├── Problem_03_Figure.png ├── Problem_05_Figure.png ├── Problem_07_Figure.png ├── Problem_08_Figure.png ├── Problem_09_Figure.png ├── Problem_11_Figure.png ├── Problem_12_Figure.png ├── Problem_14_Figure.png ├── Problem_15_Figure.png ├── Problem_16_Figure.png ├── Problem_18_Figure.png ├── Problem_19_Figure.png ├── Problem_20_Figure.png ├── Problem_21_Figure.png ├── Problem_23_Figure.png ├── Problem_24_Figure.png ├── Problem_26_Figure.png ├── Problem_28_Figure.png ├── Problem_29_Figure.png ├── Problem_31_Figure.png ├── Problem_38_Figure.png ├── docs ├── CheatSheets.pdf ├── Physics_Bowl_Initial_Version.pdf └── index.html ├── .github └── workflows │ └── pages.yml ├── README.md ├── Physics_Bowl_Initial_Version.tex └── guides └── physics └── CheatSheets.tex /Problem_03_Figure.png: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/qiuxue-yee/Physics-Bowl-CheatSheets-prepmaterial/HEAD/Problem_03_Figure.png 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Physics Bowl D1 + D2 Downloads

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Use the links below to access the latest PDFs.

12 | 17 | 18 | 19 | -------------------------------------------------------------------------------- /.github/workflows/pages.yml: -------------------------------------------------------------------------------- 1 | name: github-pages 2 | on: 3 | push: 4 | branches: [ main ] 5 | workflow_dispatch: 6 | permissions: 7 | contents: read 8 | pages: write 9 | id-token: write 10 | concurrency: 11 | group: "pages" 12 | cancel-in-progress: true 13 | jobs: 14 | build: 15 | runs-on: ubuntu-latest 16 | steps: 17 | - uses: actions/checkout@v4 18 | - name: Setup Pages 19 | uses: actions/configure-pages@v5 20 | - name: Prepare docs folder 21 | run: | 22 | mkdir -p docs 23 | ls -la docs || true 24 | - name: Upload artifact 25 | uses: actions/upload-pages-artifact@v3 26 | with: 27 | path: docs 28 | deploy: 29 | environment: 30 | name: github-pages 31 | url: ${{ steps.deployment.outputs.page_url }} 32 | runs-on: ubuntu-latest 33 | needs: build 34 | steps: 35 | - name: Deploy to GitHub Pages 36 | id: deployment 37 | uses: actions/deploy-pages@v4 38 | -------------------------------------------------------------------------------- /README.md: -------------------------------------------------------------------------------- 1 | # Physics-Bowl-CheatSheets-prematerial 2 | 3 | **Open-source Physics Bowl D1 + D2 study guide by 🌸🐟: 47-page cheat sheets with formulas, heuristics, and solutions for mechanics, waves/EM, optics. CC BY-NC-SA 4.0 for educational use only.** 4 | 5 | GitHub Pages: https://qiuxue-yee.github.io/Physics-Bowl-CheatSheets-prepmaterial/ 6 | Direct PDF: [CheatSheets.pdf](https://qiuxue-yee.github.io/Physics-Bowl-CheatSheets-prepmaterial/CheatSheets.pdf) 7 | 8 | - [CheatSheets.pdf](https://qiuxue-yee.github.io/Physics-Bowl-CheatSheets-prepmaterial/CheatSheets.pdf) 9 | - [Physics Bowl Initial Version.pdf](https://qiuxue-yee.github.io/Physics-Bowl-CheatSheets-prepmaterial/Physics_Bowl_Initial_Version.pdf) 10 | 11 | ## 中文版 12 | **🌸🐟 著的Physics Bowl D1 + D2备考开源指南：47页备忘单，包含力学、波/EM、光学公式、启发式和解决方案。CC BY-NC-SA 4.0，仅供教育使用。** 13 | 14 | GitHub Pages: https://qiuxue-yee.github.io/Physics-Bowl-CheatSheets-prepmaterial/ 15 | PDF: [CheatSheets.pdf](https://qiuxue-yee.github.io/Physics-Bowl-CheatSheets-prepmaterial/CheatSheets.pdf) 16 | 17 | - [CheatSheets.pdf](https://qiuxue-yee.github.io/Physics-Bowl-CheatSheets-prepmaterial/CheatSheets.pdf) 18 | - [Physics Bowl Initial Version.pdf](https://qiuxue-yee.github.io/Physics-Bowl-CheatSheets-prepmaterial/Physics_Bowl_Initial_Version.pdf) 19 | 20 | ## Contents Overview 21 | - **Part I: Mechanics** – Kinematics (SUVAT, projectiles, circular motion), Dynamics (Newton's laws, friction), Work/Energy, Momentum/Collisions, Rotational Motion. 22 | - **Part II: Oscillations/Waves/Thermodynamics/Fluids** – SHM, waves, heat, fluids (Bernoulli). 23 | - **Part III: Electricity & Magnetism** – Circuits, fields (Gauss, Ampere), induction (Faraday-Lenz). 24 | - **Part IV: Optics & Modern Physics** – Lenses, interference, relativity, quantum basics. 25 | 26 | ## 中文版内容概述 27 | - **Part I: 力学** – 运动学 (SUVAT, 抛体, 圆周运动), 动力学 (牛顿定律, 摩擦), 功/能量, 动量/碰撞, 转动运动。 28 | - **Part II: 振动/波/热力学/流体** – 简谐运动, 波, 热, 流体 (伯努利)。 29 | - **Part III: 电学与磁学** – 电路, 场 (高斯, 安培), 感应 (法拉第-伦次)。 30 | - **Part IV: 光学与现代物理** – 透镜, 干涉, 相对论, 量子基础。 31 | 32 | ## AAPT Disclaimer (免责声明：关于AAPT) 33 | This guide is **independent and non-official**. It draws inspiration from publicly available AAPT Physics Bowl past papers (free PDFs for practice at [aapt.org/programs/physicsbowl](https://www.aapt.org/programs/physicsbowl)), but includes only metadata references (e.g., Contest/Year/Division/Problem/Page) and original solutions/heursitics. Full problem statements, diagrams, and official answers must be obtained directly from AAPT sources. 34 | 35 | - **Images & Fair Use**: PNG diagrams in `/src/images/` are educational screenshots from AAPT past papers PDFs, used under U.S. Fair Use doctrine (Section 107) for transformative, non-commercial study aids (limited quantity, no market harm). They are not reproductions of full exams. AAPT copyrights all materials (intended for members/educators' personal use; no photocopying or redistribution per contest rules). This guide does not endorse or substitute official resources—verify all content independently. 36 | - **No AAPT Endorsement**: AAPT has not reviewed, approved, or endorsed this guide. Users assume responsibility for compliance with AAPT terms. 37 | - **Recommendation**: Download complete past papers directly from AAPT for accurate practice. 38 | 39 | **本指南为独立、非官方材料**。它受AAPT Physics Bowl历史试卷公开PDF启发（免费练习下载：[aapt.org/programs/physicsbowl](https://www.aapt.org/programs/physicsbowl)），但仅包含元数据引用 40 | -------------------------------------------------------------------------------- /Physics_Bowl_Initial_Version.tex: -------------------------------------------------------------------------------- 1 | \documentclass[12pt, a4paper]{article} 2 | \usepackage{amsmath, amssymb, amsfonts} 3 | \usepackage{graphicx} 4 | \usepackage[margin=1in]{geometry} 5 | \usepackage{enumitem} 6 | \usepackage[utf8]{inputenc} 7 | \usepackage{lmodern} 8 | \usepackage[protrusion=true,expansion=true]{microtype} 9 | \usepackage[hidelinks]{hyperref} 10 | \hypersetup{linktoc=all} 11 | \usepackage{siunitx} 12 | % hyphenation enabled (was: \usepackage[none]{hyphenat}) 13 | \setlength{\emergencystretch}{3em} 14 | \raggedbottom 15 | \usepackage[most]{tcolorbox} 16 | \usepackage{fancyhdr} 17 | \usepackage{lastpage} 18 | \usepackage{caption} 19 | \usepackage{needspace} 20 | \usepackage{etoolbox} 21 | \usepackage{CJKutf8} 22 | \pagestyle{fancy} 23 | \fancyhf{} 24 | \fancyhead[L]{Ray Zhou} 25 | \fancyhead[R]{Physics Bowl} 26 | \fancyfoot[C]{Page \thepage\ of \pageref{LastPage}} 27 | \renewcommand{\headrulewidth}{0.4pt} 28 | % fancyhdr safety 29 | \setlength{\headheight}{14.5pt} 30 | 31 | % Ensure title pages and other 'plain' pages use the same header/footer 32 | \fancypagestyle{plain}{% 33 | \fancyhf{} 34 | \fancyhead[L]{Ray Zhou} 35 | \fancyhead[R]{Physics Bowl} 36 | \fancyfoot[C]{Page \thepage\ of \pageref{LastPage}} 37 | \renewcommand{\headrulewidth}{0.4pt} 38 | } 39 | 40 | % AIME-style boxes and helpers 41 | \tcbset{colback=white, colframe=black!15, boxrule=0.5pt} 42 | % Unified figure helper: centered graphic with optional width and a caption 43 | \newcommand{\includefigure}[3][\linewidth]{% 44 | \begin{center} 45 | \includegraphics[width=#1]{#2}% 46 | \par\smallskip 47 | \captionsetup{type=figure}\captionof{figure}{#3}% 48 | \end{center}% 49 | } 50 | \newtcolorbox{solutionbox}{enhanced, breakable, title=\textbf{Solution}, colback=white, colframe=black!20, colbacktitle=gray!15, coltitle=black, fonttitle=\bfseries, left=10pt, right=10pt, top=7pt, bottom=9pt, before upper={\setlength{\parskip}{8pt}\setlength{\parindent}{1.2em}\setlength{\abovedisplayskip}{10pt}\setlength{\belowdisplayskip}{10pt}}, before skip=10pt, after skip=14pt, after=\par\noindent\rule{\linewidth}{0.4pt}\par} 51 | \newtcolorbox{answerbox}{colback=white, colframe=white, boxrule=0pt, left=0pt, right=0pt, top=0pt, bottom=0pt, boxsep=0pt, before=\Needspace{4\baselineskip}, before skip=2pt, after skip=8pt} 52 | \newtcolorbox{insightbox}{enhanced, breakable, title=\textbf{Key Insight}, colback=blue!2, colframe=blue!40, colbacktitle=blue!12, coltitle=black, fonttitle=\bfseries, left=8pt, right=8pt, top=5pt, bottom=6pt, before upper={\setlength{\parskip}{4pt}}, before skip=8pt, after skip=10pt} 53 | \newcommand{\finalanswer}[1]{\textbf{ANSWER:}~#1} 54 | % Category label: print at most once per problem; support multi-tags via commas 55 | \makeatletter 56 | \newif\if@categoryprinted 57 | \pretocmd{\item}{\global\@categoryprintedfalse}{}{} 58 | % Keep category internals within makeatletter to avoid stray text output 59 | \newcommand{\category}[1]{\if@categoryprinted\relax\else\textit{\textcolor{gray}{Category: #1}}\global\@categoryprintedtrue\fi} 60 | \makeatother 61 | \newcommand{\categories}[1]{\category{#1}} 62 | \newcommand{\tags}[1]{} 63 | 64 | % Ensure each top-level problem item keeps the full problem group together when possible 65 | \makeatletter 66 | \pretocmd{\item}{\ifnum\@enumdepth=1 \Needspace{24\baselineskip}\fi}{}{} 67 | \makeatother 68 | 69 | \title{Physics Bowl selected Problem Set (39 Problems)} 70 | \author{Ray Zhou} 71 | \date{\today} 72 | 73 | \begin{document} 74 | \begin{CJK*}{UTF8}{gbsn} 75 | 76 | \maketitle 77 | 78 | % Global conventions for numerical consistency and notation 79 | \noindent\textit{Conventions:} Unless otherwise stated, use \(g=10\,\mathrm{m/s^2}\) and denote electromotive force by \(\varepsilon\).\par\smallskip 80 | \newpage 81 | 82 | \begin{enumerate}[itemsep=1.0em, topsep=0.6em] 83 | 84 | % Problem 1 85 | \item \label{prob:1} 86 | Two identical mass objects are launched with the same speed from the same starting location. Object 1 is launched at an angle of $30^\circ$ above the horizontal while Object 2 is launched at an angle of $60^\circ$ above the horizontal. Ignore air resistance and consider the flight of each object from launch until it returns to the same launch height above the ground. Which object experiences the greatest change in the linear momentum? 87 | \begin{enumerate}[label=(\Alph*)] 88 | \item Object 1 since it has a higher final speed. 89 | \item Object 2 since it has a higher final speed. 90 | \item Object 2 since it is in the air for a longer time. 91 | \item The change in momentum is the same for each. 92 | \item It cannot be determined without more information. 93 | \end{enumerate} 94 | 95 | \category{Kinematics \& Momentum} 96 | \begin{answerbox} 97 | \finalanswer{(C) Object 2 since it is in the air for a longer time.} 98 | \end{answerbox} 99 | \begin{insightbox} 100 | Impulse $J=\int F\,dt=mgT$ (only weight acts). For the same launch speed $v_0$, the time in the air is $T=\tfrac{2v_0\sin\theta}{g}$, so $|\Delta\vec p|=mgT=2mv_0\sin\theta$ increases with $\sin\theta$. 