di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Electromagnetism


Question: Two electrons are in a uniform electric field $\mathbf{E}=E_{0} \hat{\mathbf{z}}$ where $E_{0}=10^{-11} \mathrm{~N} / \mathrm{C}$. One electron is at the origin, and another is $10 \mathrm{~m}$ above the first electron. The electron at the origin is moving at $u=10 \mathrm{~m} / \mathrm{s}$ at an angle of $30^{\circ}$ from the line connecting the electrons at $t=0$, while the other electron is at rest at $t=0$. Find the minimum distance between the electrons. You may neglect relativistic effects.

![](https://cdn.mathpix.com/cropped/2023_12_21_a93d620a5fbe0da0bef9g-1.jpg?height=605&width=1049&top_left_y=1351&top_left_x=538)

Solution:
Let $\ell=10 \mathrm{~m}$. First, switch into the reference frame accelerating at $-\frac{E q}{m} \hat{z}$. In this frame, the electrons are not affected by the electric field. Now, switch into the center of mass reference frame from here. In this frame, we have both conservation of angular momentum and conservation of energy. Both electrons in this frame are moving at $\frac{u}{2}$ initially at an angle of $\theta=30^{\circ}$. At the smallest distance, both electrons will be moving perpendicular to the line connecting them. Suppose that they both move with speed $v$ and are a distance $r$ from the center of mass. By conservation of angular momentum,
$$
\begin{gathered}
2 m \cdot \frac{u}{2} \cdot \frac{\ell}{2} \sin \theta=2 m v r \\
v r=\frac{u \ell}{4} \sin \theta
\end{gathered}
$$



Now, by conservation of energy,

$$
m v^{2}+\frac{k e^{2}}{2 r}=\frac{1}{4} m u^{2}+\frac{k e^{2}}{\ell}
$$

Now, we just solve this system of equations to determine the value of $r$. Substituting $v=\frac{u l}{4 r} \sin \theta$ into the conservation of energy equation, we can solve the ensuing quadratic to find:

$$
r=\frac{\frac{k e^{2}}{2}+\sqrt{\left(\frac{k e^{2}}{2}\right)^{2}+\left(m u^{2}+\frac{4 k e^{2}}{\ell}\right)\left(\frac{m u^{2} \ell^{2}}{16} \sin ^{2}(\theta)\right)}}{\frac{1}{2} m u^{2}+\frac{2 k e^{2}}{\ell}}
$$

Finally, remembering that the distance between the electrons is actually $2 r$, we obtain $2 r=6.84 \mathrm{~m}$ as the final answer.

Final answer: 6.84 m
Subfield: Electromagnetism


Question: Consider a long uniform conducting cylinder. First, we divide the cylinder into thirds and remove the middle third. Then, we perform the same steps on the remaining two cylinders. Again, we perform the same steps on the remaining four cylinders and continuing until there are 2048 cylinders.

We then connect the terminals of the cylinder to a battery and measure the effective capacitance to be $C_{1}$. If we continue to remove cylinders, the capacitance will reach an asymptotic value of $C_{0}$. What is $C_{1} / C_{0}$ ?

You may assume each cylindrical disk to be wide enough to be considered as an infinite plate, such that the radius $R$ of the cylinders is much larger than the $d$ between any successive cylinders.
![](https://cdn.mathpix.com/cropped/2023_12_21_09b501d99d90cb579ed8g-1.jpg?height=192&width=1006&top_left_y=1272&top_left_x=556)

Note: The diagram is not to scale.

Solution:
The capacitance is proportional to $C \propto \frac{1}{d}$, where $d$ is the distance between successive parallel plates. When we add capacitor plates in series, their effective capacitance will be:
$$
C \propto\left(\frac{1}{1 / d_{1}}+\frac{1}{1 / d_{2}}+\cdots\right)^{-1}=\frac{1}{d_{1}+d_{2}+\cdots} \Longrightarrow C \propto \frac{1}{d_{\text {total }}}
$$

