weblab-llm-competition-2025-bridge/neko-prelim-HLE_SFT_OlympiadBench huggingface dataset:
Problem: A planet has a radius of $10 \mathrm{~km}$ and a uniform density of $5 \mathrm{~g} / \mathrm{cm}^{3}$. A powerful bomb detonates at the center of the planet, releasing $8.93 \times 10^{17} \mathrm{~J}$ of energy, causing the planet to separate into three large sections each with equal masses. You may model each section as a perfect sphere of radius $r^{\prime}$. The initial and final distances between the centers of any two given sections is $2 r^{\prime}$. How long does it take for the three sections to collide again?

Solution: Due to conservation of momentum, the three masses must form an equilateral triangle at all times. Let us determine the force as a function of $r$, the distance between each mass and the center.


<img_4307>

The vector sum of the net force on any individual mass is

$$
F=\frac{G m^{2}}{d^{2}} \sqrt{2-2 \cos 120^{\circ}}=\frac{\sqrt{3} G m^{2}}{d^{2}}
$$

where $d$ is the distance between the mass and the center.

$$
d^{2}=3 r^{2}
$$

The net force is thus

$$
F=\frac{G m^{2}}{\sqrt{3} r^{2}}
$$

The system behaves as if there was a stationary mass $m^{\prime}=m / \sqrt{3}$ at the center, simplifying the problem greatly into a restricted two body system. Next, we need to figure out the height of the apoapsis. This can be done via conservation of energy.

$$
E_{\text {binding,initial }}+E=3 E_{\text {binding,final }}-\frac{G m^{2}}{\sqrt{3} \ell} \Longrightarrow \ell=-\frac{G m^{2}}{\sqrt{3}\left(-\frac{3 G M^{2}}{5 R}+8.93 \cdot 10^{17}-\left(-3 \frac{3 G m^{2}}{5 r_{f}}\right)\right)}=101,000 \mathrm{~m}
$$

If you have a stationary mass $M$ at the center. The time it takes for an object to fall into it is:

$$
T=\pi \sqrt{\frac{\ell^{3}}{8 G M}}
$$

our time will be double this, and the mass in the center will be $M=m / \sqrt{3}$. So plugging in numbers gives:

$$
t=2 \pi \sqrt{\frac{l^{3}}{8 G\left(\frac{m}{\sqrt{3}}\right)}}=138,000
$$



If we take into account a nonzero radius so final separation is $10 / \cos \left(30^{\circ}\right) \mathrm{km}$, then the answer should be:

$$
-2 \int_{l}^{\frac{10000}{\cos \left(\frac{\pi}{6}\right)}}\left(\sqrt{\frac{x l}{2 G\left(\frac{m}{\sqrt{3}}\right)(l-x)}}\right) d x=136,000 \mathrm{~s}
$$

Note if you set the upper bound to zero, you get the same answer as before. Both these answers will be accepted.

Answer: 136000
Problem: A point charge $+q$ is placed a distance $a$ away from an infinitely large conducting plate. The force of the electrostatic interaction is $F_{0}$. Then, an identical conducting plate is placed a distance $3 a$ from the charge, parallel to the first one such that the charge is "sandwiched in." The new electrostatic force the particle feels is $F^{\prime}$. What is $F^{\prime} / F_{0}$ ? Round to the nearest hundredths.

Solution: We solve this via the method of image charges. Let us first reflect the charge $+q$ across the closest wall. The force of this interaction will be:
$$
F_{0}=\frac{q^{2}}{4 \pi \epsilon_{0} a^{2}} \frac{1}{2^{2}}
$$

and this will be the only image charge we need to place if there were only one conducting plane. Since there is another conducting plane, another charge will be reflected to a distance $a+4 a$ past the other conducting plane, and thus will be $a+4 a+3 a=8 a$ away from the original charge. All these reflections cause a force that points in the same direction, which we will label as the positive + direction. Therefore:

$$
F_{+}=\frac{q^{2}}{4 \pi \epsilon_{0} a^{2}}\left(\frac{1}{2^{2}}+\frac{1}{8^{2}}+\frac{1}{10^{2}}+\frac{1}{16^{2}} \frac{1}{18^{2}}+\frac{1}{24^{2}}+\cdots\right)
$$

