# Blake Test

#### Contents

## SVG Crowbar

\[1 / \sqrt{2} \ket{\psi }\] \[\frac{1}{\sqrt{2}}\]

## Tests

There has been a great deal of discussion about this problem in the solutions. Many people have invoked an experiment performed by the Mythbusters several years back which showed that a fired bullet and a dropped bullet hit the ground at the same time. We here at Brilliant love the Mythbusters, but we love physics more. Let's break this problem down a bit:

**Quadratic Drag**

As some of the solutions have indicated, the interesting bit of physics in this problem comes from the fact that the drag force from air resistance is proportional to the squared velocity of the bullet. If we break down the velocity into its \(v_x\) and \(v_y\) components, that means the drag term is something like:

\[F_d = -k|v|^2,\]

where \(|v| = \sqrt{v_x^2 + v_y^2}\).

**Breaking down the force**

Like any other force, we can break down the drag force into components along the \(x\) and \(y\) axes. If the bullet's trajectory is off from horizontal by an angle \(\theta\), then the \(y\) component of the force is \(F_d \sin{\theta}\) and the \(x\) component of the force is \(F_d \cos{\theta}.\) To avoid any unnecessary variables, we can express these trigonometric functions using the bullet's \(x\) and \(y\) components. Using the right triangle above, we can observe that:

\[\sin{\theta} = \frac{v_y}{|v|}, \\ \cos{\theta} = \frac{v_x}{|v|}.\]

When substituted into the \(x\) and \(y\) components of the drag force, the divided factor of \(|v|\) simplifies and yields \(x\) and \(y\) components of \(-k |v| v_x\) and \(-k |v| v_y\) respectively. Even though these components of the force are orthogonal, notice that they both still depend on the total velocity \(|v|\) in addition to their velocity component.

**Equations of motion**

Now that we've broken down the drag force into the \(x\) and \(y\) components, we can construct the equations of motion according to Newton's second law:

\[ \begin{align} m \, v_x'(t) &= -k |v| v_x \\ m \, v_y'(t) &= -mg - k |v| v_y \end{align} \]

Substituting \(|v|\) as the norm of the two spatial components:

\[\begin{align} m \, v_x'(t) &= -k v_x \sqrt{v_x^2 + v_y^2} \\ m \, v_y'(t) &= -mg - k v_y \sqrt{v_x^2 + v_y^2} \end{align}\]

You might notice now that these equations are coupled through \(|v|\), and likely can't be solved explicitly. Even without solving, we can consider what effect the total velocity of the bullet \(|v| = \sqrt{v_x^2 + v_y^2}\) contributes to the drag term in both components of the bullet's velocity. If the bullet is fired from a gun, a large horizontal velocity will increase the vertical drag force that resists the pull of gravity. If a bullet is simply dropped, the horizontal velocity will be zero, and the total velocity will be much lower. This would yield a weaker drag force to resist gravity, and explains why a dropped bullet hits the ground quicker

**Finding the trajectories**

Though explicitly solving these coupled differential equations is nasty, we can use a numerical solver in Mathematica (or your computational tool of choice). Picking some physical constants for the drag coefficient \(k\) and the mass \(m\) yields the following trajectories for a dropped bullet and one fired at \(\SI{500}{\meter \per \second}\).

The fired bullet hits the ground after about 0.60 seconds, while the dropped one hits after only 0.45 seconds.

**What about linear drag?**

If we were to use a simpler linear approximation of air drag, the total velocity would no longer factor in to the drag of both components, and the equations would no longer be coupled. I leave that to the reader to construct similar equations of motion using a linear drag term like \(F_d = -k |v|\). The trajectory of such a bullet looks something like the one below, and hits the ground just like the dropped one, after about 0.45 seconds.

## Basic

The wavelength of green light is about \(\SI{500}{\nano \meter}\). As the wavelength of light gets smaller, its energy gets larger. Green light is about \(200 \times\) smaller than the width of a hair, and about \(2000 \times\) larger than most atoms.

We use visible light in microscopes to observe forensic samples of hair, and even to visualize the inside of cells, which are only about \(2 \times\) larger than the wavelength of green light. But we can't observe individual atoms with a light microscope: In fact as a rule, you can't really observe objects much smaller than the wavelength you're using to look at them. It would be like trying to track a speck of dust in the sky with a radar station.

What does this imply about subatomic particle physics?

- Subatomic particles can only be studied with very long wavelength light.
- As we probe more deeply into subatomic particles, we'll need to build higher energy accelerators.
- To observe individual subatomic particles, the experiment must be extremely cold.

## Intermediate

Hydrogen is the simplest atom, with only one proton and one electron, interacting with a Coulomb potential:

\[U = \frac{e^2}{4 \pi \epsilon_0 r}\]

where \(e\) is the elementary charge and \(\epsilon_0\) is the permitivity of free space.

If the electron were a classical charged particle, its orbit around the nucleus would decay until it came to rest at the nucleus. Since the electron is a quantum object, this doesn't happen. The spread in position and momentum of the electron is limited by the uncertainty principle:

\[\Delta x \Delta p \geq \frac{\hbar}{2}\]

where \(\hbar\) is the reduced Planck's constant.

The radius of the electron orbit **must** be larger than the spread in position, likewise the momentum of the electron **must** be larger than this spread in momentum. Otherwise, the Hydrogen atom would be violating this fundamental limit.

Treat the Hydrogen atom as a two body problem where the mass of the proton is much greater than the electron. Estimate the minimum radius of the electron orbit.

## Advanced

The founding fathers of quantum mechanics largely ignored relativity when they first applied quantum theory to chemistry:

"The general theory of quantum mechanics is now almost complete ... these give rise to difficulties only when high-speed particles are involved, and are therefore of no importance in the consideration of atomic and molecular structure and ordinary chemical reactions"- Paul Dirac

Treat an atom as a two body problem with one electron and a central nucleus with a charge \(Z\) and a mass much greater than the electron. Find the characteristic length scale of a ground state electron orbit. At what element in the periodic table would relativistic effects become important for an electron in a ground state?

*Note: Significant errors start to appear between classical and relativistic dynamics when \(\frac{v}{c} \geq 0.1\).*