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Natural Science Forum / Physics / General Physics / July 2008The opposing rockets and the box
A box is placed in orbit, On two opposite sides of the box there are two equal rockets placed that can fire in opposite directions. Rocket (A) and rocket (B). If you fire rocket (A) for 10 seconds) and then shut it down. Will it now take only 10 seconds for rocket (B) to slow the box to an "at rest" state again? :) Feel free to use any mass for the box and any propulsion force from the rockets as long as the rockets have the same propulsive force. :) A hint: The box will have an inertial force after rocket (A) fires and shuts down. Don't forget the inertial force of the box. :) Pick an answer 1) Rocket (B) will need to fire longer to stop the box 2) Yes it will only take 10 seconds of Rocket (B) firing to stop the box. 3) Rocket (A) will not fire because there is a "dead and alive" cat inside it. :)  SignatureJames M Driscoll Jr Spaceman > A box is placed in orbit, > On two opposite sides of the box there [quoted text clipped - 6 lines] > Will it now take only 10 seconds for rocket > (B) to slow the box to an "at rest" state again? No, the box will be lighter than before since the first rocket threw away some mass. It should take less time if the rocket thrust is the same. > Feel free to use any mass for the box and > any propulsion force from the rockets as long as > the rockets have the same propulsive force. Ah, they do? Then it should take less time. > A hint: > The box will have an inertial force after rocket (A) > fires and shuts down. No, inertial force is a reaction force to attempted changes in the state of motion. As long as the box is only coasting the inertial force is zero. > Don't forget the inertial force of the box. The inertial force is the reaction force that manifests when the box is accelerated, that is, when the rockets are firing. It's what the rockets are pushing against, resisting the acceleration of the box. > Pick an answer > 1) Rocket (B) will need to fire longer to stop the box > 2) Yes it will only take 10 seconds of Rocket (B) firing to stop the > box. 3) Rocket (A) will not fire because there is a "dead and alive" > cat inside it. 4) None of the above. >> A box is placed in orbit, >> On two opposite sides of the box there [quoted text clipped - 10 lines] > first rocket threw away some mass. It should take > less time if the rocket thrust is the same. Ok, You found a flaw, I will admit. Lets change it to this problem instead. 2 cannons are placed 200 ft away from a mass 1 cannon is fired at the mass and the collision causes the mass to move towards the other cannon, Will the other cannon be able to stop the mass with 1 shot? >>> A box is placed in orbit, >>> On two opposite sides of the box there [quoted text clipped - 20 lines] > Will the other cannon be able to stop the mass > with 1 shot? You need to define what happens to the cannon balls. Do they bounce off or become embedded in the mass? Also, is the mass on a frictionless surface? Or perhaps the whole lot is simply floating in deep space? >>>> A box is placed in orbit, >>>> On two opposite sides of the box there [quoted text clipped - 23 lines] > You need to define what happens to the cannon balls. > Do they bounce off or become embedded in the mass? Try the "embedded in the mass" since that is what occurs in the propulsion posts I made. Then if you wish, also try the bouncing off. :) > Also, is the mass on a frictionless surface? Or > perhaps the whole lot is simply floating in deep > space? They are all floating in space and we can ignore the motion of the cannons after they fire since both will move the same away from the firing direction. SignatureJames M Driscoll Jr Spaceman >>>>> A box is placed in orbit, >>>>> On two opposite sides of the box there [quoted text clipped - 37 lines] > the motion of the cannons after they fire since both > will move the same away from the firing direction. First cannon fires ball with mass m at velocity v (for momentum m*v) at object with mass M. Mass M is initially at rest in our chosen frame of reference, so its momentum is initially zero. Ball strikes object and is embedded. Total momentum of the the ball and object must remain the same before and after collision: m*v = (M + m)*v2 v2 is the velocity of the object+ball v2 = v*m/(M + m) So v2 is slower than the original speed of the cannon ball by the ratio m/(M + m). But the total momentum of the pair is still m*v in our frame of reference. Second cannon fires its ball. It's momentum is also m*v but directed in the opposite direction. So call it -m*v. When it merges with the object and first ball the total momentum must sum to zero ( -m*v + m*v = 0). The object and balls have zero momentum in our frame of reference, so no velocity. They come to rest. >>>>>> A box is placed in orbit, >>>>>> On two opposite sides of the box there [quoted text clipped - 54 lines] > ball by the ratio m/(M + m). But the total momentum > of the pair is still m*v in our frame of reference. Hang on Greg, You just made v*m/(M + m) = m*v It seems you have a slight problem with your "totalling". :) >>>>>>> A box is placed in orbit, >>>>>>> On two opposite sides of the box there [quoted text clipped - 59 lines] > v*m/(M + m) = m*v > It seems you have a slight problem with your "totalling". v*m/(M + m) is the velocity of the first ball and object after their collision. Their momentum is then mass times velocity, the mass being (M + m) because they're stuck together, so: (M + m) * v*m/(M + m) = m*v That is, the momentum is conserved. The momentum of the object plus ball is the same as the object of the ball was before the collision. >>>>>>>> A box is placed in orbit, >>>>>>>> On two opposite sides of the box there [quoted text clipped - 62 lines] > v*m/(M + m) is the velocity of the first ball and object > after their collision. So it is v2 > Their momentum is then mass times velocity, the mass being > (M + m) because they're stuck together, so: > > (M + m) * v*m/(M + m) = m*v If v*m/(M+m) truly equals = v2 you have (M + m) * v2 = m*v :) Again I will say this. Somehow you are getting the momentum of the larger and smaller mass together to be equal to the momentum of the smaller mass without the larger mass at all. (m*v) That is just plain wrong. You are "removing" that larger mass completely and don't even see why such would be wrong. And of course, removing the larger mass would make both smaller masses equal. Sorry you are just wrong by removing the larger mass like you are doing. SignatureJames M Driscoll Jr Spaceman >>>>>>>>> A box is placed in orbit, >>>>>>>>> On two opposite sides of the box there [quoted text clipped - 73 lines] > you have > (M + m) * v2 = m*v Yes. Plug in the expression for v2 in terms of v and you see that the (M + m) bits cancel, as I showed. > Again I will say this. > Somehow you are getting the momentum of the larger and > smaller mass together to be equal to the momentum of the smaller mass > without > the larger mass at all. (m*v) > That is just plain wrong. No, momentum is a combination of both mass and velocity. If the mass is larger, a smaller velocity is required to have the same momentum. That's why after the collision of the first ball with the object the ensemble travels slower than the ball did alone. In fact, it travelled slower by the ratio m/(M + m). > You are "removing" that larger mass completely and don't even > see why such would be wrong. > And of course, removing the larger mass would make both > smaller masses equal. > Sorry you are just wrong by removing the larger mass like > you are doing. James, at least learn a little about conservation of momentum and calculating the results of collisions. It's high school Newtonian physics. >>>>>>>>>> A box is placed in orbit, >>>>>>>>>> On two opposite sides of the box there [quoted text clipped - 76 lines] > Yes. Plug in the expression for v2 in terms of v and > you see that the (M + m) bits cancel, as I showed. No, you multiplied an extra (M+m) to do such and because you did such you completely removed the larger mass and showed you are lost and could never figure out how the opposing forces could not be the same once a relative mass increase occurs to the M. You basically left the 2 masses moving (M+m) with the same momentum the single mass had to begin with. That is a freakin joke. Why even bother doing anything. Just ignore the large mass completely and you get the same result. You are ignorant to your own "wrong" methods. > No, momentum is a combination of both mass and velocity. > If the mass is larger, a smaller velocity is required to [quoted text clipped - 13 lines] > and calculating the results of collisions. It's high school > Newtonian physics. Greg, You just proved to me that you are clueless. You removed the M for no reason at all. You made it basically dissapear from the problem without even knowing where you went wrong. For some stupid a.s reason you made m*v hit a larger mass and still stay as m*v again even though it combined with a larger mass. Are you really that dense? You have m*v = v*m/(M + m) You are truly lost. If m = 30 and M = 100 You have 10* 30 = 30*10/(100 + 30) Why you can just multiply the (M +m) for no reason to the v2 is amazing that you would try to pull such sh.t off at all.. SignatureJames M Driscoll Jr Spaceman > You removed the M for no reason at all. > You made it basically dissapear from the problem [quoted text clipped - 6 lines] > You have m*v = v*m/(M + m) > You are truly lost. That is not what I wrote. I wrote v2 = v*m/(M + m). That's the velocity of the combined masses following the collision. > If m = 30 and M = 100 > You have 10* 30 = 30*10/(100 + 30) > Why you can just multiply the (M +m) for no reason to the v2 is > amazing that you would try to pull such sh.t off at all.. So James, you want a concrete example with numbers. Okay, let's follow your suggestion. I presume in the above you meant to say that: M = 100 kg m = 30 kg v1 = 10 m/s the initial velocity of mass m Mass m impacts mass M which is initially at rest. For convenience we will assume that mass m is initially travelling along the x-axis in the positive direction in our chosen frame of reference. Mass m sticks to mass M in the collision (inelastic collision). What, in your view, is the resulting velocity of the combined masses after collision? >> You removed the M for no reason at all. >> You made it basically dissapear from the problem [quoted text clipped - 10 lines] > That's the velocity of the combined masses following > the collision. And then you said that was equal to m*v all over again. in other words, why did you even bother doing such since you came out with the same thing the small mass alone was. Sheesh Again. >> If m = 30 and M = 100 >> You have 10* 30 = 30*10/(100 + 30) [quoted text clipped - 17 lines] > What, in your view, is the resulting velocity of the > combined masses after collision? I sure don't think it is the same as the small ball alone like you seem to think when you went back to have m*v all over again. As I said, you removed the large mass completely without any real reasoning for such. And since you did that of course you will be equal with the other small mass from the other direction. Just admit it Greg, you were wrong. There is no way the large mass and the small mass both in motion can be the same m*a that you had for the small mass alone. SignatureJames M Driscoll Jr Spaceman >>> You removed the M for no reason at all. >>> You made it basically dissapear from the problem [quoted text clipped - 12 lines] > > And then you said that was equal to m*v all over again. No, I said that the momentum was conserved, and that the momentum after the collision and the bodies are stuck together is given by v2*(M + m). I then plugged in the expression for the velocity v2 into that. > in other words, why did you even bother doing such > since you came out with the same thing the small mass > alone was. Sorry, I thought you might be interested in the details. I should have guessed that you're not interested in details, only denials. >>> If m = 30 and M = 100 >>> You have 10* 30 = 30*10/(100 + 30) [quoted text clipped - 21 lines] > like you seem to think when you went back to have > m*v all over again. I provided an expression for the velocity of the combined masses previously. One of the details that you ignored. You mean to say you couldn't even retain that bit of information for the time it took to read down to here? > As I said, > you removed the large mass completely without > any real reasoning for such. You simply skipped over the reasoning, which I gratuitously provided, and then preceded to attack the result without thinking. > And since you did that of course you will be equal > with the other small mass from the other direction. > Just admit it Greg, you were wrong. I was not wrong. You simply are unable to follow trivial physics math. Not my problem. > There is no way the large mass and the small mass > both in motion can be the same m*a that you had for the > small mass alone. What's a? I never mentioned acceleration. I claimed that momentum is conserved, and must be equal to the momentum of the initial launched ball, both before and after the first collision. > No, I said that the momentum was conserved, and that > the momentum after the collision and the bodies are > stuck together is given by v2*(M + m). Sorry, that should read "the velocity after the collision", not the momentum. My bad. >>>> You removed the M for no reason at all. >>>> You made it basically dissapear from the problem [quoted text clipped - 19 lines] > I then plugged in the expression for the velocity v2 > into that. No you did not. You came out with the momentum of the larger and smaller mass as being m*v all over again. You posted this....* *> So v2 is slower than the original speed of the cannon *> ball by the ratio m/(M + m). But the total momentum *> of the pair is still m*v in our frame of reference. That is you saying the larger mass and the smaller mass is equal to just the smaller mass alone. So Greg, Now you just proved you are a lyer and never wrong even when you are. This is you being the diversional a.shole you realy are and adding the lying factor also. So. Now we all know Greg will never be wrong and will lie as much as possible to never be wrong. This means I will now stop wasting my time with you since you are basically just a f.cking lying pile of dog sh.t. So Bye Greg, the lying a.s dog sh.t. SignatureJames M Driscoll Jr Spaceman >>>>> You removed the M for no reason at all. >>>>> You made it basically dissapear from the problem [quoted text clipped - 21 lines] > > No you did not. Sorry, but I did. Also, I corrected myself with regards the above paragraph in a separate post. I meant to say that the *velocity* after the collision was given by v2 = v*m/(M + m). The momentum is thus (M + m) * v2 = (M + m) * v*m/(M + m) = m*v. That is, momentum is conserved (the same before and after the collision). > You came out with the momentum of the larger > and smaller mass as being m*v all over again. Which it is. Momentum is conserved. Numerically, the momentum before and after is equal to m*v. > You posted this....