Therefore, this essentially becomes a math problem: What is the total length of the spacing in between? Between successive 'cuts', the length of each cylinder is cut down by $1 / 3$, but the number of gaps double. Therefore, the spacing grows by a factor of $2 / 3$ each time. For $n=2^{1}$, the spacing starts off as $1 / 3$. For $n=2^{10}$, the spacing becomes:

$$
\frac{1}{C_{\text {eff }}} \propto d=\frac{1}{3}\left(\frac{1-(2 / 3)^{10}}{1-2 / 3}\right) L=0.983 L
$$

for $n \rightarrow \infty$, it is clear the total spacing will converge to $L$. Therefore:

$$
C_{1} / C_{0}=1.017
$$

Final answer: 1.017
Subfield: Mechanics


Question: A square based pyramid (that is symmetrical) is standing on top of a cube with side length $\ell=10 \mathrm{~cm}$ such that their square faces perfectly line up. The cube is initially standing still on flat ground and both objects have the same uniform density. The coefficient of friction between every surface is the same value of $\mu=0.3$. The cube is then given an initial speed $v$ in some direction parallel to the floor. What is the maximum possible value of $v$ such that the base of the pyramid will always remain parallel to the top of the cube? Answer in meters per second.

![](https://cdn.mathpix.com/cropped/2023_12_21_19adaf49165bed09665eg-1.jpg?height=1017&width=772&top_left_y=554&top_left_x=671)

Solution:
Let $x$ be the relative displacement of the two objects. Then:
$$
v_{i}^{2}=2 a x
$$

where the relative acceleration is:

$$
a=\frac{2 h}{3 \ell} g \mu
$$

work The acceleration of pyramid is $g \mu$ and the acceleration of cube is:

$$
\rho \ell^{3} a=\rho\left(\ell^{3}+\frac{\ell^{2} h}{3}\right) g \mu+\rho \frac{\ell^{2} h}{3} g \mu \Longrightarrow a=\frac{3 \ell+2 h}{3 \ell} g \mu
$$

Therefore, the relative acceleration is:

$$
a=\frac{6 \ell+2 h}{3 \ell} g \mu
$$

We want to direct the motion of the cube diagonally (such that horizontal sides of the cube form a 45 degree angle with the displacement). Initially, we may think that we need to let $x=\frac{\sqrt{2}}{2} \ell$ but it can start tipping before that. Moving into a non-inertial reference frame for the pyramid, we see that the effective gravity needs to point towards the back corner of the cube, so it needs to satisfy the criteria

$$
\frac{h_{\mathrm{cm}}}{\frac{\sqrt{2}}{2} \ell-x}=\frac{m g}{m g \mu} \Longrightarrow x=\frac{\sqrt{2}}{2} \ell-\mu \frac{1}{4} h
$$



Here I used the fact that the center of mass was $1 / 4$ of the way up. Substituting, we get:

$$
v=\sqrt{2\left(\frac{6 \ell+2 h}{3 \ell} g \mu\right)\left(\frac{\sqrt{2}}{2} l-\frac{\mu}{4} h\right)}=\sqrt{2 g \mu\left(2+\frac{2 h}{3 \ell}\right)\left(\frac{\sqrt{2}}{2} \ell-\frac{\mu}{4} h\right)}
$$

The maximum $v$ occurs at

$$
h / \ell=\frac{-3+\frac{2 \sqrt{2}}{\mu}}{2}
$$

giving a maximum $v$ of $v_{\max }=1.07 \mathrm{~m} / \mathrm{s}$.

Final answer: 1.07 m/s
Subfield: Mechanics


Question: During quarantine, the FBI has been monitoring a young physicists suspicious activities. After compiling weeks worth of evidence, the FBI finally has had enough and searches his room.

The room's door is opened with a high angular velocity about its hinge. Over a very short period of time, its angular velocity increases to $\omega=8.56 \mathrm{rad} / \mathrm{s}$ due to the force applied at the end opposite from the hinge. For simplicity, treat the door as a uniform thin rod of length $L=1.00 \mathrm{~m}$ and mass $M=9.50 \mathrm{~kg}$. The hinge (pivot) is located at one end of the rod. Ignore gravity. At what distance from the hinge of the door is the door most likely to break? Assume that the door will break at where the bending moment is largest. (Answer in metres.)

