Deriving Momentum From Newton's Second Law of Motion

[tomtomtom1]

Homework Statement

Deriving Momentum From Newton's Second Law of Motion

Relevant Equations

Force = Mass * Acceleration
Momentum = Mass * Velocity

Hello everyone

I was hoping someone could shed some light on the following:-

I am trying to derive the equation of Momentum from Newton's 2nd Law.

What I know is the following:-

I don't know how to get from Force = Mass * Acceleration TO Momentum = Mass * Velocity.

I have attempted to re-write Mass and Velocity which gets me to the following:-

But i still am struggling to derive the equation of Momentum (Mass * Velocity) from the above equations?

I am not sure what i have missed out?

Any one point me in the right direction?

Thank you.

 

[Orodruin]

You also need Newton's third law.

 

[archaic]

Newton's second law says that force is the rate of change of momentum, $F=\dfrac{dp}{dt}$. For movements with a constant acceleration, which is what you seem to be doing, it becomes $F=\dfrac{\Delta p}{\Delta t}=\dfrac{mv_f-mv_i}{\Delta t}$.
It is postulated this way.

 

[ehild]

Homework Statement:: Deriving Momentum From Newton's Second Law of Motion
Homework Equations:: Force = Mass * Acceleration
Momentum = Mass * Velocity

You know that acceleration is the time derivative of velocity, a=dv/dt.. So Newton's second law is:
F= mdv/dt. For constant mass, mdv/dt is the time derivative of (mv), called momentum, p.

 

[tomtomtom1]

You know that acceleration is the time derivative of velocity, a=dv/dt.. So Newton's second law is:
F= mdv/dt. For constant mass, mdv/dt is the time derivative of (mv), called momentum, p.

Hi ehild

Thank you for your response, to be honest i don't think i fully understand.

I re-wrote velocity out to get:-

But this is as far as i get.

I tried to re-write acceleration (as you have shown) which gets me to:-

From set 7, i have Force = Mass * derivative of velocity with respect to time.

However when i multiply Mass by dv, then dv/dt is no longer the derivative but Delta V - or have i gotton myself confused.

Never the less I'm still no closer to getting the momentum equation from Newtons law - still struggling to understand it.

Can you elaborate any more (& please be as simple as you can, i won't be offended)

Thank you.

 

[PeroK]

Never the less I'm still no closer to getting the momentum equation from Newtons law - still struggling to understand it.

What equation are you trying to derive?

$F = \frac{dp}{dt}$, where $p = mv$?

 

[archaic]

Hi ehild

Thank you for your response, to be honest i don't think i fully understand.

I re-wrote velocity out to get:-

View attachment 255265
But this is as far as i get.

I tried to re-write acceleration (as you have shown) which gets me to:-

View attachment 255266
From set 7, i have Force = Mass * derivative of velocity with respect to time.

However when i multiply Mass by dv, then dv/dt is no longer the derivative but Delta V - or have i gotton myself confused.

Never the less I'm still no closer to getting the momentum equation from Newtons law - still struggling to understand it.

Can you elaborate any more (& please be as simple as you can, i won't be offended)

Thank you.

$F=ma=m\dfrac{dv}{dt}$, but since $m$ is assumed to be constant you can take it inside the derivative operator, i.e $F=\dfrac{d(mv)}{dt}$. Now, if $F$ is the time derivative of momentum, then $mv=\int F\,dt$.
But, I don't think it is useful to derive momentum from force, as force is DEFINED by Newton as the rate of change of momentum. The quantity $mv$ has been called momentum before the concept of force sprung forth.

 

[PeroK]

$F=ma=m\dfrac{dv}{dt}$, but since $m$ is assumed to be constant you can take it inside the derivative operator, i.e $F=\dfrac{d(mv)}{dt}$. Now, if $F$ is the time derivative of momentum, then $mv=\int F\,dt$.
But, I don't think it is useful to derive momentum from force, as force is DEFINED by Newton as the rate of change of momentum. The quantity $mv$ has been called momentum before the concept of force sprung forth.

You can define force by $F = ma$, then show $F = \frac{dp}{dt}$. If you define force as $F = \frac{dp}{dt}$, then you have to show that $F = ma$. Either way, it amounts to much the same thing.