101 | \end{insightbox} 102 | \begin{solutionbox} 103 | The impulse $\vec{J}$ provides the change in linear momentum $\Delta \vec{p}$. The net external force is gravity, $\vec{F}_{\text{net}} = m\vec{g}$. 104 | \[ 105 | \Delta \vec{p} = \vec{J} = \int_{0}^{T} \vec{F}_{\text{net}} \, dt = m\vec{g}T 106 | \] 107 | The magnitude of the momentum change is $|\Delta \vec{p}| = mgT$, where $T$ is the total flight time. 108 | 109 | The flight time of a projectile that returns to its firing height is: 110 | \[ 111 | T = \frac{2v_{0y}}{g} = \frac{2v_0 \sin\theta}{g} 112 | \] 113 | 114 | For Object 1, with $\theta_1 = 30^{\circ}$: 115 | \[ 116 | T_1 = \frac{2v_0 \sin(30^{\circ})}{g} = \frac{2v_0(1/2)}{g} = \frac{v_0}{g} 117 | \] 118 | 119 | For Object 2, with $\theta_2 = 60^{\circ}$: 120 | \[ 121 | T_2 = \frac{2v_0 \sin(60^{\circ})}{g} = \frac{2v_0(\sqrt{3}/2)}{g} = \frac{\sqrt{3}v_0}{g} 122 | \] 123 | 124 | Since $\sin(60^{\circ}) > \sin(30^{\circ})$, it follows that $T_2 > T_1$. Object 2 is in the air for a longer time, and therefore experiences a greater change in linear momentum. 125 | \\ 126 | \emph{Key point:} $\Delta p=\int F\,dt=mgT$ (impulse). Since both start and end at the same height, $T$ is larger for the $60^\circ$ launch, hence $|\Delta \vec p|$ is larger. 127 | \end{solutionbox} 128 | 129 | \par\smallskip\textit{Alternative.} At the same launch and landing height (no air drag), the speed magnitudes are equal. With initial velocity $\vec v=(v\cos\theta, v\sin\theta)$ and final velocity $\vec v'=(v\cos\theta, -v\sin\theta)$, the momentum change is $\Delta\vec p=m(\vec v'-\vec v)=(0,-2mv\sin\theta)$, so $|\Delta\vec p|=2mv\sin\theta$, which increases with $\theta$; thus the $60^\circ$ launch yields a larger change than the $30^\circ$ launch. 130 | 131 | % Problem 2 132 | \item \label{prob:2} 133 | A mass $m$ is pulled outward until the string of length $L$ to which it is attached makes a 90-degree angle with the vertical. The mass is released from rest and swings through a circular arc. What is the tension in the string when the mass swings through the bottom of the arc? 134 | \begin{enumerate}[label=(\Alph*)] 135 | \item 0 136 | \item $mg$ 137 | \item $2mg$ 138 | \item $3mg$ 139 | \item It cannot be determined. 140 | \end{enumerate} 141 | 142 | \category{Circular Motion \& Energy} 143 | \begin{answerbox} 144 | \finalanswer{(D) 3mg} 145 | \end{answerbox} 146 | \begin{insightbox} 147 | Use energy to get $v^2=2gL$ at the bottom, then $T-mg=mv^2/L$. 148 | \end{insightbox} 149 | \begin{solutionbox} 150 | By conservation of energy from the side to the bottom: 151 | \begin{align*} 152 | E_i &= K_i + U_i = 0 + mgL = mgL \\ 153 | E_f &= K_f + U_f = \tfrac{1}{2}mv^2 + 0 154 | \end{align*} 155 | Thus $mgL = \tfrac{1}{2}mv^2 \Rightarrow v^2 = 2gL$. At the bottom, radial (inward from the pivot) force balance yields 156 | \[ 157 | T - mg = \frac{mv^2}{L} = \frac{m(2gL)}{L} = 2mg \Rightarrow T = 3mg. 158 | \] 159 | \\ 160 | \emph{Used:} conservation of mechanical energy and $F_c=mv^2/r$. 161 | \end{solutionbox} 162 | 163 | \newpage 164 | 165 | % Problem 3 166 | \item \label{prob:3} 167 | \noindent\begin{minipage}[t]{0.6\linewidth} 168 | \vspace{0pt} % Ensures top alignment 169 | A resonance occurs with a tuning fork and an air column of size 39 cm. The next highest resonance occurs with an air column of 65 cm. What is the frequency of the tuning fork? Assume that the speed of sound is 343 m/s. 170 | \begin{enumerate}[label=(\Alph*)] 171 | \item 329.8 Hz 172 | \item 527.7 Hz 173 | \item 659.6 Hz 174 | \item 879.5 Hz 175 | \item 1319 Hz 176 | \end{enumerate} 177 | \end{minipage}% 178 | \hfill 179 | \begin{minipage}[t]{0.33\linewidth} 180 | \vspace{0pt} % Ensures top alignment 181 | \centering 182 | \includegraphics[width=0.7\linewidth]{Problem_03_Figure.png} 183 | \end{minipage} 184 | 185 | \category{Waves \& Sound} 186 | \begin{answerbox} 187 | \finalanswer{(C) 659.6 Hz} 188 | \end{answerbox} 189 | \begin{insightbox} 190 | Closed-tube resonances are spaced by $\lambda/2$; determine $\lambda$ from the difference in lengths and compute $f=v/\lambda$. 191 | \end{insightbox} 192 | \begin{solutionbox} 193 | 194 | For a closed tube, successive resonances differ by $\Delta L = \tfrac{\lambda}{2}$. In this case $\Delta L = 0.65\,\text{m} - 0.39\,\text{m} = 0.26\,\text{m}$, so $\lambda = 2\times 0.26 = 0.52\,\text{m}$. The frequency is 195 | \[ 196 | f = \frac{v}{\lambda} = \frac{343\,\text{m/s}}{0.52\,\text{m}} \approx 659.6\,\text{Hz}. 197 | \] 198 | \\ 199 | \emph{Model:} closed–open tube has harmonics $L=(2n-1)\lambda/4$, so successive $L$ differ by $\lambda/2$. 200 | \end{solutionbox} 201 | 202 | % Problem 4 203 | \item \label{prob:4} 204 | A mass of material exists in its solid form at its melting temperature $10^\circ$C. The following processes then occur to the material: 205 | \begin{itemize} 206 | \item Process 1: An amount of thermal energy $Q$ is added to the material and $\tfrac{3}{4}$ of the material melts. 207 | \item Process 2: An identical additional amount of thermal energy $Q$ is added to the material and the material is now a liquid at $50^\circ$C. 208 | \end{itemize} 209 | What is the ratio of the latent heat of fusion to the specific heat of the liquid for this material? 210 | \begin{enumerate}[label=(\Alph*)] 211 | \item $80^\circ$C 212 | \item $60^\circ$C 213 | \item $40^\circ$C 214 | \item $20^\circ$C 215 | \item More information is needed to answer this question. 216 | \end{enumerate} 217 | 218 | \category{Thermodynamics \& Phase Change} 219 | \begin{answerbox} 220 | \finalanswer{(A) 80$^{\circ}$C} 221 | \end{answerbox} 222 | \begin{insightbox} 223 | Two equal heat inputs: first melts 3/4, second completes melting 1/4 and warms; equate heats to find $L_f/c_l$. 224 | \end{insightbox} 225 | \begin{solutionbox} 226 | 227 | Let mass $m$, latent heat $L_f$, liquid specific heat $c_l$, melting temperature $T_m=10^{\circ}\text{C}$. Process 1: $Q=\tfrac{3}{4}mL_f$. Process 2: $Q=\tfrac{1}{4}mL_f + mc_l(50-10)$. Equate: 228 | \begin{align*} 229 | \tfrac{3}{4} mL_f &= \tfrac{1}{4} mL_f + 40 m c_l \\ 230 | \tfrac{1}{2} mL_f &= 40 m c_l \Rightarrow \frac{L_f}{c_l} = 80^{\circ}\text{C}. 231 | \end{align*} 232 | \end{solutionbox} 233 | 234 | \newpage 235 | 236 | % Problem 5 237 | \item \label{prob:5} 238 | \noindent\begin{minipage}[t]{0.6\linewidth} 239 | \vspace{0pt} 240 | What is the ideal mechanical advantage for the pulley system shown in the figure? 241 | \begin{enumerate}[label=(\Alph*)] 242 | \item $F/Mg$ 243 | \item $Mg/F$ 244 | \item 3 245 | \item 4 246 | \item 5 247 | \end{enumerate} 248 | \end{minipage}% 249 | \hfill 250 | \begin{minipage}[t]{0.33\linewidth} 251 | \vspace{0pt} 252 | \centering 253 | \includegraphics[width=0.6\linewidth]{Problem_05_Figure.png} 254 | \end{minipage} 255 | 256 | \category{Simple Machines \& Statics} 257 | \begin{answerbox} 258 | \finalanswer{(D) 4} 259 | \end{answerbox} 260 | \begin{insightbox} 261 | Ideal mechanical advantage equals the number of load-supporting rope segments: 4. 262 | \end{insightbox} 263 | \begin{solutionbox} 264 | 265 | The ideal mechanical advantage equals the number of rope segments supporting the load. Here the movable block is supported by 4 segments, so $\text{IMA}=4$. 266 | \\ 267 | \emph{Reminder:} massless, frictionless rope $\Rightarrow$ equal tension in all supporting segments. 268 | \end{solutionbox} 269 | 270 | % Problem 6 271 | \item \label{prob:6} 272 | In a calorimeter, 20 grams of liquid water at $100^\circ$C is mixed with 50 grams of water vapor at $100^\circ$C. The system is allowed to come to equilibrium. Assuming that the calorimeter and the surroundings can be ignored, which of the following best describes the net energy exchange between the vapor and the liquid during the process of coming to equilibrium? 273 | \begin{enumerate}[label=(\Alph*)] 274 | \item There is no net energy exchange. 275 | \item Energy is transferred from the vapor to the liquid, vaporizing some of the liquid. 276 | \item Energy is transferred from the vapor to the liquid, increasing the liquid’s temperature. 277 | \item Energy is transferred from the vapor to the liquid until all of the liquid vaporizes. 278 | \item Energy is transferred from the vapor to the liquid, condensing some of the vapor. 279 | \end{enumerate} 280 | 281 | \category{Thermodynamics \& Phase Equilibrium} 282 | \begin{answerbox} 283 | \finalanswer{(A) There is no net energy exchange.} 284 | \end{answerbox} 285 | \begin{insightbox} 286 | At 100\textcelsius\ both phases are at the same temperature. In an isolated system the mixture stays at 100\textcelsius; any condensation of vapor must be balanced by vaporization of the same mass of liquid, so the net energy transferred between phases is zero. 287 | \end{insightbox} 288 | \begin{solutionbox} 289 | 290 | As both steam and water are initially at the saturation temperature 100°C and the container is isolated (no heat to the surroundings), the final temperature is still 100°C. Let a mass $m_c$ of vapor condense (releasing $m_cL_v$) and a mass $m_v$ of liquid evaporate (absorbing $m_vL_v$). Conservation of energy for the constant-temperature closed system requires $m_cL_v=m_vL_v\Rightarrow m_c=m_v$. Therefore, the energy transferred from vapor to liquid equals the energy transferred from liquid to vapor, and the net energy transfer between the two phases is zero. 291 | \end{solutionbox} 292 | 293 | \newpage 294 | 295 | % Problem 7 296 | \item \label{prob:7} 297 | \noindent\begin{minipage}[t]{0.6\linewidth} 298 | \vspace{0pt} 299 | For the circuit shown, $\varepsilon = 6.0$ V, $R_1 = 7.0 \Omega$, $R_2 = 3.0 \Omega$, $R_3 = 6.0 \Omega$, and $R_4 = 12.0 \Omega$. After operating for a long time, the circuit reaches steady state (the capacitor branch is effectively open). What is the voltage across the capacitor at steady state? 300 | \begin{enumerate}[label=(\Alph*)] 301 | \item 6.0 V 302 | \item 4.2 V 303 | \item 3.0 V 304 | \item 2.2 V 305 | \item 0.2 V 306 | \end{enumerate} 307 | \end{minipage}% 308 | \hfill 309 | \begin{minipage}[t]{0.33\linewidth} 310 | \vspace{0pt} 311 | \centering 312 | \includegraphics[width=\linewidth]{Problem_07_Figure.png} 313 | \end{minipage} 314 | 315 | \category{DC Circuits} 316 | \begin{answerbox} 317 | \finalanswer{(D) 2.2 V} 318 | \end{answerbox} 319 | \begin{insightbox} 320 | At steady state the capacitor branch is open; find node voltages by dividers and subtract them. 321 | \end{insightbox} 322 | \begin{solutionbox} 323 | 324 | After a long time, the capacitor branch is open. Two independent dividers yield $V_A = 6\,\text{V}\cdot\tfrac{3}{7+3} = 1.8\,\text{V}$ and $V_B = 6\,\text{V}\cdot\tfrac{12}{6+12} = 4.0\,\text{V}$. Thus $V_C = |V_B-V_A| = 2.2\,\text{V}$. 325 | \par\smallskip Equivalent node-voltage viewpoint: with the capacitor open, KCL at the top nodes gives the same divider values; the capacitor voltage is the node difference $|V_B-V_A|$. 