a theory concerning space propulsion

[ninus maximus]

hello everyone , I'm new to the forum.

 

and I have a theory of a new type of propulsion for space travel.

 

consider a rail gun that fires a 100 kg mass along a 500 meter rail.

 

then the mass enters a turnaround at the end of the 500 meter rail.

having accelerated the 500 meters , it will present a force to the turnaround.

 

the mass turns 180 degrees through the turnaround and begins traveling the opposite direction and free floats in zero g for 500 meters in the

opposite direction.

 

it then enters a second 180 degree turnaround and presents a force to the second turnaround.

the mass turns 180 degrees again in the second turnaround and now enters the rail gun a second time

and it is accelerated even faster the second time.

 

the two forces that are presented to the turnarounds by the 100 kg mass will cancel each other out and the force that was used to accelerate the 100 kg mass will propel the spacecraft.

 

 

so you end up with

2 positive forces for propulsion ;

1) the force used to accelerate the mass

2) the force presented to the second turnaround by the mass

and

1 negative force for propulsion ;

1) the force presented to the first turnaround by the mass

 

this process repeats giving the space ship the ability to accelerate as long

as power is used to accelerate the mass.

 

as the mass inside is following a path inside the space ship the momentum

of the mass in reference to the space ship is conserved.

 

and since force is not momentum the conservation of momentum

is not violated.

 

it is a closed system but matter is not crossing the boundary.

the forces that are applied to the space ship from the inside , push

the space craft the same way a mass ejection propulsion system works

by ejecting or throwing mass out of the spaceship.

 

 

the force that accelerates the mass inside the space ship is transfered to

the space ship which gives the space ship acceleration.

 

nothing needs to be ejected or thrown from the ship for propulsion.

Edited June 25, 2010 by ninus maximus

[swansont]

the two forces that are presented to the turnarounds by the 100 kg mass will cancel each other out

 

I suspect upon more careful analysis that you will find that this is not the case. I think you have assumed the speed remains constant, and/or the speed profiles are identical for the two turns. Is that really the case?

[ninus maximus]

Originally Posted by swansont: I suspect upon more careful analysis that you will find that this is not the case. I think you have assumed the speed remains constant, and/or the speed profiles are identical for the two turns. Is that really the case?

 

Hello swansont

 

proposing that a force of 500 N is used to accelerate the 100 kg mass

in the rail gun for the 500 meter distance.

 

we can assume that the direction that the rail gun accelerates the 100 kg mass

is 0 degrees in a 360 degree range.

 

since the mass will be changing direction it is best to include direction when

describing the motion of the mass using velocity , so I will include direction

in the calculations to avoid confussion.

 

the velocity of the 100 kg mass will increase from 0 m/s/0d to 70.711 m/s/0d

in 14.142 seconds.

 

Durring acceleration the mass has a average velocity of 35.3555 m/s/0d

so the mass travels a distance of

35.3555 m/s/0d * 14.1422 seconds = 500.0045521 meters/0d

in 14.142 seconds time.

 

since the masses force of momentum is the product of

its mass * its velocity P=mv

 

the momentum of the mass as it enters the first 180 degree turnaround will be

 

P=mv

 

100 kg * 70.711 m/s/0d = 7071.1 N/0d

 

this force of momentum of 7071.1 N/0d will be presented to the first turnaround

as the mass passes through the first turnaround and changes its momentum from

 

P=7071.1 N/0d to P=7071.1 N/180d

 

now the 100 kg mass free floats to the second turnaround requiring 0 additional force

as it is not being accelerated and it still has a momentum that is the product of its

mass * its new velocity , and it is met by 0 resistance as the mass free floats in zero g

to the second turnaround.

 

the mass then enters the second turnaround and presents a force to the turnaround of

100 kg * its new velocity = slightly less than 7071.1 N / 180d

 

I have not included any resistance to movement as the mass passes through the

(2) 180 degree turnarounds , however using rolling friction such as .0001 which is

common in low friction bearings available on todays market should deliver a

useable calculation for speculation purposes.

