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


Question: A quarantined physics student decides to perform an experiment to land a small box of mass $m=60 \mathrm{~g}$ onto the center of a target a distance $\Delta d$ away. The student puts the box on a top of a frictionless ramp with height $h_{2}=0.5 \mathrm{~m}$ that is angled $\theta=30^{\circ}$ to the horizontal on a table that is $h_{1}=4 \mathrm{~m}$ above the floor. If the student pushes the spring with spring constant $k=6.5 \mathrm{~N} / \mathrm{m}$ down by $\Delta x=0.3 \mathrm{~m}$ compared to its rest length and lands the box exactly on the target, what is $\Delta d$ ? Answer in meters. You may assume friction is negligible.

![](https://cdn.mathpix.com/cropped/2023_12_21_d6ba160aede69eaaf49ag-1.jpg?height=673&width=1217&top_left_y=585&top_left_x=451)

Solution:
By conservation of energy, we have that
$$
\frac{1}{2} k x^{2}=m g x \sin \theta+\frac{1}{2} m v^{2} \Longrightarrow v=\sqrt{\frac{k}{m} x^{2}-2 g x \sin \theta}
$$

By kinematic formulae for motion with constant acceleration,

$$
-4.5 \mathrm{~m}=v \sin \theta t-\frac{1}{2} g t^{2}
$$

Solving for $t$ through the quadratic, $t=1.099 \mathrm{~s}$ and the velocity is $v=2.60 \mathrm{~m} / \mathrm{s}$. Therefore, the distance, $\Delta d=v \cos \theta t=2.47 \mathrm{~m}$.

Final answer: 2.47 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 quarantined physics student decides to perform an experiment to land a small box of mass $m=60 \mathrm{~g}$ onto the center of a target a distance $\Delta d$ away. The student puts the box on a top of a frictionless ramp with height $h_{2}=0.5 \mathrm{~m}$ that is angled $\theta=30^{\circ}$ to the horizontal on a table that is $h_{1}=4 \mathrm{~m}$ above the floor. If the student pushes the spring with spring constant $k=6.5 \mathrm{~N} / \mathrm{m}$ down by $\Delta x=0.3 \mathrm{~m}$ compared to its rest length and lands the box exactly on the target, what is $\Delta d$ ? Answer in meters. You may assume friction is negligible.

<img_4280>

Solution:
By conservation of energy, we have that
$$
\frac{1}{2} k x^{2}=m g x \sin \theta+\frac{1}{2} m v^{2} \Longrightarrow v=\sqrt{\frac{k}{m} x^{2}-2 g x \sin \theta}
$$

By kinematic formulae for motion with constant acceleration,

$$
-4.5 \mathrm{~m}=v \sin \theta t-\frac{1}{2} g t^{2}
$$

Solving for $t$ through the quadratic, $t=1.099 \mathrm{~s}$ and the velocity is $v=2.60 \mathrm{~m} / \mathrm{s}$. Therefore, the distance, $\Delta d=v \cos \theta t=2.47 \mathrm{~m}$.

Final answer: 2.47 m
Subfield: Mechanics


Question: A wooden bus of mass $M=20,000 \mathrm{~kg}$ ( $M$ represents the mass excluding the wheels) is on a ramp with angle $30^{\circ}$. Each of the four wheels is composed of a ring of mass $\frac{M}{2}$ and radius $R=1 \mathrm{~m}$ and 6 evenly spaced spokes of mass $\frac{M}{6}$ and length $R$. All components of the truck have a uniform density. Find the acceleration of the bus down the ramp assuming that it rolls without slipping.

Answer in $\mathrm{m} / \mathrm{s}^{2}$.