 

[archaic]

You can define force by $F = ma$, then show $F = \frac{dp}{dt}$. If you define force as $F = \frac{dp}{dt}$, then you have to show that $F = ma$. Either way, it amounts to much the same thing.

I did what you mentioned first in my post, as you've seen, and of course the second follows from the definition. I am just saying that he should take the postulate as it is in its original form as it is more general (as the expression for momentum changes in general relativity).

 

[tomtomtom1]

$F=ma=m\dfrac{dv}{dt}$, but since $m$ is assumed to be constant you can take it inside the derivative operator, i.e $F=\dfrac{d(mv)}{dt}$. Now, if $F$ is the time derivative of momentum, then $mv=\int F\,dt$.
But, I don't think it is useful to derive momentum from force, as force is DEFINED by Newton as the rate of change of momentum. The quantity $mv$ has been called momentum before the concept of force sprung forth.

Hi Archaic

That has cleared up a lot of things for me thank you, If i may ask some final question just to clear up my doubt which are:-

1) Force can be defined as Mass * Acceleration which is the same thing as saying Force is equal to change of momentum with respect to time, which is the same thing as saying the Force acting on a body is equal to the rate of change of momentum with respect to time - would all these statements be correct and equal to each other?

2) Would it be correct to say that Force is equal to the Change in Momentum DIVIDED by Change in Time i.e Delta p / Delta t (p = momentum, t = time) - would this be a correct statement.

3) Finally, given the example:-
Car traveling at 40 m/s with a mass of 256kg
After 92 seconds the car is traveling at 45 m/s with a mass of 256kg
Initial Momentum = 40 *256 = 10240
Final Momentum = 45 * 256 = 11520
Change in time = 92 seconds
Change in momentum = 11520 - 10240 = 1280
The force the car is traveling at is = 1280 / 92 = 13.9 Newtons

Please say this is correct - this makes sense to me.

I know it is very wordy and i know that I have repeated the same statements but with different words but i get confused when different people explain things but us different words but they mean the same thing.

Your thoughts?

I thank you again for your comments.

 

[ehild]

From set 7, i have Force = Mass * derivative of velocity with respect to time.

However when i multiply Mass by dv, then dv/dt is no longer the derivative but Delta V - or have i gotton myself confused.

We speak of derivatives, so do not separate dv and dt. dv/dt is the derivative of v with respect to time, Delta v is the change of the velocity. They are different things. Also, dv/dt (derivative) is not the same as $\Delta v/\Delta t$
You know what is the derivative of a function f(t) multiplied by a constant c : d(c f(t))/dt = c df/dt.
Here the constant is the mass, m, and the function is the velocity v(t), so
m(dv/dt) = d(mv)/dt = dp/dt .
mv=p is called momentum.

So we can rewrite Newton's second law as F=ma=m(dv/dt)=d(mv)/dt=dp/dt
F=dp/dt :
The force is equal to the time derivative of the momentum.

 

[PhanthomJay]

"The force the car is traveling at is = 1280 / 92 = 13.9 Newtons"

Cars do not travel at a force; rather, the net force acting on the car is 13.9 N, in the direction of the acceleration, assumed constant.

 

[archaic]

If you do not know how vectors work, it is ok, just imagine that there are no arrows.

1) Force can be defined as Mass * Acceleration

Yes, within the folds of Newtonian physics (or Classical mechanics, if you like). If you will take advanced physics courses, you will see that $\vec F=\dfrac{d\vec p}{dt}$ won't necessarily reduce to $m\vec a$.

which is the same thing as saying Force is equal to change of momentum with respect to time, which is the same thing as saying the Force acting on a body is equal to the rate of change of momentum with respect to time

