weblab-llm-competition-2025-bridge/neko-prelim-HLE_SFT_OlympiadBench huggingface dataset:
Problem: a. Suppose you drop a block of mass $m$ vertically onto a fixed ramp with angle $\theta$ with coefficient of static and kinetic friction $\mu$. The block is dropped in such a way that it does not rotate after colliding with the ramp. Throughout this problem, assume the time of the collision is negligible.


i. Suppose the block's speed just before it hits the ramp is $v$ and the block slides down the ramp immediately after impact. What is the speed of the block right after the collision?

Solution: During the collision, the block receives impulses from the normal force, friction, and gravity. Since the collision is very short, the impulse due to gravity is negligible. Let $p_{N}$ and $p_{F}$ be the magnitudes of the impulses from the normal force and friction force.



Since the block stays on the ramp after the collision, its final momentum is parallel to the ramp. Then the normal force must completely eliminate the block's initial momentum perpendicular to the ramp, so



$$

p_{N}=m v \cos \theta

$$



Since the block still moves after the collision,



$$

p_{F}=\mu p_{N}

$$



The block's initial momentum parallel to the ramp is $m v \sin \theta$, so



$$

m v \sin \theta-p_{F}=m u

$$



where $u$ is the final speed of the block. Solving for $u$ gives



$$

u=v(\sin \theta-\mu \cos \theta)

$$

Answer: $u=v(\sin \theta-\mu \cos \theta)$
Problem: ii. What is the minimum $\mu$ such that the speed of the block right after the collision is 0 ?

Solution: We set $u=0$ to obtain



$$

\mu=\tan \theta

$$



Note that this is simply the no-slip condition for a block resting on an inclined plane! This is because in both cases, equality is achieved when the normal force and maximal friction force sum to a purely vertical force.

Answer: $\mu=\tan \theta$
Problem: b. Now suppose you drop a sphere with mass $m$, radius $R$ and moment of inertia $\beta m R^{2}$ vertically onto the same fixed ramp such that it reaches the ramp with speed $v$.


i. Suppose the sphere immediately begins to roll without slipping. What is the new speed of the sphere in this case?

Solution: If the sphere immediately begins to roll without slipping, we can calculate the frictional impulse independently of the normal impulse. We have



$$

m v \sin \theta-p_{F}=m u

$$



The frictional impulse is responsible for the sphere's rotation, so its angular momentum about its center of mass is $L=p_{F} R$. But we also know that



$$

L=\beta m R^{2} \omega=\beta m R u

$$



Then



$$

p_{F}=\beta m u \text {. }

$$



Substituting into the previous expression gives



$$

m v \sin \theta=(1+\beta) m u \Rightarrow u=\frac{v \sin \theta}{1+\beta}

$$

Answer: $u=\frac{v \sin \theta}{1+\beta}$
Problem: ii. What is the minimum coefficient of friction such that the sphere rolls without slipping immediately after the collision?

Solution: As in part (a), the normal impulse is $p_{N}=m v \cos \theta$ and the maximal frictional impulse is $p_{F}=\mu p_{N}$. From the previous part, we need



$$

p_{F}=\frac{\beta m v \sin \theta}{1+\beta}

$$



and equating these expressions gives



$$

\mu=\frac{\beta \tan \theta}{1+\beta}

$$

Answer: $\mu=\frac{\beta \tan \theta}{1+\beta}$
Problem: a. What is the electric potential at a corner of the same cube? Write your answer in terms of $\rho, a, \epsilon_{0}$, and any necessary numerical constants.

Solution: By dimensional analysis, the answer takes the form



$$

\Phi_{c}(a, \rho) \approx \frac{C \rho a^{2}}{\epsilon_{0}}

$$



for a dimensionless constant $C$. Note that a cube of side length $a$ consists of 8 cubes of side length $a / 2$, each with a corner at the center of the larger cube. Then



$$

\frac{0.1894 \rho a^{2}}{\epsilon_{0}}=8 \frac{C \rho(a / 2)^{2}}{\epsilon_{0}}

$$



so $C=0.1894 / 2=0.0947$.