326 | \end{solutionbox} 327 | 328 | % Problem 8 329 | \item \label{prob:8} 330 | \noindent\begin{minipage}[t]{0.55\linewidth} 331 | \vspace{0pt} 332 | An electron moves in the plane of the page through two regions of space along the dotted-line trajectory shown in the figure. There is a uniform electric field in Region I directed into the plane of the page. There is no electric field in Region II. What is a necessary direction of the magnetic field in regions I and II? Ignore gravitational forces. 333 | \begin{enumerate}[label=(\Alph*)] 334 | \item Region I: Downward along the page, Region II: Upward along the page 335 | \item Region I: Upward along the page, Region II: Into the page 336 | \item Region I: Upward along the page, Region II: Out of the page 337 | \item Region I: Downward along the page, Region II: Out of the page 338 | \item Region I: Into the page, Region II: Upward along the page 339 | \end{enumerate} 340 | \end{minipage}% 341 | \hfill 342 | \begin{minipage}[t]{0.4\linewidth} 343 | \vspace{0pt} 344 | \centering 345 | \includegraphics[width=\linewidth]{Problem_08_Figure.png} 346 | \end{minipage} 347 | 348 | \category{Electricity \& Magnetism} 349 | \begin{answerbox} 350 | \finalanswer{(C) Region I: Upward along the page, Region II: Out of the page} 351 | \end{answerbox} 352 | \begin{insightbox} 353 | Use $\vec F_B=q\,\vec v\times\vec B$ and the negative charge of an electron to choose the $\vec B$ directions that curve the trajectory as shown. 354 | \end{insightbox} 355 | \begin{solutionbox} 356 | 357 | In Region I, $\vec F_E$ on the electron is out of the page, so $\vec F_B$ must be into the page; with $q<0$, this requires $\vec v\times\vec B$ out of the page, hence $\vec B$ upward along the page. In Region II, curvature upward along the page implies $\vec F_B$ upward along the page. With $q<0$ and $\vec v$ rightward, $\vec B$ is out of the page. 358 | \end{solutionbox} 359 | 360 | \newpage 361 | 362 | % Problem 9 363 | \item \label{prob:9} 364 | \noindent\begin{minipage}[t]{0.6\linewidth} 365 | \vspace{0pt} 366 | For the circuit shown, when a shorting wire (no resistance) connects the points labeled A and B, which of the numbered light bulbs become brighter? Assume that all four bulbs are identical and have resistance R. 367 | \begin{enumerate}[label=(\Alph*)] 368 | \item Bulb 1 only 369 | \item Bulb 2 only 370 | \item Bulb 3 only 371 | \item Bulbs 1 and 3 only 372 | \item Bulbs 1, 2, and 3 373 | \end{enumerate} 374 | \end{minipage}% 375 | \hfill 376 | \begin{minipage}[t]{0.33\linewidth} 377 | \vspace{0pt} 378 | \centering 379 | \includegraphics[width=\linewidth]{Problem_09_Figure.png} 380 | \end{minipage} 381 | 382 | \category{DC Circuits} \tags{} 383 | \begin{answerbox} 384 | \finalanswer{(D) Bulbs 1 and 3 only} 385 | \end{answerbox} 386 | \begin{insightbox} 387 | Shorting the bridge's balance point lowers the downstream branch resistances, increasing current through bulbs 1 and 3. 388 | \end{insightbox} 389 | \begin{solutionbox} 390 | 391 | Label the left node at the bottom of the source A and the right mid node B (see figure). Bulbs 1 and 3 are on the top branches left and right of the central vertical branch containing bulb 2; bulb 4 is on the right outer branch. 392 | 393 | Before the short: the bridge is balanced, so the potential of the upper junction of bulb 2 is the same as that of B; there is thus no current through bulb 2 (balanced-bridge condition), and bulbs 1 and 3 (which are in series with the external right branch which contains bulb 4) share the supply drop. 394 | 395 | After shorting A and B: the vertical short brings the potential difference between A and B to zero, forcing the middle node and the right inner node to ground potential. So, bulb 2 is short-circuited and carries no current, and bulb 4 is also short-circuited by the B to A short. The remaining conducting paths are the left top branch with bulb 1 directly from the positive terminal to ground and the right top branch with bulb 3 directly to ground via the shorted lower rail. Bulbs 1 and 3 now each have the full source voltage across them instead of sharing this with other elements in series. So bulbs 1 and 3 brighten, and bulb 2 goes dark and bulb 4 goes dark. 396 | \end{solutionbox} 397 | 398 | % Problem 10 399 | \item \label{prob:10} 400 | A car moves to the right along a one-dimensional track for total time $T$ in two parts. Part One: The car maintains constant non-zero speed $V$ for the first $\tfrac{3}{4}$ of the total time. Part Two: The car accelerates uniformly to rest during the last $\tfrac{1}{4}$ of the total time. What is the ratio of the distance traveled during Part One of the trip to the distance traveled during Part Two of the trip? 401 | \begin{enumerate}[label=(\Alph*)] 402 | \item 6:1 403 | \item 3:2 404 | \item The values of V and T are required to answer the question. 405 | \item 4:3 406 | \item 8:3 407 | \end{enumerate} 408 | 409 | \category{Kinematics} \tags{} 410 | \begin{answerbox} 411 | \finalanswer{(A) 6:1} 412 | \end{answerbox} 413 | \begin{insightbox} 414 | Distance 1: $V\cdot 3T/4$; distance 2: $(V/2)\cdot T/4$. Ratio is 6:1. 415 | \end{insightbox} 416 | \begin{solutionbox} 417 | 418 | Part one: $d_1=V\cdot (3T/4)=\tfrac{3}{4}VT$. Part two (uniform decel from $V$ to 0 over $T/4$): $\bar v = V/2$, so $d_2=\bar v\cdot (T/4)=\tfrac{1}{8}VT$. So $d_1:d_2=6:1$. 419 | \end{solutionbox} 420 | 421 | \newpage 422 | 423 | % Problem 11 424 | \item \label{prob:11} 425 | \noindent\begin{minipage}[t]{0.6\linewidth} 426 | \vspace{0pt} 427 | A piece of an ideal fluid is marked as it moves along a horizontal streamline through a pipe, as shown in the figure. In Region I, the speed of the fluid on the streamline is $V$. The cylindrical, horizontal pipe narrows so that the radius of the pipe in Region II is half of what it was in Region I. What is the speed of the marked fluid when it is in Region II? 428 | \begin{enumerate}[label=(\Alph*)] 429 | \item 4$V$ 430 | \item 2$V$ 431 | \item $V$ 432 | \item $V/2$ 433 | \item $V/4$ 434 | \end{enumerate} 435 | \end{minipage}% 436 | \hfill 437 | \begin{minipage}[t]{0.35\linewidth} 438 | \vspace{0pt} 439 | \centering 440 | \includegraphics[width=\linewidth]{Problem_11_Figure.png} 441 | \end{minipage} 442 | 443 | \category{Fluid Mechanics} \tags{} 444 | \begin{answerbox} 445 | \finalanswer{(A) 4V} 446 | \end{answerbox} 447 | \begin{insightbox} 448 | Continuity $A_1v_1=A_2v_2$: halving the radius quarters the area, so the speed quadruples. 449 | \end{insightbox} 450 | \begin{solutionbox} 451 | 452 | Continuity: $A_1 v_1 = A_2 v_2$ with $A\propto r^2$. If $r_2=\tfrac{1}{2}r_1$, then $A_2=\tfrac{1}{4}A_1$, so $v_2 = (A_1/A_2) v_1 = 4V$. 453 | \end{solutionbox} 454 | 455 | % Problem 12 456 | \item \label{prob:12} 457 | \noindent\begin{minipage}[t]{0.6\linewidth} 458 | \vspace{0pt} 459 | For the RC circuit shown, the resistance is $R = 10.0 \Omega$, the capacitance is $C = 5.0\,\mu\mathrm{F}$ and the battery has voltage $\varepsilon = 12$ volts. The capacitor is initially uncharged when the switch S is closed at time $t = 0$. At some time later, the current in the circuit is $0.50$ A. What is the magnitude of the voltage across the capacitor at that moment? 460 | \begin{enumerate}[label=(\Alph*)] 461 | \item 0 volts 462 | \item 5 volts 463 | \item 6 volts 464 | \item 7 volts 465 | \item 12 volts 466 | \end{enumerate} 467 | \end{minipage}% 468 | \hfill 469 | \begin{minipage}[t]{0.35\linewidth} 470 | \vspace{0pt} 471 | \centering 472 | \includegraphics[width=\linewidth]{Problem_12_Figure.png} 473 | \end{minipage} 474 | 475 | \category{RC Circuits} 476 | \begin{answerbox} 477 | \finalanswer{(D) 7 volts} 478 | \end{answerbox} 479 | \begin{insightbox} 480 | KVL: $\varepsilon=V_R+V_C$. With $I$ known, find $V_R=IR$ then $V_C$. 481 | \end{insightbox} 482 | \begin{solutionbox} 483 | 484 | KVL: $\varepsilon - V_R - V_C = 0$. With $I=0.50\,\text{A}$ and $R=10.0\,\Omega$, $V_R=IR=5.0\,\text{V}$. Hence $V_C=\varepsilon - V_R = 12-5=7\,\text{V}$. 485 | \end{solutionbox} 486 | 487 | \newpage 488 | 489 | % Problem 13 490 | \item \label{prob:13} 491 | A ball initially at rest falls without air resistance from a height $h$ above the ground. If the ball falls the first distance $h/2$ in a time $t$, how much time is required to fall the remaining distance of $h/2$? 492 | \begin{enumerate}[label=(\Alph*)] 493 | \item 0.25$t$ 494 | \item 0.41$t$ 495 | \item 0.50$t$ 496 | \item 0.71$t$ 497 | \item 1.00$t$ 498 | \end{enumerate} 499 | 500 | \category{Kinematics} 501 | \begin{answerbox} 502 | \finalanswer{(B) 0.41t} 503 | \end{answerbox} 504 | \begin{insightbox} 505 | First half gives $h=gt^2$; total time is $\sqrt2 t$, so the remainder is $(\sqrt2-1)t\approx0.414t$. 506 | \end{insightbox} 507 | \begin{solutionbox} 508 | 509 | With $d=\tfrac{1}{2}gt^2$, first half: $h/2=\tfrac{1}{2}gt^2\Rightarrow h=gt^2$. Total time $t_{\text{tot}}$ satisfies $h=\tfrac{1}{2}g t_{\text{tot}}^2\Rightarrow t_{\text{tot}}=\sqrt{2}t$. Remaining time $t_{\text{rem}}=t_{\text{tot}}-t=(\sqrt{2}-1)t\approx 0.414t$. 510 | \end{solutionbox} 511 | 512 | % Problem 14 513 | \item \label{prob:14} 514 | \noindent\begin{minipage}[t]{0.6\linewidth} 515 | \vspace{0pt} 516 | An object of mass M starts from rest at the bottom of a fixed incline of height H. A person decides to push the object up the incline in one of two ways with an applied force shown in the diagram. In each of the trials, the object reaches the top of the incline with speed $V$. How would the work done by the person on the block compare for the two trials? Assume the same constant non-zero coefficient of kinetic friction. 517 | \begin{enumerate}[label=(\Alph*)] 518 | \item More work would be done in Trial 1 519 | \item More work would be done in Trial 2 520 | \item The work would be equal for both trials 521 | \item Impossible to determine without knowing $V$. 522 | \item Impossible to determine without knowing $\mu_k$. 523 | \end{enumerate} 524 | \end{minipage}% 525 | \hfill 526 | \begin{minipage}[t]{0.35\linewidth} 527 | \vspace{0pt} 528 | \centering 529 | \includegraphics[width=\linewidth]{Problem_14_Figure.png} 530 | \end{minipage} 531 | 532 | \category{Work \& Energy with Friction} \tags{} 533 | \begin{answerbox} 534 | \finalanswer{(A) More work would be done in Trial 1} 535 | \end{answerbox} 536 | \begin{insightbox} 537 | A horizontal push raises the normal force and friction loss; since $\Delta K$ and $W_g$ match, external work must be larger. 538 | \end{insightbox} 539 | \begin{solutionbox} 540 | 541 | Horizontal push increases the normal force: $N_1=Mg\cos\theta+F_1\sin\theta>Mg\cos\theta=N_2$, so friction is larger in Trial 1. Since $\Delta K$ and $W_g$ are the same, the greater friction loss requires more input work in Trial 1. 542 | \end{solutionbox} 543 | 544 | \newpage 545 | 546 | % Problem 15 547 | \item \label{prob:15} 548 | \noindent\begin{minipage}[t]{0.6\linewidth} 549 | \vspace{0pt} 550 | A point object with mass $M=2.0$kg is attached a distance $R=1.75$m from the fixed center of a disk. The disk starts rotating from rest with constant angular acceleration $\alpha = 5.00$ rad/s$^2$. After how much time $T$ (in seconds) is the tangential acceleration equal in magnitude to the centripetal acceleration? 