![](https://cdn.mathpix.com/cropped/2023_12_21_d10918c220dd12cc2515g-1.jpg?height=745&width=1008&top_left_y=560&top_left_x=556)

Solution:
The moment of inertia of each wheel can be thought of as a superposition of the six spokes and a ring. Therefore, we get:
$$
I_{\text {wheel }}=\frac{M}{2} R^{2}+6\left(\frac{1}{3} \frac{M}{6} R^{2}\right)=\frac{5}{6} M R^{2}
$$

The moment of inertia of four wheels is:

$$
I_{\mathrm{bus}}=4\left(\frac{5}{6} M R^{2}\right)=\frac{10}{3} M R^{2}
$$

The total mass of the bus is

$$
m=M+4\left(\frac{M}{2}+6 \cdot \frac{M}{6}\right)=7 M
$$

Using Newton's second law down the ramp,

$$
7 M a=7 M g \sin \theta-4 f
$$

if $f$ is the friction at each wheel, and the torque balance on each wheel is:

$$
\frac{5}{6} M R^{2} \alpha=f R
$$

Letting $a=\alpha r$ for the no slip condition, we can solve for $f$ to be:

$$
f=\frac{5}{6} M a
$$

so our force balance equation becomes:

$$
7 M a=7 M g \sin \theta-\frac{10}{3} M a \Longrightarrow a=\frac{21}{31} g \sin \theta=3.32 \mathrm{~m} / \mathrm{s}^{2}
$$

Final answer: 3.32 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: In an old coal factory, a conveyor belt will move at a constant velocity of $20.3 \mathrm{~m} / \mathrm{s}$ and can deliver a maximum power of $15 \mathrm{MW}$. Each wheel in the conveyor belt has a diameter of $2 \mathrm{~m}$. However a changing demand has pushed the coal factory to fill their coal hoppers with a different material with a certain constant specific density. These "coal" hoppers have been modified to deliver a constant $18 \mathrm{~m}^{3} \mathrm{~s}^{-1}$ of the new material to the conveyor belt. Assume that the kinetic and static friction are the same and that there is no slippage. What is the maximum density of the material?

Solution:
The maximal force the convey belt can provide to a particle is:
$$
F=\frac{P}{v}
$$

The conveyor belt must provide an impulse to the particles to have a momentum of $p=m v$, where $m$ is the mass of the particle and $v$ is the velocity.

$$
F=\frac{d p}{d t}
$$

where $\frac{d p}{d t}$ is:

$$
\rho \dot{V} v
$$

Solving for for the maximum density we get:

$$
\begin{gathered}
\rho=\frac{P}{\dot{V} v^{2}} \\
\rho=2022.2 \frac{\mathrm{kg}}{\mathrm{m}^{3}}
\end{gathered}
$$

Final answer: $2022.2$ $\frac{\mathrm{kg}}{\mathrm{m}^{3}}$
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 ball is situated at the midpoint of the bottom of a rectangular ditch with width $1 \mathrm{~m}$. It is shot at a velocity $v=5 \mathrm{~m} / \mathrm{s}$ at an angle of $30^{\circ}$ relative to the horizontal. How many times does the ball collide with the walls of the ditch until it hits the bottom of the ditch again? Assume all collisions to be elastic and that the ball never flies out of the ditch.

<img_4255>

Solution:
We use the idea of mirroring the walls of the ditch. We can then draw out the normal path of the projectile and find the number of intersections the projectile makes with the mirror walls. The total time of the projectile to travel a path is given by
$$
t=\frac{2 v_{0} \sin \theta}{g}
$$

The projectile will cover a horizontal distance

$$
\left(N+\frac{1}{2}\right) a=v \cos \theta t \Longrightarrow N=\frac{v}{a} \cos \theta t-\frac{1}{2}
$$

Taking the ceiling of this gives:

$$
N=\left\lceil\frac{R-a / 2}{a}\right\rceil=2 .
$$

Final answer: 2
Subfield: Mechanics


Question: A ball is situated at the midpoint of the bottom of a rectangular ditch with width $1 \mathrm{~m}$. It is shot at a velocity $v=5 \mathrm{~m} / \mathrm{s}$ at an angle of $30^{\circ}$ relative to the horizontal. How many times does the ball collide with the walls of the ditch until it hits the bottom of the ditch again? Assume all collisions to be elastic and that the ball never flies out of the ditch.

