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Originally Posted By: paul

red cart 6100 kg mass v +12.84 m/s
blue cart 100 kg mass v -40 m/s
results show that the blue cart bounces off of the red cart
the red cart continues in the + direction.
red cart velocity = +11.13 m/s
blue cart velocity = +63.97 m/s


OK, that's one mass turning around. Now do it for the other 49 masses. See if the red cart changes direction.

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Originally Posted By: paul
Quote:
We have to transfer -78324Ns of momentum to the pipe to stop it.


yes you do !!!
and you cant do that with only (-)4000Ns


Don't we have 50 masses, each with that much momentum?

Even if they just give all their momentum to the pipe, as you repeatedly claimed they would, then it would be enough to turn it around.

Actually I think there might be a mistake in you calculation of the -78324Ns. Did you consider that the pipe is already moving when it accelerates the 2nd mass?

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Here's a simple way to analyse any number of masses doing only one acceleration and the 1st turn. You have to agree to at least this. I realize there's still the 2nd turn or end stop that's not included. But if you don't agree on these two stages we should solve that first.

The system (only accelerator and 1st turn) is actually a rocket firing masses forward (+ve direction) as they come out of the 1st turn and no longer interact with anything.

So it's a rocket firing its engine in the +ve direction. That means it is accelerating in the -ve direction. It makes no difference how the internal details of the engine work. Any rocket will accelerate in the opposite direction to where the engine is pointed.

Right?

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Originally Posted By: paul
red cart 6100 kg mass v +12.84 m/s
blue cart 100 kg mass v -40 m/s

results show that the blue cart bounces off of the red cart
the red cart continues in the + direction.

red cart velocity = +11.13 m/s

blue cart velocity = +63.97 m/s


Hey Paul, there's a dirty little lie hidden in these numbers. That website is obviously operated by me to fool people about how momentum works:

red cart's momentum before:
6100kg * 12.84m/s = 78324 Ns

red cart's momentum after:
6100kg * 11.13m/s = 67893 Ns

Change in momentum:
67893Ns - 78324Ns = -10431Ns

But it was only hit by a cart with 100kg * -40m/s = -4000Ns of momentum!!!!

Impossible!

Can you explain why it gives those results? Do you accept them?

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Quote:
OK, that's one mass turning around. Now do it for the other 49 masses. See if the red cart changes direction.



at 60 seconds the pipe has a mass of 5000 kg
and a velocity of 20.59908 m/s so its momentum is

+102995.4Ns

results are
red cart v = +18.22264549019608 m/s
blue cart v = +78.82172549019607 m/s

now lets do 80 seconds

at 80 seconds the pipe has a mass of 3000 kg
and a velocity of 48.25031 m/s so its momentum is

+144750.934Ns

results are
red cart v = +42.556741612903224 m/s
blue cart v = +130.80705161290322 m/s

now lets do 100 seconds

at 100 seconds the pipe has a mass of 1000 kg
and a velocity of 121.7047 m/s so its momentum is

+121704.74Ns

results are
red cart v = +92.30384545454545 m/s
blue cart v = +254.00854545454547 m/s

note :
I did not subtract or add the resultant velocities of the pipe or the mass and include them for each calculation above
therefore the differences will be extremely different.

ie...the results above wont even touch the true results.
because by decelerating the mass the 4000N I did use will be much greater so the pipe will be accelerating much faster than the above.
which means that the pipe velocity will increase greatly each second.


but I will do that today.

but from the looks of it your going to need a lot of stuff on the spacecraft that uses electricity , because the blue cart has really gained speed through the turn.

perhaps some type of deflector shield and a molecular transporter would consume the extra electricity generated.

and of cource we shouldnt forget the phaser banks and photon torpedoes.

LOL

btw

the wiki web site says that when a more massive body collides with a less massive body the more massive body does not change its velocity.

and the less massive body then has twice the speed of the more massive body less its original speed.

http://en.wikipedia.org/wiki/Momentum

Quote:
Thus the more massive body does not change its velocity, and the less massive body travels at twice the velocity of the more massive body less its own original velocity. Assuming both masses were heading towards each other on impact, the less massive body is now therefore moving in the opposite direction at twice the speed of the more massive body plus its own original speed.


but surely you want something for your -4000Ns momentum influence !! to use to slow or stop the pipe.

but since were using a 100% elastic collision
you cant have anything to use !!!

