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1020 | Context Example 1:
Question:
The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or "Mountain of Gems". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle ("PT") and its variants.
Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below.
PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below.
<image>
As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question.
Clark's Triangle: If the left side of PT is replaced with consecutive multiples of 6 , starting with 0 , but the right entries (except the first) and the generating rule are left unchanged, the result is called Clark's Triangle. If the $k^{\text {th }}$ entry of the $n^{\text {th }}$ row is denoted by $\mathrm{Cl}(n, k)$, then the formal rule is:
$$
\begin{cases}\mathrm{Cl}(n, 0)=6 n & \text { for all } n \\ \mathrm{Cl}(n, n)=1 & \text { for } n \geq 1 \\ \mathrm{Cl}(n, k)=\mathrm{Cl}(n-1, k-1)+\mathrm{Cl}(n-1, k) & \text { for } n \geq 1 \text { and } 1 \leq k \leq n-1\end{cases}
$$
The first four rows of Clark's Triangle are given below.
<image>
Compute $\mathrm{Cl}(11,3)$.
Solution:
$\mathrm{Cl}(11,3)=2025$.
Answer: 2025
Context Example 2:
Question:
In the diagram, $A B C D$ is a quadrilateral in which $\angle A+\angle C=180^{\circ}$. What is the length of $C D$ ?
<image>
Solution:
In order to determine $C D$, we must determine one of the angles (or at least some information about one of the angles) in $\triangle B C D$.
To do this, we look at $\angle A$ use the fact that $\angle A+\angle C=180^{\circ}$.
<img_3524>
Using the cosine law in $\triangle A B D$, we obtain
$$
\begin{aligned}
7^{2} & =5^{2}+6^{2}-2(5)(6) \cos (\angle A) \\
49 & =61-60 \cos (\angle A) \\
\cos (\angle A) & =\frac{1}{5}
\end{aligned}
$$
Since $\cos (\angle A)=\frac{1}{5}$ and $\angle A+\angle C=180^{\circ}$, then $\cos (\angle C)=-\cos \left(180^{\circ}-\angle A\right)=-\frac{1}{5}$.
(We could have calculated the actual size of $\angle A$ using $\cos (\angle A)=\frac{1}{5}$ and then used this to calculate the size of $\angle C$, but we would introduce the possibility of rounding error by doing this.)
Then, using the cosine law in $\triangle B C D$, we obtain
$$
\begin{aligned}
7^{2} & =4^{2}+C D^{2}-2(4)(C D) \cos (\angle C) \\
49 & =16+C D^{2}-8(C D)\left(-\frac{1}{5}\right) \\
0 & =5 C D^{2}+8 C D-165 \\
0 & =(5 C D+33)(C D-5)
\end{aligned}
$$
So $C D=-\frac{33}{5}$ or $C D=5$. (We could have also determined these roots using the quadratic formula.)
Since $C D$ is a length, it must be positive, so $C D=5$.
(We could have also proceeded by using the sine law in $\triangle B C D$ to determine $\angle B D C$ and then found the size of $\angle D B C$, which would have allowed us to calculate $C D$ using the sine law. However, this would again introduce the potential of rounding error.)
Answer: 5
Context Example 3:
Question:
Large Hadron Collider
Please read the general instructions in the separate envelope before you start this problem.
In this task, the physics of the particle accelerator LHC (Large Hadron Collider) at CERN is discussed. CERN is the world's largest particle physics laboratory. Its main goal is to get insight into the fundamental laws of nature. Two beams of particles are accelerated to high energies, guided around the accelerator ring by a strong magnetic field and then made to collide with each other. The protons are not spread uniformly around the circumference of the accelerator, but they are clustered in so-called bunches. The resulting particles generated by collisions are observed with large detectors. Some parameters of the LHC can be found in table 1 .
| LHC ring | |
| :--- | :--- |
| Circumference of ring | $26659 \mathrm{~m}$ |
| Number of bunches per proton beam | 2808 |
| Number of protons per bunch | $1.15 \times 10^{11}$ |
| Proton beams | |
| Energy of protons | $7.00 \mathrm{TeV}$ |
| Centre of mass energy | $14.0 \mathrm{TeV}$ |
Table 1: Typical numerical values of relevant LHC parameters.
Particle physicists use convenient units for the energy, momentum and mass: The energy is measured in electron volts $[\mathrm{eV}]$. By definition, $1 \mathrm{eV}$ is the amount of energy gained by a particle with elementary charge, e, moved through a potential difference of one volt $\left(1 \mathrm{eV}=1.602 \cdot 10^{-19} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2}\right)$.
The momentum is measured in units of $\mathrm{eV} / c$ and the mass in units of $\mathrm{eV} / c^{2}$, where $c$ is the speed of light in vacuum. Since $1 \mathrm{eV}$ is a very small quantity of energy, particle physicists often use $\mathrm{MeV}\left(1 \mathrm{MeV}=10^{6} \mathrm{eV}\right)$, $\mathrm{GeV}\left(1 \mathrm{GeV}=10^{9} \mathrm{eV}\right)$ or $\mathrm{TeV}\left(1 \mathrm{TeV}=10^{12} \mathrm{eV}\right)$.
Part A deals with the acceleration of protons or electrons. Part B is concerned with the identification of particles produced in the collisions at CERN.
Part A.LHC accelerator
Acceleration:
Assume that the protons have been accelerated by a voltage $V$ such that their velocity is very close to the speed of light and neglect any energy loss due to radiation or collisions with other particles.
Context question:
A.1 Find the exact expression for the final velocity $v$ of the protons as a function of the accelerating voltage $V$, and physical constants.
Context answer:
\boxed{$v=c \cdot \sqrt{1-(\frac{m_{p} \cdot c^{2}}{m_{p} \cdot c^{2}+V \cdot e})^{2}}$}
Extra Supplementary Reading Materials:
A design for a future experiment at CERN plans to use the protons from the LHC and to collide them with electrons which have an energy of $60.0 \mathrm{GeV}$.
Context question:
A.2 For particles with high energy and low mass the relative deviation $\Delta=(c-v) / c$ of the final velocity $v$ from the speed of light is very small. Find a first order approximation for $\Delta$ and calculate $\Delta$ for electrons with an energy of $60.0 \mathrm{GeVusing}$ the accelerating voltage $V$ and physical constants.
Context answer:
\boxed{$\frac{1}{2}\left(\frac{m_{e} \cdot c^{2}}{m_{e} \cdot c^{2}+V \cdot e}\right)^{2}$, $\Delta=3.63 \cdot 10^{-11}$}
Extra Supplementary Reading Materials:
We now return to the protons in the LHC. Assume that the beam pipe has a circular shape.
Context question:
A.3 Derive an expression for the uniform magnetic flux density $B$ necessary to keep the proton beam on a circular track. The expression should only contain the energy of the protons $E$, the circumference $L$, fundamental constants and numbers. You may use suitable approximations if their effect is smaller than precision given by the least number of significant digits.
Calculate the magnetic flux density $B$ for a proton energy of $E=7.00 \mathrm{TeV}$, neglecting interactions between the protons.
Context answer:
\boxed{$B=\frac{2 \pi \cdot E}{e \cdot c \cdot L}$ , $B=5.50$}
Extra Supplementary Reading Materials:
Radiated Power:
An accelerated charged particle radiates energy in the form of electromagnetic waves. The radiated power $P_{\mathrm{rad}}$ of a charged particle that circulates with a constant angular velocity depends only on its acceleration $a$, its charge $q$, the speed of light $c$ and the permittivity of free space $\varepsilon_{0}$.
Context question:
A.4 Use dimensional analysis to find an expression for the radiated power $P_{\text {rad }}$.
Context answer:
\boxed{$P_{\text {rad }}=a^{\alpha} \cdot q^{\beta} \cdot c^{\gamma} \cdot \epsilon_{0}^{\delta}$}
Extra Supplementary Reading Materials:
The real formula for the radiated power contains a factor $1 /(6 \pi)$; moreover, a full relativistic derivation gives an additional multiplicative factor $\gamma^{4}$, with $\gamma=\left(1-v^{2} / c^{2}\right)^{-\frac{1}{2}}$.
Context question:
A.5 Calculate $P_{\text {tot }}$, the total radiated power of the LHC, for a proton energy of E= 7.00 $\mathrm{TeV}$ (Note table 1). You may use suitable approximations.
Context answer:
\boxed{5.13}
Extra Supplementary Reading Materials:
Linear Acceleration:
At CERN, protons at rest are accelerated by a linear accelerator of length $d=30.0 \mathrm{~m}$ through a potential difference of $V=500 \mathrm{MV}$. Assume that the electrical field is homogeneous. A linear accelerator consists of two plates as sketched in Figure 1.
A.6 Determine the time T that the protons take to pass through this field
<image>
Figure 1: Sketch of an accelerator module.
Solution:
2nd Newton's law
$$
\begin{gathered}
F=\frac{d p}{d t} \quad \text { leads to } \\
\frac{V \cdot e}{d}=\frac{p_{f}-p_{i}}{T} \text { with } p_{i}=0
\end{gathered}
$$
Conservation of energy:
$$
E_{t o t}=m \cdot c^{2}+e \cdot V
$$
Since
$$
\begin{gathered}
E_{t o t}^{2}=\left(m \cdot c^{2}\right)^{2}+\left(p_{f} \cdot c\right)^{2} \\
\rightarrow p_{f}=\frac{1}{c} \cdot \sqrt{\left(m \cdot c^{2}+e \cdot V\right)^{2}-\left(m \cdot c^{2}\right)^{2}}=\sqrt{2 e \cdot m \cdot V+\left(\frac{e \cdot V}{c}\right)^{2}} \\
\rightarrow T=\frac{d \cdot p_{f}}{V \cdot e}=\frac{d}{V \cdot e} \sqrt{2 e \cdot m_{p} \cdot V+\left(\frac{e \cdot V}{c}\right)^{2}} \\
T=218 \mathrm{~ns}
\end{gathered}
$$
2nd Newton's law
$$
\frac{V \cdot e}{d}=\frac{p_{f}-p_{i}}{T} \text { with } p_{i}=0
$$
velocity from A1 or from conservation of energy
$$
v=c \cdot \sqrt{1-\left(\frac{m_{p} \cdot c^{2}}{m_{p} \cdot c^{2}+V \cdot e}\right)^{2}}
$$
and hence for $\gamma$
$$
\begin{gathered}
\gamma=1 / \sqrt{1-\frac{v^{2}}{c^{2}}}=1+\frac{e \cdot V}{m_{p} \cdot c^{2}} \\
\rightarrow p_{f}=\gamma \cdot m_{p} \cdot v=\left(1+\frac{e \cdot V}{m_{p} \cdot c^{2}}\right) \cdot m_{p} \cdot c \cdot \sqrt{1-\left(\frac{m_{p} \cdot c^{2}}{m_{p} \cdot c^{2}+V \cdot e}\right)^{2}} \\
\rightarrow T=\frac{d \cdot p_{f}}{V \cdot e}=\frac{d \cdot m_{p} \cdot c}{V \cdot e} \cdot \sqrt{\left(\frac{m_{p} \cdot c^{2}+e \cdot V}{m_{p} \cdot c^{2}}\right)^{2}-1}=\frac{d}{V \cdot e} \sqrt{2 e \cdot m_{p} \cdot V+\left(\frac{e \cdot V}{c}\right)^{2}} \\
T=218 \mathrm{~ns}
\end{gathered}
$$
Energy increases linearly with distance $\mathrm{x}$
$$
\begin{gathered}
E(x)=\frac{e \cdot V \cdot x}{d} \\
t=\int d t=\int_{0}^{d} \frac{d x}{v(x)} \\
v(x)=c \cdot \sqrt{1-\left(\frac{m_{p} \cdot c^{2}}{m_{p} \cdot c^{2}+\frac{e \cdot V \cdot x}{d}}\right)^{2}}=c \cdot \frac{\sqrt{\left(m_{p} \cdot c^{2}+\frac{e \cdot V \cdot x}{d}\right)^{2}-\left(m_{p} \cdot c^{2}\right)^{2}}}{m_{p} \cdot c^{2}+\frac{e \cdot V \cdot x}{d}} \\
=c \cdot \frac{\sqrt{\left(1+\frac{e \cdot V \cdot x}{d \cdot m_{p} \cdot c^{2}}\right)^{2}-1}}{1+\frac{e \cdot V \cdot x}{d \cdot m_{p} \cdot c^{2}}}
\end{gathered}
$$
Substitution : $\xi=\frac{e \cdot V \cdot x}{d \cdot m_{p} \cdot c^{2}} \quad \frac{d \xi}{d x}=\frac{e \cdot V}{d \cdot m_{p} \cdot c^{2}}$
$$
\rightarrow t=\frac{1}{c} \int_{0}^{b} \frac{1+\xi}{\sqrt{(1+\xi)^{2}-1}} \frac{d \cdot m_{p} \cdot c^{2}}{e \cdot V} d \xi \quad b=\frac{e \cdot V}{m_{p} \cdot c^{2}}
$$
$$
1+\xi:=\cosh (s) \quad \frac{d \xi}{d s}=\sinh (s)
$$
0.1
$$
t=\frac{m_{p} \cdot c \cdot d}{e \cdot V} \int \frac{\cosh (s) \cdot \sinh (s) d s}{\sqrt{\cosh ^{2}(s)-1}}=\frac{m_{p} \cdot c \cdot d}{e \cdot V}[\sinh (s)]_{b_{1}}^{b_{2}}
$$
$$
\text { with } \quad b_{1}=\cosh ^{-1}(1), \quad b_{2}=\cosh ^{-1}\left(1+\frac{e \cdot V}{m_{p} \cdot c^{2}}\right)
$$
$$
T=218 \mathrm{~ns}
$$
Alternative: differential equation
$$
\begin{gathered}
\rightarrow \frac{V \cdot e}{d}=\frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{m \cdot v}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\right)=\frac{m \cdot a\left(1-\frac{v^{2}}{c^{2}}\right)+m \cdot a \frac{v^{2}}{c^{2}}}{\left(1-\frac{v^{2}}{c^{2}}\right)^{\frac{3}{2}}}=\gamma^{3} \cdot m \cdot a \\
a=\ddot{s}=\frac{V \cdot e}{d \cdot m}\left(1-\frac{\dot{s}^{2}}{c^{2}}\right)^{\frac{3}{2}}
\end{gathered}
$$
$$
\begin{gathered}
\rightarrow s(t)=\frac{c}{V \cdot e}\left(\sqrt{e^{2} \cdot V^{2} \cdot t^{2}+c^{2} \cdot m^{2} \cdot d^{2}}-c \cdot m \cdot d\right) \\
s=d \rightarrow T=\frac{d}{V \cdot e} \sqrt{\left(\frac{V \cdot e}{c}\right)^{2}+2 V \cdot e \cdot m} \\
T=218 \mathrm{~ns}
\end{gathered}
$$
Answer: 218
Context Example 4:
Question:
- 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,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$
Two astronauts, Alice and Bob, are standing inside their cylindrical spaceship, which is rotating at an angular velocity $\omega$ clockwise around its axis in order to simulate the gravitational acceleration $g$ on earth. The radius of the spaceship is $R$. For this problem, we will only consider motion in the plane perpendicular to the axis of the spaceship. Let point $O$ be the center of the spaceship. Initially, an ideal zero-length spring has one end fixed at point $O$, while the other end is connected to a mass $m$ at the "ground" of the spaceship, where the astronauts are standing (we will call this point $A$ ). From the astronauts' point of view, the mass remains motionless.
Next, Alice fixes one end of the spring at point $A$, and attaches the mass to the other end at point $O$. Bob starts at point $A$, and moves an angle $\theta$ counterclockwise to point $B$ (such that $A O B$ is an isosceles triangle). At time $t=0$, the mass at point $O$ is released. Given that the mass comes close enough for Bob to catch it, find the value of $\theta$ to the nearest tenth of a degree.
Assume that the only force acting on the mass is the spring's tension, and that the astronauts' heights are much less than $R$.
<image>
Solution:
First, we must have $\omega^{2} R=g$ and $k=m \omega^{2}$. Now, we step into the frame of point $A$, rotating around with the spaceship. We will thus have three fictitious forces: translational, centrifugal, and coriolis. Note that because centrifugal is $m \omega^{2} r$, pointing away from $A$, it cancels with the spring force. Thus, the only forces left to consider are translational and coriolis.
The translational force points "down" with a constant magnitude of $m g$, like gravity. The coriolis force
points perpendicular to the velocity with magnitude $2 m \omega v$. We recognize this setup is analogous to that of a charged particle moving in an E field and B field. It is well known that the mass will follow a cycloid shape. Writing the equation of the cycloid, and finding where the cycloid hits the circle (spaceship), we can find $\theta$. Note that we have to use numerical methods.
Specifically, the cycloid can be parametrized as $\frac{R}{4}(\alpha-\sin \alpha, 1-\cos \alpha)$, and we need to find where this intersects the circle $x^{2}+y^{2}=R^{2}$, so $(\alpha-\sin \alpha)^{2}+(1-\cos \alpha)^{2}=16$. Solving gives $\alpha=3.307$, and since $\cot \theta=\frac{1-\cos \alpha}{\alpha-\sin \alpha}$, we have $\theta=60.2^{\circ}$.
Answer: $60.2$
Context Example 5:
Question:
In the diagram, the area of $\triangle A B C$ is 1 . Trapezoid $D E F G$ is constructed so that $G$ is to the left of $F, D E$ is parallel to $B C$, $E F$ is parallel to $A B$ and $D G$ is parallel to $A C$. Determine the maximum possible area of trapezoid $D E F G$.
