School Formula Derivation

High School Physics Formula Derivation

It is common practice to get formula sheets for physics exams. Doing physics isn’t just about memorization and so these sheets remind you of the equations, but they don’t give any explanations for how they work. Yet the problem of doing physics this way is that it can lead to an axiomal fallacy.

An axiom is something that we assume to be true. The idea is that you choose as few axioms as possible and then use them as foundations to build upon. A famous example is Euclid’s five axioms from which it is possible to derive all of planar geometry. These five rules, which we cannot test and so assume to be true, contain within them all the information needed to understand geometry. It is simply up to us to reshape them into a form that makes it easier to understand.

Sometimes when we look at a formula sheet we can forget this fact. Instead, we see each equation as a separate axiom from which physics is built upon. (Or at least this is how I saw it in high school…) Instead, this could not be further from the truth and often many equations are simply checkpoints in a long derivation from one equation to another. They are phenomenally useful in a test (you don’t have enough time to do the derivation again!) but we must remember that not all equations are equal.

In New Zealand high schools (secondary schools) you are introduced to Newtonian mechanics in year 12 at Level 2 NCEA (second to last year of school). For the mechanics exam you are given a formula sheet of all the relevant formulas, however they can all be derived from Newton’s three laws of motion.

This post will go through the NCEA Level 2 Physics Formula Sheet and break down how each formula is derived focusing on the mechanics equations.

Axioms

We need to first establish some axioms to base our derivations from. Newton’s three laws are as follows

1. Everything that is in motion will continue in motion unless something acts on it
2. Force is the change in momentum over time¹
3. Every action has an equal and opposite reaction

That’s enough to jump in!

Definitions

To do some physics we need to put variables to the laws above!

Momentum

Momentum is how much a moving object wants to continue to move. A rolling bike has momentum, but a speeding car has more; both because it is heavier but also since it moves faster. As such momentum is directly proportional to mass and velocity, or expressed as an equation as $$p=mv \tag{1}$$

Where $p$ is momentum, $m$ is mass and $v$ is velocity (speed).

Velocity

Speaking of velocity, the speed that something moves is just how much its position changes in a certain amount of time. This is often phrased that velocity is the change in distance over time but it means the same thing. We could write this mathematically as $$v= \frac{\Delta x}{\Delta t} \tag{2}$$

Where $\Delta x$ is the change in distance and $\Delta t$ is the change in time. The $\Delta$ symbol simply means “change in”.

If you have ever done calculus you will know that this method of describing a quantity as the change in another quantity over time is known as a derivative. We could then imagine that the change in time is so so small that it approaches 0 and we get the velocity at that instance in time. This is represented mathematically as² $$v = \frac{dx}{dt}$$

Acceleration

We define acceleration in a similar way. No physics is being done here, just definitions. Acceleration is how much our velocity is changing and how quickly we speed up (or slow down) over time.

Its common to phrase it as the change in velocity over time and so once again we can write this as a fraction and as a derivative. Both mean the same thing, it just depends on how large your time scale is! $$a = \frac{dv}{dt} = \frac{\Delta v}{\Delta t} \tag{3}$$

Hey look at that! We’ve already knocked off three formulas from our formula sheet! That’s not bad progress.

Force

We already have a definition for force as given by Newton’s second law, but now we simply need to convert it to an equation. As a reminder, force is the change in momentum over time. This sounds an awful lot like a derivative mentioned above, which it indeed is. $$F= \frac{dp}{dt}$$

These $dp$ and $dt$ represent very very small changes in momentum and time, however we can find the average force if we extend them to a larger period by $dp \to \Delta p$. This is the opposite idea to what we just did earlier. If we do this and rearrange for momentum we get another equation except that now this is the average force. $$\Delta p = F \Delta t \tag{4}$$

We also have the definition of momentum, so we could substitute that in for $p$ into Newton’s second law. If we expand that derivative using the chain rule we then get $$F = \frac{d}{dt}(mv) = \frac{dm}{dt}v + m \frac{dv}{dt}$$

Next if we make the (common) assumption that mass is constant over time, which allows us to drop the first term as $dm/dt=0$, and notice that we defined something to be equal to the change in velocity above, acceleration, we can rewrite it as $$F = m\frac{dv}{dt} = ma \tag{5}$$

Now we have force in a form that we are often used to, yet it is just a restatement of the above axiom! In fact this common form is only true for constant mass, but we never would have found that out unless we performed this derivation.

