weblab-llm-competition-2025-bridge/neko-prelim-HLE_SFT_OlympiadBench huggingface dataset:
Problem: A rocket with mass of 563.17 (not including the mass of fuel) metric tons sits on the launchpad of the Kennedy Space Center (latitude $28^{\circ} 31^{\prime} 27^{\prime \prime} \mathrm{N}$, longitude $80^{\circ} 39^{\prime} 03^{\prime \prime} \mathrm{W}$ ), pointing directly upwards. Two solid fuel boosters, each with a mass of $68415 \mathrm{~kg}$ and providing $3421 \mathrm{kN}$ of thrust are pointed directly downwards.
The rocket also has a liquid fuel engine, that can be throttled to produce different amounts of thrust and gimbaled to point in various directions. What is the minimum amount of thrust, in $\mathrm{kN}$, that this engine needs to provide for the rocket to lift vertically (to accelerate directly upwards) off the launchpad?

Assume $G=6.674 \times 10^{-11} \frac{\mathrm{N} \cdot \mathrm{m}^{2}}{\mathrm{~s}^{3}}$, and that the Earth is a perfect sphere of radius $6370 \mathrm{~km}$ and mass $5.972 \times 10^{24} \mathrm{~kg}$ that completes one revolution every $86164 \mathrm{~s}$ and that the rocket is negligibly small compared to the Earth. Ignore buoyancy forces.

Solution: Note the additional information provided in the problem (latitude, Earth radius, revolution period), which makes it clear that the effect of the rotation of the Earth must also be considered.
We first compute local gravitational acceleration:

$$
g=\frac{G M}{R^{2}}=9.823 \frac{\mathrm{m}}{\mathrm{s}^{2}}
$$



And also acceleration due to the Earth's rotation:

$$
a=(R \cos \theta) \omega^{2}=0.02976 \frac{\mathrm{m}}{\mathrm{s}^{2}}
$$

Then vertical acceleration is:

$$
g-a \cos \theta=9.7965 \frac{\mathrm{m}}{\mathrm{s}^{2}}
$$

And horizontal acceleration:

$$
a \sin \theta=0.0142 \frac{m}{s^{2}}
$$

The total mass of the craft is 700 metric tons, so the needed forces for liftoff are:

$$
\left(F_{x}, F_{y}\right)=(6857527,9948) \mathrm{N}
$$

Subtracting out the effect of the solid fuel boosters, the force the liquid fuel engine needs to provide is:

$$
\left(F_{x}, F_{y}\right)=(15527,9948) \mathrm{N}
$$

Answer: $18.44$
Problem: A spacecraft is orbiting in a very low circular orbit at a velocity $v_{0}$ over the equator of a perfectly spherical moon with uniform density. Relative to a stationary frame, the spacecraft completes a revolution of the moon every 90 minutes, while the moon revolves in the same direction once every 24 hours. The pilot of the spacecraft would like to land on the moon using the following process:So the final answer is 

1. Start by firing the engine directly against the direction of motion.So the final answer is 

2. Orient the engine over time such that the vertical velocity of the craft remains 0 , while the horizontal speed continues to decrease.

3. Once the velocity of the craft relative to the ground is also 0 , turn off the engine.

Assume that the engine of the craft can be oriented instantly in any direction, and the craft has a TWR (thrust-to-weight ratio, where weight refers to the weight at the moon's surface) of 2, which remains constant throughout the burn. If the craft starts at $v_{0}=500 \mathrm{~m} / \mathrm{s}$, compute the delta-v expended to land, minus the initial velocity, i.e. $\Delta v-v_{0}$.

Solution: The trick in this question is to work in dimensionless units. Let $v$ be the ratio of the craft's velocity to orbital velocity. Then, if the craft has horizontal velocity $v$, the acceleration downwards is the following:

$$
a_{V}=\frac{v_{0}^{2}}{r}-\frac{\left(v_{0} v\right)^{2}}{r}=\frac{v_{0}^{2}}{r}\left(1-v^{2}\right)=g_{m}\left(1-v^{2}\right)
$$

