I believe that for any physical quantity, to know what significance it has in any situation or that it's different from other quantities, we have to first start by assigning a value to it, or some means of comparison.
For example, empirical temperature started with a lot of experiments on thermal equilibrium and so on, until we knew what it was (at least it's effect on thermal equilibrium state) and then we made devices that can measure it and so we could assign a value to compare between temperatures.
I want to know what exactly is the definition of mass in Newtonian mechanics, I'm not looking for what do we know about mass now or how do we interpret it..etc, I want to know how was mass measured and assigned a numerical value and what are the criteria for saying that two bodies have the same "mass number", do they behave the same in specific experiments or what?
Best Answer
Consider a pair of bodies $b_1$ and $b_2$ in an inertial reference frame. If the bodies $b_1$ and $b_2$ are far from the other objects of the universe and to each other, they have constant velocity. As soon as they become sufficiently close to each other accelerations take place in view of the interactions between them. However physical evidence shows that, inedpendently form the nature of the interaction, there are two strictly positive constants $m_1,m_2$ such that $$m_1 \vec{v}_1 + m_2 \vec{v}_2 = \vec{constant} \quad \mbox{in time}\tag{1}$$ even if $\vec{v}_i$ change in time.
If you replace $b_2$ for $b'_2$, you see that $m_1$ does not change, it is a property of $b_1$ only.
Furthermore, changing inertial reference frame masses do not change.
Another classical property of the mass is that if the two (or more) bodies impact and give rise to a third body $b_3$ it turns out that $m_3 = m_1+m_2$. The same happens if a body breaks down into two (or more) bodies.
(1) can ideally be exploited to measure the mass of bodies. Assume per definition that a fixed body has unit mass $1$. To measure the mass $m$ of $b$, just measure the velocities in two different instants when they are different in view of the interaction of the bodies, $$1\vec{V}(t) + m \vec{v}(t) = 1\vec{V}(t') + m \vec{v}(t')$$ and thus $$1(\vec{V}(t) -\vec{V}(t')) = m (\vec{v}(t')-\vec{v}(t))$$ this identity determines $m$ univocally.