Answer: $\Phi_{c}(a, \rho) \approx \frac{C \rho a^{2}}{\epsilon_{0}}$, $C=0.0947$
Problem: a. Find the standard deviation $\Delta r$ of the distribution for $r$, the distance from the center of the telescope mirror to the point of reflection of the photon.

Solution: The square of the standard deviation is the variance, so



$$

(\Delta r)^{2}=\left\langle r^{2}\right\rangle-\langle r\rangle^{2}

$$



Computing the average value of $r^{2}$ is mathematically the exact same thing as computing the moment of inertia of a uniform disk; we have



$$

\left\langle r^{2}\right\rangle=\frac{1}{\pi R^{2}} \int_{0}^{R} r^{2}(2 \pi r d r)=\frac{1}{2} R^{2}

$$



Similarly, the average value of $r$ is



$$

\langle r\rangle=\frac{1}{\pi R^{2}} \int_{0}^{R} r(2 \pi r d r)=\frac{2}{3} R

$$



Then we have



$$

\Delta r=\frac{R}{\sqrt{18}}

$$

Answer: $\Delta r=\frac{R}{\sqrt{18}}$
Problem: i. Given that the total power radiated from the sun is given by $L_{\odot}$, find an expression for the radiation pressure a distance $R$ from the sun.

Solution: We will assume that light from the sun is completely absorbed, and then re-radiated as blackbody isotropically. In that case,



$$

P=\frac{I}{c}=\frac{L \odot}{4 \pi R^{2} c}

$$



The relationship between the pressure $P$ and the energy density $I$ can be derived from the equation $p=E / c$ for photons, or simply postulated by dimensional analysis, since there is no other relevant speed.



Alternatively, this relation can be derived using classical electromagnetism, though this was not required. For a plane wave, the momentum density (i.e. pressure) of the electromagnetic field is $P=\epsilon_{0}|\mathbf{E} \times \mathbf{B}|$, while the energy density is $I=|\mathbf{S}| / c$ where $\mathbf{S}=\mathbf{E} \times \mathbf{B} / \mu_{0}$. Then $P=I / c$ since $c^{2}=1 / \epsilon_{0} \mu_{0}$.

Answer: $P=\frac{L \odot}{4 \pi R^{2} c}$
Problem: ii. Assuming that the particle has a density $\rho$, derive an expression for the ratio $\frac{F_{\text {radiation }}}{F_{\text {gravity }}}$ in terms of $L_{\odot}$, mass of sun $M_{\odot}, \rho$, particle radius $r$, and quality factor $Q$.

Solution: We have



$$

F_{\text {gravity }}=\frac{G M_{\odot}}{R^{2}} \frac{4}{3} \pi \rho r^{3}, \quad F_{\text {radiation }}=\frac{L_{\odot}}{4 \pi R^{2} c} Q \pi r^{2}=\frac{L_{\odot}}{4 R^{2} c} Q r^{2}

$$



which gives



$$

\frac{F_{\text {radiation }}}{F_{\text {gravity }}}=\frac{3 L_{\odot}}{16 \pi G c M_{\odot} \rho} \frac{Q}{r}

$$

Answer: $\frac{F_{\text {radiation }}}{F_{\text {gravity }}}=\frac{3 L_{\odot}}{16 \pi G c M_{\odot} \rho} \frac{Q}{r}$
Problem: i. Assume that a particle is in a circular orbit around the sun. Find the speed of the particle $v$ in terms of $M_{\odot}$, distance from sun $R$, and any other fundamental constants.

Solution: Using the circular motion equation



$$

\frac{G M_{\odot}}{R^{2}}=\frac{v^{2}}{R}

$$



we have



$$

v=\sqrt{\frac{G M_{\odot}}{R}}

$$

Answer: $v=\sqrt{\frac{G M_{\odot}}{R}}$
Problem: ii. Because the particle is moving, the radiation force is not directed directly away from the sun. Find the torque $\tau$ on the particle because of radiation pressure. You may assume that $v \ll c$.