As the TWR is 2 , the total acceleration the engine provides is $a_{0}=2 g_{m}$, where $g_{m}$ is the surface gravity of the moon. As this total acceleration is the sum of horizontal and vertical components, and the vertical component cancels out the downwards acceleration:

$$
a_{H}=\sqrt{\left(2 g_{m}\right)^{2}-\left(g_{m}\left(1-v^{2}\right)\right)^{2}}=g_{m} \sqrt{2^{2}-\left(1-v^{2}\right)^{2}}
$$

And $a_{H}$ is related to $v$ (which is dimensionless!) by the following relation:

$$
\begin{gathered}
\frac{\mathrm{d}\left(v_{0} v\right)}{\mathrm{d} t}=-a_{H} \\
\frac{\mathrm{d} v}{\mathrm{~d} t}=-\frac{g_{m}}{v_{0}} \sqrt{2^{2}-\left(1-v^{2}\right)^{2}} \\
\frac{\mathrm{d} v}{\sqrt{2^{2}-\left(1-v^{2}\right)^{2}}}=-\frac{g_{m}}{v_{0}} \mathrm{~d} t
\end{gathered}
$$

At the start, $v=1$. However, at landing, velocity of the craft is 0 relative to the surface, not a stationary frame! Therefore, we use the orbital periods data to determine the final $v$ to be $\frac{1.5}{24}=\frac{1}{16}$. Then integrating:

$$
\int_{v=1}^{v=1 / 16} \frac{\mathrm{d} v}{\sqrt{2^{2}-\left(1-v^{2}\right)^{2}}}=-\frac{g_{m}}{v_{0}} t
$$

And as delta-v is related to time by $\Delta v=a_{0} t$ :

$$
\Delta v=a_{0} t=a_{0} \frac{v_{0}}{g_{m}} \int_{1 / 16}^{1} \frac{\mathrm{d} v}{\sqrt{2^{2}-\left(1-v^{2}\right)^{2}}}=a_{0} \frac{v_{0}}{a_{0}} 2(0.503)=500 \cdot 1.006=503.06 \frac{\mathrm{m}}{\mathrm{s}}
$$

And subtracting:

$$
\Delta v-v_{0}=3.06 \frac{\mathrm{m}}{\mathrm{s}}
$$

Answer: $3.06$
Problem: Consider two points $S$ and $S^{\prime}$ randomly placed inside a $D$-dimensional hyper-rectangular room with walls that are perfect-reflecting $(D-1)$-dimensional hyper-plane mirrors. How many different

light-rays that start from $S$, reflect $N$ times on one of the walls and $N-1$ times on each of the rest, then go to $S^{\prime}$ ? Use $D=7$ and $N=3$.

Solution: Using the hyper-rectangular room as the fundamental unit-cell of an infinite hyper-grid in space, then find all possible positions of $S^{\prime}$-images through reflections: we can realize that there is only one position of $S^{\prime}$-image inside every unit-cell.

(A)
<img_4257>
<img_4259>

Consider two opposite hyper-plane mirrors, since the rest of the mirrors are perpendicular to them, the numbers of reflections on them for any light-path traveled from point $S$ and point $S^{\prime}$ can only different by 1 or less. If the numbers are both equal and non-zero, then the available positions $S^{\prime}$ image the light-path from $S$ should reach are two unit-cell hyper-rows that parallel to the mirrors. If the numbers are not equal, then the available positions $S^{\prime}$-image the light-path from $S$ should reach are one unit-cell hyper-rows that that parallel to the mirrors.

Say, without loss of generality, pick one mirror to be reflected $N$ times and the rest to be reflected $N-1$ times each, then the number of light-rays for that pick should equal to the number of unit-cell intersections between all relevant unit-cell hyper-rows, which is half the total number of vertices a hyper-rectangular has, thus $2^{D-1}$. There are $2 D$ walls, thus the total number of light-rays that satisfies the task given is $2 D \times 2^{D-1}=D 2^{D}$, independent of $N$ for all values $N>1$.



