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Since my early studies, I have learned the principle of Newton for the law of dynamics as follows:

$$\sum \vec{F} = m \vec{a}\tag{1}$$ with $\sum \vec{F}$ all the forces applied on the object, $m$ its mass and $\vec{a}$ its acceleration.

Later, after this first version, I have learned another version:

$$\sum \vec{F} = \dfrac{d\vec{p}}{dt}\tag{2}$$ with $\vec{p}$ the momentum which is equal to $\vec{p}=m\,\vec{v}$

With this second form, we have a derivate both on mass and on speed.

I know the first version is not demonstrable (except with Euler-Lagrange equations or maybe Hamiltonian formalism).

But I would like to know how to get the expression of the second version. I ask this because the famous Einstein's formula $E=mc^2$ can be obtained from the second version, i.e $\sum \vec{F} = \dfrac{d\vec{p}}{dt}$ where mass $m$ may not be constant.

Finally, in which context do I have to use the first or the second one?

UPDATE :

Thanks for your answer. However, I don't know how to prove the general form, i.e $\vec{F}=\dfrac{\text{d}\vec{p}}{\text{d}t}$ with Euler-Lagrange equations. With Hamiltonian formalism, I can prove that momentum $\dot p=-\dfrac{\partial H}{\partial q} \equiv -\dfrac{\partial E_p}{\partial q}\equiv F$ (If $\vec{F}$ is only a conservative force).

Then I get the general form : $\dot p = F$, don't I ?

Is it the only way to prove it ?

Regards

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  • $\begingroup$ Isn't that the definition of a force? $\endgroup$ May 27, 2017 at 1:00
  • $\begingroup$ This isn't a full answer to your question, but you should always use $\textbf{F}=\mathrm{d}\textbf{p}/\mathrm{d}t$. In the case of constant mass, this simply becomes $\textbf{F}=m\textbf{a}$. The change in momentum is much more general and allows you to deal with changing masses. "The second one," as you put it, is the fundamental equation, while "the first one" is simply a special case. $\endgroup$ May 27, 2017 at 23:08

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Acceleration is a change in velocity over time, which can be written $$\vec a = \frac {d \vec v}{dt}$$

So, rearranging your first formula we can get $$\sum \vec F = m \frac{d \vec v}{dt}$$

Now considering that $ \vec p = m \vec v$, if $m$ is constant, we can re-write equation 2 like this $$\sum \vec F = \frac {d \vec p}{dt} = \frac {d (m \vec v)}{dt}=m\frac {d \vec v}{dt} = m \vec a$$ This should help illustrate that a change in velocity is the same as an acceleration. Since momentum is the product of velocity and mass, and force is the product of acceleration and mass, change of momentum over time is also equal to force.

The momentum version makes a more general equation that doesn't assume mass is constant.

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  • $\begingroup$ I was going to answer but this sums it up perfectly. $\endgroup$
    – Steve
    May 27, 2017 at 1:01
  • $\begingroup$ Could add that the term $\propto\frac{dm}{dt}$, which is neglected becomes very relevant for rockets. $\endgroup$ May 27, 2017 at 22:07

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