Q1 Is $F=\frac{d(mv)}{dt} just a definition or is there something "more" behind the formula for force?
$\vec{p} = m\vec{v}$ is a definition! $\vec{F} = m\vec{a}$ is not! Pratically speaking, you can measure both Force (using an elastic material and its deformation) and acceleration (double derivative of space respect to time), <PERSON> discovered they're proportional so, since they're definite indipendently, $\vec{F} = m\vec{a}$ is the relation between $\vec{F}$ and $\vec{a}$ not the definition of force! You can't measure momentum, so you'll need to define it! The definition is $\vec{p} = m\vec{v}$ and the reasons are perfectly explained on <PERSON> books.
$$\vec{F} = m\vec{a} = m\frac{d\vec{v}}{dt} = \frac{d(m\vec{v})}{dt}$$
Q2 Is W=∫F⃗ ⋅ds⃗ just a definition or is there something more behind? I mean, can you derive the formula for work not by taking the formula for the kinetic energy as given.
$$W = \int \vec{F} \cdot d\vec{s}$$ is the definition of work done by a force $\vec{F}$ along a path, you do not need kinetic energy to define work. Energy is the possible work doable by a moving body.
Q3 How to derive the formula for the kinetic energy and work only form the conservation of momentum Σmivi→=const.?
You know that $\vec{F} = \frac{d(m\vec{v})}{dt} = \frac{d\vec{p}}{dt}$ but since $\vec{p}$ is costant, is derivative will be zero. 512
In case it is moving with costant speed $\vec{v_0}$. Now, imagine a costant force applied to the mass, you'll have that $v^2 - v_{0}^2 = 2as$ (from cinematics) $Fs = W = \Delta E$, $mas = W \Rightarrow$ $$v^2 - v_0^2 = 2W/m$$ $$W = \frac{1}{2}mv^2 - \frac{1}{2}mv_0^2$$ The applied force is going to stop the mass, so the mass reacts with an opposite force, and since energy is the possible work done by the moving mass. Since it is opposed and $v= 0$ we obtain: $$ W = \frac{1}{2}mv_0^2$$ Energy depends only on the initial velocity because momentum (before the costant force applies) is conservating. 512