## Impulse and Momentum Conservation and the Third law

A conserved quantity in physics is one that is unchanged during a time interval—typically a “before” and “after” some event. These statements are called “ConservationLaws.”

Let’s work out the most important one of these and get to know it. It’s inventor? Why, Mr Newton, which should be a surprise to no one. But Mr Huygens was ahead of Newton in many ways and when it came to collisions, he was king. So let's give Mr Huygens a short, one-clap round of applause. Like here:
A one-clap applause for Huygens. At the top, we aim both hands toward one another and in the bottom, they collide.
In each case, we carefully manage our clap so that our identical hands have the same speeds, but oppositely-directed. Since our hands are identical (?) they have the same mass. So each has a momentum value of $$p(\text{hand}) = m(\text{hand})\times v(\text{hand})$$ But since they are oppositely directed and since $v$ is a vector and so $p$ is a vector, they are negatives of one another and our diagram shows that I've chosen left as the negative direction. Now what happens: you brought your hands together with enough force to make a sound (I mean, why applaud if there's no sound?) and then what happened? They stopped dead...right in front of you. What happed to the momentum of the hands? Stopping dead is just a colorful way to say that they both lost their velocities and so they both lost their momenta. Look at this example in order to see how we construct an important argument based on this lauditory expression of appreciation for our Dutchman. It's based on an analysis of the impluse and out of it falls...Newton's third law. ```{admonition}   Please study Example 1: :class: warning Applause for Huygens ``` The bottom line is that the total momentum at the beginning of this hand-collision is: $$ \begin{align*} p(\text{total}) &=p(\text{left}) + p(\text{right}) \\ &= p - p \\ &= 0 \end{align*} $$ And by just looking at your hands *after the collision*, the total momentum is also $0$. They're the same. >The momentum of the whole system (two hands) *at the beginning of this event* is the same as the momentum of the whole system (still, two hands) *at the end of the event*. This is always the case! The total momentum of a system never changes! Let's imagine that we have two objects colliding, each with it's own momentum. Say, like Collision 2, which I'll redraw here:
TIME GOES FROM TOP TO BOTTOM IN THREE STEPS: Collision 2: $\\mathbf{a}+\\mathbf{B}\\to \\mathbf{a}+\\mathbf{B}$
Our little $a$ and bigger $B$ don't necessarily have the same momenta...they could be recoiling with different speeds and we'l look at some examples. But what we can say---AND WILL ALWAYS SAY---is that the total before is equal to the total after. Here's what this means in symbols. Admire it: $$ \begin{align} p(\text{total, before}) &= p_{a,0} + p_{B,0} \nonumber \\ p(\text{total, after}) &= p_{a} + p_{B} \nonumber \\ \text{and, since } p(\text{total, before}) &=p(\text{total, after}) \nonumber \\ p_{a,0} + p_{B,0} &= p_{a} + p_{B} (\#eq:momcons) \end{align} $$

Take a deep breath and remember where you were at this moment. Seriously.

Equation \ref{eq:momcons} is a really important relationship. First, it's the statement of “momentum conservation.” Second, it is what's behind the otherwise, bumper-stickery, out-of-the-blue-sounding "For every action there's an equal and opposite reaction." Newton's third law? Just a memorable and wordy way to state momentum conservation…where the rubber really meets the road, so to speak. Here’s one of the apparently true things about the universe:

Momentum is always conserved.

...drops mic…which bounces, according to momentum conservation.