You’re right. That was supposed to be m1u1+m2u2=(m+m)v.
Because in the case of your couch and carpet, the carpet (being connected to your house, which is in turn connected to the ground) is exerting a force on your couch.
However, there is no external force acting to bend the metal of our two colliding cars. All of the forces are internal.
You’re right. In considering what the velocity is of our two cars after colliding, it doesn’t matter what they are made of (so long as they stick together, which makes our collision perfectly inelastic).
It comes at the expense of energy, not momentum. (Specifically, it comes from the initial kinetic energy of the colliding cars, which is not conserved.) Again, momentum is always conserved.
Indeed, that’s pretty much the system we are describing during our car collisions. What external force do you think is acting on our two car system?
robby is right. Kinetic energy is not conserved and is what is “lost” when the metal is bent. The momentum, however, is conserved.
Tell you what, let’s just cut to the chase and you tell me why you get to make that assumption. We all agree what would happen if this were the case, but I’m unconvinced why this would be the case. I don’t understand why the Coefficient of Restitution would be assumed to be zero.
This is incorrect. From the theoretical textbook example, the cases are identical. The energy dissipated in the collision is the same. The momentum is the same. Ergo, identical.
The thing is, each car is experiencing a 50 km/h collision, but two cars are involved. In the 2 cars, one of which is at 100 km/h and one is at 0 km/hr, the result has two cars involved as well. Each car absorbs some of the damage from the first moving car.
For a non-textbook case, you have to consider whether the “parked” car is in neutral or with parking brake on or with drive train engaged, and how all that balances out resistance-wise.
A solid wall is different than a parked car. The solid wall acted as a giant immovable object that was essentially undamaged afterwards. All the energy was dissipated through damage to the single car. Two cars impacting each other had two cars to absorb the same energy. The single car 50 mph into wall was very similar to 2 cars 50 mph each head to head, because the second car in the reverse direction was essentially the same as a brick wall. But if the car going 100 mph hit a car that is sitting at zero and can (a) collapse/get damaged, and (b) move backwards – **two things that the wall did not do **-- then the situation for the single 100 mph car will look like an impact of a single car at 50 mph hitting a wall. Each car sees the impact as if it is going 50 mph and hitting an immovable object.
This is wrong. You are assuming that the two cars are at rest immediately after the collision. This is false. The second car will start moving due to the impact from the first car. The textbook example shows that the momentum of state prior to crash is 1 car at 100 km/hr and one car at 0 km/h, the momentum of the state after crash is 2 cars at 50 km/hr. The still car does not have any way to resist moving, it is not bolted to the ground, tied down, concreted in place. It is sitting there, and can roll or slide. It’s resistance to motion is either internal friction of engine/brakes, or external friction of tires to road.
Your case is the equivalent of hitting a car at rest that is parked against a giant wall. 2 cars absorb the energy, but all of the original KE is dissipated through the collision, rather than only half of the KE being dissipated.
No, it was a simple conceptual demo. Two cars lightly bumping are two cars lightly buming, whether one car is at rest or both cars are moving at near light speed. The delta V is the significant contributor if the masses are the same.
No, some of the original KE from one car is now KE in two cars moving away from the crash. The second car that was initially parked is now moving. That KE came from somewhere.
In 1, why don’t the energies cancel? Yes, I’m serious. I know that they don’t, I just don’t know why.
There will be some non-zero coefficient of restitution, in that the bent metal and plastic will have a slight give. But you will see that most of the damage to each car remains after the crash. You end up with two crumpled heaps, not two pristine cars. Ergo, the restitution is small, and the collision is effectively inelastic.
How much restitution? Is it 5% is it 0.5% Where is the significant cut off for the purposes of the conversation between assuming perfectly inelastic and saying that assumption isn’t valid? From an engineer’s perspective, it’s close enough that the car is totaled. Who cares if the metal unflexed a half a percent? You still have to tow that mess away.
First off, realize that momentum is always conserved, whether or not the collision is elastic or inelastic.
We know that the collision is not elastic, because that is an unrealistic ideal condition in which kinetic energy is conserved, like when perfectly rigid billiard balls collide.
