Just finished reading Poor Charlie’s Almanack. It was good, entertaining and had a few nice pieces that I want to come back to. Note to myself — skip anything added by Peter D. Kaufman as it often sounds as personality cult following.

I might have been hard on Google as Apple showed their Liquid Glass design. Now, they are taking the crown of the stupidest design system from Google. More readability and accessibility issues, more bubbafication, more stupid animations.

There are no adults left. Zero.

From Better, Easier, Emotional UX: The research behind Google’s bold new direction for design

Material 3 Expressive is the most researched update to Google’s design system, ever. Here, Material researchers share the data behind the designs and new insights into users’ preference for emotion-driven UX.

Google released Material 3. Now, this might be the stupidest and ugliest design system out there.

From Building a slow web:

The internet I came to love was quieter. Smaller – which is not to say small. It was still vast, but it was a vast collection of small sites instead of a small collection of vast sites.

Ditching big platforms and using my personal blog instead was the best decision I did in the last year. I still post on Instagram once or twice a year and visit a few subreddits, but my RSS reader is now my primary source for content.

Feynman’s Lectures Exercises 8.2 via the Center-of-Mass Frame

First, calculate the velocity of the CM frame:

${\displaystyle {\mathbf{v}}_{\text{cm}}=\frac{m\mathbf{v}+m\cdot 0}{m+m}=\frac{\mathbf{v}}{2}}$

Now, calculate velocities of particles in the CM frame:

${\displaystyle {\mathbf{v}}_1^{\text{cm}}=\mathbf{v}-{\mathbf{v}}_{\text{cm}}=\frac{\mathbf{v}}{2}}$

${\displaystyle {\mathbf{v}}_2^{\text{cm}}=0-{\mathbf{v}}_{\text{cm}}=-\frac{\mathbf{v}}{2}}$

In the CM frame, particles will have the same speeds after elastic collision, meaning:

${\displaystyle \mathrm{\mid }{\mathbf{u}}_1\mathrm{\mid }=\mathrm{\mid }{\mathbf{u}}_2\mathrm{\mid }=\mathrm{\mid }\mathbf{u}\mathrm{\mid }=\frac{v}{2}}$

The directions of these velocities will be opposite, that is:

${\displaystyle \begin{array}{rl}{\displaystyle {\mathbf{u}}_1}& {\displaystyle =\mathbf{u}}\\ {\displaystyle {\mathbf{u}}_2}& {\displaystyle =-\mathbf{u}}\end{array}}$

Now, let’s move back to the lab frame:

${\displaystyle \begin{array}{rl}{\displaystyle {\mathbf{v}}_1}& {\displaystyle =\mathbf{u}+\frac{\mathbf{v}}{2}}\\ {\displaystyle {\mathbf{v}}_2}& {\displaystyle =-\mathbf{u}+\frac{\mathbf{v}}{2}}\end{array}}$

Let’s calculate the dot product:

${\displaystyle \begin{array}{rl}{\displaystyle {\mathbf{v}}_1\cdot {\mathbf{v}}_2}& {\displaystyle =(\mathbf{u}+\frac{\mathbf{v}}{2})\cdot (-\mathbf{u}+\frac{\mathbf{v}}{2})}\\ {\displaystyle }& {\displaystyle =-\mathrm{\mid }\mathbf{u}{\mathrm{\mid }}^2+\mathbf{u}\cdot \frac{\mathbf{v}}{2}-\mathbf{u}\cdot \frac{\mathbf{v}}{2}+\frac{\mathbf{v}}{2}\cdot \frac{\mathbf{v}}{2}}\\ {\displaystyle }& {\displaystyle =-{\left(\frac{v}{2}\right)}^2+{\left(\frac{v}{2}\right)}^2=0}\end{array}}$

Therefore, the final velocities are orthogonal:

${\displaystyle {\mathbf{v}}_1\perp {\mathbf{v}}_2}$

Previous solution, which doesn’t use the center-of-mass frame, can be found at A more elegant solution can be found in Feynman’s Lectures Exercises 8.2.

Feynman’s Lectures Exercises 8.2

‌A moving particle collides perfectly elastically with an equally massive particle initially at rest. Show that the two particles move at right angles to one another after the collision.