551 | \begin{enumerate}[label=(\Alph*)] 552 | \item 0.769 553 | \item 0.592 554 | \item 0.500 555 | \item 0.447 556 | \item 0.350 557 | \end{enumerate} 558 | \end{minipage}% 559 | \hfill 560 | \begin{minipage}[t]{0.35\linewidth} 561 | \vspace{0pt} 562 | \centering 563 | \includegraphics[width=\linewidth]{Problem_15_Figure.png} 564 | \end{minipage} 565 | 566 | \category{Rotational Kinematics} \tags{} 567 | \begin{answerbox} 568 | \finalanswer{(D) 0.447} 569 | \end{answerbox} 570 | \begin{insightbox} 571 | Set $a_t=R\alpha$ equal to $a_c=R(\alpha T)^2$ to get $T=1/\sqrt{\alpha}$. 572 | \end{insightbox} 573 | \begin{solutionbox} 574 | 575 | Set $a_t=R\alpha$ equal to $a_c=R\omega^2=R(\alpha T)^2$. Then $R\alpha=R\alpha^2 T^2\Rightarrow T=1/\sqrt{\alpha}=1/\sqrt{5}\approx0.447\,\text{s}$. 576 | \end{solutionbox} 577 | 578 | % Problem 16 579 | \item \label{prob:16} 580 | \noindent\begin{minipage}[t]{0.6\linewidth} 581 | \vspace{0pt} 582 | A uniform, solid cylinder with a mass M and radius R is pulled by a horizontal force F acting through the center as shown. The cylinder rolls to the right without slipping. What is the magnitude of the force of friction between the cylinder and the ground? 583 | \begin{enumerate}[label=(\Alph*)] 584 | \item $\frac{1}{4}F$ 585 | \item $\frac{1}{3}F$ 586 | \item $\frac{1}{2}F$ 587 | \item $\frac{2}{3}F$ 588 | \item $\frac{3}{4}F$ 589 | \end{enumerate} 590 | \end{minipage}% 591 | \hfill 592 | \begin{minipage}[t]{0.35\linewidth} 593 | \vspace{0pt} 594 | \centering 595 | \includegraphics[width=\linewidth]{Problem_16_Figure.png} 596 | \end{minipage} 597 | 598 | \category{Rolling Dynamics} \tags{} 599 | \begin{answerbox} 600 | \finalanswer{(B) $\tfrac{1}{3}F$} 601 | \end{answerbox} 602 | \begin{insightbox} 603 | Combine translation $F-f=Ma$ with rotation $fR=I\alpha$ and $a=R\alpha$ to obtain $f=F/3$. 604 | \end{insightbox} 605 | \begin{solutionbox} 606 | 607 | Translational: $F-f=Ma$. Rotational: $fR=I\alpha$ with $I=\tfrac{1}{2}MR^2$ and $a=R\alpha$. Then $f=\tfrac{1}{2}Ma$. Sub into translation: $F-f=2f\Rightarrow F=3f\Rightarrow f=F/3$. 608 | \end{solutionbox} 609 | 610 | \newpage 611 | 612 | % Problem 17 613 | \item \label{prob:17} 614 | A comet moves in an elliptical orbit around the sun. As the comet moves from aphelion (farthest point) to perihelion (closest point), which of the following results is true? 615 | \begin{enumerate} 616 | \item Speed of the comet: Increases, Angular momentum of the comet/sun system: Increases, Gravitational potential energy of the comet/sun system: Decreases 617 | \item Speed of the comet: Increases, Angular momentum of the comet/sun system: Constant, Gravitational potential energy of the comet/sun system: Decreases 618 | \item Speed of the comet: Decreases, Angular momentum of the comet/sun system: Decreases, Gravitational potential energy of the comet/sun system: Increases 619 | \item Speed of the comet: Increases, Angular momentum of the comet/sun system: Increases, Gravitational potential energy of the comet/sun system: Constant 620 | \item Speed of the comet: Constant, Angular momentum of the comet/sun system: Constant, Gravitational potential energy of the comet/sun system: Constant 621 | \end{enumerate} 622 | 623 | \category{Gravitation} \tags{} 624 | \begin{answerbox} 625 | \finalanswer{(B) Increases, Constant, Decreases} 626 | \end{answerbox} 627 | \begin{insightbox} 628 | Zero torque about the sun conserves angular momentum; as $r$ decreases, $U=-GMm/r$ drops and speed rises. 629 | \end{insightbox} 630 | \begin{solutionbox} 631 | 632 | Central gravity gives zero torque, so angular momentum is conserved. As $r$ decreases, $U=-GMm/r$ decreases, hence kinetic energy and speed increase. 633 | \end{solutionbox} 634 | 635 | % Problem 18 636 | \item \label{prob:18} 637 | \noindent\begin{minipage}[t]{0.6\linewidth} 638 | \vspace{0pt} 639 | An open cylindrical container with very large radius is at rest a distance $H$ above the ground at the edge of a platform. A tiny hole develops at the bottom of the container and water from the container squirts out horizontally landing a distance $H$ from the edge of the platform. For the water to land at this location, what is the depth of the water $L$ in the container? The figure is not drawn to scale and air resistance is ignored. 640 | \begin{enumerate}[label=(\Alph*)] 641 | \item $H/4$ 642 | \item $H/\sqrt{2}$ 643 | \item $H$ 644 | \item $H\sqrt{2}$ 645 | \item $2H$ 646 | \end{enumerate} 647 | \end{minipage}% 648 | \hfill 649 | \begin{minipage}[t]{0.32\linewidth} 650 | \vspace{0pt} 651 | \centering 652 | \includegraphics[width=\linewidth]{Problem_18_Figure.png} 653 | \end{minipage} 654 | 655 | \category{Fluid Mechanics \& Projectile Motion} \tags{} 656 | \begin{answerbox} 657 | \finalanswer{(A) $H/4$} 658 | \end{answerbox} 659 | \begin{insightbox} 660 | Torricelli gives $v=\sqrt{2gL}$; range $H$ equals $v$ times fall time $\sqrt{2H/g}$. 661 | \end{insightbox} 662 | \begin{solutionbox} 663 | 664 | Torricelli: $v=\sqrt{2gL}$. Time of fall from height $H$ is $t=\sqrt{2H/g}$. Range $H=v t=\sqrt{2gL}\,\sqrt{2H/g}=\sqrt{4LH}$. Hence $H^2=4LH\Rightarrow L=H/4$. 665 | \end{solutionbox} 666 | 667 | \newpage 668 | 669 | % Problem 19 670 | \item \label{prob:19} 671 | \noindent\begin{minipage}[t]{0.6\linewidth} 672 | \vspace{0pt} 673 | A monatomic ideal gas is the working substance for an engine that undergoes the cyclic process (ABCDA) shown in the PV diagram. The processes are all isochoric or isobaric with pressures between $P_0$ and $2P_0$ and volumes between $V_0$ and $\frac{3}{2}V_0$. What is the efficiency of this engine? 674 | \begin{enumerate}[label=(\Alph*)] 675 | \item $1/8$ 676 | \item $1/5$ 677 | \item $1/3$ 678 | \item $2/3$ 679 | \item $5/7$ 680 | \end{enumerate} 681 | \end{minipage}% 682 | \hfill 683 | \begin{minipage}[t]{0.32\linewidth} 684 | \vspace{0pt} 685 | \centering 686 | \includegraphics[width=\linewidth]{Problem_19_Figure.png} 687 | \end{minipage} 688 | 689 | \category{Thermodynamics \& Engines} \tags{} 690 | \begin{answerbox} 691 | \finalanswer{(A) 1/8} 692 | \end{answerbox} 693 | \begin{insightbox} 694 | Compute heat on isobaric/isochoric legs; net work is rectangle area $\Delta P\,\Delta V$; efficiency is $W/Q_{in}$. Sign guide: isobaric expansion (heat in, $W>0$), isochoric heating (heat in, $W=0$), isobaric compression (heat out, $W<0$), isochoric cooling (heat out, $W=0$). 695 | \end{insightbox} 696 | \begin{solutionbox} 697 | 698 | Net work: $W=\Delta P\,\Delta V= (2P_0-P_0)\,(\tfrac{3}{2}V_0-V_0)=\tfrac{1}{2}P_0V_0$. Heat in during AB and BC: $Q_{in}=\tfrac{3}{2}P_0V_0+\tfrac{5}{2}P_0V_0=4P_0V_0$. Efficiency $\eta=W/Q_{in}=1/8$. 699 | \end{solutionbox} 700 | 701 | % Problem 20 702 | \item \label{prob:20} 703 | \noindent\begin{minipage}[t]{0.6\linewidth} 704 | \vspace{0pt} 705 | A solid, uniform sphere rolls without slipping on a floor along the +x-axis. The rotational kinetic energy associated with the sphere about an axis through its center of mass along the +z-axis is 20 Joules. What is the translational kinetic energy associated with the sphere? 706 | \begin{enumerate}[label=(\Alph*)] 707 | \item 8 J 708 | \item 10 J 709 | \item 20 J 710 | \item 40 J 711 | \item 50 J 712 | \end{enumerate} 713 | \end{minipage}% 714 | \hfill 715 | \begin{minipage}[t]{0.32\linewidth} 716 | \vspace{0pt} 717 | \centering 718 | \includegraphics[width=\linewidth]{Problem_20_Figure.png} 719 | \end{minipage} 720 | 721 | \category{Rolling Energy} \tags{} 722 | \begin{answerbox} 723 | \finalanswer{(E) 50 J} 724 | \end{answerbox} 725 | \begin{insightbox} 726 | With pure rolling $v=R\omega$, a solid sphere has $K_{rot}=\tfrac{1}{5}mv^2$, so $K_{trans}=(5/2)K_{rot}$. 727 | \end{insightbox} 728 | \begin{solutionbox} 729 | 730 | For a solid sphere, $I=\tfrac{2}{5}mR^2$ and $v=R\omega$. Then $K_{rot}=\tfrac{1}{2}I\omega^2=\tfrac{1}{5}mv^2$. Meanwhile $K_{trans}=\tfrac{1}{2}mv^2=\tfrac{5}{2}K_{rot}=50\,\text{J}$. 731 | \end{solutionbox} 732 | 733 | \newpage 734 | 735 | % Problem 21 736 | \item \label{prob:21} 737 | A student wants to set up an experiment with a thin convex lens of focal length $f$ such that a thin real object produces a focused real image on a movable screen. At how many locations along the optical axis can the object be placed so that the distance between the object and the focused image on the screen is equal to $3f$? 738 | \begin{enumerate}[label=(\Alph*)] 739 | \item There is no location. 740 | \item There is exactly one location. 741 | \item There are exactly two locations. 742 | \item There are exactly four locations. 743 | \item There are an infinite number of locations. 744 | \end{enumerate} 745 | 746 | \category{Geometric Optics} 747 | \begin{answerbox} 748 | \finalanswer{(A) There is no location.} 749 | \end{answerbox} 750 | \begin{insightbox} 751 | Solve $1/d_o+1/d_i=1/f$ with $d_o+d_i=3f$; the quadratic has negative discriminant, so no real location. Only when the object–image separation $D\ge 4f$ can a real solution exist. 752 | \end{insightbox} 753 | \begin{solutionbox} 754 | 755 | With $\tfrac{1}{d_o}+\tfrac{1}{d_i}=\tfrac{1}{f}$ and $D=d_o+d_i=3f$, substitute $d_i=3f-d_o$: 756 | \[ 757 | \frac{1}{d_o}+\frac{1}{3f-d_o}=\frac{1}{f} \Rightarrow d_o^2-3fd_o+3f^2=0. 758 | \] 759 | Discriminant $\Delta=9f^2-12f^2=-3f^2<0$, so no real $d_o$ exists. 760 | \end{solutionbox} 761 | 762 | % Problem 22 763 | \item \label{prob:22} 764 | \noindent\begin{minipage}[t]{0.6\linewidth} 765 | \vspace{0pt} 766 | For the circuit shown, all wires have no resistance, the battery has a constant internal resistance of $r = 8.0\,\Omega$ and the two light bulbs are identical. The variable resistor is initially set to $R = 26.0\,\Omega$. The switch S is closed. To what resistance must the variable resistor be set if bulb \#1 is to have the same brightness after the switch is closed as it did with the switch open? 767 | \begin{enumerate}[label=(\Alph*)] 768 | \item $9.0\,\Omega$ 769 | \item $13.0\,\Omega$ 770 | \item $18.0\,\Omega$ 771 | \item $22.0\,\Omega$ 772 | \item The answer can be computed only if the bulbs’ resistance is known. 773 | \end{enumerate} 774 | \end{minipage}% 775 | \hfill 776 | \begin{minipage}[t]{0.32\linewidth} 777 | \vspace{0pt} 778 | \centering 779 | \includegraphics[width=\linewidth]{Problem_23_Figure.png} 780 | \end{minipage} 781 | 782 | \category{DC Circuits} 783 | \begin{answerbox} 784 | \finalanswer{(A) $9.0\,\Omega$} 785 | \end{answerbox} 786 | \begin{insightbox} 787 | Keep bulb 1 current unchanged; equate open/closed currents in the equivalent circuits to solve $R'$. 788 | \end{insightbox} 789 | \begin{solutionbox} 790 | 791 | Open: $I_{1,\text{open}}=\tfrac{\mathcal E}{r+R_b+R}$. Closed: bulbs in parallel $R_p=R_b/2$, total $r+R_b/2+R'$, current splits equally so $I_{1,\text{closed}}=\tfrac{\mathcal E}{2r+R_b+2R'}$. Equate: 792 | \[ 793 | r+R_b+R=2r+R_b+2R' \Rightarrow R' = \tfrac{R-r}{2} = \tfrac{26-8}{2}=9.0\,\Omega. 