# What is momentum?

Momentum tells you the mass of the object and how fast it is going right? So if I have a 2 kg ball moving at 2 m/s, then the ball has 4 kg⋅m/s of momentum. My question is why do we multiply mass and velocity to get momentum. (From the example above) Why cant we just say the ball is 2 kg moving at a speed of 2 m/s and that is momentum. Why do we have to multiply it?

• Momentum tells you the mass of the object and how fast it is going right? No. Knowing the momentum does not tell you either the mass or the velocity. If I tell you “the product of two numbers is 24”, how would you know what the two numbers are? – G. Smith Apr 11 at 6:15
• Of course you could say 2 kg moving at a speed of 2m/s and if you did that, your audience would need to work out what you meant, by doing the multiplication for you. If you had particular reason for singling out mass or speed, that would not be the same question. – Robbie Goodwin Apr 11 at 20:23
• If you tried to solve even one trivial "homework" type problem using p=mv it should be completely obvious why. – J... Apr 12 at 11:45
• This video posted almost two weeks ago (cough, cough) did a nice job not explaining what momentum is at all. – Arthur Apr 12 at 22:09

The deep reason for introducing the momentum is that momentum is a quantity that in some circumstances can be conserved, while this is not the case for the velocity. The case of one particle is not really enlightening, but as soon as we move to systems of more than one particle the advantage of introducing momentum is clear.

The special role of momentum can be fully understood only using the full conceptual framework of analytic mechanics. However, some hint about the importance of this concept can come from a few facts:

1. it turns out that it is possible to define and assign a value to the momentum of massless entities like electromagnetic radiations or photons;
2. Relativity introduces a deep revision of the concepts of classical mechanics, but momentum, although defined in a different way, still plays the same role as in classical mechanics;
3. Quantum mechanics does not allow a meaningful definition of velocity (the concept of trajectory becomes meaningless), still, it is possible to define the momentum of a particle, playing a key role in the theory.
• "Quantum mechanics does not allow a meaningful definition of velocity" That is wrong. While it is true that velocity cannot be defined as the time derivative of the position in quantum mechanics, the velocity operator is still meaningfully defined as p/m. See e.g. physics.stackexchange.com/questions/430118/… – oliver Apr 11 at 15:26
• Can you give an example where momentum (minus leakage in and out of the system under consideration) is not preserved? – Peter - Reinstate Monica Apr 11 at 18:14
• @Peter: Maybe an accelerating reference frame fails to properly preserve it? I haven't actually done the math, though. – Kevin Apr 11 at 18:53
• @Peter-ReinstateMonica in general relativity, momentum is not, in general, conserved. For example, consider light in an expanding spacetime; the expansion causes redshift of the light, which decreases its energy and momentum. – John Dumancic Apr 11 at 18:56
• @oliver by meaningful definition I meant something analogous to the definition in classical mechanics. Your comment shows even more that momentum plays a role in QM independent of the velocity. – GiorgioP Apr 11 at 19:23

Why cant we just say the ball is 2 kg moving at a speed of 2m/s and that is momentum

If you did do that, you can’t call it “momentum”. The mass and the velocity are two seperate quantities and looking at them separately does not give you the quantity momentum.

By definition$$^1$$, their product gives you the (linear) momentum $$\tag 1 p=mv$$

Why do we have to multiply it?

Your question is more or less asking what is important about the quantity $$mv$$. It is therefore important to note that momentum or the quantity $$mv$$ is a conserved quantity and the fact that this product is indeed conserved, is an experimentally determined fact with fundamental implications in physics.

Historically one could have termed the quantity $$m\sqrt{v}$$ (or any other similar arbitrary quantity involving mass and velocity) “momentum” but such quantities are not conserved (except of course kinetic energy).

The quantity momentum in equation 1 was formulated by Newton in his Principia and did so because this quantity is conserved and therefore has important and fundamental implications in physics. In fact, I believe that Newton was working with the quantity $$mv$$ and realised its physical significance before he$$^2$$ named it “momentum”.