Torque

For systems that rotate we can extend the idea of force to torque. Torque is similar to force except for the fact that we can use leverage to increase it.

When you screw in a bolt you use a wrench so that the distance from the pivot is increased. By doing this, the same force applies greater torque as it is further away from the pivot. Mathematically speaking torque increases directly proportional to force and the radial distance which means we can write it as $$\tau = Fr \tag{6}$$

This is a similar way to how we wrote momentum. Unfortunately this is just a definition and only useful to define what $\tau$ means.

Movement

There are four formulas which are all used for translational motion in one dimension. Students commonly use these to solve projectile motion questions or acceleration questions as they act on the assumption that acceleration is constant. They are the following²

$$\begin{align*} x &=v_it + \frac 12 at^2 \\ v_f &= at + v_i \\ v_f^2 &= v_i^2 + 2ax \\ x &= \frac{v_f + v_i}{2}t \\ \end{align*}$$

We can derive these from our definitions above. Specifically lets focus on the definition of acceleration and velocity. $$a = \frac{dv}{dt} \quad\text{ and }\quad v=\frac{dx}{dt}$$

If we substitute the velocity equation into the acceleration equation we gain acceleration in terms of distance and time. In words this means that acceleration is the change in the change of distance over time. It’s how fast the change is occurring is happening! In mathematics this is represented by the second derivative.

$$a = \frac{d^2 x}{dt^2}$$

Now we want to find an equation for $x$ that will satisfy this second derivative. If we assume that acceleration is constant with time, then we just need to find an equation that when differentiated twice is constant. A simple solution can be guessed to be a quadratic with some arbitrary constants.³

$$x = \alpha t^2 + \beta t + \gamma $$

If we differentiate this twice we then get $a = 2\alpha$ so this means that $\alpha = a/2$ to fix our constants correctly. Our starting position is also arbitrary, so we can set $\gamma = 0$ since the laws of motion are the same no matter where we are. This then leaves our equation with one unknown $\beta$. $$x = \frac a2 t^2+ \beta t$$

To find what $\beta$ is we can substitute a point that we know. For instance lets say that the starting velocity is $v_i$ where $i$ is meaning the initial velocity. We can differentiate the above equation to get $$v = at + \beta$$

and then substitute in that $v=v_i$ when $t=0$ to get $\beta = v_i$. This is actually a useful equation to keep around so we’ll just rearrange it for time and leave it for later $$t = \frac{v-v_i}{a} \tag{A}$$

If we plug in our value of $\beta$ into the above equation, that was missing it, we get our first desired equation! $$x = \frac a2 t^2 + v_i t \tag{7}$$

Our useful equation above will then help us get two more equations. Firstly we need to note that the final velocity $v_f$ is just the same as the velocity given from the formula. Its the same variable but wearing different clothes so $v=v_f$. This means that we can rearrange $(\textrm{A})$ into $$v_f = v_i + at \tag{8}$$

This equation makes sense as your final velocity should be the sum of your starting velocity as well as the length of time that you were allowed to accelerate for.

We can also substitute $(\textrm{A})$ into $(7)$ by replacing $t$. This results in a nasty expression $$x = v_i \left(\frac{v_f-v_i}{a}\right) + \frac a2 \left( \frac{v_f-v_i}{a} \right)^2$$

But if we chug through and expand the brackets then combine the fractions together, a nice form results $$x = \frac{v_iv_f - v_i^2 }{a} + \frac{v_f^2 - 2v_fv_i + v_i^2}{2a} = \frac{v_f^2 - v_i^2}{2a}$$ We can then rearrange this to get the form that we know and love $$v_f^2 = v_i^2 + 2ax \tag{9}$$

We can also rearrange $(\textrm{A})$ to give us acceleration and then substitute this into $(7)$ in a similar way to what we did above. This gives us $$x = v_it + \frac 12 \left( \frac{v_f-v_i}{t} \right) t^2$$

which we can simply rearrange into the final form of $$x = \frac{v_f + v_i}{2}t \tag{10}$$

The useful thing about these equations is that they all share five variables $x,a,t,v_f,v_i$ yet each equation is missing one of them. This allows us to be more efficient when using them as we don’t need to calculate variables that we don’t need. However, we can always convert between the forms if we need since they are simply a rexpression of the same fundamental equation.