Now let us look at what happens if we originally reflect the charge $+q$ across the other wall. Repeating the steps above, we see that through subsequent reflections, each force will point in the negative - direction. Therefore:

$$
F_{-}=\frac{-q^{2}}{4 \pi \epsilon_{0} a^{2}}\left(\frac{1}{6^{2}}+\frac{1}{8^{2}}+\frac{1}{14^{2}}+\frac{1}{16^{2}}+\frac{1}{22^{2}}+\frac{1}{24^{2}}+\cdots\right)
$$

The net force is a result of the superposition of these two forces, giving us:

$$
\begin{aligned}
F^{\prime}=\frac{q^{2}}{4 \pi \epsilon_{0} a^{2}} & \left(\frac{1}{2^{2}}+\frac{1}{8^{2}}+\frac{1}{10^{2}}+\frac{1}{16^{2}} \frac{1}{18^{2}}+\frac{1}{24^{2}}+\cdots\right. \\
& \left.-\frac{1}{6^{2}}-\frac{1}{8^{2}}-\frac{1}{14^{2}}-\frac{1}{16^{2}}-\frac{1}{22^{2}}-\frac{1}{24^{2}}-\cdots\right)
\end{aligned}
$$

Even terms can be cancelled out to give:

$$
\begin{aligned}
F^{\prime} & =\frac{q^{2}}{4 \pi \epsilon_{0} a^{2}}\left(\frac{1}{2^{2}}-\frac{1}{6^{2}}+\frac{1}{10^{2}}-\frac{1}{14^{2}}+\frac{1}{18^{2}}-\frac{1}{22^{2}}+\cdots\right) \\
& =\frac{q^{2}}{4 \pi \epsilon_{0} a^{2}} \frac{1}{4}\left(\frac{1}{1^{2}}-\frac{1}{3^{2}}+\frac{1}{5^{2}}-\frac{1}{7^{2}}+\frac{1}{9^{2}}-\frac{1}{11^{2}}+\cdots\right)
\end{aligned}
$$

You may recognize the infinite series inside the parentheses to be Catalan's constant $G \approx 0.916$. Alternatively, you can use a calculator and evaluate the first seven terms to get a rough answer (but will still be correct since we asked for it to be rounded). Therefore:

$$
F^{\prime} / F=\frac{1}{1^{2}}-\frac{1}{3^{2}}+\frac{1}{5^{2}}-\frac{1}{7^{2}}+\frac{1}{9^{2}}-\frac{1}{11^{2}}+\cdots=G \approx 0.916
$$

Answer: 0.916
Problem: Jerry spots a truckload of his favourite golden yellow Swiss cheese being transported on a cart moving at a constant velocity $v_{0}=5 \mathrm{~m} / \mathrm{s} \hat{i}$ along the x-axis, which is initially placed at $(0,0)$. Jerry, driven by desire immediately starts pursuing the cheese-truck in such a way that his velocity vector always points towards the cheese-truck; however, Jerry is smart and knows that he must maintain a constant distance $\ell=10 \mathrm{~m}$ from the truck to avoid being caught by anyone, no matter what. Note that Jerry starts at coordinates $(0, \ell)$.


Let the magnitude of velocity (in $\mathrm{m} / \mathrm{s}$ ) and acceleration (in $\mathrm{m} / \mathrm{s}^{2}$ ) of Jerry at the moment when the (acute) angle between the two velocity vectors is $\theta=60^{\circ}$ be $\alpha$ and $\beta$ respectively. Compute $\alpha^{2}+\beta^{2}$.

