Merged DDWFTTW in a “Real” Wind

[humber]

I knew I has read something recently about collissions and I found it. From out buddy Rhett who was wring about ddwfttw.

http://scienceblogs.com/dotphysics/2010/05/mythbusters_and_double_the_spe.php

So he is wrong about this as well.

Jamie said:
"Two cars crashing head on at 50 mph is the same as one car crashing into a wall at 100 mph"
Jamie was wrong, but that is not what I am going to talk about. Instead, I am going to talk about Adam's small scale test of this situation.
( about the experiment, not the original error)


Which is what I said about the two missiles. Rhett is right, Jamie is wrong.

ETA:
Jamie's response to the original error.
“….that was a mistake. You know what? I am ok with that. That is how you learn stuff”
http://www.wired.com/wiredscience/tag/mythbusters/page/2/

 

[3point14]

Nothing travels at the speed of the medium.

Not even the medium?

 

[humber]

Not even the medium?

Doh.

 

[3point14]

I stole it from someone else in the thread.

Edit: From 3point14 to be exact and now he has gone back to correct it!! I will edit mine to match.



It scares me that I am not surprised. Let me guess, his explanation involved RMS?

Thanks!

Working out the smaller/larger thing was, for some reason, really tricky for me!

 

[3point14]

Doh.

Well, that told me.


It told me that you don't get why the question is significant.

 

[jduffy77]

While Googling for Rhett I came across another link which I hope one of you will critique.

http://warp.povusers.org/grrr/collisionmath.html

A quote from this link happens to be exactly how I initially thought about the collisions.

"So we have two forces: Vehicle 1 applies the equivalent of 50 km/h times its mass to vehicle 2, and vehicle 2 applies an equal force to vehicle 1. This causes both vehicles to stop right there.

Where would the additional 50 km/h times the mass of the vehicle come from? Vehicle 1 cannot apply that force to itself. It's applying it to vehicle 2. So where is it coming from?

The answer is that it's not coming from anywhere because the force applied to vehicle 1 is not 100 km/h times the mass, but only 50 km/h times the mass. The same as with the rock wall."

 

[humber]

Thanks!

Working out the smaller/larger thing was, for some reason, really tricky for me!

Well, that told me.


It told me that you don't get why the question is significant.

You don't understand RMS or fluid dynamics, but unlike Jamie, are unwilling to accept that knowledge counts, and have no interest in learning.

 

[Modified]

I knew I has read something recently about collisions and I found it. From our buddy Rhett who was wrong about ddwfttw.

http://scienceblogs.com/dotphysics/2010/05/mythbusters_and_double_the_spe.php

So he is wrong about this as well.

He's not wrong about Jamie being wrong.

Two cars crashing head on at 50 mph is the same as one car crashing into a wall at 100 mph

is obviously wrong, as it equates a car and a wall. Two cars crashing head on at 50 mph is the same as one car crashing into another at 100 mph. One car crashing into a wall at 100 mph is the same as a car and a wall crashing head on at 50 mph. More or less, that is; the speed of the road relative to the cars will have some minor effect.

 

[Jack by the hedge]

The total is larger, and perhaps the heat of collision will vaporizer all the mass, so the result is a cloud hot gas containing 40000J, or the other way, no smoke and thunder, but two crumpled missiles, one with with 0J and the other. 40000J. That can't happen when both have only 10000J each.

Can't happen at all.

Energy is not the only property which is conserved; momentum is conserved too. After their inelastic collision, the missiles (or at least the centre of gravity of their debris) is not moving in the 600 vs 600 mph example but it is moving (to the left, at 300mph) in the 0 vs 1,200 mph example.

So the latter example retains half its 40,000J of KE as KE. Both collisions have the same amount of KE available to contribute to smoke and thunder. Both examples are in fact identical apart from the frame of reference which, when you consider the two missiles as a closed system, gives the second example 20,000 J of KE merely because the combined COG of the system is moving relative to the FOR.

(As ArmillarySphere already explained rather clearly.)

 

[3point14]

You don't understand RMS or fluid dynamics, but unlike Jamie, are unwilling to accept that knowledge counts, and have no interest in learning.

So, what you're saying is that (and for all I know, this might work) if you have a liquid made up of two different elements/compounds, and you pump them down a long enough hose, eventually you'll split them perfectly with one moving faster down the pipe than the other?

That's a pretty nifty trick if it works.