To illustrate the above explanation, let's take a look at the simple case of $D=2$ and $N=2>1$. Consider a rectangular room, with random points $S, S^{\prime}$ and $2 D=4$ walls $W_{1}, W_{2}, W_{3}, W_{4}$ (see Fig. A). Without loss of generality, we want to find light-rays that go from $S$, reflect $N=2$ times on $W_{1}$ and $N-1=1$ times on $W_{2}, W_{3}, W_{4}$ then come to $S^{\prime}$. Each image of $S^{\prime}$ is an unique point in a unit-cell generated by the room (see Fig. B). Note that every light-ray from $S$ that reach $S^{\prime}$-images in the $\rightarrow W_{2} \rightarrow W_{4} \rightarrow$ and $\rightarrow W_{4} \rightarrow W_{2} \rightarrow$ unit-cell green-rows will satisfy the requirement of one reflection on each of $W_{2}$ and $W_{4}$, every light-ray from $S$ that reach $S^{\prime}$-images in the $\rightarrow W_{1} \rightarrow W_{3} \rightarrow W_{1} \rightarrow$ unit-cell blue-rows will satisfy the requirement of two reflection on $W_{1}$ and one reflection on $W_{3}$ (see Fig. C). The intersection of these rows are two unit-cells, corresponds two possible images thuss two possible light-rays that satisfies the requirement (see Fig. D).

For $D=7$ and $N=3>1$, we get 896 light-rays.

Answer: 895
Problem: Two concentric isolated rings of radius $a=1 \mathrm{~m}$ and $b=2 \mathrm{~m}$ of mass $m_{a}=1 \mathrm{~kg}$ and $m_{b}=2 \mathrm{~kg}$ are kept in a gravity free region. A soap film of surface tension $\sigma=0.05 \mathrm{Nm}^{-1}$ with negligible mass is spread over the rings such that it occupies the region between the rings. The smaller ring is pulled slightly along the axis of the rings. Find the time period of small oscillation in seconds.

Solution: The force on the two rings when they are a distance $L$ apart follows as
$$
F=4 \pi r \sigma \sin \theta
$$

In small displacements, the change in $\theta$ is small. Therefore,

$$
\begin{aligned}
F & =4 \pi r \sigma \theta \\
\frac{F}{4 \pi r \sigma} & =\frac{\mathrm{d} y}{\mathrm{~d} r} \\
\int_{a}^{b} \frac{F}{4 \pi \sigma r} \mathrm{~d} r & =\int_{0}^{L} \mathrm{~d} y \\
\frac{F}{4 \pi \sigma} \ln (b / a) & =L
\end{aligned}
$$

Now let $a_{1}$ and $a_{2}$ be acceleration of a and $\mathrm{b}$ respectively. We have that

$$

a_{\text {net }}=a_{1}+a_{2}
\tag{12}
$$
$$
a_{\text {net }}=F\left(\frac{m_{1}+m_{2}}{m_{1} m_{2}}\right)
\tag{13}
$$
$$
L \omega^{2}=\frac{4 \pi \sigma}{\ln (b / a)} L\left(\frac{m_{1}+m_{2}}{m_{1} m_{2}}\right)
\tag{14}
$$
$$
T=2 \pi \sqrt{\frac{\ln (b / a) m_{1} m_{2}}{4 \pi \sigma\left(m_{1}+m_{2}\right)}}
\tag{15}
$$
$$
T=2 \pi \sqrt{\frac{10 \ln (2)}{3 \pi}}=5.388 \mathrm{~s}
\tag{16}
$$

Answer: $2 \pi \sqrt{\frac{10 \ln (2)}{3 \pi}}$
Problem: A table of unknown material has a mass $M=100 \mathrm{~kg}$, width $w=4 \mathrm{~m}$, length $\ell=3 \mathrm{~m}$, and 4 legs of length $L=0.5 \mathrm{~m}$ with a Young's modulus of $Y=1.02 \mathrm{MPa}$ at each of the corners. The cross-sectional area of a table leg is approximately $A=1 \mathrm{~cm}^{2}$. The surface of the table has a coefficient of friction of $\mu=0.1$. A point body with the same mass as the table is put at some position from the geometric center of the table. What is the minimum distance the body must be placed from the center such that it slips on the table surface immediately after? Report your answer in centimeters.
The table surface and floor are non-deformable.