* > *> So v2 is slower than the original speed of the cannon [quoted text clipped - 3 lines] > That is you saying the larger mass and the smaller mass > is equal to just the smaller mass alone. No it is far from saying that. Momentum involves two components: mass *and* velocity. If the mass is greater, the velocity is smaller to produce the same momentum magnitude. > So Greg, > Now you just proved you are a lyer and never wrong > even when you are. Nope. You're just confused over mass versus momentum. > This is you being the diversional a.shole you realy are > and adding the lying factor also. [quoted text clipped - 6 lines] > So > Bye Greg, the lying a.s dog sh.t. I love you too. Kisses. >>>>>> You removed the M for no reason at all. >>>>>> You made it basically dissapear from the problem [quoted text clipped - 30 lines] > momentum is conserved (the same before and after the > collision). See, you did it again and can't even tell. You started with the small mass moving (m*v) And then you played all around and came up yet again with (m*v) all over again Greg. You might as well ignore the larger mass completely. You would get the same answer of m*v. Sheesh SignatureJames M Driscoll Jr Spaceman >>>>>>> You removed the M for no reason at all. >>>>>>> You made it basically dissapear from the problem [quoted text clipped - 35 lines] > And then you played all around and came up yet again > with (m*v) all over again Greg. Yes, of course! The momentum is conserved so it's numerical value, after the collision, is equal to the value it had prior to the collision. That value is m*v. It's also (M + m)*v2. They are equal. The same. I figured why carry around the more complicated expression into the next stage of the analysis when m*v will do nicely. > You might as well ignore the larger mass completely. > You would get the same answer of m*v. If you were to think about it, James, you'd realize that when considering momentum the amount of mass is not critical; It's the product of mass and velocity that determiness the momentum. A baseball and a cannon ball can have the same momentum with appropriate velocities for each, and both will 'deliver' the same momentum to a target if they become embedded. >>>>>>>> You removed the M for no reason at all. >>>>>>>> You made it basically dissapear from the problem [quoted text clipped - 40 lines] > prior to the collision. That value is m*v. It's also > (M + m)*v2. They are equal. The same. Wow. see again you are stating (M + m)*v2 = m*v LOL Poor Greg. so stuck in conservation of momentum he can't see the math of what he is stating is wrong unless of course the larger mass of M = 0. Poor Greg "still ignoring the larger mass completely to come up with his "conservation of momentum" wrong answer. LOL SignatureJames M Driscoll Jr Spaceman >>>>>>>>> You removed the M for no reason at all. >>>>>>>>> You made it basically dissapear from the problem [quoted text clipped - 45 lines] > again you are stating > (M + m)*v2 = m*v Yes. You do realize, don't you, that v2 is not equal to v? >>>>>>>>>> You removed the M for no reason at all. >>>>>>>>>> You made it basically dissapear from the problem [quoted text clipped - 47 lines] > > Yes. You do realize, don't you, that v2 is not equal to v? You do realize that when you came up with the answer of the 2 boxes together being m*v , you removed that difference? You said it was equal and the you used the m*v without any considereation of the actual difference. You still basically ignored the larger mass completely to "conserve" your momentum. You seem to not want to see the difference because that difference will make the other mass hitting the new combined mass not actually stop the mass. Poor Greg. :) SignatureJames M Driscoll Jr Spaceman >>>>>>>>>>> You removed the M for no reason at all. >>>>>>>>>>> You made it basically dissapear from the problem [quoted text clipped - 58 lines] > that difference will make the other mass hitting the new > combined mass not actually stop the mass. So if I read you correctly, what you're saying is that you don't think that momentum is conserved and that Newtonian physics is wrong. Does that about sum it up? > So if I read you correctly, what you're saying is that > you don't think that momentum is conserved and that > Newtonian physics is wrong. Does that about sum it up? No, I am saying you are not finding the actual reason the momentum is conserved and yet motion occurs. You seem to wish to ignore the momentum of the larger mass that will have greater KE than it had when it was "at rest". :) SignatureJames M Driscoll Jr Spaceman >> So if I read you correctly, what you're saying is that >> you don't think that momentum is conserved and that [quoted text clipped - 4 lines] > the momentum is conserved and yet motion > occurs. Um, James, momentum is a "quantity" of motion. > You seem to wish to ignore the momentum of the > larger mass that will have greater KE than it had when > it was "at rest". KE is not conserved and so is not as useful a property as momentum for doing these sorts of calculations. You can always determine the KE using the formula KE = (1/2)*m*v^2 But what do you plan to do with the value? It's not like you can write a balanced equation for pre-collision versus post-collision kinetic energy; kinetic energy isn't a conserved quantity! > KE is not conserved and so is not as useful a property > as momentum for doing these sorts of calculations. You > can always determine the KE using the formula It is what proves the final motion even after the second ball tries to stop it, occurs Greg. I see you are affraid of it because of such a fact. :) SignatureJames M Driscoll Jr Spaceman >> KE is not conserved and so is not as useful a property >> as momentum for doing these sorts of calculations. You [quoted text clipped - 3 lines] > second ball tries to stop it, occurs Greg. > I see you are affraid of it because of such a fact. In what way does it prove it? You'll have to show your math. >>> KE is not conserved and so is not as useful a property >>> as momentum for doing these sorts of calculations. You [quoted text clipped - 6 lines] > In what way does it prove it? You'll have to show > your math. And yet again, you won't do such math so you will never prove yourself wrong. ROFLOL! SignatureJames M Driscoll Jr Spaceman >>>> KE is not conserved and so is not as useful a property >>>> as momentum for doing these sorts of calculations. You [quoted text clipped - 9 lines] > And yet again, you won't do such math so you will > never prove yourself wrong. Hey, you do the math and prove me wrong. Wouldn't that be more satisfying? > Hey, you do the math and prove me wrong. Wouldn't > that be more satisfying? I like explanations, not math. The KE of the large mass is increased from 0 to greater than 0, The smaller mass will now need more than the original KE that moved the larger mass to begin with simply because it is no longer a 0 KE. What can't you grasp about that simple fact Greg? I can see why you are affraid to do the "actual" math because of course you would find out the larger mass will still move and the same KE used to move it will not stop it now. Poor Greg, I am trying to let him do his own math to see where he is wrong but of course. he refuses. :) SignatureJames M Driscoll Jr Spaceman >> Hey, you do the math and prove me wrong. Wouldn't >> that be more satisfying? [quoted text clipped - 4 lines] > KE that moved the larger mass to begin with simply because it > is no longer a 0 KE. You haven't explained why that should be. "Because I Said So" is not an explanation. Point to or produce the equations that back up your contention. Or, if you prefer, state the equations in prose. Either way, you need to unambiguously state the relationship you're inviking. > What can't you grasp about that simple fact Greg? > I can see why you are affraid to do the "actual" math > because of course you would find out > the larger mass will still move and the same > KE used to move it will not stop it now. An arbitrary KE can be used to stop it. What matters is the momentum. As an extreme case, consider a very, very massive wall (say it's immovably bolted to the Earth) already at rest. It has zero kinetic energy in the at rest frame. Your mass runs into the wall and stops. There, it was stopped using an object with no kinetic energy at all. > Poor Greg, > I am trying