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I have read the answer about the energy and momentum conservation question but I have a great question now! Imagine we have two masses which one of them is at rest and another move and collide to the rest one. Now we know that the momentum remain always constant even if the collision was inelastic but the kinetic energy convert to heat. So in this case the energy converted to heat and radiate to space because of the (heat radiation nature) so we will receive that energy and then as we know the momentum will conserved there and lead that the second mass start to move with the same velocity as the firs mass before collide. Imagine again there were a chain much more masses spaced with same divided far from each other. One mass will collide to the next and radiate heat and the the second mass start to move with the same speed and collide to third mass and radiate heat and then start to move toward the forth mass and .... see? we were receiving heat from naught? While the system lost no velocity amount of energy freed. This is impossible. I have test this problem by MSC ADAMS. ADAMS shows that the masses velocity remain same. So where is the sure of this heat energy?

In the image you will see that the two mass collide together inelastically fore ever because the momentum conserved the system never loss velocity, but we have receive heat from their collision

In this fig you will see that the two mass collide together inelastically fore ever because the momentum conserved the system never loss velocity, but we have receive heat from their collision

enter image description here

AMIN EJLALI
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2 Answers2

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... so we will receive that energy and then as we know the momentum will conserved there and lead that the second mass start to move with the same velocity as the firs mass before collide.

But you said that it was an inelastic collision. But if it were truly inelastic, then not all of the kinetic energy of the first ball that was rolling would have been transferred to the second one that you said was at rest. Hence, the second ball can't be rolling with the same speed.

... In the image you will see that the two mass collide together inelastically fore ever because the momentum conserved the system never loss velocity, but we have receive heat from their collision

This is a violation of the conservation of energy, that followed because of the error you made while using the momentum formulas.

TO POINT OUT (from the comments you posted): It is not true that the velocity will remain the same with each collision. Because the collisions are inelastic, some amount of kinetic energy will be lost. So, you have to change the values of velocity that you input in the momentum conservation formula with each collision (you are right in saying that momentum is conserved in inelastic collisions).

Karthik
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  • thanks for your answer, actually I don't persist on the first mass stop after collision. I was focused on the momentum conservation rather the first mass stop or reverse or continue with less velocity. For second answer: in the figure the wall is rigid and so there is no energy lost from mass collision to the walls but of course the K energy will lost in the masses collision because of inelastically being that. But according to the answers from the pasted link we will never lost momentum even if the collision was inelastic and this means while the masses are constant the velocity will remain – AMIN EJLALI Jan 08 '19 at 15:59
  • If you feel this answers your question, you can click on that tick sign to accept the answer. – Karthik Jan 08 '19 at 17:48
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You ask whether how the collision between a single moving particle and a single stationary particle can be inelastic if the moving particle comes to rest. The short answer is: it can't. If the collision is inelastic then the initially moving particle must keep moving.

For simplicity take the two particles to have equal mass $m$ and the collision to be in 1D. Then momentum conservation requires the initial speed $u$ to equal the sum of the two final speeds $v_1, v_2$ of the two particles. But we have:

$$u^2 = (v_1 + v_2)^2 = v_1^2 + v_2^2 + 2v_1v_2 \geq v_1^2+ v_2^2$$

This implies that the initial kinetic energy is larger than the final, except in the case where equality holds, which is one of the particles ends up at rest.

So in the inelastic case, the first particle does not come to rest, but keeps moving somewhat. The most inelastic case has the two particles congeal together to form a single mass $2m$.

jacob1729
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  • Thanks for your answer. I understood this proof and confirm it. Now according to your proof this is my question. The two particle continue with the same velocity because of the momentum conservation. At next this particles collide to a rigid wall and reverse. Again two particle collide together and lost K energy but not momentum. {to be continued in next the comment...} – AMIN EJLALI Jan 08 '19 at 17:36
  • According to this theorem colliding to a rigid wall and reversion will not absorb their momentum so they come back with constant velocity and collide together again. saying the momentum conserved in this occurrence the masses never loss velocity and by being a rigid wall in opposite (fig at the commence question) the masses will continue to collide occasionally forever and they never stop. Do you confirm that the two masses never stop? – AMIN EJLALI Jan 08 '19 at 17:36
  • @AMINEJLALI this sounds like a new question, but unlike I'm misunderstanding your situation the masses still don't stop. With each inelastic collision between masses they keep getting closer to having half the initial speed each. – jacob1729 Jan 08 '19 at 17:42
  • @AMINEJLALI when you said somewhere in the beginning that the two blocks with one at rest, collided, the one at rest would move at the same speed as the first one. And then the first one would come to rest. But in an inelastic collision, because some kinetic energy is lost, it does not make any sense for the block at rest to move with the same speed as the first one. Why? Because some kinetic energy is lost. It could have gone as heat or sound or whatever. Block 1 may or may not have come to rest. – Karthik Jan 08 '19 at 18:03
  • @KarthikV block 1 can't come to rest and block 2 can't move off at the same speed as the unitial speed $u$. The final velocities $v_1,v_2$ have to obey the strict inequalities $0 < v_1 < v_2 < u $ for a 1D inelastic collision. – jacob1729 Jan 08 '19 at 19:53
  • @jacob1729 Ya that's what I meant. He thought that the block at rest will get the same speed in an inelastic collision. Since it cannot be, he cannot get unlimited heat energy out of his system. – Karthik Jan 08 '19 at 21:36
  • No I did not persist in the No 1 stopping or moving. I meant how it is possible that the masses never stop while they generate heat in every collision and fir ever – AMIN EJLALI Jan 09 '19 at 18:18
  • @jacob1729 Are you tellin me that v_1+v_2=u? or not – AMIN EJLALI Jan 09 '19 at 18:19
  • @jacob1729 is this possible?? : v1+v2<u – AMIN EJLALI Jan 09 '19 at 18:20
  • My uni proff was UCLA PHD top student at his time, He told me that in the real world the masses stop someday. he said ADAMS is defined so that take the momentum always conserved unless you define some code in which the momentum decrease: v1+v2<u even in Meriam dynamic book he did not noted clearly that the momentum always and anyway will conserved, he just said that the momentum conserved but not clear meant about elastic or inelastic collision – AMIN EJLALI Jan 09 '19 at 18:24