From conservation of momentum, we write the equations for the $x$ and $y$ axes:

${\displaystyle \begin{array}{rl}{\displaystyle v_1}& {\displaystyle =v_1^{\mathrm{\prime }}\cos \alpha +v_2^{\mathrm{\prime }}\cos \beta }\\ {\displaystyle 0}& {\displaystyle =v_1^{\mathrm{\prime }}\sin \alpha -v_2^{\mathrm{\prime }}\sin \beta }\end{array}}$

From the second equation:

${\displaystyle v_1^{\mathrm{\prime }}\sin \alpha =v_2^{\mathrm{\prime }}\sin \beta }$

This gives:

${\displaystyle \frac{v_1^{\mathrm{\prime }}}{v_2^{\mathrm{\prime }}}=\frac{\sin \beta }{\sin \alpha }}$

Now, from conservation of kinetic energy:

${\displaystyle \begin{array}{rl}{\displaystyle \frac{1}{2}m{v}_1^2}& {\displaystyle =\frac{1}{2}m{v}_1^{\mathrm{\prime }2}+\frac{1}{2}m{v}_2^{\mathrm{\prime }2}}\\ {\displaystyle v_1^2}& {\displaystyle =v_1^{\mathrm{\prime }2}+v_2^{\mathrm{\prime }2}}\end{array}}$

Square the momentum equation and substitute $v_1^2$:

${\displaystyle \begin{array}{rl}{\displaystyle (v_1^{\mathrm{\prime }}\cos \alpha +v_2^{\mathrm{\prime }}\cos \beta {)}^2}& {\displaystyle =v_1^2}\\ {\displaystyle v_1^{\mathrm{\prime }2}{\cos }^2\alpha +v_2^{\mathrm{\prime }2}{\cos }^2\beta +2v_1^{\mathrm{\prime }}{v}_2^{\mathrm{\prime }}\cos \alpha \cos \beta }& {\displaystyle =v_1^{\mathrm{\prime }2}+v_2^{\mathrm{\prime }2}}\end{array}}$

Rearranging:

${\displaystyle \begin{array}{rl}{\displaystyle 2v_1^{\mathrm{\prime }}{v}_2^{\mathrm{\prime }}\cos \alpha \cos \beta }& {\displaystyle =v_1^{\mathrm{\prime }2}(1-{\cos }^2\alpha )+v_2^{\mathrm{\prime }2}(1-{\cos }^2\beta )}\\ {\displaystyle 2v_1^{\mathrm{\prime }}{v}_2^{\mathrm{\prime }}\cos \alpha \cos \beta }& {\displaystyle =v_1^{\mathrm{\prime }2}{\sin }^2\alpha +v_2^{\mathrm{\prime }2}{\sin }^2\beta }\end{array}}$

Divide both sides by $v_1' v_2'$:

${\displaystyle 2\cos \alpha \cos \beta =\frac{v_1^{\mathrm{\prime }}}{v_2^{\mathrm{\prime }}}{\sin }^2\alpha +\frac{v_2^{\mathrm{\prime }}}{v_1^{\mathrm{\prime }}}{\sin }^2\beta }$

Use the earlier ratio:

${\displaystyle \frac{v_1^{\mathrm{\prime }}}{v_2^{\mathrm{\prime }}}=\frac{\sin \beta }{\sin \alpha }}$

Substitute:

${\displaystyle 2\cos \alpha \cos \beta =2\sin \alpha \sin \beta }$

So:

${\displaystyle \cos \alpha \cos \beta -\sin \alpha \sin \beta =0}$

This gives:

${\displaystyle \cos (\alpha +\beta )=0}$

Therefore:

${\displaystyle \alpha +\beta =90^{\circ }}$

The two particles move at right angles after the collision.

A more elegant solution can be found in Feynman’s Lectures Exercises 8.2 via the Center-of-Mass Frame.

Replaced my old IKEA desk lamp with Artemide Tolomeo.

It’s a simple design. Some screws are visible, and you can tighten them when needed. My version is aluminum, so there’s no chance of paint chipping. You can use any bulb up to 70W. The lamp and its mount are sold separately, giving you plenty of options—you can even turn it into a reading lamp with a floor stand. Artemide sells all the spare parts, so you can fix it if anything breaks.

The philosophy of Artemide is that the product lasts for generations. Starting from the design, the quality of materials and manufacturing, our lamps are made to last over time and designed to be repaired if any component is damaged by accidental events during use.

This lamp is the definition of good design.