 

so we can say that there will be a .0001 N loss of the momentum

that was the result of the 500 N force * 14.142 seconds used to accelerate the mass initially.

 

for instance the loss of momentum of the mass as it travels through the first turnaround

would be 7071.1 N * .0001 N = 0.70711 N

 

Knowing that the .70711 N momentum will be lost as the mass passes through the first

180 degree turnaround , and that there will also be a loss of momentum as the mass passes

through the second 180 degree turnaround we can establish that the mass will loose momentum passing

through both turnarounds.

 

0.70711 N * 2 = 1.41422 N

 

and we can establish that the new momentum of the mass after it has passed through both

turnarounds will be the product of its mass * its new velocity.

 

to provide for a easier to follow topic we can say that we have established that the mass

looses a momentum of 2 N through the entire cycle.

 

we have not established a mass for the space craft yet , so we can assign a mass to the space craft.

 

since we do not yet know the new velocity that the mass attains while passing through the 2 turnarounds

we can only give a estimation of the space crafts movement.

 

suppose the space craft has a mass of 5000 kg

 

the force applied to the space craft is 500 N

 

its acceleration is 0.1 m/s/s

 

its initial velocity is 0 m/s/180d

its average acceleration is 0.70711 m/s/s/180d

its final velocity is 1.41422 m/s/180d

 

the space craft moves a distance and direction of apx 10.000091042 meters/180d in 14.1422 seconds

 

we can establish that the space craft has gained a momentum of apx 7071.1 N/180d

 

therefore your question is certainly one that needs addressing as this will have a grave impact on determining

the velocity and distance that the space craft will travel.

Edited June 25, 2010 by ninus maximus

[swansont]

the momentum of the mass as it enters the first 180 degree turnaround will be

 

P=mv

 

100 kg * 70.711 m/s/0d = 7071.1 N/0d

 

Terminology of your vector notation (+ and - are fine for this) and units (N is not the unit of momentum) aside, OK. This is is not the problem.

 

this force of momentum of 7071.1 N/0d will be presented to the first turnaround

as the mass passes through the first turnaround and changes its momentum from

 

P=7071.1 N/0d to P=7071.1 N/180d

 

 

No. This is an internal force to the system, and you will exchange momentum with the railgun+track. The projectile will have less momentum after it changes direction. You have to treat this as you would a collision, though you may treat it as an elastic collision in the ideal case.

 

Your analysis is true only in the limiting case where the railgun has an infinite mass, in which case it will never move.

[ninus maximus]

No. This is an internal force to the system, and you will exchange momentum with the railgun+track. The projectile will have less momentum after it changes direction. You have to treat this as you would a collision, though you may treat it as an elastic collision in the ideal case.

will exchange momentum with the railgun+track.

yes , and the railgun and track are attached to the spacecraft !

 

and that exchange of momentum is the equal and opposite reaction of the spacecrafts momentum due to the force applied in accelerating the mass.

 

I am treating this as a collision , in fact 2 collisions.

 

the mass turning around in the turnaround will push the spacecraft , just as if the mass were compressing a spring.

although the spring completely stops the mass and then accelerates the mass in the opposite direction , the spring also applies

the compression force to the spacecraft.

 

the only real difference is the mass does not stop as it passes through the turnaround in fact its speed does not greatly reduce.

 

Your analysis is true only in the limiting case where the railgun has an infinite mass , in which case it will never move.

 

thats taking it to an extreme wouldnt you think , I sure hope that you are not simply a nay sayer that will go to any expence just

to try and prove a point.

 

I have found no viable reason why this would not work , and I have looked , yet you pick something infinite from thin air and use it.

 

if you have found a reason why it would not work would you mind posting the reason , also I am looking for a viable reason so if you dont mind , please avoid posting violations to physics laws , as simply posting a physical law without the inclusion of your reason the physical law appears to be violated will never prove or disprove the viability of any concept.

 

ie ...

 

F=ma

0=1000kg*1000m/s/s

 

the above shows a physical law violation...

 

vs

 

that wont work because it violates newtons laws of motion.

 

the above does not show a physical law violation.

