Author: zachary

Solution to Gangbuster Geometry

Diagram of pond after being hit by rock. waves are just for illustration purposes and are not part of the problem.

We’re trying to find the amount of time it takes for a ball to resurface after hitting water. The simplest way to do this is to make a simulation of all the forces acting on the ball. The buoyant forces, given by the second equation require you to calculate the amount of water the ball has displaced, so you must calculate the volume of a sphere cap. The volume of a sphere cap is given by the fourth equation. The third equation, for drag, requires the frontal area of the sphere cap, which is given by the fifth equation. We put these together to get a simulation. The answer is the time in between a peak and a trough. This time actually fluctuates a bit. Our answer is that the average frequency is about one ripple every 0.3969 seconds.

\vec F_G=1kg g\ down\\[8pt] \vec F_B=\frac{998.2kg gV_w}{m^3}\ up\\[8pt] \vec F_d=\frac{469.2kgA_f\vec V^2}{m^3}\ up\\[16pt] V_w=\pi rh^2-\frac{\pi h^3}3\\[8pt] A_f={\pi r^2\over(r-d)^2}-\pi

Gangbuster Geometry

Geometry, trigonometry, and calculus are required for this challenge.

Diagram of pond after being hit by rock. waves are just for illustration purposes and are not part of the problem.

We left an unsolved puzzle at the end of Fascinating Frequencies 3. We want to find the time it takes for a stone to stop after hitting water. The stone has an initial velocity of 1 m/s, a mass of 1 kilogram, and radius 10cm. The force acting on it from gravity is 1kg*g down. The buoyant force acting on it is the weight of the water displaced by the stone. The water has density 0.9982 kg/m3. The friction from the water in any direction acts perpendicular to the surface, inwards, with a strength equal to the sine of the angle multiplied by the density of the water.

\vec F_G=1kg g\ down\\[8pt] \vec F_B=\frac{998.2kg gV_w}{m^3}\ up\\[8pt] \vec F_d=\frac{469.2kgA_f\vec V^2}{m^3}\ up
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Fascinating Frequencies 5

You will need to use physics for this Fascinating Frequencies challenge.

An interesting device can be created when you have a wheel of gears. Of course, this wheel will have an even number of gears. If the device is anchored and you spin one gear, the others will all also spin at the same frequencies. But if you anchor the gear you spin, the others can spin around it. Try building this device yourself, it’s very cool. Say you make one of these devices with six gears. The radius of each gear is 10 centimetres, and the centripetal force felt by the gear opposite to the anchored one you are spinning is 1 metre per second squared. The challenge is to find what frequency the gear is spinning.

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Solution to Didactic Dice

We are trying to find the average number of moves of a game. The game is big enough that we can approximate the probabilities with a bell curve. We could also use a simulation. Both give an accurate enough result. We do not need the exact answer. The average roll moves you 3.5 spaces. The probability that you land on the star on any given pass is thus 2/7, and miss it, 5/7. So the average number of passes you make is 5/(7-5) = 5/2 = 2.5 passes. 2.5*24 equals 60 squares. The average number of rolls will then be 60/3.5 equals 17 and a seventh rolls. The standard deviation for this is calculated by the formula below, and it gives about 7.071. What we do next is surprisingly simple: subtract the average number of turns it takes to move standard deviation from the average. In equation form, 17.143-7.071/3=Our answer is that the game takes an average of about 14.79 turns.

D=\sqrt{\frac{qty\times(s^2-1)}{12}}\\[8pt] D=5\sqrt2\approx7.071

Fascinating Frequencies 4

A physics simulation is needed to solve this challenge.

Diagram of Baffling Bungee Jump.

Bob the Bungee jumper from Baffling Bungee is upset about his bungee jumping experience. If you attempt the problem, you will soon see that Bob’s bungee jump will take too long. Far too long. This is because C5H8 is very springy. In fact, Bob’s bungee jump would take about 2 months … if you were thinking of a mathematical model. You see, the diagram shows a cliff behind Bob. The rubber is so springy he would bash into it, cutting his jump short. But set that problem aside for now. This challenge does not concern those. In bob’s jumps, the ups and downs would gradually, very gradually, become smaller and smaller. But they would still occur at the same frequency. Can you guess today’s challenge? Calculate that frequency.

