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


Question: Colliding Conducting Slab A thin conducting square slab with side length $s=5 \mathrm{~cm}$, initial charge $q=0.1 \mu \mathrm{C}$, and mass $m=100 \mathrm{~g}$ is given a kick and sent bouncing between two infinite conducting plates separated by a distance $d=0.5 \mathrm{~cm} \ll s$ and with surface charge density $\pm \sigma= \pm 50 \mu \mathrm{C} / \mathrm{m}^{2}$. After a long time it is observed exactly in the middle of the two plates to be traveling with velocity of magnitude $v=3 \mathrm{~m} / \mathrm{s}$ and direction $\theta=30^{\circ}$ with respect to the horizontal line parallel to the plates. How many collisions occur after it has traveled a distance $L=15 \mathrm{~m}$ horizontally from when it was last observed? Assume that all collisions are elastic, and neglect induced charges. Note that the setup is horizontal so gravity does not need to be accounted for.
![](https://cdn.mathpix.com/cropped/2023_12_21_b6b6c3c8a654288caf6bg-1.jpg?height=280&width=710&top_left_y=1676&top_left_x=705)

Solution:
After the first collision, the charge approaches a constant magnitude. Let's look at a collision with the plate at charge $q$ hitting the $-\sigma$ plate. Since the slab is thin, to keep the conducting surface an equipotential, the charge on the conduting slab has to become $-\sigma s^{2}$. The initial charge doesn't matter because the plates are infinite. So the charge is $-\sigma s^{2}$ after leaving the negative plate and $+\sigma s^{2}$ after leaving the positive plate.
Now the acceleration always has a constant magnitude, which is directed toward the plate it is traveling to, which we can find using $F=m a$,

$$
\left(\sigma s^{2}\right) \frac{\sigma}{\epsilon_{0}}=m a \Longrightarrow a=\frac{\sigma^{2} s^{2}}{\epsilon_{0} m}=7.06 \mathrm{~m} / \mathrm{s}^{2}
$$



Then here we can use the trick of reflecting over the plates, and now it's a basic kinematics problem. The time to travel $L$ is $t=\frac{L}{v \cos \theta}$, and then the distance it travels vertically is

$$
y=v \sin \theta \frac{L}{v \cos \theta}+\frac{1}{2} a\left(\frac{L}{v \cos \theta}\right)^{2}=126.36 \mathrm{~m}
$$

And since $25272 d+d / 2 \approx 87.129$, the number of collisions is 25273 .

Final answer: 25273
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 player throws two tennis balls on a level ground at $v=20 \mathrm{~m} / \mathrm{s}$ in the same direction, once at an angle of $\alpha=35^{\circ}$ and once at an angle $\beta=55^{\circ}$ to the horizontal. The distance between the landing spots of the two balls is $d$. Find $d$ in meters.
Assume the height of the player is negligble and ignore air resistance.

Solution:
The range of a projectile is proportional as $R \propto \sin 2 \theta$, or $R \propto \cos \theta \sin \theta$. As $\cos (90-\theta)=\sin \theta$, and $\alpha+\beta=90$, the distance travelled by both projectiles are the same.
$0 \mathrm{~m}$

Final answer: 0 m
Subfield: Mechanics


Question: Consider the following simple model of a bow and arrow. An ideal elastic string has a spring constant $k=10 \mathrm{~N} / \mathrm{m}$ and relaxed length $L=1 \mathrm{~m}$ which is attached to the ends of an inflexible fixed steel rod of the same length $L$ as shown below. A small ball of mass $m=2 \mathrm{~kg}$ and the thread are pulled by its midpoint away from the rod until each individual part of the thread have the same length of the rod, as shown below. What is the speed of the ball in meters per seconds right after it stops accelerating? Assume the whole setup is carried out in zero gravity.
![](https://cdn.mathpix.com/cropped/2023_12_21_78a63d65ef2bfc8b2947g-1.jpg?height=220&width=273&top_left_y=1164&top_left_x=926)