Therefore the collision will either be inelastic or perfectly inelastic. A perfectly inelastic collision is one in which the two colliding objects stick togeter. It is also a simpler case to consider, and is not unrealistic. I’m sure you can picture two cars getting mangled together and sticking together after a collision.
Remember, the original point of the OP was to compare two cases. Considering an inelastic (but not perfectly inelastic) just makes the problem needlessly complex, like bringing air resistance into introductory parabolic trajectory problems.
Nars, in the first case, each car has it’s own equivalent share of momentum and energy. Each car effectively dissipates it’s own energy, and the momentum cancels. Momentum is a vector, direction matters, but kinetic energy is a scalar, direction is irrelevant.
In the second case, one car starts with all the KE and momentum, then both cars end with some KE and momentum. The momentum gained by the stationary car is momentum removed from the initially moving car. Thus the initially moving car sees exactly the same situation - a delta 50 km/h velocity change and and a momentum transfer of 50m. The initially stationary car also sees a 50km/h velocity change and a momentum transfer of 50m. It’s just in the reverse direction.
Because energy (unlike momentum) is a scalar quantity. Therefore there is no such thing as negative energy (in the non-quantum macroscopic world), and therefore there is no such thing as energy “cancelling.”
Also, in inelastic collisions, note that kinetic energy is NOT conserved.
OK, well it’s quittin’ time at work, so let me just register my disagreement with “significant” and “needlessly”. Seeing as how we agreed on page one about the OP’s question and we agree on the physics anyhow, I see no reason to continue quibbling about whether we should un-assume the spherical cow.
OP, the collisions are equivalent. Peace out. 
How about this: There are two answer to the question. (I haven’t taught physics, but I do have a degree in it, and actually do some physics for my work, so I think I’m a ‘physicist’).
In the big picture, robby is absolutely right. If the collision took place on a perfectly smooth and frictionless sheet of ice or something, the two collisions would be identical, except in the 100/0 case the two cars would be sliding along at 50mph, while in the 50/50 case, they’d be (more or less) at rest. In fact (again if the ice is physics-problem-land perfectly featureless) you couldn’t tell whether you saw a 50/50 crash from a stationary position, or a 100/0 crash from a separate third vehicle coasting along at 50mph – in each case you’d see two cars approaching one from each side at 50 mph relative to you, an earth-shattering kaboom, and then a mangled two-car mashup not moving relative to you.
Now, in the real-world, there’s going to be some difference. Depending on how the stationary car is pointed, what kind of surface it’s sitting on, whether it has brakes set, how soon the crash locks up the wheels or pushes parts of the car into the ground, there’s going to be some kind of friction. So in the 100/0 crash the two cars will be sliding away at something less than 50mph, and they won’t be sliding harmlessly forever either. Some of that frictional force could cause more damage, even if a lot of it will dissipate more or less harmlessly as heat. So, yes I think the 100/0 crash will probably – though it all depends on the details – do some amount more damage than the 50/50 crash. But, in the big picture, not much more. I’m sure a 75/0 crash will do less overall damage than a 50/50, for instance
(Of course we’re assuming that the two cars sliding/rolling away in the aftermath of the 100/0 crash won’t hit anything else, which could certainly cause more damage. But that is I think not the point of the OP).
It’s been said before, but its worth repeating. Except for the frame of reference, these are the same. The two vehicles collide with a closing speed of 100km/h.
If anybody cares to disagree, I would be interested in why you think so.
The wheels. A stationary car’s wheels aren’t spinning, whereas a moving cars is.
I hope we can agree that as the thought experiment this is, the effects of spinning wheels are negligible.
Scary that so many do not even know simple high school physics. Doubling velocity (speed) does not double energy. It quadruples energy. Everyone here should know that. Otherwise scammers and politicians make money and get elected because the public is uneducated and naive. These fundamental facts are taught to everyone in high school to make dishonesty and scams difficult.
Energy of one car at 100 km/hr is equal to the energy of two cars at 70.7 km/hr. Basic physics. Obviously. Posting again is what everyone should have known when the OP first asked his question:
Energy = one half mass times velocity squared.