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Didactic Dice

Combinatorics is needed to solve today’s challenge.

Shown in the diagram to the right is the board for a dice game. The rules are as follows: start at the point. On your turn, roll 2 six sided dice. Move forward thet many spaces allong the path. The path is marked by the arrow and has 24 squares. If you reach the black square and have more moves, move back to the dot. You do not continue moving; your turn ends immediately. The first player to end exactly on the black square wins. You play this game with a friend. The challenge is to find the average number of turns the game will take. After 1 player wins, the other does not continue. This is very important.

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Solution to Fascinating Frequencies 3

Diagram of pond after being hit by rock. waves are just for illustration purposes and are not part of the problem.

We’re trying to find the frequency of ripples on water at 20 °C from a stone. When the stone hits the water, it has a kinetic energy of 0.5 m2kg/s2. We get this from the equation KE=mv2/2. The water it passes through has a density of ≈0.9982 kg/m3. At T=0, the ripple will begin. When the stone is completely submerged, the first ripple will end. The time between is the frequency of the ripples. But how long will it take the ball to submerge? The deeper it is, the more water it must push. But the slower it goes, the less energy required to push it. There is always the force of gravity on the ball. The ball weighs 1kg and will need to displace ≈1333π cm3 of water. The volume of a sphere is 4/3πr3. This actually weighs ≈4.1802 kilograms, far more than the stone. The stone will float! The buoyant forces acting on the rock will push it upwards, with a force equal to the weight of the water displaced. The water will slow it down according to the drag equation, Cd*p*A*v2/2, where Cd is the drag coefficient, p is the density of the fluid, A is the frontal area, and v is the velocity relative to the fluid. The drag coefficient for a sphere is 0.47. The density of water is 0.9982 kg/m3. So drag is taken care of. Finally, we have the force of gravity, pushing down with the weight of the rock. Calculating the time it takes for the rock’s velocity to fall to 0 will be another challenge, Gangbuster Geometry.

Solution to Ubiquitous Units

We’re trying to solve the system of equations left in Fascinating Frequencies 2. To start, factor out v in the second equation. We do this by multiplying both sides by ms/4πd. Then we combine the 2 equations. Next, substitute in the gravitational constant G and factor out d to find the distance. We factor out d by multiplying both sides by 72000s*d3/π and taking the fourth root of both sides. Substitute d into the first equation to solve for v. Our answer is approximately 8,032 metres per second.

\vec v=\frac{2\pi d}{72\,000s}\\[8pt] \frac{4\pi\vec vd}{ms}=\frac{Gm_1}{d^2}\\[8pt] G=\frac{6.674m^3}{10^{11}s^2kg}\\[8pt] \vec v=\frac{Gmm_1s}{4\pi d^3}=\frac{\pi d}{36\,000s}\\[8pt] \vec d\approx\sqrt[4]{\frac{2.403m^4m_1s^2}{4\times10^5\pi^2s^2kg}-1}\\[8pt] \vec v\approx\sqrt[4]{\frac{\pi^2m^4m_1s^2}{11.99 \times10^8\pi^4s^6kg}-1}\\[8pt] \vec v\approx\sqrt[4]{\frac{4.162\times10^{15}m^4}{\pi^2s^4}}=8.032\times10^3\frac ms

Fascinating Frequencies 3

Some chemistry principles are required to solve this challenge.

Diagram of pond after being hit by rock. waves are just for illustration purposes and are not part of the problem.

In this Fascinating Frequencies challenge, we will cover fluid dynamics. A perfect homogeneous sphere with radius 10 cm and mass 1 kg hits the middle of a pond of pure water on earth. It’s velocity is 1 m/s downward. The temperature of the pond is a uniform 20° celsius. The pond will begin to ripple. There are no animals in the pond to interfere with this. The challenge is to calculate the frequency of the ripples.

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