When you examine the modern interpretation of Newton’s Laws of motion, texts often refer to a discussion of forces. In truth, Newton’s laws are talking about Inertia in motion.

Lex 1: Every body perseveres in its state of rest, or of uniform motion in a right line, unless it is compelled to change that state by forces impressed thereon.

Lex 2: The alteration of motion is ever proportional to the motive force impressed ; and is made in the direction of the right line in which that force

is impressed.

The language of the time has a somewhat different meaning than today as much of the terminology was new. Newton is talking about inertia in motion, which in modern times we call momentum. We define the momentum **p**, in terms of inertia and velocity.

**p**=m**v**

Note that momentum and velocity are in bold type because they are vectors, they have a magnitude and a direction. Mass is a scalar and thus has no direction. So when Newton refers to the *alteration of motion*, he is actually talking about a *change in momentum*.

Newton did not originate the ideas of momentum, but merely summed up the ideas so elegantly in the Principia that the concept truly entered the mainstream. Descartes laid the grounding for the idea of momentum by talking about the “*quantity of motion*.” However, Descartes was talking about mass and speed, and as we will find out later, this distinction to include the vector nature is critically important. Galileo discussed a similar idea calling it “*impetus*.” John Wallis, another British philosopher can be given the credit for coining the term *momentum*. In fact, most of the credit for Newton’s first two laws should actually be given to Wallis.

Example: What is the momentum of a 5.0 *kg* bowling ball rolling forward at a speed of 3 *m/s*?

m = 5.0 *kg *

* ***v*** = *3.0 *m/s *

* ***p*** = ??? *

**p** = m**v** = (5.0 *kg)(*3.0 *m/s) = *15 *kg·m/s
*

You will find that momentum does not have its own unit and is defined in terms of other units.

Coming back to Newton’s Second Law, we examine the phrase “*motive force impressed*.” This actually refers to a force exerted for a duration of time, which we formally call ** impulse**. In modern language, we could say that an impulse,

**J**, causes a change in momentum. This can be formally written as

**J = F**Δt=Δ**p**

where F is the force causing the change in momentum, t is the duration of time the force is exerted for, and Δ**p** is the change in momentum. This is sometimes known as the **Force-Impulse Relationship**. In most textbook cases, when we refer to a change in momentum, we are also referring to a change in velocity.

Δ**p = **mΔ**v**

However, a change in momentum could also be due to a change in mass. A rocket engine burning fuel is a great example of a change in mass. In this course, we will limit our calculations to changes in velocity. Using this definition of Newton’s second law, we can quickly see where our modern force base interpretation comes from.

**J = F**Δt=Δ**p**

Substituting in the change in momentum we get

**F**Δt=** **Δ**p = **mΔ**v**

Now, if we divide both sides by ** **Δt

**F** **= **m (Δ**v/**Δt) = m**a**

Where we arrive at our familiar equation for Newton’s Second Law. At this point, I want to also point out another common set of units for momentum. Looking at the definition of impulse, you can see that the *Newton-second* or *N·s* is equivalent to the *kg·m/s.*

As mentioned above, momentum is a vector and has a direction. This because critical when examining changes in momentum involved with interactions such as an accelerating force, or even simple collisions. Consider a bouncing ball. A ball will have the opposite momentum after it bounces off of a wall than just before it hits (assuming no change in speed). This can be accounted for when we define momentum as a vector.

Δp = p_{f} – p_{i}

If the speed is the same, then the change in momentum is actually twice the original.

Example: A 1 *kg* dodge-ball hits the wall at a speed of 4 *m/s*. It rebounds from the wall at the same speed. What is its change in momentum of the dodge-ball?

m = 2 *kg *

v_{i}= 4 *m/s*

v_{f} = -4 *m/s *

Δp = ?????

* *Δp = p_{f} – p_{i }= (1 *kg)(*-4 *m/s**) – *(1 *kg)(*4 *m/s**) *= 8 *kg·m/s*