# Comparing the Impulse-Momentum and Work-Energy theorems

#### Prerequisites

- Restating Newton's 2nd law: momentum
- System Schema Introduction
- Kinetic energy and the work-energy theorem

Newton's second law (N2),

$$\overrightarrow{a}_A= \frac{\overrightarrow{F}^{net}_A}{m_A}$$

is the key result in our study of motion that tells us how objects (defined however we want to define them -- even as parts of things) respond to the forces they feel. Since there can be lots of forces and lots of objects, and the resulting motions can be quite complicated, we have manipulated N2 in a variety of ways to allow us to see the content it expresses in a variety of different ways.

One way is by defining momentum. This allows us to see a force acting on an object for a certain amount of time adding something to the object that represents its motion (the momentum). The amount of momentum a force adds to an object equals the force times the time it acts (or, better, the integral of the force over the time).

$$\Delta\overrightarrow{p}_A = \int_{t_i}^{t_f}\overrightarrow{F}^{net}_A dt$$

A second way is by defining kinetic energy. This allows us to see a force acting on an object over a certain distance as adding something to the object that represents its speed (the kinetic energy). The amount of kinetic energy a force adds to an object equals the force in the direction of motion times the distance the object moves (or, better, the integral of the dot product of the force with the displacement).

$$\Delta \big({1 \over 2} \; m_A v_A^2 \big) = \int \overrightarrow{F}^{net}_A \cdot \overrightarrow{dr}$$

Something interesting happened to the momentum when we looked at a system with just two objects that interact with each other.

Recall that each interaction between a pair of objects is a force that when considered on one of the objects is in one direction and when considered on the other object is an equal force in the opposite direction. (Newton's 3rd law) The result was that if we add together the impulse momentum theorems for the two objects, we get

$$\Delta (m_A\overrightarrow{v}_A) = \overrightarrow{F}_{B \rightarrow A} \Delta t$$

$$\Delta (m_B \overrightarrow{v}_B) = \overrightarrow{F}_{A \rightarrow B} \Delta t$$

$$\Delta (m_A\overrightarrow{v}_A + m_B \overrightarrow{v}_B) = \overrightarrow{F}_{B \rightarrow A} \Delta t + \overrightarrow{F}_{A \rightarrow B} \Delta t = (\overrightarrow{F}_{B \rightarrow A} + \overrightarrow{F}_{A \rightarrow B}) \Delta t = 0$$

That 0 happens because the two objects interact over the same time interval, so Deltat can be factored out, and then N3 results in the forces cancelling. The result is momentum conservation: when two objects interact, they change each other's momentum in opposite ways so that the total change is 0.

What if we try to do something similar for the work energy theorems for a pair of interacting objects? Again, something interesting happens – but we no longer get a simple cancellation.Here are the Work-Energy theorems for two interacting objects moving a small distance in a time Δ*t*.

$$\Delta \big({1 \over 2} \; m_A v_A^2 \big) = \overrightarrow{F}_{B \rightarrow A} \cdot \overrightarrow{\Delta r}_A$$

$$\Delta \big({1 \over 2} \; m_B v_B^2 \big) = \overrightarrow{F}_{A \rightarrow B} \cdot \overrightarrow{\Delta r}_B$$

If we add them together and try to use N3 to simplify, we find that they don't cancel like they did in the momentum case.

$$\Delta \big({1 \over 2} \; m_A v_A^2 + {1 \over 2} \; m_B v_B^2\big) = \overrightarrow{F}_{B \rightarrow A} \cdot \overrightarrow{\Delta r}_A + \overrightarrow{F}_{A \rightarrow B} \cdot \overrightarrow{\Delta r}_B$$

$$\Delta \big({1 \over 2} \; m_A v_A^2 + {1 \over 2} \; m_B v_B^2\big) = \overrightarrow{F}_{B \rightarrow A} \cdot (\overrightarrow{\Delta r}_A - \overrightarrow{\Delta r}_B)$$

$$\Delta \big({1 \over 2} \; m_A v_A^2 + {1 \over 2} \; m_B v_B^2\big) = \overrightarrow{F}_{B \rightarrow A} \cdot \overrightarrow{\Delta r}_{AB}$$

where $\Delta r_{AB}$ means the change in the relative position of the two objects.

In the momentum case, because the times of interactions were the same, we could factor it out from the impulse terms and the forces cancelled by N3. In the kinetic energy case, the two objects could each be moving – so in the same time intervals they might each move different distances. As a result, if we use N3 and factor out the force it doesn't cancel. Instead it is multiplied (with the dot product to get the right part of the force for changing the speed) by the change in the relative position of the two objects.

This means that we no longer get a conservation theorem: the change in the kinetic energy of two interacting objects is not 0: it is equal to the change of a new quantity, *the potential energy of interaction*:

$$ \Delta U_{AB} = \overrightarrow{F}_{A \rightarrow B} \cdot \overrightarrow{\Delta r}_{AB}$$

(The signs are a little tricky here. We'll work them out carefully in a later page.)

What this shows is that when two objects interact with the right kind of force* the total change in there kinetic energy is balanced by an opposing change in a new kind of energy. We can still get a conservation theorem, but it is a little trickier than momentum conservation.

This also shows that the potential energy belongs to the interaction of the two objects, not to one or the other of the objects. This can sometimes be confusing, because one example of potential energy is gravitational. In this case, because we only care about the gravity from the earth (since gravity is such a weak force that we need at least one huge object to see it), and since the earth doesn't respond much to the forces from other objects (since its mass is so big), we can treat the gravitational PE as if it belongs to the small object and forget about the earth. This is less appropriate in the case of electrical forces (though we can sometimes get away with it).

* The action at a distance forces of gravity and electricity work to give us a potential energy, as do spring forces, but resistive forces do not. The work-energy term is OK, but you can't write it as a change in some potential function, U.

Joe Redish 11/19/13