![](https://cdn.mathpix.com/cropped/2023_12_21_949eb8cca07484578c58g-1.jpg?height=212&width=605&top_left_y=1225&top_left_x=760)

Solution:
Let $N$ be the force from the pivot and $F$ be the applied force at the end. Let $\alpha$ be the angular acceleration. Writing the torque equation and Newton's 2nd law for the whole door, we get:
$$
\begin{aligned}
& F \cdot L=\frac{1}{3} M L^{2} \alpha \\
& N+F=\frac{1}{2} M L \alpha
\end{aligned}
$$

Solving, we get $F=\frac{1}{3} M L \alpha$ and $N=\frac{1}{6} M L \alpha$. Now, we consider the part of the door with length $x$ attached to the pivot. The rest of the door applies a torque $\tau$ and shear force $f$ on our system. (There is also tension force). Let $\lambda=\frac{M}{L}$. We can write the torque equation and Newton's 2 nd law for our system:

$$
\begin{aligned}
& \tau+f x=\frac{1}{3} \lambda x^{3} \alpha \\
& N+f=\lambda x \cdot \frac{x}{2} \alpha
\end{aligned}
$$

Solving, we get

$$
\tau=\frac{1}{6} \lambda x \alpha\left(L^{2}-x^{2}\right)
$$

and

$$
f=\frac{1}{6} \lambda \alpha\left(3 x^{2}-L^{2}\right)
$$

We maximize $\tau$ (which is equivalent to maximizing bending moment) to get $x=\frac{L}{\sqrt{3}}=0.577 \mathrm{~m}$

Final answer: $\frac{\sqrt{3}}{3}$ m
Subfield: Mechanics


Question: A solid half-disc of mass $m=1 \mathrm{~kg}$ in the shape of a semi-circle of radius $R=1 \mathrm{~m}$ is kept at rest on a smooth horizontal table. QiLin starts applying a constant force of magnitude $F=10 \mathrm{~N}$ at point A as shown, parallel to its straight edge. What is the initial linear acceleration of point B? (Answer in $\mathrm{m} / \mathrm{s}^{2}$ )

![](https://cdn.mathpix.com/cropped/2023_12_21_325b4a9088b5394ee833g-1.jpg?height=409&width=1004&top_left_y=438&top_left_x=558)

Note: the diagram above is a top down view.

Solution:
Let C denote the location of the centre of mass of the disc. It is well known that $O C=\frac{4 R}{3 \pi}$. Note that the initial angular velocity of the disc about its centre of mass is 0 , and the linear acceleration is simply
$$
a_{\mathrm{CM}}=\frac{F}{m} \hat{i}
$$

Now we use the $\tau=\vec{r} \times \vec{F}=I_{\mathrm{CM}} \alpha_{\mathrm{CM}}$ about the centre of mass of the disc

$$
\left(\frac{4 R}{3 \pi}\right) F=I_{\mathrm{CM}} \alpha_{\mathrm{CM}}
$$

To compute the moment of inertia of the disc about its centre of mass, we use Steiner's theorem:

$$
I_{\mathrm{CM}}=\frac{M R^{2}}{2}-M\left(\frac{4 R}{3 \pi}\right)^{2}
$$

so the $\tau=I \alpha$ equation becomes

$$
\left(\frac{4 R}{3 \pi}\right) F=\left[\frac{M R^{2}}{2}-M\left(\frac{4 R}{3 \pi}\right)^{2}\right] \alpha_{\mathrm{CM}} \Rightarrow \alpha_{\mathrm{CM}}=\frac{\frac{4 F R}{3 \pi}}{M R^{2}\left(\frac{1}{2}-\frac{16}{9 \pi^{2}}\right)}
$$

Using kinematics equation for rotational motion, we have

$$
\vec{a}_{B}=\vec{a}_{\mathrm{CM}}+\vec{\alpha}_{\mathrm{CM}} \times \overrightarrow{\mathrm{CB}}
$$

Substituting the values of $\overrightarrow{\mathrm{CB}}=R \hat{i}-\frac{4 R}{3 \pi} \hat{j}$ and $\alpha_{\mathrm{CM}}=\frac{\frac{4 F R}{3 \pi}}{M R^{2}\left(\frac{1}{2}-\frac{16}{9 \pi^{2}}\right)} \hat{k}$ we get

$$
a_{B}=\frac{F}{m}\left[\left(1+\frac{1}{-1+\frac{9 \pi^{2}}{32}}\right) \hat{i}+\frac{2}{3 \pi\left(\frac{1}{2}+\frac{16}{9 \pi^{2}}\right)} \hat{j}\right]=15.9395 \mathrm{~m} / \mathrm{s}^{2}
$$

and we are done.

