# What is the definition of linear momentum?

Every where and book I search I get that the definition of linear momentum is the amount of speed (quantity of motion) contained in it or simply it is mass $\times$velocity? So, what is an appropriate definition of linear momentum? What did Newton think when he discovered it? He certainly did not think it as the amount of speed in a body.

• It is very important to distinguish between speed and velocity. Momentum is never defined as amount of speed at the link you provided. And what is ridiculous about $mass\ *\ velocity$? – Bill N Nov 4 '15 at 15:37

Newton thought of momentum as "Quantity of motion" - as we can see in the translated version of 'Principia'. Particularly, he defined momentum in the following words:

The quantity of motion is the measure of the same, arise from the velocity and quantity of matter conjointly.

So yeah, that is the definition of momentum. The question why we defined the momentum the way we defined is the actual question you have in mind, I think. Well, the answer can be thought of like this.

In physics, we actually try to find some combinations of some directly measured quantities of the objects whose appropriate summation remains constant in time - no matter whatever process the system is going through. There exists different such groups of terms of such a nature that summation of terms in group remain constant in time, but individual terms do not remain contstant in time, in general. We address each term of a particular group by one name - i.e. by the name of that group. Also we assign further names to individual terms of the group to further identify them.

Like there exists a group of terms $1/2 mv^2 + mgh + ...$ which remains constant in time. We call each term of it to be the energy and then further associate different identification with each term, like Kinetic Energy, Gravitational Potential Energy, etc.

A similar other group of terms is known and we call each term of it to be some momentum. Classically we had just summation of $m$v in this group. But later on, we found that there were some other terms also in this group and all of them together remain constant in time. For example the momentum of electromagnetic fields.

So we don't priori know how and what we should define as momentum. But we observe some traits in nature and to keep a track of these patterns we define some things and associate some intuition with it. (particularly while naming. Like Newton named it as quantity of motion.)

• I'm not sure you're using the word 'legacy' correctly. Does the dictionary definition agree with what you had in mind? – Emilio Pisanty Nov 5 '15 at 17:32
• @Emilio Thank you for notifying. You are correct. Legacy doesn't really describe what I meant. I think group/sequence can be proper words. Any other suggestions? – Dvij Mankad Nov 5 '15 at 17:44
• The current formulation is much better. – Emilio Pisanty Nov 5 '15 at 18:53

Newton (if I recall correctly) typically referred to the concept of inertia, which was an objects resistance to changes in velocity when subjected to external forces. You are right about him not thinking about it as just the speed of the object, because this is where the mass term comes in. Many people think of Newton's second law as being written as $F = ma$, and while this is true, I think that Newton liked to write it in terms of momentum as

$$F = \frac{dp}{dt}$$

Obviously, not with this notation since that came later. So, this can be thought of in the following way. The change in an objects momentum is equal to the force applied and the time interval over which the force is applied. Or,

$$\Delta p = F\Delta t$$

Starting from rest, this will give you the total momentum of the object, $p$. Considering inertia, objects with a higher mass will exhibit more resistance to velocity change. If the object doesn't change mass, then $\Delta p = m\Delta v$. Starting from rest, where $v_0 = 0$ and $p_0 = 0$, you end up with $p=mv$ for the total linear momentum of an object.

What you're looking for is an intuitive explanation or how you could visualize momentum. You can think of momentum as the quantity/amount of motion or "how much would I not want be in the path of this body."

I'm going to try and provide some intuition through a few examples:

A car of mass 1000 kg moving at 5 m/s would have the same "quantity/amount of motion" as a truck of mass 5000 kg moving at 1 m/s or a bicycle of mass 20 kg moving at 250 m/s.

If a rocket of mass 500 kg moving at at 250 m/s, uses up 100 kg of stored fuel to accelerate the rocket upto 312.5 m/s, although the rocket is moving faster, the mass has decreased by 100 kg and hence the "quantity of motion" has remained the same or "how much would I not want be in the path of this rocket" has remained the same.

Therefore, Netwon defined the force as something that changes the "amount of motion" that a body has in a certain amount of time, or:

$F = \frac{dp}{dt}$

Where p represents momentum and t represents time

I suppose Newton may have devised the momentum equation to numerically express how objects of exact speeds (but different densities) would create different effects upon impact and perhaps how much energy would be needed to move such objects to a given speed. Consider the following:

1. a wood ball (25g, 33.5 cc) hurled at a sheet metal target at an average speed of 9 m/s. p = 225
2. a lead sphere (380g, 33.5 cc) hurled at a sheet metal target at an average speed of 9 m/s. p = 3,420

Most people would probably intuitively know which of those objects would cause the greatest impact. We can calculate that the lead sphere needs over 15 times the momentum (force) to get to the same average velocity as the wood ball--which also means more force or energy is needed to move the lead to that average speed. It also means it's impact force is greater.