Solution: If the distance between Jerry and the cheese truck is constant, then Jerry moves in circle of radius $\ell$ in the reference frame of the cheese truck. There is no radial component of Jerry's velocity in this reference frame, so we must have $\alpha=v_{0} \cos \theta=\frac{5}{2}$. In this case, the tangential velocity is $v_{0} \sin \theta$. Furthermore, the radial acceleration in this frame is given by the centripetal acceleration which is $\frac{\left(v_{0} \sin \theta\right)^{2}}{\ell}=\frac{v_{0}^{2} \sin ^{2} \theta}{\ell}$. The tangential acceleration is

$$
\frac{d}{d t}\left(v_{0} \sin \theta\right)=v_{0} \cos \theta \cdot \frac{d \theta}{d t}=v_{0} \cos \theta \cdot \frac{-v_{0} \sin \theta}{\ell}=-\frac{v_{0}^{2} \sin \theta \cos \theta}{\ell}
$$

The vector sum of these accelerations has magnitude

$$
\beta=\sqrt{\left(\frac{v_{0}^{2} \sin ^{2} \theta}{\ell}\right)^{2}+\left(\frac{v_{0}^{2} \sin \theta \cos \theta}{\ell}\right)^{2}}=\frac{v_{0}^{2} \sin \theta}{\ell}=\frac{5 \sqrt{3}}{4} .
$$

The final answer is $\alpha^{2}+\beta^{2}=10.9375$.

Answer: 10.9375
Problem: What is the smallest number of $1 \Omega$ resistors needed such that when arranged in a certain arrangement involving only series and parallel connections, that the equivalent resistance is $\frac{7}{6} \Omega$ ?

Solution: We can write:
$$
\frac{7}{6}=\frac{1}{2}+\frac{2}{3}
\tag{2}
$$

It takes two resistors (connected in parallel) to create a $\frac{1}{2} \Omega$ resistor. If we write $\frac{2}{3}=\frac{2 \cdot 1}{2+1}$, then it takes three resistors to create a $\frac{2}{3} \Omega$ (this is accomplished by connecting a 2 resistor in parallel with a 1 resistor.

Combining the $\frac{1}{2} \Omega$ element in series with the $\frac{2}{3} \Omega$ element gives us our desired amount. To prove this is the minimum, we can easily check all possible combinations using 4 or fewer resistors.

Answer: 5
Problem: A coaxial cable is cylindrically symmetric and consists of a solid inner cylinder of radius $a=2 \mathrm{~cm}$ and an outer cylindrical shell of inner radius $b=5 \mathrm{~cm}$ and outer radius $c=7 \mathrm{~cm}$. A uniformly distributed current of total magnitude $I=5 \mathrm{~A}$ is flowing in the inner cylinder and a uniformly distributed current of the same magnitude but opposite direction flows in the outer shell. Find the magnitude $B(r)$ of the magnetic field $B$ as a function of distance $r$ from the axis of the cable. As the final result, submit $\int_{0}^{\infty} B(r) \mathrm{d} r$. In case this is infinite, submit 42 .

Solution: Ampere's law $\int B \cdot \mathrm{d} l=\mu_{0} I$ is all we need. For every point on the wire, we can write the magnetic field as a function of the distance from its center $r$. Thus,
$$
B(r)= \begin{cases}\frac{5 \mu_{0} r}{8 \pi} & r \leq 2 \\ \frac{5 \mu_{0}}{2 \pi r} & 2<r<5 \\ \frac{5 \mu_{0}\left(-r^{2}+49\right)}{48 \pi r} & 5 \leq r \leq 7 \\ 0 & r>7\end{cases}
$$

Now we just sum each integral from each interval, or in othe words

$$
\int_{0}^{\infty} B(r) \mathrm{d} r=\int_{0}^{2} B(r) \mathrm{d} r+\int_{2}^{5} B(r) \mathrm{d} r+\int_{5}^{7} B(r) \mathrm{d} r
$$

This is now straightforward integration.

