Restating Newton's 2nd law: momentum


Defining momentum

Newton's 2nd law tells us that an object's velocity changes in response to unbalanced forces:

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

Consider, for example, a ball thrown straight upwards. Let's assume that for the speeds and distances involved we can ignore the effect of resistive forces on the ball. What our Newton's 0th law (object egotism) tells us is that once the ball leaves our hand, the hand can exert no more force on the ball. Nothing is touching it (we are ignoring the effect of the air) so the only force it feels is its weight — the gravitational force of the earth pulling it down. But the hand is what made it move! What did it give to the ball? 

Newton 2 doesn't formulate it in that way. It just talks about the acceleration of the object — how its velocity changes. But if we re-write N2 in a slightly different way we can create a description that identifies a moving object as "carrying something".

Let's write N2 for the object A by expressing it in a way that shows the force as changing something belonging to the object.

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

$$\overrightarrow{F}^{net}= m_A \frac{d\overrightarrow{v}_A}{dt} = \frac{d(m_A \overrightarrow{v}_A)}{dt}$$

Since the object's mass is a constant, it's OK to bring it inside the derivative. When we write it this way, it shows us that a net force changes something belonging to the object — its mass times its velocity. This is a rather nice quantity to carry information about the the "oomph" of an object — the intensity of its motion; something you would have to get rid of (using a net force) in order to stop it.

Momentum has the following nice properties:

  • It's bigger for a bigger mass moving with the same velocity.
  • It's bigger for the same mass moving with a faster velocity.
  • It has a direction (is a vector).
  • It's what the net force changes directly with no intervening factors.

Let's give this combination a name: the object's linear momentum.

$$\overrightarrow{p}_A = m_A \overrightarrow{v}_A$$

With this, N2 can be reformulated as:

$$\overrightarrow{F}^{net}_A= \frac{d \overrightarrow{p}_A}{dt}$$

The Impulse-Momentum Theorem

We can integrate this, and express the result in a variety of ways. It we assume that the net force on the object is constant, we can take a (not very small) time interval and express the change in momentum produced by a force acting for a time interval:

$$\overrightarrow{F}^{net}_A= \frac{\Delta \overrightarrow{p}_A}{\Delta t}$$

$$\Delta \overrightarrow{p}_A = \overrightarrow{F}^{net}_A \Delta t$$

If the force is a function of time and is changing over the time interval we want to pay attention to, we can't treat it as a constant. We have to take small time intervals and add them up — do an integral.

$$d\overrightarrow{p}_A = \overrightarrow{F}^{net}_A dt$$

$$\int_{p_A^i}^{p_A^f} d\overrightarrow{p}_A = \int_{t_i}^{t_f}\overrightarrow{F}^{net}_A dt$$

$$\overrightarrow{p}_A^f - \overrightarrow{p}_A^i = \int_{t_i}^{t_f}\overrightarrow{F}^{net}_A dt$$

where the superscripts i and f mean initial and final since the integrals are done from the starting to the ending conditions. The integral of dp is just of p evaluated at the endpoints, we just get the change in the momentum. We can therefore write even the integral form this way:

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

The change in momentum is equal to the force times the time step added up over time.

Since the force times the time is what we have called the impulse these results are known as The Impulse-Momentum Theorem.

In this form, the story of the ball thrown straight up takes a nice form.

As the hand starts to move up, it acts over time adding momentum to the ball. When it releases the ball it has given the ball a substantial amount of upward momentum. The ball continues upward as a result of that momentum, but now gravity is an unbalanced force on the ball pointing down so, acting over time, gravity reduces the upward momentum until it has acted for enough time to take the momentum all the way down to 0. At that instant, the ball has 0 momentum and therefore 0 velocity — it's at the top. But gravity is still unbalanced and keeps acting. Now it adds downward momentum and the more time it acts the more downward momentum it gives the ball.

We now can see the force as being the rate of change of the momentum and the confusion between force and momentum as a failure to distinguish between a quantity and its rate of change -- a quite common error.

Dangerous bend! Momentum is defined as a vector so it has a direction. To change an object's direction of motion requires a net force, just as any other change to its velocity. In my experience, students beginning to deal with momentum often forget this and treat momentum as if it were a scalar. This can easily lead to incorrect conclusions!

Joe Redish 10/18/11


Article 396
Last Modified: February 6, 2019