Solution:
We can use conservation of energy. The bow string has its potential increased as
$$
E_{p}=\frac{1}{2} k(2 L-L)^{2}=\frac{1}{2} k L^{2}
$$

This all turns into the kinetic energy of the ball $E_{k}=\frac{1}{2} m v^{2}$, so

$$
E_{p}=E_{k} \Longrightarrow \frac{1}{2} k L^{2}=\frac{1}{2} m v^{2} \Longrightarrow v=L \sqrt{\frac{k}{m}}
$$

Final answer: $2.23$ m/s
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: Consider the following simple model of a bow and arrow. An ideal elastic string has a spring constant $k=10 \mathrm{~N} / \mathrm{m}$ and relaxed length $L=1 \mathrm{~m}$ which is attached to the ends of an inflexible fixed steel rod of the same length $L$ as shown below. A small ball of mass $m=2 \mathrm{~kg}$ and the thread are pulled by its midpoint away from the rod until each individual part of the thread have the same length of the rod, as shown below. What is the speed of the ball in meters per seconds right after it stops accelerating? Assume the whole setup is carried out in zero gravity.
<img_4282>

Solution:
We can use conservation of energy. The bow string has its potential increased as
$$
E_{p}=\frac{1}{2} k(2 L-L)^{2}=\frac{1}{2} k L^{2}
$$

This all turns into the kinetic energy of the ball $E_{k}=\frac{1}{2} m v^{2}$, so

$$
E_{p}=E_{k} \Longrightarrow \frac{1}{2} k L^{2}=\frac{1}{2} m v^{2} \Longrightarrow v=L \sqrt{\frac{k}{m}}
$$

Final answer: $2.23$ m/s
Subfield: Mechanics


Question: A truck (denoted by $S$ ) is driving at a speed $v=2 \mathrm{~m} / \mathrm{s}$ in the opposite direction of a car driving at a speed $u=3 \mathrm{~m} / \mathrm{s}$, which is equipped with a rear-view mirror. Both $v$ and $u$ are measured from an observer on the ground. Relative to this observer, what is the speed (in $\mathrm{m} / \mathrm{s}$ ) of the truck's image $S^{\prime}$ through the car's mirror? Car's mirror is a plane mirror.

![](https://cdn.mathpix.com/cropped/2023_12_21_b2b497998ee2d73231a3g-1.jpg?height=488&width=615&top_left_y=192&top_left_x=752)

Solution:
In the mirror's frame of reference, the source speed and the image speed is both $u+v$ but in opposite direction. Now, go back the the observer's frame of reference, the image speed becomes $v^{\prime}$ :
$$
v^{\prime}=(u+v)+u=2 u+v=8 \mathrm{~m} / \mathrm{s}
\tag{1}
$$

Final answer: $8$ m/s
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 truck (denoted by $S$ ) is driving at a speed $v=2 \mathrm{~m} / \mathrm{s}$ in the opposite direction of a car driving at a speed $u=3 \mathrm{~m} / \mathrm{s}$, which is equipped with a rear-view mirror. Both $v$ and $u$ are measured from an observer on the ground. Relative to this observer, what is the speed (in $\mathrm{m} / \mathrm{s}$ ) of the truck's image $S^{\prime}$ through the car's mirror? Car's mirror is a plane mirror.

<img_4297>

Solution:
In the mirror's frame of reference, the source speed and the image speed is both $u+v$ but in opposite direction. Now, go back the the observer's frame of reference, the image speed becomes $v^{\prime}$ :
$$
v^{\prime}=(u+v)+u=2 u+v=8 \mathrm{~m} / \mathrm{s}
\tag{1}
$$

Final answer: $8$ m/s
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: For this problem, assume the Earth moves in a perfect circle around the sun in the $x y$ plane, with a radius of $r=1.496 \times 10^{11} \mathrm{~m}$, and the Earth has a mass $m=5.972 \times 10^{24} \mathrm{~kg}$. An alien stands far away from our solar system on the $x$ axis such that it appears the Earth is moving along a one dimensional line, as if there was a zero-length spring connecting the Earth and the Sun.
For the alien at this location, it is impossible to tell just from the motion if it's 2D motion via gravity or 1D motion via a spring. Let $U_{g}$ be the gravitational potential energy ignoring its self energy if Earth moves via gravity, taking potential energy at infinity to be 0 and $U_{s}$ be the maximum spring potential energy if Earth moves in $1 \mathrm{D}$ via a spring. Compute $U_{g} / U_{s}$.