Answer: $1.6 \times 10^{-8}$
Problem: A train of length $100 \mathrm{~m}$ and mass $10^{5} \mathrm{~kg}$ is travelling at $20 \mathrm{~m} / \mathrm{s}$ along a straight track. The driver engages the brakes and the train starts deccelerating at a constant rate, coming to a stop after travelling a distance $d=2000 \mathrm{~m}$. As the train decelerates, energy released as heat from the brakes goes into the tracks, which have a linear heat capacity of $5000 \mathrm{~J} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$. Assume the rate of heat generation and transfer is uniform across the length of the train at any given moment.
If the tracks start at an ambient temperature of $20^{\circ} \mathrm{C}$, there is a function $T(x)$ that describes the temperature (in Celsius) of the tracks at each point $x$, where the rear of where the train starts is at $x=0$. Assume (unrealistically) that $100 \%$ of the original kinetic energy of the train is transferred to the tracks (the train does not absorb any energy), that there is no conduction of heat along the tracks, and that heat transfer between the tracks and the surroundings is negligible.

Compute $T(20)+T(500)+T(2021)$ in degrees celsius.

Solution: Consider a small element of the tracks at position $x$ with width $\mathrm{d} x$. Since the rate of heat generation is uniform along the length of the train $L$, we have that the rate of heat given to the track element is mav $\frac{\mathrm{d} x}{L}$, where $m$ is the train's mass, $a$ is the train's deceleration, and $v$ is the trains speed. Integrating over time gives the total heat given to the track element: $\mathrm{d} Q=m a \frac{\mathrm{d} x}{L} \Delta x$, where $\Delta x$ is the total distance the train slips on the track element. Combining with $\mathrm{d} Q=c \mathrm{~d} x \cdot \Delta T$, we get $T(x)=T_{0}+\frac{m a}{c L} \Delta x$, where $c$ is the linear heat capacity. Now we split into 3 cases:
- $0<x<L$,
- $L<x<d$,
- $d<x<d+L$

where $d$ is the total distance. If $0<x<L$, the train slips a distance $\Delta x=x$ on the track element, so we have $T(x)=T_{0}+\frac{m a}{c L} x$ (linear). If $L<x<d$, the train slips a distance $\Delta x=L$ on the track element, so we have $T(x)=T_{0}+\frac{m a}{c}$ (constant). If $d<x<d+L$, the train slips a distance $\Delta x=d+L-x$, so we have $T(x)=T 0+\frac{m a}{c L}(d+L-x)$ (linear). Note that $a=\frac{v_{0}^{2}}{2 d}$ where $v_{0}$ is the train's initial velocity.

$63.98^{\circ} \mathrm{C}$

Answer: $63.98$
Problem: Consider a gas of mysterious particles called nieons that all travel at the same speed, $v$. They are enclosed in a cubical box, and there are $\rho$ nieons per unit volume. A very small hole of area $A$ is punched in the side of the box. The number of nieons that escape the box per unit time is given by
$$
\alpha v^{\beta} A^{\gamma} \rho^{\delta}
\tag{10}
$$

where $\alpha, \beta, \gamma$, and $\delta$ are all dimensionless constants. Calculate $\alpha+\beta+\gamma+\delta$.

Solution: The main idea is this: if a nieon is to escape in time $\Delta t$, then it must be traveling towards the hole and be within a hemisphere of radius $v \Delta t$, centered on the hole.
(Note that we will ignore all collisions between nieons. As with many calculations in these kinds of problems, collisions will not affect our answer. Even if they did, we can assume that the hole is so small that we can take $v \Delta t \rightarrow 0$, making the radius of the hemisphere much much smaller than the mean free path.)

First, we will break up this hemisphere into volume elements. We will calculate how many nieons are inside each volume element. Then we will find out how many of these nieons are traveling in the direction of the hole.

We define a spherical coordinate system $(r, \theta, \phi)$. The volume of each differential element is $r^{2} \sin \theta d r d \theta d \phi$, which means that there are $\rho r^{2} \sin \theta d r d \theta d \phi$ nieons inside this volume element.

Within this volume element, how many of these nieons are going towards the hole? We must think about what each nieon "sees." Visualize a sphere of radius $r$ around the volume element. The surface of the sphere will pass through the small hole in the wall. Each nieon is equally likely to go in any direction and is equally likely to end up going through any patch of this spherical surface. Thus, each nieon within this volume element has a probability $A^{\prime} / 4 \pi r^{2}$ of going through the hole, where $A^{\prime}$ is the perceived area of the hole from where the nieon is situated.