Energy

Next we’ll focus on equations that deal with energy.

Work

A useful quantity is how much energy was used to force an object a specific distance. This is commonly known as how much work was done. If you have to do work, you must have put energy into the system (Also if you get energy, then the system did work, like a petrol motor for instance).

The units of energy are defined to be newton meters $Nm$ which are commonly referred to as joules to confused the unwary. So to find the total energy done we need to add up all the forces as the object is moved.

If we give this definition a mathematical framework we must use an integral of force over the distance moved. $$\delta W = \int_a^b F \, dx$$

Where $\delta W$ is the change in work done⁴, $F$ is the force, $dx$ a very small unit of distance and $a$ and $b$ the starting and ending distances respectively. The integral then adds up all the forces as the object is moved by an amount $dx$.

If the force $F$ is constant everywhere we can simply integrate. In more complex situations this often is not true. $$\delta W = Fx \bigg\rvert_a^b = F(b-a)$$

but the quantity $b-a$ is just the distance that the object is moved. So if we call the distance $d=b-a$ we get the standard form of work. $$W=Fd \tag{11}$$

Power

Some objects are able to output more work than others in the same time. Consider turning a crank shaft. You and a car could both do it, but the car would be able to put much more energy into turning it in the same amount of time. This is as the car engine can output more work than you can per second.

We call this output of work per unit time power. A car has a higher power than you have (probably). As before this is represented as a derivative since power is the change in work over the change in time, written as $$P = \frac{dW}{dt}$$

If we want the average power done, we can extend the time period $dt \to t$ and get a simple fraction form that we see on the formula sheet. $$P = \frac{W}{t} \tag{12}$$

Kinetic Energy

Every moving object has energy associated with its movement. It takes energy to start something moving, but you can get some of it back if you slow it down again. Yet how much energy is exactly associated with a moving object?

As work is energy we can start with the same integral definition as above, but let’s swap force with momentum using Newton’s second law.

$$\delta W = \int F \, dx = \int \frac{dp}{dt} \, dx$$

Now we can use a trick of mathematics to swap the derivatives around and make the integral a subject of momentum instead $$\int \frac{dp}{dt}\, dx =\int \frac{dx}{dt}\, dp $$

Next we can notice that the derivative of position is velocity $v=dx/dt$ (which is how quickly position changes over some change in time) and that we can break momentum down into mass and velocity by $p=mv$. This allows us to reexpress it as $$\int \frac{dx}{dt}\, dp = \int vm\, dv = m \int v\, dv$$

Where $m$ is removed from the integral as it is assumed to be constant, with respect to velocity (a reasonable assumption but not always valid such as in a rocket burning fuel).

This integral can be evaluated to give us the form of kinetic energy shown. To represent that it is kinetic energy we rewrite work $\delta W \to E_k$ by convention, but they both mean the same thing fundamentally. $$E_k = \frac 12 mv^2 \tag{13}$$

It is remarkable how much a form like this shows up in physics.

Gravitational Potential Energy

Potential wells also have a similar situation as you can store energy in them. For instance lifting a ball up against gravity takes energy but if you release it that energy is returned (in the form of kinetic energy). This energy is stored as gravitational potential energy.

To calculate it we need only to look at the force the earth has on an object. As we derived above we can write force as $F=ma$ and on earth we have an acceleration of $9.8ms^{-1}$ downwards. By convention, we write this acceleration as $g$, but it means the same thing as $a$, so we could rewrite the force due to gravity as $$F=mg$$

This can then be substitued in our derivation for work above (as work is energy) to give us $$\delta W = Fd = mgd$$

Now with a simple rewrite of the variables we are done. To represent that it is potential energy we write the work as $\delta W \to \Delta E_p$ and to make it clear that $d$ is the distance moved and not the height above the earth or something, we can rewrite it as the change in height $d \to \Delta h$. This then gives us the desired form $$\Delta E_p = mg\Delta h \tag{14}$$

It’s important to note that this isn’t the total gravitational potential energy; instead it is only the change that occurred moving the object by $\Delta h$. This is a result of $\delta W$ not being the total work, but just the change in work that occurred moving this object.