Irishman - if you know that is wrong, then you have posted numbers and equations that says so. Subjective claim or denial suggests outright lying or lack of basic high school education. Honest answers always include the numbers. So, where are your numbers? The always required numbers when you insist someone is wrong.
Explain how two cars at 50 km/hr is energy equal to one car at 100 km/hr? Explain means posting the numbers.
Energy in a 100 Km/hr car is one half times its mass times 100 squared. Energy in each 50 km/hr car is one half times the same mass times 50 squared. Please show everyone how that energy is equal. You claimed it is so – but only subjectively. Prove it. A point I make often because scammers and politicians can only exists when numbers are replaced by subjective denials (also called hearsay or junk science).
Why must every adult know this? Because a car doing 100 km/hr (or 70 mph) requires twice as much braking to stop. As compared to a car only doing 70 .7 km/hr (50 mph). But then every high school graduate is required to know this. Otherwise jobs must go overseas to those who did learn what is critically important – the numbers. Show me your numbers or withdrawal subjective (empty) denials.
Your conservation of momentum exists only when a collision is elastic. Car crashes are not elastic. M1 * V1 = M2 * V2 only when kinetic energy before the collision is exactly same as energy after the collision. Obviously that does not exist for two crashing autos. Crash zones are about converting that kinetic energy into scrap metal. To not conserve momentum. To make the collision as inelastic as possible. So that momentum is not conserved. And so that less energy appears inside the passenger compartment.
Again, I make the point because it is why we need all citizens educated. To not know this stuff subverts the economy by empowering scammers and corrupt politicians.
Energy equivalent of one car at 100 km/hr is two cars at 70.7 km/hr. Again, the basic energy equation that must be in you every conclusion:
Energy = one half times mass times velocity squared.
Conservation of energy (not momentum) applies.
Energy is always conserved. Two cars at 50 km/hr or one car at 100 km/hr. When the collision is finished, that energy defined by velocity is somewhere else. Both cars are at 0 km/hr. Where did that energy go? Exactly why crumple zones are created.
No way around what is taught to everyone in high school physics. And fundamental to answering the OP’s question IF that is what the OP meant by a vague term: “detrimental”.
If you’d prefer, we could assume some other degree of elasticity for the collisions, or leave it as a free parameter. Assuming that it’s perfectly inelastic is just a simplification to make the calculations easier, but we could use any other elasticity at all. And as long as the elasticity is the same for both cases (no reason it shouldn’t be), the collisions will be equivalent.
Let me make it a bit more general, even: A head-on collision of cars of mass M each going at speed v, vs. a car of mass M going at 2*v hitting a car at rest.
Case 1: Each car has an initial kinetic energy of 0.5Mv^2. Both cars end up at rest, for 0 final kinetic energy. All of the kinetic energy is therefore absorbed and converted to heat by the crumple zones, and the crumple zone of each car has absorbed an energy of 0.5Mv^2.
Case 2: The moving car has an initial kinetic energy of 0.5M(2v)^2, or 2Mv^2, while the initially-stationary car has no kinetic energy initially. After the the collision, from conservation of momentum, we find that both cars are moving at speed v, half the speed of the initially-moving car. The final kinetic energy of the two cars is therefore 0.5(2M)v^2, or Mv^2. Thus, the kinetic energy has decreased by Mv^2, which means that much energy must have been absorbed by the crumple zones. This energy is divided between the two cars, which means that each car’s crumple zone absorbed an energy of 0.5Mv^2.
Note that in both cases, each car absorbs an energy of 0.5Mv^2, so the damage is equivalent in both cases.
Since everyone does know that this isn’t really helping. What you are missing is that after the collision both cars are moving at 50 km/hr. That’s where the extra energy is that you keep ignoring. At the end of the collision with the stationary car all that extra energy is now moving the crumpled cars.
This is not correct. Momentum is conserved always, in every form of interaction known to science, from subatomic particles to billion-solar-mass black holes. There is never any situation in which momentum is not conserved.