What is momentum? Well, like energy, it doesn't really have a good definition outside of a mathematical one. The best definition I can think of is that momentum describes how hard an object is to stop, but there are several problems with this definition. Most sources just cite the mathematical definition. We will only deal with linear momentum for now, so here's its definition:
$$ \vec{p} = m \vec{v} $$ This equation is very unassuming. Now, this isn't very useful unless we can connect it to previous concepts, which is where impulse ($J$) comes in. Impulse is defined as the product of the force and the time over which it acts. It has a unit of newton-seconds ($\textrm{N}\cdot \textrm{s}$). If you pay close attention, these units can be converted into $\textrm{kg~m/s}$ , which are the units of momentum. Wonder what that could mean?
$$ J = F \Delta t $$ For a varying force, the impulse can be calculated with an integral. You can probably guess that it's the integral of the force over the time period.
$$ J = \int_{t_1}^{t_b} F~dt $$ If the graph is simple enough, it may be easier just to calculate the area under the force v. time graph with geometry. For a force that changes with time, the impulse can still be found as the area under the force versus time graph.
Now, this formula isn't very useful. After all, what even is impulse? What does it do? Why do we even care?
Figure 1: You are very cool if you get the reference.
Well, Minos Prime (that's the guy's name) isn't exactly correct here. You see, there's something called the impulse-momentum theorem that connects the concepts of impulse and momentum. It's actually an extremely simple relation.
The impulse-momentum theorem: $$ J = \Delta p $$
Yep. That's all there is to it. Impulse is merely the change in momentum. Of course, when calculating, it is good to note that this is the impulse exerted by the net force. Also, the change in momentum is usually $m\Delta v$, but not always. You'll see what I mean.
Now we can do a short practice problem to hammer home the concept.
How much impulse does gravity exert on a 2.0 kg object as it slides down a 30 degree inclined plane in three seconds? Using this, how much does the velocity change?
Hopefully, you realize that only the portion of gravity pointed down the incline is exerting an impulse. This gives us a force of $F = mg\sin \theta = \frac{mg}{2}$. We can now easily find the impulse.
$$ J = F t = \dfrac{mg}{2}t = \bbox[3px, border: 0.5px solid white]{29.4 ~\textrm{N} \cdot \textrm{s} }$$ After we have found this, we can use the impulse-momentum theorem to calculate the velocity change:
$$ J = \Delta p = m \Delta v $$ $$ \Delta v = \dfrac{J}{m} = \bbox[3px, border: 0.5px solid white] {14.7 ~\textrm{m/s} } $$ We get the same result if we use kinematics techniques, letting us know that our answer is correct.
The second thing I'd like to talk about here is Newton's Second Law. You might think that was a typo or that I've completely gone insane, but that was intentional. (I'm not going to comment about insanity.) The form we gave for Newton's Second Law is a generalized form for the cases where mass is constant.
The form of this new second-law variant involves a derivative, as well as the concept of momentum.
$$ F = \dfrac{dp}{dt} $$ This was actually Newton's preferred form of his second law, since it's more general and can deal with more cases. We can also write this as an average force:
$$ F_{av} = \dfrac{\Delta p}{\Delta t} $$ The exact form is one that uses calculus, but essentially it states that the force on an object is equal to the rate of change of its momentum. We can write this as an average value instead to avoid the calculus here:
$$ F_{av} = \dfrac{\Delta p}{\Delta t} $$ Now, think about how this translates back to the regular form of Newton's Second Law. I said that the form we used earlier assumed that mass was constant, so we can simply put this assumption back in and see if we can make any substititions. (Hint, hint: we can.)
Since mass is constant, we can rewrite $\Delta p$ as $m \Delta v$. This lets us substitute this into our equation for average force:
$$ F_{av} = \dfrac{m \Delta v}{\Delta t} $$ Now, something should look familiar here. The term $\Delta v / \Delta t$ is equal to the average acceleration!
$$ F_{av} = m a_{av} $$ We can drop the average symbols. The exact reasoning here is a little sketchy, but my idea is that both values are averages, so you can make them both instantaneous. In any case, the calculus version of this proof makes much more sense and is much more logically rigorous. However, this is probably good enough for a basic understanding. Now, we can go back to our familiar form of the Second Law:
$$ F = ma $$ We can prove this more rigorously using calculus, by following similar logic. We first start by defining $dp = m dv$ and then use identical logic to arrive back at the same result. I'll leave the exact details for you to do.
In any case, the result has significance in that it's not restricted to the case where mass is constant. This allows us to handle problems we previously were not able to. I'll give you one very common and illustrative example on the next page. This can be easily shown to be equivalent to our previous statement of Newton's Second Law. If you're interested in all the details of the math, you can see our Algebra-based level. The basic idea behind it is that if we assume that mass is constant, then the formula simplifies to the familiar $F=ma$ result. First, we can write:
$$\Delta p = m \Delta v$$ This is only true if mass is constant. The term for change in momentum can technically have both a change in mass and a change in velocity, so we have to know mass is constant in order to write it in this form. Most of the time, we've dealt solely with unchanging mass, but this can change.
