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

## Real Expansion



Consider a "real" monatomic gas consisting of $N$ atoms of negligible volume and mass $m$ in equilibrium inside a closed cubical container of volume $V$. In this "real" gas, the attractive forces between atoms is small but not negligible. Because these atoms have negligible volume, you can assume that the atoms do not collide with each other for the entirety of the problem.
Context question:
a. Consider an atom in the interior of this container of volume $V$. Suppose the potential energy of the interaction is given by


$$

u(r)= \begin{cases}0 & r<d \\ -\epsilon\left(\frac{d}{r}\right)^{6} & r \geq d\end{cases}

$$



where $d \ll V^{1 / 3}$ is the minimum allowed distance between two atoms. Assume the gas is uniformly distributed within the container, what is the average potential energy of this atom? Write your answer in terms of $a^{\prime}=\frac{2 \pi d^{3} \epsilon}{3}, N$, and $V$.
Context answer:
\boxed{$-2 a^{\prime} N / V$}


Question: b. What is the average potential energy of an atom near the boundary of the box? Assume that there is no interaction between atoms near the boundary and the box itself.

Solution:
Now only half of the shell of radius $r$ is full of gas, and the other half is outside of the box. This mean that the potential energy is lessened by a factor of two, to $-a^{\prime} N / V$.

Final answer: $-a^{\prime} N / V$
Subfield: Thermodynamics

## Real Expansion



Consider a "real" monatomic gas consisting of $N$ atoms of negligible volume and mass $m$ in equilibrium inside a closed cubical container of volume $V$. In this "real" gas, the attractive forces between atoms is small but not negligible. Because these atoms have negligible volume, you can assume that the atoms do not collide with each other for the entirety of the problem.
Context question:
a. Consider an atom in the interior of this container of volume $V$. Suppose the potential energy of the interaction is given by


$$

u(r)= \begin{cases}0 & r<d \\ -\epsilon\left(\frac{d}{r}\right)^{6} & r \geq d\end{cases}

$$



where $d \ll V^{1 / 3}$ is the minimum allowed distance between two atoms. Assume the gas is uniformly distributed within the container, what is the average potential energy of this atom? Write your answer in terms of $a^{\prime}=\frac{2 \pi d^{3} \epsilon}{3}, N$, and $V$.
Context answer:
\boxed{$-2 a^{\prime} N / V$}


Context question:
b. What is the average potential energy of an atom near the boundary of the box? Assume that there is no interaction between atoms near the boundary and the box itself.
Context answer:
\boxed{$-a^{\prime} N / V$}


Question: c. Using Bernoulli's law $P+U+\rho v^{2} / 2=$ constant, with pressure $P$, potential energy density $U$, mass density $\rho$ and fluid velocity $v$, what is the pressure at the boundary of the box? Assume the interior pressure is given by the ideal gas law.

Solution:
The potential energy density difference is $-a^{\prime} \frac{N^{2}}{V^{2}}$. Since there is no velocity difference, this is also the pressure difference. If the pressure on the interior is $\frac{N k T}{V}$, then the pressure on the box is $\frac{N k T}{V}-a^{\prime} \frac{N^{2}}{V^{2}}$

Final answer: $\frac{N k T}{V}-a^{\prime} \frac{N^{2}}{V^{2}}$
Subfield: Thermodynamics

## Real Expansion



Consider a "real" monatomic gas consisting of $N$ atoms of negligible volume and mass $m$ in equilibrium inside a closed cubical container of volume $V$. In this "real" gas, the attractive forces between atoms is small but not negligible. Because these atoms have negligible volume, you can assume that the atoms do not collide with each other for the entirety of the problem.
Context question:
a. Consider an atom in the interior of this container of volume $V$. Suppose the potential energy of the interaction is given by


$$

u(r)= \begin{cases}0 & r<d \\ -\epsilon\left(\frac{d}{r}\right)^{6} & r \geq d\end{cases}

$$



where $d \ll V^{1 / 3}$ is the minimum allowed distance between two atoms. Assume the gas is uniformly distributed within the container, what is the average potential energy of this atom? Write your answer in terms of $a^{\prime}=\frac{2 \pi d^{3} \epsilon}{3}, N$, and $V$.
Context answer:
\boxed{$-2 a^{\prime} N / V$}