Final answer: $15.9395$ $\mathrm{m} / \mathrm{s}^{2}$
Subfield: Mechanics


Question: A regular tetrahedron of mass $m=1 \mathrm{~g}$ and unknown side length is balancing on top of a hemisphere of mass $M=100 \mathrm{~kg}$ and radius $R=100 \mathrm{~m}$. The hemisphere is placed on a flat surface such that it is at its lowest potential. For a certain value of the length of the regular tetrahedron, the oscillations become unstable. What is this side length of the tetrahedron?

![](https://cdn.mathpix.com/cropped/2023_12_21_ac3dcf6b8fd2f77ccf4bg-1.jpg?height=382&width=403&top_left_y=481&top_left_x=861)

Solution:
For stable equilibrium the height resultant from a slight displacement must be greater than the original height of the center of mass of the cube. A small displacement from the original position can be modeled as an $\mathrm{x}$ displacement of
$$
R d \theta
$$

which raises the height by

$$
R d \theta \sin (d \theta)
$$

added to the height

$$
s \cos (d \theta)
$$

This must be greater than $s$, where $s$ is the distance from the center of mass to the point of contact to the sphere. Approximating to the second degree of $\theta$ using Taylor series, we solve the inequality and get that $s=R$. The altitude of the tetrahedron is therefore $4 s=400$

$$
l^{2}-(l \sqrt{3} / 3)^{2}=400^{2}
$$

Therefore, $l=\sqrt{400^{2} \cdot 3 / 2}=490 \mathrm{~m}$

Final answer: $200\sqrt{6}$ m
Subfield: Mechanics

- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,

$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$

- Solar Mass

$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$

- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,

$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$

- Planck's constant,

$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$

- Permittivity of free space,

$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$

- Coulomb's law constant,

$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$

- Permeability of free space,

$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$

- Magnetic constant,

$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$

- 1 atmospheric pressure,

$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$

- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,

$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$

Question: A regular tetrahedron of mass $m=1 \mathrm{~g}$ and unknown side length is balancing on top of a hemisphere of mass $M=100 \mathrm{~kg}$ and radius $R=100 \mathrm{~m}$. The hemisphere is placed on a flat surface such that it is at its lowest potential. For a certain value of the length of the regular tetrahedron, the oscillations become unstable. What is this side length of the tetrahedron?

<img_4267>

Solution:
For stable equilibrium the height resultant from a slight displacement must be greater than the original height of the center of mass of the cube. A small displacement from the original position can be modeled as an $\mathrm{x}$ displacement of
$$
R d \theta
$$

which raises the height by

$$
R d \theta \sin (d \theta)
$$

added to the height

$$
s \cos (d \theta)
$$

This must be greater than $s$, where $s$ is the distance from the center of mass to the point of contact to the sphere. Approximating to the second degree of $\theta$ using Taylor series, we solve the inequality and get that $s=R$. The altitude of the tetrahedron is therefore $4 s=400$

$$
l^{2}-(l \sqrt{3} / 3)^{2}=400^{2}
$$

Therefore, $l=\sqrt{400^{2} \cdot 3 / 2}=490 \mathrm{~m}$

Final answer: $200\sqrt{6}$ m
Subfield: Mechanics


Question: Two astronauts, Alice and Bob, are standing inside their cylindrical spaceship, which is rotating at an angular velocity $\omega$ clockwise around its axis in order to simulate the gravitational acceleration $g$ on earth. The radius of the spaceship is $R$. For this problem, we will only consider motion in the plane perpendicular to the axis of the spaceship. Let point $O$ be the center of the spaceship. Initially, an ideal zero-length spring has one end fixed at point $O$, while the other end is connected to a mass $m$ at the "ground" of the spaceship, where the astronauts are standing (we will call this point $A$ ). From the astronauts' point of view, the mass remains motionless.

