What I find scary is that you don’t know simple high school physics and yet accuse others of it.
People explained to you, multiple times, some with words, some with equations, that what’s important is not the initial energy, but the difference between initial energy (pre-collision) and final energy (after collision). And that difference is the same in both cases.
True. But it has nothing to do with the question in the OP.
It means that the crashing car speed is not 0. Energy is being disipated by the cars rolling and slowing. In the head to head crash all energy is disipated by the two cars chrunching.
So all collisions are elastic. So every car crash is still moving down the road? Of course not. Each crash is two cars not moving. What happened to that energy? Crumple zones. No moving vehicles. And more energy dissipated as tire skid marks.
The quadrupled energy means four times more crumpled metal, tire wear, burned breakes, etc.
The numbers that so many others have also posted. You did read every post before replying - as expected.
A car at 100 km/hr is one hald mass times velocity squared. 100 sqaured times one half means energy is 5000 times mass.
A parked car is one half times mass times zero squared or zero energy.
A car at 50 km/hr is 50 square times one half or 1250 times mass.
The energy of a 100 km/hr car into a parked car is 5000 times mass plus 0 times mass. Or 5000 times mass.
Then eerngy of two 50 km/hr cars is 1250 time mass plus 1250 times mass. Or 2500 times mass. Again, this was posted by others multiple times.
Show me how those numbers are wrong. Your second equation of
0.5*(2*M)v^2, or Mv^2. for v=50 is 2500 times mass.
Your first equation of 0.5Mv^2 for v=100 is 5000 times mass.
So, if both cars absorb energy equally, then the 50 km/hr example means each car absorbs energy at 1250 time mass. A 100 km/hr car crashing into a park car means both cars absorb 2500 times mass. How does 1250 equal 2500.
BTW, this is the classic question used in SATs to separate those who learned their high school physics from those who foolishly assume the collisions are equal energy.
Even your own posted equations agree with me.
First off, all collisions are not elastic.
However, what I think you are missing is that you are looking at the second case after the mangled wreck of two cars has slid to a stop. You need to look at the situation immediately after the collision, in which the car traveling at 100 km/hr has struck the parked car. By conservation of momentum, the two cars after the collision will have a resultant velocity of 50 km/hr immediately after the collision. They will then slide to a halt. During the slide, the KE of the two mangled cars will be dissipated as frictional heat, but this does not necessarily cause harm to the occupants.
If it helps, consider the ideal case where the collision takes place on a frictionless surface (idealized as a sheet of ice). There is no question here that the two cars after the collision will move off with a speed of 50 km/hr, which will then NOT dissipate (because of the frictionless surface). Yet there was still a horrific collision that caused damage when the vehicle traveling 100 km/hr struck the motionless vehicle.
Because, as I stated back in Post #24, in the first case, the initial total KE of the two vehicles is 2500m. The final KE is zero, so the KE dissipated in the actual collision is 2500m.
In the second case, the initial KE of car A is 5000m, but the KE of the two mangled vehicles immediately after the collision is 2500m. (The rest of the KE will then be dissipated as friction as the two mangled vehicles slide to a stop). So the KE dissipated in the collision will again be 2500m.
:rolleyes:
From the New York Times regarding Robert Goddard and the impossibility of a rocket functioning in a vacuum:
Needless to say, like westom here, the Times was dead wrong, too. The Times, at least, published a retraction 49 years later following the launch of Apollo 11.
I’ve repeatedly given both explanations as well as numbers in my posts.
So two cars at 50 Km/hr crash. And keep moving at 50 km/hr after the collision? Nonsense. Must be true for momentum to be conserved. If the collision were elastic (ie between pool table balls). But a collision between cars is inelastic. When done, both cars are at 0 km/hr. Momentum is not conserved. Energy is. When done, all cars are at 0 km/hr.
Where did energy go? Into crumple zones, burned tires, brakes, etc. Energy defined by 50 km/hr is now found in scrap metal. An inelastic crash.
OP’s question defines inelastic collisions. When done, a 100 km/hr car, the parked car, and the 50 km/hr cars are all doing zero km/hr. No car is still moving at 50 km/hr.
Energy of one car at 100 km/hr is equal to two cars at 70.7 km/hr. Not two cars at 50 km/hr as so many try to claim subjectively. Subjective claims imply lying - as scammers and politicians love to do. Make bogus claims without numbers. Numbers - to be energy equal, both cars must be at 70.7; not at 50.
BTW, this has long been a typical question on SATs to separate those who think subjectively from those who learned their physics and do the numbers.