 

I personaly cannot find a violation of any law of physics in this concept.

Edited June 25, 2010 by ninus maximus

[swansont]

will exchange momentum with the railgun+track.

yes , and the railgun and track are attached to the spacecraft !

 

and that exchange of momentum is the equal and opposite reaction of the spacecrafts momentum due to the force applied in accelerating the mass.

 

I am treating this as a collision , in fact 2 collisions.

 

the mass turning around in the turnaround will push the spacecraft , just as if the mass were compressing a spring.

although the spring completely stops the mass and then accelerates the mass in the opposite direction , the spring also applies

the compression force to the spacecraft.

 

the only real difference is the mass does not stop as it passes through the turnaround in fact its speed does not greatly reduce.

 

But the point is that the speed will reduce, and your numbers do not agree with that statement. Your numbers violate the conservation of energy, because the KE of the projectile is the same, and yet there must be a transfer of energy to the craft, since there is a transfer of momentum to the craft.

 

 

thats taking it to an extreme wouldnt you think , I sure hope that you are not simply a nay sayer that will go to any expence just

to try and prove a point.

 

I have found no viable reason why this would not work , and I have looked , yet you pick something infinite from thin air and use it.

 

Not at all. Looking at limiting behavior in solutions is a common analysis to see if the answer is correct; one does this when one finds a general solution to a problem. If you look to the side of the post you'll see the title "Physics Expert." That means I am not simply a nay sayer. I say your proposal is flawed because from a physics standpoint, your analysis is flawed. You can only get the answer you gave if the mass of the craft is infinite. How do I know this? I've solved the elastic collision problem, hundreds of times.

 

[math]v_{1f} = \frac{m_1-m_2}{m_1+m_2}v_{1i}[/math]

 

The only way for the speeds to have equal magnitudes is for m2 to be infinite

 

 

if you have found a reason why it would not work would you mind posting the reason , also I am looking for a viable reason so if you dont mind , please avoid posting violations to physics laws , as simply posting a physical law without the inclusion of your reason the physical law appears to be violated will never prove or disprove the viability of any concept.

 

ie ...

 

F=ma

0=1000kg*1000m/s/s

 

the above shows a physical law violation...

 

vs

 

that wont work because it violates newtons laws of motion.

 

the above does not show a physical law violation.

 

I personaly cannot find a violation of any law of physics in this concept.

 

Your concept violates conservation of energy and/or conservation of momentum. Which violation depends on what part of your proposal you look at. I've given an example above which shows the violation of conservation of energy. In my previous post I mentioned how you violated conservation of momentum.

[ninus maximus]

just what I thought , you are a simple nay sayer that will go to great extents

just to prove a point , normaly these nay sayers will choose to accuse a

concept as violating a physical law as you have , yet nay sayers offer

nothing viable as a reason.

 

they will type hundreds or even thousands of words so that the readers will

think that they are correct , but they only offer lip service as a reason.

 

when a single calculation can prove them right if they are right.

thats why they dont use calculations , they arent right , they are simply nay saying.

 

they dont even show where a physical law is suspected as being violated.

 

notice how he took plenty of time telling me that my math was wrong , that I used wrong units.

but did he exchange the units for the correct units and then work the calculation , no.

 

he cant prove it wont work using the math side of physics , and its not like it would

take a rocket scientist or a "expert nay sayer " to do that.

 

but maybe a expert physicist could use math.

 

they simply say they have been violated.

 

until you can offer something other than lip service as in a mathematical reason my concept stands.

 

and your lip service stands.

Edited June 26, 2010 by ninus maximus

[Bignose]

just what I thought , you are a simple nay sayer that will go to great extents

just to prove a point , normaly these nay sayers will choose to accuse a

concept as violating a physical law as you have , yet nay sayers offer

nothing viable as a reason.

 

give me a break.

 

He offered a "viable" reason your analysis is incorrect. It violates conservation of energy.

 

Do you realize, that in EVERY SINGLE situation EVER observed, conservation of energy has been followed? NEVER has there been a situation found where conversation of energy has been violated. Many times it has been thought to have been broken, but in every instance, the error was found and in the end, conservation of energy held every time. It may very well be THE most tested physical law ever, and it has NEVER been broken yet.

