Momentum

In real life, it seems obvious that stopping a baseball traveling at high speed is much easier than, say, a refrigerator hurdling towards you at high speed.  The mass, or weight, of the refrigerator is much greater than the baseball!  Everyone knows that it becomes increasingly difficult to move or change something as it becomes heavier, but how are we to describe this phenomenon using physics?

In order to represent the combination of an object's mass ($m$) and velocity ($\vec{v}$), a vector quantity called momentum, $\vec{p}$, is defined.  Notice that mass is a scalar and velocity a vector.  In our visible, daily lives, this equation exists simply as $\vec{p} = m \times \vec{v}$.  However there is another important factor that we must account for.

Certain experiments have shown that as particles travel closer and closer to the speed of light ($c = 3 \times 10^8$ m/s), the amount of interaction required in order to create an increase in velocity becomes increasingly large.  This is Einstein's "relativistic" definition of momentum, defined below.

In order to model the disparity between low speed velocity and velocity that approaches the speed of light, we introduce the scalar proportionality factor, gamma $\gamma$, which is equal to:  $$\gamma = \frac{1}{\sqrt{1-(\frac{|\vec{v}|}{c})^2}}$$
Using this proportionality factor, we can now write the correct equation to model momentum at all speeds:  $$\vec{p} = \frac{1}{\sqrt{1-(\frac{|\vec{v}|}{c})^2}} m \vec{v}$$ $$\vec{p} = \gamma \times m \vec{v}$$

Generally, gamma is approximately exactly 1 at low speeds, and is only effective at very high speeds.  We can simplify our momentum equations in low speed situations for the sake of ease down to:  $$\vec{p} = 1 \times m \vec{v} = m \vec{v}$$

Momentum is the integral of the net force ($\vec{F}_{net}$)$$\vec{p} = \int \vec{F}_{net}$$Using the definition of acceleration and its relation to velocity ($\vec{a} = \frac{\delta \vec{v}}{\delta t}$), we can define the approximate rate of change of momentum as$$\frac{\delta \vec{p}}{\delta t} = m\vec{a}$$$$\vec{F}_{net} = (\frac{\delta \vec{p}}{\delta t} = m\frac{\delta \vec{v}}{\delta t}) = m\vec{a}$$