Solution: Work in the reference frame of the particle. In this frame, the radiation hits the particle at an angle $\theta=v / c$ from the radial direction. The particle then re-emits the radiation isotropically, contributing no additional radiation pressure. (We ignore relativistic effects because they occur at second order in $v / c$, while the effect we care about is first order.) The tangential component of the force is



so



$$

F=\frac{v}{c} \frac{L \odot}{4 \pi R^{2} c} Q \pi r^{2}=\frac{v}{c} \frac{L_{\odot}}{4 R^{2} c} Q r^{2}

$$



$$

\tau=-\frac{v}{c} \frac{L \odot}{4 R c} Q r^{2}

$$


where the negative sign is because this tends to decrease the angular momentum.



The problem can also be solved in the reference frame of the Sun. In this frame, the radiation hits the particle radially, but the particle does not re-emit the radiation isotropically since it is moving; instead the radiation is Doppler shifted. This eventually leads to the same result, after a much more complicated calculation.

Answer: $\tau=-\frac{v}{c} \frac{L \odot}{4 R c} Q r^{2}$
Problem: iii. Since $\tau=d L / d t$, the angular momentum $L$ of the particle changes with time. As such, develop a differential equation to find $d R / d t$, the rate of change of the radial location of the particle. You may assume the orbit is always quasi circular.

Solution: The angular momentum is



$$

L=m v R=\frac{4}{3} \pi \rho r^{3} \sqrt{\frac{G M_{\odot}}{R}} R=\frac{4}{3} \pi \rho r^{3} \sqrt{G M_{\odot} R}

$$



Differentiating both sides with respect to time,



$$

-\frac{v}{c} \frac{L_{\odot}}{4 R c} Q r^{2}=\frac{4}{3} \pi \rho r^{3} \sqrt{\frac{G M_{\odot}}{R}} \frac{1}{2} \frac{d R}{d t}

$$



which simplifies to



$$

-\frac{1}{c^{2}} \frac{L_{\odot}}{R} Q=\frac{8}{3} \pi \rho r \frac{d R}{d t}

$$

Answer: $-\frac{1}{c^{2}} \frac{L_{\odot}}{R} Q=\frac{8}{3} \pi \rho r \frac{d R}{d t}$
Problem: iv. Develop an expression for the time required to remove particles of size $r \approx 1 \mathrm{~cm}$ and density $\rho \approx 1000 \mathrm{~kg} / \mathrm{m}^{3}$ originally in circular orbits at a distance $R=R_{\text {earth }}$, and use the numbers below to simplify your expression.

Solution: Integrating both sides,



$$

T \frac{L_{\odot} Q}{c^{2}}=\frac{4}{3} \pi \rho r R^{2}

$$



Since $r \gg \lambda$, we have $Q=1$ and



$$

T=\frac{4}{3} \frac{\pi c^{2}}{L \odot} \rho r R^{2} \approx 2 \times 10^{14} \mathrm{~s} \approx 7 \times 10^{6} \mathrm{y}

$$

Answer: $2 \times 10^{14}$
Problem: 1.  First consider an infinitely long line with charge per unit length $\lambda$ in vacuum. Calculate the electric field $\mathbf{E}(\mathbf{r})$ due to the line.

Solution: Using Gauss law

$$
\oint \mathbf{E} \cdot \mathbf{d} \mathbf{A}=\frac{q}{\epsilon_{0}} .
\tag{1}
$$

From symetry we know that the electric field only has radial component. Choose a cylinder (with a line charge as the axis) as the Gaussian surface, we obtain

$$
E .2 \pi r l=\frac{\lambda l}{\epsilon_{0}}
$$

Simplify to obtain

$$
\mathbf{E}=\hat{r} \frac{\lambda}{2 \pi \epsilon_{0} r}
\tag{2}
$$



Answer: $\mathbf{E}=\hat{r} \frac{\lambda}{2 \pi \epsilon_{0} r}$
Problem: 2.  The potential due to the line charge could be written as

$$
V(r)=f(r)+K,
$$

where $K$ is a constant. Determine $f(r)$.