1. Note that this is the definition of classical momentum, and in the late 1800s to early 1900s was found to be not exact in certain situations, and so very soon after, this definition was reformulated to $$p=\gamma mv$$ where $$\gamma = \frac{1}{\sqrt{1- \frac{v^2}{c^2}}}$$ thanks to Einstein and his theory of Special Relativity.

2. I’m not sure who named it “momentum” :)

The significant question then would be

Why we define the momentum the way we do?

There are different level to understand this but I'll just give the significance as (considering that OP is a High school student) Newton's second law:

The second law states that the rate of change of momentum of a body over time is directly proportional to the force applied, and occurs in the same direction as the applied force.

$$\mathbf{F}=\frac{d\mathbf{p}}{dt}$$ where $$\mathbf{p}$$ is the momentum of body. The momentum defined as $$\mathbf{p}=m\mathbf{v}.$$

So we defined a quantity because it plays a significant role in a different place. You can still write $$m\mathbf{v}$$ in all the places in Newtonian mechanics and that makes no difference but you find it good to say momentum instead of multiplication of mass and velocity.

As you go along you find more significance in reference to symmetry and all.

One thing a beginning physics student will learn is that the things you define are mostly just a matter of personal taste, not one of authority. Therefore, your question is a very good one.

You can always define the quantities you want. However, you have two important responsibilities:

1. choose definitions so that you can express your observations in the most economic way, because memory (either brain or computer or paper) is limited
2. accepting given definitions in order to be able to communicate with others efficiently

With respect to list item 2, you could of course report mass and velocity separately, if the receiver of the information is interested in it. You could also say "garble" instead of "momentum", but then others would not understand you, and hence, would hesitate to listen to you.

With respect to item 1, the most economic description of nature has to account for the observation that momentum is a conserved quantity. For a closed system, this economizes your description in that you just have to determine momentum once, and you would know that it possesses the same value forever. You might agree that writing down one number (or three, if you work in 3D space, or four in special relativity) is much more economic than writing down millions and millions at many different instants in time (and still missing a lot of changes inbetween these arbitrarily chosen instants).

Another very important property of momentum is that it is an extensive quantity. This means that you can add the momenta of different bodies together to obtain the compound momentum. This is already not that obvious: as long as the bodies move at the same velocity, the summation property follows from mass (if you have already accepted classical mass conservation). But what if the bodies move at different velocities? Hence, being able to add momenta is non-trivial. From momentum the concept of the center of gravity/mass directly derives.

Finally, that extensive quantities even exist is non-trivial: just take temperature: if you have a body at 270 Kelvin and one at 300 Kelvin, will the both of them have 570 Kelvin? No, not at all, you will get some value inbetween, that depends on their masses and specific heat capacity and so on. Temperature is an example of an intensive quantity. You will probably learn what that means in thermodynamics.

• Brushing everything under "economic" is underselling the power and insight of definitions. – Passer By Apr 11 at 20:15
• @PasserBy: Honestly, I don't know what you want to say by this. It is beyond a shadow of a doubt, that the laws of physics in general are the most ingenious form of data compression known to man. If calling this "economic" means underselling, then I wonder what form of respect you want to hear me saying. – oliver Apr 12 at 18:44
• In my head (being a Chemical Engineer and having taken too much thermodynamics way back when), I see tight relationship between momentum (well, more inertia than momentum) and the specific-heat/temperature of system. In your example, the system consists of two bodies, likely of different masses, perhaps of different compositions, and of different temperatures. If you calculate the "heats" of each body (from their masses, their specific heats and their temperatures) then you can get a "heat" for the system (where "heat" isn't "heat = Energy"). We regularly called this the "thermal inertia" – Flydog57 Apr 13 at 0:14

It's a matter of equating and comparing. When dealing with momentum, we're generally not interested what the specific number is, but rather how that number (and thus the object's momentum) changes (increase/decrease) or doesn't (equality).

And when discussing equality, it's much easier to have to compare one specific value (per comparable element) instead of multiple.