to let him do his own math to see where > he is wrong but of course. he refuses. That's a laugh. You're 'letting' me do the math because you can't do it yourself. If you could you would be backing up your position with the mathematical ammunition to do so. >>> Hey, you do the math and prove me wrong. Wouldn't >>> that be more satisfying? [quoted text clipped - 8 lines] > Said So" is not an explanation. Point to or produce the > equations that back up your contention. Poor Greg. He is affraid to do the equations that prove him wrong and knows I will not just for fun, so he keeps parroting the same thing. :) You truly think the larger mass will still have a 0 KE even when moving huh Greg? LOL SignatureJames M Driscoll Jr Spaceman >>>> Hey, you do the math and prove me wrong. Wouldn't >>>> that be more satisfying? [quoted text clipped - 15 lines] > You truly think the larger mass will still have a 0 KE even > when moving huh Greg? You're not reading. Or you are reading and not absorbing. I gave you the expression for the kinetic energy. Can you remember what it was? >>>>> Hey, you do the math and prove me wrong. Wouldn't >>>>> that be more satisfying? [quoted text clipped - 19 lines] > absorbing. I gave you the expression for the > kinetic energy. Can you remember what it was? Greg, I have been trying to get you to show a freakin KE difference and you are a stubborn a.shole not to do such. If you really wanted to prove me wrong, you could have done such long ago. But of course. You only prove you are an a.shole that lets someone be wrong without actually using some proof they would agree upon. The truth is. I know the KE would not be different. and I have been trying to get you to show that simple fact instead of your twising a.s arrogant methods you use instead. but of course. you prove to me yet again that you are just an a.shole that proves things only your way and refuse to prove them any other way other than such. So. To stop this bullshit and to be the better man. I admit I have been pulling your leg the whole time about the "KE being different and I admit I have been trolling you the entire time but only trying to get you to just once show such stuff using classical physics (which you still seem to lack to do such) I will also point out to all you are a big a.s for not just showing some simple KE equations that would have proved such long ago. But the most funniest thing is. You can't post such simple facts to prove the devices I proposed in the other threads will not work. simply because they are not "simple" KE problems nor "simple momentum stuff either" And that is probably why you have not taken them on the way they were stated. :) so. Greg has proven he lost "classical physics" and can't use such to figure out the propulsion devices nor can he show they will work or will not work. HA HA HA HA HA HA HA HA :) SignatureJames M Driscoll Jr Spaceman >>>>>> Hey, you do the math and prove me wrong. Wouldn't >>>>>> that be more satisfying? [quoted text clipped - 22 lines] > Greg, I have been trying to get you to show a freakin KE > difference and you are a stubborn a.shole not to do such. Why do you think that anyone here is obliged to jump through your hoops? > If you really wanted to prove me wrong, you could > have done such long ago. > But of course. You only prove you are an a.shole > that lets someone be wrong without actually > using some proof they would agree upon. History shows that you simply deny evidence that doesn't suit your preconceived notions. So unless you cough up the numbers yourself so that your method can be critiqued, what good would it do? > The truth is. > I know the KE would not be different. > and I have been trying to get you to show that > simple fact instead of your twising a.s arrogant methods > you use instead. Of course you realize that the kinetic energy will actually be *less* after the collision, rather than "not be different", right? > but of course. you prove to me yet again that you > are just an a.shole that proves things only your way > and refuse to prove them any other way other than such. So why don't you just prove it your way? Let us see what you can do? > So. > To stop this bullshit and to be the better man. [quoted text clipped - 4 lines] > using classical physics (which you still seem to lack > to do such) So you think that conservation of momentum is not classical physics? Newton would be shocked. > I will also point out to all you are a big > a.s for not just showing some simple KE equations > that would have proved such long ago. What was stopping you from typing them? > But the most funniest thing is. "Most funniest"? > You can't post such simple facts to prove > the devices I proposed in the other threads [quoted text clipped - 3 lines] > And that is probably why you have > not taken them on the way they were stated. Your scenarios are horribly vague in their statement and allow for too many misinterpretations to occur. It seems that any attempt to pin down the specifics leads you into a rant about "twisting" and "diverting". What's up with that? Can't you formulate an airtight example? > :) > so. > Greg has proven he lost "classical physics" > and can't use such to figure out the propulsion > devices nor can he show they will work or will not work. > HA HA HA HA HA HA HA HA We await the patenting and demonstration of your self contained propulsion device with great anticipation. It should go well with your perpetual motion machine, infinite energy device, and Relativity-free clock. > Why do you think that anyone here is obliged to jump > through your hoops? I don't, I just place the hoops and the smartest stupid people jump through them with just a simple cracker that the parrots want. ROFLOL > History shows that you simply deny evidence that > doesn't suit your preconceived notions. So unless > you cough up the numbers yourself so that your > method can be critiqued, what good would it do? No Greg. I have always admitted when I was wrong. It is you that denies simple stuff like a clock malfunction. It is you and other moronic Relativists that still can not think out of the box at all and are stuck in your rubber sheet, rubber ruler, malfunctoning clocks, multiple standards of measurement universe that was all created from supposedly nothing and then a Big Bang occured. LOL > Of course you realize that the kinetic energy will > actually be *less* after the collision, rather > than "not be different", right? The kinetic energy will actually be less because of changes into heat but of course some would also say that "heat" is also a kinetic energy so the total will not lose any and will only be changed into another form that is still basically kinetic in nature. > So you think that conservation of momentum is not > classical physics? Newton would be shocked. No, He would most likely be glad I showed he is still 100% correct. :) To bad you don't get such about him anymore. You still think he is wrong in some areas. But when you need him, he seems to pop up just in "time". LOL > What was stopping you from typing them? As I said. I was seeing if you ever would. Finally I got sick of waiting and you proved you were never going to. So I gave up and ended the "experiment" that you were the guinea pig for. LOL > Your scenarios are horribly vague in their statement > and allow for too many misinterpretations to occur. > It seems that any attempt to pin down the specifics > leads you into a rant about "twisting" and "diverting". > What's up with that? Can't you formulate an airtight > example? Like The clock malfunctioned? You still hate that one so that is why you troll me huh? http://www.hyperdeath.co.uk/spaceman/message.html LOL Airtight and you hate it. also. Like the fact that you use multiple standards for time and distance in relativity. Air tight again but you hate that one also. LOL > We await the