Context question:
b. What is the average potential energy of an atom near the boundary of the box? Assume that there is no interaction between atoms near the boundary and the box itself.
Context answer:
\boxed{$-a^{\prime} N / V$}


Context question:
c. Using Bernoulli's law $P+U+\rho v^{2} / 2=$ constant, with pressure $P$, potential energy density $U$, mass density $\rho$ and fluid velocity $v$, what is the pressure at the boundary of the box? Assume the interior pressure is given by the ideal gas law.
Context answer:
\boxed{$\frac{N k T}{V}-a^{\prime} \frac{N^{2}}{V^{2}}$}


Question: d. Assuming most atoms are in the interior of the box, what is the total energy of the atoms in the box?

Solution:
The total kinetic energy is $\frac{3}{2} N k T$. The total potential energy is $-a^{\prime} N^{2} / V$ (we drop a factor of two to avoid double-counting). So the total energy is $\frac{3}{2} N k T-a^{\prime} N^{2} / V$.

Final answer: $\frac{3}{2} N k T-a^{\prime} N^{2} / V$
Subfield: Thermodynamics

## Real Expansion



Consider a "real" monatomic gas consisting of $N$ atoms of negligible volume and mass $m$ in equilibrium inside a closed cubical container of volume $V$. In this "real" gas, the attractive forces between atoms is small but not negligible. Because these atoms have negligible volume, you can assume that the atoms do not collide with each other for the entirety of the problem.
Context question:
a. Consider an atom in the interior of this container of volume $V$. Suppose the potential energy of the interaction is given by


$$

u(r)= \begin{cases}0 & r<d \\ -\epsilon\left(\frac{d}{r}\right)^{6} & r \geq d\end{cases}

$$



where $d \ll V^{1 / 3}$ is the minimum allowed distance between two atoms. Assume the gas is uniformly distributed within the container, what is the average potential energy of this atom? Write your answer in terms of $a^{\prime}=\frac{2 \pi d^{3} \epsilon}{3}, N$, and $V$.
Context answer:
\boxed{$-2 a^{\prime} N / V$}


Context question:
b. What is the average potential energy of an atom near the boundary of the box? Assume that there is no interaction between atoms near the boundary and the box itself.
Context answer:
\boxed{$-a^{\prime} N / V$}


Context question:
c. Using Bernoulli's law $P+U+\rho v^{2} / 2=$ constant, with pressure $P$, potential energy density $U$, mass density $\rho$ and fluid velocity $v$, what is the pressure at the boundary of the box? Assume the interior pressure is given by the ideal gas law.
Context answer:
\boxed{$\frac{N k T}{V}-a^{\prime} \frac{N^{2}}{V^{2}}$}


Context question:
d. Assuming most atoms are in the interior of the box, what is the total energy of the atoms in the box?
Context answer:
\boxed{$\frac{3}{2} N k T-a^{\prime} N^{2} / V$}


Extra Supplementary Reading Materials:

Now consider an insulated partitioned container with two sections, each of volume $V$. We fill one side of the container with $N$ atoms of this "real" gas at temperature $T$, which the other side being a vacuum. We then quickly remove the partition and let the gas expand to fill the entirety of the partitioned container. During this expansion, the energy of the gas remains unchanged.

Question: e. What is the final temperature of the gas after the expansion?

Solution:
Naively, we might say that the total potential energy of the gas is $-2 a^{\prime} N^{2} / V$, but to avoid double-counting, we divide by 2 and instead arrive at $-a^{\prime} N^{2} / V$. Then, the quantity



$$

E=\frac{3}{2} N k_{B} T-\frac{a^{\prime} N^{2}}{V}

$$



is conserved. Therefore,



$$

T^{\prime}=T-\frac{a^{\prime} N}{3 k_{B} V}

$$

Final answer: $T^{\prime}=T-\frac{a^{\prime} N}{3 k_{B} V}$
Subfield: Thermodynamics