Solution: This problem requires some 3 dimensional reasoning. Suppose $\mathbf{s}=\left(s_{x}, s_{y}\right)$ is the gradient of the table. We can use this to calculate the additional torque from the displacement of the mass. The forces from each table leg are
$$
F_{i}=\frac{Y A}{L}\left( \pm s_{x} \frac{\ell}{2} \pm s_{y} \frac{w}{2}\right)
$$

Taking the cross product as $\boldsymbol{\tau}=\mathbf{r} \times \mathbf{F}_{\mathbf{i}}$ shows that torque is given as

$$
\boldsymbol{\tau}=\frac{Y A}{L}\left(\begin{array}{c}
-s_{y} w^{2} \\
s_{x} \ell^{2}
\end{array}\right)
$$

which must balance out the torque $M g d$ from a point mass. Hence, rewriting yields

$$
d=\frac{Y A}{M g L} \sqrt{s_{y}^{2} w^{4}+s_{x}^{2} \ell^{4}}
$$

Furthermore, note that the angle required from slipping is given from a force analysis as

$$
m g \sin \theta=\mu m g \cos \theta \Longrightarrow \mu=\tan \theta=|\mathbf{s}|=\sqrt{s_{x}^{2}+s_{y}^{2}}
$$

When $w>\ell$, we can rewrite

$$
s_{y}^{2} w^{4}+s_{x}^{2} \ell^{4}=s_{y}^{2}\left(w^{4}-\ell^{4}\right)+\ell^{4} \mu^{2}
$$

which is minimized to $\ell^{4} \mu^{2}$ when $s_{y}=0$. Hence, we obtain

$$
d=\frac{\mu \ell^{2} Y A}{M g L}
$$

Answer: $18.71$
Problem: A uniform spherical metallic ball of mass $m$, resistivity $\rho$, and radius $R$ is kept on a smooth friction-less horizontal ground. A horizontal uniform, constant magnetic field $B$ exists in the space parallel to the surface of ground. The ball was suddenly given an impulse perpendicular to magnetic field such that ball begin to move with velocity $v$ without losing the contact with ground. Find the time in seconds required to reduce its velocity by half.
Numerical Quantities: $m=2 \mathrm{~kg}, 4 \pi \epsilon_{0} R^{3} B^{2}=3 \mathrm{~kg}, \rho=10^{9} \Omega \mathrm{m}, v=\pi \mathrm{m} / \mathrm{s}$.

Solution: WLOG assuming magnetic field to be into the plane (negative z-axis) and velocity of the block along x-axis. Let any time $\mathrm{t}$ ball has velocity $\mathrm{v}$ and surface charge $\sigma \cos \theta$, where $\theta$ is measured from $\mathrm{Y}$-axis. As the ball is moving it will also generate and electric field $\mathrm{E}=\mathrm{vB}$ along positive Y-axis. Which will be opposed by the electric field of ball. Also we know the electric filed generated by the ball is $\frac{\sigma}{3 \epsilon_{0}}$ in negative $\mathrm{Z}$-axis. As this charge distribution will arise form a vertical electric field and the subsequent induced charges will also produce only a vertical electric field only thus our assumption about charge distribution and net electric field must be true.
Now at any point on the surface of the sphere rate of increase in surface charge density is given by-



$$
\begin{gathered}
J d A \cos \theta=\frac{d(\sigma \cos \theta)}{d t} d A \\
J=\frac{d \sigma}{d t} \\
E-\frac{\sigma}{3 \epsilon_{0}}=\rho \frac{d \sigma}{d t} \\
v B-\frac{\sigma}{3 \epsilon_{0}}=\rho \frac{d \sigma}{d t}
\end{gathered}
\tag{44}
$$

As the magnetic field is uniform to calculate force path of the current will not matter. Hence assuming it to be straight line between two points located at $\theta$ and $-\theta$. So the force can be written as -

$$
\begin{gathered}
d F=B \times d I \times l \\
d F=B(2 \pi(R \sin \theta) \times R d \theta \times J)(2 R \cos (\theta)) \\
F=4 \pi R^{3} B \frac{d \sigma}{d t} \int_{0}^{\pi / 2} \sin \theta \cos ^{2} \theta d \theta \\
F=\frac{4}{3} \pi R^{3} B \frac{d \sigma}{d t}
\end{gathered}
\tag{45}
$$

Now writing force equation on the sphere we have

$$
\begin{gathered}
F=-m \frac{d v}{d t} \\
\frac{4}{3} \pi R^{3} B \frac{d \sigma}{d t}=-m \frac{d v}{d t} \\
\frac{4}{3} \pi R^{3} B \int_{0}^{\sigma} d \sigma=-m \int_{v_{0}}^{v} d v \\
\frac{4}{3} \pi R^{3} B \sigma=m v_{0}-m v
\end{gathered}
\tag{46}
$$