Solution: The potential is given by

$$
\begin{aligned}
V & =-\int_{\mathrm{ref}}^{r} \mathbf{E} \cdot \mathbf{d} \mathbf{l} \\
& =-\int_{\mathrm{ref}}^{r} E \cdot d r \\
V & =-\frac{\lambda}{2 \pi \epsilon_{0}} \ln r+K,
\end{aligned}
\tag{3}
$$

so $f(r)=-\frac{\lambda}{2 \pi \epsilon_{0}} \ln r$. where $K$ is a constant.

Answer: $f(r)=-\frac{\lambda}{2 \pi \epsilon_{0}} \ln r$
Problem: 1.  When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame).

Solution: The equation of motion is given by

$$
\begin{aligned}
F & =\frac{d}{d t}(\gamma m v) \\
& =\frac{m c \dot{\beta}}{\left(1-\beta^{2}\right)^{\frac{3}{2}}}
\end{aligned}
\tag{1}
$$

$$
F=\gamma^{3} m a
\tag{2}
$$

where $\gamma=\frac{1}{\sqrt{1-\beta^{2}}}$ and $\beta=\frac{v}{c}$. So the acceleration is given by

$$
a=\frac{F}{\gamma^{3} m}
\tag{3}
$$

Answer: $a=\frac{F}{\gamma^{3} m}$
Problem: 2.  Calculate the velocity of the particle $\beta(t)=\frac{v(t)}{c}$ at time $t$ (in rest frame), in terms of $F, m, t$ and $c$.

Solution: Eq.(3) can be rewritten as

$$
\begin{aligned}
c \frac{d \beta}{d t} & =\frac{F}{\gamma^{3} m} \\
\int_{0}^{\beta} \frac{d \beta}{\left(1-\beta^{2}\right)^{\frac{3}{2}}} & =\frac{F}{m c} \int_{0}^{t} d t \\
\frac{\beta}{\sqrt{1-\beta^{2}}} & =\frac{F t}{m c}
\end{aligned}
\tag{4}
$$

$$
\beta=\frac{\frac{F t}{m c}}{\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}}
\tag{5}
$$

Answer: $\beta=\frac{\frac{F t}{m c}}{\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}}$
Problem: 3.  Calculate the position of the particle $x(t)$ at time $t$, in term of $F, m, t$ and $c$.

Solution: Using Eq.(5), we get

$$
\begin{aligned}
\int_{0}^{x} d x & =\int_{0}^{t} \frac{F t d t}{m \sqrt{1+\left(\frac{F t}{m c}\right)^{2}}} \\
x & =\frac{m c^{2}}{F}\left(\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}-1\right) .
\end{aligned}
\tag{6}
$$

Answer: $x=\frac{m c^{2}}{F}\left(\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}-1\right)$
Problem: 5.  Calculate the velocity of the particle $\beta(\tau)$, when the time as experienced by the particle is $\tau$. Express the answer in $g, \tau$, and $c$.

Solution: Eq.(3) can also be rewritten as

$$
c \frac{d \beta}{\gamma d \tau}=\frac{g}{\gamma^{3}}
\tag{14}
$$

$$
\int_{0}^{\beta} \frac{d \beta}{1-\beta^{2}} =\frac{g}{c} \int_{0}^{\tau} d \tau
$$
$$
\ln \left(\frac{1}{\sqrt{1-\beta^{2}}}+\frac{\beta}{\sqrt{1-\beta^{2}}}\right) =\frac{g \tau}{c}
\tag{15}
$$
$$
\sqrt{\frac{1+\beta}{1-\beta}} =e^{\frac{g \tau}{c}}
$$

$$
\beta\left(e^{\frac{g \tau}{c}}+e^{-\frac{g \tau}{c}}\right) =e^{\frac{g \tau}{c}}-e^{-\frac{g \tau}{c}}
$$

$$
\beta =\tanh \frac{g \tau}{c} .
\tag{16}
$$

Answer: $\beta=\tanh \frac{g \tau}{c}$
Problem: 6. ( $\mathbf{0 . 4} \mathbf{~ p t s )}$ Also calculate the time $t$ in the rest frame in terms of $g, \tau$, and $c$.