LOL


resistance is futile
you will be assimilated

Originally Posted By: paul
but I will do that today.


I guess all I have to do is make the 100 kg mass speed twice the speed of the pipe and add its original speed to it as it enters the decelerator.

before each 1 second calculation after the 51st second of course.

and add the extra momentum that the pipe will receive due to the extra speed that the 100 kg mass has after passing through the turn.

thats what I'll do today.



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as far as I can tell , with the use of physics math
the 100 kg mass that has a original speed of 40 m/s will inherit 13.49721 m/s x 2 = 26.99442

for a grand total of 66.99442 m/s speed

so its velocity would be +66.99442 m/s

the following formula is used to find force using
distance , time , momentum
http://en.wikipedia.org/wiki/Force



distance = 1000 meters
time = 50 seconds
momentum = 100 kg x +66.99442 m/s


momentum = 100 kg x 66.99442 m/s = 6699.442Ns

distance x momentum = 1000 meters x 6699.442Ns = 6699442

distance x time = 1000 meters x 50 seconds = 50,000


F = 6699442 / 50000 = 133.98884N

so during the time period between 51 seconds to 52 seconds the pipe will have a force of
+134N and a force of +4000N causing it to accelerate.

after this second has passed the pipes momentum will become higher than the previous second.

which should cause the next 100 kg mass to have a even higher speed as it leaves the turn.

etc etc etc ,,, bla bla bla...

I'll finish it later , Im trying to put this in a program.




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Originally Posted By: paul
btw

the wiki web site says that when a more massive body collides with a less massive body the more massive body does not change its velocity.

and the less massive body then has twice the speed of the more massive body less its original speed.

http://en.wikipedia.org/wiki/Momentum

Quote:
Thus the more massive body does not change its velocity, and the less massive body travels at twice the velocity of the more massive body less its own original velocity. Assuming both masses were heading towards each other on impact, the less massive body is now therefore moving in the opposite direction at twice the speed of the more massive body plus its own original speed.


but surely you want something for your -4000Ns momentum influence !! to use to slow or stop the pipe.

but since were using a 100% elastic collision
you cant have anything to use !!!


No, Paul. Read the preceding line:
"When the first body is much more massive than the other (that is, m1 » m2), the final velocities are approximately given by"

You can use the two equations immediately above that line. Those are exact not approximate. They're what I already used in my program, which you've seen.

Last edited by kallog; 09/27/11 02:05 AM.
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Originally Posted By: paul
the 100 kg mass that has a original speed of 40 m/s will inherit 13.49721 m/s x 2 = 26.99442

distance x momentum = 1000 meters x 6699.442Ns = 6699442

distance x time = 1000 meters x 50 seconds = 50,000

F = 6699442 / 50000 = 133.98884N

I'll finish it later , Im trying to put this in a program.


No, no, no. All completely wrong. Don't waste your time programming it. We had this issue before. Instead of guessing, find where on the same page, the meaning of "d" is defined as "distance".

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your right , it does say differential above the formula.

so I'll use a different means that has the same result.

I'll change it around below.

-------------------------------------------------------

as far as I can tell , with the use of physics math
the 100 kg mass that has a original speed of 40 m/s will inherit 13.49721 m/s x 2 = 26.99442

for a grand total of 66.99442 m/s speed

so its velocity would be +66.99442 m/s

the 100 kg mass accelerated from 40 m/s to 67 m/s
in 1 second.

so the acceleration from 40 m/s to 67 m/s
in 1 second was 27 m/s/s

if I decelerate the 100 kg mass from 67 m/s to 0 m/s
the pipe will feel a force each second of

first lets check if this will work
v = vi x (a x t)

v = 67 m/s = 0m/s X ( 1.34 x 50 seconds)

so we decelerate it at a rate of -1.34 m/s/s
lets check it.

f=ma
134N = 100 kg x 1.34




so during the time period between 51 seconds to 52 seconds the pipe will have a force of
+134N and a force of +4000N causing it to accelerate.

after this second has passed the pipes momentum will become higher than the previous second.

which should cause the next 100 kg mass to have a even higher speed as it leaves the turn.

etc etc etc ,,, bla bla bla...