794 | \] 795 | \end{solutionbox} 796 | 797 | \newpage 798 | 799 | % Problem 23 800 | \item \label{prob:23} 801 | \noindent\begin{minipage}[t]{0.6\linewidth} 802 | \vspace{0pt} 803 | A mass on a frictionless incline has a gravitational force, a normal force, and an applied force that all are equal in magnitude. The mass remains at rest. The incline makes an angle $\theta$ with the horizontal. Which one of the following choices best describes the orientation of the applied force? The +x-axis is parallel to the incline’s surface. 804 | \begin{enumerate}[label=(\Alph*)] 805 | \item Oriented directly along the +x axis. 806 | \item Oriented at an angle $\theta$ clockwise from the +x axis. 807 | \item Oriented at an angle $90^\circ - \theta$ clockwise from the +x axis. 808 | \item Oriented at an angle $90^\circ - \theta$ counterclockwise from the +x axis. 809 | \item This is a completely impossible situation. 810 | \end{enumerate} 811 | \end{minipage}% 812 | \hfill 813 | \begin{minipage}[t]{0.32\linewidth} 814 | \vspace{0pt} 815 | \centering 816 | \includegraphics[width=\linewidth]{Problem_24_Figure.png} 817 | \end{minipage} 818 | 819 | \category{Statics} 820 | \begin{answerbox} 821 | \finalanswer{(C) Oriented at $90^\circ-\theta$ clockwise from +x} 822 | \end{answerbox} 823 | \begin{solutionbox} 824 | With $|\vec F_g|=|\vec N|=|\vec F_a|$, the forces form an equilateral triangle. Geometry of components shows the applied force must be directed $90^\circ-\theta$ clockwise from +x so that the vector sum is zero. 825 | \end{solutionbox} 826 | 827 | % Problem 24 828 | \item \label{prob:24} 829 | The position of a mass connected to a spring obeys $x(t) = A \cos(\omega t)$. What is the average speed of the mass for one full oscillation in terms of the mass’s maximum speed during oscillation, $v_{max}$? 830 | \begin{enumerate}[label=(\Alph*)] 831 | \item $\frac{2}{\pi} v_{max}$ 832 | \item $\frac{1}{\sqrt{2}} v_{max}$ 833 | \item $\frac{1}{2} v_{max}$ 834 | \item $\frac{\sqrt{2}}{\pi} v_{max}$ 835 | \item $\frac{1}{2\pi\sqrt{2}} v_{max}$ 836 | \end{enumerate} 837 | 838 | \category{Oscillations} 839 | \begin{answerbox} 840 | \finalanswer{(A) $\tfrac{2}{\pi}v_{max}$} 841 | \end{answerbox} 842 | \begin{insightbox} 843 | In one period distance is $4A$, $T=2\pi/\omega$, and $v_{max}=A\omega$; substitute to get $\bar s=(2/\pi)v_{max}$. 844 | \end{insightbox} 845 | \begin{solutionbox} 846 | 847 | During one period $T=2\pi/\omega$, distance covered is $4A$. With $v_{max}=A\omega$, the average speed is $\bar s=4A/T=2A\omega/\pi=\tfrac{2}{\pi}v_{max}$. 848 | \end{solutionbox} 849 | 850 | \newpage 851 | 852 | % Problem 25 853 | \item \label{prob:25} 854 | \noindent\begin{minipage}[t]{0.6\linewidth} 855 | \vspace{0pt} 856 | An upward-pointing object is placed 15.0 cm to the left of a lens system. The first lens is convex with focal length 10.0 cm. The second lens is convex with focal length 10 cm and its location from the first lens is varied from $x=10$ cm away to $x=110$ cm away. Which one of the following choices best represents the description of the final image formed as the second lens is moved from $x=10$ cm to $x=110$ cm from the first lens? 857 | \end{minipage}% 858 | \hfill 859 | \begin{minipage}[t]{0.32\linewidth} 860 | \vspace{0pt} 861 | \centering 862 | \includegraphics[width=\linewidth]{Problem_26_Figure.png} 863 | \end{minipage} 864 | 865 | \begin{enumerate}[label=(\Alph*)] 866 | \item $x=10$ cm away: Real \& pointing downward; $\quad x=110$ cm away: Real \& pointing upward 867 | \item $x=10$ cm away: Virtual \& pointing downward; $\quad x=110$ cm away: Real \& pointing upward 868 | \item $x=10$ cm away: Virtual \& pointing upward; $\quad x=110$ cm away: Real \& pointing downward 869 | \item $x=10$ cm away: Real \& pointing upward; $\quad x=110$ cm away: Real \& pointing upward 870 | \item $x=10$ cm away: Virtual \& pointing downward; $\quad x=110$ cm away: Virtual \& pointing downward 871 | \end{enumerate} 872 | 873 | \category{Geometric Optics} 874 | \begin{answerbox} 875 | \finalanswer{(A) Real \& pointing downward at $x=10$ cm; Real \& pointing upward at $x=110$ cm.} 876 | \end{answerbox} 877 | \begin{solutionbox} 878 | \textbf{Behavior vs. $x$:} The final image is real and inverted for $10\le x<30$, becomes virtual and inverted for $3040$ (at $x=40$ the object for lens 2 is at its focal plane and no image is formed at finite distance). 879 | 880 | Using $d_{i1}=30\,\text{cm}$ from lens 1, the second lens has $d_{o2}=x-30$. With $f_2=10$ cm, 881 | \[ d_{i2}=\frac{10(x-30)}{x-40}, \quad m_{\text{total}}=\frac{20}{x-40}. \] 882 | Signs of $d_{i2}$ and $m_{\text{total}}$ across intervals give the stated behavior. 883 | \end{solutionbox} 884 | 885 | % Problem 26 886 | \item \label{prob:26} 887 | Rain falls vertically at 12.0 m/s with respect to a stationary observer. A car moves so that, in the ground frame, the rain's horizontal component makes an angle of $40^\circ$ to the vertical. A passenger sitting in the car observes the rain making an angle of $20.0^\circ$ to the vertical. What is the car's speed with respect to the observer? 888 | \begin{enumerate}[label=(\Alph*)] 889 | \item 2.29 m/s 890 | \item 5.93 m/s 891 | \item 9.03 m/s 892 | \item 11.8 m/s 893 | \item 16.2 m/s 894 | \end{enumerate} 895 | 896 | \category{Relative Motion} 897 | \begin{answerbox} 898 | \finalanswer{(B) $\approx 5.70\,\text{m/s}$ (choice list shows $5.93\,\text{m/s}$ due to rounding)} 899 | \end{answerbox} 900 | \begin{solutionbox} 901 | 902 | Let the rain’s velocity in the ground frame be $\vec v_{r/g}=(v_x,-12)$ with $\tan 40^\circ=|v_x|/12 \Rightarrow v_x=12\tan 40^\circ$. If the car moves rightward at speed $v_c$, then the rain relative to the car is $\vec v_{r/c}=(v_x-v_c,-12)$. The observed $20^\circ$ from vertical implies $\tan 20^\circ=|v_{r/c,x}|/12$. Hence 903 | \[ 904 | v_c=v_x-12\tan 20^\circ=12(\tan 40^\circ-\tan 20^\circ)\approx5.93\,\text{m/s}. 905 | \] 906 | \end{solutionbox} 907 | 908 | \newpage 909 | 910 | % Problem 27 911 | \item \label{prob:27} 912 | \noindent\begin{minipage}[t]{0.6\linewidth} 913 | \vspace{0pt} 914 | A 2.0 kg mass is connected to the end of string and moves about the string’s fixed end in a conical motion with a constant speed of 4.0 m/s. The string has a length of 2.50 m and forms an angle of $\theta$ with the vertical. What is the tension in the string? 915 | \begin{enumerate}[label=(\Alph*)] 916 | \item 20.0 N 917 | \item 23.7 N 918 | \item 27.4 N 919 | \item 29.8 N 920 | \item 32.5 N 921 | \end{enumerate} 922 | \end{minipage}% 923 | \hfill 924 | \begin{minipage}[t]{0.35\linewidth} 925 | \vspace{0pt} 926 | \centering 927 | \includegraphics[width=\linewidth]{Problem_28_Figure.png} 928 | \end{minipage} 929 | 930 | \category{Circular Motion} 931 | \begin{answerbox} 932 | \finalanswer{(C) 27.4 N} 933 | \end{answerbox} 934 | \begin{solutionbox} 935 | 936 | From $T\cos\theta=mg$ and $T\sin\theta=mv^2/r$ with $r=L\sin\theta$, eliminate $\theta$ to get $\cos\theta\approx0.725$ and $T=mg/\cos\theta\approx27.0\,\text{N}$ (agrees with choice 27.4 N). 937 | \end{solutionbox} 938 | 939 | % Problem 28 940 | \item \label{prob:28} 941 | \noindent\begin{minipage}[t]{0.6\linewidth} 942 | \vspace{0pt} 943 | A box slides with uniform acceleration up an incline. The box has an initial speed of 9.0 m/s and rises vertically 2.60 m before coming to rest. If the angle of the incline is $30^\circ$, what is the coefficient of kinetic friction between the box and the incline? 944 | \begin{enumerate}[label=(\Alph*)] 945 | \item 0.298 946 | \item 0.322 947 | \item 0.372 948 | \item 0.483 949 | \item 0.557 950 | \end{enumerate} 951 | \end{minipage}% 952 | \hfill 953 | \begin{minipage}[t]{0.35\linewidth} 954 | \vspace{0pt} 955 | \centering 956 | \includegraphics[width=\linewidth]{Problem_29_Figure.png} 957 | \end{minipage} 958 | 959 | \category{Work \& Energy with Friction} 960 | \begin{answerbox} 961 | \finalanswer{(B) 0.322} 962 | \end{answerbox} 963 | \begin{solutionbox} 964 | 965 | $\Delta K+\Delta U+W_f=0$: $-\tfrac{1}{2}m9^2 + mg(2.60) - \mu_k mg\cos30^\circ\,d=0$ with $d=2.60/\sin30^\circ=5.20$ m gives $\mu_k\approx0.322$ (using $g\approx10\,\text{m/s}^2$). Using $g=9.8\,\text{m/s}^2$ yields $\mu_k\approx0.340$. 966 | \end{solutionbox} 967 | 968 | \newpage 969 | 970 | % Problem 29 971 | \item \label{prob:29} 972 | \noindent\begin{minipage}[t]{0.6\linewidth} 973 | \vspace{0pt} 974 | A 9.20 m long uniform plank rests on a frictionless ice pond. A 52 kg box rests on the plank’s left end while a 71 kg person stands at the plank’s right end. After the person walks to the left on the plank and stands at the same location as the box, the plank has slid 3.84 m to the right relative to the pond’s shore. Which one of the following choices best represents the mass of the plank? 975 | \begin{enumerate}[label=(\Alph*)] 976 | \item 123 kg 977 | \item 61.5 kg 978 | \item 47.1 kg 979 | \item 36.5 kg 980 | \item 31.2 kg 981 | \end{enumerate} 982 | \end{minipage}% 983 | \hfill 984 | \begin{minipage}[t]{0.35\linewidth} 985 | \vspace{0pt} 986 | \centering 987 | \includegraphics[width=\linewidth]{Problem_31_Figure.png} 988 | \end{minipage} 989 | 990 | \category{Center of Mass} 991 | \begin{answerbox} 992 | \finalanswer{(C) 47.1 kg} 993 | \end{answerbox} 994 | \begin{solutionbox} 995 | 996 | Stationary center of mass: $M_p\Delta x_p+M_b\Delta x_b+M_{pers}\Delta x_{pers}=0$. With given displacements, $3.84M_p+199.68-380.56=0 \Rightarrow M_p\approx47.1\,\text{kg}$. 997 | \end{solutionbox} 998 | 999 | % Problem 30 1000 | \item \label{prob:30} 1001 | At the top of a high cliff, a small rock is dropped from rest. A ball is launched straight downward with an initial speed of 36.0 m/s at a time of 2.10 s after the rock was dropped. When the ball has fallen 28.0 m further than the initially dropped rock, what is the speed of the ball relative to the rock? 1002 | \begin{enumerate}[label=(\Alph*)] 1003 | \item 15.0 m/s 1004 | \item 16.0 m/s 1005 | \item 20.0 m/s 1006 | \item 21.0 m/s 1007 | \item 36.0 m/s 1008 | \end{enumerate} 1009 | 1010 | \category{Kinematics} 1011 | \begin{answerbox} 1012 | \finalanswer{(A) 15.0 m/s} 1013 | \end{answerbox} 1014 | \begin{solutionbox} 1015 | 1016 | Downward speeds: $v_r=gt$, $v_b=36+g(t-2.10)$ for $t\ge2.10$. Relative speed $v_{rel}=v_b-v_r=36-g\cdot2.10\approx15.0\,\text{m/s}$ (using $g\approx10\,\text{m/s}^2$). 1017 | \end{solutionbox} 1018 | 1019 | \newpage 1020 | 1021 | % Problem 31 1022 | \item \label{prob:31} 1023 | A mass that is in simple harmonic motion obeys the following position versus time equation: $y(t)=0.50\,\mathrm{m}\,\sin\big[(\pi/2)\,t\big]$ where $t$ is in seconds. What is the period of vibration of this mass? 1024 | \begin{enumerate}[label=(\Alph*)] 1025 | \item 1.0 s 1026 | \item 2.0 s 1027 | \item 3.0 s 1028 | \item 4.0 s 1029 | \item 5.0 s 1030 | \end{enumerate} 1031 | 1032 | \category{Oscillations} \tags{} 1033 | \begin{answerbox} 1034 | \finalanswer{(D) 4.0 s} 1035 | \end{answerbox} 1036 | \begin{insightbox} 1037 | With $\omega=\pi/2$, $T=2\pi/\omega=4$ s. 1038 | \end{insightbox} 1039 | \begin{solutionbox} 1040 | 1041 | Identify $\omega=\pi/2$. Then $T=2\pi/\omega=4.0\,\text{s}$. 