<image>
Solution:
Suppose that $A B=c, A C=b$ and $B C=a$.
Since $D G$ is parallel to $A C, \angle B D G=\angle B A C$ and $\angle D G B=\angle A C B$, so $\triangle D G B$ is similar to $\triangle A C B$.
(Similarly, $\triangle A E D$ and $\triangle E C F$ are also both similar to $\triangle A B C$.)
Suppose next that $D B=k c$, with $0<k<1$.
Then the ratio of the side lengths of $\triangle D G B$ to those of $\triangle A C B$ will be $k: 1$, so $B G=k a$ and $D G=k b$.
Since the ratio of the side lengths of $\triangle D G B$ to $\triangle A C B$ is $k: 1$, then the ratio of their areas will be $k^{2}: 1$, so the area of $\triangle D G B$ is $k^{2}$ (since the area of $\triangle A C B$ is 1 ).
Since $A B=c$ and $D B=k c$, then $A D=(1-k) c$, so using similar triangles as before, $D E=(1-k) a$ and $A E=(1-k) b$. Also, the area of $\triangle A D E$ is $(1-k)^{2}$.
Since $A C=b$ and $A E=(1-k) b$, then $E C=k b$, so again using similar triangles, $E F=k c$, $F C=k a$ and the area of $\triangle E C F$ is $k^{2}$.
Now the area of trapezoid $D E F G$ is the area of the large triangle minus the combined areas of the small triangles, or $1-k^{2}-k^{2}-(1-k)^{2}=2 k-3 k^{2}$.
We know that $k \geq 0$ by its definition. Also, since $G$ is to the left of $F$, then $B G+F C \leq B C$ or $k a+k a \leq a$ or $2 k a \leq a$ or $k \leq \frac{1}{2}$.
Let $f(k)=2 k-3 k^{2}$.
Since $f(k)=-3 k^{2}+2 k+0$ is a parabola opening downwards, its maximum occurs at its vertex, whose $k$-coordinate is $k=-\frac{2}{2(-3)}=\frac{1}{3}$ (which lies in the admissible range for $k$ ).
Note that $f\left(\frac{1}{3}\right)=\frac{2}{3}-3\left(\frac{1}{9}\right)=\frac{1}{3}$.
Therefore, the maximum area of the trapezoid is $\frac{1}{3}$.
Answer: $\frac{1}{3}$
Context Example 6:
Question:
In the diagram, $O A=15, O P=9$ and $P B=4$. Determine the equation of the line through $A$ and $B$. Explain how you got your answer.
<image>
Solution:
Since $O P=9$, then the coordinates of $P$ are $(9,0)$.
Since $O P=9$ and $O A=15$, then by the Pythagorean Theorem,
$$
A P^{2}=O A^{2}-O P^{2}=15^{2}-9^{2}=144
$$
so $A P=12$.
Since $P$ has coordinates $(9,0)$ and $A$ is 12 units directly above $P$, then $A$ has coordinates $(9,12)$.
Since $P B=4$, then $B$ has coordinates $(13,0)$.
The line through $A(9,12)$ and $B(13,0)$ has slope $\frac{12-0}{9-13}=-3$ so, using the point-slope form, has equation $y-0=-3(x-13)$ or $y=-3 x+39$.
Answer: $y=-3 x+39$
Context Example 7:
Question:
In the diagram, points $P(p, 4), B(10,0)$, and $O(0,0)$ are shown. If $\triangle O P B$ is right-angled at $P$, determine all possible values of $p$.
<image>
Solution:
Since $\angle O P B=90^{\circ}$, then $O P$ and $P B$ are perpendicular, so the product of their slopes is -1 .
The slope of $O P$ is $\frac{4-0}{p-0}=\frac{4}{p}$ and the slope of $P B$ is $\frac{4-0}{p-10}=\frac{4}{p-10}$.
Therefore, we need
$$
\begin{aligned}
\frac{4}{p} \cdot \frac{4}{p-10} & =-1 \\
16 & =-p(p-10) \\
p^{2}-10 p+16 & =0 \\
(p-2)(p-8) & =0
\end{aligned}
$$
and so $p=2$ or $p=8$. Since each these steps is reversible, then $\triangle O P B$ is right-angled precisely when $p=2$ and $p=8$.
Since $\triangle O P B$ is right-angled at $P$, then $O P^{2}+P B^{2}=O B^{2}$ by the Pythagorean Theorem. Note that $O B=10$ since $O$ has coordinates $(0,0)$ and $B$ has coordinates $(10,0)$.
Also, $O P^{2}=(p-0)^{2}+(4-0)^{2}=p^{2}+16$ and $P B^{2}=(10-p)^{2}+(4-0)^{2}=p^{2}-20 p+116$. Therefore,
$$
\begin{aligned}
\left(p^{2}+16\right)+\left(p^{2}-20 p+116\right) & =10^{2} \\
2 p^{2}-20 p+32 & =0 \\
p^{2}-10 p+16 & =0
\end{aligned}
$$
and so $(p-2)(p-8)=0$, or $p=2$ or $p=8$. Since each these steps is reversible, then $\triangle O P B$ is right-angled precisely when $p=2$ and $p=8$.
Answer: 2,8
Context Example 8:
Question:
Five distinct integers are to be chosen from the set $\{1,2,3,4,5,6,7,8\}$ and placed in some order in the top row of boxes in the diagram. Each box that is not in the top row then contains the product of the integers in the two boxes connected to it in the row directly above. Determine the number of ways in which the integers can be chosen and placed in the top row so that the integer in the bottom box is 9953280000 .
<image>
Solution:
Suppose that the integers in the first row are, in order, $a, b, c, d, e$.
Using these, we calculate the integer in each of the boxes below the top row in terms of these variables, using the rule that each integer is the product of the integers in the two boxes above:
$a$
| $b$ | $c$ | $c$ | | $d$ |
| :---: | :---: | :---: | :---: | :---: |
| $a b^{2} c$ | $b c$ | $c d$ | $c$ | |
| | $a b^{3} c^{3} d$ | $b c^{2} d$ | $c d^{2} e$ | |$\quad d e$
Therefore, $a b^{4} c^{6} d^{4} e=9953280000$.
Next, we determine the prime factorization of the integer 9953280000 :
$$
\begin{aligned}
9953280000 & =10^{4} \cdot 995328 \\
& =2^{4} \cdot 5^{4} \cdot 2^{3} \cdot 124416 \\
& =2^{7} \cdot 5^{4} \cdot 2^{3} \cdot 15552 \\
& =2^{10} \cdot 5^{4} \cdot 2^{3} \cdot 1944 \\
& =2^{13} \cdot 5^{4} \cdot 2^{3} \cdot 243 \\
& =2^{16} \cdot 5^{4} \cdot 3^{5} \\
& =2^{16} \cdot 3^{5} \cdot 5^{4}
\end{aligned}
$$
Thus, $a b^{4} c^{6} d^{4} e=2^{16} \cdot 3^{5} \cdot 5^{4}$.
Since the right side is not divisible by 7 , none of $a, b, c, d$, $e$ can equal 7 .
Thus, $a, b, c, d, e$ are five distinct integers chosen from $\{1,2,3,4,5,6,8\}$.
The only one of these integers divisible by 5 is 5 itself.
Since $2^{16} \cdot 3^{5} \cdot 5^{4}$ includes exactly 4 factors of 5 , then either $b=5$ or $d=5$. No other placement of the 5 can give exactly 4 factors of 5 .
Case 1: $b=5$
Here, $a c^{6} d^{4} e=2^{16} \cdot 3^{5}$ and $a, c, d, e$ are four distinct integers chosen from $\{1,2,3,4,6,8\}$. Since $a c^{6} d^{4} e$ includes exactly 5 factors of 3 and the possible values of $a, c, d$, e that are divisible by 3 are 3 and 6 , then either $d=3$ and one of $a$ and $e$ is 6 , or $d=6$ and one of $a$ and $e$ is 3 . No other placements of the multiples of 3 can give exactly 5 factors of 3 .
Case 1a: $b=5, d=3, a=6$
Here, $a \cdot c^{6} \cdot d^{4} \cdot e=6 \cdot c^{6} \cdot 3^{4} \cdot e=2 \cdot 3^{5} \cdot c^{6} \cdot e$.
This gives $c^{6} e=2^{15}$ and $c$ and $e$ are distinct integers from $\{1,2,4,8\}$.
Trying the four possible values of $c$ shows that $c=4$ and $e=8$ is the only solution in this case. Here, $(a, b, c, d, e)=(6,5,4,3,8)$.
Case 1b: $b=5, d=3, e=6$ We obtain $(a, b, c, d, e)=(8,5,4,3,6)$.
Case 1c: $b=5, d=6, a=3$
Here, $a \cdot c^{6} \cdot d^{4} \cdot e=3 \cdot c^{6} \cdot 6^{4} \cdot e=2^{4} \cdot 3^{5} \cdot c^{6} \cdot e$.
This gives $c^{6} e=2^{12}$ and $c$ and $e$ are distinct integers from $\{1,2,4,8\}$.
Trying the four possible values of $c$ shows that $c=4$ and $e=1$ is the only solution in this case. Here, $(a, b, c, d, e)=(3,5,4,6,1)$.
Case 1d: $b=5, d=6, e=3$ We obtain $(a, b, c, d, e)=(1,5,4,6,3)$.
Case 2: $d=5$ : A similar analysis leads to 4 further quintuples $(a, b, c, d, e)$.
Therefore, there are 8 ways in which the integers can be chosen and placed in the top row to obtain the desired integer in the bottom box.
Answer: 8
Context Example 9:
Question:
- 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,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$
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 mass $M=5 \mathrm{~kg}$.
<image>
The coefficient of restitution for this collision is given as $e=\frac{1}{2}$. Now consider a different alternative. Now let the circular loop have a uniform coefficient of friction $\mu=0.6$, while the rest of the path is still frictionless. Assume that the balls can once again collide with a restitution coefficient of $e=\frac{1}{2}$. Considering the balls to be point masses, find the minimum value of $h$ such that the ball of mass $M$ would be able to move all the way around the loop. Both balls can be considered as point masses.
Solution:
Let the angle formed by $M$ at any moment of time be angle $\theta$ with the negative y-axis. The normal force experienced by $M$ is just
$$
N=M g \cos \theta+M \frac{v(\theta)^{2}}{R}
$$
by balancing the radial forces at this moment. Now, applying the work energy theorem, we have
$$
\begin{gathered}
\int-\mu\left[M g \cos \theta+M \frac{v(\theta)^{2}}{R}\right] R \mathrm{~d} \theta=\frac{1}{2} M v(\theta)^{2}-\frac{1}{2} M v_{0}^{2}+M g R(1-\cos \theta) \\
\Rightarrow-\mu\left[M g \cos \theta+M \frac{v(\theta)^{2}}{R}\right] R=\frac{M}{2} \frac{\mathrm{d}\left(v(\theta)^{2}\right)}{\mathrm{d} \theta}+M g R \sin \theta
\end{gathered}
$$
Rearranging, we have
$$
\frac{\left.\mathrm{d}\left(v(\theta)^{2}\right)\right)}{\mathrm{d} \theta}+2 \mu v(\theta)^{2}=-2 g R(\sin \theta+\mu \cos \theta)
$$
Let $v^{2}(\theta)=y$. Thus we have a first order linear ODE of the form
$$
\frac{\mathrm{d} y}{\mathrm{~d} \theta}+P(\theta) y=Q(\theta)
$$
This is easily solvable using the integrating factor $e^{\int P(\theta) \mathrm{d} \theta}$. Here the integrating factor is
$$
e^{\int 2 \mu \mathrm{d} \theta}=e^{2 \mu \theta}
$$
So multiplying by the integrating factor, we get
$$
\begin{gathered}
\int \mathrm{d}\left(e^{2 \mu \theta} y\right)=\int-2 g R(\sin \theta+\mu \cos \theta) e^{2 \mu \theta} \mathrm{d} \theta \\
\Rightarrow y=\frac{\int-2 g R(\sin \theta+\mu \cos \theta) e^{2 \mu \theta} \mathrm{d} \theta}{e^{2 \mu \theta}}
\end{gathered}
$$
Now we use the well known integrals
$$
\begin{aligned}
& \int e^{a x} \sin x \mathrm{~d} x=\frac{e^{a x}}{1+a^{2}}(a \sin x-\cos x) \\
& \int e^{a x} \cos x \mathrm{~d} x=\frac{e^{a x}}{1+a^{2}}(a \cos x+\sin x)
\end{aligned}
$$
(These integrals can be computed using integration by parts.) Thus, plugging and chugging these integration formulas into our expression for $y$ and integrating from $\theta=0$ to $\theta=\phi$, we have upon solving
$$
v^{2}(\phi)-v_{0}^{2}=\frac{-2 g R}{1+4 \mu^{2}}\left[\left(3 \mu \sin \phi+\left(2 \mu^{2}-1\right) \cos \phi-\left(2 \mu^{2}-1\right) e^{-2 \mu \phi}\right]\right.
$$
where $v_{0}$ is the velocity at $\phi=0$. Solving gives us the velocity as a function of angle covered
$$
v(\phi)=\sqrt{v_{0}^{2}-\frac{2 g R}{1+4 \mu^{2}}\left[\left(3 \mu \sin \phi+\left(2 \mu^{2}-1\right) \cos \phi-\left(2 \mu^{2}-1\right) e^{-2 \mu \phi}\right]\right.}
$$
But to cover a complete circle, at the top most point
$$
N=m g-\frac{m v^{2}(\pi)}{R} \geq 0 \Rightarrow v(\pi) \leq \sqrt{g R}
$$
Thus
$$
v_{0} \leq \sqrt{g R\left[1+\frac{2\left(1-2 \mu^{2}\right)}{1+4 \mu^{2}}\left(1+e^{-2 \mu \pi}\right)\right]}
$$
From the previous expression,
$$
v_{0}=\frac{m(1+e) \sqrt{2 g h}}{M+m} \geq \sqrt{g R\left[1+\frac{2\left(1-2 \mu^{2}\right)}{1+4 \mu^{2}}\left(1+e^{-2 \mu \pi}\right)\right]}
$$
Hence
$$
h \geq \frac{R(M+m)^{2}}{2 m^{2}(1+e)^{2}}\left[1+\frac{2\left(1-2 \mu^{2}\right)}{1+4 \mu^{2}}\left(1+e^{-2 \mu \pi}\right)\right]
$$
We get $h \geq 72.902 \mathrm{~m}$ and we are done.
Answer: $72.902$
Context Example 10:
Question:
A triangle has vertices $A(0,3), B(4,0)$, $C(k, 5)$, where $0<k<4$. If the area of the triangle is 8 , determine the value of $k$.
<image>
Solution:
We "complete the rectangle" by drawing a horizontal line through $C$ which meets the $y$-axis at $P$ and the vertical line through $B$ at $Q$.
<img_3215>
Since $C$ has $y$-coordinate 5 , then $P$ has $y$-coordinate 5 ; thus the coordinates of $P$ are $(0,5)$.
Since $B$ has $x$-coordinate 4 , then $Q$ has $x$-coordinate 4 .
Since $C$ has $y$-coordinate 5 , then $Q$ has $y$-coordinate 5 .
Therefore, the coordinates of $Q$ are $(4,5)$, and so rectangle $O P Q B$ is 4 by 5 and so has area $4 \times 5=20$.
Now rectangle $O P Q B$ is made up of four smaller triangles, and so the sum of the areas of these triangles must be 20 .
Let us examine each of these triangles:
- $\triangle A B C$ has area 8 (given information)
- $\triangle A O B$ is right-angled at $O$, has height $A O=3$ and base $O B=4$, and so has area $\frac{1}{2} \times 4 \times 3=6$.
- $\triangle A P C$ is right-angled at $P$, has height $A P=5-3=2$ and base $P C=k-0=k$, and so has area $\frac{1}{2} \times k \times 2=k$.
- $\triangle C Q B$ is right-angled at $Q$, has height $Q B=5-0=5$ and base $C Q=4-k$, and so has area $\frac{1}{2} \times(4-k) \times 5=10-\frac{5}{2} k$.
Since the sum of the areas of these triangles is 20 , then $8+6+k+10-\frac{5}{2} k=20$ or $4=\frac{3}{2} k$ and so $k=\frac{8}{3}$.
Answer: $\frac{8}{3}$
Current Question:
3.1 Calculate the electric field intensity $\vec{E}_{p}$ at a distance $r$ from an ideal electric dipole $\vec{p}$ at the origin $\mathrm{O}$ along the direction of $\vec{p}$ in Figure 2.
$p=2 a q, \quad r \gg a$
<image_1>
FIGURE 2 | [
"$E_{p}=\\frac{2 p}{4 \\pi \\varepsilon_{0} r^{3}}$"
] | [
"Using Coulomb's Law, we write the electric field at a distance $r$ is given by\n\n$$\n\\begin{aligned}\n& E_{p}=\\frac{q}{4 \\pi \\varepsilon_{0}(r-a)^{2}}-\\frac{q}{4 \\pi \\varepsilon_{0}(r+a)^{2}} \\\\\n& E_{p}=\\frac{q}{4 \\pi \\varepsilon_{0} r^{2}}\\left(\\frac{1}{\\left(1-\\frac{a}{r}\\right)^{2}}-\\frac{1}{\\left(1+\\frac{a}{r}\\right)^{2}}\\right)\n\\end{aligned}\n\\tag{1}\n$$\n\nUsing binomial expansion for small $a$,\n\n$$\n\\begin{aligned}\nE_{p} & =\\frac{q}{4 \\pi \\varepsilon_{0} r^{2}}\\left(1+\\frac{2 a}{r}-1+\\frac{2 a}{r}\\right) \\\\\n& =+\\frac{4 q a}{4 \\pi \\varepsilon_{0} r^{3}}=+\\frac{q a}{\\pi \\varepsilon_{0} r^{3}} \\\\\n& =\\frac{2 p}{4 \\pi \\varepsilon_{0} r^{3}}\n\\end{aligned}\n\\tag{2}\n$$"
] | false | null | Expression | null | OE_MM_physics_en_COMP | Modern Physics | 3. To Commemorate the Centenary of Rutherford's Atomic Nucleus: the Scattering of an Ion by a Neutral Atom
<img_4421>
An ion of mass $m$, charge $Q$, is moving with an initial non-relativistic speed $v_{0}$ from a great distance towards the vicinity of a neutral atom of mass $M>>m$ and of electrical polarisability $\alpha$. The impact parameter is $b$ as shown in Figure 1.
