Does the law of conservation of momentum also hold for position and acceleration?

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.

• 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? Jun 13 '20 at 2:56
– Dale
Jun 13 '20 at 3:00
• @Dale I seemed to have written it incorrectly. Thanks for the edit. Jun 13 '20 at 3:07

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.