Next, Alice fixes one end of the spring at point $A$, and attaches the mass to the other end at point $O$. Bob starts at point $A$, and moves an angle $\theta$ counterclockwise to point $B$ (such that $A O B$ is an isosceles triangle). At time $t=0$, the mass at point $O$ is released. Given that the mass comes close enough for Bob to catch it, find the value of $\theta$ to the nearest tenth of a degree.

Assume that the only force acting on the mass is the spring's tension, and that the astronauts' heights are much less than $R$.

![](https://cdn.mathpix.com/cropped/2023_12_21_60ae367721568530fa75g-1.jpg?height=1206&width=1217&top_left_y=964&top_left_x=454)

Solution:
First, we must have $\omega^{2} R=g$ and $k=m \omega^{2}$. Now, we step into the frame of point $A$, rotating around with the spaceship. We will thus have three fictitious forces: translational, centrifugal, and coriolis. Note that because centrifugal is $m \omega^{2} r$, pointing away from $A$, it cancels with the spring force. Thus, the only forces left to consider are translational and coriolis.
The translational force points "down" with a constant magnitude of $m g$, like gravity. The coriolis force



points perpendicular to the velocity with magnitude $2 m \omega v$. We recognize this setup is analogous to that of a charged particle moving in an E field and B field. It is well known that the mass will follow a cycloid shape. Writing the equation of the cycloid, and finding where the cycloid hits the circle (spaceship), we can find $\theta$. Note that we have to use numerical methods.

Specifically, the cycloid can be parametrized as $\frac{R}{4}(\alpha-\sin \alpha, 1-\cos \alpha)$, and we need to find where this intersects the circle $x^{2}+y^{2}=R^{2}$, so $(\alpha-\sin \alpha)^{2}+(1-\cos \alpha)^{2}=16$. Solving gives $\alpha=3.307$, and since $\cot \theta=\frac{1-\cos \alpha}{\alpha-\sin \alpha}$, we have $\theta=60.2^{\circ}$.

Final answer: $60.2$ $^{\circ}$
Subfield: Mechanics


Question: A bicycle wheel of mass $M=2.8 \mathrm{~kg}$ and radius $R=0.3 \mathrm{~m}$ is spinning with angular velocity $\omega=5 \mathrm{rad} / \mathrm{s}$ around its axis in outer space, and its center is motionless. Assume that it has all of its mass uniformly concentrated on the rim. A long, massless axle is attached to its center, extending out along its axis. A ball of mass $m=1.0 \mathrm{~kg}$ moves at velocity $v=2 \mathrm{~m} / \mathrm{s}$ parallel to the plane of the wheel and hits the axle at a distance $h=0.5 \mathrm{~m}$ from the center of the wheel. Assume that the collision is elastic and instantaneous, and that the ball's trajectory (before and after the collision) lies on a straight line.

![](https://cdn.mathpix.com/cropped/2023_12_21_ace597fee09961022231g-1.jpg?height=1207&width=1130&top_left_y=562&top_left_x=492)

Find the time it takes for the axle to return to its original orientation. Answer in seconds and round to three significant figures.

Solution:
After the collision, let the wheel have speed $v_{1}$ and the ball have speed $v_{2}$. Conserving momentum, energy, and angular momentum gives:
$$
\begin{gathered}
m v=M v_{1}+m v_{2} \\
\frac{1}{2} m v^{2}+\frac{1}{2} M R^{2} \omega^{2}=\frac{1}{2} M v_{1}^{2}+\frac{1}{2} m v_{2}^{2}+\frac{1}{2} M R^{2} \omega^{2}+\frac{1}{2} \cdot \frac{1}{2} M R^{2} \omega_{1}^{2} \\
m\left(v-v_{2}\right) h=\frac{1}{2} M R^{2} \omega_{1}
\end{gathered}
$$
where $\omega_{1}$ is the angular velocity (after collision) of the wheel in the direction perp. to the axis and the velocity of the ball.