If by "change of momentum with respect to time" you are referring to $\dfrac{\Delta \vec p}{\Delta t}$ and by "rate of change of momentum with respect to time" you mean $\dfrac{d\vec p}{dt}$ then not necessarily!
The first is used when the acceleration is constant:
$$\begin{align*}
\dfrac{\Delta \vec p}{\Delta t}&=\dfrac{m\vec v(t+\Delta t)-m\vec v(t)}{\Delta t}\text{ where }\vec v(t)=\vec at+\vec v_0\text{ and }\vec v_0\text{ is the initial velocity vector}\\
&=\dfrac{m\vec a*(t+\Delta t)+m\vec v_0-m\vec a*(t)-m\vec v_0}{\Delta t}\\
&=\dfrac{m\vec a*\Delta t}{\Delta t}\\
&= m\vec a
\end{align*}$$
In the second case, i.e the acceleration vector is not constant, the quantity $\dfrac{\Delta \vec p}{\Delta t}$ represents an "average" net force. To get the "instantaneous" net force, however, you will need to take the time derivative of momentum, $\dfrac{d\vec p}{dt}=\lim_{\Delta t\to0}\dfrac{\vec p(t+\Delta t)-\vec p(t)}{\Delta t}$ (ignore this if you do not know what it means, for now).
On another note, when the acceleration is constant $\dfrac{d\vec p}{dt}=\dfrac{\Delta \vec p}{\Delta t}$

2) Would it be correct to say that Force is equal to the Change in Momentum DIVIDED by Change in Time i.e Delta p / Delta t (p = momentum, t = time) - would this be a correct statement.

I guess you can answer that yourself from what you should have just read as a response to what has been mentioned :)

3) Finally, given the example:-
Car traveling at 40 m/s with a mass of 256kg
After 92 seconds the car is traveling at 45 m/s with a mass of 256kg
Initial Momentum = 40 *256 = 10240
Final Momentum = 45 * 256 = 11520
Change in time = 92 seconds
Change in momentum = 11520 - 10240 = 1280
The force the car is traveling at is = 1280 / 92 = 13.9 Newtons

Dr. @PhanthomJay answered this for you ;) Note, however, that the force should be considered as an average if the acceleration in your example is not constant.

 

[ehild]

Newton's second law says that force is the rate of change of momentum, $F=\dfrac{dp}{dt}$.

Does it?

LAW II​

The alteration of motion is ever proportional to the motive force impressed; and is made in the direction of the right line in which that force is impressed.

https://jamesbrennan.org/physics/notes/Force/axioms.htm

I wonder what he meant by "alteration of motion"

 

[PeroK]

I wonder what he meant by "alteration of motion"

That's a translation from Latin!

 

[archaic]

Does it?https://jamesbrennan.org/physics/notes/Force/axioms.htm

I wonder what he meant by "alteration of motion"

Wikipedia https://en.wikipedia.org/wiki/Newton's_laws_of_motion#Newton's_second_law

 

[ehild]

That's a translation from Latin!

I know. Mutationem motus - change of motion.
Newton did not mention velocity and acceleration in his laws. He spoke about motion, and change of motion. I wonder why is it said that the original wording of the Second Law meant that "the change of momentum is equal to the force". It can be understood as "the change of velocity is proportional to the force".

 

[ehild]

Wikipedia https://en.wikipedia.org/wiki/Newton's_laws_of_motion#Newton's_second_law

I thought of Newton's original laws.

 

[Orodruin]

I thought of Newton's original laws.

I think what the original laws said is less relevant to teaching classical mechanics than the current preferred definition. If not, we would all be teaching from original publications, ignoring years (or in this case, centuries) of development and didactics.

 

[ehild]

Wikipedia https://en.wikipedia.org/wiki/Newton's_laws_of_motion#Newton's_second_law

In classical case, F=dp/dt is not valid when the mass changes. Imagine a container, with liquid, moving with constant velocity. Somehow a hole is made and the leaking container losses mass at a certain rate. The water drops just fall out of the container. Will the velocity of the container change? There is no external force along the motion.
According to dp/dt=0, m'v+mv'=0, so v' = -m'v/m. As m' is negative, the container will accelerate. Is it true? Of course, not.
The momentum changes when the system losses some material with non-zero velocity. And loosing some material is not force.

 

[ehild]

I think what the original laws said is less relevant to teaching classical mechanics than the current preferred definition. If not, we would all be teaching from original publications, ignoring years (or in this case, centuries) of development and didactics.

Yes, but it is said usually that Newton's original statement was that the change of momentum is proportional to the force. I do not think that he meant that when he wrote about change of motion.