Solving equation 1 and 3 gives us

$$
\left(m+4 \pi R^{3} B^{2} \epsilon_{0}\right) v-m v_{0}=-3 m \rho \epsilon_{0} \frac{d v}{d t}
$$

Integrating it from $v_{0}$ to $\frac{v_{0}}{2}$ gives

$$

t=\frac{3 m \rho \epsilon_{0}}{\left(m+4 \pi R^{3} B^{2} \epsilon_{0}\right)} \ln \left(\frac{8 \pi R^{3} B^{2} \epsilon_{0}}{4 \pi R^{3} B^{2} \epsilon_{0}-m}\right)
\tag{47}
$$
$$
t=\frac{6 \ln (6)}{5} \rho \epsilon_{0}=0.019 \mathrm{~s}
\tag{48}
$$

Answer: $0.019$
Problem: Zed is trying to model the repulsive interaction between 2 objects, $A$ and $B$ (with masses $m_{A}$ and $m_{B}$, respectively), in a relativistic setting. He knows that in relativity, forces cannot act at a distance, so he models the repulsive force with a small particle of mass $m$ that bounces elastically between $A$ and $B$. Throughout this problem, assume everything moves on the x-axis. Suppose that initially, $A$ and $B$ have positions and velocities $x_{A}, v_{A}$ and $x_{B}, v_{B}$, respectively, where $x_{A}<x_{B}$ and $v_{A}>v_{B}$. The particle has an initial (relativistic) speed $v$.
For simplicity, assume that the system has no total momentum. You may also assume that $v_{A}, v_{B} \ll v$, and that $p_{m} \ll p_{A}, p_{B}$, where $p_{m}, p_{A}, p_{B}$ are the momenta of the particle, $A$, and $B$, respectively. Do NOT assume $v \ll c$, where $c$ is the speed of light.

Find the position (in $\mathrm{m}$ ) of $A$ when its velocity is 0 , given that $m_{A}=1 \mathrm{~kg}, m_{B}=2 \mathrm{~kg}, v_{A}=0.001 c$, $m=1 \times 10^{-6} \mathrm{~kg}, v=0.6 c, x_{A}=0 \mathrm{~m}, x_{B}=1000 \mathrm{~m}$.

Note: Answers will be tolerated within $0.5 \%$, unlike other problems.

Solution: Since total momentum is 0 , we have $m_{A} v_{A}+m_{B} v_{B}=0$, so $v_{B}=-0.0005 c$ By conservation of energy:

$$
\frac{1}{2} m_{A} v_{A}^{2}+\frac{1}{2} m_{B} v_{B}^{2}+\gamma_{0} m c^{2}=\gamma m c^{2}
$$

where we define $\gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$ to correspond to the final state of the particle and $\gamma_{0}$ to the initial state. This equation allows us to solve for $\gamma$. Since the speed of the particle is much larger than the speed of the masses $A$ and $B$, we can imagine the particle moving in a infinite well potential where the walls are slowly moving. Applying the adiabatic theorem, we get that the adiabatic invariant $p x$ is conserved, where $p$ is the particle's momentum, and $x$ is the distance between $A$ and $B$. Thus, $\gamma v x$ is conserved, so

$$
\gamma v x=\gamma_{0} v_{0}\left(x_{B}-x_{A}\right)
$$

We can solve for $x$, since we know $\gamma$ and $v$. Finally, we realize that the center of mass stays at $x_{c} m=\frac{m_{A} x_{A}+m_{B} x_{B}}{m_{A}+m_{B}}$, so the final position of $A$ is simply

$$
x_{c} m-\frac{m_{B}}{m_{A}+m_{B}} x=378 \mathrm{~m}
$$

Answer: 378
Problem: Consider an infinite square grid of equal resistors where the nodes are exactly the lattice points in the 2D Cartesian plane. A current $I=2.7 \mathrm{~A}$ enters the grid at the origin $(0,0)$. Find the current in Amps through the resistor connecting the nodes $(N, 0)$ and $(N, 1)$, where $N=38$ can be assumed to be much larger than 1.