1042 | \end{solutionbox} 1043 | 1044 | % Problem 32 1045 | \item \label{prob:32} 1046 | A particle has a total energy of 500 MeV and a linear momentum of 300 MeV/c. What is the mass of the particle? 1047 | \begin{enumerate}[label=(\Alph*)] 1048 | \item 800 MeV/c$^2$ 1049 | \item 583 MeV/c$^2$ 1050 | \item 400 MeV/c$^2$ 1051 | \item 267 MeV/c$^2$ 1052 | \item 200 MeV/c$^2$ 1053 | \end{enumerate} 1054 | 1055 | \category{Modern Physics} 1056 | \begin{answerbox} 1057 | \finalanswer{(C) 400 MeV/c$^2$} 1058 | \end{answerbox} 1059 | \begin{solutionbox} 1060 | 1061 | $E^2=(pc)^2+(m_0c^2)^2$: $(500)^2=(300)^2+(m_0c^2)^2 \Rightarrow m_0c^2=\sqrt{160000}=400\,\text{MeV}$. 1062 | \end{solutionbox} 1063 | 1064 | \newpage 1065 | 1066 | % Problem 33 1067 | \item \label{prob:33} 1068 | Water flows out of a horizontal drainpipe at the rate of 120 kg per minute. Its initial vertical velocity is zero and it falls 3.20 m to the ground. What is the average force it exerts when it hits the ground? 1069 | \begin{enumerate}[label=(\Alph*)] 1070 | \item 6.0 N 1071 | \item 10.0 N 1072 | \item 12.0 N 1073 | \item 16.0 N 1074 | \item 20.0 N 1075 | \end{enumerate} 1076 | 1077 | \category{Momentum \& Impulse} 1078 | \begin{answerbox} 1079 | \finalanswer{(D) 16.0 N} 1080 | \end{answerbox} 1081 | \begin{solutionbox} 1082 | 1083 | Assume an inelastic impact where the water comes to rest instantaneously upon hitting the ground. Mass flow rate $\dot m=120/60=2.0\,\text{kg/s}$. Speed from $v=\sqrt{2gh}=8.0\,\text{m/s}$. Average force $F=\dot m\,v=16.0\,\text{N}$. 1084 | \end{solutionbox} 1085 | 1086 | % Problem 34 1087 | \item \label{prob:34} 1088 | Let M represent the magnification of an image. For which of the following arrangements of an object and an optical device would $-1 < M < 0$? 1089 | \begin{enumerate}[label=(\Alph*)] 1090 | \item The object is placed less than one focal length in front of a converging mirror. 1091 | \item The object is placed between one focal length and two focal lengths in front of a diverging mirror. 1092 | \item The object is placed less than one focal length in front of a diverging lens. 1093 | \item The object is placed more than two focal lengths in front of a converging lens. 1094 | \item The object is placed between one focal length and two focal lengths in front of a plane mirror. 1095 | \end{enumerate} 1096 | 1097 | \category{Geometric Optics} \tags{} 1098 | \begin{answerbox} 1099 | \finalanswer{(D) Converging lens with $d_o>2f$} 1100 | \end{answerbox} 1101 | \begin{solutionbox} 1102 | 1103 | This produces a real, inverted, reduced image: $M=-d_i/d_o$ with $f0$ means work done on the system). 193 | \item \textbf{Sign conventions}: Choose axes first and maintain consistency. For Doppler in a stationary medium, prefer the rule "approach increases frequency; recession decreases" and employ a sign diagram; see the Waves unit for the formula being given in detail. For KVL, direction of tracing loop; voltage rises are positive. 194 | \end{itemize} 195 | 196 | % --- Disclaimer (moved under Reading Instructions as a subsection) --- 197 | \subsection*{Disclaimer} 198 | \addcontentsline{toc}{subsection}{Disclaimer} 199 | \begin{disclaimerbox} 200 | This \textit{"Physics Bowl D1 + D2 Cheat Sheets"} (Version 1.7 by Ray Zhou) is an independent, non-official study guide created for educational purposes only. It summarizes key concepts, formulas, heuristics, and original solutions based on standard physics principles and publicly available Physics Bowl metadata (e.g., Contest/Year/Division/Problem/Page).\par\medskip 201 | Full problem statements are sourced from official AAPT materials---please visit the American Association of Physics Teachers (AAPT) website at \href{https://www.aapt.org/programs/physicsbowl}{aapt.org/programs/physicsbowl} for complete exams and answers.\par\medskip 202 | This guide is provided "as is" without warranties of accuracy or completeness. It is not endorsed by AAPT and should not be used as a substitute for official resources. Users should verify information independently. Distributed under CC BY-NC-SA 4.0 license for non-commercial, educational sharing.\par\medskip 203 | For feedback or questions, contact \href{mailto:zhouxinrui2025@163.com}{zhouxinrui2025@163.com}, or visit the project on GitHub: \url{https://qiuxue-yee.github.io/Physics-Bowl-CheatSheets-prepmaterial/}. 204 | \end{disclaimerbox} 205 | 206 | % Ensure the Table of Contents starts on a fresh page 207 | \clearpage 208 | 209 | % --- Table of Contents (header suppressed during TOC pages) --- 210 | \begingroup 211 | \pagestyle{plain}% suppress fancy header/footer on TOC pages 212 | \setcounter{tocdepth}{2}% up to subsections in TOC 213 | \tableofcontents 214 | \endgroup 215 | \clearpage 216 | \pagenumbering{arabic}% main matter 217 | \pagestyle{fancy} 218 | 219 | % ===================== Part I: Mechanics ===================== 220 | \parttheme{mech} 221 | \section*{Part I: Mechanics} 222 | \addcontentsline{toc}{section}{Part I: Mechanics} 223 | 224 | % -------- Unit 1: Kinematics -------- 225 | \Unit[Vectors vs scalars; constant-acceleration (SUVAT); projectile motion (same-level range/time/height); uniform circular motion ($a_c=v^2/r$, $T=2\pi r/v$); relative motion; calculus forms $v=dx/dt$, $a=dv/dt$ {\DTwo}; areas under $v$--$t$ and $a$--$t$ graphs {\DTwo}]{Unit 1: Kinematics} 226 | 227 | \begin{KnowledgePoint}{Vectors \& Components \DOne} 228 | \KPFormulas 229 | \begin{formulabox} 230 | \textbf{Concept explanation:} Vectors have both magnitude and direction. The decomposition of a vector into orthogonal components translates geometric relations into algebraic relations such that calculations are simple component arithmetic. 231 | 232 | \textbf{Core formulas:} 233 | \[ 234 | \left\{\begin{aligned} 235 | &\vec v=(v_x,v_y)\ (\text{2D}),\quad v=\|\vec v\|=\sqrt{v_x^2+v_y^2},\\ 236 | &\text{Unit vectors: }\hat\imath=(1,0),\ \hat\jmath=(0,1),\ \vec v=v_x\hat\imath+v_y\hat\jmath. 237 | \end{aligned}\right. 238 | \] 239 | 240 | \textbf{Variable definitions:} $v_x,v_y$ scalar components; $v$ speed (magnitude of velocity); $\hat\imath,\hat\jmath$ orthonormal basis. 241 | 242 | 243 | \textbf{Prerequisites \& scope:} Axes must be orthogonal for Pythagorean magnitude; extend to 3D with $\hat k$ and $v=\sqrt{v_x^2+v_y^2+v_z^2}$. 244 | \end{formulabox} 245 | 246 | \KPHeuristics 247 | \begin{heuristicsbox} 248 | \begin{itemize}[leftmargin=*] 249 | \item Align axes with directions of steepest slopes and launch angles to reduce components. 250 | \item Separate vectors before writing equations (Newton's laws, kinematics) to prevent late trigonometry. 251 | \end{itemize} 252 | \vspace{0.4em} 253 | \begin{itemize}[leftmargin=*] 254 | \item Mixing magnitude and component equations. Solution: write separate equations for $x$ and $y$ and add if applicable. 255 | \end{itemize} 256 | \end{heuristicsbox} 257 | 258 | \KPProblems 259 | \begin{cheatproblem} 260 | Resolve velocity vector $\vec v=(30\,\text{m/s},40\,\text{m/s})$ into magnitude and direction. 261 | \begin{solutionbox} 262 | $v=\sqrt{30^2+40^2}=50\,\text{m/s}$. Direction $\theta=\arctan(40/30)\approx53.1^\circ$ above the horizontal. 263 | \end{solutionbox} 264 | \end{cheatproblem} 265 | \begin{cheatproblem} 266 | A force of $100\,\text{N}$ acts at $30^\circ$ to the horizontal. Find its horizontal and vertical components. 267 | \begin{solutionbox} 268 | $F_x=100\cos30^\circ=50\sqrt3\,\text{N}$, $F_y=100\sin30^\circ=50\,\text{N}$. 269 | \end{solutionbox} 270 | \end{cheatproblem} 271 | \end{KnowledgePoint} 272 | 273 | \begin{KnowledgePoint}{Constant Acceleration (SUVAT) \DOne} 274 | \KPFormulas 275 | \begin{formulabox} 276 | \textbf{Concept explanation:} Position and velocity are quadratic/linear in time with constant acceleration, providing closed-form relations (SUVAT equations). 277 | 278 | \textbf{Core formulas:} 279 | \[ 280 | \left\{\begin{aligned} 281 | &x=x_0+v_0 t+\tfrac12 a t^2,\\ 282 | &v=v_0+at,\\ 283 | &v^2=v_0^2+2a\,(x-x_0). 284 | \end{aligned}\right. 285 | \] 286 | 287 | \textbf{Variable definitions:} $x$ position; $v$ velocity; $a$ constant acceleration; subscript $0$ initial value; $t$ time; $\Delta x=x-x_0$. 288 | 289 | 290 | \textbf{Prerequisites \& scope:} Acceleration must be constant over the interval; otherwise use calculus forms. 291 | \end{formulabox} 292 | 293 | \KPHeuristics 294 | \begin{heuristicsbox} 295 | \begin{itemize}[leftmargin=*] 296 | \item Choose the equation that excludes the unknown you lack (e.g., use $v^2$-form when time is absent). 297 | \item Work symbolically until the end to avoid compounding rounding errors. 298 | \end{itemize} 299 | \vspace{0.4em} 300 | \begin{itemize}[leftmargin=*] 301 | \item Applying constant-$a$ formulas when $a=a(t)$ or $a=a(v)$. Fix: switch to calculus forms or energy methods. 302 | \end{itemize} 303 | \end{heuristicsbox} 304 | 305 | \KPProblems 306 | \begin{cheatproblem} 307 | The velocity of a cart is given by the piecewise function $v(t)=\begin{cases} a t, & 0\le t\le T/2,\\ aT/2, & T/2< t\le T,\end{cases}$ with constant $a>0$. Find the total displacement in time $T$. 308 | \begin{solutionbox} 309 | Area under $v$--$t$: first triangle area $\tfrac12(a\,T/2)(T/2)=\tfrac{aT^2}{8}$; second rectangle area $(aT/2)(T/2)=\tfrac{aT^2}{4}$. Total $\Delta x=\tfrac{3aT^2}{8}$. 310 | \end{solutionbox} 311 | \end{cheatproblem} 312 | \begin{cheatproblem} 313 | A car accelerates uniformly from rest to $20\,\text{m/s}$ in $4\,\text{s}$. Find the acceleration and distance traveled. 314 | \begin{solutionbox} 315 | $a=\dfrac{v-v_0}{t}=\dfrac{20-0}{4}=5\,\text{m/s}^2$. Distance $\Delta x=v_0 t+\tfrac12 a t^2=0+\tfrac12\cdot5\cdot16=40\,\text{m}$. 316 | \end{solutionbox} 317 | \end{cheatproblem} 318 | \end{KnowledgePoint} 319 | 320 | \begin{KnowledgePoint}{Projectile (Equal Heights) \DOne} 321 | \KPFormulas 322 | \begin{formulabox} 323 | \textbf{Concept explanation:} Horizontal and vertical motions independent in uniform gravity. Decomposition into components provides closed-form solutions for flight time, range, and peak height at the same launch/landing height. 324 | 325 | \textbf{Core formulas:} 326 | \[ 327 | \left\{\begin{aligned} 328 | &T=\frac{2v_0\sin\theta}{g},\\ 329 | &R=\frac{v_0^2\sin2\theta}{g},\\ 330 | &H=\frac{v_0^2\sin^2\theta}{2g}. 331 | \end{aligned}\right. 332 | \] 333 | 334 | \textbf{Variable definitions:} $v_0$ launch speed; $\theta$ launch angle; $g$ gravitational acceleration; $T$ flight time; $R$ range; $H$ apex height. 335 | 336 | \textbf{Prerequisites \& scope:} Launch/landing at equal heights; neglect air resistance; for unequal heights, solve quadratic in $t$. 337 | \end{formulabox} 338 | 339 | \KPHeuristics 340 | \begin{heuristicsbox} 341 | % (title removed per style unification) 342 | \begin{itemize}[leftmargin=*] 343 | \item Separate $x$ and $y$ equations; eliminate $t$ or use symmetry about the apex for time splits. 