Solution: The time dilation relation is

$$
d t=\gamma d \tau
\tag{17}
$$

From eq.(16), we have

$$
\gamma=\frac{1}{\sqrt{1-\beta^{2}}}=\cosh \frac{g \tau}{c}
\tag{18}
$$

Combining this equations, we get

$$
\begin{aligned}
\int_{0}^{t} d t & =\int_{0}^{\tau} d \tau \cosh \frac{g \tau}{c} \\
t & =\frac{c}{g} \sinh \frac{g \tau}{c} .
\end{aligned}
\tag{19}
$$

Answer: $t=\frac{c}{g} \sinh \frac{g \tau}{c}$
Problem: 2.  A uniform magnetic field $\boldsymbol{B}$ exists and it makes an angle $\phi$ with a particle's magnetic moment $\boldsymbol{\mu}$. Due to the torque by the magnetic field, the magnetic moment $\boldsymbol{\mu}$ rotates around the field $\boldsymbol{B}$, which is also known as Larmor precession. Determine the Larmor precession frequency $\omega_{0}$ of the magnetic moment with respect to $\boldsymbol{B}=B_{0} \boldsymbol{k}$.

Solution: For a magnetic moment making an angle of $\phi$ with $\mathbf{B}$,

$$
\begin{aligned}
\frac{d \boldsymbol{\mu}}{d t} & =-\gamma \boldsymbol{\mu} \times \mathbf{B} \\
|\boldsymbol{\mu}| \sin \phi \frac{d \theta}{d t} & =\gamma|\boldsymbol{\mu}| B_{0} \sin \phi \\
\omega_{0}=\frac{d \theta}{d t} & =\gamma B_{0}
\end{aligned}
\tag{5}
$$

Answer: $\omega_{0}=\gamma B_{0}$
Problem: 2.  For $\boldsymbol{B}=B_{0} \boldsymbol{k}$, what is the new precession frequency $\Delta$ in terms of $\omega_{0}$ and $\omega$ ?

Solution: The new precession frequency as viewed on the rotating frame $S^{\prime}$ is

$$
\begin{aligned}
\vec{\Delta} & =\left(\omega_{0}-\omega\right) \mathbf{k}^{\prime} \\
\Delta & =\gamma B_{0}-\omega .
\end{aligned}
\tag{7}
$$

Answer: $\Delta =\gamma B_{0}-\omega$
Problem: 4.  Instead of applying the field $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}+\sin \omega t \boldsymbol{j})$, now we apply $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}-\sin \omega t \boldsymbol{j})$, which rotates in the opposite direction and hence $\boldsymbol{B}=B_{0} \boldsymbol{k}+b(\cos \omega t \boldsymbol{i}-\sin \omega t \boldsymbol{j})$. What is the effective magnetic field $\boldsymbol{B}_{\text {eff }}$ for this case (in terms of the unit vectors $\boldsymbol{i}^{\prime}, \boldsymbol{j}^{\prime}, \boldsymbol{k}^{\prime}$ )? What is its time average, $\overline{\boldsymbol{B}_{\text {eff }}}$ (recall that $\overline{\cos 2 \pi t / T}=\overline{\sin 2 \pi t / T}=0$ )?

Solution: In this case, the effective magnetic field becomes

$$
\begin{aligned}
\mathbf{B}_{\mathrm{eff}} & =\mathbf{B}-\omega / \gamma \mathbf{k}^{\prime} \\
& =\left(B_{0}-\frac{\omega}{\gamma}\right) \mathbf{k}^{\prime}+b\left(\cos 2 \omega t \mathbf{i}^{\prime}-\sin 2 \omega t \mathbf{j}^{\prime}\right)
\end{aligned}
\tag{9}
$$

which has a time average of $\overline{\mathbf{B}_{\mathrm{eff}}}=\left(B_{0}-\frac{\omega}{\gamma}\right) \mathbf{k}^{\prime}$.

