I have published this in my secondary school’s mathematics journal here.

A lot of the formulae given to you in HKDSE Physics, as it turns out, can be derived from the calculus taught in M2. In this series of posts we’re going to go through deriving some of them. For a more detailed treatise on this topic, I would highly recommend checking out the dynamics lecture notes, which is a course for first year mathematics at Oxford.

We would look at rectilinear motion in part I, projectile / circular motion in part II and waves in part III.

Rectilinear motion

Rectilinear motion is one-dimensional motion along a straight line. Due to it only having one dimension, all properties about the system could be represented by one variable only. We wouldn’t need to deal with coordinates.

Consider some point particle with constant mass \(m\). As we’ve seen in M2, we can respectively let displacement, velocity and acceleration as functions of time

\[\begin{align*} \text{Displacement } &= r(t) \\ \text{Velocity } &= v(t) = \frac{dr}{dt} \\ \text{Acceleration } &= a(t) = \frac{d^2r}{dt^2} \end{align*}\]

Under this language, we can reframe Newton’s First law as

\[\text{Momentum } = p(t) = mv(t) = m\frac{dr}{dt}\]

and Newton’s second law as

\[\text{Force } = F(t) = \frac{dp}{dt} = m\frac{dv}{dt} = ma(t)\]

Introducing assumptions

To get any further, we need to introduce some assumptions in DSE physics. In rectilinear motion we assume that

Force is constant (e.g. gravitational force)

This means that acceleration is constant! We would now write \(a(t)\) as \(a\) as it’s just a constant. This is crucial as it means that

\[\begin{align*} a(t) = \frac{d^2 r}{ dt^2} &= a \\ v(t) = \frac{dr}{ dt} &= at + C_1 \\ r(t) &= \frac{1}{2}at^2 + C_1 t + C_2 \end{align*}\]

by repeated indefinite integration for some constants \(C_1, C_2\). Naturally, we ask what those constant are. We can see that

\[\begin{align*} v(0) &= a \cdot 0 + C_1 = C_1 \\ r(0) &= \frac{1}{2}a \cdot 0 + C_1 \cdot 0 + C_2 = C_2 \\ \end{align*}\]

So \(C_1\) is the velocity at \(t=0\). \(C_2\) is the displacement at \(t=0\), which is generally taken to be \(0\).

Finally putting it all together we have

\[\begin{align*} v(t) &= at + v(0) \\ r(t) &= \frac{1}{2}at^2 + v(0)t + r(0) \end{align*}\]

Does this look familiar?

Conservation of energy

To see why energy is conserved, we must first define the kinetic energy of a point particle at time \(t\) to be

\[T(t) = \frac{1}{2} m \bigg( \frac{dr}{dt} \bigg)^2\]

and the potential energy for a point particle with displacement \(r\) (under constant force) to be

\[V(r) = -mar\]

From DSE physics, we know that energy is conserved. I.e. \(T+V\) is kept constant. However this is rather unobvious. Note how kinetic energy is with respect to time, but potential energy is with respect to displacement. In general, why would something with respect to time be related to something with respect to displacement?

It turns out that for energy to be conserved, the force needs to be conservative. In the one dimensional case, this means that there must exist a potential energy function \(V(r)\) such that \(F(r)=-\frac{d}{dr}V(r)\). This also means that the force is dependent on displacement only: If you are at the same displacement at different times, the force experienced is still the same.

For the case of DSE physics, as the acceleration / force is kept constant we could have \(V(r) = -mar\), so the force is conservative. Note how we can add any constant to \(V(r)\) and it would still be a valid potential function. Refer to the lecture notes for a more general analysis on conservative forces.

Now how do we show conservation of energy for this specific case? There’s two ways of doing it. Either we expand all the terms as follows

\[\begin{align*} &\phantom{=} (T+V) \\ &= \bigg[ \frac{1}{2} m \bigg( \frac{dr}{dt} \bigg)^2 -mar \bigg] \\ &= m \bigg[ \frac{1}{2} \bigg( at + v(0) \bigg)^2 -a (\frac{1}{2}at^2 +v(0)t +r(0)) \bigg] \\ &= m \bigg[ \frac{1}{2}a^2t^2 + v(0)at + \frac{1}{2}v(0)^2 - \frac{1}{2}a^2t^2 - av(0)t - ar(0)\bigg] \\ &= \frac{1}{2}m v(0)^2 - mar(0)\\ \end{align*}\]

Or we can do it more abstractly by considering the derivative of \(T+V\)

\[\begin{align*} &\phantom{=} \frac{d}{dt}(T+V) \\ &= \frac{d}{dt} \bigg[ \frac{1}{2} m \bigg( \frac{dr}{dt} \bigg)^2 + V(r)\bigg] \\ &= \frac{1}{2}m \cdot 2 \frac{d^2r}{dt^2} \cdot \frac{dr}{dt} + \frac{dV}{dr} \frac{dr}{dt} & \text{ product and chain rule} \\ &= m \frac{d^2r}{dt^2} \cdot \frac{dr}{dt} - m \frac{d^2r}{dt^2} \cdot \frac{dr}{dt} & \frac{dV}{dr} = -F(r) = -m\frac{d^2r}{dt^2}\\ &= 0 \end{align*}\]

So \(T+V\) is constant.

In particular, this means that \(\frac{1}{2}mv(t)^2 - mar(t) = \frac{1}{2}mv(0)^2 - mar(0)\). Does this look familiar?