The atom is instantaneously polarised by the electric field $\vec{E}$ of the in-coming (approaching) ion. The resulting electric dipole moment of the atom is $\vec{p}=\alpha \vec{E}$. Ignore any radiative losses in this problem. |
|
1028 | Context Example 1:
Question:
- 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,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$
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 $M=5 \mathrm{~kg}$.
<image>
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,
$$
\alpha=\frac{k^{2}}{h^{2}}=2.37 \mathrm{~m}^{2}
$$
Answer: $2.37$
Context Example 2:
Question:
- 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,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$
Two astronauts, Alice and Bob, are standing inside their cylindrical spaceship, which is rotating at an angular velocity $\omega$ clockwise around its axis in order to simulate the gravitational acceleration $g$ on earth. The radius of the spaceship is $R$. For this problem, we will only consider motion in the plane perpendicular to the axis of the spaceship. Let point $O$ be the center of the spaceship. Initially, an ideal zero-length spring has one end fixed at point $O$, while the other end is connected to a mass $m$ at the "ground" of the spaceship, where the astronauts are standing (we will call this point $A$ ). From the astronauts' point of view, the mass remains motionless.
Next, Alice fixes one end of the spring at point $A$, and attaches the mass to the other end at point $O$. Bob starts at point $A$, and moves an angle $\theta$ counterclockwise to point $B$ (such that $A O B$ is an isosceles triangle). At time $t=0$, the mass at point $O$ is released. Given that the mass comes close enough for Bob to catch it, find the value of $\theta$ to the nearest tenth of a degree.
Assume that the only force acting on the mass is the spring's tension, and that the astronauts' heights are much less than $R$.
<image>
Solution:
First, we must have $\omega^{2} R=g$ and $k=m \omega^{2}$. Now, we step into the frame of point $A$, rotating around with the spaceship. We will thus have three fictitious forces: translational, centrifugal, and coriolis. Note that because centrifugal is $m \omega^{2} r$, pointing away from $A$, it cancels with the spring force. Thus, the only forces left to consider are translational and coriolis.
The translational force points "down" with a constant magnitude of $m g$, like gravity. The coriolis force
points perpendicular to the velocity with magnitude $2 m \omega v$. We recognize this setup is analogous to that of a charged particle moving in an E field and B field. It is well known that the mass will follow a cycloid shape. Writing the equation of the cycloid, and finding where the cycloid hits the circle (spaceship), we can find $\theta$. Note that we have to use numerical methods.
Specifically, the cycloid can be parametrized as $\frac{R}{4}(\alpha-\sin \alpha, 1-\cos \alpha)$, and we need to find where this intersects the circle $x^{2}+y^{2}=R^{2}$, so $(\alpha-\sin \alpha)^{2}+(1-\cos \alpha)^{2}=16$. Solving gives $\alpha=3.307$, and since $\cot \theta=\frac{1-\cos \alpha}{\alpha-\sin \alpha}$, we have $\theta=60.2^{\circ}$.
Answer: $60.2$
Context Example 3:
Question:
1. Restore equilibrium
A wooden plank of length $1 \mathrm{~m}$ and uniform cross-section is hinged at one end to the bottom of a tank as shown in figure. The tank is filled with water up to a height $0.5 \mathrm{~m}$. The specific gravity of the plank (or ratio of plank density to water density) is 0.5 . Find the angle $\theta$ that the plank makes with the vertical in the equilibrium position (exclude the case $\theta=0^{\circ}$ )
<image>
Solution:
<img_4519>
One could find the net buoyant force by integrating about the point of reference. Or one can directly use the fact this upthrust force acts at the center of portion submerged in water.
A sample solution:
Three forces will act on the plank-
1. Weight which will act at centre of plank.
2. Upthrust which will act at centre of submerged portion.
3. Force from the hinge at $\mathrm{O}$.
Notation used: Submerged length $=0.5 \sec \theta, \mathrm{F}=$ Upthrust, $w=$ Weight
Taking moments of all three forces about point O. Moment of hinge force will be zero.
$$
(A \lg \rho) \frac{l}{2} \sin \theta=A(0.5 \sec \theta)\left(\rho_{w}\right)(g)\left(\frac{0.5 \sec \theta}{2}\right) \sin \theta
\tag{1}
$$
Solve to get $\theta=45^{\circ}$
Answer: $\theta=45$
Context Example 4:
Question:
An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:
$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$
Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write
$$
S_{3}[263415]=(132,312,231,213)
$$
More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.
In this power question, you will be asked to analyze some of the properties of labels and signatures.
We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :
<image>
A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:
<image>
Consider the shape below:
<image>
Find the 2-signature that corresponds to this shape.
Solution:
The first pair indicates an increase; the next three are decreases, and the last pair is an increase. So the 2-signature is $(12,21,21,21,12)$.
Answer: $(12,21,21,21,12)$
Context Example 5:
Question:
In triangle $A B C, A B=B C=25$ and $A C=30$. The circle with diameter $B C$ intersects $A B$ at $X$ and $A C$ at $Y$. Determine the length of $X Y$.
<image>
Solution:
Join $B Y$. Since $B C$ is a diameter, then $\angle B Y C=90^{\circ}$. Since $A B=B C, \triangle A B C$ is isosceles and $B Y$ is an altitude in $\triangle A B C$, then $A Y=Y C=15$.
Let $\angle B A C=\theta$.
Since $\triangle A B C$ is isosceles, $\angle B C A=\theta$.
Since $B C Y X$ is cyclic, $\angle B X Y=180-\theta$ and so $\angle A X Y=\theta$.
<img_3475>
Thus $\triangle A X Y$ is isosceles and so $X Y=A Y=15$.
Therefore $X Y=15$.
Join $B Y . \angle B Y C=90^{\circ}$ since it is inscribed in a semicircle.
$\triangle B A C$ is isosceles, so altitude $B Y$ bisects the base.
Therefore $B Y=\sqrt{25^{2}-15^{2}}=20$.
Join $C X . \angle C X B=90^{\circ}$ since it is also inscribed in a semicircle.
<img_3739>
The area of $\triangle A B C$ is
$$
\begin{aligned}
\frac{1}{2}(A C)(B Y) & =\frac{1}{2}(A B)(C X) \\
\frac{1}{2}(30)(20) & =\frac{1}{2}(25)(C X) \\
C X & =\frac{600}{25}=24 .
\end{aligned}
$$
From $\triangle A B Y$ we conclude that $\cos \angle A B Y=\frac{B Y}{A B}=\frac{20}{25}=\frac{4}{5}$.
In $\Delta B X Y$, applying the Law of Cosines we get $(X Y)^{2}=(B X)^{2}+(B Y)^{2}-2(B X)(B Y) \cos \angle X B Y$.
Now (by Pythagoras $\triangle B X C$ ),
$$
\begin{aligned}
B X^{2} & =B C^{2}-C X^{2} \\
& =25^{2}-24^{2} \\
& =49 \\
B X & =7 .
\end{aligned}
$$
Therefore $X Y^{2}=7^{2}+20^{2}-2(7)(20) \frac{4}{5}$
$$
=49+400-224
$$
$$
=225 \text {. }
$$
Therefore $X Y=15$.
Answer: 15
Context Example 6:
Question:
Let $T=24$. A regular $n$-gon is inscribed in a circle; $P$ and $Q$ are consecutive vertices of the polygon, and $A$ is another vertex of the polygon as shown. If $\mathrm{m} \angle A P Q=\mathrm{m} \angle A Q P=T \cdot \mathrm{m} \angle Q A P$, compute the value of $n$.
<image>
Solution:
Let $\mathrm{m} \angle A=x$. Then $\mathrm{m} \angle P=\mathrm{m} \angle Q=T x$, and $(2 T+1) x=180^{\circ}$, so $x=\frac{180^{\circ}}{2 T+1}$. Let $O$ be the center of the circle, as shown below.
<img_3423>
Then $\mathrm{m} \angle P O Q=2 \mathrm{~m} \angle P A Q=2\left(\frac{180^{\circ}}{2 T+1}\right)=\frac{360^{\circ}}{2 T+1}$. Because $\mathrm{m} \angle P O Q=\frac{360^{\circ}}{n}$, the denominators must be equal: $n=2 T+1$. Substitute $T=24$ to find $n=\mathbf{4 9}$.
Answer: 49
Context Example 7:
Question:
In the diagram, $\triangle A B C$ is right-angled at $C$. Also, $2 \sin B=3 \tan A$. Determine the measure of angle $A$.
<image>
Solution:
Since $\triangle A B C$ is right-angled at $C$, then $\sin B=\cos A$.
Therefore, $2 \cos A=3 \tan A=\frac{3 \sin A}{\cos A}$ or $2 \cos ^{2} A=3 \sin A$.
Using the fact that $\cos ^{2} A=1-\sin ^{2} A$, this becomes $2-2 \sin ^{2} A=3 \sin A$
or $2 \sin ^{2} A+3 \sin A-2=0$ or $(2 \sin A-1)(\sin A+2)=0$.
Since $\sin A$ is between -1 and 1 , then $\sin A=\frac{1}{2}$.
Since $A$ is an acute angle, then $A=30^{\circ}$.
Since $\triangle A B C$ is right-angled at $C$, then $\sin B=\frac{b}{c}$ and $\tan A=\frac{a}{b}$.
Thus, the given equation is $\frac{2 b}{c}=\frac{3 a}{b}$ or $2 b^{2}=3 a c$.
Using the Pythagorean Theorem, $b^{2}=c^{2}-a^{2}$ and so we obtain $2 c^{2}-2 a^{2}=3 a c$ or $2 c^{2}-3 a c-2 a^{2}=0$.
Factoring, we obtain $(c-2 a)(2 c+a)=0$.
Since $a$ and $c$ must both be positive, then $c=2 a$.
Since $\triangle A B C$ is right-angled, the relation $c=2 a$ means that $\triangle A B C$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle, with $A=30^{\circ}$.
Answer: $30^{\circ}$
Context Example 8:
Question:
## Disk Jockey
A disk of uniform mass density, mass $M$, and radius $R$ sits at rest on a frictionless floor. The disk is attached to the floor by a frictionless pivot at its center, which keeps the center of the disk in place, but allows the disk to rotate freely. An ant of mass $m \ll M$ is initially standing on the edge of the disk; you may give your answers to leading order in $m / M$.
a. The ant walks an angular displacement $\theta$ along the edge of the disk. Then it walks radially inward by a distance $h \ll R$, tangentially through an angular displacement $-\theta$, then back to its starting point on the disk. Assume the ant walks with constant speed $v$.
<image>
Through what net angle does the disk rotate throughout this process, to leading order in $h / R$ ?
Solution:
During the first leg of the trip, the disk has angular velocity
$$
\omega=-\frac{2 m v}{M R}
$$
by conservation of angular momentum. Thus, the disk rotates through an angle
$$
\phi_{1}=-\frac{2 m v}{M R} \frac{\theta R}{v}=-\frac{2 m \theta}{M}
$$
to leading order in $m / M$. (Here we have neglected the fact that the disk rotates under the ant as it is walking, somewhat reducing the distance it has to walk; this changes the answer only to higher order in $m / M$. The exact answer is a more complicated function of $m / M$. By going to "leading order", we mean we are expanding that exact answer in a series in $m / M$, such as with the binomial theorem, and keeping only the first nonzero term.)
When the ant is moving radially, $\omega=0$, so no rotation occurs. On the last leg of the trip, the disk has angular velocity
$$
\omega=\frac{2 m v(R-h)}{M R^{2}}
$$
which means the disk rotates through an angle
$$
\phi_{2}=\frac{2 m v(R-h)}{M R^{2}} \frac{\theta(R-h)}{v}=\frac{2 m \theta}{M}\left(1-\frac{h}{R}\right)^{2}
$$
The net rotation is
$$
\phi_{1}+\phi_{2}=\frac{2 m \theta}{M}\left(\left(1-\frac{h}{R}\right)^{2}-1\right) \approx-\frac{4 m}{M} \frac{h \theta}{R}
$$
The sign is not important since it is convention-dependent. (Solutions that were not fully approximated were also accepted; however, not approximating early dramatically increases the amount of work you have to do.)
Incidentally, you might have thought the answer had to be zero, by angular momentum conservation. After all, when a system has zero total linear momentum, its center of mass can't move. But this problem shows that systems with zero total angular momentum can perform net rotations, which is the reason, e.g. that a falling cat can always land on its feet. In more advanced physics, this would be described by saying the constraint on the disc's motion coming from angular momentum conservation is not holonomic.
Answer: $\frac{2 m \theta}{M}\left(\left(1-\frac{h}{R}\right)^{2}-1\right)$
Context Example 9:
Question:
Consider two liquids A and B insoluble in each other. The pressures $p_{i}(i=\mathrm{A}$ or $\mathrm{B})$ of their saturated vapors obey, to a good approximation, the formula:
$$
\ln \left(p_{i} / p_{o}\right)=\frac{\alpha_{i}}{T}+\beta_{i} ; \quad i=\mathrm{A} \text { or } \mathrm{B}
$$
where $p_{o}$ denotes the normal atmospheric pressure, $T$ - the absolute temperature of the vapor, and $\alpha_{i}$ and $\beta_{i}$ ( $i=\mathrm{A}$ or $\left.\mathrm{B}\right)$ - certain constants depending on the liquid. (The symbol $\ln$ denotes the natural logarithm, i.e. logarithm with base $e=2.7182818 \ldots)$
The values of the ratio $p_{i} / p_{0}$ for the liquids $\mathrm{A}$ and $\mathrm{B}$ at the temperature $40^{\circ} \mathrm{C}$ and $90^{\circ} \mathrm{C}$ are given in Tab. 1.1.
Table 1.1
| $t\left[{ }^{\circ} \mathrm{C}\right]$ | $i=\mathrm{A}$ | $p_{i} / p_{0}$ |
| :---: | :---: | :---: |
| | 0.284 | 0.07278 |
| 40 | 1.476 | 0.6918 |
| 90 | | |
The errors of these values are negligible.
Context question:
A. Determine the boiling temperatures of the liquids A and B under the pressure $p_{0}$.
Context answer:
\boxed{$76.8$ , $100$}
B. The liquids A and B were poured into a vessel in which the layers shown in Fig. 1.1 were formed. The surface of the liquid B has been covered with a thin layer of a non-volatile liquid $\mathrm{C}$, which is insoluble in the liquids $\mathrm{A}$ and $\mathrm{B}$ and vice versa, thereby preventing any free evaporation from the upper surface of the liquid $B$, The ratio of molecular masses of the liquids A and B (in the gaseous phase) is:
$$
\gamma=\mu_{A} / \mu_{B}=8
$$
<image>
Fig. 1.1
<image>
Fig. 1.2
The masses of the liquids A and B were initially the same, each equal to $m=100 \mathrm{~g}$. The heights of the layers of the liquids in the vessel and the densities of the liquids are small enough to make the assumption that the pressure in any point in the vessel is practically equal to the normal atmospheric pressure $p_{0}$.
The system of liquids in the vessel is slowly, but continuously and uniformly, heated. It was established that the temperature $t$ of the liquids changed with time $\tau$ as shown schematically in the Fig. 1.2.
Determine the temperatures $t_{1}$ and $t_{2}$ corresponding to the horizontal parts of the diagram and the masses of the liquids $\mathrm{A}$ and $\mathrm{B}$ at the time $\tau_{1}$. The temperatures should be rounded to the nearest degree (in ${ }^{\circ} \mathrm{C}$ ) and the masses of the liquids should be determined to one-tenth of gram.
REMARK: Assume that the vapors of the liquids, to a good approximation,
(1) obey the Dalton law stating that the pressure of a mixture of gases is equal to the sum of the partial pressures of the gases forming the mixture and
(2) can be treated as perfect gases up to the pressures corresponding to the saturated vapors
Solution:
PART B
As the liquids are in thermal contact with each other, their temperatures increase in time in the same way.
At the beginning of the heating, what corresponds to the left sloped part of the diagram, no evaporation can occur. The free evaporation from the upper surface of the liquid B cannot occur - it is impossible due to the layer of the non-volatile liquid $\mathrm{C}$. The evaporation from the inside of the system is considered below.
Let us consider a bubble formed in the liquid $\mathrm{A}$ or in the liquid $\mathrm{B}$ or on the surface that separates these liquids. Such a bubble can be formed due to fluctuations or for many other reasons, which will not be analyzed here.
The bubble can get out of the system only when the pressure inside it equals to the external pressure $p_{0}$ (or when it is a little bit higher than $p_{0}$ ). Otherwise, the bubble will collapse.