 

This of course does not mean that it is bullet proof and perfect. But, an awful lot of extraordinary evidence will have to be found to show a case when it has been broken. If you can provide this evidence, then let's see it.

 

Otherwise, calling an analysis that violates the conservation of energy incorrect is perfectly justified, because no situation has ever been found that yet violates it.

 

So, what else can you present to change minds instead of petty name calling and and whining about the "big bad naysayers"?

[ninus maximus]

point to where it violates the conservation of energy then big nose "Maths Expert".

 

show the readers how much of an expert you are.

 

just saying it violates something , isnt saying much.

 

and having "Maths Expert" underneath your name only means you have posted alot here in this forum.

 

something anyone can accomplish , by posting meaningless replies such as you do.

 

the momentum of the mass as it enters the first 180 degree turnaround will be

 

P=mv

 

100 kg * 70.711 m/s/0d = 7071.1 N/0d

 

were not talking about acceleration as the mass enters the first turnaround.

 

momentum is given as kgm/s

 

1 newton = 1 kgm/s/s

 

so I should have used 7071.1 kgm/s/0d

 

that unit discrepancy can in no way prove the overall concept is flawed.

 

however I immediately followed with

 

this force of momentum of 7071.1 N/0d will be presented to the first turnaround

as the mass passes through the first turnaround and changes its momentum from

 

P=7071.1 N/0d to P=7071.1 N/180d

 

you cannot present a momentum to a object.

momentum is not a force.

 

the force that an moving object presents to another object is quantified by its mass * its acceleration.

so when I used P=mv

I was quantifying the masses force due to its momentum.

 

Momentum measures the 'motion content' of an object, and is based on the product of an object's mass and velocity. Momentum doubles, for example, when velocity doubles. Similarly, if two objects are moving with the same velocity, one with twice the mass of the other also has twice the momentum.

 

Force, on the other hand, is the push or pull that is applied to an object to CHANGE its momentum. Newton's second law of motion defines force as the product of mass times ACCELERATION (vs. velocity). Since acceleration is the change in velocity divided by time, you can connect the two concepts with the following relationship:

 

force = mass x (velocity / time) = (mass x velocity) / time = momentum / time

 

Multiplying both sides of this equation by time:

 

force x time = momentum

 

 

if he was refering to impulse measure in N*s then he was wrong in assumming

that the mass would be experiencing an impulse or change in momentum as it enters the turnaround.

 

after it enters the turnaround it will experience a change in momentum however

and I clearly showed that change.

But the point is that the speed will reduce, and your numbers do not agree with that statement.

 

would you mind telling the readers how much the speed reduces?

 

obviously you have compared my numbers to your numbers , wouldnt you think

that the readers would like to know the speed reduction also.

 

and with your proof using your numbers then you can have a basis for your assumption that the concept would not work.

 

Im not sure why exactly you dont think it would work.

 

suppose you and big nose were both in separate boats that weigh the same.

and you both weigh the same , one of you two pushes the other ones boat

with a force of 100 N

 

one boat will go one way because of the 50 N force that is applied to it

and the other will go the other way because of the 50 N force applied to it.

 

if just 1 of you were in a boat and push off from the pier with a 100 N force

you would travel further because the pier is not elastic.

Edited June 26, 2010 by ninus maximus
Consecutive posts merged.

[Bignose]

point to where it violates the conservation of energy then big nose "Maths Expert".

 

show the readers how much of an expert you are.

 

just saying it violates something , isnt saying much.

 

and having "Maths Expert" underneath your name only means you have posted alot here in this forum.

 

 

Because in an isolated system (and a spacecraft away from anything else in deep space is isolated), the total amount of energy will remain constant. If the ship expends some energy to accelerate a slug of mass on a track, at the very, very best, that slug of mass can return only the exact same amount of energy to the ship (via rail, etc.) And, since you included friction, it will actually deliver significantly less than originally used to accelerate it.

