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Let's take two solids of mass $M_1$ and $M_2$.

I link them by a cable that is not under tension initially. The mass of the rope is negligible.

I pull the solid $M_2$ with a force $F$.

Is the law of Newton $\sum F = m a$ valid on the ensemble ($M_1$,$M_2$) ?

The thing that is confusing me is only because the rope is not under tension so basically when I will start pulling, only $M_2$ will move at the beginning.

But as we talk about center of inertia when we use Newton law, maybe the center of inertia of both solid will still follow the motion predicted by Newton law, I just would like to check.

[edit] : actually it is not even clear for me where would be the center of inertia in this ensemble as the two solids with the rope don't really make a "solid" ensemble.

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But as we talk about center of gravity when we use Newton law, maybe the center of gravity of both solid will still follow the motion predicted by Newton law, I just would like to check.

Yes, the center of mass of the two bodies will still move according to the Newton's second law, whether the two bodies are connected by a loose string or not.

The acceleration of the COM will be $\vec a_{COM}=\vec a \frac {M_2}{M_1+M_2}$. This is easy to see, if we take into consideration that, with $M_1$ at rest, the displacement $\vec {dr}$ of $M_2$ will translate to the displacement $\vec {dr} \frac {M_2}{M_1+M_2}$ of the COM.

So, if $\vec F=M_2\vec a$, then $\vec F=M_2 \vec a_{COM} \frac {M_1+M_2} {M_2}=(M_1+M_2) \vec a_{COM}=M_{COM} \vec a_{COM}$, i.e., the Newton's second law holds for the center of mass of the two bodies.

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