Show me one post (with equations and numbers) that dispute what I have posted. Please do not mock everyone by citing a post number. Quote the exact numbers and put them into the relevant equation. Otheriwise it is just anothger useless subjective post.
For example, Telemark’s equations and numbers agreed with me. Energy of 1250 times mass (for each car) is not same as 2500 times mass. Those numbers from his own equations. Or do you know something different? Good. Then your next post shows those equations with the appropriate numbers inserted.
If someone posted numbers that say otherwise., then you quote each number in your post. Honesty demands that. Do not just cite a post number. If you have knowledge, your post puts each number into the appropriate equation. The equation is this simple:
Energy = 0.5 times mass timed velocity squared.
Insert the numbers. Show me how total energy of 2500 times mass is the same as 5000 times mass. Telemark made that claim even though numbers in his own equations contradicted him.
Snnipe 70E - the crash is done when all cars are at 0 km/hr. But again, confusion created by the OP’s vague question: define “detrimental”. One car at 100 km/hr or two cars at 50 km/hr. The crash is done when all cars are at 0 km/hr.
So you evidently missed that question. :rolleyes:
This thread went from frustrating to comical, and now it has entered the pathetic stage. It is certainly a testament to the motto of the board - “it’s taking longer than we thought.”
I applaud anyone who has the patience to continue. Hopefully those individuals who were confused have gained some insight from multiple knowledgeable posters.
To the OP, this is not a difficult problem. Please disregard the ignorance of the vocal minority. I hope it is apparent who that is.
No. They start out moving at 50 km/hr after the collision. Then, due to friction, they slow down to 0.
Momentum is, in fact, conserved. After the collision and after the cars slowed down, the earth is spinning just a little bit faster.
Why burned tires? Let’s say the tires are not locked - they will just keep rolling (unless crimpled metal obstructs them) Same with brakes.
A car starts at 100 miles an hour. Then you put the car into neutral. It keeps rolling, and eventually stops. No crumple zones. No burned tires. No burned brakes. Is this equivalent to a collision with a brick wall at 100 mph? After all, according to your logic, the same energy is involved in both.
Wrong.
Chronos did a pretty good job.
Momentum is always conserved.
Wrong. In the second case, immediately after the collision, the two cars must be moving together at 50 km/hr, by conservation of momentum (assuming they stick together). They then slide to a stop, but this takes place AFTER the collision.
True, but irrelevant here, because it assumes that both situations will end up with the cars at rest immediately after the collision. They don’t.
I and others have posted (with equations and numbers) the correct answer repeatedly. I’ve only been referencing post numbers because it’s getting exhausting repeating myself.
Situation after a collision is all cars stopped. You are now changing the definition of ‘detrimental’ to justify your conclusion.
What happens as each slides to a halt? “Detrimental” continues. “Detrimental” does not end until all cars have halted by scrapping pavement, tearing tires, burning brakes, rolling and killing passengers, etc. That is when the crash is done as defined by the word “detrimental”.
If the cars continue rolling 3 miles down the road and into a bridge, that is still part of ‘detrimental’ - the event called a crash.
You also posted virtually no numbers. This equation you post with your numbers in it:.
Energy = .5 times mass times velocity squared.
Show me how one car at 100 km/hr is energy equivalent to two cars at 50 km/hr? Not a rhetorical question. One that demands your post include (if necessary) numbers you claim to have already posted.
Again the not so subtle hint. Two cars must be at 70.7 km/hr to be energy equivalent. To create a collision with energy at 5000 times mass.
It’s easy. 0.5 times 70.7 squared times mass is what? 2500 mass. One for each car. Total of 5000 times mass. Don’t take my word for it. Honesty also demands you confirm the arithmetic.
Easy. 0.5 times 100 squared times mass is 5000 times mass.
Both cars must be doing 70.7 km/hr to be energy equivalent to one car at 100 km/hr. Two cars at 50 km/hr is a much less violent crash. That will always confuse those who only think subjectively. Honest means you always do and post the math – so as to not be subjective like scammers and corrupt politicians.
Show me your numbers. Show me where those above numbers are wrong. Please post each numbers with specific corrections. You know those numbers are wrong? Then post the numbers. And then show which number is wrong.
In the interest of fighting ignorance, here goes:
Assume that both cars are identical and have a mass of 1000 kg. To simplify matters further, we will assume that the collision takes place on a frictionless surface. (If you find this too abstract, consider the case of train cars on a train track). We will assume that the cars stick together in both collisions. (In the case of train cars, assume that the cars lock together.)