 

And, if no mass leaves the ship, the entire system can do whatever it wants inside the ship, but it won't change the outside of the system.

 

I suggest you build a prototype of your device, if you are so sure it will work. Go out on a lake, and have a boat with a circular rail. So long as the rail and slug of mass running on it doesn't interact with the water, it will be just like being in space. I predict that you can spin that mass however you want, but you won't move that boat (be sure to neglect any wind or current effects!). But, go do that experiment yourself and prove us all wrong.

 

If you bring back objective indisputable proof, I guarantee that I will publicly admit I was wrong and you are right.

 

And, I actually am not a very prolific poster at all, compared to many others here. That title was given to me by the site administrated, just as swansont's was. Sure, I fully admit it doesn't "mean" anything -- nor did I or anyone ever claim that it did in the first place.

 

Physical law doesn't give a hoot if the person citing them is titled with "Maths Expert", or "Physics Expert", or "Kindergarten Expert" or "Banana Expert". Physical law just is. And, your claim on being able to propel a spacecraft just by spinning a slug of mass inside of it goes against physical law as we understand it today. So, again, go make your device and prove us wrong. Quit wasting your time here.

[swansont]

just what I thought , you are a simple nay sayer that will go to great extents

just to prove a point , normaly these nay sayers will choose to accuse a

concept as violating a physical law as you have , yet nay sayers offer

nothing viable as a reason.

 

they will type hundreds or even thousands of words so that the readers will

think that they are correct , but they only offer lip service as a reason.

 

when a single calculation can prove them right if they are right.

thats why they dont use calculations , they arent right , they are simply nay saying.

 

they dont even show where a physical law is suspected as being violated.

 

I explained where conservation of energy was violated.

 

I gave an equation for the projectile speed after an elastic collision. (In my world, equations count as math)

 

I invite you to go back and reread my previous posts.

[ninus maximus]

I gave an equation for the projectile speed after an elastic collision. (In my world, equations count as math)

 

the below equation is the only equation you have provided.

 

 

Not at all. Looking at limiting behavior in solutions is a common analysis to see if the answer is correct; one does this when one finds a general solution to a problem. If you look to the side of the post you'll see the title "Physics Expert." That means I am not simply a nay sayer. I say your proposal is flawed because from a physics standpoint, your analysis is flawed. You can only get the answer you gave if the mass of the craft is infinite. How do I know this? I've solved the elastic collision problem, hundreds of times.

 

 

 

 

 

 

The only way for the speeds to have equal magnitudes is for m2 to be infinite

 

is the above equation that uses two seperate masses where you gave a equation?

 

because it is the only equation you have gave!

 

V1f = M1-M2/M1+M2 * V1i

 

WE ARE NOT CHANGING THE MASS

WE ARE ONLY CHANGING THE VELOCITY OF THE MASS

 

you said my math violates the laws of physics , then you use an impossibility such

as a infinite mass to give a reason why the concept wouldnt work.

 

since mass does not change durring acceleration , it would be more factual to say

that your math violates the laws of physics.

 

as I said earlier a simple nay sayer will go to great extents

just to prove a point.

 

But the point is that the speed will reduce, and your numbers do not agree with that statement. Your numbers violate the conservation of energy, because the KE of the projectile is the same, and yet there must be a transfer of energy to the craft, since there is a transfer of momentum to the craft.

 

tie a string to a mass , then drop the mass from a height , does the mass stop as it gets closest to the ground.

or does the mass continue to swing until it reaches its apx dropped from height?

 

the speed is zero before you drop the mass.

the speed is zero after it swings to the other side and stops , before it swings again.

 

it did not loose a great amount of KE just as the mass in my concept does not loose a great amount of KE

passing through the turnarounds.

 

in the above example of a mass tied to a string the mass is propelled by gravity and it is met by the same exact force

as it swings upward again.

 

the mass in my concept is propelled by a force and is not met with a opposing force as the mass and string example.

 

no resistance = no loss in KE.

little resistance = little loss in KE.

 

 

perhaps there is someone in this forum who knows physics that would like to discuss this concept , as it stands no one that knows physics and how to apply physics has stepped forward.

