Why is momentum defined as mass times velocity?

Question

Why is momentum defined as mass times velocity? I asked this question because everywhere people try to answer this by saying that $F=ma$ and if we integrate it with respect to ${\rm d}t$ we can get $p=mv$. But that's not true.

Originally, Newton put forth the second law as $F={\rm d}p/{\rm d}t$ and then using $p=mv$ we got $F=ma$. So its a stupid thing to use $F=ma$ and answer the question that I put forth.

So the real question is why Sir Issac Newton defined momentum as mass times velocity? It could have been velocity squared or mass squared or anything like that. Before we go forward in physics we need to check if the roots of physics were correct.

Answer 1

Newton's definition of momentum $$p=mv$$ is useful, because only then the total momentum is conserved: $$\sum p= \text{const},$$ as can be verified experimentally. Remember that the purpose of physical theories is to be able to predict the result of experiments.

Other definitions (e.g. $p=mv^3$) would certainly be possible, but would be completely useless.

_{Historical side-note: 200 years after Newton it turned out that momentum conservation based on the definition $p=mv$ actually is only an approximation (valid for not too high velocities $v$). But it is not the exact law. In order to preserve the conservation law of total momentum, momentum needed to be redefined as $$p=\frac{mv}{\sqrt{1-v^2/c^2}}$$ where $c$ is the speed of light.}

Answer 2

Newton did not just make up his laws of motion - he based them on extensive experiments and qualitative observations.

In Newton's Principia, the Scholium that follows the statements of the laws of motion describes in detail how he started with Galileo's discoveries concerning projectiles. Then, working with his collaborators such as Wren and Huygens, he carried out a series of very detailed experiments using pendulums released from various heights, to determine the dynamics of both elastic and inelastic collisions. It was the results of these experiments that led him to recognise the importance of what he called "the quantity of motion", which we now call momentum.

Answer 3

At the time Newton and Leibniz were working, both were trying to understand what was happening when mechanical systems changed speeds as a result of forces, that is, trying to find models that would fit with the observations. As a general rule, when trying to understand a novel system, identifying an invariant is tremendously useful.

Newton thought mv was a useful conserved quantity.

Leibniz thought mv² was a useful conserved quantity.

If you read the history, you'll find there was much discussion, rivalry, and even bad blood as each pushed the benefits of their particular view. Each thought that their quantity was more fundamental, or more important.

Now, we see that both are useful, just in different contexts.

I'm sure somebody briefly toyed with the expressions mv³ and maybe even m²v before quickly finding that they didn't stay constant under any reasonable set of constraints, so had no predictive power. That's why they're not named, or used for anything.

So why has the quantity mv been given a name? Because it's useful, it's conserved, and it allows us to make predictions about some parameters of a mechanical system as it undergoes interactions with other systems.

Answer 4

So the real question is why Sir Issac Newton defined momentum as mass times velocity? It could have been velocity squared or mass squared or anything like that. Before we go forward in physics we need to check if the roots of physics were correct.

You have it backwards here. It's not like Newton was thinking "Ah yes! Momentum! What should its definition be?" Quantities in physics come about because they are useful in describing the universe and the world around us. What is much more likely is that Newton realized that the product of mass and velocity was a useful quantity, so it became focused on, given a name, etc. You can't be correct or incorrect with a definition. It's just a definition.

Things like $mv^2$ or $m^2v$ or even $\alpha m^3-\beta \log(v^{1/2}/\gamma)$ wouldn't be incorrect to use either. If these values did turn out to be useful for explaining a variety of phenomena, then we would give it a more permanent definition, and people would focus on it.

Therefore, suggesting a quantity could have been defined incorrectly just isn't solid here. All you can suggest is if 1)a definition is useful or not, and 2) if that definition of being applied correctly or not in some area of physics.

All of the answers saying "well what if momentum was "this" instead are also missing the point, unless one is specifically asking "what do we take the time derivative of to get $ma$?", but that's a trivial question where a "proof by contradiction" is not needed.

Answer 5

As @gandalf61 said , Newton made his laws on the basis of several experiments.

I would like to tell you that momentum means quantity of motion ( this is what I read on many websites and is easier to understand).

Case 1: Suppose there are two moving bodies say of same mass but one with twice the speed of the other. Then to stop both of them we will have to apply a greater force to the one which has a greater velocity and thus this means that it has more quantity of motion. So from here , we simply notice that your quantity of motion depends on your velocity .