Case 1 (Two cars hit head-on, each traveling at 50 km/hr):
The initial momentum of car A is m[sub]A[/sub]v[sub]A[/sub] = (1000 kg)(50 km/hr) = 50,000 kg km/hr. The initial momentum of car B is m[sub]B[/sub]v[sub]B[/sub] = (1000 kg)(-50 km/hr) = -50,000 kg km/hr. You will notice that the initial total momentum of the combined two-car system being considered is zero, when you add them together.
The initial KE of car A is 0.5m[sub]A[/sub]v[sub]A[/sub][sup]2[/sup] = 0.5(1000 kg)(50 km/hr)[sup]2[/sup] = 1,250,000 kg km[sup]2[/sup]/hr[sup]2[/sup]. The initial KE of car B is 0.5m[sub]B[/sub]v[sub]B[/sub][sup]2[/sup] = 0.5(1000 kg)(-50 km/hr)[sup]2[/sup] = 1,250,000 kg km[sup]2[/sup]/hr[sup]2[/sup]. The total initial KE is 2,500,000 kg km[sup]2[/sup]/hr[sup]2[/sup].
After the collision, for momentum to be conserved, and because the initial total momentum was zero, both cars must have zero final velocity. In other words, both cars are at rest after the collision.
Because the cars are now at rest, the final kinetic energy of the two cars is also zero. Where did this energy go? It went into heat, noise, and/or the crumpling of metal. Note that kinetic energy was not conserved. Indeed, the change in KE is equal to 2,500,000 kg km[sup]2[/sup]/hr[sup]2[/sup].
Case 2 (A car traveling at 100 km/hr strikes a car at rest):
The initial momentum of car A is m[sub]A[/sub]v[sub]A[/sub] = (1000 kg)(100 km/hr) = 100,000 kg km/hr. The initial momentum of car B is m[sub]B[/sub]v[sub]B[/sub] = (1000 kg)(0 km/hr) = zero. You will notice that the initial total momentum of the combined two-car system being considered is 100,000 kg km/hr, when you add them together.
The initial KE of car A is 0.5m[sub]A[/sub]v[sub]A[/sub][sup]2[/sup] = 0.5(1000 kg)(100 km/hr)[sup]2[/sup] = 5,000,000 kg km[sup]2[/sup]/hr[sup]2[/sup]. The initial KE of car B is 0.5m[sub]B[/sub]v[sub]B[/sub][sup]2[/sup] = 0.5(1000 kg)(0 km/hr)[sup]2[/sup] = zero. The total initial KE is 5,000,000 kg km[sup]2[/sup]/hr[sup]2[/sup].
After the collision, for momentum to be conserved, the final momentum after the collision must be 100,000 kg km/hr. It cannot be zero, unless an outside force acts on our cars. With a frictionless surface, no outside force can act on our cars in the direction of motion. So the final momentum is equal to 100,000 kg km/hr = m[sub]AB[/sub]v[sub]AB[/sub] = (1000 kg + 1000 kg)v[sub]AB[/sub]. Solving for v[sub]AB[/sub], gives v[sub]AB[/sub] = 50 km/hr. This is the final velocity of the two cars after the collision. On a frictionless surface, this velocity would remain unchanged. On a road, friction would quickly slow the cars down, but this takes place AFTER THE COLLISION.
So the final kinetic energy of the two cars is 0.5m[sub]AB[/sub]v[sub]AB[/sub][sup]2[/sup] = 0.5(1000 kg + 1000 kg)(50 km/hr)[sup]2[/sup] = 1,250,000 kg km[sup]2[/sup]/hr[sup]2[/sup] = 2,500,000 kg km[sup]2[/sup]/hr[sup]2[/sup]. You will notice that the change in KE is equal to 5,000,000 kg km[sup]2[/sup]/hr[sup]2[/sup] minus 2,500,000 kg km[sup]2[/sup]/hr[sup]2[/sup] = 2,500,000 kg km[sup]2[/sup]/hr[sup]2[/sup]. You will also notice that the change in kinetic energy is equivalent to Case 1.
Q.E.D.
No, that is another, subsequent collision. We are only considering the first collision here.
The first collision ends when all cars stop moving. You cannot change the problem to suit your needs.
If momentum is conserved, then no energy is lost in crumple zones. You cannot have it both ways.
Momentum is only conserved if the collision is elastic. If crumple zones do not crumple; do not absorb energy. Only then can MV before a collision equal MV after the collision.
