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Newton defined rate of transfer of momentum from one particle to another as force. Can we predict motion using F=mvs or something similiar instead of $F=ma$?

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Why $F=ma$, why not $F=mvs$

Because it works. Try to do it in another (nonsimilar) way, and it won't work. It seems like a more than fair reason to me. This is Newton's Flaming Laser Sword after all.

Can we predict motion using $p=mv$ instead of $F=ma$?

Yes we can: one can use $F=\frac{dp}{dt}$ and get the same answer (and it even works with more general cases, like time-dependent mass!).

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  1. It is because an object with constant velocity can't be distinguished from a stationary object if one moves into the frame of reference of the object.
  2. The bare minimum required to see the motion of a body in its own frame of reference is change in velocity.
  3. Hence, force should at least depend on the second derivative of displacement.
  4. There are forces that depend on the third derivative as well like Abraham–Lorentz force.
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I am assuming that "$s$" in your super cool new force $F:=mvs$ means the position of the mass (along a preferred direction).

Generally, the quantities you define will not restrict the things you can predict with them, that is, unless you define quantities that lose some information (for example, $v^2$ is not a suitable state variable of a mechanical system, because it lacks the sign of $v$).

However, the most important question in physics is, and has always been, what is the simplest formulation of the equations that allow you to predict something. Given your definition of the force, and additionally defining old-school grandmummy's force $F_0:=ma$, we know for example, that Hooke's law of elasticity gives $$F_0 = -k\cdot s$$ When you try to relate this to your "next gen force", you first notice that (I assume constant mass for the sake of clarity) $$F_0 = m\frac{d}{dt}(v)=\frac{d}{dt}\left(\frac{F}{s}\right)=\frac{1}{s} \frac{dF}{dt}-\frac{F}{s^2}\frac{ds}{dt}$$ and when you substitute this into Hooke's law, you might be inclined to multiply the whole equation by $s^2$ in order to get rid of the state variable $s$ in the denominator, which would be a little problematic. So you finally arrive at $$s\frac{dF}{dt}-F\frac{ds}{dt}=-ks^3$$ While you are still writing down the equation, just before you think about what the heck this equation means and how you could solve it, and like 30 minutes before you fetch your 1000 pages ordinary differential equations handbook and trying to figure out in what category this equation falls, all your buddies using grandma's force have already solved the equation, namely the well-known harmonic oscillator.

This might still not be enough to convince you that Newton's force is the best choice, and this attitude is okay, because you might find something new, even among those things that hundreds of thousands of physicists have been thinking about over and over again.

In fact, the geocentric model was the accepted standard in astronomy from the times of Ptolemy (born 100 AD) until the late middle ages, when it was replaced by the heliocentric model (by Copernicus, born 1473). The latter succeded, because it was much easier to handle, although at first it did not predict anything that the geocentric model wasn't capable of. If Copernicus had not dared to ask, why one preferred to consider everything rotating around the earth, we would still wallow in the mire of naive intuition about astronomy.

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So, you are correct that this is a definition rather than an assertion about the world. As a definition, it cannot be invalidated by any particular experiment, it lives at the “theory level” which cannot be meaningfully questioned, as opposed to the “models” we make which can correspond to reality.

So what actually happens is that certain choices of theory make our models much simpler, and people prefer to use those simpler models. In that respect science in general proceeds via a sort of evolution by natural selection, where the genes are in the theories and their reproductive success is how many papers are written using those theories, as those papers have a sort of genealogy where this paper led to that one and so on.

That is a really abstract idea so let me connect it back to something concrete: Galilean relativity. (I don't actually know if Galileo observed this or if it is just named after him.) The idea is, as long as the windows are closed and the track is smooth, I can juggle on a train without having to adjust anything from how I would juggle on the ground. Obviously if it crashes or the track curves or some other deviations from uniform motion in a straight line happen, my juggling will be affected. Even if you take the roof off of the train my juggling will be affected, it will be like I'm trying to juggle in a windstorm. But, there is an ideal circumstance that we can point to where the train moves at a non-zero uniform velocity, and the physics inside the train car is the same as the physics inside and equivalent room on the ground, not in motion.

The mathematical manifestation of this, is that every particle in the train car has had a certain velocity added to it, but the physics has remained invariant.

As a result, theories will intrinsically be simpler if they only make reference to changes in velocity, not absolute velocities. A theory with absolute velocities cannot explain my ability to juggle on a train by saying “it's the same as juggling on the ground.”

This definition by Newton divides a difference in velocity by a difference in time. This is handy because in classical mechanics everybody agrees on time as well. That you need a ratio of differences comes back to Newton’s larger agenda which is to apply calculus: a change by itself cannot be associated with any particular time, but a ratio of two changes can, because there is a limit as the two times approach. That the one difference is a change in velocity, is very handy because that is a Galilean relativistic invariant, so I don't need to give you a rule for how to calculate forces in a moving frame; they're exactly the same as the forces in a stationary frame. If you use a change in position or an absolute position or so, you no longer get this reference frame invariance!

So that is the simplification that this definition is making: it is that as a force starts to move an object, I do not need to worry about “translating the force into the frame of the object” or any such thing, because I have defined forces as the invariant. Conversely, when I discover that there are things like light where I cannot add velocity to them, and we transition into special relativity, then this notion of force needs to be redefined. But Newtonian mechanics works extremely well in part because Galileian relativity works so darn well.

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Yes.

$\displaystyle F=\frac{dp}{dt}=\frac{d}{dt}(mv)$

If mass $m$ is constant this simplifies to the more familiar

$\displaystyle F=m\frac{dv}{dt}=ma$

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  • $\begingroup$ mass is always a constant $\endgroup$ Jun 14 at 16:36
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    $\begingroup$ @brucesmitherson Not always. There are problems in dynamics where the mass of the system is not constant e.g. sand falling on a moving conveyor belt, or a bucket of water with a leak in it. To solve these problems you have to use $F = \frac{dp }{dt}$ rather than $F=ma$. $\endgroup$
    – gandalf61
    Jun 14 at 17:49
  • $\begingroup$ In most cases the mass moves in or out of the system, and you cannot use F=v dm\dt+am. Example, a rocket. $\endgroup$ Jun 14 at 18:56
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Lets look at this simple example

enter image description here

with:

$$a=\frac{dv}{dt}~,v=\frac{ds}{dt}\\F=-k\,s$$

you obtain the equation of motion

I) $~m\,a=F$

$$m\,\ddot s+k\,s=0$$ and with $s(0)=s_0~,\dot{s}(0)=0$

$$s \left( t \right) =s_{{0}}\cos \left( {\frac {\sqrt {k}t}{\sqrt {m}}} \right) $$

II) $~m\,v\,s=F$

$$m\,\dot s \,s+k\,s=0$$ thus with $~s(0)=s_0$

$$s \left( t \right) =-{\frac {kt}{m}}+s_{{0}}$$

but this result is not what we expecting form the physic, so the second ansatz is wrong and Newton ansatz is the correct one.

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