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Re: Calculating Newtons in Joules and Joules/s Re: Calculating Newtons in Joules and Joules/s Source: http://sci.tech−archive.net/Archive/sci.physics/2006−07/msg00705.html • From: "Dennis B" <Utopian@xxxxxxxx> • Date: 6 Jul 2006 22:50:12 −0700 Dennis B wrote: Randy Poe wrote: Dennis B wrote: PD wrote: Dennis B wrote: Randy Poe wrote: Dennis B wrote: You are blind if you cannot see that force is energy. Saying it over and over won't make it so. Force can be Re: Calculating Newtons in Joules and Joules/s 1 Re: Calculating Newtons in Joules and Joules/s measured in J/m. Yes, and energy can't. The distinctions between momentum, energy, work, and force are that they are different measurements of one thing: Energy. Incorrect. For everything is energy. You cannot argue with that. Yes I can. − Randy I don't have the time or energy at the present Re: Calculating Newtons in Joules and Joules/s 2 Re: Calculating Newtons in Joules and Joules/s moment to respond to all of the messages in detail. Not that I don't intend to. In the interim, perhaps following evidence will prove to you that momentum, force, work, and energy are all unified: Momentum = mv Force = (delta)mv (Kinetic) Energy = 1/2mv^2 work = 1/2mv^2 In other words, momenum, force, kinetic energy, and work are all measurements of mv... Perhaps now you will begin to understand my perspective? A couple of comments: These formulas you have here are NOT universal definitions, despite what you may have read. They work sometimes, under some circumstances, for some kinds of things. Just because m and v appear in all of them does NOT mean that they are the same thing. For example, if momentum and energy were the same thing, then if the momentum for an object A happened to be the same as the momentum for object B, then you would expect the energies to be the same, too. But they're not. Let's take an example: Object A: m = 2 kg, v = 4 m/s Object B: m = 4 kg, v = 2 m/s Here momentum of A = (2 kg)(4 m/s) = 8 kg*m/s And momentum of B = (4 kg)(2 m/s) = 8 kg*m/s. They have the same momentum. So if momentum and energy are the same thing, then if these Re: Calculating Newtons in Joules and Joules/s 3 Re: Calculating Newtons in Joules and Joules/s two momenta for A and B are equal, then so should be the energies. Here energy of A = 1/2(2 kg)(4 m/s)^2 = 16 J But energy of B = 1/2(4 kg)(2 m/s)^2 = 8 J. Not the same. So the momenta of A and B are equal, but the energies of A and B are NOT equal. −Dennis I have a question for you: If the 2kg and 4kg masses have the same momentum, what would happen if they each collided with a 3 kg mass (each having identical properties) and transferred their momentum? In collisions, typically one mass doesn't "transfer it's momentum" to another. Most of the time both masses end up with some momentum and KE after the collision. The amount of momentum and KE transferred will be different for the 2 kg and 4 kg masses if you assume perfectly elastic or perfectly inelastic collisions. Since the 2kg mass and the 4kg mass have the same momentum, wouldn't both of the 3 kg masses have the same momentum as well? Momentum and energy are separately conserved. Due to those requirements, your hypothetical momentum transfers can't both happen. If the 3 kg mass were to get all of the momentum of the 2 kg mass, it would be moving at 2/3 of the velocity and would therefore have (3/2)*(4/9) = 2/3 of the KE that the 2 kg mass originally had. That is possible in the right inelastic collision, with the remaining 1/3 being lost to heat. What does the (4/9) represent? Re: Calculating Newtons in Joules and Joules/s 4 Re: Calculating Newtons in Joules and Joules/s If the 3 kg mass were to get all of the momentum of the 4 kg mass, it would be moving at 4/3 of the velocity and would have (3/4)*(16/9) = 4/3 of the KE that the 4 kg mass originally had. That is impossible. What does the (16/9) represent? It is not possible for a 4 kg mass to transfer all of its momentum to a 3 kg mass in a collision. − Randy Make it two 4 kilogram masses instead of two 3 kg masses. And make the collision totally inelastic, so that all of the momentum is transferred. Then what would the results of a transfer of momentum be? I don't see why the velocities would be any different. Although the 2kg and 4kg masses differ, the 2kg mass travels at a greater velocity to make up for the difference so that the momentum of the two is the same. Therefore, I would expect that the 2kg mass should have just as much of an impact as the 4 kg mass. −Dennis Perhaps I should ask what the results of a perfectly elastic collision would be as opposed to an inelastic collision. The reason being that while studying momentum I was astonished to find that in an inelastic collision, the colliding masses supposedly stick together, presumably such that both then travel together. Yet, I ask...is it not possible for the mass at rest to absorb all of the energy of the colliding mass and accelerate to the same velocity of the colliding mass which, having given up all of it's kinetic energy, then effectively absorbs the zero velocity "momentum" of the mass at rest such that it stays put while the other mass carries on where the other left off. That would be a true transfer of momentum. An inelastic collision wherein both masses continue onward together after the collision does NOT represent a complete transfer of momentum in my opinion, since the colliding mass retains some of it's momentum. −Dennis . Re: Calculating Newtons in Joules and Joules/s 5