We know 1250 times mass (for each 50 km/hr car) is reduced because energy dissipates in crumple zones. Or do you also deny that? For example, ‘1000 times mass’ energy dissipates in the crumple zone. Leaving ‘250 times mass’ to keep moving the car. The equation is simple.
E = .5 times mass times velocity squared.
A remaining 250 times mass = .5 times mass times velocity squared.
What is the new velocity as cars separate? About 11 km/hr.
So momentum is conserved - you say. MV=MV? Mass times 50 km/hr equals mass times 11 km/hr?
Why does the math not work? Because momentum is never conserved when a collision is inelastic. When crumple zones exist. But then I am only repeating - and again with numbers - high school physics.
Energy is always conserved. Or do you also deny that? Not a rhetorical question. You answer it – but you won’t.
Provided are equations. Rather than post subjectively, post your equations with your “corrected” numbers. Because honesty demands your every post do so. Honesty says you explain how a car moving less than 50 km/hr after a crash is conservation of momentum. Explain how conservation of momentum exists when energy dissipates in crumple zones.
BTW, good luck changing the laws of physics.
This is incorrect, as others have tried to explain. Momentum is always conserved, including during inelastic collisions.
Kinetic energy is not conserved during an inelastic collision (because of the crumpling), but MOMENTUM IS STILL CONSERVED.
If you don’t believe me, look it up in ANY PHYSICS TEXTBOOK EVER PUBLISHED.
Momentum is not the same thing as energy.
Momentum is always conserved. Energy is always conserved. Kinetic energy is only conserved in ideal, perfectly elastic collisions. In inelastic collisions, kinetic energy is not conserved.
Wrong. Wrong. WRONG. :rolleyes:
Momentum is always conserved. It is conserved in elastic collisions as well as inelastic collisions.
What’s wrong here is not the law of conservation of linear momentum, but your apparent confusion between energy and momentum, your apparent inability to comprehend velocity vectors, your idea that momentum is dissipated in crumple zones, and your erroneous conclusion that the final velocity of the cars is “11 km/hr.” In other words, a bunch of gibberish resulted in an erroneous conclusion.
The math doesn’t work because you set up the problem wrong. And if your high school physics teacher read your nonsensical posts, they would either weep or be rolling over in their grave.
Sure I will. I agree that energy is always conserved. Note that kinetic energy is not always conserved, though.
I have tried to explain, but I can’t do much about willful ignorance.
Speak for yourself. :rolleyes:
Again, momentum is ALWAYS conserved.
Two cases:
Is (1) more or less “detrimental” than (2)? Note that EXACTLY the same kinetic energy is dissipated in both cases.
Feel free to ignore answering this if you feel uncomfortable doing so.
It is also possible to show that the change in momentum for each vehicle is the same in both cases:
Case 1 (Two cars hit head-on, each traveling at 50 km/hr):
The initial momentum of car A is m[sub]A[/sub]v[sub]A[/sub] = (1000 kg)(50 km/hr) = 50,000 kg km/hr. The initial momentum of car B is m[sub]B[/sub]v[sub]B[/sub] = (1000 kg)(-50 km/hr) = -50,000 kg km/hr.
After the collision, for momentum to be conserved, and because the initial total momentum was zero, both cars must have zero final velocity. In other words, both cars are at rest after the collision.
For both vehicles, the change in momentum before and after the collision is 50,000 kg km/hr.
Case 2 (A car traveling at 100 km/hr strikes a car at rest):
The initial momentum of car A is m[sub]A[/sub]v[sub]A[/sub] = (1000 kg)(100 km/hr) = 100,000 kg km/hr. The initial momentum of car B is m[sub]B[/sub]v[sub]B[/sub] = (1000 kg)(0 km/hr) = zero.
After the collision, by the analysis presented in the quoted section above, the velocity of the two vehicles is 50 km/hr. The momentum of car A is m[sub]A[/sub]v[sub]A[/sub] = (1000 kg)(50 km/hr) = 50,000 kg km/hr. The momentum of car B is m[sub]B[/sub]v[sub]B[/sub] = (1000 kg)(50 km/hr) = 50,000 kg km/hr.
Note that the change in momentum for car A is 50,000 kg km/hr. The change in momentum for car B is also 50,000 kg km/hr. This is equivalent to Case 1.
Q.E.D.
This conclusion is significant because the true measure of damage to the vehicles is the force experienced by the vehicles over the course of the collision.
The force experienced by the vehicles is equal to the change in momentum (i.e. impulse) divided by the time interval over which the force is applied. If we assume that the time interval is the same in both collisions, all we need to look at then is the change in momentum.