Solution:
One naive idea is to directly compute $U_{g}$ and $U_{s}$, but we can use the fact that their frequencies are the same, or:
$$
\omega^{2}=\frac{k}{m}=\frac{G M}{r^{3}} \Longrightarrow k r^{2}=\frac{G M}{r}
$$

Then,

$$
U_{g}=-\frac{G M m}{r}=-k r^{2}
$$

and

$$
U_{s}=\frac{1}{2} k r^{2}
$$

Therefore,

$$
U_{g} / U_{s}=-2
$$

Final answer: -2
Subfield: Optics


Question: These days, there are so many stylish rectangular home-designs (see figure A). It is possible from the outline of those houses in their picture to estimate with good precision where the camera was. Consider an outline in one photograph of a rectangular house which has height $H=3$ meters (see figure B for square-grid coordinates). Assume that the camera size is negligible, how high above the ground (in meters) was the camera at the moment this picture was taken?![](https://cdn.mathpix.com/cropped/2023_12_21_6ad9a16bf503154c2401g-1.jpg?height=990&width=892&top_left_y=1260&top_left_x=622)

Solution:
The formation of the house's image seen in the picture is due to pinhole principle, and note that the fish-eye effect here is weak (straight-lines stays straight). Define points $A, B, C, A^{\prime}, B^{\prime}, C^{\prime}$ as in the attached Fig., since $A A^{\prime}, B B^{\prime}, C C^{\prime}$ stays parallel we know that the camera looked horizontally



at the time this picture is taken.

![](https://cdn.mathpix.com/cropped/2023_12_21_479ef03f149db178c6b9g-1.jpg?height=632&width=1480&top_left_y=313&top_left_x=317)

To determine the height of the camera at the very same moment, we need to know the where is the horizontal plane passing through the camera in the picture which is collapsed into a line. That can be found by finding the intersection $M$ of $A B \cap A^{\prime} B^{\prime}$ and the intersection $N$ of $B C \cap B^{\prime} C^{\prime}$, then $M N$ is the line of interests. $M N$ intersects $B B^{\prime}$ at $P$, the position of $P$ can be calculated too be $(22,0.9)$, therefore the height of the camera is the length-ratio $P B^{\prime} / B B^{\prime}$ times $3 \mathrm{~m}$, which equals to $0.9 \mathrm{~m}$.

Final answer: $0.9$ m
Subfield: Electromagnetism


Question: Consider a thin rigid wire-frame MNPP'N'M' in which MNN'M' and NPP'N' are two squares of side $L$ with resistance per unit-length $\lambda$ and their planes are perpendicular. The frame is rotated with a constant angular velocity $\omega$ around an axis passing through $\mathrm{NN}$ ' and put in a region with constant magnetic field $B$ pointing perpendicular to $\mathrm{NN}^{\prime}$. What is the total heat released on the frame per revolution (in Joules)? Use $L=1 \mathrm{~m}, \lambda=1 \Omega / \mathrm{m}, \omega=2 \pi \mathrm{rad} / \mathrm{s}$ and $B=1 \mathrm{~T}$.
![](https://cdn.mathpix.com/cropped/2023_12_21_479ef03f149db178c6b9g-1.jpg?height=651&width=681&top_left_y=1710&top_left_x=711)

