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.

I’ve been working on a weather forecast app that presents information in a more compact form.

It’s easy to see the difference in perception when you compare the two approach side by side. You can see the entire day’s temperatures at a glance, without scrolling or tapping. The current temperature is visible but doesn’t take half the screen. A full week’s forecast is accessible with just one or two scrolls.

It’s currently available at weather.simonuvarov.com, but works for Dublin only.

It’s a Progressive Web App (PWA), meaning you can install it on your phone’s home screen. Once installed, it behaves almost like a native app, including the app icon (shown at the beginning of this post).

There’s still work to be done, but it’s so much nicer to use already.

Watched Brutalist. This might be the best movie I’m going to watch in 2025.

Feynman’s Lectures Exercises 6.9

Before firing, the total momentum is:

${\displaystyle (M+m_b)V_0=0}$

After firing one bullet, the platform moves with velocity $V_p$. By momentum conservation, total momentum remains zero:

${\displaystyle M{V}_p+m_b{V}_b=0}$

Solving for $V_p$:

${\displaystyle V_p=-\frac{m_b{V}_b}{M}=-0.005\text{ m/s}}$

A negative sign indicates the platform moves opposite to the bullet’s direction.

A bullet travels 5 m at 500 m/s, thus its travel time is:

${\displaystyle t=\frac{5}{500}=\frac{1}{100}\text{ s}}$

Since bullets are fired every $\frac{1}{10}\text{ s}$, only one bullet is airborne at a time.

Approach 1: Considering 10 bullets per second

Each bullet provides a velocity of $0.005$ m/s to the platform, then it flies for some time and then it stops. In total, the platform moves for due to 10 bullets per seconds:

${\displaystyle t_{\text{total}}=nt=10\times \frac{1}{100}s=\frac{1}{10}s}$

Thus, the effective average velocity is:

${\displaystyle V=0.005\times \frac{1}{10}\times 10=5\times 10^{-4}\text{ m/s}}$

Approach 2: Using total distance traveled

The distance traveled by the platform per bullet fired is:

${\displaystyle s_p=V_{p}t=0.005\times \frac{1}{100}=5\times 10^{-5}\text{ m}}$

With 10 bullets fired each second, the effective velocity is:

${\displaystyle V=5\times 10^{-5}\times 10=5\times 10^{-4}\text{ m/s}}$

Feynman’s Lectures Exercises 6.9

Before firing, the total momentum is:

${\displaystyle (M+m_b)V_0=0}$

After firing one bullet, the platform moves with velocity $V_p$. By momentum conservation, total momentum remains zero:

${\displaystyle M{V}_p+m_b{V}_b=0}$

Solving for $V_p$:

${\displaystyle V_p=-\frac{m_b{V}_b}{M}=-0.005\text{ m/s}}$

A negative sign indicates the platform moves opposite to the bullet’s direction.

A bullet travels 5 m at 500 m/s, thus its travel time is:

${\displaystyle t=\frac{5}{500}=\frac{1}{100}\text{ s}}$

Since bullets are fired every $\frac{1}{10}\text{ s}$, only one bullet is airborne at a time.

Approach 1: Considering 10 bullets per second

Each bullet provides a velocity of $0.005$ m/s to the platform, then it flies for some time and then it stops. In total, the platform moves for due to 10 bullets per seconds:

${\displaystyle t_{\text{total}}=nt=10\times \frac{1}{100}s=\frac{1}{10}s}$

Thus, the effective average velocity is:

${\displaystyle V=0.005\times \frac{1}{10}\times 10=5\times 10^{-4}\text{ m/s}}$

Approach 2: Using total distance traveled

The distance traveled by the platform per bullet fired is:

${\displaystyle s_p=V_{p}t=0.005\times \frac{1}{100}=5\times 10^{-5}\text{ m}}$

With 10 bullets fired each second, the effective velocity is:

${\displaystyle V=5\times 10^{-5}\times 10=5\times 10^{-4}\text{ m/s}}$