The pressure inside the bubble formed in the volume of the liquid A or in the volume of the liquid B equals to the pressure of the saturated vapor of the liquid A or B, respectively. However, the pressure inside the bubble formed on the surface separating the liquids A and B is equal to the sum of the pressures of the saturated vapors of both these liquids, as then the bubble is in a contact with the liquids $\mathrm{A}$ and $\mathrm{B}$ at the same time. In the case considered the pressure inside the bubble is greater than the pressures of the saturated vapors of each of the liquids $\mathrm{A}$ and $\mathrm{B}$ (at the same temperature).
Therefore, when the system is heated, the pressure $p_{0}$ is reached first in the bubbles that were formed on the surface separating the liquids. Thus, the temperature $t_{1}$ corresponds to a kind of common boiling of both liquids that occurs in the region of their direct contact. The temperature $t_{1}$ is for sure lower than the boiling temperatures of the liquids $\mathrm{A}$ and $\mathrm{B}$ as then the pressures of the saturated vapors of the liquids $\mathrm{A}$ and $\mathrm{B}$ are less then $p_{0}$ (their sum equals to $p_{0}$ and each of them is greater than zero).
In order to determine the value of $t_{1}$ with required accuracy, we can calculate the values of the sum of the saturated vapors of the liquids A and B for several values of the temperature $t$ and look when one gets the value $p_{0}$.
From the formula given in the text of the problem, we have:
$$
\begin{aligned}
& \frac{p_{A}}{p_{0}}=e^{\frac{\alpha_{A}}{T}+\beta_{A}}, \\
& \frac{p_{B}}{p_{0}}=e^{\frac{\alpha_{B}}{T}+\beta_{B}} .
\end{aligned}
$$
$p_{A}+p_{B}$ equals to $p_{0}$ if
$$
\frac{p_{A}}{p_{0}}+\frac{p_{B}}{p_{0}}=1
$$
Thus, we have to calculate the values of the following function:
$$
y(x)=e^{\frac{\alpha_{A}}{t+t_{0}}+\beta_{A}}+e^{\frac{\alpha_{B}}{t+t_{0}}+\beta_{B}}
$$
(where $t_{0}=273.15^{\circ} \mathrm{C}$ ) and to determine the temperature $t=t_{1}$, at which $y(t)$ equals to 1 . When calculating the values of the function $y(t)$ we can divide the intervals of the temperatures $t$ by 2 (approximately) and look whether the results are greater or less than 1 .
We have:
Table 1.2
| $t$ | $y(t)$ |
| :--- | :--- |
| $40^{\circ} \mathrm{C}$ | $<1($ see Tab. 1.1$)$ |
| $77^{\circ} \mathrm{C}$ | $>1\left(\right.$ as $t_{1}$ is less than $\left.t_{b A}\right)$ |
| $59^{\circ} \mathrm{C}$ | $0.749<1$ |
| $70^{\circ} \mathrm{C}$ | $1.113>1$ |
| $66^{\circ} \mathrm{C}$ | $0.966<1$ |
| $67^{\circ} \mathrm{C}$ | $1.001>1$ |
| $66.5^{\circ} \mathrm{C}$ | $0.983<1$ |
Therefore, $t_{1} \approx 67^{\circ} \mathrm{C}$ (with required accuracy).
Now we calculate the pressures of the saturated vapors of the liquids $A$ and $B$ at the temperature $t_{1} \approx 67^{\circ} \mathrm{C}$, i.e. the pressures of the saturated vapors of the liquids $\mathrm{A}$ and $\mathrm{B}$ in each bubble formed on the surface separating the liquids. From the equations (1) and (2), we get:
$$
\begin{aligned}
& p_{A} \approx 0.734 p_{0}, \\
& p_{B} \approx 0.267 p_{0}, \\
& \left(p_{A}+p_{B}=1.001 p_{0} \approx p_{0}\right) .
\end{aligned}
$$
These pressures depend only on the temperature and, therefore, they remain constant during the motion of the bubbles through the liquid $\mathrm{B}$. The volume of the bubbles during this motion also cannot be changed without violation of the relation $p_{A}+p_{B}=p_{0}$. It follows from the above remarks that the mass ratio of the saturated vapors of the liquids A and B in each bubble is the same. This conclusion remains valid as long as both liquids are in the system. After total evaporation of one of the liquids the temperature of the system will increase again (second sloped part of the diagram). Then, however, the mass of the system remains constant until the temperature reaches the value $t_{2}$ at which the boiling of the liquid (remained in the vessel) starts. Therefore, the temperature $t_{2}$ (the higher horizontal part of the diagram) corresponds to the boiling of the liquid remained in the vessel.
The mass ratio $m_{A} / m_{B}$ of the saturated vapors of the liquids $A$ and $B$ in each bubble leaving the system at the temperature $t_{1}$ is equal to the ratio of the densities of these vapors $\rho_{A} / \rho_{B}$. According to the assumption 2, stating that the vapors can be treated as ideal gases, the last ratio equals to the ratio of the products of the pressures of the saturated vapors by the molecular masses:
$$
\frac{m_{A}}{m_{B}}=\frac{\rho_{A}}{\rho_{B}}=\frac{p_{A} \mu_{A}}{p_{B} \mu_{B}}=\frac{p_{A}}{p_{B}} \mu
$$
Thus,
$$
\frac{m_{A}}{m_{B}} \approx 22.0
$$
We see that the liquid A evaporates 22 times faster than the liquid B. The evaporation of $100 \mathrm{~g}$ of the liquid A during the "surface boiling" at the temperature $t_{1}$ is associated with the
evaporation of $100 \mathrm{~g} / 22 \approx 4.5 \mathrm{~g}$ of the liquid B. Thus, at the time $\tau_{1}$ the vessel contains 95.5 $\mathrm{g}$ of the liquid $\mathrm{B}$ (and no liquid $\mathrm{A}$ ). The temperature $t_{2}$ is equal to the boiling temperature of the liquid $\mathrm{B}: t_{2}=100^{\circ} \mathrm{C}$.
Answer: $67$ , $100$
Context Example 10:
Question:
Points $P$ and $Q$ are located inside the square $A B C D$ such that $D P$ is parallel to $Q B$ and $D P=Q B=P Q$. Determine the minimum possible value of $\angle A D P$.
<image>
Solution:
Placing the information on the coordinate axes, the diagram is indicated to the right.
We note that $P$ has coordinates $(a, b)$.
By symmetry (or congruency) we can label lengths $a$ and $b$ as shown. Thus $Q$ has coordinates $(2-a, 2-b)$.
Since $P D=P Q, a^{2}+b^{2}=(2-2 a)^{2}+(2-2 b)^{2}$
$$
\begin{aligned}
& 3 a^{2}+3 b^{2}-8 a-8 b+8=0 \\
& \left(a-\frac{4}{3}\right)^{2}+\left(b-\frac{4}{3}\right)^{2}=\frac{8}{9}
\end{aligned}
$$
<img_3483>
$P$ is on a circle with centre $O\left(\frac{4}{3}, \frac{4}{3}\right)$ with $r=\frac{2}{3} \sqrt{2}$.
The minimum angle for $\theta$ occurs when $D P$ is tangent to the circle.
So we have the diagram noted to the right.
Since $O D$ makes an angle of $45^{\circ}$ with the $x$-axis then $\angle P D O=45-\theta$ and $O D=\frac{4}{3} \sqrt{2}$.
Therefore $\sin (45-\theta)=\frac{\frac{2}{3} \sqrt{2}}{\frac{4}{3} \sqrt{2}}=\frac{1}{2}$ which means $45^{\circ}-\theta=30^{\circ}$ or $\theta=15^{\circ}$.
Thus the minimum value for $\theta$ is $15^{\circ}$.
<img_3977>
Let $A B=B C=C D=D A=1$.
Join $D$ to $B$. Let $\angle A D P=\theta$. Therefore, $\angle P D B=45-\theta$.
Let $P D=a$ and $P B=b$ and $P Q=\frac{a}{2}$.
We now establish a relationship between $a$ and $b$.
In $\triangle P D B, b^{2}=a^{2}+2-2(a)(\sqrt{2}) \cos (45-\theta)$
$$
\text { or, } \quad \cos (45-\theta)=\frac{a^{2}-b^{2}+2}{2 \sqrt{2} a}
$$
<img_3307>
In $\triangle P D R,\left(\frac{a}{2}\right)^{2}=a^{2}+\left(\frac{\sqrt{2}}{2}\right)^{2}-2 a \frac{\sqrt{2}}{2} \cos (45-\theta)$
or, $\cos (45-\theta)=\frac{\frac{3}{4} a^{2}+\frac{1}{2}}{a \sqrt{2}}$
Comparing (1) and (2) gives, $\frac{a^{2}-b^{2}+2}{2 \sqrt{2} a}=\frac{\frac{3}{4} a^{2}+\frac{1}{2}}{a \sqrt{2}}$.
Simplifying this, $a^{2}+2 b^{2}=2$
$$
\text { or, } \quad b^{2}=\frac{2-a^{2}}{2}
$$
Now $\cos (45-\theta)=\frac{a^{2}+2-\left(\frac{2-a^{2}}{2}\right)}{2 a \sqrt{2}}=\frac{1}{4 \sqrt{2}}\left(3 a+\frac{2}{a}\right)$.
Now considering $3 a+\frac{2}{a}$, we know $\left(\sqrt{3 a}-\sqrt{\frac{2}{a}}\right)^{2} \geq 0$
$$
\text { or, } \quad 3 a+\frac{2}{a} \geq 2 \sqrt{6}
$$
Thus, $\cos (45-\theta) \geq \frac{1}{4 \sqrt{2}}(2 \sqrt{6})=\frac{\sqrt{3}}{2}$
$$
\cos (45-\theta) \geq \frac{\sqrt{3}}{2}
$$
$\cos (45-\theta)$ has a minimum value for $45^{\circ}-\theta=30^{\circ}$ or $\theta=15^{\circ}$.
Join $B D$. Let $B D$ meet $P Q$ at $M$. Let $\angle A D P=\theta$.
By interior alternate angles, $\angle P=\angle Q$ and $\angle P D M=\angle Q B M$.
Thus $\triangle P D M \cong \triangle Q B M$ by A.S.A., so $P M=Q M$ and $D M=B M$.
So $M$ is the midpoint of $B D$ and the centre of the square.
Without loss of generality, let $P M=1$. Then $P D=2$.
Since $\theta+\alpha=45^{\circ}$ (see diagram), $\theta$ will be minimized when $\alpha$ is maximized.
<img_3749>
Consider $\triangle P M D$.
Using the sine law, $\frac{\sin \alpha}{1}=\frac{\sin (\angle P M D)}{2}$.
To maximize $\alpha$, we maximize $\sin \alpha$.
But $\sin \alpha=\frac{\sin (\angle P M D)}{2}$, so it is maximized when $\sin (\angle P M D)=1$.
In this case, $\sin \alpha=\frac{1}{2}$, so $\alpha=30^{\circ}$.
Therefore, $\theta=45^{\circ}-30^{\circ}=15^{\circ}$, and so the minimum value of $\theta$ is $15^{\circ}$.
We place the diagram on a coordinate grid, with $D(0,0)$, $C(1,0), B(0,1), A(1,1)$.
Let $P D=P Q=Q B=a$, and $\angle A D P=\theta$.
Drop a perpendicular from $P$ to $A D$, meeting $A D$ at $X$.
Then $P X=a \sin \theta, D X=a \cos \theta$.
Therefore the coordinates of $P$ are $(a \sin \theta, a \cos \theta)$.
Since $P D \| B Q$, then $\angle Q B C=\theta$.
So by a similar argument (or by using the fact that $P Q$ are symmetric through the centre of the square), the coordinates of $Q$ are $(1-a \sin \theta, 1+a \cos \theta)$.
<img_3706>
Now $(P Q)^{2}=a^{2}$, so $(1-2 a \sin \theta)^{2}+(1-2 a \cos \theta)^{2}=a^{2}$
$$
2+4 a^{2} \sin ^{2} \theta+4 a^{2} \cos ^{2} \theta-4 a(\sin \theta+\cos \theta)=a^{2}
$$
$$
\begin{aligned}
2+4 a^{2}-a^{2} & =4 a(\sin \theta+\cos \theta) \\
\frac{2+3 a^{2}}{4 a} & =\sin \theta+\cos \theta \\
\frac{2+3 a^{2}}{4 \sqrt{2} a} & =\frac{1}{\sqrt{2}} \sin \theta+\frac{1}{\sqrt{2}} \cos \theta=\cos \left(45^{\circ}\right) \sin \theta+\sin \left(45^{\circ}\right) \cos \theta \\
\frac{2+3 a^{2}}{4 \sqrt{2} a} & =\sin \left(\theta+45^{\circ}\right)
\end{aligned}
$$
$$
\text { Now } \begin{aligned}
\left(a-\sqrt{\frac{2}{3}}\right)^{2} & \geq 0 \\
a^{2}-2 a \sqrt{\frac{2}{3}}+\frac{2}{3} & \geq 0 \\
3 a^{2}-2 a \sqrt{6}+2 & \geq 0 \\
3 a^{2}+2 & \geq 2 a \sqrt{6} \\
\frac{3 a^{2}+2}{4 \sqrt{2} a} & \geq \frac{\sqrt{3}}{2}
\end{aligned}
$$
and equality occurs when $a=\sqrt{\frac{2}{3}}$.
So $\sin \left(\theta+45^{\circ}\right) \geq \frac{\sqrt{3}}{2}$ and thus since $0^{\circ} \leq \theta \leq 90^{\circ}$, then $\theta+45^{\circ} \geq 60^{\circ}$ or $\theta \geq 15^{\circ}$.
Therefore the minimum possible value of $\angle A D P$ is $15^{\circ}$.
Answer: $15^{\circ}$
Current Question:
<image_1>
In certain lakes there is a strange phenomenon called "seiching" which is an oscillation of the water. Lakes in which you can see this phenomenon are normally long compared with the depth and also narrow. It is natural to see waves in a lake but not something like the seiching, where the entire water volume oscillates, like the coffee in a cup that you carry to a waiting guest.
In order to create a model of the seiching we look at water in a rectangular container. The length of the container is $L$ and the depth of the water is $h$. Assume that the surface of the water to begin with makes a small angle with the horizontal. The seiching will then start, and we assume that the water surface continues to be plane but oscillates around an axis in the horizontal plane and located in the middle of the container.
Create a model of the movement of the water and derive a formula for the oscillation period $T$. The starting conditions are given in figure above.
Assume that $\xi<<h$. The table below shows experimental oscillation periods for different water depths in two containers of different lengths. Check in some reasonable way how well the formula that you have derived agrees with the experimental data. Give your opinion on the quality of your model.
Table 1. $L=479 \mathrm{~mm}$
| $h / m m$ | 30 | 50 | 69 | 88 | 107 | 124 | 142 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $T / s$ | 1.78 | 1.40 | 1.18 | 1.08 | 1.00 | 0.91 | 0.82 |
Table 2. $L=143 \mathrm{~mm}$
| $h / m m$ | 31 | 38 | 58 | 67 | 124 |
| :---: | :---: | :---: | :---: | :---: | ---: |
| $T / s$ | 0.52 | 0.52 | 0.43 | 0.35 | 0.28 |
The graph below shows results from measurements in lake Vättern in Sweden. This lake has a length of $123 \mathrm{~km}$ and a mean depth of $50 \mathrm{~m}$. What is the time scale in the graph?
<image_2>
The water surface level in Bastudalen (northern end of lake Vättern) and Jönköping (southern end). | [
"$\\frac{\\pi L}{\\sqrt{3 h}}$"
] | [
"In the coordinate system of the figure, we have for the centre of mass coordinates of the two triangular parts of the water\n\n$$\n\\left(x_{1}, y_{1}\\right)=(L / 3, h / 2+\\xi / 3) \\quad\\left(x_{2}, y_{2}\\right)=(-L / 3, h / 2-\\xi / 3) .\n$$\n\nFor the entire water mass the centre of mass coordinates will then be\n\n$$\n\\left(x_{C O M}, y_{C O M}\\right)=\\left(\\frac{\\xi L}{6 h}, \\frac{\\xi^{2}}{6 h}\\right)\n$$\n\nDue to that the $y$ component is quadratic in $\\xi$ will be much much smaller than the $x$ component.\n\nThe velocities of the water mass are\n\n$$\n\\left(v_{x}, v_{y}\\right)=\\left(\\frac{g_{L}}{6 h}, \\frac{g_{\\xi}}{3 h}\\right)\n$$\n\nand again the vertical component is much smaller the the horizontal one.\n\nWe now in our model neglect the vertical components. The total energy (kinetic + potential) will then be\n\n$$\nW=W_{K}+W_{P}=\\frac{1}{2} M \\frac{\\xi^{2} L^{2}}{36 h^{2}}+M g \\frac{\\xi^{2}}{6 h^{2}}\n$$\n\nFor a harmonic oscillator we have\n\n$$\nW=W_{K}+W_{P}=\\frac{1}{2} m x^{2}+\\frac{1}{2} m \\omega^{2} x^{2}\n$$\n\nIdentifying gives\n\n$$\n\\omega=\\sqrt{\\frac{12 g h}{L}} \\text { or } T_{\\text {model }}=\\frac{\\pi L}{\\sqrt{3 h}} \\text {. }\n$$\n\nComparing with the experimental data we find $T_{\\text {experiment }} \\approx 1.1 \\cdot T_{\\text {model }}$ our model gives a slight underestimation of the oscillation period.\n\nApplying our corrected model on the Vättern data we have that the oscillation period of the seiching is about 3 hours."