Answer: $\mathbf{B}_{\mathrm{eff}}=\left(B_{0}-\frac{\omega}{\gamma}\right) \mathbf{k}^{\prime}+b\left(\cos 2 \omega t \mathbf{i}^{\prime}-\sin 2 \omega t \mathbf{j}^{\prime}\right)$ , $\overline{\mathbf{B}_{\mathrm{eff}}}=\left(B_{0}-\frac{\omega}{\gamma}\right) \mathbf{k}^{\prime}$
Problem: 1.  In the rotating frame $S^{\prime}$, show that the effective field can be approximated by

$$
\boldsymbol{B}_{\text {eff }} \approx b \boldsymbol{i}^{\prime},
$$

which is commonly known as rotating wave approximation. What is the precession frequency $\Omega$ in frame $S^{\prime}$ ?

Solution: The oscillating field can be considered as a superposition of two oppositely rotating field:

$$
2 b \cos \omega_{0} t \mathbf{i}=b\left(\cos \omega_{0} t \mathbf{i}+\sin \omega_{0} t \mathbf{j}\right)+b\left(\cos \omega_{0} t \mathbf{i}-\sin \omega_{0} t \mathbf{j}\right)
$$

which gives an effective field of (with $\omega=\omega_{0}=\gamma B_{0}$ ):

$$
\mathbf{B}_{\mathrm{eff}}=\left(B_{0}-\frac{\omega}{\gamma}\right) \mathbf{k}^{\prime}+b \mathbf{i}^{\prime}+b\left(\cos 2 \omega_{0} t \mathbf{i}^{\prime}-\sin 2 \omega_{0} t \mathbf{j}^{\prime}\right)
$$

Since $\omega_{0} \gg \gamma b$, the rotation of the term $b\left(\cos 2 \omega_{0} t \mathbf{i}^{\prime}-\sin 2 \omega_{0} t \mathbf{j}^{\prime}\right)$ is so fast compared to the frequency $\gamma b$. This means that we can take the approximation

$$
\mathbf{B}_{\mathrm{eff}} \approx\left(B_{0}-\frac{\omega}{\gamma}\right) \mathbf{k}^{\prime}+b \mathbf{i}^{\prime}=b \mathbf{i}^{\prime}
\tag{10}
$$

where the magnetic moment precesses with frequency $\Omega=\gamma b$.

As $\Omega=\gamma b \ll \omega_{0}$, the magnetic moment does not "feel" the rotating term $b\left(\cos 2 \omega_{0} t \mathbf{i}^{\prime}-\sin 2 \omega_{0} t \mathbf{j}^{\prime}\right)$ which averaged to zero.

Answer: $\Omega=\gamma b$
Problem: 3.  Under the application of magnetic field described above, determine the fractional population of each spin up $P_{\uparrow}=N_{\uparrow} / N$ and spin down $P_{\downarrow}=N_{\downarrow} / N$ as a function of time. Plot $P_{\uparrow}(t)$ and $P_{\downarrow}(t)$ on the same graph vs. time $t$. The alternating spin up and spin down population as a function of time is called Rabi oscillation.

Solution: From the relations

$$
\begin{aligned}
P_{\uparrow}-P_{\downarrow} & =\frac{\mu_{z}}{\mu}=\cos \theta, \\
P_{\uparrow}+P_{\downarrow} & =1,
\end{aligned}
$$



we obtain the results $\left(\omega=\omega_{0}\right)$

$$
\begin{aligned}
P_{\downarrow} & =\frac{1-\cos \theta}{2} \\
& =\frac{1-\cos ^{2} \alpha-\sin ^{2} \alpha \cos \Omega t}{2} \\
& =\sin ^{2} \alpha \frac{1-\cos \Omega t}{2} \\
& =\frac{b^{2}}{\left(B_{0}-\frac{\omega}{\gamma}\right)^{2}+b^{2}} \sin ^{2} \frac{\Omega t}{2} \\
& =\sin ^{2} \frac{\Omega t}{2},
\end{aligned}
\tag{13}
$$

and

$$
P_{\uparrow}=\frac{b^{2}}{\left(B_{0}-\frac{\omega}{\gamma}\right)^{2}+b^{2}} \cos ^{2} \frac{\Omega t}{2}=\cos ^{2} \frac{\Omega t}{2}
\tag{14}
$$

<img_4555>

Answer: $P_{\downarrow}=\sin ^{2} \frac{\Omega t}{2}$ , $P_{\uparrow}=\cos ^{2} \frac{\Omega t}{2}$