Energy of a 100 km/hr car completely dissipates in the scrap metal once called two cars. All kinetic energy dissipates in both cars - that are now at 0 km/hr.
Two 50 km/hr cars have one half that energy. All kinetic energy dissipates in both cars - that are now at 0 km/hr. The concept is well understood – conservation of energy.
If ‘detriment’ is defined by energy dissipation, then a 100 km/hr collision is doubly detrimental. Obviously. Notice – every post defines the term ‘detriment’. You do not.
All energy from moving cars must be conserved. All that energy dissipates in damage to both cars because all cars are now at 0 km/hr. Conservation of energy always exists. Always (despite someone earlier who denied it).
A collision of one car at 100 km/hr means double energy dissipates in both cars. No way around reality.
Energy change is obvious. One car at 100 km/hr is now 0 km/hr. Two cars at 50 km/hr are now at 0 km/hr. All energy once defined by mass and speed is now measured by damage.
BTW, your NASCAR example is bogus. Speed of both cars after the 0.05 mph collision is not defined. So no answer is possible. Obviously, without mass, another reason why no answer is possible. Please present your answer with numbers. Subjective reasoning is also called junk science.
Obviously initial energy matters. It is called conservation of energy. Why is that so difficult? All energy in once moving cars is now dissipated in what is called scrap metal.
I think everyone else is saying that the extra energy is present in the system in the form of both cars moving at 50 km/hr after the collision. That energy isn’t going to crumple the cars.
The energy that gets dissipated need not result in damage to the cars, it very much depends HOW it is dissipated. Example, the cars gently sideswipe each other, and the drivers use their brakes to stop both cars. Same energy dissipated, but much different levels of damage.
In these cases, the energy dissipated during the collision is the same, what differs is how much energy is left over post-collision to be dissipated in another way.
So again you post without defining the word ‘detrimental’. Energy dissipated in scrap metal or brakes is still the same energy conserved. At least I define detirimental. Energy in each vehicle. Burn brakes - conservatioin of energy - is still defined by detriment.
westom, you are thoroughly confusing yourself (and possibly others) by only considering energy.
Energy dissipated as noise or friction or the crumpling of metal does not necessarily hurt the occupants of the vehicles.
Instead, as I have stated repeatedly, the detriment to the occupants in the vehicles would be due to the peak forces experienced by the occupants. Period.
Also, nobody has denied that energy is not conserved. I did state earlier that **kinetic **energy is not conserved. Which it isn’t in this type of inelastic collison, by definition.
robby, I think you are forgetting the limits on the material used to construct car. I agree with you that in case 2 immediately contact that there will be crumpling to both car in some degree then both car would move. That’s why I said “Though even system 2 have twice the kinetic energy it might have longer collision time so the occupant would experience less G force.”
But I am just not sure both of them would them move at around 50 km/h immediately after contact then slide to a stop.
As car A slams into a stationary car B some of the 5000m of energy is going to be dissipated in the crumpling of both car but depending on the geometry and material used the whole car might just collapse and pancaking the passengers inside. It would take special engineering to assume half of the energy would be absorbed by materials and rest into friction with road. Different rigidity, materials would all give you different results in case 2.
I think this is what the entire thread digests down to. If you change your frame of reference to put one of the cars in the center and keep it there, then both scenarios result in the other car smashing into it a 100 kph. Hence, they are the same, in the ideal case.
In the real world, there would be some additional forces in the stationary car scenario due to the frictional forces that must be overcome to get the car moving.
