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# 7.1: (Linear) Momentum Conservation Model

[ "article:topic", "authorname:ucd7" ]

### Overview of the Model

As you begin this chapter, listening in lecture and working in DL, it must seem, at least at first, that you are being introduced to a lot of new concepts. The representation of the motion of an object and the forces acting on an object are necessary ideas to understand before we can fully understand this new conserved quantity called momentum. One goal of this (and the next) chapter is to understand the effects of forces on motion and we begin to do this in this chapter through a discussion of momentum and transfers of momentum.

We introduce two concepts which are completely new: momentum and impulse. However, we are taking great pains to help you see how these concepts play roles very similarly to energy and work. So, yes, you have to memorize that momentum, $$p$$, is the product of mass and velocity $$(p = mv$$). And you have to be careful to not forget that momentum has vector properties, just as velocity does. But, impulse is not an isolated construct you file away in your brain somewhere. Rather, you should really strive to understand impulse in analogy to work. A transfer of energy as work changes the energy of a physical system. Similarly, a transfer of momentum as an impulse changes the momentum of a system. Energy is conserved. Momentum is also conserved. Of course, there are differences between momentum and energy and between impulse and work. As you work in DL, as you study this text, as you work the FNT’s, and as you interact with other students and with instructors as you mentally struggle with this material, try to understand these new concepts in relation to what you already know, rather than as simply some more isolated facts that you memorize.

### Review and Extension of the “Before and After” Interaction Approach

In Chapters 1 and 2 we focused on changes in the energy of a physical system. Energy has meaning for one particle or 1023 particles, for objects as small as the nucleus of an atom and as large as a galaxy. It really is a universal concept that applies to any physical system. It turns out that there are two other concepts that are like energy in that they are universally applicable, are transferred among systems as a result of interactions, and the amount transferred gives very useful information that does not depend on the details of the interaction. These are the concepts of momentum and angular momentum.

#### Integrating “the Agent” of Interactions

We have called force the “agent of interactions”. Interactions occur between objects as they exert forces on each other. Objects experience changes in energy when other objects exert forces on them and do work on them. We recall that the amount of energy change caused by a force is the integral of the force over a distance. This integral is called the $$work$$ done on a system. The only component that contributes to the work, however, is the component of the force that is parallel to the motion. We usually indicated this component with the symbol $$F_{||}$$:

$W = \int _{x_i} ^{x_f} F_{||}(x) dx = E_f - E_i = \Delta E \tag{7.1.1}$

Or, if $$F$$ is constant, or we define an average force $$F_{avg}$$, we can write

$W =F_{avg|| }\Delta x = E_f - E_i = \Delta E \tag{7.1.2}$

In other words, the parallel component of force integrated over the path of the motion is the work, and this work equals the amount of energy transferred to the system due to the application of the force by an object outside the system.

A similar integral of the force is equal to the change in momentum of the system. But instead of integrating over distance, we now integrate over time. This integral is called the impulse of the force, $$F$$. We represent the impulse with the symbol $$J$$.

$J = \int _{t_i} ^{t_f} F_{||}(t) dx = p_f - p_i = \Delta p \tag{7.1.3}$

Or, if $$F$$ is constant, or we define an average force, $$F_{avg}$$, then

$J = F_{avg} \Delta t = p_f - p_i = \Delta p \tag{7.1.4}$

Impulse is a vector quantity and causes a change in a vector property of a system: specifically, a change in the linear momentum, $$\Delta p$$. The change in momentum, is of course, independent of what Galilean reference frame we choose to measure the momenta in.

Note on units: Force has SI units of newtons, of course. Impulse must therefore have units of newton seconds, $$N \times s$$. Momentum, the product of mass and velocity, must have SI units of kilogram meter per second, kg m/s. Since these two quantities are equated, these units must be equivalent, as you can show using the relation $$N = kg~ m/s^2$$.

### Linear Momentum

The linear momentum of an object is simply the product of the object’s mass and velocity: $$p = mv$$ Linear momentum incorporates the notion of inertia, expressed as mass, as well as the speed and direction of motion. In some ways it is similar to kinetic energy, 1/2 mv2, but an obvious difference is that momentum has a direction; it is described as a vector. (Often, the word “momentum” is used without the modifier “linear,” when talking about linear momentum. Later, however, the modifier “angular” is always used when talking about angular momentum.)

Example 7.1.1: Calculating Momentum: A Football Player and a Football

(a) Calculate the momentum of a 110-kg football player running at 8.00 m/s. (b) Compare the player’s momentum with the momentum of a hard-thrown 0.410-kg football that has a speed of 25.0 m/s.

Strategy

No information is given regarding direction, and so we can calculate only the magnitude of the momentum, $$p$$ (As usual, a symbol that is in italics is a magnitude, whereas one that is italicized, boldfaced, and has an arrow is a vector.) In both parts of this example, the magnitude of momentum can be calculated directly from the definition of momentum given in the equation, which becomes

$p = mv$

when only magnitudes are considered.