And as I have demonstrated here, the change in momentum during the actual collision is the same in both cases.
You first said the initial momentum was 50,000 kg km/hr. For a 1000 kg car, that means initial velocity is 50 km/hr.
Later you change things to suit your needs. Now your initial total momentum changes to zero? When did sleight of hand become part of physics?
MV before the collision. 50,000 kg km/hr. That means 1,250,000 kg km2/hr2 of energy - your numbers. Cars collide. You said, crumple zones, et al absorb energy.
1,250,000 must be smaller because you said energy is absorbed. Any number - the crumple zone absorbs 1,000,000. Now the car only has 250.000 kg km2/hr2 remaining.
E / .5 / mass = velocity squared.
250,000 / .5 / 1000 = 125 = velocity squared.
Due to energy absorbed in crumple zones, the new velocity is less. In this example, 11 km/hr. Where is your conservation of momentum?
MV for 50 km/hr is not equal to MV for 11 km/hr. Why? Because collisions are not elastic. Because crumple zones absorb energy. Since a crumple zone absorbed only 1,000,000 kg km2/hr2 of energy, the new velocity goes from 50 km/hr to 11 km/hr. Where is conservation of momentum?
The collision was not elastic. Cars are intentionally designed (for human safety) to be the least elastic. Conservation of momentum can only exist in elastic collisions. Your own numbers demonstrate that conservation of momentum does not exist.
Then you also invented an initial velocity of zero when your own numbers started with an initial velocity of 50.
I aggressively asked for numbers in every post from every poster. Now that you did, identified is where your numbers only make sense when conservation of momentum is disposed.
Using your own numbers, if a crash absorbs energy, then momentum is not conserved. I used your numbers. Crumple zones reduce velocity from 50 o any other numbers down to zero (ie 11 in the example). If velocity decreases, then conservation of momentum does not exist. We intentionally design cars to make inelastic collisions. To not conserve momentum.
Finally using your own numbers. First collision: each car at 50 km/hr has an energy content of 1,250,000 kg km2/hr2. Second collision: a car moving at 100 km/hr has an energy content of 5,000,000 kg km2/hr2. For a parked car: 0 kg km2/hr2
Total energy in the first collision: 2,500,000. Total energy in a 100 km/hr crash: 5,000,000. Energy numbers do not lie. A 100 km/hr car crash is twice as violent - your numbers.
And yet somehow you then spin those numbers to make 2,500,000 equal to 5,000,000. How? You ignored conservation of energy. You assumed a crash and crumpling zone absorbs no energy. Assume conservation of momentum exists. Let’s use your numbers.
Second collisioni: what happens when 5,000,000 kg km2/hr2 energy is applied to two 1000 kg cars? Do the numbers. Both cars are now moving at 70.7 km/hr.
E = .5 times mass of both cars times the velocity of both squared.
5,000,000 = .5 * 2000 Kg * velocity squared.= 5000
Velocity is 70.7 km/hr
What happened to your conservation of momentum? When did 1000 kg times 100 kg/hr become 2000 times 70.7 km/hr? It didn’t. Why. Conservation of momentum here is bogus. Conservation of energy says so.
But again, using your numbers. Everything works as long as conservation of momentum is bogus. You cannot have it both ways. Either you believe in conservation of energy. Or you believe in conservation of momentum. Both cannot exist – as demonstrated by your numbers.
How do you get 2000 kg times 50 km/hr to equal 1000 kg times 100 km/hr? Somehow you must change initial energy from 5,000,000 kg km2/hr2 to 2,500,000. How do you do that? Change initial velocity from 100 km/hr to 70.7 km/hr.
Changing initial conditions is the only way to make conservation of momentum work. But the initial conditions are 100 km/hr – not 70.7. You cannot change facts to justify a myth - ‘conservation of momentum’. Why? Because conservation of momentum does not apply to inelastic collisions. And because your own numbers said the incoming 100 km/hr car is 5,000,000 kg km2/hr2. Not 2,500,000 as your conservation of momentum claims.
Now that you have provided the numbers - finally - I can use your numbers to expose mistakes. Your own numbers says the energy of a 100 km/hr car is not equivalent to a collision of two 50 km/hr cars. And yet using a myth called conservation of momentum, you magically changed 5,000,000 kg km2/hr2 into 2,500,000 kg km2/hr2.
Again, either you have conservation of momentum OR conservation of energy. Your own numbers say both cannot exist. Your own numbers say the two collisions have vastly different energy. You even changed initial conditions from 50 km/hr to zero km/hr just to make a bogus ‘conservation of momentum’ work.