## A kind of fun high school physics collision problem, generalized slightly.

fig. 1. The collision problem.

Karl’s studying for his grade 12 physics final, and I picked out some problems from his text [1] for him to work on. Here’s one, fig. 1, that he made a numerical error with.

I solved this two ways, the first was quick and dirty using Mathematica, so he could check his answer against a number, and then while he was working on it, I also tried it on paper. I found the specific numeric values annoying to work with, so tackled the slightly more general problem of an object of mass $$m_1$$ colliding with an object of mass $$m_2$$ initially at rest, and determined the final velocities of both.

If we want to solve this, we start with a plain old conservation of energy relationship, with initial potential energy, equal to pre-collision kinetic energy
\label{eqn:collisionproblem:20}
m_1 g h = \inv{2} m_1 v^2,

where for this problem $$h = 3 – 3 \cos(\pi/3) = 1.5 \,\textrm{m}$$, and $$m_1 = 4 \,\textrm{kg}$$. This gives us big ball’s pre-collision velocity
\label{eqn:collisionproblem:40}
v = \sqrt{2 g h}.

For the collision part of the problem, we have energy and momentum balance equations
\label{eqn:collisionproblem:60}
\begin{aligned}
\inv{2} m_1 v^2 &= \inv{2} m_1 v_1^2 + \inv{2} m_2 v_2^2 \\
m_1 v &= m_1 v_1 + m_2 v_2.
\end{aligned}

Clearly, the ratio of masses is more interesting than the masses themselves, so let’s write
\label{eqn:collisionproblem:80}
\mu = \frac{m_1}{m_2}.

For the specific problem at hand, this is a value of $$\mu = 2$$, but let’s not plug that in now, instead writing
\label{eqn:collisionproblem:100}
\begin{aligned}
\mu v^2 &= \mu v_1^2 + v_2^2 \\
\mu \lr{ v – v_1 } &= v_2,
\end{aligned}

so
\label{eqn:collisionproblem:120}
v^2 = v_1^2 + \mu \lr{ v – v_1 }^2,

or
\label{eqn:collisionproblem:140}
v_1^2 \lr{ 1 + \mu } – 2 \mu v v_1 = v^2 \lr{ 1 – \mu }.

Completing the square gives

\label{eqn:collisionproblem:160}
\lr{ v_1 – \frac{\mu}{1 + \mu} v }^2 = \frac{\mu^2}{(1 + \mu)^2} v^2 + v^2 \frac{ 1 – \mu }{1 + \mu},

or
\label{eqn:collisionproblem:180}
\begin{aligned}
\frac{v_1}{v}
&= \frac{\mu}{1 + \mu} \pm \inv{1 + \mu} \sqrt{ \mu^2 + 1 – \mu^2 } \\
&= \frac{\mu \pm 1}{1 + \mu}.
\end{aligned}

Our second velocity, relative to the initial, is
\label{eqn:collisionproblem:200}
\begin{aligned}
\frac{v_2}{v}
&= \mu \lr{ 1 – \frac{v_1}{v} } \\
&= \mu \lr{ 1 – \frac{\mu \pm 1}{1 + \mu} } \\
&= \mu \frac{ 1 + \mu – \mu \mp 1 }{1 + \mu} \\
&= \mu \frac{ 1 \mp 1 }{1 + \mu}.
\end{aligned}

The post collision velocities are
\label{eqn:collisionproblem:220}
\begin{aligned}
v_1 &= \frac{\mu \pm 1}{1 + \mu} v \\
v_2 &= \mu v \frac{ 1 \mp 1 }{1 + \mu},
\end{aligned}

but we see the equations describe one scenario that doesn’t make sense physically, because the positive case, describes the first mass teleporting through and past the second mass, and continuing merrily on its way with its initial velocity. That means that our final solution is
\label{eqn:collisionproblem:240}
\begin{aligned}
v_1 &= \frac{\mu – 1}{1 + \mu} v \\
v_2 &= 2 \frac{ \mu }{1 + \mu} v,
\end{aligned}

For the original problem, that is $$v_1 = 2 v / 3$$ and $$v_2 = 4 v /3$$, where $$v = \sqrt{ 2(9.8) 1.5 } \,\textrm{m/s}$$.

For the post-collision heights part of the question, we have
\label{eqn:collisionproblem:260}
\begin{aligned}
\inv{2} m_1 \lr{ \frac{2 v}{3} }^2 &= m_1 g h_1 \\
\inv{2} m_2 \lr{ \frac{4 v}{3} }^2 &= m_2 g h_1,
\end{aligned}

or
\label{eqn:collisionproblem:280}
\begin{aligned}
h_1 &= \frac{2}{9} \frac{v^2}{g} = \frac{4}{9} h \\
h_2 &= \frac{8}{9} \frac{v^2}{g} = \frac{16}{9} h,
\end{aligned}

where $$h = 1.5 \,\textrm{m}$$.

The original question doesn’t ask for the second, or Nth, collision. That would be a bit more fun to try.

# References

[1] Bruni, Dick, Speijer, and Stewart. Physics 12, University Preparation. Nelson, 2012.

## New video: Velocity and angular momentum with geometric algebra

In this video, we compute velocity in a radial representation $$\mathbf{x} = r \mathbf{\hat{r}}$$.

We use a scalar radial coordinate $$r$$, and leave all the angular dependence implicitly encoded in a radial unit vector $$\mathbf{\hat{r}}$$.

We find the geometric algebra structure of the $$\mathbf{\hat{r}}’$$ in two different ways, to find

$$\mathbf{\hat{r}}’ = \frac{\mathbf{\hat{r}}}{r} \left( \mathbf{\hat{r}} \wedge \mathbf{\hat{x}}’ \right),$$

then derive the conventional triple vector cross product equivalent for reference:

$$\mathbf{\hat{r}}’ = \left( \mathbf{\hat{r}} \times \mathbf{\hat{x}}’ \right) \times \frac{\mathbf{\hat{r}}}{r}.$$

We then compute kinetic energy in this representation, and show how a bivector-valued angular momentum $$L = \mathbf{x} \wedge \mathbf{p}$$, falls naturally from that computation, where we have

$$\frac{m}{2} \mathbf{v}^2 = \frac{1}{2 m} {(m r’)}^2 – \frac{1}{2 m r^2 } L^2.$$

Prerequisites: calculus (derivatives and chain rule), and geometric algebra basics (vector multiplication, commutation relationships for vectors and bivectors in a plane, wedge and cross product equivalencies, …)

Errata: at around 4:12 I used $$\mathbf{r}$$ instead of $$\mathbf{x}$$, then kept doing so every time after that when the value for $$L$$ was stated.

As well as being posted to Google’s censorship-tube, this video can also be found on odysee.

## Video: Spherical basis vectors in geometric algebra

I’ve made a new manim-based video with a geometric algebra application.

In the video, the geometric algebra form for the spherical unit vectors are derived, then unpacked to find the conventional vector algebra form. We will then use our new tools to find the expression for the kinetic energy of a particle in spherical coordinates.

Prerequisites: calculus (derivatives and chain rule), complex numbers (exponential polar form), and geometric algebra basics (single sided rotations, vector multiplication, vector commutation sign changes, …)

You can find the video on Google’s censorship-tube, or on odysee.