If the center of gravity moves, it moves with the velocity v. So it will be in the position r+v*t after some time t. Now it’s at the position r. A different position of gravity (mass) center means different position. For the whatever finite t.
In the case when the gravity center doesn’t move, you can divide the composition into two sub-compositions, where both gravity centers do move. If only one had moved, then the combined gravity center would move and we would have the solved case above.
But if both gravity centers move, they can either move apart and never collide—in which case they will both have different position vectors lately - or they will collide. In that case, they will reverse their directions after the elastic collision and we have a solved case then.
But if both gravity centers move, they can either move apart and never collide—in which case they will both have different position vectors lately—or they will collide. In that case, they will reverse their directions after the elastic collision and we have a solved case then.
I’m not convinced by this bit. Usually we can calculate the results of an elastic collision by using both conservation of energy and conservation of momentum. But we can’t know the energy of the sub-compositions based just on the velocity of their centres of mass. They will also have some internal energy. So we can’t calculate the results of the collision.
Then, both gravity centers travel with a certain velocity each and will collide. How can they return back here? If they will reverse both velocities after the collision. Then, they can return. But with the opposite velocities, and therefore this will not be the same state.
Since they always move in lines, there will be no another collision. And therefore no return to the present state.
Doesn’t matter how many collisions will happen, the momentum conservation will hold. Even if only two small balls of each gravity center will collide, the sum of momentums of that two gravity centers will remain the same. Doesn’t matter which partition of balls we chose, only that is the same before and after the collision.
After some collisions have happened and the two parts are heading away from each other the two parts could still overlap and then some more of their balls could collide. This could lead to the two parts heading back together.
I’m assuming no gravity (and that at least one ball is moving). Do you have a proof for your assertion?
Sure.
If the center of gravity moves, it moves with the velocity v. So it will be in the position r+v*t after some time t. Now it’s at the position r. A different position of gravity (mass) center means different position. For the whatever finite t.
In the case when the gravity center doesn’t move, you can divide the composition into two sub-compositions, where both gravity centers do move. If only one had moved, then the combined gravity center would move and we would have the solved case above.
But if both gravity centers move, they can either move apart and never collide—in which case they will both have different position vectors lately - or they will collide. In that case, they will reverse their directions after the elastic collision and we have a solved case then.
Well, that’s an approximate proof.
I’m not convinced by this bit. Usually we can calculate the results of an elastic collision by using both conservation of energy and conservation of momentum. But we can’t know the energy of the sub-compositions based just on the velocity of their centres of mass. They will also have some internal energy. So we can’t calculate the results of the collision.
Do we agree, that if there will be no collision, and both gravity centers move, that they will never return to the present position?
Yes.
Then, both gravity centers travel with a certain velocity each and will collide. How can they return back here? If they will reverse both velocities after the collision. Then, they can return. But with the opposite velocities, and therefore this will not be the same state.
Since they always move in lines, there will be no another collision. And therefore no return to the present state.
What do you mean by “the” collision? If each part has several balls then there will be multiple collisions.
Doesn’t matter how many collisions will happen, the momentum conservation will hold. Even if only two small balls of each gravity center will collide, the sum of momentums of that two gravity centers will remain the same. Doesn’t matter which partition of balls we chose, only that is the same before and after the collision.
After some collisions have happened and the two parts are heading away from each other the two parts could still overlap and then some more of their balls could collide. This could lead to the two parts heading back together.
No, that’s impossible. However you choose to divide this set of balls and however they later collide, both impulses are still conserved.