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Conservation of momentum can be derived during collisions by using Newton's laws of motion. But in other cases, do we simply take it like an axiom ?

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    $\begingroup$ Newton's laws apply whether there is a collision or not. $\endgroup$
    – Paul
    Commented Jan 11, 2016 at 14:15
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    $\begingroup$ Conservation of momentum comes from Noether's theorem and shift symmetry so it isn't an axiom. There surely must already be a duplicate question discussing this, but I must admit that a quick search has failed to find one ... $\endgroup$ Commented Jan 11, 2016 at 14:22
  • $\begingroup$ Conservation of momentum ca be derived using the third law of newton $\endgroup$
    – David 2000
    Commented Jan 11, 2016 at 14:27

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We don't take conservation of momentum as an assumption, and neither do we take Newtonian mechanics as an assumption. Instead the fundamental assumption we make is that the systems we study in classical mechanics obey the principle of least action.

It is hard to overstate just how important this principle is. From it we obtain Lagrangian mechanics and Newtonian mechanics, but it gives us a lot more. For example the relevance to your question is that for a system described by a Lagrangian Noether's theorem tells us that conservation laws are related to symmetries (of the action). If the system has translational symmetry, i.e. action is invariant under translations in space, then momentum must be conserved.

The principle of least action is a somewhat abstract approach for students beginning their study of physics, and you normally won't be introduced to it until you start university. However if you are looking for the fundamental assumptions involved in mechanics then this is it.

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The second Newton law can be written in its general formulation as \begin{equation} \sum\vec{F} = \frac{d\vec{p}}{dt}. \end{equation} The formula $\sum\vec{F} = m \vec{a}$ is implied only under the condition of a closed system (a system which does not loose or gain mass) and, by considering $\vec{p}=m\vec{v}$, you can easily show that \begin{equation} \sum\vec{F} = \underbrace{\frac{dm}{dt}}_{=0}\vec{v} + m \frac{d\vec{v}}{dt} = m \vec{a}. \end{equation} Then, you can understand that momentum conservation is implied by $\sum F =0$ since \begin{equation} \sum F = \frac{d\vec{p}}{dt} = 0 \qquad \Rightarrow \qquad \vec{p} = \text{constant}. \end{equation} Summarizing, you cannot consider momentum conservation as an axiom since it is not! This particular conservation property is implied only if your system is not subjected to a resulting force, like a collision of two particles on a plane without friction forces.

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Conservation of linear momentum is not just an assumed axiom, its a deep fact of space, the laws of physics are the same throughout all space. The connection between them is Noether's theorem. To see that you will have to get a book in classical mechanics. I recommend you Leonard Susskind's lectures in classical mechanics, they are on youtube and are very understandable for begginers. You can also check wikipedia. Here is an extract:

"Conservation of momentum is a mathematical consequence of the homogeneity (shift symmetry) of space (position in space is the canonical conjugate quantity to momentum). That is, conservation of momentum is a consequence of the fact that the laws of physics do not depend on position; this is a special case of Noether's theorem."

Noether's theorem:https://en.wikipedia.org/wiki/Momentum#Symmetry_and_conservation"

Leonard Susskind lectures: http://theoreticalminimum.com/courses/classical-mechanics/2011/fall

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