I'll finish it later , Im trying to put this in a program.

---------------------------------------------

it looks like the result was the same.




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Quote:
much more massive


how would you determine when a mass is much more massive?

is there a much more massive formula that is used to determine when a object is much more massive than another object?

is 6000kg much more massive than 100 kg?

or is it merely more massive?




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Originally Posted By: paul
Quote:
much more massive


how would you determine when a mass is much more massive?

is there a much more massive formula that is used?

is 6000kg much more massive than 100 kg?

or is it merely more massive?




That's why you can't use that formula for something that needs to be precise. There's no point at which it changes from wrong to right. It's only an approximation. Anyone using it has to decide if their large mass is large enough for the approximation to be good enough for their application. For us it is never good enough because we have a cumulative effect of many masses hitting it again and again, so the little error gets worse and worse.

The two equations immediately before it at the exact ones. Use them instead. They give correct results for large or small masses.

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on the force page?
http://en.wikipedia.org/wiki/Force


where are the two your talking about?



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Originally Posted By: paul
so the acceleration from 40 m/s to 67 m/s
in 1 second was 27 m/s/s


Huh? Where does a mass have a velocity of 40m/s? How did it get into that state?

The animation app showed one changing from -40m/s to 69m/s (or similar).

-40m/s is not the same as 40m/s.

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Originally Posted By: paul
where are the two your talking about?

section on momentum page

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Quote:
section on momentum page


you should just post the formula you prefer that way I dont use the wrong one.


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let me rephrase that just for you , so that you can comprehend it.

Quote:
so the 100 kg mass passing through the turn that is experiencing a speed increase from 40 m/s to 67 m/s
in 1 second experiences an acceleration of 27 m/s/s


therefore

a = (sf - si)/t

where
a = acceleration m/s/s
s = speed
t = time in seconds
f = final
i = initial

a = 67 m/s - 40 m/s / 1 second = 27 m/s/s



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did you know that.

if I was stationary in space in zero g

and the pipe had a velocity of +40 m/s

I would see the pipe moving at +40 m/s.

and if the pipe were transparent I would see the mass moving at +40 m/s just before it is launched in the (-) direction toward the turn.

after the 100 kg mass has traveled the 1000 meter distance to the turn and has been accelerated to -40 m/s during the 50 second acceleration time.

the 100 kg mass would not have a velocity !!!
and it would not have momentum !!!

in fact it has been decelerated from +40 m/s to 0m/s







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Im replying to my post because Im having a really hard time with the below from wiki.

Quote:
Thus the more massive body does not change its velocity, and the less massive body travels at twice the velocity of the more massive body less its own original velocity. Assuming both masses were heading towards each other on impact, the less massive body is now therefore moving in the opposite direction at twice the speed of the more massive body plus its own original speed.


Thus the more massive body does not change its velocity
that doesnt seem like it belongs in physics.

and the less massive body travels at twice the velocity of the more massive body less its own original velocity

that makes absolutely no sense.

twice the velocity !

no matter what the momentum is of the 2 bodies as long as one of them is more massive than the other.

thats like saying a car hits a train head on and the trains speed does not decrease.

train mass = 5 million kg
train velocity = +10 m/s
50 million Ns

what if the car has a speed of -5 million m/s
or -500 million m/s?

I say if the car weighs only 1 kg and it has enough speed it can not only stop the train but reverse its direction.

I dont think I'll use that method of determining the speed of the mass as it passes through the turn.

because this is the year 2011 and physics should have advanced beyond that type of crap.

its like there applying speed limits to math.
like they did to light.
like they own it.
or trying to control it would be the better choice of words.


so since physics math is not sure of what happens in a collision I dont think we should treat the turn as a collision.

we have too much troubles with the mass having to stop
and then reverse direction.

but the mass in our turn does not stop.




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Originally Posted By: paul
Quote:
section on momentum page


you should just post the formula you prefer that way I dont use the wrong one.



You can use any correct formulas. If it's not suitable then it won't produce a result. You might find others that you prefer.

But my preference is the first 2 formulas on that link.

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