After plugging this into our previous equations, we can simplifiy our result using the basic kinematics relation that $a = \Delta v/ \Delta t$ to arrive at our familiar form of Newton's Second Law:
$$F = ma $$ This new expression is not really all that useful for most cases since 90% of the time, mass will be constant. However, it is more versatile and allows us to tackle problems involving a changing mass.
A cart of mass $m$ slides along a frictionless rail with some velocity $v$. At a certain point in the track, the cart starts being filled up with sand that falls vertically into the cart at a constant rate $R$, where $R$ describes the mass of the sand that is added to the cart per time interval. What is the force that must be exerted on the cart to keep it going at a constant velocity?
Figure 2: Magic pocket sand! Now, you might be confused why we would need to exert a force to keep the cart moving, since there's no friction. Doesn't Newton's First Law state that an object will remain in its state of motion if no net force acts on it? Therefore, we wouldn't need to exert any force...right?
Well, no. First off, the cart remaining at the same speed would violate the conservation of energy, since the mass is increasing which means the kinetic energy is increasing without anything else changing. Therefore, we would need to exert a net force to maintain the cart's motion. The cart also does experience a net force in the form of the sand hitting the back of the cart since it falls in without any horizontal velocity. We need to use the new form of Newton's Second Law to solve.
There is no velocity change, but there is a mass change. Therefore, we can write the change in momentum as:
$$ \Delta p = \Delta m v$$ Now, we can use our force formula, which we don't have to take the average of because the force required is constant.
$$ F = \dfrac{\Delta p} {\Delta t} = \dfrac{\Delta m} {\Delta t} v $$ Now, we recognize that the rate of change of mass $R$ is equal to $\Delta m/ \Delta t $, which lets us write our final answer:
$$\bbox[3px, border: 0.5px solid white]{F = R v} $$ There we go! The logic and reasoning required can be a bit tricky at first, but the actual calculations are very easy once you understand what's going on. Mass isn't always constant, and knowing this technique lets you expand your horizons.
Here's an interesting follow-up question for you. Now, what would happen if sand was leaking from the cart instead?
Well, you might jump to conclude that the opposite would just be true, that the cart would speed up. However, this isn't the case, as you might have realized by thinking of real-life examples! A car leaking oil doesn't suddenly go faster.
The sand is leaking from the cart, which means that it will fall out with zero relative velocity to the cart. This means it takes kinetic energy away from the cart, causing it to remain at the same speed despite its mass having decreased. It would just continue forward at the same speed as before.
That being said, it is possible for this to speed up the cart. The sand would have to be expelled from the cart with some relative velocity to the cart, meaning we would do something like launching it out the back. This is the principle of action of rocket engines, which launch hot gases at high velocities in order to propel themselves forward. Who knew we would glimpse into rocket science here? Let's keep it conceptual here. A cart slides along a frictionless rail in a straight line. At a certain point, sand begins to fall vertically into the box at some constant rate. What would happen to the cart?
Figure 2: Magic pocket sand! You might assume that nothing would happen, since according to Newton's First Law an object will tend to remain in its state of motion unless acted on by a net force, and there seems to be no obvious net force here. However, consider it this way. The cart would be gaining kinetic energy out of nowhere if it kept moving at the same speed, since its mass is increasing. Therefore, something has to be happening.
The key is to analyze the following equation to see what would happen. Since all rates and such are constant, we can assume the force would be constant too.
$$ F = \dfrac{\Delta p}{\Delta t} $$ What exactly does this do? Well, it tells us that some net force is being exerted on the cart because its momentum is changing. More accurately, its mass is changing, which means a change in momentum. We can combine this result with the fact that the total energy of the cart must remain constant to conclude that the cart experiences a backward force and will slow down if nothing else is done to it. This result is not entirely unexpected, since if you weigh something down it's harder to move!
This result is quite useful. We can actually modify the problem to solve for the required force such that the cart remains at a constant velocity, which is what we've done in the higher difficulty levels. Knowing this result helps you deal with problems where mass is not constant, which admittedly are not numerous but still definitely present. And, it's an interesting result!
What happens if both mass and velocity change? Well, this requires some tricky calculus and isn't common, but can be dealt with. You are not strictly required to know this method, but it can come in handy if you're aiming to do physics at a higher level.
Take a look back at our new expression for Newton's Second Law. $$ F = \dfrac{dP}{dt}$$ Where:
$$ p = mv $$ In all previous scenarios, either mass or velocity was constant, so we could have treated them as such and didn't have to differentiate them. However, if both are changing, then we have to write a more general form. The key is to use the product rule.
$$F = \dfrac{dp}{dt} = m \dfrac{dv}{dt} + \dfrac{dm}{dt} v $$ This might not look exactly like the product rule you're used to, but if you take a closer look you will find that this indeed does follow the formula. Indeed, this method can be applied generally and not just to this formula if necessary.
Now, you have a way to find the net force on any object at any point in time, provided you know certain pieces of information.
This lesson was short because it was introductory. The next few lessons will cover more serious and difficult topics relating to conservation and problems involving momentum. They are really the representative lessons of momentum, and this one usually gets left in the dust as an unappealing addendum to the unit. However, this lesson is important in its own right, and you'll do well to remeber its results, especially our brand-new form of Newton's Second Law.
If you're ready to face the concept of conservation again after not talking about it for one whole lesson, let's move on!
The AntiMatter Lab 