 

[Modified]

The physics forum will alert you to how common your misunderstanding is, Modified.

It was your misunderstanding, not mine. I was following your logic, and you now confirm that you were wrong.

And you still have not answered the question about two objects colliding in space. Since according to you, only KE measured in the reference frame of the ground is meaningful, how do we determine the strength of the collision? Do we need to find the nearest point to the surface of the earth, even if it is billions of miles away?

 

[humber]

He's not wrong about Jamie being wrong.

So you are.

Two cars crashing head on at 50 mph is the same as one car crashing into a wall at 100 mph is obviously wrong, as it equates a car and a wall. Two cars crashing head on at 50 mph is the same as one car crashing into another at 100 mph.

No it is not. The difference lies in the total kinetic energy, all of which are in the car or cars. And the example of the missiles, is two missiles not a wall and a missile.

One car crashing into a wall at 100 mph is the same as a car and a wall crashing head on at 50 mph. More or less, that is; the speed of the road relative to the cars will have some minor effect.

Then you need to take into account the KE of the moving wall to equate to two moving cars. The stationary wall has no KE. Jamie had the good sense not to post spurious and contradictory arguments in defense of his error.

 

[Modified]

So, what you're saying is that (and for all I know, this might work) if you have a liquid made up of two different elements/compounds, and you pump them down a long enough hose, eventually you'll split them perfectly with one moving faster down the pipe than the other?

That's a pretty nifty trick if it works.

Nah, it only works if one of the liquids is balloons.

 

[3point14]

Nah, it only works if one of the liquids is balloons.

Rats. I was thinking desalination.

 

[Jack by the hedge]

The difference lies in the total kinetic energy, all of which are in the car or cars. And the example of the missiles, is two missiles not a wall and a missile.


Then you need to take into account the KE of the moving wall. The stationary wall has no KE. Jamie had the good sense not to post spurious and contradictory arguments in defense of his error.

You need to take into account the conservation of momentum and the velocity of the centre of gravity of the whole system.

 

[humber]

It was your misunderstanding, not mine. I was following your logic, and you now confirm that you were wrong.

And you still have not answered the question about two objects colliding in space. Since according to you, only KE measured in the reference frame of the ground is meaningful, how do we determine the strength of the collision? Do we need to find the nearest point to the surface of the earth, even if it is billions of miles away?

Physics Professor Rhett, and Jamie, a keen and adroit experimenter holding, I understand, a degree in Russian, both say you are wrong. Spork is wrong.

Theory wrong, TM wrong. Experiments and trials wrong.

 

[Modified]

He's not wrong about Jamie being wrong. is obviously wrong, as it equates a car and a wall. Two cars crashing head on at 50 mph is the same as one car crashing into another at 100 mph. One car crashing into a wall at 100 mph is the same as a car and a wall crashing head on at 50 mph. More or less, that is; the speed of the road relative to the cars will have some minor effect.

To be a bit more precise, now that I think about it, the stationary wall would have to be on the same "wheels" as the moving wall. Or, the wall could be big and heavy enough that it doesn't matter much either way.

 

[EHocking]

Correct. It won't.

So a balloon won't reach windspeed.

If there is a dog in that basket. Does it wag its tail?

As per usual, you obfuscate when backed into a corner.
Your post was:

Originally Posted by humber
a helium balloon ...moves at the speed of the wind in the direction of the wind.

I create a new gif for just this type of posting.

The perceived velocity depends on the position within the slip stream.

Colloquially, the air over the top an airfoil, moves faster than the air on the underside.

Dodge, dodge, weave, weave.
Your post was describing a helium balloon not an airfoil.

<snip obfuscation about airfoils>
If the air in the basket is close to the free stream velocity, then the air over the balloon itself may be faster or slower.

What a load of guff. The subject was your post about a helium balloon, which you stated DOES move at WS.

I'll quote you again to remind you of the fact.

Originally Posted by humber
a helium balloon ...moves at the speed of the wind in the direction of the wind.

TRY to keep your story straight.
No wonder the threads you get in tend to be so prolonged, you even argue with yourself, let alone other posters.

 

[Dr. Keith]

Thanks!

Working out the smaller/larger thing was, for some reason, really tricky for me!

I thought it was great and quoted the wrong version of it!! So, it wasn't intuitive for me either . . .

 

[Dr. Keith]

Doh.

I'm sorry, is this you realizing how wrong you were or not understanding the questions?