patenting and demonstration of your > self contained propulsion device with great > anticipation. It should go well with your > perpetual motion machine, infinite energy device, > and Relativity-free clock. No worries here Greg. It is you that is stuck in the box with the malfunctioning clock and rubber rulers. Not I. :) SignatureJames M Driscoll Jr Spaceman >> Why do you think that anyone here is obliged to jump >> through your hoops? [quoted text clipped - 11 lines] > No Greg. > I have always admitted when I was wrong. That wasn't the point. The point was having you provide your own mathematical (or simpy numerical) results. There's no wishy-washying around with hard numbers or equations. > It is you that denies simple stuff like a clock > malfunction. [quoted text clipped - 4 lines] > created from supposedly nothing and then a Big Bang > occured. There we go, back onto James' hobby horse. A fanatic is one who cannot change his mind, nor the subject. >> Of course you realize that the kinetic energy will >> actually be *less* after the collision, rather [quoted text clipped - 6 lines] > changed into another form that is still basically > kinetic in nature. Yes, yes, some energy is lost in real collisions due to the nature of real materials. But beyond this, for even theoretically perfect materials and perfect collisions where no energy is lost as heat or sound or otherwise, you do know that that kinetic energy is not conserved, right? >> So you think that conservation of momentum is not >> classical physics? Newton would be shocked. > > No, > He would most likely be glad I showed he is > still 100% correct. Huh? Then why are you so seemingly mystified by the law of conservation of momentum? You can't have it both ways. > To bad you don't get such about him anymore. > You still think he is wrong in some areas. [quoted text clipped - 6 lines] > As I said. > I was seeing if you ever would. Very lame excuse noted. > Finally I got sick of waiting and you proved > you were never going to. Oh dear, imagine that. > So I gave up and ended the "experiment" that you > were the guinea pig for. Riiiight. So you got up and walked out because the material was too difficult for you. >> Your scenarios are horribly vague in their statement >> and allow for too many misinterpretations to occur. [quoted text clipped - 8 lines] > LOL > Airtight and you hate it. And there he goes again. It's almost like clockwork... > also. > Like the fact that you use multiple standards for time > and distance in relativity. > Air tight again but you hate that one also. And again. Old faithfull. >> We await the patenting and demonstration of your >> self contained propulsion device with great [quoted text clipped - 6 lines] > and rubber rulers. > Not I. Nice finish. Too bad it doesn't add content. > That wasn't the point. The point was having you > provide your own mathematical (or simpy numerical) > results. There's no wishy-washying around with > hard numbers or equations. I have used hard numbers. I use 1 + 1 = 2 all the time and you came up with some math that supposedly proves such is wrong yet you used the same math base I did so your proof itself was wrong if my math was wrong at all. Poor Greg He still thinks that c+c does not equal 2c and he has the math that proves it also! LOL > Yes, yes, some energy is lost in real collisions due > to the nature of real materials. But beyond this, > for even theoretically perfect materials and perfect > collisions where no energy is lost as heat or sound > or otherwise, you do know that that kinetic energy is > not conserved, right? I asked you to show me why. I still have not seen "real" proof it is not conserved. You said you showed scuh already but I do not see where. I asked you to show me again and I said pretty please and everything so I am still waiting. :) > Huh? Then why are you so seemingly mystified by > the law of conservation of momentum? You can't have > it both ways. What do you mean You say I can't have Newton be correct and the conservation of momentum correct also? That is kinda silly. > And there he goes again. It's almost like clockwork... LOL clockwork, Like you know anything about that! LOL SignatureJames M Driscoll Jr Spaceman >> That wasn't the point. The point was having you >> provide your own mathematical (or simpy numerical) [quoted text clipped - 24 lines] > I asked you to show me again and I said pretty please > and everything so I am still waiting. Continue to wait. You've lost your begging rights with your recent outbursts. You can earn back your credibility by making an effort. You'll forgive me I hope if I simply point out where you're wrong. >> Huh? Then why are you so seemingly mystified by >> the law of conservation of momentum? You can't have [quoted text clipped - 3 lines] > You say I can't have Newton be correct > and the conservation of momentum correct also? More reading comprehension deficit noted. Considering that it was Newton that showed the conservation law, your statement is ludicrous and illiterate. > That is kinda silly. Yes, indeed. You are. > >>>> Hey, you do the math and prove me wrong. Wouldn't > >>>> that be more satisfying? [quoted text clipped - 15 lines] > > You truly think the larger mass will still have a 0 KE even > > when moving huh Greg? > You're not reading. Or you are reading and not > absorbing. I gave you the expression for the > kinetic energy. Can you remember what it was? sh.t, Spaceman doesn't remember where his a.s is and can't find it without a set of instructions written at the third grade level with big print and a flashlight that someone else turned on for him. Why do you bother as he isn't a particularly amusing drooling crackpot, just brain damaged and easily replaced by an emulator? http://www.hyperdeath.co.uk/spaceman/ in case you didn't know. SignatureJim Pennino Remove .spam.sux to reply. > sh.t, Spaceman doesn't remember where his a.s is and can't find it > without a set of instructions written at the third grade level with [quoted text clipped - 4 lines] > > http://www.hyperdeath.co.uk/spaceman/ in case you didn't know. In case others did not know This entire post was made to show the a.ses that visit this group. Jim Pennino has just proven he is part of the a.s patrol. :) Hey Jim, Read this. http://www.hyperdeath.co.uk/spaceman/message.html Show me where it is wrong please. :) SignatureJames M Driscoll Jr Spaceman >> You're not reading. Or you are reading and not >> absorbing. I gave you the expression for the [quoted text clipped - 8 lines] > > http://www.hyperdeath.co.uk/spaceman/ in case you didn't know. Oh, I knew all right. I just enjoy provoking him into a froth while correcting his atrocious physics. Witness the most recent tirades. > <j...