Returned from Montenegro, where we were sailing for a week.

Not going back to the Balkans anytime soon. The airport in Tivat is the worst I’ve ever been to. Prices are as high as in Ireland, and Ireland is one of the most expensive countries in Europe.

The contrast between widespread poverty and the luxury in marinas is revolting.

There’s no real national cuisine. Compared to the food in Turkey or Cyprus — let alone France or Italy — it’s disappointing.

Far fewer marinas than in Turkey, and Turkey also had many more small anchorages.

This trip pushed us to start working on getting a skipper license ourselves. Shouldn’t take more than a few months.

Some good places:

• Marina Lazure, Montenegro. Very good showers, helpful mariners, and a nice promenade to Herceg Novi that’s good for running.
• Lustica Bay, Montenegro. Ugly and unauthentic, but had good showers and one shop.
• Saint Nicholas Island, a good swimming spot.
• Climbing the old Kotor Fort trail. On our way back we were charged 15 euros.
• Mon Bistro Cafétéria. Unexpectedly good coffee and croissants.

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.