First of all you cannot separate linear from angular momentum. They work together just like linear and angular velocities do (or forces and torques).

I am going to answer your question from the perspective of geometry. The quantity of momentum is not so important as the geometrical construction that momentum implies. Let's see if you can follow:

All possible movements of a rigid body can be idealized as a rotation about some axis (finite or at infinity). This axis is actually a 3D line perfectly defined from the 6 motion components of a rigid body.

The motion causes the rigid body to have linear and angular momentum at the center of mass. This combination is actually a manifestation of an instantaneous linear momentum along another 3D line. This line is called The Axis of Percussion of the rigid body for the said rotation (also known as the sweet spot). This axis is perpendicular to the rotation axis and at a distance $r = \frac{\rho^2}{c}$ away from the center of mass. The pivot to center of mass distance is $c$ and the radius of gyration about the rotation $\rho$.

The geometrical interpretation of momentum is the axis by which if you apply an equal and opposite impulse the body is going to instantaneously stop rotating.

Read this answer for more details on how to find the axis of pair of free and line vectors like linear and angular momenta.

Essentially if you have a moving body with linear momentum $\boldsymbol{p} = m \boldsymbol{v}_{cm}$ and angular momentum at the center of mass $\boldsymbol{L}_{cm}=\mathrm{I}_{cm} \boldsymbol{\omega}$.

• The direction of the percussion axis is $$\boldsymbol{n} = \frac{\boldsymbol{p}}{\|\boldsymbol{p}\|}$$
• The location of the percussion axis relative to the center of mass is $$\boldsymbol{r} = \frac{\boldsymbol{p} \times \boldsymbol{L}_{cm} }{\| \boldsymbol{p} \|^2}$$

NOTE: $\times$ is the vector cross product.

A simple example is a thin rod of length $\ell$ (like a baseball bat) and mass $m$ rotating about one end. The center of mass is half way along the rod at $c=\frac{\ell}{2}$. Place the rod horizontally (along the x-axis) and rotate it about the z-axis with $\boldsymbol{\omega} = (0,0,\dot{\theta})$.

1. The linear momentum of this configuration is $\boldsymbol{p}=(0,m\, v_{cm},0) = (0,m \frac{\ell}{2} \dot{\theta},0)$
2. The angular momentum about the center of mass is $\boldsymbol{L}_{cm} = I_{cm} \boldsymbol{\omega} = (0,0,\frac{\ell^2}{12} m \dot{\theta})$ since the mass moment of inertia of a thin rod about the center is $m \frac{\ell^2}{12}$. Note that the radius of gyration about the center is $\rho =\sqrt{\frac{I_{cm}}{m}}= \frac{1}{\sqrt{12}} \ell$
3. The direction of the axis of percussion is along the y-axis $$\boldsymbol{n} = \frac{(0,m \frac{\ell}{2} \dot{\theta},0)}{\|(0,\frac{\ell}{2} \dot{\theta},0)\|} = (0,1,0)$$
4. The position of the percussion axis from the center is along the rod (x axis) $$\boldsymbol{r} = \frac{(0,m \frac{\ell}{2} \dot{\theta},0) \times (0,0,\frac{\ell^2}{12} m \dot{\theta})}{\|(0,m \frac{\ell}{2} \dot{\theta},0)\|} = \frac{(\frac{m^2 \ell^3 \dot{\theta}^2}{24},0,0)}{\frac{m^2 \ell^2 \dot{\theta}^2}{4}} = (\frac{\ell}{6},0,0)$$
5. The distance of the percussion axis (CoP) from the pivot is then $\frac{\ell}{2} + \frac{\ell}{6} = \frac{2}{3} \ell$ (same value as in slide 12 of http://www.iitg.ac.in/asil/Lecture-12.pdf).
6. You can compare the distance to the CoP with the expression $$\frac{\rho^2}{c} = \frac{ \frac{m}{12} \ell^2}{\frac{\ell}{2}} = \frac{\ell}{6}$$

So you can arrive at the center of percussion either from the shorthand expression $r=\frac{\rho^2}{c}$ or by the combination of the momentum vectors $$\boxed{ \boldsymbol{r} = \frac{\boldsymbol{p} \times \boldsymbol{L}_{cm} }{\| \boldsymbol{p} \|^2} }$$

For me momentum always describes the axis of percussion.

## protected by Qmechanic♦Nov 4 '15 at 14:40

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