![](https://cdn.mathpix.com/cropped/2023_12_21_ceba72362d0faa7a8dc8g-1.jpg?height=567&width=813&top_left_y=459&top_left_x=648)

Solution:
We use the idea of mirroring the walls of the ditch. We can then draw out the normal path of the projectile and find the number of intersections the projectile makes with the mirror walls. The total time of the projectile to travel a path is given by
$$
t=\frac{2 v_{0} \sin \theta}{g}
$$

The projectile will cover a horizontal distance

$$
\left(N+\frac{1}{2}\right) a=v \cos \theta t \Longrightarrow N=\frac{v}{a} \cos \theta t-\frac{1}{2}
$$

Taking the ceiling of this gives:

$$
N=\left\lceil\frac{R-a / 2}{a}\right\rceil=2 .
$$

Final answer: 2
Subfield: Mechanics


Question: A frictionless track contains a loop of radius $R=0.5 \mathrm{~m}$. Situated on top of the track lies a small ball of mass $m=2 \mathrm{~kg}$ at a height $h$. It is then dropped and collides with another ball (of negligible size) of mass $M=5 \mathrm{~kg}$.

![](https://cdn.mathpix.com/cropped/2023_12_21_4d9f6a3f09e5d7bf56fag-1.jpg?height=357&width=797&top_left_y=407&top_left_x=661)

Let $h$ be the minimum height that $m$ was dropped such that $M$ would be able to move all the way around the loop. The coefficient of restitution for this collision is given as $e=\frac{1}{2}$.

Now consider a different scenario. Assume that the balls can now collide perfectly inelastically, which means that they stick to each other instantaneously after collision for the rest of the motion. If $m$ was dropped from a height $3 R$, find the minimum value of $\frac{m}{M}$ such that the combined mass can fully move all the way around the loop. Let this minimum value be $k$. Compute $\alpha=\frac{k^{2}}{h^{2}}$. (Note that this question is only asking for $\alpha$ but you need to find $h$ to find $\alpha$ ). Assume the balls are point masses (neglect rotational effects).

Solution:
The velocity of $m$ when it gets to the bottom of the track will be given by
$$
v_{b}=\sqrt{2 g h} .
$$

Claim: The velocity of $M$ after collision will be given by $v_{M}=\frac{(1+e) m}{m+M} \sqrt{2 g h}$.

Proof: Conservation of momentum before and after the collision is expressed by:

$$
m v_{b}=m v_{m}+M v_{M}
$$

By coefficient of restitution,

$$
v_{M}-v_{m}=e v_{b}
$$

These equations may be solved directly to find $v_{m}, v_{M}$ to give

$$
v_{M}=\frac{(1+e) m}{m+M} v_{b}
$$

Next, when the objects get to the top of the loop, conservation of energy gives the speed of $M$ when it gets to the top as

$$
\frac{1}{2} v_{M}^{2}=\frac{1}{2} v_{t}^{2}+2 g R .
$$

At the top of the loop, $M$ must at least have an acceleration $g$ to maintain circular motion, and thus

$$
\frac{v_{t}^{2}}{R}=g \Longrightarrow v_{t}=\sqrt{g R}
$$



Substituting these results into our conservation equation gives us

$$
\begin{aligned}
\frac{(1+e)^{2} m^{2}}{(m+M)^{2}}(\sqrt{2 g h})^{2} & =g R+4 g R \\
\frac{2(1+e)^{2} m^{2} g h}{(m+M)^{2}} & =5 g R \\
h & =\frac{5(m+M)^{2} R}{2(1+e)^{2} m^{2}} \\
& =6.805 \mathrm{~m}
\end{aligned}
$$

In the second scenario, conservation of momentum before and after the collision gives:

$$
m v_{b}=(M+m) v_{f} \Longrightarrow v_{f}=\frac{m v_{b}}{M+m}
$$

The same conservation of energy formula as in the first scenario yields

$$
\begin{aligned}
\frac{m^{2}}{(M+m)^{2}}(\sqrt{2 g h})^{2} & =5 g R \\
\frac{2 m^{2} g h}{(M+m)^{2}} & =5 g R \\
\frac{6 m^{2} g R}{(M+m)^{2}} & =5 g R \\
\left(\frac{k}{k+1}\right)^{2} & =\frac{5}{6} \\
\frac{k}{k+1} & =\sqrt{\frac{5}{6}} \\
k & =\frac{\sqrt{5}}{\sqrt{6}-\sqrt{5}} \\
& =10.477
\end{aligned}
$$

Thus,

$$
\alpha=\frac{k^{2}}{h^{2}}=2.37 \mathrm{~m}^{2}
$$

Final answer: $2.37$ $\mathrm{~m}^{2}$
Subfield: Electromagnetism


Question: The following diagram depicts a single wire that is bent into the shape below. The circuit is placed in a magnetic field pointing out of the page, uniformly increasing at the rate $\frac{d B}{d t}=2.34 \mathrm{~T} / \mathrm{s}$. Calculate the magnitude of induced electromotive force in the wire, in terms of the following labelled areas $\left(\mathrm{m}^{2}\right)$. Note that $B$ is non-inclusive of $C$ and that $A=4.23, B=2.74$, and $C=0.34$.