344 | \item For maximum range at equal heights, use $\theta=45^\circ$. 345 | \end{itemize} 346 | \vspace{0.4em} 347 | \begin{itemize}[leftmargin=*] 348 | \item Using $R$ formula when launch and landing heights differ. Fix: solve general quadratic and then compute $x(T)$. 349 | \end{itemize} 350 | \end{heuristicsbox} 351 | 352 | \KPProblems 353 | \begin{cheatproblem} 354 | A projectile is launched at speed $v_0$ and angle $\theta$ from flat ground and lands at the same height. Express the time of flight $T$, the maximum height $H$, and the range $R$ in terms of $v_0,\theta,g$. 355 | \begin{solutionbox} 356 | $T=\dfrac{2v_0\sin\theta}{g}$, $H=\dfrac{v_0^2\sin^2\theta}{2g}$, $R=\dfrac{v_0^2\sin2\theta}{g}$, by separating vertical and horizontal components and using constant-$g$ kinematics. 357 | \end{solutionbox} 358 | \end{cheatproblem} 359 | 360 | \begin{cheatproblem} 361 | Two projectiles launched with the same $v_0$ at angles $\theta$ and $90^\circ-\theta$ have equal ranges. What is the ratio of their maximum heights $H_\theta/H_{90^\circ-\theta}$? 362 | \begin{solutionbox} 363 | $H=\dfrac{v_0^2\sin^2\theta}{2g}$. Hence $\dfrac{H_\theta}{H_{90^\circ-\theta}}=\dfrac{\sin^2\theta}{\cos^2\theta}=\tan^2\theta$. 364 | \end{solutionbox} 365 | \end{cheatproblem} 366 | \end{KnowledgePoint} 367 | 368 | \begin{KnowledgePoint}{Uniform Circular Motion \DOne} 369 | \KPFormulas 370 | \begin{formulabox} 371 | \textbf{Concept description:} In uniform speed circular motion, the acceleration is centripetal (centerwards) with magnitude $v^2/r$; period and speed are related by circumference. 372 | 373 | \textbf{Core formulas:} 374 | \[ 375 | \left\{\begin{aligned} 376 | &a_c=\frac{v^2}{r}=\frac{4\pi^2 r}{T^2},\\ 377 | &T=\frac{2\pi r}{v}. 378 | \end{aligned}\right. 379 | \] 380 | 381 | \textbf{Variable definitions:} $r$ radius; $v$ speed; $T$ period; $a_c$ centripetal acceleration. 382 | 383 | \textbf{Prerequisites \& scope:} Speed constant; acceleration direction changes; for non-uniform circular motion add tangential component. 384 | \end{formulabox} 385 | 386 | \KPHeuristics 387 | \begin{heuristicsbox} 388 | % (title removed per style unification) 389 | \begin{itemize}[leftmargin=*] 390 | \item Draw radial/tangential components explicitly; set $a_t=0$ for UCM. 391 | \item Vertical circle normals: at the top, $T+mg=mv^2/r$; at the bottom, $T-mg=mv^2/r$. 392 | \item Minimum speed at the top for a taut string (no slack): $v_{\text{top}}\ge \sqrt{gr}$ (else $T=0$ at the top). 393 | \end{itemize} 394 | \vspace{0.4em} 395 | \begin{itemize}[leftmargin=*] 396 | \item Treating centripetal force as an extra force. Fix: centripetal is the net radial component of existing forces. 397 | \end{itemize} 398 | \end{heuristicsbox} 399 | 400 | \KPProblems 401 | \begin{cheatproblem} 402 | A bead moves in a circle of radius $r$ with period $T$. Compute its speed and centripetal acceleration. 403 | \begin{solutionbox} 404 | $v=\dfrac{2\pi r}{T}$ and $a_c=\dfrac{v^2}{r}=\dfrac{4\pi^2 r}{T^2}$. 405 | \end{solutionbox} 406 | \end{cheatproblem} 407 | \end{KnowledgePoint} 408 | 409 | \begin{KnowledgePoint}{Relative Motion \DOne} 410 | \KPFormulas 411 | \begin{formulabox} 412 | \textbf{Concept explanation:} Relative velocity cancels observer motion: the velocity of $A$ relative to $B$ is $\vec v_A-\vec v_B$. 413 | 414 | \textbf{Core formula:} 415 | \[ 416 | \vec v_{A/B}=\vec v_A-\vec v_B. 417 | \] 418 | 419 | \textbf{Variable definitions:} $\vec v_{A/B}$ velocity of $A$ relative to $B$; $\vec v_A,\vec v_B$ velocities in an inertial frame. 420 | 421 | \textbf{Prerequisites \& scope:} Within Galilean (non-relativistic) regime; in 2D/3D apply component-wise. 422 | \end{formulabox} 423 | 424 | \KPHeuristics 425 | \begin{heuristicsbox} 426 | \begin{itemize}[leftmargin=*] 427 | \item Move to the target's rest frame to align directions and simplify timing. 428 | \item For winds/rivers, remove drift by aiming to oppose the current component. 429 | \item Boundary condition: if $v\le u$, landing directly across is not possible; you will be swept downstream and must land downstream. 430 | \end{itemize} 431 | \vspace{0.4em} 432 | \begin{itemize}[leftmargin=*] 433 | \item Superposing speeds scalarly when directions are different. Remedy: take difference of vectors component-wise. 434 | \end{itemize} 435 | \end{heuristicsbox} 436 | 437 | \KPProblems 438 | \begin{cheatproblem} 439 | A river of width $W$ flows east at speed $u$. A boat of speed $v$ relative to water aims at angle $\alpha$ north of west to land directly across. Find $\alpha$ and the crossing time, assuming $v>u$. 440 | \begin{solutionbox} 441 | Require zero east drift: west component equals current, so $v\cos\alpha=u$ and $\alpha=\arccos(u/v)$. North component is $v\sin\alpha=\sqrt{v^2-u^2}$, so time $t=\dfrac{W}{\sqrt{v^2-u^2}}$. 442 | \end{solutionbox} 443 | \end{cheatproblem} 444 | \end{KnowledgePoint} 445 | 446 | \begin{KnowledgePoint}{Calculus Forms \& Graph Areas \DTwo} 447 | \KPFormulas 448 | \begin{formulabox} 449 | \textbf{Concept explanation:} With acceleration varying, kinematics is found by integrating velocity and acceleration; graph areas are summaries of change. 450 | 451 | \textbf{Core formulas:} 452 | \[ 453 | \left\{\begin{aligned} 454 | &v=\frac{dx}{dt},\quad a=\frac{dv}{dt}=\frac{d^2x}{dt^2}.\\ 455 | &v(t)=v(t_0)+\int_{t_0}^t a(\tau)\,d\tau,\quad x(t)=x(t_0)+\int_{t_0}^t v(\tau)\,d\tau.\\ 456 | &\Delta x=\int v\,dt\ (\text{area under }v\text{--}t),\quad \Delta v=\int a\,dt\ (\text{area under }a\text{--}t).\\ 457 | &\text{Chain rule: } a=\frac{dv}{dt}=\frac{dv}{dx}\,v\ \Rightarrow\ v\,dv=a(x)\,dx.\\ 458 | &\text{If }a=\text{const: } v^2=v_0^2+2a\,(x-x_0). 459 | \end{aligned}\right. 460 | \] 461 | 462 | \textbf{Variable definitions:} $x,v,a$ position/velocity/acceleration; $t$ time; integrals are definite over the time interval. 463 | 464 | \textbf{Prerequisites \& scope:} Differentiability over interval; interpret signed areas for direction-sensitive quantities. 465 | \end{formulabox} 466 | 467 | \KPHeuristics 468 | \begin{heuristicsbox} 469 | \begin{itemize}[leftmargin=*] 470 | \item Read slopes of $x$--$t$ as $v$ and slopes of $v$--$t$ as $a$; use areas for accumulated change. 471 | \item When $a=a(v)$ or $a=a(x)$, separate variables via $v\,dv=a\,dx$ to avoid time explicitly. 472 | \item If motion crosses turning points where $v=0$, integrate piecewise and track signs to avoid taking an incorrect branch. 473 | \end{itemize} 474 | \vspace{0.4em} 475 | \begin{itemize}[leftmargin=*] 476 | \item Confusing displacement with distance on $v$--$t$ when $v$ changes sign. Fix: integrate absolute value for distance. 477 | \end{itemize} 478 | \end{heuristicsbox} 479 | 480 | \KPProblems 481 | \begin{cheatproblem} 482 | An object has acceleration $a(t)=a_0+bt$ with constants $a_0,b$. If $x(0)=0$ and $v(0)=v_0$, find $x(t)$. 483 | \begin{solutionbox} 484 | Integrate: $v(t)=v_0+\int_0^t(a_0+b\tau)\,d\tau=v_0+a_0 t+\tfrac12 b t^2$. Then $x(t)=\int_0^t v(\tau)\,d\tau=v_0 t+\tfrac12 a_0 t^2+\tfrac16 b t^3$. 485 | \end{solutionbox} 486 | \end{cheatproblem} 487 | 488 | \begin{cheatproblem} 489 | A particle obeys $\dfrac{dx}{dt}=k\sqrt{x}$ with $k>0$ and $x(0)=0$. Find $x(t)$. 490 | \begin{solutionbox} 491 | Separate: $\dfrac{dx}{\sqrt{x}}=k\,dt\Rightarrow 2\sqrt{x}=kt+C$. With $x(0)=0$, $C=0$. Hence $x(t)=\dfrac{k^2 t^2}{4}$ and $v(t)=\dfrac{k^2 t}{2}$. 492 | \end{solutionbox} 493 | \end{cheatproblem} 494 | \end{KnowledgePoint} 495 | 496 | % (Removed outdated Practice Pointers for Part I Unit 1) 497 | 498 | % -------- Placeholders for remaining Mechanics Units (skeletons only) -------- 499 | \Unit[Newton's laws; common forces (weight, spring, friction, normal); free-body diagrams; non-uniform circular motion; variable force and impulse]{Unit 2: Newtonian Dynamics} 500 | 501 | \begin{KnowledgePoint}{Newton's Laws and Free-Body Diagrams \DOne} 502 | \KPFormulas 503 | \begin{formulabox} 504 | \[ 505 | \left\{\begin{aligned} 506 | &\sum \vec F=m\vec a\ \text{; action--reaction pairs on separate bodies}. 507 | \end{aligned}\right. 508 | \] 509 | \end{formulabox} 510 | 511 | \KPHeuristics 512 | \begin{heuristicsbox} 513 | \begin{itemize}[leftmargin=*] 514 | \item Split each body; sketch clean FBDs and project along convenient directions. 515 | \item Don't place action--reaction pairs on the same drawing; they're on separate bodies and cancel only at the system level. 516 | \item Use radial/tangential axes for curves or inclined axes on ramps; use friction direction as unknown and solve for its sign. 517 | \item For connected bodies (strings/pulleys), apply kinematic constraints (e.g., equal string lengths imply proportional accelerations) and apply one tension per ideal massless string. 518 | \item In non-inertial frames (elevators, speeding cars), add $-m\,\vec a_{\text{frame}}$ only if you change frames explicitly. 519 | \end{itemize} 520 | \end{heuristicsbox} 521 | 522 | \KPProblems 523 | \begin{cheatproblem} 524 | A $5\,\text{kg}$ block rests on a horizontal surface. A horizontal force $F=20\,\text{N}$ is applied. Find its acceleration if friction is negligible. 525 | \begin{solutionbox} 526 | $\sum F=ma\Rightarrow 20=5a\Rightarrow a=4\,\text{m/s}^2$. 527 | \end{solutionbox} 528 | \end{cheatproblem} 529 | \begin{cheatproblem} 530 | A $2\,\text{kg}$ block hangs from a string. Find the tension in the string when the block is at rest. 531 | \begin{solutionbox} 532 | $\sum F_y=0\Rightarrow T-mg=0\Rightarrow T=2\cdot9.8=19.6\,\text{N}$. 533 | \end{solutionbox} 534 | \end{cheatproblem} 535 | \end{KnowledgePoint} 536 | 537 | \begin{KnowledgePoint}{Friction and Springs \DOne} 538 | \KPFormulas 539 | \begin{formulabox} 540 | \[ 541 | \left\{\begin{aligned} 542 | &\text{Friction: } f_s\le \mu_s N\ \text{(variable up to max)},\quad f_k=\mu_k N,\\ 543 | &\text{Spring force: } F=-kx\ \text{(Hooke) within elastic limit}. 544 | \end{aligned}\right. 545 | \] 546 | \end{formulabox} 547 | 548 | \KPHeuristics 549 | \begin{heuristicsbox} 550 | \begin{itemize}[leftmargin=*] 551 | \item Assume the direction of static friction to be unknown; solve and interpret its sign from the solution. 552 | \item Apply $f_s\le \mu_s N$; on the verge of sliding, take $f_{s,\max}=\mu_s N$. For steady sliding, use $f_k=\mu_k N$. 553 | \item For springs in series, use $k_{\text{eq}}$: in series $\tfrac{1}{k_{\text{eq}}}=\sum \tfrac{1}{k_i}$, in parallel $k_{\text{eq}}=\sum k_i$. 554 | \end{itemize} 555 | \end{heuristicsbox} 556 | 557 | \KPProblems 558 | \begin{cheatproblem} 559 | A $3\,\text{kg}$ block on a horizontal surface has $\mu_k=0.2$. Find the kinetic friction force when the block slides. 560 | \begin{solutionbox} 561 | $N=mg=3\cdot9.8=29.4\,\text{N}$. $f_k=\mu_k N=0.2\cdot29.4=5.88\,\text{N}$. 562 | \end{solutionbox} 563 | \end{cheatproblem} 564 | \begin{cheatproblem} 565 | A spring with constant $k=200\,\text{N/m}$ is compressed by $x=0.1\,\text{m}$. Find the restoring force. 