Solution for (a)

To determine the momentum of the player, substitute the known values for the player’s mass and speed into the equation.

$p_{player} = (110 \space kg)(8.00 \space m/s) = 880 \space kg \cdot m/s$

Solution for (b)

To determine the momentum of the ball, substitute the known values for the ball’s mass and speed into the equation.

$p_{ball} = (0.410 \space kg)(25.0 \space m/s) = 10.3 \space kg \cdot m/s$

The ratio of the player’s momentum to that of the ball is

$\dfrac{p_{player}}{p_{ball}} = \dfrac{880}{10.3} = 85.0$

Discussion

Although the ball has greater velocity, the player has a much greater mass. Thus the momentum of the player is much greater than the momentum of the football, as you might guess. As a result, the player’s motion is only slightly affected if he catches the ball. We shall quantify what happens in such collisions in terms of momentum in later sections.

#### Temporary restriction to non-rotating objects and center of mass

Until we consider rotation of objects in the next model/approach, Angular Momentum Conservation, we will consider phenomena in which extended objects act only like point particles. A useful construct that will become much more meaningful when we consider rotation, is center of mass. Right now we can simply consider that any extended object acts like a single particle whose mass is equal to the mass of the object, located at the special point, the center of mass.

### The Construct of Net Force and net Impulse

Be sure to review the discussion of Net Force in Chapter 6. The central point is that the effect of all forces acting on an object can be represented by a single vector construct called the unbalanced force or the net force, $$\sum F$$. When we use the concept of impulse, it is sometimes useful to consider the impulse, $$J$$, due to a particular force; we would write this as:

$J_A = F_A \Delta t \tag{7.1.5}$

$$(\text{The subscript “A” tells us that the impulse is due to particular force exerted by the object “A”})$$

However, the power of the construct of impulse comes into play when we consider the impulse of the net force; we write this as

$J_{net} = J = \sum F \Delta t = \Delta p \tag{7.1.6}$

$$(\text{J without a subscript usually means the impulse due to the net force}.)$$

In words, $$\text{the net impulse is equal to the change in the linear momentum of the system}$$. We explore this relationship further below.

### Momentum of a System of Particles

The momentum of a single object is simply the product of its mass and velocity. Suppose we define a physical system that contains several particles which move with different velocities. The total linear momentum of this physical system is the vector sum of the individual linear momenta.

$p_{system} = p_1+p_2+p_3+...=\sum p_i \tag{7.1.7}$

If the particles in our system interact with each other, they exert forces on each other, and there will be an impulse associated with each of these forces. Newton’s 3rd law tells us, however, that the impulse that particle a, for example, exerts on particle b is equal in magnitude and opposite in direction to the impulse exerted by particle b on particle a. And using the relation that the impulse is equal to a change in momentum of a particle, we see that the change in momenta of particles $$a~and ~b$$ due to their interaction will be equal in magnitude, but opposite in direction.

Generalizing the above argument to interactions between any of the particles within the system, we see that if the momentum of one particle changes a certain amount, another particle’s momentum changes the same amount in the opposite direction. Thus, when we sum over all the momenta of the system, the total momentum of the system does not change in response to interactions among the particles within the system.

However, if the particles of our system interact with particles (objects) outside the system, then the total momentum of the system might change. The figure below shows some of the forces that might be acting on the particles of the system. Some, labeled int (for internal) don’t change the total momentum of the system. The forces labeled ext (for external) do change the momentum of the system.

Figure 7.1.1

Of the various impulses shown in the figure 7.1.1, only the impulse caused by $$F_{ext ~on ~c}$$ causes a change in momentum of the system of particles.

### Statement of Conservation of Momentum

So, for a system of particles (objects) it is useful to write the impulse/momentum relation in a way that emphasizes the external interaction:

$Net ~Impulse_{ext} = J_{ext} = \int \sum F_{ext}(t) dt = p_f - p_i = \Delta p_{system} \tag{7.1.8}$

A system acted on by external forces undergoes a change in total linear momentum equal to the net impulse (total impulse) of the external forces. We can rephrase the relationship stated above as a conservation principle for the total momentum of a system of particles:

Note: Conservation of Linear Momentum

If the net external impulse acting on a system is zero, then there is no change in the total linear momentum of that system; otherwise, the change in momentum is equal to the net external impulse.

This statement is an expression of Conservation of Linear Momentum. The total linear momentum of a system of objects remains constant as long as there is no net impulse due to forces that arise from interactions with objects outside the system. It does not matter that the objects of the system interact with each other and exert impulses on each other. These internal impulses cause changes in the individual momenta of the objects, but not the sum or total momentum of the systemof objects.

We can rephrase this discussion in terms of open and closed systems:

1. Closed system - A closed system does not interact with its environment so there is no net external impulse. The total momentum of a closed system is conserved. That is, the total momentum of the system remains constant.
2. Open system - An open system interacts with its environment, so that it can exchange both energy and momentum with the environment. For an open system the change in the total momentum is equal to the net impulse added from the environment–from objects outside the system.