Solution:
In this setting, for every orientation during rotation the total magnetic flux passing though MNPP'N'M' is tthe same as through MPP'M', which has area $S=\sqrt{2} L^{2}$.
The magnetic flux is:

$$
\Phi(t)=B S \sin (\omega t)=\sqrt{2} B L^{2} \sin (\omega t) .
\tag{3}
$$

The emf running around the wire-frame is:

$$
E(t)=\frac{d}{d t} \Phi(t)=\sqrt{2} B L^{2} \omega \cos (\omega t)
\tag{4}
$$

The electrical current running around the wire-frame is:

$$
I(t)=\frac{E(t)}{6 \lambda L}=\frac{B L \omega \cos (\omega t)}{3 \sqrt{2} \lambda}
\tag{5}
$$

The heat released power is:

$$
\frac{d}{d t} Q(t)=I^{2}(t) \times 6 \lambda L=\frac{B^{2} L^{3} \omega^{2} \cos ^{2}(\omega t)}{3 \lambda}
\tag{6}
$$

Thus the total heat released per revolution is:

$$
Q=\int_{0}^{2 \pi / \omega} d t \frac{d}{d t} Q(t)=\frac{B^{2} L^{3} \omega^{2} \int_{0}^{2 \pi / \omega} d t \cos ^{2}(\omega t)}{3 \lambda}=\frac{\pi B^{2} L^{3} \omega}{3 \lambda} \approx 6.58 \mathrm{~J}
\tag{7}
$$

Final answer: $6.58$ J
Subfield: Mechanics


Question: A scale of uniform mass $M=3 \mathrm{~kg}$ of length $L=4 \mathrm{~m}$ is kept on a rough table (infinite friction) with $l=1 \mathrm{~m}$ hanging out of the table as shown in the figure below. A small ball of mass $m=1 \mathrm{~kg}$ is released from rest from a height of $h=5 \mathrm{~m}$ above the end of the scale. Find the maximum angle (in degrees) that the scale rotates by in the subsequent motion if ball sticks to the scale after collision. Take gravity $g=10 \mathrm{~m} / \mathrm{s}^{2}$.

![](https://cdn.mathpix.com/cropped/2023_12_21_4f8039b1096042486648g-1.jpg?height=523&width=612&top_left_y=186&top_left_x=754)

Solution:
The ball falls with velocity $\sqrt{2 g h}=10$. Applying conservation of angular momentum about the end point of the table.
$$
\begin{gathered}
L_{i}=m v x \\
=10 \\
L_{f}=I_{1} \omega \\
I_{1}=4+3+1=8 \\
\Longrightarrow \omega=1.25
\end{gathered}
$$

Now applying energy conservation

$$
\begin{gathered}
E_{i}=\frac{1}{2} I \omega^{2}=6.25 \\
E_{f}=M_{\text {total }} g x_{c o m} \sin (\theta)=20 \sin \theta \\
\theta=18.21^{\circ}
\end{gathered}
$$

Final answer: $18.21$ $^{\circ}$
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 scale of uniform mass $M=3 \mathrm{~kg}$ of length $L=4 \mathrm{~m}$ is kept on a rough table (infinite friction) with $l=1 \mathrm{~m}$ hanging out of the table as shown in the figure below. A small ball of mass $m=1 \mathrm{~kg}$ is released from rest from a height of $h=5 \mathrm{~m}$ above the end of the scale. Find the maximum angle (in degrees) that the scale rotates by in the subsequent motion if ball sticks to the scale after collision. Take gravity $g=10 \mathrm{~m} / \mathrm{s}^{2}$.

<img_4317>

Solution:
The ball falls with velocity $\sqrt{2 g h}=10$. Applying conservation of angular momentum about the end point of the table.
$$
\begin{gathered}
L_{i}=m v x \\
=10 \\
L_{f}=I_{1} \omega \\
I_{1}=4+3+1=8 \\
\Longrightarrow \omega=1.25
\end{gathered}
$$

Now applying energy conservation

$$
\begin{gathered}
E_{i}=\frac{1}{2} I \omega^{2}=6.25 \\
E_{f}=M_{\text {total }} g x_{c o m} \sin (\theta)=20 \sin \theta \\
\theta=18.21^{\circ}
\end{gathered}
$$

Final answer: $18.21$ $^{\circ}$