Edited June 26, 2010 by ninus maximus

[insane_alien]

it may be the only equation he gave, but its the only equation he needed to give to prove his point.

 

seriously, you're making some basic mistakes about newtonian physics. swansont knows what he is talking about when it comes to physics.

[ninus maximus]

it may be the only equation he gave, but its the only equation he needed to give to prove his point.

 

seriously, you're making some basic mistakes about newtonian physics. swansont knows what he is talking about when it comes to physics.

 

he used the wrong equation and if he knows physics , and thats a strong if , he knows he used the wrong equation.

 

he has lost trust in his ability to perform.

he can gain trust only by posting the correct equation , and by posting correct results obtained by a correct equation

that pertains to the issue.

 

 

from what I have seen so far he doesnt know diddly , squat.

 

sorry but thats just the way he presents his knowlege of physics.

 

perhaps there is someone in this forum who knows physics that would like to discuss this concept , as it stands no one that knows physics and how to apply physics has stepped forward.

Edited June 26, 2010 by ninus maximus

[swansont]

You have two masses. The mass of the projectile and the mass of the craft, i.e. all the rest of the mass in the system. In the equation I gave, m2 would be the mass of the rest of the system.

 

Your value of the momentum after the projectile has changed direction was stated, not derived. This is your proposal, so the burden of proof is upon you to show how you got that number. My equation (derived from conservation of KE and momentum) shows that the only way to get v1f = -v1i is to let m2 go to infinity, which I seem to recall having stated already.

 

If you want to take the position that you conserve momentum, then your proposal violates conservation of energy. If you want to take the position that you conserve energy, then you violate conservation of momentum.

 

Using your numbers of 100 kg for the projectile and 5000 kg for the craft, and the initial velocity of 70.7 m/s in the + direction, we see that when the projectile has turned around it will have a velocity of -67.9 m/s. The speeds are unequal. Your assertion is wrong.

 

In addition to not working to propel a craft, you have the added problem that the craft will end up rotating in this scenario.

[ninus maximus]

You have two masses. The mass of the projectile and the mass of the craft, i.e. all the rest of the mass in the system. In the equation I gave, m2 would be the mass of the rest of the system.

 

wow , i was right you dont have a clue.

 

none of the mass in the entire system changes.

only the velocity of the masses changes.

 

the equation you used deals in changing masses , not changes in velocity.

 

ie............... the changes that would occur in velocity if the mass changes in magnatude durring acceleration.

 

My equation (derived from conservation of KE and momentum) shows that the only way to get v1f = -v1i is to let m2 go to infinity, which I seem to recall having stated already.

 

the only way to get V1f = -V1i

 

momentum ---- your talking about momentum ------ momentum ----- is not a force.

 

Your value of the momentum after the projectile has changed direction was stated, not derived. This is your proposal, so the burden of proof is upon you to show how you got that number.

 

if I drive my car south at 50 mph then turn the car around and drive it north at 50 mph

I get V1f=-V1i

 

if you will take the time to read the post and comprehend them before you reply to them with your expert opinion you might save a little face.

 

I stated that I had not included resistance yet , as the mass passes through the two turnarounds as follows

I have not included any resistance to movement as the mass passes through the

(2) 180 degree turnarounds , however using rolling friction such as .0001 which is

common in low friction bearings available on todays market should deliver a

useable calculation for speculation purposes.

Edited June 26, 2010 by ninus maximus

[insane_alien]

he used the wrong equation and if he knows physics , and thats a strong if , he knows he used the wrong equation.

 

if he did not know this physics (and much much more beyond it) then he wouldn't be employed. his job as a physicist does tend to be highly dependant on him having physics knowledge.

 

my job also requires a good grasp of physics (i'm a chemical engineer) especially when things are going round corners such as material in pipes. i can confirm he is correct and using the appropriate equation.

 

edit: from your most recent post you said that none of the masses change. this is true, swansont never claimed this at all. don't claim people said stuff they didn't, it only makes you look stupid.

