\(\def \u#1{\,\mathrm{#1}}\) \(\def \abs#1{\left|#1\right|}\) \(\def \ast{*}\) \(\def \deg{^{\circ}}\) \(\def \ten#1{\times 10^{#1}}\) \(\def \redcancel#1{{\color{red}\cancel{#1}}}\) \(\def \BLUE#1{{\color{blue} #1}}\) \(\def \RED#1{{\color{red} #1}}\) \(\def \PURPLE#1{{\color{purple} #1}}\) \(\def \th#1,#2{#1,\!#2}\) \(\def \lshift#1#2{\underset{\Leftarrow\atop{#2}}#1}}\) \(\def \rshift#1#2{\underset{\Rightarrow\atop{#2}}#1}}\) \(\def \dotspot{{\color{lightgray}{\circ}}}\)
Chapter 5: Impulse and Momentum
2.

Momentum

Momentum
\(\vec p\)
Ns
The momentum of an object is the product of its mass and its velocity, and is represented by the letter $\vec p$:
$$\vec p=m\vec v$$

It has the same SI units as impulse does (Ns, which can also be written as $\u{kg\cdot m/s}$). The momentum is a vector which points in the direction that the object is moving, and there is a relationship between the momentum and the impulse: when an impulse acts on an object, it changes its momentum according to

$\vec J=\Delta\vec p=\vec p_f-\vec p_i$
$\text{or }\vec p_f = \vec p_i + \vec J$

This is incredibly useful, as we will see in the following pages.

We can actually prove this relationship, if you're interested:

According to Newton's Second Law, $$\RED{\vec F_{avg}} = \RED{m\vec a_{avg}}$$
The average acceleration is related to the change in velocity $$\begin{align} \vec a_{avg}&=\frac{\vec{v}_f-\vec v_i}{\Delta t}\\ \implies \BLUE{\vec a_{avg}\Delta t} &=\BLUE{\vec v_f-\vec v_i}\\ \end{align}$$
The Impulse is equal to $$\begin{align} \vec J&=\RED{\vec F_{avg}}\Delta t\\ &=(\RED{m\vec a_{avg}})\Delta t\\ &=m(\BLUE{\vec a_{avg}\Delta t})\\ &=m(\BLUE{\vec v_f-\vec v_i})\\ &=m\vec v_f-m\vec v_i\\ \end{align}$$
And since momentum is $\vec p=m\vec v$, $$\vec J=\vec p_f-\vec p_i$$