Relationship between mass, momentum and kinetic energy

Question

Is there any fundamental reason why (at least mathematically) momentum is the integral of mass wrt velocity and kinetic energy the integral of momentum also wrt velocity? ie $$p= \int m \ dv = mv$$ $$E = \int p \ dv = \int mv \ dv = \frac 12 mv^2$$ If so, does this work for higher order integrals and what would be the associated physical parameters?

Answer 1

On the relation between momentum and kinetic energy:

Momentum

I regard momentum $mv$ as the integral of $ma$ (mass times acceleration) with respect to time, from starting time coordinate $t_0$ to final time coordinate $t$.

$$ \int_{t_0}^t m \ a \ dt \tag{1} $$

For the purpose of evaluating (1) we can restate $a=\tfrac{dv}{dt}$ as: $a \ dt = dv$

Using that to change from $dt$ to $dv$, with corresponding change of limits:

$$ \int_{v_0}^v m \ dv = mv - mv_0 \tag{2} $$

Which of course simplifies to $mv$ when the initial velocity is zero.

Kinetic energy

Kinetic energy is the integral of $ma$ with respect to position coordinate, from starting position $s_0$ to final position $s$.

$$ \int_{s_0}^s m \ a \ ds \tag{3} $$

Use $ds = v \ dt$, with corresponding change of limits:

$$ \int_{t_0}^t m \ a \ v \ dt \tag{4} $$

Use $a \ dt = dv$, with corresponding change of limits:

$$ \int_{v_0}^v m \ v \ dv = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \tag{5} $$