$A^{\prime}$ can be calculated as follows. $\hat{\mathbf{j}}$ is the unit vector pointing perpendicular to the hole. Let $\hat{\mathbf{r}}$ be the



unit vector pointing from the hole to the volume element. Then it can be verified that $A^{\prime}=A \cos \psi$, where $\psi$ is the angle between $\hat{\mathbf{j}}$ and $\hat{\mathbf{r}}$.

Expressing $\hat{\mathbf{r}}$ in terms of the other unit vectors, it can be shown that $A^{\prime}=A \sin \theta \sin \phi$. Hence, the fraction of nieons inside the volume element that are going towards the hole is

$$
\frac{A \sin \theta \sin \phi}{4 \pi r^{2}}
$$

Integrating over the entire hemisphere, the number of nieons escaping through the hole in a time $\Delta t$ is

$$
\int_{0}^{v \Delta t} d r \int_{0}^{\pi} d \phi \int_{0}^{\pi} d \theta\left(\frac{A}{4 \pi r^{2}} \sin \theta \sin \phi\right)\left(\rho r^{2} \sin \theta\right)=\frac{1}{4} v A \rho \cdot \Delta t
$$

Answer: 3.25
Problem: Now, Poncho has encountered a different Pico-Pico game that uses the same shaped frictionless track, but lays it horizontally on a table with friction and coefficient of friction $\mu=0.8$. In addition, the ball, which can once again be considered a point mass, is placed on the other side of the track as the ball in part 1. Finally, a buzzer on the other side of the track requires the mass to hit with at least velocity $v_{f}=2 \mathrm{~m} / \mathrm{s}$ in order to trigger the buzzer and win the game. Find the minimum velocity $v_{0}$ required for the ball to reach the end of the track with a velocity of at least $v_{f}$. The initial velocity must be directed along the track.

Solution: We simply need to find the work done by friction and we can finish with conservation of energy. To get the work done by friction, since the force is constant, we just need to find the arc length of the track. That is
$$
\ell=\int_{-\sqrt{5 / 2}}^{\sqrt{5 / 2}} \sqrt{1+h^{\prime}(x)^{2}} \mathrm{~d} x=\int_{-\sqrt{5 / 2}}^{\sqrt{5 / 2}} \sqrt{1+16 x^{2}} \mathrm{~d} x=10.76 \mathrm{~m}
$$

Then by conservation of energy,

$$
\frac{1}{2} v_{0}^{2}-\mu g \ell=\frac{1}{2} v_{f}^{2} \Longrightarrow v_{0}=\sqrt{v_{f}^{2}+2 \mu g \ell}
$$

We simply plug in the numbers to finish.

Answer: $13.1$
Problem: Let $A B C$ be a solid right triangle $(A B=5 s, A C=12 s$, and $B C=13 s$ ) with uniform charge density $\sigma$. Let $D$ be the midpoint of $B C$. We denote the electric potential of a point $P$ by $\phi(P)$. The electric potential at infinity is 0 . If $\phi(B)+\phi(C)+\phi(D)=\frac{k \sigma s}{\epsilon_{0}}$ where $k$ is a dimensionless constant, determine $k$.

Solution: If we put two of these right triangles together, we can form a rectangle with side lengths $5 s$ and $12 s$. Let $V$ be the potential at the center of this rectangle. By superposition, $\phi(D)=\frac{V}{2}$. Consider the potential at the corner. It can be decomposed into the potential from each of the right triangles, which is precisely $\phi(B)+\phi(C)$. Now, also note that if we put 4 of these rectangles side by side, to make a larger rectangle with side lengths $10 s$ and $24 s$, the potential is scaled by a factor of 2 due to dimensional analysis arguments. Thus, the potential at the center of this larger rectangle is $2 \mathrm{~V}$, but this can be decomposed into the sum of the potentials at the


corners of the 4 smaller rectangles. Thus, the potential in the corner of the smaller rectangle is $\frac{V}{2}=\phi(B)+\phi(C)$. Thus, we obtain $\phi(B)+\phi(C)+\phi(D)=V$.