Take the following example. We have three objects:

• A 4 kg ball moving at 3.6 m/s
• A 7 kg ball moving at 1.9 m/s
• A 3 kg ball moving at 4.8 m/s

Which one would hurt you the least when it hit you?

It's hard to figure out, because your brain can't balance the two sets of numbers and easily compare the ratios. And keep in mind that this is even just simple multiplication, let alone more complicated relations between properties, e.g. if I had given you the weight of the ball and the height I drop it from.

But if I had said this:

• A ball with 14.4 kgm/s momentum
• A ball with 13.3 kgm/s momentum
• A ball with 14.4 kgm/s momentum

It's a lot easier to spot the answer, right?

This comes with a drawback though, you can no longer figure out the individual speed or weight. Notice how examples 1 and 3 have the same momentum, yet a different speed and weight. When dealing with momentum, the two have become indistinguishable. So what you said here is incorrect:

Momentum tells you the mass of the object and how fast it is going right?

Momentum doesn't tell you both things, momentum expresses a relation between the two things.

It's easier to consider [ momentum, speed, weight ] as a grab bag. If you know two of them, you can inherently figure out the third. But if you only know one of them, then there are a whole range of possibilities that could fit that scenario (such as my examples 1 and 3 from before, I could find literally thousands of examples of balls with the same momentum).

However, in most cases where momentum is being observed, the specific speed and weight doesn't really matter all that much.

For example, let's say you're going to paint your room. What you need to know is the surface area you're going to be painting. While you did in fact calculate the surface area from the width and height of your walls (and how many walls there are), those values don't really matter much themselves. Two people with wildly different shapes of rooms but with the same surface area need to buy the same amount of paint.

Inside the paint shop, rooms of equal surface area are considered equal and specific width/height don't matter. And for the same reason in most momentum discussions the specific speed/weight don't matter either.

• "Which one would hurt you the least when it hit you?" none of the balls would leave any possibility to ever feel pain again, I'm afraid :) – scristalli Apr 12 at 23:01
• @scristalli: In a sense, I just unwittingly proved my own pudding, as I had no concept of the absolute values, only the comparative nature ;) Brought everything down by a factor of 10 to make it more reasonable. – Flater Apr 13 at 9:20

A formula like $$p=mv$$ is the mathematical expression of the definition of a physical property or change. A formula consists of variables which affect the quantity that it defines. And the formula also shows the relation between those variables.

The concept of momentum is the result of some observations in universe. Think collision between two objects. For example, a rock and a car. When a large rock (large mass) which is moving at a low speed hits the car, it will dent the car's body. And when a small rock (small mass) which is moving at a high speed hits the car, it will dent the car's body in the same way.

This is a very simplified example, but when we analyze it, we see that there is a relation between the mass and velocity of the rock and the result of the collision. And we see (via experiments) that there is a specific relation between mass and velocity here. When one of them halves the other doubles; we see this when we compare two different collisions with same results. Like this situation, when the masses of the two objects are same and the velocity of one of them is twice the other, its effect in the collision is more. How much? 2 times. This means that these two variables are related in a way that when the value of one of them increases twofold, the observed effect increases twofold too. This means that the increase in velocity or mass does not "add" to the result of the collision, it "multiplies" it.

So we can define a quantity now, which is momentum in this case, to explain the properties of the collision (in other words, the observed relation between quantities in an event) and express it with a formula in which there are two variables that are multiplied.

$$p=mv$$

Then we observe that momentum is conserved in a closed system.

In the context of classical mechanics, and especially rigid body motions momentum can have the following interpretation. The statements below might seems a bit circular, but they work.

Momentum is the quantity needed to completely remove all movement from a rigid body or a point mass.

Specifically, momentum is a vector quantity applied along an infinite line in space (the axis of percussion). An impulse (a short lived force) of equal and opposite magnitude and direction, but along the same line applied on a moving body will instantaneously stop all motion(s). There are specially cases than more than one impulses needed to be applied for this to happen, and in those cases momentum is defined from the net impulse acting on the body.