$g=9.8 \mathrm{~N} / \mathrm{kg}$



$k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$



$c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$



$N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$



$\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$



$1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$



$m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$



$m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$



$\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$



$$

\begin{aligned}

& G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\

& k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\

& k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\

& R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\

& e=1.602 \times 10^{-19} \mathrm{C} \\

& h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\

& (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\

& \ln (1+x) \approx x \text { for }|x| \ll 1 \\

& \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1

\end{aligned}

$$

$$

\begin{array}{lrl}

\text { velocity parameter } & \beta & =\frac{v}{c} \\

\text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\

\text { relativistic momentum } & p & =\gamma \beta m c \\

\text { relativistic energy } & E & =\gamma m c^{2} \\

\text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}}

\end{array}

$$

Extra Supplementary Reading Materials:

Consider two objects with equal heat capacities $C$ and initial temperatures $T_{1}$ and $T_{2}$. A Carnot engine is run using these objects as its hot and cold reservoirs until they are at equal temperatures. Assume that the temperature changes of both the hot and cold reservoirs is very small compared to the temperature during any one cycle of the Carnot engine.

Question: a. Find the final temperature $T_{f}$ of the two objects, and the total work $W$ done by the engine.

Solution:
Since a Carnot engine is reversible, it produces no entropy,



$$

d S_{1}+d S_{2}=\frac{d Q_{1}}{T_{1}}+\frac{d Q_{2}}{T_{2}}=0

$$


By the definition of heat capacity, $d Q_{i}=C d T_{i}$, so



$$

\frac{d T_{1}}{T_{1}}=-\frac{d T_{2}}{T_{2}}

$$



Integrating this equation shows that $T_{1} T_{2}$ is constant, so the final temperature is



$$

T_{f}=\sqrt{T_{1} T_{2}}

$$



The change in thermal energy of the objects is



$$

C\left(T_{f}-T_{1}\right)+C\left(T_{f}-T_{2}\right)=C\left[2 \sqrt{T_{1} T_{2}}-T_{1}-T_{2}\right]

$$



By the First Law of Thermodynamics, the missing energy has been used to do work, so



$$

W=C\left[T_{1}+T_{2}-2 \sqrt{T_{1} T_{2}}\right]

$$

Final answer: $T_{f}=\sqrt{T_{1} T_{2}}$, $W=C\left[T_{1}+T_{2}-2 \sqrt{T_{1} T_{2}}\right]$
Subfield: Modern Physics

$g=9.8 \mathrm{~N} / \mathrm{kg}$



$k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$



$c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$



$N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$



$\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$



$1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$



$m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$



$m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$



$\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$



$$

\begin{aligned}

& G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\

& k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\

& k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\

& R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\

& e=1.602 \times 10^{-19} \mathrm{C} \\

& h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\

& (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\

& \ln (1+x) \approx x \text { for }|x| \ll 1 \\

& \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1

\end{aligned}

$$

$$

\begin{array}{lrl}

\text { velocity parameter } & \beta & =\frac{v}{c} \\

\text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\

\text { relativistic momentum } & p & =\gamma \beta m c \\

\text { relativistic energy } & E & =\gamma m c^{2} \\

\text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}}

\end{array}

$$

Extra Supplementary Reading Materials:

Relativistic particles obey the mass energy relation



$$

E^{2}=(p c)^{2}+\left(m c^{2}\right)^{2}

$$



where $E$ is the relativistic energy of the particle, $p$ is the relativistic momentum, $m$ is the mass, and $c$ is the speed of light.



A proton with mass $m_{p}$ and energy $E_{p}$ collides head on with a photon which is massless and has energy $E_{b}$. The two combine and form a new particle with mass $m_{\Delta}$ called $\Delta$, or "delta". It is a one dimensional collision that conserves both relativistic energy and relativistic momentum.
Context question:
a. Determine $E_{p}$ in terms of $m_{p}, m_{\Delta}$, and $E_{b}$. You may assume that $E_{b}$ is small.
Context answer:
\boxed{$\frac{m_{\Delta}^{2}-m_{p}^{2}}{4 E_{b}}$}


Question: b. In this case, the photon energy $E_{b}$ is that of the cosmic background radiation, which is an EM wave with wavelength $1.06 \mathrm{~mm}$. Determine the energy of the photons, writing your answer in electron volts.

