I am going to add a few assumptions here so we don’t get into another plane on treadmill mess.
The cars are identical models.
They collide squarely head-on.
One car does not go flying over the other.
One half second after the initial contact, the “detriment” to cars and occupants would be the same in both cases.
In case #1, I would assume both cars have come to a complete stop, and no further damage can occur.
In case#2, both cars are now moving at 50kmh. In a real life scenario, it is entirely possible that the cars could could flip and roll, causing additional “detriment.” It is also possible that they skid harmlessly to a stop.
This is not correct. After the collision in scenario #2, both cars are moving in the same direction at 50 km/hr. If you doubt this, go watch some 2-vehicle crash tests on YouTube.
After the collision in scenario #1 (both cars headed toward each other at 50 km/hr), both cars are at rest, so in that case yes, all of the initial kinetic energy is dissipated in the collision.
scenario 1:
Initial total kinetic energy of both vehicles: E1 = 2*(0.5m50^2) = 2500m (m = mass of one car)
Final total kinetic energy of both vehicles: E2 = 0 Energy dissipated in scenario 1 impact: E1 - E2 = 2500m scenario 2:
Initial kinetic energy of both vehicles: E1 = .5m100^2 = 5000m
Final kinetic energy of both vehicles: E2 = 2*(.5m50^2) = 2500m Energy dissipated in scenario 2 impact: E1-E2 = 2500m
Either way, the energy dissipated into the crumple zones is the same, so both scenarios present equally severe impacts for the vehicle occupants.
None of this really matters, because the fact that the cars have to be moving away from the collision at 50 km/hr comes from conservation of momentum (assuming they are stuck together, which is not a bad assumption in a case where the objects colliding tend to crumple up when they collide).
Also, assuming the cars hit head-on in both cases, there should not be any difference in the time of collision.
Both cases are equivalent. The single car going 100 km/hr has more energy, but not all of that energy will go into damaging the cars: Some of it will go into making the other car move. This is different from hitting a proverbial brick wall, because the brick wall can’t move, so the energy has no place to go but into damage. Hitting a brick wall at 50 km/hr is equivalent to a head-on collision between two cars each going 50 km/hr.
Westom’s definition of “detriment” makes my brain hurt. By this definition, a car slamming into a brick wall at 100kmh undergoes the same detriment as a similar car initially moving at 100kmh coasting to stop without hitting anything.
So you are talking about oranges while others discuss pears. If detriment is what you have defined, then every number you have posted is useless. We need mass, velocity, etc for the occupants. We need to know the elasticity of their collision. We need to know if the occupants are restrained. The 50 and 100 km/hr numbers tell us nothing useful for your definition of detriment. Worse, your posts are only subjective.
As first noted, what you are calling ‘detrimental’ cannot even be discussed due to too many hypotheticals. Due to missing numbers. Due to facts not found in the OP’s original question. Irrelevant even to what the myth busters demonstrated.
Other answers are addressing a completely different “detrimental”. Oranges and pears. Your detriment cannot be defined by anything in the OP’s original post - a 100 km/hr car or two 50 km/hr cars. We can only say the car and occupant as a unit are confronted by twice the energy when one car does 100 km/hr. Nothing more.
To say more means you must define “where” energy dissipates in the car and in its occupants - with numbers and vectors. I posted “where” in at the start. Your question in post #57 was repeatedly answered by “where”.
newme - it’s simple. Is detriment a relationship between the occupant and the dashboard? Or detriment defined by the energy that turned metal into scrap? Even the now moving car dissipates energy ‘detrimentally’ as tire rubber is ripped off by the pavement. “Where” does energy dissipate? robby says he is concerned with physics inside the passenger compartment. But then subjectively argues about energy that converts metal to scrap. Why is he concerned about energy dissipated outside when his ‘detriment’ is should be concerned with how hard the head hits a dashboard?
Why do politicians so easily manipulate the naive? The naive forget to define a problem before answering the question - as we have here. Or have brain pain rather than learn how to do the work. Always define a problem long before trying to answer it. Always. Otherwise scammers and politicians seek you out.
For most, ‘detrimental’ is the energy expended destructively into both vehicles. That includes any brake wear used to stop a once parked vehicle. When is that energy highest – most detrimental? When one vehicle at 100 km/hr strikes a parked vehicle.
It doesn’t matter what the OP means by detriment because both collisions are the same for all intents and purposes and hence will have the same “detriment” however you want to define it.