@specsol.spam.sux.com> wrote in message > [quoted text clipped - 15 lines] > Oh, I knew all right. I just enjoy provoking him into > a froth You see, Spaceman? Look! There are three people talking about you right now, discussing how easy and enjoyable it is to provoke you into a froth. And yet you walk right into it, thinking you can dish it as well as it's given to you, and not noticing that the only one here with a bloody nose is you, and that's because you just swung and hit yourself in the face. Ever consider not continuing to do that? > while correcting his atrocious physics. Witness > the most recent tirades. > You see, Spaceman? Look! There are three people talking about you > right now, discussing how easy and enjoyable it is to provoke you into > a froth. And yet you walk right into it, thinking you can dish it as > well as it's given to you, and not noticing that the only one here > with a bloody nose is you, and that's because you just swung and hit > yourself in the face. Ever consider not continuing to do that? And yet more proof of the "conservation of a.sholes" LOL I have no "bloody nose" PD. I did not hit myself in the face but you have hit yourself pretty hard and just did such again proving you are nothing more than an arrogant a.shole for the umteenth time. :) SignatureJames M Driscoll Jr Spaceman > > >> You're not reading. Or you are reading and not [quoted text clipped - 9 lines] > > > > http://www.hyperdeath.co.uk/spaceman/ in case you didn't know. > Oh, I knew all right. I just enjoy provoking him into > a froth while correcting his atrocious physics. Witness > the most recent tirades. I understand. I kill filled the drooling, babbling idiot long ago as totally uninteresting and boring and don't see anything he posts unless it is quoted in a reply. Habshi is my drooling whipping bitch; everyone needs one. Have fun. SignatureJim Pennino Remove .spam.sux to reply. > I understand. > > I kill filled the drooling, babbling idiot long ago as totally > uninteresting and boring and don't see anything he posts unless it is > quoted in a reply. Poor Jim killfiled me when he could not prove the clock did not malfunction! LOL SignatureJames M Driscoll Jr Spaceman >>>> Also, I corrected myself with regards the above paragraph in >>>> a separate post. I meant to say that the *velocity* after the [quoted text clipped - 12 lines] >> prior to the collision. That value is m*v. It's also >> (M + m)*v2. They are equal. The same. >Wow. >see [quoted text clipped - 8 lines] >to come up with his "conservation of momentum" wrong answer. >LOL Wow is right. Spaceshit, it's rather strange to see you rail on why relativity is wrong when you don't even seem to know high school physics. Think about it. When that cannonball gets fired into that lump of clay stationary in space, does the cannonball/clay lump move: 1) not at all 2) slower than the cannonball before impact 3) the same speed as the cannonball 4) faster than the cannonball More important is Why? That's what Greg was explaining mathematically. Maybe if I use a different description that even a tire salesman could understand. A 9 ton truck is stopped in neutral on a level road with its parking brake off. Assume the road, tires, bearings etc. are frictionless. A 1 ton compact car, also with frictionless tires etc. rolls into the truck at 10 mph and sticks to it on impact. Does the truck/car combo move? At what speed? Why? > Wow is right. Spaceshit, it's rather strange to see you rail on why > relativity is wrong when you don't even seem to know high school [quoted text clipped - 5 lines] > 3) the same speed as the cannonball > 4) faster than the cannonball Of course it would be 2 Mike. But the problem is Greg is ignoring the actual difference when he gets back to m*v all over again. The small ball was m*v alone and now he has the larger and small ball being m*v again. He might as well have not done any math to come up with m*v all over again. Sheesh. > More important is Why? That's what Greg was explaining > mathematically. > > Maybe if I use a different description that even a tire salesman could > understand. f.ck Off. It seems you also have no clue what I actually am. You have fallen for the lemming description of my job. LOL > A 9 ton truck is stopped in neutral on a level road with its parking > brake off. Assume the road, tires, bearings etc. are frictionless. > A 1 ton compact car, also with frictionless tires etc. rolls into the > truck at 10 mph and sticks to it on impact. Does the truck/car combo > move? At what speed? Why? Maybe you should think about such. Do you think once the truck is moving, it will take the same opposite force impact to stop the truck again? IF you do. you have created a wonderful problem with ignoring the new KE of the truck. LOL SignatureJames M Driscoll Jr Spaceman >> A 9 ton truck is stopped in neutral on a level road with its parking >> brake off. Assume the road, tires, bearings etc. are frictionless. [quoted text clipped - 7 lines] > IF you do. you have created a wonderful problem > with ignoring the new KE of the truck. I don't understand why you're advocating the breaking of the law of energy conservation. >>> A 9 ton truck is stopped in neutral on a level road with its parking >>> brake off. Assume the road, tires, bearings etc. are frictionless. [quoted text clipped - 10 lines] > I don't understand why you're advocating the breaking > of the law of energy conservation. I am not Greg. How much kinetic energy will it take to stop the now moving truck? How much energy did it take to move the truck to begin with? >>>> A 9 ton truck is stopped in neutral on a level road with its >>>> parking brake off. Assume the road, tires, bearings etc. are [quoted text clipped - 15 lines] > truck? > How much energy did it take to move the truck to begin with? (1/2)*m*v^2 is the kinetic energy of the truck. The *work* required to halt it will be the same. This work can be done by applying a small force over an extended time, or by smashing another compact car into it... > (1/2)*m*v^2 is the kinetic energy of the truck. The > *work* required to halt it will be the same. The same as the truck in motion. and that is not the same as what made the truck move. Poor Greg. He will start to twist and use a different experiment soon so he will remove the given experiment so he can be correct while also being wrong about the previous experiment. LOL >This work > can be done by applying a small force over an extended > time, or by smashing another compact car into it... Smashing another compact car that has the same KE that move the truck will not stop it. Apparently you love to ignore the new kinetic energy the truck did not have when it was "at rest". LOL SignatureJames M Driscoll Jr Spaceman >> (1/2)*m*v^2 is the kinetic energy of the truck. The >> *work* required to halt it will be the same. > > The same as the truck in motion. > and that is not the same as what made the truck move. Sure it is. Can you show otherwise? > Poor Greg. > He will start to twist and use a different experiment soon [quoted text clipped - 8 lines] > Smashing another compact car that has the same > KE that move the truck will not stop it. It will if it happens to have the same mass as the truck (as unlikely as that may seem)! Then their kinetic energies and momenta will both be equal. > Apparently you love to ignore the new kinetic energy > the truck did not have when it was "at rest". As I have been trying to point out (rather ineffectively apparently) it's the momentum and not the kinetic energy that is the salient factor for determining whether or not the larger mass can be brought to a halt by the smaller. If the momentum of a system sums to zero then all parts can, through suitable interaction, be brought to rest. >>> (1/2)*m*v^2 is the kinetic energy of the truck. The >>> *work* required to halt it will be the same. [quoted text clipped - 3 lines] > > Sure it is. Can you show otherwise? The original KE of the box is 0 the new KE of the box is greater than 0 You are just to ignorant to find such out. :) > It will if it happens to have the same mass as the > truck (as unlikely as that may seem)! Then their > kinetic energies and momenta will both be equal. And now you need to change the experiment yet again just so you dont admit you are wrong. LOL > As I have been trying to point out (rather ineffectively > apparently) it's the momentum and not the kinetic [quoted text clipped - 3 lines] > sums to zero then all parts can, through suitable > interaction, be brought to rest. As I have shown you, You must completely ignore the facts about the different KE of the box once moving so you can divert yourself from being wrong about the motion actually occuring. LOL SignatureJames M Driscoll Jr Spaceman >>>> (1/2)*m*v^2 is the kinetic energy of the truck. The >>>> *work* required to halt it will be the same. [quoted text clipped - 7 lines] > the new KE of the box is greater than 0 > You are just to ignorant to find such out. So