 

[Orodruin]

In classical case, F=dp/dt is not valid when the mass changes. Imagine a container, with liquid, moving with constant velocity. Somehow a hole is made and the leaking container losses mass at a certain rate. The water drops just fall out of the container. Will the velocity of the container change? There is no external force along the motion.
According to dp/dt=0, m'v+mv'=0, so v' = -m'v/m. As m' is negative, the container will accelerate. Is it true? Of course, not.
The momentum changes when the system losses some material with non-zero velocity. And loosing some material is not force.

This is not entirely correct. The mass loss corresponds to a momentum transfer out of your system so the net force on the system is negative, just as it needs to be to result in zero acceleration. The definition F = dp/dt is a perfectly fine definition of force and the one that most readily generalises to relativistic mechanics. However, you need to carefully consider all sources of momentum transfer from a system. From the correct interpretation of F = dp/dm, it readily follows that ma = m’vrel + Fext, where Fext is due to external forces not due to mass transfer and vrel is the relative velocity between the object and the ejecta.

I do not think that he meant that when he wrote about change of motion.

What we think he meant is irrelevant. In particular if we do not want to get into a discussion about history and I think we should avoid that.

 

[archaic]

I know. Mutationem motus - change of motion.
Newton did not mention velocity and acceleration in his laws. He spoke about motion, and change of motion. I wonder why is it said that the original wording of the Second Law meant that "the change of momentum is equal to the force". It can be understood as "the change of velocity is proportional to the force".
View attachment 255282

In french, $mv$ is called quantité de mouvement or quantity of motion in english (or motion's quantity), it was the expression used by Descartes, and it should have reached Newton as such. Knowing that, I think that this is the reason why it Newton's law is taken like that, i.e alteration of motion = alteration of $mv$.
https://www.mathpages.com/home/kmath106/kmath106.htm

 

[ehild]

In french, $mv$ is called quantité de mouvement or quantity of motion in english (or motion's quantity), it was the expression used by Descartes, and it should have reached Newton as such. Knowing that, I think that this is the reason why it Newton's law is taken like that, i.e alteration of motion = alteration in $mv$.
https://www.mathpages.com/home/kmath106/kmath106.htm

Thank you for the link!

 

[archaic]

Thank you for the link!

You are welcome :)

 

[PeroK]

In classical case, F=dp/dt is not valid when the mass changes. Imagine a container, with liquid, moving with constant velocity. Somehow a hole is made and the leaking container losses mass at a certain rate. The water drops just fall out of the container. Will the velocity of the container change? There is no external force along the motion.
According to dp/dt=0, m'v+mv'=0, so v' = -m'v/m. As m' is negative, the container will accelerate. Is it true? Of course, not.
The momentum changes when the system losses some material with non-zero velocity. And loosing some material is not force.

Effectively all you are doing is redefining your system. In this case the momentum changed because you decided to ignore particles and mass that had been part of your system. In principle, they don't even have to leak out. You can simply decide you don't want to count their momentum any longer.

Imagine, for example, that you have a system of objects that can light up. Perhaps balls that are normally white, but can light up and become red.

If you measure "the momentum of the red balls", then that can change with no regard to Newton's laws at all.

In the classical case, mass is conserved, so the mass of a system can only change by inclusion or exclusion of particles, which isn't really a change of momentum at all. It's always a redefinition of the system.

 

[ehild]

Effectively all you are doing is redefining your system. In this case the momentum changed because you decided to ignore particles and mass that had been part of your system.

When I first met with the problem of rocket motion I wanted to use dp/dt = 0 and it was wrong. Since then, I have learned to take the rocket and the expelled fuel as a single system, and apply conservation of momentum. But is looks a bit weird. I am interested in the motion of the rocket, the rocket should be my system.
Expelling fuel changes the momentum of the rocket, and that transfer has to be taken into account. That can be done by using the equation dp_r/dt=dp_t/dt + F where dp_t/dt means the rate of momentum change caused by mass transfer.
In an other problem, a raindrop falling through a cloud and growing, as vapor condenses on it, there is mass transfer, but no transfer of momentum, so dp/dt=F holds.

 

[Orodruin]

When I first met with the problem of rocket motion I wanted to use dp/dt = 0 and it was wrong.

It is not wrong if you interpret it correctly. That is the point.