] | false | null | Expression | null | OE_MM_physics_en_COMP | Optics | ||
1082 | Context Example 1:
Question:
The parabola $y=-x^{2}+4$ has vertex $P$ and intersects the $x$-axis at $A$ and $B$. The parabola is translated from its original position so that its vertex moves along the line $y=x+4$ to the point $Q$. In this position, the parabola intersects the $x$-axis at $B$ and $C$. Determine the coordinates of $C$.
<image>
Solution:
The parabola $y=-x^{2}+4$ has vertex $P(0,4)$ and intersects the $x$-axis at $A(-2,0)$ and $B(2,0)$. The intercept $B(2,0)$ has its pre-image, $B^{\prime}$ on the parabola $y=-x^{2}+4$. To find $B^{\prime}$, we find the point of intersection of the line passing through $B(2,0)$, with slope 1 , and the parabola $y=-x^{2}+4$.
The equation of the line is $y=x-2$.
Intersection points, $x-2=-x^{2}+4$
$$
\begin{array}{r}
x^{2}+x-6=0 \\
(x+3)(x-2)=0 .
\end{array}
$$
Therefore, $x=-3$ or $x=2$.
For $x=-3, y=-3-2=-5$. Thus $B^{\prime}$ has coordinates $(-3,-5)$.
If $(-3,-5) \rightarrow(2,0)$ then the required general translation mapping $y=-x^{2}+4$ onto the parabola with vertex $Q$ is $(x, y) \rightarrow(x+5, y+5)$.
Possibility 1
Using the general translation, we find the coordinates of $Q$ to be, $P(0,4) \rightarrow Q(0+5,4+5)=Q(5,9)$.
If $C$ is the reflection of $B$ in the axis of symmetry of the parabola, i.e. $x=5, C$ has coordinates $(8,0)$.
Possibility 2
If $B^{\prime}$ has coordinates $(-3,-5)$ then $C^{\prime}$ is the reflection of $B^{\prime}$ in the $y$-axis. Thus $C^{\prime}$ has coordinates $(3,-5)$.
If we apply the general translation then $C$ has coordinates $(3+5,-5+5)$ or $(8,0)$.
Thus $C$ has coordinates $(8,0)$.
Possibility 3
Using the general translation, we find the coordinates of $Q$ to be, $P(0,4) \rightarrow Q(0+5,4+5)=Q(5,9)$.
The equation of the image parabola is $y=-(x-5)^{2}+9$.
To find its intercepts, $-(x-5)^{2}+9=0$
$$
\begin{aligned}
(x-5)^{2} & =9 \\
x-5 & = \pm 3 .
\end{aligned}
$$
Therefore $x=8$ or $x=2$.
Thus $C$ has coordinates $(8,0)$.
The translation moving the parabola with equation $y=-x^{2}+4$ onto the parabola with vertex $Q$ is $T(t, t)$ because the slope of the line $y=x+4$ is 1 .
The pre-image of $B^{\prime}$ is $(2-t,-t)$.
Since $B^{\prime}$ is on the parabola with vertex $P$, we have
$$
\begin{aligned}
-t & =-(2-t)^{2}+4 \\
-t & =-4+4 t-t^{2}+4 \\
t^{2}-5 t & =0 \\
t(t-5) & =0
\end{aligned}
$$
Therefore, $t=0$ or $t=5$.
Thus $B^{\prime}$ is $(-3,-5)$.
Let $C$ have coordinates $(c, 0)$.
The pre-image of $C$ is $(c-5,-5)$.
Therefore, $-5=-(c-5)^{2}+4$.
Or, $(c-5)^{2}=9$.
Therefore $c-5=3$ or $c-5=-3$.
$$
c=8 \text { or } \quad c=2
$$
Thus $C$ has coordinates $(8,0)$.
Answer: $(8,0)$
Context Example 2:
Question:
In rectangle $A B C D$, point $E$ is on side $D C$. Line segments $A E$ and $B D$ are perpendicular and intersect at $F$. If $A F=4$ and $D F=2$, determine the area of quadrilateral $B C E F$.
<image>
Solution:
Since $\triangle A F D$ is right-angled at $F$, then by the Pythagorean Theorem,
$$
A D=\sqrt{A F^{2}+F D^{2}}=\sqrt{4^{2}+2^{2}}=\sqrt{20}=2 \sqrt{5}
$$
since $A D>0$.
Let $\angle F A D=\beta$.
Since $A B C D$ is a rectangle, then $\angle B A F=90^{\circ}-\beta$.
Since $\triangle A F D$ is right-angled at $F$, then $\angle A D F=90^{\circ}-\beta$.
Since $A B C D$ is a rectangle, then $\angle B D C=90^{\circ}-\left(90^{\circ}-\beta\right)=\beta$.
<img_3979>
Therefore, $\triangle B F A, \triangle A F D$, and $\triangle D F E$ are all similar as each is right-angled and has either an angle of $\beta$ or an angle of $90^{\circ}-\beta$ (and hence both of these angles).
Therefore, $\frac{A B}{A F}=\frac{D A}{D F}$ and so $A B=\frac{4(2 \sqrt{5})}{2}=4 \sqrt{5}$.
Also, $\frac{F E}{F D}=\frac{F D}{F A}$ and so $F E=\frac{2(2)}{4}=1$.
Since $A B C D$ is a rectangle, then $B C=A D=2 \sqrt{5}$, and $D C=A B=4 \sqrt{5}$.
Finally, the area of quadrilateral $B C E F$ equals the area of $\triangle D C B$ minus the area $\triangle D F E$. Thus, the required area is
$$
\frac{1}{2}(D C)(C B)-\frac{1}{2}(D F)(F E)=\frac{1}{2}(4 \sqrt{5})(2 \sqrt{5})-\frac{1}{2}(2)(1)=20-1=19
$$
Since $\triangle A F D$ is right-angled at $F$, then by the Pythagorean Theorem,
$$
A D=\sqrt{A F^{2}+F D^{2}}=\sqrt{4^{2}+2^{2}}=\sqrt{20}=2 \sqrt{5}
$$
since $A D>0$.
Let $\angle F A D=\beta$.
Since $A B C D$ is a rectangle, then $\angle B A F=90^{\circ}-\beta$. Since $\triangle B A F$ is right-angled at $F$, then $\angle A B F=\beta$.
Since $\triangle A F D$ is right-angled at $F$, then $\angle A D F=90^{\circ}-\beta$.
Since $A B C D$ is a rectangle, then $\angle B D C=90^{\circ}-\left(90^{\circ}-\beta\right)=\beta$.
<img_3412>
Looking at $\triangle A F D$, we see that $\sin \beta=\frac{F D}{A D}=\frac{2}{2 \sqrt{5}}=\frac{1}{\sqrt{5}}, \cos \beta=\frac{A F}{A D}=\frac{4}{2 \sqrt{5}}=\frac{2}{\sqrt{5}}$, and $\tan \beta=\frac{F D}{A F}=\frac{2}{4}=\frac{1}{2}$.
Since $A F=4$ and $\angle A B F=\beta$, then $A B=\frac{A F}{\sin \beta}=\frac{4}{\frac{1}{\sqrt{5}}}=4 \sqrt{5}$.
Since $F D=2$ and $\angle F D E=\beta$, then $F E=F D \tan \beta=2 \cdot \frac{1}{2}=1$.
Since $A B C D$ is a rectangle, then $B C=A D=2 \sqrt{5}$, and $D C=A B=4 \sqrt{5}$.
Finally, the area of quadrilateral $E F B C$ equals the area of $\triangle D C B$ minus the area $\triangle D F E$. Thus, the required area is
$$
\frac{1}{2}(D C)(C B)-\frac{1}{2}(D F)(F E)=\frac{1}{2}(4 \sqrt{5})(2 \sqrt{5})-\frac{1}{2}(2)(1)=20-1=19
$$
Answer: 19
Context Example 3:
Question:
Square $A B C D$ has side length 22. Points $G$ and $H$ lie on $\overline{A B}$ so that $A H=B G=5$. Points $E$ and $F$ lie outside square $A B C D$ so that $E F G H$ is a square. Compute the area of hexagon $A E F B C D$.
<image>
Solution:
Note that $G H=A B-A H-B G=22-5-5=12$. Thus
$$
\begin{aligned}
{[A E F B C D] } & =[A B C D]+[E F G H]+[A E H]+[B F G] \\
& =22^{2}+12^{2}+\frac{1}{2} \cdot 5 \cdot 12+\frac{1}{2} \cdot 5 \cdot 12 \\
& =484+144+30+30 \\
& =\mathbf{6 8 8} .
\end{aligned}
$$
Answer: 688
Context Example 4:
Question:
A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.
This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.
Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.
<image>
Circle $A$ has curvature 2; circle $B$ has curvature 1 .
<image>
Circle $A$ has curvature 2; circle $B$ has curvature -1 .
Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then
$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$
where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Determine the total number of plots sold up to and including day $n$.
Solution:
The total number of plots sold up to and including day $n$ is
$$
\begin{aligned}
2+\sum_{k=1}^{n} X_{k} & =2+2 \sum_{k=1}^{n} 3^{k-1} \\
& =2+2 \cdot\left(1+3+3^{2}+\ldots+3^{n-1}\right) \\
& =3^{n}+1
\end{aligned}
$$
Alternatively, proceed by induction: on day 0 , there are $2=3^{0}+1$ plots sold, and for $n \geq 0$,
$$
\begin{aligned}
\left(3^{n}+1\right)+X_{n+1} & =\left(3^{n}+1\right)+2 \cdot 3^{n} \\
& =3 \cdot 3^{n}+1 \\
& =3^{n+1}+1 .
\end{aligned}
$$
Answer: $3^{n}+1$
Context Example 5:
Question:
Large Hadron Collider
Please read the general instructions in the separate envelope before you start this problem.
In this task, the physics of the particle accelerator LHC (Large Hadron Collider) at CERN is discussed. CERN is the world's largest particle physics laboratory. Its main goal is to get insight into the fundamental laws of nature. Two beams of particles are accelerated to high energies, guided around the accelerator ring by a strong magnetic field and then made to collide with each other. The protons are not spread uniformly around the circumference of the accelerator, but they are clustered in so-called bunches. The resulting particles generated by collisions are observed with large detectors. Some parameters of the LHC can be found in table 1 .
| LHC ring | |
| :--- | :--- |
| Circumference of ring | $26659 \mathrm{~m}$ |
| Number of bunches per proton beam | 2808 |
| Number of protons per bunch | $1.15 \times 10^{11}$ |
| Proton beams | |
| Energy of protons | $7.00 \mathrm{TeV}$ |
| Centre of mass energy | $14.0 \mathrm{TeV}$ |
Table 1: Typical numerical values of relevant LHC parameters.
Particle physicists use convenient units for the energy, momentum and mass: The energy is measured in electron volts $[\mathrm{eV}]$. By definition, $1 \mathrm{eV}$ is the amount of energy gained by a particle with elementary charge, e, moved through a potential difference of one volt $\left(1 \mathrm{eV}=1.602 \cdot 10^{-19} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2}\right)$.
The momentum is measured in units of $\mathrm{eV} / c$ and the mass in units of $\mathrm{eV} / c^{2}$, where $c$ is the speed of light in vacuum. Since $1 \mathrm{eV}$ is a very small quantity of energy, particle physicists often use $\mathrm{MeV}\left(1 \mathrm{MeV}=10^{6} \mathrm{eV}\right)$, $\mathrm{GeV}\left(1 \mathrm{GeV}=10^{9} \mathrm{eV}\right)$ or $\mathrm{TeV}\left(1 \mathrm{TeV}=10^{12} \mathrm{eV}\right)$.
Part A deals with the acceleration of protons or electrons. Part B is concerned with the identification of particles produced in the collisions at CERN.
Part A.LHC accelerator
Acceleration:
Assume that the protons have been accelerated by a voltage $V$ such that their velocity is very close to the speed of light and neglect any energy loss due to radiation or collisions with other particles.
Context question:
A.1 Find the exact expression for the final velocity $v$ of the protons as a function of the accelerating voltage $V$, and physical constants.
Context answer:
\boxed{$v=c \cdot \sqrt{1-(\frac{m_{p} \cdot c^{2}}{m_{p} \cdot c^{2}+V \cdot e})^{2}}$}
Extra Supplementary Reading Materials:
A design for a future experiment at CERN plans to use the protons from the LHC and to collide them with electrons which have an energy of $60.0 \mathrm{GeV}$.
Context question:
A.2 For particles with high energy and low mass the relative deviation $\Delta=(c-v) / c$ of the final velocity $v$ from the speed of light is very small. Find a first order approximation for $\Delta$ and calculate $\Delta$ for electrons with an energy of $60.0 \mathrm{GeVusing}$ the accelerating voltage $V$ and physical constants.
Context answer:
\boxed{$\frac{1}{2}\left(\frac{m_{e} \cdot c^{2}}{m_{e} \cdot c^{2}+V \cdot e}\right)^{2}$, $\Delta=3.63 \cdot 10^{-11}$}
Extra Supplementary Reading Materials:
We now return to the protons in the LHC. Assume that the beam pipe has a circular shape.
Context question:
A.3 Derive an expression for the uniform magnetic flux density $B$ necessary to keep the proton beam on a circular track. The expression should only contain the energy of the protons $E$, the circumference $L$, fundamental constants and numbers. You may use suitable approximations if their effect is smaller than precision given by the least number of significant digits.
Calculate the magnetic flux density $B$ for a proton energy of $E=7.00 \mathrm{TeV}$, neglecting interactions between the protons.
Context answer:
\boxed{$B=\frac{2 \pi \cdot E}{e \cdot c \cdot L}$ , $B=5.50$}
Extra Supplementary Reading Materials:
Radiated Power:
An accelerated charged particle radiates energy in the form of electromagnetic waves. The radiated power $P_{\mathrm{rad}}$ of a charged particle that circulates with a constant angular velocity depends only on its acceleration $a$, its charge $q$, the speed of light $c$ and the permittivity of free space $\varepsilon_{0}$.
Context question:
A.4 Use dimensional analysis to find an expression for the radiated power $P_{\text {rad }}$.
Context answer:
\boxed{$P_{\text {rad }}=a^{\alpha} \cdot q^{\beta} \cdot c^{\gamma} \cdot \epsilon_{0}^{\delta}$}
Extra Supplementary Reading Materials:
The real formula for the radiated power contains a factor $1 /(6 \pi)$; moreover, a full relativistic derivation gives an additional multiplicative factor $\gamma^{4}$, with $\gamma=\left(1-v^{2} / c^{2}\right)^{-\frac{1}{2}}$.
Context question:
A.5 Calculate $P_{\text {tot }}$, the total radiated power of the LHC, for a proton energy of E= 7.00 $\mathrm{TeV}$ (Note table 1). You may use suitable approximations.
Context answer:
\boxed{5.13}
Extra Supplementary Reading Materials:
Linear Acceleration:
At CERN, protons at rest are accelerated by a linear accelerator of length $d=30.0 \mathrm{~m}$ through a potential difference of $V=500 \mathrm{MV}$. Assume that the electrical field is homogeneous. A linear accelerator consists of two plates as sketched in Figure 1.
A.6 Determine the time T that the protons take to pass through this field
<image>
Figure 1: Sketch of an accelerator module.