![](https://cdn.mathpix.com/cropped/2023_12_21_b30a714cc20c5e96a317g-1.jpg?height=374&width=612&top_left_y=2144&top_left_x=754)

Solution:
Without loss of generality, let the current around $A$ flow in the counterclockwise direction and let the flux through $A$ be positive. Note that the current will flow in the clockwise direction in $C$ and around $B$, the area enclosed by the loop is $B+C$. The flux will be negative here. Therefore, the total flux is then proportional to:
$$
\Phi \propto A-B-2 C=0.81
$$

and the magnitude of the induced electromotive force is:

$$
\varepsilon=(A-B-2 C) \frac{d B}{d t}=1.9 \mathrm{Tm}^{2} \mathrm{~s}^{-1} .
$$

Final answer: $1.9$ $\mathrm{Tm}^{2} \mathrm{~s}^{-1}$
Subfield: Electromagnetism

- 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: The following diagram depicts a single wire that is bent into the shape below. The circuit is placed in a magnetic field pointing out of the page, uniformly increasing at the rate $\frac{d B}{d t}=2.34 \mathrm{~T} / \mathrm{s}$. Calculate the magnitude of induced electromotive force in the wire, in terms of the following labelled areas $\left(\mathrm{m}^{2}\right)$. Note that $B$ is non-inclusive of $C$ and that $A=4.23, B=2.74$, and $C=0.34$.

<img_4260>

Solution:
Without loss of generality, let the current around $A$ flow in the counterclockwise direction and let the flux through $A$ be positive. Note that the current will flow in the clockwise direction in $C$ and around $B$, the area enclosed by the loop is $B+C$. The flux will be negative here. Therefore, the total flux is then proportional to:
$$
\Phi \propto A-B-2 C=0.81
$$

and the magnitude of the induced electromotive force is:

$$
\varepsilon=(A-B-2 C) \frac{d B}{d t}=1.9 \mathrm{Tm}^{2} \mathrm{~s}^{-1} .
$$

Final answer: $1.9$ $\mathrm{Tm}^{2} \mathrm{~s}^{-1}$
Subfield: Thermodynamics

- 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: Inside a laboratory at room temperature, a steel tuning fork in the shape of a $\mathrm{U}$ is struck and begins to vibrate at $f=426 \mathrm{~Hz}$. The tuning fork is then brought outside where it is $10^{\circ} \mathrm{C}$ hotter and the experiment is performed again. What is the change in frequency, $\Delta f$ of the tuning fork? (A positive value will indicate an increase in frequency, and a negative value will indicate a decrease.)

Note: The linear thermal coefficient of expansion for steel is $\alpha=1.5 \times 10^{-5} \mathrm{~K}^{-1}$ and you may assume the expansion is isotropic and linear. When the steel bends, there is a restoring torque $\tau=-\kappa \theta$ such that $\kappa \equiv G J$ where $G=77 \mathrm{GPa}$ is constant and $J$ depends on the geometry and dimensions of the cross-sectional area.

Solution:
Note that $\kappa$ has units of torque so dimensionally, $J$ must be proportional to $L^{3}$. Therefore, we have:
$$
\beta M L^{2} \alpha \propto-L^{3} \theta \Longrightarrow f \propto \sqrt{L}
$$

Therefore, we have:

$$
\frac{\Delta f}{f}=\frac{\Delta L}{2 L}
$$

Since $\frac{\Delta L}{L}=\alpha \Delta T$, this gives us:

$$
\Delta f=\frac{1}{2} f \alpha \Delta T=0.0320 \mathrm{~Hz}
$$

Final answer: 0.0320 Hz