Solving for $\omega_{1}$, we get
$$
\omega_{1}=\frac{4 h m v}{m\left(R^{2}+2 h^{2}\right)+M R^{2}}
$$
Now, we realize that the angular momentum of the wheel is given by $I_{x} \omega \hat{x}+I_{y} \omega_{1} \hat{y}$ where the wheel's axis is the $\mathrm{x}$-axis and the $\mathrm{y}$-axis is in the direction of $\omega_{1}$. Since angular momentum is conserved, the wheel must precess about its angular momentum vector. Let $\hat{L}$ represent the direction of the angular momentum vector. To find the rate of precession, we can decompose the angular velocity vector $\omega \hat{x}+\omega_{1} \hat{y}$ into a $\hat{L}$ component and a $\hat{x}$ component. Since $I_{x}=2 I_{y}$, the $\hat{L}$ component is $\sqrt{(2 \omega)^{2}+\omega_{1}^{2}}$, resulting in a precession period of
$$
T=\frac{\pi}{\sqrt{\omega^{2}+\frac{\omega_{1}^{2}}{4}}}=0.568 s
$$

Final answer: 0.568 s
Subfield: Mechanics


Question: A small toy car rolls down three ramps with the same height and horizontal length, but different shapes, starting from rest. The car stays in contact with the ramp at all times and no energy is lost. Order the ramps from the fastest to slowest time it takes for the toy car to drop the full $1 \mathrm{~m}$. For example, if ramp 1 is the fastest and ramp 3 is the slowest, then enter 123 as your answer choice.

Figure 1

![](https://cdn.mathpix.com/cropped/2023_12_21_45699e63605c39917edcg-1.jpg?height=439&width=480&top_left_y=645&top_left_x=275)

Figure 2

![](https://cdn.mathpix.com/cropped/2023_12_21_45699e63605c39917edcg-1.jpg?height=442&width=471&top_left_y=641&top_left_x=835)

Figure 3

![](https://cdn.mathpix.com/cropped/2023_12_21_45699e63605c39917edcg-1.jpg?height=439&width=477&top_left_y=645&top_left_x=1385)

Solution:
One can solve this by finding the time it takes for each ramp. For ramp 1:
$$
\begin{aligned}
\sqrt{2} & =\frac{1}{2} g \sin \left(45^{\circ}\right) t^{2} \\
\Longrightarrow t & =0.639 \mathrm{~s}
\end{aligned}
$$

For ramps 2, let the length of the dashed region be $x$. Then:

$$
x+x / \tan \left(30^{\circ}\right)=1 \Longrightarrow x=0.366 \mathrm{~m}
\tag{1}
$$

Due to symmetry, both the steep and shallow regions of both ramps 2 and 3 have a length of $x / \cos \left(60^{\circ}\right)=0.732 \mathrm{~m}$. This results in a time for ramp 2 as:

$$
\begin{aligned}
& 0.732=\frac{1}{2} g \sin \left(60^{\circ}\right) t_{1}^{2} \Longrightarrow t_{1}=0.415 \mathrm{~s} \\
& 0.732=\sqrt{2 g(1-x)} t_{2}+\frac{1}{2} g \sin \left(30^{\circ}\right) t_{2}^{2} \Longrightarrow t_{2}=0.184 \mathrm{~s}
\end{aligned}
$$

for a total time of $t_{1}+t_{2}=0.599 \mathrm{~s}$. For ramp 3 ,

$$
\begin{aligned}
& 0.732=\frac{1}{2} g \sin \left(30^{\circ}\right) t_{1}^{2} \Longrightarrow t_{1}=0.546 \mathrm{~s} \\
& 0.732=\sqrt{2 g(1-x)} t_{2}+\frac{1}{2} g \sin \left(60^{\circ}\right) t_{2}^{2} \Longrightarrow t_{2}=0.206 \mathrm{~s}
\end{aligned}
$$

which gives a total time of $t_{1}+t_{2}=0.752 \mathrm{~s}$. From fastest to slowest, the answer becomes 213 . Note that this answer is easily guessable via intuition.

Final answer: 213