what? You have yet to convince anyone but yourself that the kinetic energy is relevant to the issue. >> It will if it happens to have the same mass as the >> truck (as unlikely as that may seem)! Then their >> kinetic energies and momenta will both be equal. > > And now you need to change the experiment yet again > just so you dont admit you are wrong. I'm just pointing out the one situation in which what you are saying about kinetic energy is true; If the masses involved are both equal, then the kinetic energies will balance when the momenta do. >> As I have been trying to point out (rather ineffectively >> apparently) it's the momentum and not the kinetic [quoted text clipped - 9 lines] > you can divert yourself from being wrong about > the motion actually occuring. I've also ignored their paint color and the given name of their owner and his pet poodle, and for good reason: it is irrelevant! Can you provide any sort of calculation to back up your claim that it is otherwise? > So what? You have yet to convince anyone but yourself > that the kinetic energy is relevant to the issue. This one statement alone proves you are beyond help. You refuse to use the kinetic energy thoughts about such because such thoughts prove the motion will indeed occur. You have proven now that you are simply an ignorant rubber ruler worshipper that never looks "outside" the box. LOL SignatureJames M Driscoll Jr Spaceman >> So what? You have yet to convince anyone but yourself >> that the kinetic energy is relevant to the issue. > > This one statement alone proves you are beyond help. > You refuse to use the kinetic energy thoughts about such > because such thoughts prove the motion will indeed occur. No, it won't. The kinetic energy of the projectile that stops the mass is arbitrary! You cannot draw any relevant conclusions from a comparison of the kinetic energies. > You have proven now that you are simply an ignorant > rubber ruler worshipper that never looks "outside" the box. James, you don't even know what's in the box, let alone outside of it. >>> So what? You have yet to convince anyone but yourself >>> that the kinetic energy is relevant to the issue. [quoted text clipped - 5 lines] > No, it won't. The kinetic energy of the projectile > that stops the mass is arbitrary! Bullshit. The KE difference is what causes the motion to actually move even after the smaller mass tries to stop it. SignatureJames M Driscoll Jr Spaceman >>>> So what? You have yet to convince anyone but yourself >>>> that the kinetic energy is relevant to the issue. [quoted text clipped - 10 lines] > actually move even after the smaller mass tries to > stop it. Sorry, no. Conservation of momentum dictates the outcome. The kinetic energy left over after the second impact is zero. > Sorry, no. Conservation of momentum dictates the > outcome. The kinetic energy left over after the > second impact is zero. See this is the sh.t that proves you are nothing but an a.shole and help pretty much nobody here. I pratically begged you to show KE stuff that proved I was wrong. (and I waited because I actually knew it would) But of course. you only pulled the momentum blah blah blah bullshit. and the "You do the math" arrogant fuckface sh.t you always pull. You have proven you are not even worth listening to now Greg. Because you have nothing to contribute except to prove you are so perfect there is nothing you can do wrong ever because you are the best ROM memory ever created! LOL Bye Greg. SignatureJames M Driscoll Jr Spaceman >> Sorry, no. Conservation of momentum dictates the >> outcome. The kinetic energy left over after the [quoted text clipped - 4 lines] > I pratically begged you to show KE stuff > that proved I was wrong. You didn't ask nicely. No one here is obliged to provide you an education. You're begging on a steet corner and *demanding* twenties. > (and I waited because I actually knew it would) > But of course. you only pulled the momentum blah blah blah > bullshit. and the "You do the math" > arrogant fuckface sh.t you always pull. Hey, you're the one that seems to think that everyone else should do your heavy lifting. So now you get all bent out of shape when you're asked to do it yourself? Ha! > You have proven you are not even worth listening to > now Greg. > Because you have nothing to contribute except > to prove you are so perfect there is nothing you > can do wrong ever because you are the best ROM memory > ever created! Sorry to disappoint James, but in this thread I'm not providing any insights into physics that haven't been known for several hundred years. It's all highschool level stuff that was (when I was in school) thoroughly backed up with hands-on labs to confirm it all. If you don't want to take advantage of other's *free* advice and experience here, then why are you generating these threads? And if it's just to troll for annoyance, why do you get so emotionally involved? > Bye Greg. Promises, promises. > You didn't ask nicely. Actually i did, I even said "please do" somewhere in all this mess. But of course you only wished to prove how much an a.s you were but did not even know that was being proven. Bye a.s! :) >> Wow is right. Spaceshit, it's rather strange to see you rail on why >> relativity is wrong when you don't even seem to know high school [quoted text clipped - 5 lines] >> 3) the same speed as the cannonball >> 4) faster than the cannonball >Of course it would be 2 Mike. So tell us more (if you can). If you know the mass of the cannonball, and its velocity, and you know the mass of the clay lump and know its velocity is initially 0, what is the velocity of the combined mass after impact? >But the problem is Greg is ignoring the actual >difference when he gets back to m*v all over again. >The small ball was m*v alone and now he has the >larger and small ball being m*v again. >He might as well have not done any math to >come up with m*v all over again. Can't you follow his math? This is high school level. >> Maybe if I use a different description that even a tire salesman could >> understand. >f.ck Off. >It seems you also have no clue what I actually am. >You have fallen for the lemming description of my job. OK, show that you're smarter than a tire salesman, and answer the question. >> A 9 ton truck is stopped in neutral on a level road with its parking >> brake off. Assume the road, tires, bearings etc. are frictionless. >> A 1 ton compact car, also with frictionless tires etc. rolls into the >> truck at 10 mph and sticks to it on impact. Does the truck/car combo >> move? At what speed? Why? How fast is the stuck-together car/truck moving? Why is it moving at that speed? >Maybe you should think about such. >Do you think once the truck is moving, it will take >the same opposite force impact to stop the truck again? OK, you tell us. After the first impact, a second 1 ton subcompact travelling the opposite direction at 10 mph crashes into the stuck- together car-truck, and sticks. How fast is the resulting car-truck-car mass moving, and why is it moving at that speed? >IF you do. you have created a wonderful problem >with ignoring the new KE of the truck. Do you even know the difference between energy and momentum? Why should *anyone* believe your complaints about relativity if you can't even solve a simple high school physics problem? >>> Wow is right. Spaceshit, it's rather strange to see you rail on why >>> relativity is wrong when you don't even seem to know high school [quoted text clipped - 13 lines] > and know its velocity is initially 0, what is the velocity of the > combined mass after impact? It would of course be greater than 0 along with it's new kinetic energy being greater than 0. so.. with a greater than 0 KE there is no way you will get the original KE that started the motion to stop the greater KE now occuring to the larger mass. >> But the problem is Greg