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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.

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  • $\begingroup$ 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 '20 at 2:56
  • $\begingroup$ I made some edits I believe reflect your intention. Please review $\endgroup$
    – Dale
    Jun 13 '20 at 3:00
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    $\begingroup$ @Dale I seemed to have written it incorrectly. Thanks for the edit. $\endgroup$ Jun 13 '20 at 3:07
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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.

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  • $\begingroup$ 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 '20 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 '20 at 3:04

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