Solution:
2nd Newton's law
$$
\begin{gathered}
F=\frac{d p}{d t} \quad \text { leads to } \\
\frac{V \cdot e}{d}=\frac{p_{f}-p_{i}}{T} \text { with } p_{i}=0
\end{gathered}
$$
Conservation of energy:
$$
E_{t o t}=m \cdot c^{2}+e \cdot V
$$
Since
$$
\begin{gathered}
E_{t o t}^{2}=\left(m \cdot c^{2}\right)^{2}+\left(p_{f} \cdot c\right)^{2} \\
\rightarrow p_{f}=\frac{1}{c} \cdot \sqrt{\left(m \cdot c^{2}+e \cdot V\right)^{2}-\left(m \cdot c^{2}\right)^{2}}=\sqrt{2 e \cdot m \cdot V+\left(\frac{e \cdot V}{c}\right)^{2}} \\
\rightarrow T=\frac{d \cdot p_{f}}{V \cdot e}=\frac{d}{V \cdot e} \sqrt{2 e \cdot m_{p} \cdot V+\left(\frac{e \cdot V}{c}\right)^{2}} \\
T=218 \mathrm{~ns}
\end{gathered}
$$
2nd Newton's law
$$
\frac{V \cdot e}{d}=\frac{p_{f}-p_{i}}{T} \text { with } p_{i}=0
$$
velocity from A1 or from conservation of energy
$$
v=c \cdot \sqrt{1-\left(\frac{m_{p} \cdot c^{2}}{m_{p} \cdot c^{2}+V \cdot e}\right)^{2}}
$$
and hence for $\gamma$
$$
\begin{gathered}
\gamma=1 / \sqrt{1-\frac{v^{2}}{c^{2}}}=1+\frac{e \cdot V}{m_{p} \cdot c^{2}} \\
\rightarrow p_{f}=\gamma \cdot m_{p} \cdot v=\left(1+\frac{e \cdot V}{m_{p} \cdot c^{2}}\right) \cdot m_{p} \cdot c \cdot \sqrt{1-\left(\frac{m_{p} \cdot c^{2}}{m_{p} \cdot c^{2}+V \cdot e}\right)^{2}} \\
\rightarrow T=\frac{d \cdot p_{f}}{V \cdot e}=\frac{d \cdot m_{p} \cdot c}{V \cdot e} \cdot \sqrt{\left(\frac{m_{p} \cdot c^{2}+e \cdot V}{m_{p} \cdot c^{2}}\right)^{2}-1}=\frac{d}{V \cdot e} \sqrt{2 e \cdot m_{p} \cdot V+\left(\frac{e \cdot V}{c}\right)^{2}} \\
T=218 \mathrm{~ns}
\end{gathered}
$$
Energy increases linearly with distance $\mathrm{x}$
$$
\begin{gathered}
E(x)=\frac{e \cdot V \cdot x}{d} \\
t=\int d t=\int_{0}^{d} \frac{d x}{v(x)} \\
v(x)=c \cdot \sqrt{1-\left(\frac{m_{p} \cdot c^{2}}{m_{p} \cdot c^{2}+\frac{e \cdot V \cdot x}{d}}\right)^{2}}=c \cdot \frac{\sqrt{\left(m_{p} \cdot c^{2}+\frac{e \cdot V \cdot x}{d}\right)^{2}-\left(m_{p} \cdot c^{2}\right)^{2}}}{m_{p} \cdot c^{2}+\frac{e \cdot V \cdot x}{d}} \\
=c \cdot \frac{\sqrt{\left(1+\frac{e \cdot V \cdot x}{d \cdot m_{p} \cdot c^{2}}\right)^{2}-1}}{1+\frac{e \cdot V \cdot x}{d \cdot m_{p} \cdot c^{2}}}
\end{gathered}
$$
Substitution : $\xi=\frac{e \cdot V \cdot x}{d \cdot m_{p} \cdot c^{2}} \quad \frac{d \xi}{d x}=\frac{e \cdot V}{d \cdot m_{p} \cdot c^{2}}$
$$
\rightarrow t=\frac{1}{c} \int_{0}^{b} \frac{1+\xi}{\sqrt{(1+\xi)^{2}-1}} \frac{d \cdot m_{p} \cdot c^{2}}{e \cdot V} d \xi \quad b=\frac{e \cdot V}{m_{p} \cdot c^{2}}
$$
$$
1+\xi:=\cosh (s) \quad \frac{d \xi}{d s}=\sinh (s)
$$
0.1
$$
t=\frac{m_{p} \cdot c \cdot d}{e \cdot V} \int \frac{\cosh (s) \cdot \sinh (s) d s}{\sqrt{\cosh ^{2}(s)-1}}=\frac{m_{p} \cdot c \cdot d}{e \cdot V}[\sinh (s)]_{b_{1}}^{b_{2}}
$$
$$
\text { with } \quad b_{1}=\cosh ^{-1}(1), \quad b_{2}=\cosh ^{-1}\left(1+\frac{e \cdot V}{m_{p} \cdot c^{2}}\right)
$$
$$
T=218 \mathrm{~ns}
$$
Alternative: differential equation
$$
\begin{gathered}
\rightarrow \frac{V \cdot e}{d}=\frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{m \cdot v}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\right)=\frac{m \cdot a\left(1-\frac{v^{2}}{c^{2}}\right)+m \cdot a \frac{v^{2}}{c^{2}}}{\left(1-\frac{v^{2}}{c^{2}}\right)^{\frac{3}{2}}}=\gamma^{3} \cdot m \cdot a \\
a=\ddot{s}=\frac{V \cdot e}{d \cdot m}\left(1-\frac{\dot{s}^{2}}{c^{2}}\right)^{\frac{3}{2}}
\end{gathered}
$$
$$
\begin{gathered}
\rightarrow s(t)=\frac{c}{V \cdot e}\left(\sqrt{e^{2} \cdot V^{2} \cdot t^{2}+c^{2} \cdot m^{2} \cdot d^{2}}-c \cdot m \cdot d\right) \\
s=d \rightarrow T=\frac{d}{V \cdot e} \sqrt{\left(\frac{V \cdot e}{c}\right)^{2}+2 V \cdot e \cdot m} \\
T=218 \mathrm{~ns}
\end{gathered}
$$
Answer: 218
Context Example 6:
Question:
An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:
$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$
Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write
$$
S_{3}[263415]=(132,312,231,213)
$$
More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.
In this power question, you will be asked to analyze some of the properties of labels and signatures.
We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :
<image>
A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:
<image>
Determine the number of 9-labels with 2-signature
$$
(12,21,12,21,12,21,12,21) \text {. }
$$
Justify your answer.
Solution:
The answer is 7936. The shape of this 2-signature has four peaks and three intermediate valleys:
<img_3473>
We will solve this problem by building up from smaller examples. Let $f_{n}$ equal the number of $(2 n+1)$-labels whose 2 -signature consists of $n$ peaks and $n-1$ intermediate valleys. In part (b) we showed that $f_{2}=16$. In the case where we have one peak, $f_{1}=2$. For the trivial case (no peaks), we get $f_{0}=1$. These cases are shown below.
1
<img_3650>
<img_3227>
Suppose we know the peak on which the largest number, $2 n+1$, is placed. Then that splits our picture into two shapes with fewer peaks. Once we choose which numbers from $1,2, \ldots, 2 n$ to place each shape, we can compute the number of arrangements of the numbers on each shape, and then take the product. For example, if we place the 9 at the second peak, as shown below, we get a 1-peak shape on the left and a 2-peak shape on the right.
<img_4030>
For the above shape, there are $\left(\begin{array}{l}8 \\ 3\end{array}\right)$ ways to pick the three numbers to place on the left-hand side, $f_{1}=2$ ways to place them, and $f_{2}=16$ ways to place the remaining five numbers on the right.
This argument works for any $n>1$, so we have shown the following:
$$
f_{n}=\sum_{k=1}^{n}\left(\begin{array}{c}
2 n \\
2 k-1
\end{array}\right) f_{k-1} f_{n-k}
$$
So we have:
$$
\begin{aligned}
& f_{1}=\left(\begin{array}{l}
2 \\
1
\end{array}\right) f_{0}^{2}=2 \\
& f_{2}=\left(\begin{array}{l}
4 \\
1
\end{array}\right) f_{0} f_{1}+\left(\begin{array}{l}
4 \\
3
\end{array}\right) f_{1} f_{0}=16 \\
& f_{3}=\left(\begin{array}{l}
6 \\
1
\end{array}\right) f_{0} f_{2}+\left(\begin{array}{l}
6 \\
3
\end{array}\right) f_{1}^{2}+\left(\begin{array}{l}
6 \\
5
\end{array}\right) f_{2} f_{0}=272 \\
& f_{4}=\left(\begin{array}{l}
8 \\
1
\end{array}\right) f_{0} f_{3}+\left(\begin{array}{l}
8 \\
3
\end{array}\right) f_{1} f_{2}+\left(\begin{array}{l}
8 \\
5
\end{array}\right) f_{2} f_{1}+\left(\begin{array}{l}
8 \\
7
\end{array}\right) f_{3} f_{0}=7936 .
\end{aligned}
$$
Answer: 7936
Context Example 7:
Question:
In the diagram, $\angle A C B=\angle A D E=90^{\circ}$. If $A B=75, B C=21, A D=20$, and $C E=47$, determine the exact length of $B D$.
<image>
Solution:
We use the cosine law in $\triangle A B D$ to determine the length of $B D$ :
$$
B D^{2}=A B^{2}+A D^{2}-2(A B)(A D) \cos (\angle B A D)
$$
We are given that $A B=75$ and $A D=20$, so we need to determine $\cos (\angle B A D)$.
Now
$$
\begin{aligned}
\cos (\angle B A D) & =\cos (\angle B A C+\angle E A D) \\
& =\cos (\angle B A C) \cos (\angle E A D)-\sin (\angle B A C) \sin (\angle E A D) \\
& =\frac{A C}{A B} \frac{A D}{A E}-\frac{B C}{A B} \frac{E D}{A E}
\end{aligned}
$$
since $\triangle A B C$ and $\triangle A D E$ are right-angled.
Since $A B=75$ and $B C=21$, then by the Pythagorean Theorem,
$$
A C=\sqrt{A B^{2}-B C^{2}}=\sqrt{75^{2}-21^{2}}=\sqrt{5625-441}=\sqrt{5184}=72
$$
since $A C>0$.
Since $A C=72$ and $C E=47$, then $A E=A C-C E=25$.
Since $A E=25$ and $A D=20$, then by the Pythagorean Theorem,
$$
E D=\sqrt{A E^{2}-A D^{2}}=\sqrt{25^{2}-20^{2}}=\sqrt{625-400}=\sqrt{225}=15
$$
since $E D>0$.
Therefore,
$$
\cos (\angle B A D)=\frac{A C}{A B} \frac{A D}{A E}-\frac{B C}{A B} \frac{E D}{A E}=\frac{72}{75} \frac{20}{25}-\frac{21}{75} \frac{15}{25}=\frac{1440-315}{75(25)}=\frac{1125}{75(25)}=\frac{45}{75}=\frac{3}{5}
$$
Finally,
$$
\begin{aligned}
B D^{2} & =A B^{2}+A D^{2}-2(A B)(A D) \cos (\angle B A D) \\
& =75^{2}+20^{2}-2(75)(20)\left(\frac{3}{5}\right) \\
& =5625+400-1800 \\
& =4225
\end{aligned}
$$
Since $B D>0$, then $B D=\sqrt{4225}=65$, as required.
Answer: 65
Context Example 8:
Question:
- 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,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$
Consider a long uniform conducting cylinder. First, we divide the cylinder into thirds and remove the middle third. Then, we perform the same steps on the remaining two cylinders. Again, we perform the same steps on the remaining four cylinders and continuing until there are 2048 cylinders.
We then connect the terminals of the cylinder to a battery and measure the effective capacitance to be $C_{1}$. If we continue to remove cylinders, the capacitance will reach an asymptotic value of $C_{0}$. What is $C_{1} / C_{0}$ ?
You may assume each cylindrical disk to be wide enough to be considered as an infinite plate, such that the radius $R$ of the cylinders is much larger than the $d$ between any successive cylinders.
<image>
Note: The diagram is not to scale.
Solution:
The capacitance is proportional to $C \propto \frac{1}{d}$, where $d$ is the distance between successive parallel plates. When we add capacitor plates in series, their effective capacitance will be:
$$
C \propto\left(\frac{1}{1 / d_{1}}+\frac{1}{1 / d_{2}}+\cdots\right)^{-1}=\frac{1}{d_{1}+d_{2}+\cdots} \Longrightarrow C \propto \frac{1}{d_{\text {total }}}
$$
Therefore, this essentially becomes a math problem: What is the total length of the spacing in between? Between successive 'cuts', the length of each cylinder is cut down by $1 / 3$, but the number of gaps double. Therefore, the spacing grows by a factor of $2 / 3$ each time. For $n=2^{1}$, the spacing starts off as $1 / 3$. For $n=2^{10}$, the spacing becomes:
$$
\frac{1}{C_{\text {eff }}} \propto d=\frac{1}{3}\left(\frac{1-(2 / 3)^{10}}{1-2 / 3}\right) L=0.983 L
$$
for $n \rightarrow \infty$, it is clear the total spacing will converge to $L$. Therefore:
$$
C_{1} / C_{0}=1.017
$$
Answer: 1.017
Context Example 9:
Question:
Leibniz's Harmonic Triangle: Consider the triangle formed by the rule
$$
\begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases}
$$
This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below.
<image>
For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$.
Compute $\sum_{n=1}^{2011} \operatorname{Le}(n, 1)$.
Solution:
Because $\operatorname{Le}(n, 1)=\frac{1}{n}-\frac{1}{n+1}$,
$$
\begin{aligned}
\sum_{i=1}^{2011} \operatorname{Le}(i, 1) & =\sum_{i=1}^{2011}\left(\frac{1}{n}-\frac{1}{n+1}\right) \\
& =\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{2010}-\frac{1}{2011}\right)+\left(\frac{1}{2011}-\frac{1}{2012}\right) \\
& =1-\frac{1}{2012} \\
& =\frac{2011}{2012} .
\end{aligned}
$$
Answer: $\frac{2011}{2012}$
Context Example 10:
Question:
- 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,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$
An engineer has access to a tetrahedron building block with side length $\ell=10 \mathrm{~cm}$. The body is made of a thermal insulator but the edges are wrapped with a thin copper wiring with cross sectional area $S=2 \mathrm{~cm}^{2}$. The thermal conductivity of copper is $385.0 \mathrm{~W} /(\mathrm{m} \mathrm{K})$. He stacks these tetrahedrons (all facing the same direction) to form a large lattice such that the copper wires are all in contact. In the diagram, only the front row of a small section is coloured. Assume that the lattice formed is infinitely large.
At some location in the tetrahedral building block, the temperature difference between two adjacent points is $1^{\circ} \mathrm{C}$. What is the heat flow across these two points? Answer in Watts.
Note: Two adjacent points refer to two adjacent points on the tetrahedron.
<image>
Solution:
There are many ways to solve this problem. We first identify that this is exactly the same as an infinite lattice resistor problem. To solve these, we can imagine injecting a current at a node and seeing how this current spreads out. However, a faster approach is by applying Foster's Theorem on this lattice.
The resistance of a single wire is:
$$
R=\frac{\ell}{k S}=1.299 \mathrm{~W} / \mathrm{K}
$$
Foster's theorem tells us that
$$
E R=V-1
$$
where $V$ is number of vertices and $E$ is edges. Taking the limit as $E, V \rightarrow \infty$, we get: $E=6 \mathrm{~V}$ (since each vertex is connected to 12 edges, but each edge is shared by two vertices). Therefore:
$$
R_{\mathrm{eff}}=\frac{1}{6} R=0.2165 \mathrm{~W} / \mathrm{K}
$$
From Fourier's Law, we have:
$$
\dot{Q}=\frac{\Delta T}{R_{\mathrm{eff}}}=4.62 \mathrm{~W}
$$
Answer: 4.62
Current Question:
C.2 Recall the problem from part A where the particle traverses from $\mathrm{O}$ to $\mathrm{P}$ (see Fig. 4). Let an opaque partition be placed at the boundary $\mathrm{AB}$ between the two regions. There is a small opening $\mathrm{CD}$ of width $d$ in $\mathrm{AB}$ such that $d \ll\left(x_{0}-x_{1}\right)$ and $d \ll x_{1}$.
Consider two extreme paths OCP and ODP such that OCP lies on the classical trajectory discussed in part A. Obtain the phase difference $\Delta \phi_{C D}$ between the two paths to first order.
<image_1>
Figure 4 | [
"$\\Delta \\phi_{C D}=0$"
] | [
"<img_4442>\n\nConsider the extreme trajectories $O C P$ and $O D P$ of $(\\mathrm{C} 1)$\n\nThe geometrical path difference is $E D$ in region I and $C F$ in region II.\n\nThis implies (note: $d \\ll\\left(x_{0}-x_{1}\\right)$ and $d \\ll x_{1}$ )\n\n$$\n\\begin{gathered}\n\\Delta \\phi_{C D}=\\frac{2 \\pi d \\sin \\theta_{1}}{\\lambda_{1}}-\\frac{2 \\pi d \\sin \\theta_{2}}{\\lambda_{2}} \\\\\n\\Delta \\phi_{C D}=\\frac{2 \\pi m v_{1} d \\sin \\theta_{1}}{h}-\\frac{2 \\pi m v_{2} d \\sin \\theta_{2}}{h} \\\\\n=2 \\pi \\frac{m d}{h}\\left(v_{1} \\sin \\theta_{1}-v_{2} \\sin \\theta_{2}\\right) \\\\\n=0 \\quad(\\text { from } \\mathrm{A} 2 \\text { or } \\mathrm{B} 1)\n\\end{gathered}\n$$\n\nThus near the clasical path there is invariably constructive interference.\n\n"
] | false | null | Numerical | 0 | OE_MM_physics_en_COMP | Mechanics | A The Extremum Principle in Mechanics
Consider a horizontal frictionless $x-y$ plane shown in Fig. 1. It is divided into two regions, I and II, by a line $\mathrm{AB}$ satisfying the equation $x=x_{1}$. The potential energy of a point particle of mass $m$ in region I is $V=0$ while it is $V=V_{0}$ in region II. The particle is sent from the origin $\mathrm{O}$ with speed $v_{1}$ along a line making an angle $\theta_{1}$ with the $x$-axis. It reaches point $\mathrm{P}$ in region II traveling with speed $v_{2}$ along a line that makes an angle $\theta_{2}$ with the $x$-axis. Ignore gravity and relativistic effects in this entire task T-2 (all parts).
<img_4336>
Figure 1
Context question:
A.1 Obtain an expression for $v_{2}$ in terms of $m, v_{1}$ and $V_{0}$.
Context answer:
\boxed{$v_{2}=(v_{1}^{2}-\frac{2 V_{0}}{m})^{1 / 2}$}
Context question:
A.2 Express $v_{2}$ in terms of $v_{1}, \theta_{1}$ and $\theta_{2}$.
Context answer:
\boxed{$v_2=\frac{v_1 \sin\theta_1}{\sin \theta_2}$}
Extra Supplementary Reading Materials:
We define a quantity called action $A=m \int v(s) d s$, where $d s$ is the infinitesimal length along the trajectory of a particle of mass $m$ moving with speed $v(s)$. The integral is taken over the path. As an example, for a particle moving with constant speed $v$ on a circular path of radius $R$, the action $A$ for one revolution will be $2 \pi m R v$. For a particle with constant energy $E$, it can be shown that of all the possible trajectories between two fixed points, the actual trajectory is the one on which $A$ defined above is an extremum (minimum or maximum). Historically this is known as the Principle of Least Action (PLA).
