The following is the law of conservation of momentum (in terms of velocity):

$$m_1\mathbf{v_1} + m_2 \mathbf{v_2} = m_1 \mathbf{v_1}^\prime + m_2 \mathbf{v_2}^\prime.$$

Does the law of conservation of momentum also hold for position and acceleration? Since position and acceleration are the $0$th and $2nd$ derivatives (of position), respectively, I suspect that it does. If so, then, putting the law of conservation in terms of position, we get

$$m_1 \mathbf{r_1} + m_2 \mathbf{r_2} = m_1 \mathbf{r_1}^\prime + m_2 \mathbf{r_2}^\prime.$$

I would greatly appreciate it if someone would please take the time to clarify this.

  • $\begingroup$ What you have written doesn't say alot about convervation of linear momentum and it is only true if $v=v^{\prime}$. You have written $(m_1+m_2)v = (m_1+m_2)v^{\prime}$. Does this look like a demonstration of conservation of linear momentum? $\endgroup$
    – K7PEH
    Jun 13, 2020 at 2:56
  • $\begingroup$ I made some edits I believe reflect your intention. Please review $\endgroup$
    – Dale
    Jun 13, 2020 at 3:00
  • 1
    $\begingroup$ @Dale I seemed to have written it incorrectly. Thanks for the edit. $\endgroup$ Jun 13, 2020 at 3:07

1 Answer 1


Conservation of position is not valid. Consider an isolated point particle of mass $m_1$ moving inertially at some non-zero velocity. For such a particle $m_1 \mathbf r_1 \ne m_1 \mathbf r_1’$ and since it is isolated $m_2=0$. So the proposed conservation equation does not hold.

In general it does not hold whenever the center of mass of the system is moving.

  • $\begingroup$ What about in the case of this problem physics.stackexchange.com/q/559023/141502 ? If you refer to that problem, you’ll see why I asked this question and be able to infer what I’m trying to do here. I didn’t want to add too many details to this question and make it unfocused, so I left out all of this stuff and just asked the question in a focused and isolated way. $\endgroup$ Jun 13, 2020 at 3:00
  • $\begingroup$ Sorry, I saw that question earlier but didn’t have a good answer. But regardless of that question, the conservation of position isn’t valid. $\endgroup$
    – Dale
    Jun 13, 2020 at 3:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.