Solution: WLOG, let each resistor have unit resistance.

Kirchoff's current law says that the total current entering the node $(x, y) \neq(0,0)$ is zero:

$$
\begin{array}{r}
{[U(x+1, y)-U(x, y)]+[U(x-1, y)-U(x, y)]+[U(x, y+1)-U(x, y)]+[U(x, y-1)-U(x, y)]=0} \\
{[U(x+1, y)-2 U(x, y)+U(x-1, y)]+[U(x, y+1)-2 U(x, y)+U(x, y-1)]=0}
\end{array}
$$

This is an approximation of the equation

$$
\frac{\partial^{2} U}{\partial x^{2}}+\frac{\partial^{2} U}{\partial y^{2}}=0
$$

which is Laplace's equation $\nabla^{2} U=0$ in 2D. Given $U \rightarrow 0$ as $r \rightarrow \infty$, this implies that the potential



field approximates that of a point charge at $O$ in 2D. This approximation is valid far from the origin where changes in $U$ over unit length are small.

The electric field corresponding to this potential field is

$$
\mathbf{E}=-\nabla U=\left(-\partial_{x} U,-\partial_{y} U\right) \approx(U(x-1, y)-U(x, y), U(x, y-1)-U(x, y))=\left(i_{x}(x, y), i_{y}(x, y)\right)
$$

far from the origin, where $i_{x}(x, y), i_{y}(x, y)$ are the horizontal and vertical currents passing through node $(x, y)$. (Note that the current horizontal current is different to the left vs. to the right of the node, and similarly for the vertical current, but the difference is negligible for $N \gg 1$.)

The current $i=\sqrt{i_{x}^{2}+i_{y}^{2}}$ at a distance $r \gg 1$ away from the origin is given by Gauss's law:

$$
2 \pi r i(r) \approx I
$$

so at $(N, 0)$ we have

$$
i_{x}(N, 0)=i(N) \approx \frac{I}{2 \pi N}
$$

The difference between the entering and exiting horizontal currents at $(N, 0)$ is approximately

$$
-\partial_{x} i_{x}(N, 0)=\frac{I}{2 \pi N^{2}}
$$

This difference is directed equally into the vertical resistors adjacent to $(N, 0)$, so the final answer is

$$
\frac{I}{4 \pi N^{2}}=1.488 \times 10^{-4} \mathrm{~A}
$$

Answer: $1.488 \times 10^{-4}$
Problem: Suppose we have a non-ideal gas, and in a certain volume range and temperature range, it is found to satisfy the state relation
$$
p=A V^{\alpha} T^{\beta}
$$

where $A$ is a constant, $\alpha=-\frac{4}{5}$ and $\beta=\frac{3}{2}$, and the other variables have their usual meanings. Throughout the problem, we will assume to be always in that volume and temperature range.

Assume that $\gamma=\frac{C_{p}}{C_{V}}$ is found to be constant for this gas ( $\gamma$ is independent of the state of the gas), where $C_{p}$ and $C_{v}$ are the heat capacities at constant pressure and volume, respectively. What is the minimum possible value for $\gamma$ ?

Solution: We claim that the conditions given uniquely determine $\gamma$.

The fundamental thermodynamic relation gives:

$$
\mathrm{d} U=T \mathrm{~d} S-p \mathrm{~d} V
$$

So

$$
\left(\frac{\partial U}{\partial V}\right)_{T}=T\left(\frac{\partial S}{\partial V}\right)_{T}-p=T\left(\frac{\partial p}{\partial T}\right)_{V}-p
$$

where we have used a Maxwell relation.



$$
\mathrm{d} U=\left(\frac{\partial U}{\partial T}\right)_{V} \mathrm{~d} T+\left(\frac{\partial U}{\partial V}\right)_{T} \mathrm{~d} V=C_{V} \mathrm{~d} T+\left(T\left(\frac{\partial p}{\partial T}\right)_{V}-p\right) \mathrm{d} V
$$

We have

$$
C_{p}=\left(\frac{\partial U}{\partial T}\right)_{p}+p\left(\frac{\partial V}{\partial T}\right)_{p}=C_{V}+T\left(\frac{\partial p}{\partial T}\right)_{V}\left(\frac{\partial V}{\partial T}\right)_{p}
$$