Context question:
A.3 PLA implies that the trajectory of a particle moving between two fixed points in a region of constant <br> potential will be a straight line. Let the two fixed points $\mathrm{O}$ and $\mathrm{P}$ in Fig. 1 have coordinates $(0,0)$ and <br> $\left(x_{0}, y_{0}\right)$ respectively and the boundary point where the particle transits from region I to region II have <br> coordinates $\left(x_{1}, \alpha\right)$. Note that $x_{1}$ is fixed and the action depends on the coordinate $\alpha$ only. State the <br> expression for the action $A(\alpha)$. Use PLA to obtain the relationship between $v_{1} / v_{2}$ and these coordinates.
Context answer:
\boxed{$\frac{v_{1}}{v_{2}}=\frac{(y_{0}-\alpha)(x_{1}^{2}+\alpha^{2})^{1 / 2}}{\alpha[(x_{0}-x_{1})^{2}+(y_{0}-\alpha)^{2}]^{1 / 2}}$}
Extra Supplementary Reading Materials:
B The Extremum Principle in Optics
A light ray travels from medium I to medium II with refractive indices $n_{1}$ and $n_{2}$ respectively. The two media are separated by a line parallel to the $x$-axis. The light ray makes an angle $i_{1}$ with the $y$-axis in medium I and $i_{2}$ in medium II (see Fig. 2). To obtain the trajectory of the ray, we make use of another extremum (minimum or maximum) principle known as Fermat's principle of least time.
<img_4508>
Figure 2
Context question:
B.1 The principle states that between two fixed points, a light ray moves along a path such that time taken between the two points is an extremum. Derive the relation between $\sin i_{1}$ and $\sin i_{2}$ on the basis of Fermat's principle.
Context answer:
\boxed{$n_{1} \sin i_{1}=n_{2} \sin i_{2}$}
Extra Supplementary Reading Materials:
Shown in Fig. 3 is a schematic sketch of the path of a laser beam incident horizontally on a solution of sugar in which the concentration of sugar decreases with height. As a consequence, the refractive index of the solution also decreases with height.
<img_4478>
Figure 3: Tank of Sugar Solution
Context question:
B.2 Assume that the refractive index $n(y)$ depends only on $y$. Use the equation obtained in B1 to obtain the expression for the slope $d y / d x$ of the beam's path in terms of refractive index $n_{0}$ at $y=0$ and $n(y)$.
Context answer:
\boxed{$\frac{d y}{d x}=-\sqrt{(\frac{n(y)}{n_{0} \sin i_{0}})^{2}-1}$}
Context question:
B.3 The laser beam is directed horizontally from the origin $(0,0)$ into the sugar solution at a height $y_{0}$ from the Obtain an expression for $x$ in terms of $y$ and related quantities for the actual trajectory of the laser beam.[^0]
You may use: $\int \sec \theta d \theta=\ln (\sec \theta+\tan \theta)+$ constant, where $\sec \theta=1 / \cos \theta$ or <br> $\int \frac{d x}{\sqrt{x^{2}-1}}=\ln \left(x+\sqrt{x^{2}-1}\right)+$ constant
Context answer:
\boxed{$x=\frac{n_{0}}{k} \ln (\frac{n_{0}-k y}{n_{0}}+\sqrt{(\frac{n_{0}-k y}{n_{0}})^{2}-1})$}
Context question:
B.4 Obtain the value of $x_{0}$, the point where the beam meets the bottom of the tank. Take $y_{0}=10.0 \mathrm{~cm}$, <br> $n_{0}=1.50, k=0.050 \mathrm{~cm}^{-1}\left(1 \mathrm{~cm}=10^{-2} \mathrm{~m}\right)$.
Context answer:
\boxed{24.0}
Extra Supplementary Reading Materials:
C The Extremum Principle and the Wave Nature of Matter
We now explore the connection between the PLA and the wave nature of a moving particle. For this we assume that a particle moving from $\mathrm{O}$ to $\mathrm{P}$ can take all possible trajectories and we will seek a trajectory that depends on the constructive interference of de Broglie waves.
Context question:
C.1 As the particle moves along its trajectory by an infinitesimal distance $\Delta s$, relate the change $\Delta \varphi$ in the phase <br> of its de Broglie wave to the change $\Delta A$ in the action and the Planck constant.
Context answer:
\boxed{$\Delta \phi=\frac{2 \pi \Delta A}{h}$} |
|
1100 | Context Example 1:
Question:
An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:
$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$
Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write
$$
S_{3}[263415]=(132,312,231,213)
$$
More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.
In this power question, you will be asked to analyze some of the properties of labels and signatures.
We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :
<image>
A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:
<image>
Consider the shape below:
<image>
Find the 2-signature that corresponds to this shape.
Solution:
The first pair indicates an increase; the next three are decreases, and the last pair is an increase. So the 2-signature is $(12,21,21,21,12)$.
Answer: $(12,21,21,21,12)$
Context Example 2:
Question:
In rectangle $A B C D, F$ is on diagonal $B D$ so that $A F$ is perpendicular to $B D$. Also, $B C=30, C D=40$ and $A F=x$. Determine the value of $x$.
<image>
Solution:
Since $A B C D$ is a rectangle, then $A B=C D=40$ and $A D=B C=30$.
By the Pythagorean Theorem, $B D^{2}=A D^{2}+A B^{2}$ and since $B D>0$, then
$$
B D=\sqrt{30^{2}+40^{2}}=\sqrt{900+1600}=\sqrt{2500}=50
$$
We calculate the area of $\triangle A D B$ is two different ways.
First, using $A B$ as base and $A D$ as height, we obtain an area of $\frac{1}{2}(40)(30)=600$.
Next, using $D B$ as base and $A F$ as height, we obtain an area of $\frac{1}{2}(50) x=25 x$.
We must have $25 x=600$ and so $x=\frac{600}{25}=24$.
Since $A B C D$ is a rectangle, then $A B=C D=40$ and $A D=B C=30$.
By the Pythagorean Theorem, $B D^{2}=A D^{2}+A B^{2}$ and since $B D>0$, then
$$
B D=\sqrt{30^{2}+40^{2}}=\sqrt{900+1600}=\sqrt{2500}=50
$$
Since $\triangle D A B$ is right-angled at $A$, then $\sin (\angle A D B)=\frac{A B}{B D}=\frac{40}{50}=\frac{4}{5}$.
But $\triangle A D F$ is right-angled at $F$ and $\angle A D F=\angle A D B$.
Therefore, $\sin (\angle A D F)=\frac{A F}{A D}=\frac{x}{30}$.
Thus, $\frac{x}{30}=\frac{4}{5}$ and so $x=\frac{4}{5}(30)=24$.
Solution 3
Since $A B C D$ is a rectangle, then $A B=C D=40$ and $A D=B C=30$.
By the Pythagorean Theorem, $B D^{2}=A D^{2}+A B^{2}$ and since $B D>0$, then
$$
B D=\sqrt{30^{2}+40^{2}}=\sqrt{900+1600}=\sqrt{2500}=50
$$
Note that $\triangle B F A$ is similar to $\triangle B A D$, since each is right-angled and they share a common angle at $B$.
Thus, $\frac{A F}{A B}=\frac{A D}{B D}$ and so $\frac{x}{30}=\frac{40}{50}$ which gives $x=\frac{30(40)}{50}=24$.
Answer: 24
Context Example 3:
Question:
In the isosceles trapezoid $A B C D$, $A B=C D=x$. The area of the trapezoid is 80 and the circle with centre $O$ and radius 4 is tangent to the four sides of the trapezoid. Determine the value of $x$.
<image>
Solution:
Using the tangent properties of a circle, the lengths of line segments are as shown on the diagram.
Area of trapezoid $A B C D=\frac{1}{2}(8)(B C+A D)$
$$
\begin{aligned}
& =4(2 b+2 x-2 b) \\
& =8 x .
\end{aligned}
$$
<img_3854>
Thus, $8 x=80$.
Therefore, $x=10$.
Answer: 10
Context Example 4:
Question:
A regular pentagon covers part of another regular polygon, as shown. This regular polygon has $n$ sides, five of which are completely or partially visible. In the diagram, the sum of the measures of the angles marked $a^{\circ}$ and $b^{\circ}$ is $88^{\circ}$. Determine the value of $n$.
(The side lengths of a regular polygon are all equal, as are the measures of its interior angles.)
<image>
Solution:
The angles in a polygon with $n$ sides have a sum of $(n-2) \cdot 180^{\circ}$.
This means that the angles in a pentagon have a sum of $3 \cdot 180^{\circ}$ or $540^{\circ}$, which means that each interior angle in a regular pentagon equals $\frac{1}{5} \cdot 540^{\circ}$ or $108^{\circ}$.
Also, each interior angle in a regular polygon with $n$ sides equals $\frac{n-2}{n} \cdot 180^{\circ}$. (This is the general version of the statement in the previous sentence.)
Consider the portion of the regular polygon with $n$ sides that lies outside the pentagon and join the points from which the angles that measure $a^{\circ}$ and $b^{\circ}$ emanate to form a hexagon.
<img_3516>
This polygon has 6 sides, and so the sum of its 6 angles is $4 \cdot 180^{\circ}$.
Four of its angles are the original angles from the $n$-sided polygon, so each equals $\frac{n-2}{n} \cdot 180^{\circ}$.
The remaining two angles have measures $a^{\circ}+c^{\circ}$ and $b^{\circ}+d^{\circ}$.
We are told that $a^{\circ}+b^{\circ}=88^{\circ}$.
Also, the angles that measure $c^{\circ}$ and $d^{\circ}$ are two angles in a triangle whose third angle is $108^{\circ}$.
Thus, $c^{\circ}+d^{\circ}=180^{\circ}-108^{\circ}=72^{\circ}$.
Therefore,
$$
\begin{aligned}
4 \cdot \frac{n-2}{n} \cdot 180^{\circ}+88^{\circ}+72^{\circ} & =4 \cdot 180^{\circ} \\
160^{\circ} & =\left(4-\frac{4(n-2)}{n}\right) \cdot 180^{\circ} \\
160^{\circ} & =\frac{4 n-(4 n-8)}{n} \cdot 180^{\circ} \\
\frac{160^{\circ}}{180^{\circ}} & =\frac{8}{n} \\
\frac{8}{9} & =\frac{8}{n}
\end{aligned}
$$
and so the value of $n$ is 9 .
The angles in a polygon with $n$ sides have a sum of $(n-2) \cdot 180^{\circ}$.
This means that the angles in a pentagon have a sum of $3 \cdot 180^{\circ}$ or $540^{\circ}$, which means that each interior angle in a regular pentagon equals $\frac{1}{5} \cdot 540^{\circ}$ or $108^{\circ}$.
Also, each interior angle in a regular polygon with $n$ sides equals $\frac{n-2}{n} \cdot 180^{\circ}$. (This is the general version of the statement in the previous sentence.)
Consider the portion of the regular polygon with $n$ sides that lies outside the pentagon.
<img_3345>
This polygon has 7 sides, and so the sum of its 7 angles is $5 \cdot 180^{\circ}$.
Four of its angles are the original angles from the $n$-sided polygon, so each equals $\frac{n-2}{n} \cdot 180^{\circ}$.
Two of its angles are the angles equal to $a^{\circ}$ and $b^{\circ}$, whose sum is $88^{\circ}$.
Its seventh angle is the reflex angle corresponding to the pentagon's angle of $108^{\circ}$, which equals $360^{\circ}-108^{\circ}$ or $252^{\circ}$.
Therefore,
$$
\begin{aligned}
4 \cdot \frac{n-2}{n} \cdot 180^{\circ}+88^{\circ}+252^{\circ} & =5 \cdot 180^{\circ} \\
340^{\circ} & =\left(5-\frac{4(n-2)}{n}\right) \cdot 180^{\circ} \\
340^{\circ} & =\frac{5 n-(4 n-8)}{n} \cdot 180^{\circ} \\
\frac{340^{\circ}}{180^{\circ}} & =\frac{n+8}{n} \\
\frac{17}{9} & =\frac{n+8}{n} \\
17 n & =9(n+8) \\
17 n & =9 n+72 \\
8 n & =72
\end{aligned}
$$
and so the value of $n$ is 9 .
Answer: 9
Context Example 5:
Question:
Let $T=80$. In circle $O$, diagrammed at right, minor arc $\widehat{A B}$ measures $\frac{T}{4}$ degrees. If $\mathrm{m} \angle O A C=10^{\circ}$ and $\mathrm{m} \angle O B D=5^{\circ}$, compute the degree measure of $\angle A E B$. Just pass the number without the units.
<image>
Solution:
Note that $\mathrm{m} \angle A E B=\frac{1}{2}(\mathrm{~m} \widehat{A B}-m \widehat{C D})=\frac{1}{2}(\mathrm{~m} \widehat{A B}-\mathrm{m} \angle C O D)$. Also note that $\mathrm{m} \angle C O D=$ $360^{\circ}-(\mathrm{m} \angle A O C+\mathrm{m} \angle B O D+\mathrm{m} \angle A O B)=360^{\circ}-\left(180^{\circ}-2 \mathrm{~m} \angle O A C\right)-\left(180^{\circ}-2 \mathrm{~m} \angle O B D\right)-$ $\mathrm{m} \widehat{A B}=2(\mathrm{~m} \angle O A C+\mathrm{m} \angle O B D)-\mathrm{m} \widehat{A B}$. Thus $\mathrm{m} \angle A E B=\mathrm{m} \widehat{A B}-\mathrm{m} \angle O A C-\mathrm{m} \angle O B D=$ $\frac{T}{4}-10^{\circ}-5^{\circ}$, and with $T=80$, the answer is 5 .
Answer: 5
Context Example 6:
Question:
In the diagram, $\triangle P Q R$ has $P Q=a, Q R=b, P R=21$, and $\angle P Q R=60^{\circ}$. Also, $\triangle S T U$ has $S T=a, T U=b, \angle T S U=30^{\circ}$, and $\sin (\angle T U S)=\frac{4}{5}$. Determine the values of $a$ and $b$.
<image>
Solution:
Using the cosine law in $\triangle P Q R$,
$$
\begin{aligned}
P R^{2} & =P Q^{2}+Q R^{2}-2 \cdot P Q \cdot Q R \cdot \cos (\angle P Q R) \\
21^{2} & =a^{2}+b^{2}-2 a b \cos \left(60^{\circ}\right) \\
441 & =a^{2}+b^{2}-2 a b \cdot \frac{1}{2} \\
441 & =a^{2}+b^{2}-a b
\end{aligned}
$$
Using the sine law in $\triangle S T U$, we obtain $\frac{S T}{\sin (\angle T U S)}=\frac{T U}{\sin (\angle T S U)}$ and so $\frac{a}{4 / 5}=\frac{b}{\sin \left(30^{\circ}\right)}$. Therefore, $\frac{a}{4 / 5}=\frac{b}{1 / 2}$ and so $a=\frac{4}{5} \cdot 2 b=\frac{8}{5} b$.
Substituting into the previous equation,
$$
\begin{aligned}
& 441=\left(\frac{8}{5} b\right)^{2}+b^{2}-\left(\frac{8}{5} b\right) b \\
& 441=\frac{64}{25} b^{2}+b^{2}-\frac{8}{5} b^{2} \\
& 441=\frac{64}{25} b^{2}+\frac{25}{25} b^{2}-\frac{40}{25} b^{2} \\
& 441=\frac{49}{25} b^{2} \\
& 225=b^{2}
\end{aligned}
$$
Since $b>0$, then $b=15$ and so $a=\frac{8}{5} b=\frac{8}{5} \cdot 15=24$.
Answer: $24,15$
Context Example 7:
Question:
- 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,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$
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?<image>
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.
<img_4322>
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 $0.9 \mathrm{~m}$.
Answer: $0.9$
Context Example 8:
Question:
- 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,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$
A boy is riding a tricycle across along a sidewalk that is parallel to the $x$-axis. This tricycle contains three identical wheels with radius $0.5 \mathrm{~m}$. The front wheel is free to rotate while the last two wheels are parallel to each other and to the main body of the tricycle. See the diagram.
<image>
The front wheel is rotating at a constant angular speed of $\omega=3 \mathrm{rad} / \mathrm{s}$. The child is controlling the tricycle such that the front wheel is making an angle of $\theta(t)=0.15 \sin ((0.1 \mathrm{rad} / \mathrm{s}) t)$ with the main body of the tricycle. Determine the maximum lateral acceleration in $\mathrm{m} / \mathrm{s}^{2}$. Assume a massless frame. The marked plus sign implies CoM. The degree is in radians.
Solution:
Assuming the wheels roll without slipping, the cart will instantaneously rotate about a fixed point. This fixed point can be constructed by drawing perpendicular lines from all the wheels and seeing where they intersect. Consider the below diagram,
<img_4299>
We have:
$$
\sin \theta_{s}=\frac{L}{R}, \quad \quad \sin \theta_{s}^{\prime}=\frac{L}{2 R^{\prime}}
\tag{6}
$$
The angular frequency of every point on the robot is the same $\omega_{n}=\frac{V_{0}}{R}$. Therefore,
$$
\begin{aligned}
V_{C M} & =\omega_{n} R^{\prime} \\
& =\frac{V_{0}}{R} \frac{L}{2 \sin \theta_{s}^{\prime}} \\
& =\frac{V \sin \theta_{s}}{2 \sin \theta_{s}^{\prime}} .