[ninus maximus]

if he did not know this physics (and much much more beyond it) then he wouldn't be employed. his job as a physicist does tend to be highly dependant on him having physics knowledge.

 

my job also requires a good grasp of physics (i'm a chemical engineer) especially when things are going round corners such as material in pipes. i can confirm he is correct and using the appropriate equation.

 

edit: from your most recent post you said that none of the masses change. this is true, swansont never claimed this at all. don't claim people said stuff they didn't, it only makes you look stupid.

 

having a job as a physicist does not prove you have a working knowlege of physics , if you have a job as a physicisist and dont comprehend physics you can kill people.

 

 

having a job as a physicist and not knowing physics and how to apply physics only sudgest you used cheat sheets in college.

[insane_alien]

swansont has been employed for years and will have collborated with many physicists over the course of his career.

 

you can't fake it at his level.

 

besides, this is very very basic high school level stuff.

 

this also shows he is correct http://en.wikipedia.org/wiki/Collision

 

its a basic conservation of momentum equation. this is very much the most appropriate equation to use.

[ninus maximus]

besides, this is very very basic high school level stuff.

 

yes , it is.

thats why I question his ability to apply physics equations to find solutions to problems.

 

L2mdAvdPhT4

 

Collisions involve forces (there is a change in velocity).

Collisions can be elastic, meaning they conserve energy and momentum.

 

in my concept the mass passing through a turnaround is a elastic collision.

the mass conserves its energy and its momentum because it is not met by an opposing force.

 

other than a slight almost negligible resistance force as it presses against the turnaround

while passing through a turnaround.

 

swansont has been employed for years and will have collborated with many physicists over the course of his career.

 

you can't fake it at his level.

 

you cant?

Edited June 26, 2010 by ninus maximus

[insane_alien]

so you're saying momentum is not conserved? because thats what swansont is saying.

 

and conservation of momentum is one of the most tested things in physics because momentum being conserved is one of the major assumptions of nearly every single physics theory.

[ninus maximus]

so you're saying momentum is not conserved? because thats what swansont is saying.

 

and conservation of momentum is one of the most tested things in physics because momentum being conserved is one of the major assumptions of nearly every single physics theory.

 

absolutely not !!! Im showing conservation of momentum.

 

point to where Im not!

 

the item he questioned shows conservation of momentum.

 

he said , she said , why not just stay out of it.

and let swansont mind his buisness , and you can mind yours.

Edited June 26, 2010 by ninus maximus

[insane_alien]

the video shows conservation of energy.

 

energy and momentum are two very different things(although they are both conserved)

 

the difference being, in a collision like scenario(like the one you are presenting) energy can be lost to heat and other things while momentum remains strictly in the motion of the objects. how it is distributed is determined by the equation swansont posted.

 

to say it is the wrong equation to use is to deny the conservation of momentum.

[ninus maximus]

the video shows conservation of energy.

 

energy and momentum are two very different things(although they are both conserved)

 

the difference being, in a collision like scenario(like the one you are presenting) energy can be lost to heat and other things while momentum remains strictly in the motion of the objects. how it is distributed is determined by the equation swansont posted.

 

to say it is the wrong equation to use is to deny the conservation of momentum.

 

well then insane allen

 

why dont you put some numbers into the equation and give it a try , let me help you.

 

V1f=M1-M2/M1+M2 * V1i

 

V1i = 70.711 m/s

M1 = 100 kg

 

there , now solve for V1f

 

I want to see what you will use for M2

 

 

 

and when your through explain the following.

 

The law of conservation of energy is an empirical law of physics. It states that the total amount of energy in an isolated system remains constant over time (is said to be conserved over time).

 

I think it is said that our universe is an isolated system.

but there are none within our universe.

 

perhaps you can show me a isolated system.

Edited June 26, 2010 by ninus maximus

[Moontanman]

Ninus, think for a second, no math is required to see this is false, your hypothetical space craft would at best wobble back and forth and go no where.

 

No matter the sped of the slug drawing it back to it's starting position would negate the original acceleration. It's as obvious to me as falling off a building results in hitting the ground...