Now, we will find the potential at the center of the rectangle. Note that it suffices to find the potential at the vertex of an isosceles triangle because we can connect the center of the rectangle to the 4 corners to create 4 isosceles triangles. Suppose we have an isosceles triangle with base $2 x$ and height $y$. The potential at the vertex is

$$
\iint \frac{\sigma}{4 \pi \epsilon_{0} r}(r d r d \theta)=\frac{\sigma}{4 \pi \epsilon_{0}} \int_{-\tan ^{-1}\left(\frac{x}{y}\right)}^{\tan ^{-1}\left(\frac{x}{y}\right)} \int_{0}^{\frac{y}{\cos \theta}} d r d \theta=\frac{\sigma y}{2 \pi \epsilon_{0}} \log \left(\frac{x+\sqrt{x^{2}+y^{2}}}{y}\right)
$$

If the sides of the rectangle are $a$ and $b$, we then obtain the potential at the center is

$$
V=\frac{\sigma}{2 \pi \epsilon_{0}}\left(a \log \left(\frac{b+\sqrt{a^{2}+b^{2}}}{a}\right)+b \log \left(\frac{a+\sqrt{a^{2}+b^{2}}}{b}\right)\right)
$$

In this case,

$$
k=\frac{5}{2 \pi} \ln 5+\frac{6}{\pi} \ln \frac{3}{2}
$$

Answer: 2.055
Problem: A straight ladder $A B$ of mass $m=1 \mathrm{~kg}$ is positioned almost vertically such that point $B$ is in contact with the ground with a coefficient of friction $\mu=0.15$. It is given an infinitesimal kick at the point $A$ so that the ladder begins rotating about point $B$. Find the value $\phi_{m}$ of angle $\phi$ of the ladder with the vertical at which the lower end $B$ starts slipping on the ground.

Solution: By conservation of energy, we have $\frac{1}{2} I \omega_{m}^{2}=m g \frac{L}{2}\left(1-\cos \phi_{m}\right)$ where $I=\frac{1}{3} m L^{2}$. Thus,
$$
\omega_{m}=\sqrt{\frac{3 g\left(1-\cos \phi_{m}\right)}{L}}
$$

Also, by torque analysis about B, we have $\tau=m g \frac{L}{2} \sin \phi_{m}=I \alpha_{m}$ which means

$$
\alpha_{m}=\frac{3 g}{2 L} \sin \phi_{m}
$$

Thus, the centripetal and tangential accelerations of the ladder are $a_{c}=\omega_{m}^{2} \frac{L}{2}=\frac{3}{2} g\left(1-\cos \phi_{m}\right)$ and $a_{t}=\alpha_{m} \frac{L}{2}=\frac{3}{4} g \sin \phi_{m}$ respectively. The normal force is thus $N=m g-m a_{c} \cos \phi-m a_{t} \sin \phi$, so

$$
\frac{N}{m g}=1-\frac{3}{2} \cos \phi_{m}\left(1-\cos \phi_{m}\right)-\frac{3}{4} \sin ^{2} \phi_{m}
$$

The frictional force is thus $f=m a_{t} \cos \phi_{m}-m a_{c} \sin \phi_{m}$ so

$$
\frac{f}{m g}=\frac{3}{4} \sin \phi_{m} \cos \phi_{m}-\frac{3}{2} \sin \phi_{m}\left(1-\cos \phi_{m}\right)
$$

Setting $\frac{f}{N}=\mu$, we have $6 \sin \phi_{m}\left(1-\cos \phi_{m}\right)-3 \sin \phi_{m} \cos \phi_{m}=-4 \mu+6 \mu \cos \phi_{m}\left(1-\cos \phi_{m}\right)+$ $3 \mu \sin ^{2} \phi_{m}$. Simplifying

$$
6 \sin \phi_{m}-9 \sin \phi_{m} \cos \phi_{m}+9 \mu \cos ^{2} \phi_{m}-6 \mu \cos \phi_{m}=-\mu
$$

We then can solve for $\phi_{m}$ numerically.