\end{aligned}
$$
The naive idea here is to say the lateral velocity is $V_{C M} \sin \theta_{s}^{\prime}$, but note that the robot could already be rotated. Let the angle of rotation with respect to the horizontal line be $\alpha$, so we have:
$$
V_{C M, l a t}=V_{C M} \sin \left(\theta_{s}^{\prime}+\alpha\right)
$$
Using the sine addition formula, we can simplify this to
$$
\begin{aligned}
V_{C M, l a t} & =V_{C M} \sin \theta_{s}^{\prime} \cos \alpha+V_{C M} \cos \theta_{s}^{\prime} \sin \alpha \\
& =\frac{V_{0}}{2} \sin \theta_{s} \cos \alpha+\frac{V_{0} \sin \theta_{s}}{2 \tan \theta_{s}^{\prime}} \sin \alpha
\end{aligned}
$$
To simplify this, note that we can write:
$$
\frac{\sin \theta_{s}}{\tan \theta_{s}^{\prime}}=2
$$
where we used the fact that
$$
\sin \theta_{s} \approx \tan \theta_{s}=2 \tan \theta_{s}^{\prime} \approx 2 \sin \theta_{s}^{\prime}
$$
Then,
$$
\dot{y}_{C M}=\frac{V_{0}}{2} \theta_{s}+V_{0} \alpha
$$
At the center of the two back wheels (marked with an X), the speed is
$$
V_{b a c k}=R \cos \theta_{s} \omega_{n}=V_{0} \cos \theta_{s} \approx V_{0}
$$
so its lateral velocity is
$$
\dot{y}_{b a c k}=V_{0} \sin \alpha \approx V_{0} \alpha .
$$
Note that the angle $\alpha$ can be written as
$$
\alpha \approx \sin \alpha=\frac{y_{C M}-y_{b a c k}}{(L / 2)} \Longrightarrow \dot{\alpha}=\frac{2}{L}\left(\dot{y}_{C M}-\dot{y}_{b a c k}\right)=\frac{V_{0}}{L} \theta_{s}
$$
Taking higher derivatives of $\dot{y}_{C M}$, we have
$$
\begin{aligned}
\ddot{y}_{C M} & =\frac{V_{0}}{2} \dot{\theta}_{s}+V_{0} \dot{\alpha} \\
& =\frac{V_{0}}{2} \dot{\theta}_{s}+\frac{V_{0}^{2}}{L} \theta_{s} .
\end{aligned}
$$
Plugging in $\theta(t)=0.15 \sin (0.1 t)$ and $V_{0}=1.5 \mathrm{~m} / \mathrm{s}$, we can optimize $\ddot{y}_{\mathrm{CM}}$ to get $0.16912 \mathrm{~m} / \mathrm{s}$.
Answer: $0.16912$
Context Example 9:
Question:
<image>
In certain lakes there is a strange phenomenon called "seiching" which is an oscillation of the water. Lakes in which you can see this phenomenon are normally long compared with the depth and also narrow. It is natural to see waves in a lake but not something like the seiching, where the entire water volume oscillates, like the coffee in a cup that you carry to a waiting guest.
In order to create a model of the seiching we look at water in a rectangular container. The length of the container is $L$ and the depth of the water is $h$. Assume that the surface of the water to begin with makes a small angle with the horizontal. The seiching will then start, and we assume that the water surface continues to be plane but oscillates around an axis in the horizontal plane and located in the middle of the container.
Create a model of the movement of the water and derive a formula for the oscillation period $T$. The starting conditions are given in figure above.
Assume that $\xi<<h$. The table below shows experimental oscillation periods for different water depths in two containers of different lengths. Check in some reasonable way how well the formula that you have derived agrees with the experimental data. Give your opinion on the quality of your model.
Table 1. $L=479 \mathrm{~mm}$
| $h / m m$ | 30 | 50 | 69 | 88 | 107 | 124 | 142 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $T / s$ | 1.78 | 1.40 | 1.18 | 1.08 | 1.00 | 0.91 | 0.82 |
Table 2. $L=143 \mathrm{~mm}$
| $h / m m$ | 31 | 38 | 58 | 67 | 124 |
| :---: | :---: | :---: | :---: | :---: | ---: |
| $T / s$ | 0.52 | 0.52 | 0.43 | 0.35 | 0.28 |
The graph below shows results from measurements in lake Vättern in Sweden. This lake has a length of $123 \mathrm{~km}$ and a mean depth of $50 \mathrm{~m}$. What is the time scale in the graph?
<image>
The water surface level in Bastudalen (northern end of lake Vättern) and Jönköping (southern end).
Solution:
In the coordinate system of the figure, we have for the centre of mass coordinates of the two triangular parts of the water
$$
\left(x_{1}, y_{1}\right)=(L / 3, h / 2+\xi / 3) \quad\left(x_{2}, y_{2}\right)=(-L / 3, h / 2-\xi / 3) .
$$
For the entire water mass the centre of mass coordinates will then be
$$
\left(x_{C O M}, y_{C O M}\right)=\left(\frac{\xi L}{6 h}, \frac{\xi^{2}}{6 h}\right)
$$
Due to that the $y$ component is quadratic in $\xi$ will be much much smaller than the $x$ component.
The velocities of the water mass are
$$
\left(v_{x}, v_{y}\right)=\left(\frac{g_{L}}{6 h}, \frac{g_{\xi}}{3 h}\right)
$$
and again the vertical component is much smaller the the horizontal one.
We now in our model neglect the vertical components. The total energy (kinetic + potential) will then be
$$
W=W_{K}+W_{P}=\frac{1}{2} M \frac{\xi^{2} L^{2}}{36 h^{2}}+M g \frac{\xi^{2}}{6 h^{2}}
$$
For a harmonic oscillator we have
$$
W=W_{K}+W_{P}=\frac{1}{2} m x^{2}+\frac{1}{2} m \omega^{2} x^{2}
$$
Identifying gives
$$
\omega=\sqrt{\frac{12 g h}{L}} \text { or } T_{\text {model }}=\frac{\pi L}{\sqrt{3 h}} \text {. }
$$
Comparing with the experimental data we find $T_{\text {experiment }} \approx 1.1 \cdot T_{\text {model }}$ our model gives a slight underestimation of the oscillation period.
Applying our corrected model on the Vättern data we have that the oscillation period of the seiching is about 3 hours.
Answer: $\frac{\pi L}{\sqrt{3 h}}$
Context Example 10:
Question:
- 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,
$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$
A small block of mass $m$ and charge $Q$ is placed at rest on an inclined plane with a slope $\alpha=40^{\circ}$. The coefficient of friction between them is $\mu=0.3$. A homogenous magnetic field of magnitude $B_{0}$ is applied perpendicular to the slope. The speed of the block after a very long time is given by $v=\beta \frac{m g}{Q B_{0}}$. Determine $\beta$. Do not neglect the effects of gravity.
<image>
Solution:
Create a free body diagram. The direction of the magnetic field (into or out of the page) does not matter as we only need to know the magnitude of the terminal velocity. In dynamic equilibrium, we have three forces along the plane: the component of gravity along the plane, friction, and the magnetic force. The component of gravity along the plane is $m g \sin \alpha$. Friction has the magnitude $f=\mu m g \cos \alpha$. The magnetic force has magnitude $F_{B}=Q v B_{0}$. At the terminal velocity, these forces are balanced. To finish, note that friction and the magnetic force are perpendicular so the magnitude of their vector sum is equal to the component of gravity along the plane. By the Pythagorean Theorem,
$$
F_{B}^{2}+f^{2}=(m g \sin \alpha)^{2} \Longrightarrow v=\frac{m g}{Q B_{0}} \sqrt{\sin ^{2} \alpha-\mu^{2} \cos ^{2} \alpha}
$$
Answer: 0.6
Current Question:
Initially, a switch $S$ is unshorted in the circuit shown in the figure on the right, a capacitor of capacitance $2 C$ carries the electric charge $q_{0}$, a capacitor of capacitance $C$ is uncharged, and there are no electric currents in both coils of inductance $L$ and $2 L$, respectively. The capacitor starts to discharge and at the moment when the current in the coils reaches its maximum value, the switch $S$ is instantly shorted. Find the maximum current $I_{\max }$ through the switch $S$ thereafter.
<image_1> | [
"$I_{\\max }=\\frac{q_{0}}{\\sqrt{2 L C}}$"
] | [
"At the moment when the current in the coils is a maximum, the total voltage across the coils is equal to zero, so the capacitor voltages must be equal in magnitude and opposite in polarity. Let $U$ be a voltage on the capacitors at the time moment just mentioned and $I_{0}$ be that maximum current. According to the law of charge conservation\n\nthus,\n\n$$\nq_{0}=2 C U+C U\n\\tag{C1.1}\n$$\n\n$$\nU=\\frac{q_{0}}{3 C}\n\\tag{C1.2}\n$$\n\nThen, from the energy conservation law\n\n$$\n\\frac{q_{0}^{2}}{2 \\cdot 2 C}=\\frac{L I_{0}^{2}}{2}+\\frac{2 L I_{0}^{2}}{2}+\\frac{C U^{2}}{2}+\\frac{2 C U^{2}}{2}\n\\tag{C1.3}\n$$\n\nthe maximum current is found as\n\n$$\nI_{0}=\\frac{q_{0}}{3 \\sqrt{2 L C}}\n\\tag{C1.4}\n$$\n\nAfter the key $K$ is shortened there will be independent oscillations in both circuits with the frequency\n\n$$\n\\omega=\\frac{1}{\\sqrt{2 L C}}\n\\tag{C1.5}\n$$\n\nand their amplitudes are obtained from the corresponding energy conservation laws written as\n\n$$\n\\frac{2 C U^{2}}{2}+\\frac{L I_{0}^{2}}{2}=\\frac{L J_{1}^{2}}{2}\n\\tag{C1.6}\n$$\n$$\n\\frac{C U^{2}}{2}+\\frac{2 L I_{0}^{2}}{2}=\\frac{2 L J_{2}^{2}}{2} .\n\\tag{C1.7}\n$$\n\nHence, the corresponding amplitudes are found as\n\n$$\nJ_{1} =\\sqrt{5} I_{0},\n\\tag{C1.8}\n$$\n$$\nJ_{2} =\\sqrt{2} I_{0} .\n\\tag{C1.9}\n$$\n\nChoose the positive directions of the currents in the circuits as shown in the figure on the right. Then, the current flowing through the key is written as follows\n\n$$\nI=I_{1}-I_{2}\n\\tag{C1.10}\n$$\n\nThe currents depend on time as\n\n$$\nI_{1}(t)=A \\cos \\omega t+B \\sin \\omega t\n\\tag{C1.11}\n$$\n$$\nI_{2}(t)=D \\cos \\omega t+F \\sin \\omega t\n\\tag{C1.12}\n$$\n\n<img_4418>\n\nThe constants $A, B, D, F$ can be determined from the initial values of the currents and their amplitudes by putting down the following set of equations\n\n$$\nI_{1}(0)=A=I_{0},\n\\tag{C1.13}\n$$\n$$\nA^{2}+B^{2}=J_{1}^{2}, \n\\tag{C1.14}\n$$\n$$\nI_{2}(0)=D=I_{0}, \n\\tag{C1.15}\n$$\n$$\nD^{2}+F^{2}=J_{2}^{2}\n\\tag{C1.16}\n$$\n\nSolving (C1.13)-(C1.16) it is found that\n\n$$\nB=2 I_{0}, \n\\tag{C1.17}\n$$\n$$\nF=-I_{0},\n\\tag{C1.18}\n$$\n\nThe sign in $F$ is chosen negative, since at the time moment of the key shortening the current in the coil $2 L$ decreases.\n\nThus, the dependence of the currents on time takes the following form\n\n$$\nI_{1}(t)=I_{0}(\\cos \\omega t+2 \\sin \\omega t),\n\\tag{C1.19}\n$$\n$$\nI_{2}(t)=I_{0}(\\cos \\omega t-\\sin \\omega t) .\n\\tag{C1.20}\n$$\n\nIn accordance with (C1.10), the current in the key is dependent on time according to\n\n$$\nI(t)=I_{1}(t)-I_{2}(t)=3 I_{0} \\sin \\omega t .\n\\tag{C1.21}\n$$\n\nHence, the amplitude of the current in the key is obtained as\n\n$$\nI_{\\max }=3 I_{0}=\\omega q_{0}=\\frac{q_{0}}{\\sqrt{2 L C}}\n\\tag{C1.22}\n$$",
"Instead of determining the coefficients $A, B, D, F$ the vector diagram shown in the figure on the right can be used. The segment $A C$ represents the current sought and its projection on the current axis is zero at the time of the key shortening. The current $I_{1}$ in the coil of inductance $L$ grows at the same time moment because the capacitor $2 C$ continues to discharge, thus, this current is depicted in the figure by the segment $O A$. The current $I_{2}$ in the coil of inductance $2 L$ decreases at the time of the key shortening since it continues to charge the capacitor $2 C$, that is why this current is depicted in the figure by the segment $O C$.\n\nIt is known for above that $O B=I_{0}, O A=\\sqrt{5} I_{0}, O C=\\sqrt{2} I_{0}$. Hence, it is found from the Pythagorean theorem that\n\n$$\nA B=\\sqrt{O A^{2}-O B^{2}}=2 I_{0}\n\\tag{C2.1}\n$$\n$$\nB C=\\sqrt{O C^{2}-O B^{2}}=I_{0}\n\\tag{C2.2}\n$$\n\n<img_4426>\n\nThus, the current sought is found as\n\n$$\nI_{\\max }=A C=A B+B C=3 I_{0}=\\omega q_{0}=\\frac{q_{0}}{\\sqrt{2 L C}}\n\\tag{C2.3}\n$$\n\nMethod 3. Heuristic approach\n\nIt is clear that the current through the key performs harmonic oscillations with the frequency\n\n$$\n\\omega=\\frac{1}{\\sqrt{2 L C}}\n\\tag{C3.1}\n$$\n\nand it is equal to zero at the time of the key shortening, i.e.\n\n$$\nI(t)=I_{\\max } \\sin \\omega t\n\\tag{C3.2}\n$$\n\nSince the current is equal to zero at the time of the key shortening, then the current amplitude is equal to the current derivative at this time moment divided by the oscillation frequency. Let us find that current derivative. Let the capacitor of capacitance $2 C$ have the charge $q_{1}$. Then the charge on the capacitor of capacitance $C$ is found from the charge conservation law as\n\n$$\nq_{2}=q_{0}-q_{1} .\n\\tag{C3.3}\n$$\n\nAfter shortening the key the rate of current change in the coil of inductance $L$ is obtained as\n\n$$\n\\dot{I}_{1}=\\frac{q_{1}}{2 L C}\n\\tag{C3.4}\n$$\n\nwhereas in the coil of inductance $2 L$ it is equal to\n\n$$\n\\dot{I}_{2}=-\\frac{q_{0}-q_{1}}{2 L C}\n\\tag{C3.5}\n$$\n\nSince the voltage polarity on the capacitors are opposite, then the current derivative with respect to time finally takes the form\n\n$$\n\\dot{I}=\\dot{I}_{1}-\\dot{I}_{2}=\\frac{q_{0}}{2 L C}=\\omega^{2} q_{0}\n\\tag{C3.6}\n$$\n\nNote that this derivative is independent of the time of the key shortening!\n\nHence, the maximum current is found as\n\n$$\nI_{\\max }=\\frac{\\dot{I}}{\\omega}=\\omega q_{0}=\\frac{q_{0}}{\\sqrt{2 L C}}\n\\tag{C3.7}\n$$\n\nand it is independent of the time of the key shortening!"
] | false | null | Expression | null | OE_MM_physics_en_COMP | Electromagnetism | ||
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1164 | "Context Example 1:\nQuestion:\n\nA triangle has vertices $A(0,3), B(4,0)$, $C(k, 5)$, where $0<k<4$(...TRUNCATED) | [{"src":"https://datasets-server.huggingface.co/assets/yuanshengni/OlympiadBench-OE-CoT-num10/--/{da(...TRUNCATED) | [
"$v_{\\min }=3 \\sqrt{\\frac{g R}{2}}$"
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1166 | "Context Example 1:\nQuestion:\n## String Cheese\nContext question:\na. When a faucet is turned on, (...TRUNCATED) | [{"src":"https://datasets-server.huggingface.co/assets/yuanshengni/OlympiadBench-OE-CoT-num10/--/{da(...TRUNCATED) | [
"$F=\\frac{4-\\sqrt{2}}{8 \\pi \\mu_{0}} \\frac{\\Phi^{2}}{l^{2}}$"
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1218 | "Context Example 1:\nQuestion:\n\nThe arrangement of numbers known as Pascal's Triangle has fascinat(...TRUNCATED) | [{"src":"https://datasets-server.huggingface.co/assets/yuanshengni/OlympiadBench-OE-CoT-num10/--/{da(...TRUNCATED) | [
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1322 | "Context Example 1:\nQuestion:\n\n$\\quad$ Let $T=12$. As shown, three circles are mutually external(...TRUNCATED) | [{"src":"https://datasets-server.huggingface.co/assets/yuanshengni/OlympiadBench-OE-CoT-num10/--/{da(...TRUNCATED) | [
"$67$ , $100$"
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1398 | "Context Example 1:\nQuestion:\n\nIn the diagram, $\\triangle P Q R$ has $P Q=a, Q R=b, P R=21$, and(...TRUNCATED) | [{"src":"https://datasets-server.huggingface.co/assets/yuanshengni/OlympiadBench-OE-CoT-num10/--/{da(...TRUNCATED) | ["$\\frac{2 m \\theta}{M}\\left(\\left(1-\\frac{h}{R}\\right)^{2}-1\\right)$","$-\\frac{4 m}{M} \\fr(...TRUNCATED) | ["During the first leg of the trip, the disk has angular velocity\n\n\n\n$$\n\n\\omega=-\\frac{2 m v(...TRUNCATED) | false | null | Expression | null | OE_MM_physics_en_COMP | Mechanics | "## Disk Jockey\n\n\n\nA disk of uniform mass density, mass $M$, and radius $R$ sits at rest on a fr(...TRUNCATED) |
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