Sure there is. It’s in the car components that break and snap and bend. Those are “external” in the sense that they’re not motion-in-motion-out. I realize you might reject this notion at first by saying that the car parts are in the car and thus part of the system, but that’s not true. If you ignore those breakable components of the car, you’ll get your 50 km/hr output, but the cars won’t be the total wrecks that we’re talking about.
In the intro textbooks, they assume that the cows are spherical, that cars don’t break, that the ice is frictionless, that the rubber band is perfectly elastic, etc. And that’s all well and good, but the topic of the thread requires that we talk about breakable cars and cow-shaped cows.
If the cars don’t deform, then you get the typical “this skater grabs that skater at rest, how fast do they move?” problem. But these cars do deform, and that takes energy/momentum out of the system.
And lest you confuse me with other nay-sayers in this thread, remember that I’m agreeing that the situations are equivalent. I’m just saying they’re not coming out at the typical “.5mv[sup]2[/sup] + .5mv[sup]2[/sup] = 2mv” result you’d get from the textbook.
Chessic Sense, I’m sorry, but you’re just plain wrong here. None of the forces you are discussing would be considered to be “external” forces. They would be considered to be “internal” to our system under consideration.
And when you are discussing conservation of momentum, you don’t need to worry about internal forces, even the ones that that bend car parts.
We can safely assume that in the situation under discussion here, there are no external forces acting on the cars during the collision.
Therefore the linear momentum of the two cars are conserved, and in the case where a car traveling at 100 km/hr strikes a parked car, the two cars (assuming they stick together) will move off at an initial speed after the collision of 50 km/hr.
Indeed, assuming the cars stick together, it doesn’t matter whether or not the cars are unharmed or are a mangled wreck. Conservation of momentum holds in either case.
I think you’re right about that, as doing the KE calculation on 64kph and 55 kph the numbers don’t add up. A car travelling at 64kph does not have twice the KE of one travelling at 55 kph. Looking at the pictures, the barrier has a thin deformable layer on the front and a solid structure behind it.
Chessic Sense, I can’t make sense of that post. I see no conceivable way that the components of the cars breaking and snapping can be considered “external”: They’re pretty much the textbook example of internal forces. Remember, energy can “hide”, being converted to hard-to-track forms like heat, but momentum can’t hide. You can’t take momentum out of the cars without putting it into something that isn’t the cars, and that something will move (or at least, change its motion) as a result.
And the equation you have in your last line, “.5mv[sup]2[/sup] + .5mv[sup]2[/sup] = 2mv” isn’t even dimensionally consistent, and thus can’t make sense.
I am convinced that for the most part there is no difference between 1 car going 100 into a stationary car and two going 50 head on into each other. Just fix your frame of reference to the front bumper of one of the cars and the colision looks the same.
The only difference is the rotation of the wheels. In the case of the one car going 100, its wheels are spinning twice as fast as the wheels in each car in the other situation. Since the rotational kinetic energy is proportional to the square of the angular velocity, the total rotational kinetic energy of the two cars is twice as much in the case of one car going 100. This shouldn’t matter a whole lot, but I believe it does introduce greater energy regardless of your frame of reference. The wheels will still want to continue to spin after the collision which will cause the car to get a little more push from the ground. I believe that this total push is double in the first case.
This change would be very minor, but as a thought experiment, think of lightweight cars with big, heavy tires and it makes a bigger difference.
Oh! Why didn’t you just say so in the first place?
Would be. Are. I’m not sure why you think the friction between my couch and carpet is external but the stress on the metal isn’t.
If you could, then it wouldn’t matter if the car was made of metal, cheese, gold, or unobtanium. But it does, because that’s the crux of the problem.
You’ve got to be kidding me. You honestly think it doesn’t matter how the car bends and stresses? You think all that distortion comes for free? Have you ever bent a metal rod back and forth and felt it get hot? Why do you think that is? You don’t get to bend metal for free.
Conservation of momentum holds when you have a closed system, i.e. where the only force is that Newton’s-Third force that causes one object to slow down and the other to speed up.
You don’t need to know any of these things if you assume that both collisions are perfectly inelastic and that the cars, occupants, restraints, etc. are identical in both cases.
:rolleyes: It’s also possible to needlessly obfuscate a simple physics problem by trying to bring in irrelevant factors and considerations.