Answer: $11.5$
Problem: Two Ladders Two straight ladders $A B$ and $C D$, each with length $1 \mathrm{~m}$, are symmetrically placed on smooth ground, leaning on each other, such that they are touching with their ends $B$ and $C$, ends $A$ and $D$ are touching the floor. The friction at any two surfaces is negligible. Initially both ladders are almost parallel and vertical. Find the distance $A D$ when the points $B$ and $C$ lose contact.

Solution: The center of mass of both of the ladders moves in a circle, centered at the point on the ground directly beneath $B / C$. So we find when the required normal force between the two ladders is 0 . That is, when the total net force on one of the ladders is when the two ladders lose contact. Let $2 r=\ell$. Now by conservation of energy,
$$
\frac{1}{2} m v^{2}+\frac{1}{2} \frac{m r^{2}}{3} \frac{v^{2}}{r^{2}}=m g r(1-\cos \theta)
$$

where $\theta$ is defined as the angle the ladder makes witht he vertical. So we have

$$
v^{2}\left(1+\frac{1}{3}\right)=2 g r(1-\cos \theta) \Longrightarrow v^{2}=\frac{3}{2} g r(1-\cos \theta)
$$

So the centripetal acceleration is

$$
a_{c}=\frac{v^{2}}{r}=\frac{3}{2} g(1-\cos \theta)
$$



And the tangential acceleration is

$$
\frac{d v}{d t}=\sqrt{\frac{3 g r}{2}} \frac{\sin \theta}{2 \sqrt{1-\cos \theta}} \frac{d \theta}{d t}
$$

And $\frac{d \theta}{d t}=\frac{v}{r}$, so

$$
a_{\theta}=\frac{d v}{d t}=\frac{3 g}{2} \frac{\sin \theta}{2}
$$

Now for the total acceleration to be vertical, we need

$$
a_{c} \tan \theta=a_{\theta} \text {, }
$$

so

$$
1-\cos \theta=\frac{\cos \theta}{2}
$$

This simplifies to

$$
2=3 \cos \theta
$$

So the distance between the two ends is $4 r \sin \theta=4 r \sqrt{1-\frac{4}{9}}=\frac{4 r \sqrt{5}}{3}$.

Answer: $\frac{2\sqrt{5}}{3}$
Problem: An evil gamma photon of energy $E_{\gamma 1}=200 \mathrm{keV}$ is heading towards a spaceship. The commander's only choice is shooting another photon in the direction of the gamma photon such that they 'collide' head on and produce an electron-positron pair (both have mass $m_{e}$ ). Find the lower bound on the energy $E_{\gamma 2}$ of the photon as imposed by the principles of special relativity such that this occurs. Answer in keV.

Solution: The key claim is that energy is minimized when both particles are moving at the same velocity after the collision. This can be proved by transforming into the frame where the total momentum is 0 .
This idea is sufficient because it implies both the electron and positron have the same momentum and energy after the collision. Let them both have momentum $p$. We then have

$$
2 p=\frac{E_{\gamma_{1}}-E_{\gamma_{2}}}{c} \Longrightarrow p c=\frac{E_{\gamma_{1}}-E_{\gamma_{2}}}{2}
\tag{21}
$$

By energy conservation, both the electron and positron have energy $\frac{E_{\gamma_{1}}+E_{\gamma_{2}}}{2}$. Using the result $E=(p c)^{2}+\left(m c^{2}\right)^{2}$, we obtain,

$$
\left(\frac{E_{\gamma_{1}}+E_{\gamma_{2}}}{2}\right)^{2}=(p c)^{2}+\left(m_{e} c^{2}\right)^{2}
\tag{22}
$$

Combining equation 21 and equation 22 , we get

$$
\left(\frac{E_{\gamma_{1}}+E_{\gamma_{2}}}{2}\right)^{2}=\left(\frac{E_{\gamma_{1}}-E_{\gamma_{2}}}{2}\right)^{2}+\left(m_{e} c^{2}\right)^{2} \Longrightarrow E_{\gamma_{2}}=\frac{m_{e}^{2} c^{4}}{E_{\gamma_{1}}}
$$

Answer: 1306