In this context the units of momentum are Newton-second [$$\mathrm{N\,s}$$] representing a force action (force over time).

The resulting change in velocity $$\Delta \vec{v}$$ due to an impulse $$\vec{J}$$ is $$\Delta \vec{v} = \frac{1}{m} \vec{J}$$ This is true for a point mass, or for the center of mass of a rigid body. The mechanics of how an impulse $$\vec{J} = \int \vec{F} \;{\rm d}t$$ is defined does not matter here.

You can think of this backwards also. Take a stationary object and apply an impulse equivalent to the desired momentum, in magnitude, direction and location and the body with move after the application in said way.

Example(s):

1. A point mass of $$m=2\,\text{kg}$$ is moving with velocity (vector) $$\vec{v} = \pmatrix{5 & 0 & 0}\;\text{m/s}$$

An impulse of $$\vec{J}=\pmatrix{-10 & 0 & 0}\;\text{Ns}$$ causes the following step change in velocity $$\Delta \vec{v} = \tfrac{1}{m} \vec{J} = \pmatrix{-5 & 0 & 0} \; \text{m/s}$$ The final velocity is thus $$\vec{v}^\star = \vec{v} + \Delta \vec{v} = \pmatrix{0 & 0 & 0}$$ If the final velocity is zero, it means the impulse applied must have been equal and opposite to the momentum of the body. So the momentum of the body was $$\vec{p} = -m \Delta{v} = - \vec{J} = \pmatrix{10 & 0 & 0}\;\text{Ns}$$

1. A solid disk (rigid body) of mass $$m=2\;\text{kg}$$ is rolling on a horizontal plane with the velocity of the center $$\vec{v} = \pmatrix{5 & 0 & 0}\;\text{m/s}$$. The radius of the disk is $$R=1 \;\text{m}$$ mass moment of inertia of the disk is $$I = \tfrac{m}{2} R^2 = 1\;\mathrm{kg\, m^2}$$. The rotation of the disk is $$\vec{\omega} = \pmatrix{0 & 0 & -5}\,\text{rad/s}$$. The motion response of the disk that rolls must maintain that $$\Delta \vec{v} = \Delta \vec{\omega} \times \pmatrix{0 & R/2 & 0}$$.

An impulse of $$\vec{J} = \pmatrix{-10 & 0 & 0}\;\text{Ns}$$ applied at a distance $$b = \frac{I}{m R}= \tfrac{1}{2}R = 0.5\;\text{m}$$ above the center of mass has the following affect in the motion

\begin{aligned} \Delta \vec{v} & = \frac{1}{m} \vec{J} & \Delta \vec{\omega} & = I^{-1} ( \vec{b} \times \vec{J} ) \\ & = \pmatrix{-5 & 0 & 0}\;\text{m/s} & & = \pmatrix{0 & 0 & 5} \; \text{rad/s} \end{aligned}

with the final motion all zero for both rotation and translation.

So the momentum of the rolling disk is $$\vec{p} =-m \Delta \vec{v} = -\vec{J} = \pmatrix{10 & 0 & 0}\;\text{Ns}$$ and the percussion axis being horizontal through a point $$\vec{b} = \pmatrix{0 & 0.5 & 0}\;\text{m}$$ above the center of mass.

I think we should let the Original Gangster speak when talking about intuitive definitions of momentum. From Newton's Principia:

"The quantity of motion is the measure of the same, arising from the velocity and quantity of matter conjunctly.

The motion of the whole is the sum of the motions of all the parts; and therefore in a body double in quantity, with equal velocity, the motion is double; with twice the velocity, it is quadruple."

Since conjunctly means as a product. We can paraphrase this:

Momentum is the quantity of motion. Motion can double in quantity by having 2 things of equal size move at the same rate, or by having something go twice as fast. The two are equivalent.

These ideas have been updated by modern theory to some degree, but it is still safe to call momentum a measure of the quantity of motion.