Springs

Next we’ll tackle equations that deal with springs.

A long time ago Mr Hooke discovered that the longer you pulled a string, the greater the force there was trying to retract it and vice versa when he wrote “as the extension, so the force”. He noticed that this was a linear relationship meaning that doubling the length, doubled the force retracting it. Mathematically we write it down as follows $$F=-kx \tag{15}$$

Unfortunately, this is impossible to derive from Newton’s laws. It is an empirical law which means that we discovered it from plotting many points and simply finding the best line that fits. In fact, we know that it isn’t quite right, but it’s pretty good that not many people notice the difference.⁵

Regardless, we can do the same thing we did for potential energy by substituting this force into the work equation. We’ll drop the negative sign as that just indicates the direction of the force and has no affect of the total energy of the system. $$\delta W = \int kx\, dx$$

Simply integrating and changing the symbol for energy from work to potential energy $\delta W \to E_p$ then gives us the energy contained in a spring. It’s quite clear that this increases the more a string is stretched (increasing $x$) so it passes the sanity check too. $$E_p = \frac 12 kx^2 \tag{16}$$

(Notice how similar the form is to kinetic energy! It appears everywhere)

Rotational Motion

And now we arrive at the actually difficult derivation of rotational motion. Rotational motion is complicated as there is a dilemma between it being easier to calculate using angles, but easier to think about using distances. It doesn’t help that vector calculus occludes much of what is happening.

However, we can do a little trick if we assume that the speed of the object is constant. Thankfully this is valid for our situation, so we’ll step through the derivation that way.⁶

Lets assume that we have a ball moving around a circle of radius $r$ at a constant speed $v$. While the speed is constant the velocity is not constant due to the ball having to change direction to trace out a circle.

How much does the velocity change by for one revolution? Well if we imagine the vector fixed in place then the tip of the vector will trace a circle around its self. So the tip of the velocity vector moves by $2\pi v$ for one revolution.

We also need the time taken to complete one revolution. As the ball moves at a continuous speed of $v$ around a circumference of $2\pi r$ it will then take $2\pi r/v$ long to complete a revolution.

Combining this we can find the change in velocity during a change in time; commonly known as acceleration. Importantly this is only valid if the speed is constant! Otherwise, this would only be the average centripetal acceleration. This is done by diving the change in velocity of one revolution by the time taken. $$a_c = \frac{2\pi v}{\frac{2\pi r}{v}}$$

A quick simplification yields what we expect $$a_c = \frac{v^2}{r} \tag{17}$$

We can also gain the centripetal force by substituting this acceleration into Newton’s second law $F=ma$ to derive centripetal acceleration. $$F_c = \frac{mv^2}{r} \tag{18}$$

Afterword

That’s it! With that behind us we’ve managed to derive the whole formula sheet from just a few basic axioms. This is the power of physics as only a few assumptions are needed for us to build an entire field on its foundations. Even complex fields like Quantum Mechanics and General Relativity a fabricated from a set of axioms to build from.

You may have noticed that it seemed we only needed to use the second law and some logic forgoing the first and third laws completely. This isn’t the case and it is the third law that leads to conservation laws such as the conservation of energy and momentum⁷ (which are left off the formula sheet so I don’t have to prove them!). And we actually used the first law implicitly the whole time. The whole point of derivivng these equations is that we assume that once an object is doing its thing it’ll stay doing its thing until we tell it otherwise. If it didn’t then there’d be not much point in predicting what it could do. The first law makes this assumption concrete and without it our laws would not match reality.

Knowing the equations does little good if you’re unable to apply them to solving problems. Physics is not a spectator sport; you can’t get good by looking at equations from afar, you need to get your hands dirty.