Rearranging, gives

$$
C_{V}=\frac{T\left(\frac{\partial p}{\partial T}\right)_{V}\left(\frac{\partial V}{\partial T}\right)_{p}}{\gamma-1}
$$

From the symmetry of mixed second partial derivatives, we know

$$
\left(\frac{\partial C_{V}}{\partial V}\right)_{T}=\left(\frac{\partial^{2} U}{\partial T \partial V}\right)=\left(\frac{\partial}{\partial T}\left(\frac{\partial U}{\partial V}\right)_{T}\right)_{V}=\frac{\partial}{\partial T}\left(T\left(\frac{\partial p}{\partial T}\right)_{V}-p\right)=\left(\frac{\partial^{2} p}{\partial T^{2}}\right)_{V}
$$

Plugging our expression for $C_{V}$ into here, and plugging in the equation of state, we can solve for $\gamma$ to get

$$
\gamma=\frac{\alpha+\beta}{\alpha(1-\beta)}=\frac{7}{4}
$$

Answer: $\frac{7}{4}$
Problem: The coin flip has long been recognized as a simple and unbiased method to randomly determine the outcome of an event. In the case of an ideal coin, it is well-established that each flip has an equal $50 \%$ chance of landing as either heads or tails.
However, coin flips are not entirely random. They appear random to us because we lack sufficient information about the coin's initial conditions. If we possessed this information, we would always be able to predict the outcome without needing to flip the coin. For an intriguing discussion on why this observation is significant, watch this video by Vsauce.

Now, consider a scenario where a coin with uniform density and negligible width is tossed directly upward from a height of $h=0.75 \mathrm{~m}$ above the ground. The coin starts with its heads facing upward and is given an initial vertical velocity of $v_{y}=49 \mathrm{~m} / \mathrm{s}$ and a positive angular velocity of $\omega=\pi \mathrm{rad} / \mathrm{s}$. What face does the coin display upon hitting the ground? Submit $\mathbf{0}$ for heads and $\mathbf{1}$ for tails. You only have one attempt for this problem. Assume the floor is padded and it absorbs all of the coin's energy upon contact. The radius of the coin is negligible.

Solution: We have the following quadratic:
$$
\begin{aligned}
x & =x_{0}+v_{0} t+\frac{1}{2} a t^{2} \\
0 & =0.75+49 t-4.9 t^{2} \\
t & =-0.01,10.01
\end{aligned}
$$

The first solution is extraneous so $t=10.01$ is correct. Now, $\theta=\omega t \approx 10 \pi$. As one full rotation is $\phi=2 \pi$, then the coin performs 5 full rotations before landing on the ground. This means the answer is 0 , or heads.

Answer: 0
Problem: Suppose all cars on a (single-lane) highway are identical. Their length is $l=4 \mathrm{~m}$, their wheels have coefficients of friction $\mu=0.7$, and they all travel at speed $v_{0}$. Find the $v_{0}$ which maximizes the flow rate of cars (i.e. how many cars travel across an imaginary line per minute). Assume that they need to be able to stop in time if the car in front instantaneously stops. Disregard reaction time.

Solution: Suppose the maximum speed is $v^{\prime}$. Notice that $a=\mu g$ so it takes a time $t^{\prime}=\frac{v^{\prime}}{\mu g}$ amount of time to stop. This means the car will travel a distance of
$$
d=\frac{1}{2} \mu g\left(\frac{v^{\prime}}{\mu g}\right)^{2}=\frac{v^{\prime 2}}{2 \mu g}
$$

Thus for every distance $d+l=l+\frac{v^{\prime 2}}{2 \mu g}$ there is a car. This means in unit time $t$, there will be $N=\frac{d+l}{v}$ cars that passes through the line. To maximize the flow rate, we need to maximize

$$
\begin{gathered}
f\left(v^{\prime}\right)=\frac{l}{v^{\prime}}+\frac{v^{\prime}}{2 \mu g} \\
f^{\prime}=\frac{-l}{v^{\prime 2}}+\frac{1}{2 \mu g}=0 \Rightarrow v^{\prime}=\sqrt{2 \mu g l}=7.41 \mathrm{~m} / \mathrm{s}
\end{gathered}
$$

Answer: $7.41$