Solution:
Plugging in the numbers gives



$$

E=\frac{h c}{\lambda}=0.00112 \mathrm{eV}

$$

Final answer: $0.00112$ eV
Subfield: Mechanics


Question: d. A row of toppling dominoes can be considered to have a propagation speed of the length $l+t$ divided by the time between successive collisions. When the first domino is given a minimal push just large enough to topple and start a chain reaction of toppling dominoes, the speed increases with each domino, but approaches an asymptotic speed $v$.



![](https://cdn.mathpix.com/cropped/2023_12_21_ccfe1ffced15d55565eag-1.jpg?height=246&width=358&top_left_y=1449&top_left_x=905)



Suppose there is a row of dominoes on another planet. These dominoes have the same density as the dominoes previously considered, but are twice as tall, wide, and thick, and placed with a spacing of $2 l$ between them. If this row of dominoes topples with the same asymptotic speed $v$ previously found, what is the local gravitational acceleration on this planet?

Solution:
This part is independent of the others and requires only dimensional analysis. The speed $v$ can depend on $g, h, w, l$. To get a quantity with dimensions $\left[L T^{-1}\right]$, we must take



$$

v=c \sqrt{g L}

$$



where $L$ is some length made from $h, w$, and $l$. On the new planet, $v$ is the same and $L$ is twice as much, so $g$ must be half as great on the new planet.

Final answer: So the final answer is $g$ must be half as great on the new planet
Subfield: Electromagnetism


Question: d. The proton beam enters the electromagnet and is deflected by an angle $\theta=10^{\circ}$. Determine the magnetic field strength.



![](https://cdn.mathpix.com/cropped/2023_12_21_3059736d5ab3a83051d2g-1.jpg?height=342&width=659&top_left_y=363&top_left_x=836)

Solution:
Start with $F=q v B$ where $F$ is the force on the protons, and $v$ the velocity. The protons are non-relativistic, so



$$

\frac{1}{2} m v^{2}=q V

$$



They move in a circle of radius $r$ inside the field, given by



$$

\frac{m v^{2}}{r}=q v B

$$



Combining these equations gives



$$

r=\frac{m v}{q B}=\frac{m}{q B} \sqrt{\frac{2 q V}{m}}=\frac{1}{B} \sqrt{\frac{2 m V}{q}}

$$



Solving for the magnetic field strength,



$$

B=\frac{1}{r} \sqrt{\frac{2 m V}{q}}

$$



To relate this to the angle, we need to do some geometry. Sketch two circles, one of radius $b$, the other of radius $r$, that intersect perpendicular to each other, as in the diagram above. Then drawing a triangle gives



$$

\tan \frac{\theta}{2}=\frac{b}{r} .

$$



Combining our results, the answer is



$$

B=\frac{\tan \theta / 2}{b} \sqrt{\frac{2 m V}{q}}=0.0894 \mathrm{~T}

$$

Final answer: 0.0894 T
Subfield: Electromagnetism


Question: e. The electromagnet is composed of layers of spiral wound copper pipe; the pipe has inner diameter $d_{i}=0.40 \mathrm{~cm}$ and outer diameter $d_{o}=0.50 \mathrm{~cm}$. The copper pipe is wound into this flat spiral that has an inner diameter $D_{i}=20 \mathrm{~cm}$ and outer diameter $D_{o}=50 \mathrm{~cm}$. Assuming the pipe almost touches in the spiral winding, determine the length $L$ in one spiral.



Copyright (c)2017 American Association of Physics Teachers







![](https://cdn.mathpix.com/cropped/2023_12_21_b3dd0e9beea33cd68ffdg-1.jpg?height=379&width=306&top_left_y=220&top_left_x=950)

Solution:
Treat the problem as two dimensional. The area of the spiral is



$$

A=\frac{\pi}{4}\left(D_{o}^{2}-D_{i}^{2}\right)

$$



The area of the pipe is



$$

A=L d_{o}

$$



Equating and solving,



$$

L=\frac{\pi\left(D_{o}^{2}-D_{i}^{2}\right)}{4 d_{o}}=33 \mathrm{~m}

$$

Final answer: 33 m
Subfield: Mechanics

$$

\begin{array}{ll}

g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\

k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\

c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\

N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\

\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\

1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\

m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\

\sin \theta \approx \theta-\theta^{3} / 6 \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\theta^{2} / 2 \text { for }|\theta| \ll 1

\end{array}

$$

Extra Supplementary Reading Materials:

Collision Course



Two blocks, $A$ and $B$, of the same mass are on a fixed inclined plane, which makes a $30^{\circ}$ angle with the horizontal. At time $t=0, A$ is a distance $\ell=5 \mathrm{~cm}$ along the incline above $B$, and both blocks are at rest. Suppose the coefficients of static and kinetic friction between the blocks and the incline are



$$

\mu_{A}=\frac{\sqrt{3}}{6}, \quad \mu_{B}=\frac{\sqrt{3}}{3}

$$



and that the blocks collide perfectly elastically. Let $v_{A}(t)$ and $v_{B}(t)$ be the speeds of the blocks down the incline. For this problem, use $g=10 \mathrm{~m} / \mathrm{s}^{2}$, assume both blocks stay on the incline for the entire time, and neglect the sizes of the blocks.
Context question:
a. Graph the functions $v_{A}(t)$ and $v_{B}(t)$ for $t$ from 0 to 1 second on the provided answer sheet, with a solid and dashed line respectively. Mark the times at which collisions occur.
Context answer:
<img_4458>


Question: b. Derive an expression for the total distance block $A$ has moved from its original position right after its $n^{\text {th }}$ collision, in terms of $\ell$ and $n$.

Solution:
The easiest way to calculate the total distance after $n$ collisions is through the graph, namely the area enclosed by the blue line: When $n=1$, the distance traveled is $d_{A}(1)=\ell$, for $n=2, d_{A}(2)=d_{A}(1)+4 \ell$, and so on, so



$$

d_{A}(n)=\ell+4 \ell+8 \ell+\ldots+(n-1) \times 4 \ell=\ell\left(2 n^{2}-2 n+1\right) .

$$

Final answer: $d_{A}(n)=\ell \left(2 n^{2}-2 n+1\right)$
Subfield: Mechanics

$$

\begin{array}{ll}

g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\

k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\

c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\

N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\

\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\

1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\

m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\

\sin \theta \approx \theta-\theta^{3} / 6 \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\theta^{2} / 2 \text { for }|\theta| \ll 1

\end{array}

$$

Extra Supplementary Reading Materials:

Collision Course



Two blocks, $A$ and $B$, of the same mass are on a fixed inclined plane, which makes a $30^{\circ}$ angle with the horizontal. At time $t=0, A$ is a distance $\ell=5 \mathrm{~cm}$ along the incline above $B$, and both blocks are at rest. Suppose the coefficients of static and kinetic friction between the blocks and the incline are



$$

\mu_{A}=\frac{\sqrt{3}}{6}, \quad \mu_{B}=\frac{\sqrt{3}}{3}

$$



and that the blocks collide perfectly elastically. Let $v_{A}(t)$ and $v_{B}(t)$ be the speeds of the blocks down the incline. For this problem, use $g=10 \mathrm{~m} / \mathrm{s}^{2}$, assume both blocks stay on the incline for the entire time, and neglect the sizes of the blocks.
Context question:
a. Graph the functions $v_{A}(t)$ and $v_{B}(t)$ for $t$ from 0 to 1 second on the provided answer sheet, with a solid and dashed line respectively. Mark the times at which collisions occur.
Context answer:
<img_4458>


Context question:
b. Derive an expression for the total distance block $A$ has moved from its original position right after its $n^{\text {th }}$ collision, in terms of $\ell$ and $n$.
Context answer:
\boxed{$d_{A}(n)=\ell \left(2 n^{2}-2 n+1\right)$}


Extra Supplementary Reading Materials:

Now suppose that the coefficient of block $B$ is instead $\mu_{B}=\sqrt{3} / 2$, while $\mu_{A}=\sqrt{3} / 6$ remains the same.
Context question:
Again, graph the functions $v_{A}(t)$ and $v_{B}(t)$ for $t$ from 0 to 1 second on the provided answer sheet, with a solid and dashed line respectively. Mark the times at which collisions occur.
Context answer:
<img_4347>


Question: At time $t=1 \mathrm{~s}$, how far has block $A$ moved from its original position?

Solution:
Again, using the graph, it is easy to see that at $t=1 \mathrm{~s}, A$ just finished the 5 th collision, and the total distance it moved is: $5 \ell=25 \mathrm{~cm}$.

Final answer: $25$ cm