Case 2: This time let two bodies are moving with the same velocity but one of them has a mass twice the other. So to stop both of them at the same time we will need to apply a greater force to the one with a greater mass . This means that a massive body also possess greater quantity of motion. So from here , we simply notice that momentum is dependent on mass.

Summing up the two cases , we get that momentum (the quantity of motion) is dependent on both mass and velocity.

Other answers have shown why it can't be mass times velocity squared or any other .

And I think that's why it is defined as mass times velocity (which is a simple relationship between mass and velocity with momentum ).

Hope it helps ☺️.

Answer 6

The definition of momentum actually comes from the definition of mass. Once you've defined mass, conservation of momentum is right around the corner. In fact, they're connected to the extent that they're basically the same idea. The only reason mass even enters Newtonian mechanics, the only reason all objects can be ascribed this unchanging constant called 'mass', is because momentum is conserved.

In an isolated system of $n$ particles evolving over time, the following identity holds for some constants $c_i$:

$$\sum c_i v_i(t)=constant$$

The above law can be taken to define the mass of the $i^{th}$ particle. The unique constant $c_i$ which satisfies the above equation is defined to be the mass of the $i^{th}$ particle.

The traditional way mass is defined in textbooks is different from the above, but it's equivalent to the above. The traditional way mass is defined is : 'Mass is an object's resistance to change in velocity'. Or more precisely, the changes in velocities of two particles in an isolated system over time is inversely proportional to their masses: $\frac{dv_1}{dv_2}=\frac{-m_2}{m1}$, where $dv1$ and $dv2$ are the changes in velocities of the two particles. The minus sign indicates that the changes are in opposite directions.

The way we defined this mass above naturally gives us another really convenient quantity to work with, called momentum:

Consider the quantity $m1v1+m2v2$ for a system of two particles before collision as well as after collision. Before collision, this quantity's value is $m_1u_1+m_2u_2$, where $u_1,u_2$ are the initial velocities. After collision, its value is $m_1(u_1+du_1)+m_2(u_2+du_2)=m_1u_1+m_2u_2+m_1du_1+m_2du_2=m_1u_1+m_2u_2+0=m_1u_1+m_2u_2$

The quantity $m_1du_1+m_2du_2$ is zero because of the definition of mass. As change in velocity is in the inverse ratio of masses (mass resists change in velocity), $\frac{du_1}{du_2}=-\frac{m_2}{m_1}$, implying, $m_1du_1=-m_2du_2$,.

This means that the momentum of a system of particles is conserved as long as only internal interactions are involved (no net interactions from outside the system). This quantity $mv$ is like a currency which simply gets exchanged in 2-particle interactions. The conservation of momentum holds for n-particle systems too, as the interactions between $n$ particles simply comprise of a bunch of two-particle interactions (all of which are just momentum exchanges).

So the reason it's natural to work with $mv$ instead of $m^2v$ is because $mv$ is where $m$ comes from in the first place.

Answer 7

Proof by contradiction :

Lets say you define momentum $p = mv^2$,

Then $$\frac{dp}{dt}=\frac{d(mv^2)}{dt}$$

Applying multiplication rule gives :

$$ m\left(v\frac{dv}{dt}+v\frac{dv}{dt}\right) $$

Which is

$$ m\left(va+va\right) $$

Or

$$ F^{~\prime} = 2v~ma$$

Thus, proved, that what we've got here is not a Newton Law.

Answer 8

I take for granted that you accept that force must be defined as being a function of a mass accelerating. If you double the mass which you are accelerating it is reasonable to assume you will require double the force e.g. if you accelerate one bowling ball and then add another it would be surprising that that the force needed to accelerate the two bowling balls together would be different then the sum of the forces of accelerating the two bowling balls each separately. Similarly for acceleration though less obviously so at least to my paltry mind. However you must admit that multiplying mass and acceleration to measure force is the simplest and perhaps therefor the most logical formation. By suggesting momentum be anything else is to introduce complexity to the force equation.

Answer 9

Simply put,

Let's have an example, two space ships. One is one Kilogram, and One weighs a tonne. Since we're in space (lets just say a void), while we are moving, no friction is applied. Since an object with greater mass will accelerate faster, we can see that this is an amazing way to describe momentum.

Real life examples would include pushing a large box and a small box at the same time on ice. Try it! See first hand the marvels of life!