is ignoring the actual >> difference when he gets back to m*v all over again. [quoted text clipped - 4 lines] > > Can't you follow his math? This is high school level. His math is ignoring the greater KE that larger mass would have. > How fast is the stuck-together car/truck moving? Why is it moving at > that speed? Again, It is moving faster than it was originally (0 speed) and because such it now also has a non zero KE and it will take more KE to stop it than it took to move it. Sheesh Mike, this is all classical pre-highschool stuff. :) > OK, you tell us. After the first impact, a second 1 ton subcompact > travelling the opposite direction at 10 mph crashes into the stuck- > together car-truck, and sticks. How fast is the resulting > car-truck-car mass moving, and why is it moving at that speed? It won't be stopped. You don't have enough KE coming from the new impact to do such. SignatureJames M Driscoll Jr Spaceman >>>> Wow is right. Spaceshit, it's rather strange to see you rail on >>>> why relativity is wrong when you don't even seem to know high [quoted text clipped - 19 lines] > way you will get the original KE that started the motion > to stop the greater KE now occuring to the larger mass. Amazing. James has discovered an infinite energy source and perpetual motion! How could it be that 2000 years of tinkering with colliding masses has not produced one other person who has noticed this! >>> But the problem is Greg is ignoring the actual >>> difference when he gets back to m*v all over again. [quoted text clipped - 7 lines] > His math is ignoring the greater KE that larger mass > would have. But it wouldn't have a larger KE. In fact its KE would be much lower than that of the original projectile. KE involves the mass multiplied by the square of the velocity. The square function increases much more rapidly than a simple multiplication. The velocity of the combined blob and canon ball is much less than the speed that the cannon ball had before impact. >> How fast is the stuck-together car/truck moving? Why is it moving at >> that speed? [quoted text clipped - 3 lines] > and because such it now also has a non zero KE and > it will take more KE to stop it than it took to move it. Why? What do you think is so special about kinetic energy that it seems to multiply the energy put into something? > Sheesh Mike, > this is all classical pre-highschool stuff. Actually, your version is classical baloney. >> OK, you tell us. After the first impact, a second 1 ton subcompact >> travelling the opposite direction at 10 mph crashes into the stuck- [quoted text clipped - 4 lines] > You don't have enough KE coming from the new impact to > do such. That's not an answer. He asked how fast. Let's see you do the numbers. > Amazing. James has discovered an infinite energy source > and perpetual motion! How could it be that 2000 years of > tinkering with colliding masses has not produced one other > person who has noticed this! See, Still no simple proof and just arrogant a.shole bullshit twisting instead of a simple answer for the KE I asked him to prove. > But it wouldn't have a larger KE. In fact its KE would > be much lower than that of the original projectile. So why didn't you show such math Greg? It would have been very simple to do. But instead you sat on your highhorse blabbing about momentum blah blah blah. Sad Greg. I figured you would prove yourself an a.shole and you did it very well. You let me go on and on to make yourself feel even more smarter than ever. To bad you were actually being tested and failed. You are an a.s. I was trying to see if you were not the big a.s I thought you were, but you proved you are. Thanks for the proof a.s! BTW a.shole. I did the numbers before I posted this. And I knew the whole time the KE would not be larger for real because there is no way a collision will "gain" KE. but of course. you showed how much an a.s you are. Thanks again.. a.s! LOL SignatureJames M Driscoll Jr Spaceman >> Amazing. James has discovered an infinite energy source >> and perpetual motion! How could it be that 2000 years of [quoted text clipped - 24 lines] > I thought you were, but you proved you are. > Thanks for the proof a.s! So, you're now claiming to be conducting clandestine personality testing here in the science newsgroups? Well, that should win you a lot of friends. Good luck with that. > BTW a.shole. > I did the numbers before I posted this. Sure. Sure you did. Again, you claim without demonstration. > And I knew the whole time the KE would > not be larger for real because there is no > way a collision will "gain" KE. > but of course. you showed how much an a.s you are. > Thanks again.. a.s! So you admit that you've just been trolling all along. That speaks volumes. > So, you're now claiming to be conducting clandestine > personality testing here in the science newsgroups? > Well, that should win you a lot of friends. Good > luck with that. Like I want idiots that think the shortest distance between two points is a curve as my friends. ROFLOL Like I want idiots that think clocks do more than simply count in a periodic rate as friends.. ROFLOL I would never want you as a "friend" Greg. And you just proved it 100%. a.sholes for friends are not my idea of "real" friends at all. > So you admit that you've just been trolling all along. > That speaks volumes. I only troll a.sholes like you and PD mostly Greg since you basically troll me most of the time also. so you are still fair game for the fishing as long as you swim into my posts.. The more funny part is you are a fish that can not understand he uses multiple standards of time and distance to support his church of fish parrots. LOL http://www.hyperdeath.co.uk/spaceman/message.html SignatureJames M Driscoll Jr Spaceman On Jul 22, 11:52 pm, "Spaceman" <space...@yourclockmalfunctioned.duh> wrote: > I only troll a.sholes like you and PD mostly Greg > since you basically troll me most of the time also. > so you are still fair game for the fishing as long as > you swim into my posts.. Has it never occurred to you not to enter a battle of wits unarmed? Has it never occurred to you that in these confrontations, you don't end up looking so good, and that this is not even because of anything that I say in particular, but rather because you make YOURSELF look bad? Has it never occurred to you that people will provoke a belligerent but harmless animal because the animal's response is *funny*? (Gotta love "fish parrots".) Has it never occurred to you that if you change your approach, you might get a different response? PD > The more funny part is you are a fish that can not > understand he uses multiple standards of time and [quoted text clipped - 4 lines] > James M Driscoll Jr > Spaceman > Has it never occurred to you not to enter a battle of wits unarmed? ROFLOL! This from an idiot that thinks the "shortest physical distance" between two points is not a straight line! LOL SignatureJames M Driscoll Jr Spaceman On Jul 22, 11:56 pm, "Spaceman" <space...@yourclockmalfunctioned.duh> wrote: > > Amazing. James has discovered an infinite energy source > > and perpetual motion! How could it be that 2000 years of [quoted text clipped - 33 lines] > Thanks again.. a.s! > LOL I actually prepared this for your ball-in-a-box example. Let's see if we can come up with the result predicted by conservation of momentum without actually using it as a law. First, some definitions, per high-school physics (no relativity here, only Sir Isaac Newton and friends). Fundamentals: Time is a separation between two events; its units are seconds (s). Position is a vector describing where something is in relation to a reference point; its units are metres (m). Mass is a quantity of matter; its units are kg (for convenience). Then: Velocity is a change in position over time; its un |
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