566 | \begin{solutionbox} 567 | $F=-kx=-200\cdot0.1=-20\,\text{N}$ (opposite to compression). 568 | \end{solutionbox} 569 | \end{cheatproblem} 570 | \end{KnowledgePoint} 571 | 572 | \begin{KnowledgePoint}{Non-uniform Circular Motion \DOne} 573 | \KPFormulas 574 | \begin{formulabox} 575 | \[ 576 | \left\{\begin{aligned} 577 | &\text{Radial: } \sum F_r=m v^2/r\ ;\ \text{tangential: } \sum F_t=m a_t. 578 | \end{aligned}\right. 579 | \] 580 | \end{formulabox} 581 | 582 | \KPHeuristics 583 | \begin{heuristicsbox} 584 | Split radial and tangential components explicitly; centripetal is net radial. 585 | \end{heuristicsbox} 586 | 587 | \KPProblems 588 | \begin{cheatproblem} 589 | A car of mass $m$ rounds a curve of radius $r=50\,\text{m}$ at $v=20\,\text{m/s}$. Find the required centripetal force. 590 | \begin{solutionbox} 591 | $F_c=\dfrac{mv^2}{r}=m\dfrac{400}{50}=8m\,\text{N}$. 592 | \end{solutionbox} 593 | \end{cheatproblem} 594 | \begin{cheatproblem} 595 | A ball on a string swings in a vertical circle of radius $r$. At the top, the tension is $T$ and speed is $v$. Write the centripetal force equation. 596 | \begin{solutionbox} 597 | At the top, $T+mg=\dfrac{mv^2}{r}$ (both point inward). 598 | \end{solutionbox} 599 | \end{cheatproblem} 600 | \end{KnowledgePoint} 601 | 602 | \begin{KnowledgePoint}{Impulse and Variable Forces \DTwo} 603 | \KPFormulas 604 | \begin{formulabox} 605 | \[ 606 | \left\{\begin{aligned} 607 | &\text{Impulse: } \vec J=\int \vec F\,dt=\Delta \vec p. 608 | \end{aligned}\right. 609 | \] 610 | \end{formulabox} 611 | 612 | \KPHeuristics 613 | \begin{heuristicsbox} 614 | Extrapolate force-time profiles for varying forces; examine external impulses for system imparted momentum changes. 615 | \end{heuristicsbox} 616 | 617 | \KPProblems 618 | \begin{cheatproblem} 619 | A constant force $F=10\,\text{N}$ acts on a mass for $\Delta t=3\,\text{s}$. Find the impulse delivered. 620 | \begin{solutionbox} 621 | $J=F\Delta t=10\cdot3=30\,\text{N}\cdot\text{s}$. 622 | \end{solutionbox} 623 | \end{cheatproblem} 624 | \begin{cheatproblem} 625 | A particle's momentum changes from $\vec p_i=(5,0)\,\text{kg·m/s}$ to $\vec p_f=(5,10)\,\text{kg·m/s}$ in $2\,\text{s}$. Find the average force. 626 | \begin{solutionbox} 627 | $\vec J=\Delta\vec p=(0,10)\,\text{kg·m/s}$. $\vec F_{\text{avg}}=\dfrac{\vec J}{\Delta t}=(0,5)\,\text{N}$. 628 | \end{solutionbox} 629 | \end{cheatproblem} 630 | \end{KnowledgePoint} 631 | 632 | \Unit[Work $W=\int \vec F\cdot d\vec r$; Work--Energy theorem; potential energies (gravity, spring); mechanical energy conservation; power (instantaneous vs average); conservative fields ($F_x=-\,dU/dx$)]{Unit 3: Work, Energy, Power} 633 | 634 | \begin{KnowledgePoint}{Work and the Work--Energy Theorem \DOne} 635 | \KPFormulas 636 | \begin{formulabox} 637 | \[ 638 | \left\{\begin{aligned} 639 | &\text{For constant }F\text{ and displacement }d\text{ at angle }\theta:\ \ W=Fd\cos\theta,\\ 640 | &\text{Work--Energy: }\Delta K=W_{\text{net}},\\ 641 | &\text{Power: instantaneous }P=\vec F\cdot\vec v,\quad \text{average over }\Delta t:\ \bar P=\dfrac{\Delta W}{\Delta t}. 642 | \end{aligned}\right. 643 | \] 644 | \end{formulabox} 645 | 646 | \KPHeuristics 647 | \begin{heuristicsbox} 648 | \begin{itemize}[leftmargin=*] 649 | \item Choose the system to eliminate internal forces; only external work should be considered. 650 | \item Include nonconservative work explicitly (friction, thrust). For rolling without slipping, static friction often does no work on the rolling body. 651 | \item Check for indications by comparing the endpoints: $K_i+U_i$ vs. $K_f+U_f$. 652 | \end{itemize} 653 | \end{heuristicsbox} 654 | 655 | \KPProblems 656 | \begin{cheatproblem} 657 | A $10\,\text{kg}$ block is pushed $5\,\text{m}$ horizontally by a constant force $F=30\,\text{N}$. Find the work done. 658 | \begin{solutionbox} 659 | $W=Fd\cos\theta=30\cdot5\cdot\cos0=150\,\text{J}$. 660 | \end{solutionbox} 661 | \end{cheatproblem} 662 | \begin{cheatproblem} 663 | A net force of $50\,\text{N}$ accelerates a $5\,\text{kg}$ mass from rest over $10\,\text{m}$. Find the final kinetic energy and speed. 664 | \begin{solutionbox} 665 | $W_{\text{net}}=50\cdot10=500\,\text{J}$. By work-energy theorem, $\Delta K=W_{\text{net}}=500\,\text{J}$. Then $\tfrac12mv^2=500\Rightarrow v=\sqrt{200}=10\sqrt2\,\text{m/s}$. 666 | \end{solutionbox} 667 | \end{cheatproblem} 668 | \end{KnowledgePoint} 669 | 670 | \begin{KnowledgePoint}{Potential Energy and Conservation \DOne} 671 | \KPFormulas 672 | \begin{formulabox} 673 | \[ 674 | \left\{\begin{aligned} 675 | &U_g=mgh\ \text{(near-Earth)},\quad U_s=\tfrac12 kx^2,\\ 676 | &\text{Conservation (when nonconservative work zero): }K_i+U_i=K_f+U_f. 677 | \end{aligned}\right. 678 | \] 679 | \end{formulabox} 680 | 681 | \KPHeuristics 682 | \begin{heuristicsbox} 683 | Use reference $U=0$ conveniently; only differences matter. 684 | \end{heuristicsbox} 685 | 686 | \KPProblems 687 | \begin{cheatproblem} 688 | A $2\,\text{kg}$ block falls $h=5\,\text{m}$ from rest. Find its speed at the bottom using energy conservation. 689 | \begin{solutionbox} 690 | $K_i+U_i=K_f+U_f\Rightarrow0+mgh=\tfrac12mv^2+0\Rightarrow v=\sqrt{2gh}=\sqrt{2\cdot9.8\cdot5}=\sqrt{98}\approx9.9\,\text{m/s}$. 691 | \end{solutionbox} 692 | \end{cheatproblem} 693 | \begin{cheatproblem} 694 | A spring ($k=100\,\text{N/m}$) is compressed by $x=0.2\,\text{m}$. Compute the stored elastic potential energy. 695 | \begin{solutionbox} 696 | $U_s=\tfrac12 kx^2=\tfrac12\cdot100\cdot0.04=2\,\text{J}$. 697 | \end{solutionbox} 698 | \end{cheatproblem} 699 | \end{KnowledgePoint} 700 | 701 | \begin{KnowledgePoint}{Conservative Fields \DTwo} 702 | \KPFormulas 703 | \begin{formulabox} 704 | \[ 705 | F_x=-\\,\\dfrac{dU}{dx} 706 | \] 707 | Path independence in conservative fields; use energy methods when applicable. 708 | \end{formulabox} 709 | 710 | \KPHeuristics 711 | \begin{heuristicsbox} 712 | Use energy methods if forces are conservative; verify curl-free areas. 713 | \end{heuristicsbox} 714 | 715 | \KPProblems 716 | \begin{cheatproblem} 717 | A mass moves in a potential $U(x)=\tfrac12 kx^2$. Find the force acting on it. 718 | \begin{solutionbox} 719 | $\vec F=-\dfrac{dU}{dx}\,\hat\imath=-kx\,\hat\imath$ (Hooke's law in 1D). 720 | \end{solutionbox} 721 | \end{cheatproblem} 722 | \begin{cheatproblem} 723 | Show that gravitational potential $U=-\dfrac{GMm}{r}$ gives force $F=-\dfrac{GMm}{r^2}$ (radial). 724 | \begin{solutionbox} 725 | First compute $\dfrac{dU}{dr}=\dfrac{d}{dr}\Big(-GMm\,r^{-1}\Big)=+\dfrac{GMm}{r^2}$. Therefore the radial force is $\vec F=-\dfrac{dU}{dr}\,\hat{\mathbf r}=-\dfrac{GMm}{r^2}\,\hat{\mathbf r}$ (attractive, inward). 726 | \end{solutionbox} 727 | \end{cheatproblem} 728 | \end{KnowledgePoint} 729 | 730 | \Unit[Momentum $\vec p=m\vec v$; impulse; momentum conservation; elastic/inelastic/fully inelastic collisions; center of mass; rocket equation]{Unit 4: Momentum \& Collisions} 731 | 732 | \begin{KnowledgePoint}{Momentum and Impulse \DOne} 733 | \KPFormulas 734 | \begin{formulabox} 735 | \textbf{Concept explanation:} Momentum measures motion; impulse is the total effect of force over time and is equal to the change in momentum. 736 | 737 | \textbf{Core formulas:} 738 | \[ 739 | \left\{\begin{aligned} 740 | &\vec p=m\,\vec v,\\ 741 | &\vec J=\int_{t_1}^{t_2} \vec F\,dt=\Delta\vec p,\\ 742 | &\text{System: }\ \vec P=\sum_i m_i\vec v_i,\quad \Delta\vec P=\vec J_{\text{ext}}.\ (\text{If }\vec J_{\text{ext}}=0,\ \vec P\ \text{conserved}) 743 | \end{aligned}\right. 744 | \] 745 | 746 | \textbf{Variable definitions:} $\vec p$ momentum; $\vec P$ total momentum; $\vec J$ impulse; $\vec F$ external force; $m$ mass; $\vec v$ velocity. 747 | 748 | 749 | 750 | \textbf{Prerequisites \& scope:} Valid for Newtonian mechanics; for varying mass systems, take care with momentum flux (see rockets). 751 | \end{formulabox} 752 | 753 | \KPHeuristics 754 | \begin{heuristicsbox} 755 | \begin{itemize}[leftmargin=*] 756 | \item Identify a closed system (no external impulse) to use momentum conservation immediately. 757 | \item For short-duration large forces (collisions), use impulse–momentum rather than force–time details. 758 | \end{itemize} 759 | \vspace{0.4em} 760 | \begin{itemize}[leftmargin=*] 761 | \item Treating internal forces as external. Fix: define the system to include interacting bodies so internal forces cancel. 762 | \end{itemize} 763 | \end{heuristicsbox} 764 | 765 | \KPProblems 766 | \begin{cheatproblem} 767 | A $4\,\text{kg}$ object moves with velocity $\vec v=(6,8)\,\text{m/s}$. Find its momentum vector and magnitude. 768 | \begin{solutionbox} 769 | $\vec p=m\vec v=4(6,8)=(24,32)\,\text{kg·m/s}$, so $|\vec p|=\sqrt{24^2+32^2}=40\,\text{kg·m/s}$ and the direction matches $\vec v$. 770 | \end{solutionbox} 771 | \end{cheatproblem} 772 | \begin{cheatproblem} 773 | A force $F(t)=10t\,\text{N}$ acts on a particle from $t=0$ to $t=2\,\text{s}$. Find the impulse delivered. 774 | \begin{solutionbox} 775 | $J=\int_0^2 10t\,dt=\left[5t^2\right]_0^2=20\,\text{N}\cdot\text{s}$. 776 | \end{solutionbox} 777 | \end{cheatproblem} 778 | \end{KnowledgePoint} 779 | 780 | \begin{KnowledgePoint}{Elastic and Inelastic Collisions \DOne} 781 | \KPFormulas 782 | \begin{formulabox} 783 | \textbf{Concept explanation:} Collisions conserve total momentum; elastic ones also conserve kinetic energy. The center-of-mass (COM) frame makes algebra simpler. 784 | 785 | \textbf{Core formulas (1D):} 786 | \[ 787 | \left\{\begin{aligned} 788 | &\text{Momentum: } m_1u_1+m_2u_2=m_1v_1+m_2v_2,\\ 789 | &\text{Elastic energy: } \tfrac12 m_1u_1^2+\tfrac12 m_2u_2^2=\tfrac12 m_1v_1^2+\tfrac12 m_2v_2^2,\\ 790 | &\text{Result (elastic): } v_1=\frac{(m_1-m_2)u_1+2m_2u_2}{m_1+m_2},\ \ v_2=\frac{2m_1u_1+(m_2-m_1)u_2}{m_1+m_2},\\ 791 | &\text{Completely inelastic: stick }(v_1=v_2=v)=\frac{m_1u_1+m_2u_2}{m_1+m_2}.\\ 792 | \end{aligned}\right. 793 | \] 794 | 795 | \textbf{Variable definitions:} $u_i$ initial, $v_i$ final velocities; $m_i$ masses; all along one line (1D). 796 | 797 | 798 | 799 | \textbf{Prerequisites \& scope:} For oblique/2D, conserve vector momentum and use geometry; kinetic energy changes via deformation/heat in inelastic cases. 800 | \end{formulabox} 801 | 802 | \KPHeuristics 803 | \begin{heuristicsbox} 804 | \begin{itemize}[leftmargin=*] 805 | \item Switch to the COM frame where total momentum is zero; velocities reverse in elastic 1D collisions. 806 | \item For 2D glancing collisions, conserve components along orthogonal axes; use restitution or geometry for angles. 807 | \end{itemize} 808 | \vspace{0.4em} 809 | \begin{itemize}[leftmargin=*] 810 | \item Enforcing kinetic energy conservation in inelastic impact. Fix: only momentum is guaranteed; account for energy loss. 811 | \end{itemize} 812 | \end{heuristicsbox} 813 | 814 | \KPProblems 815 | \begin{cheatproblem} 816 | Two carts collide elastically in 1D. Mass $m_1=2\,\text{kg}$ at $u_1=5\,\text{m/s}$ hits mass $m_2=3\,\text{kg}$ at rest. Find $v_1$ and $v_2$. 817 | \begin{solutionbox} 818 | $v_1=\dfrac{(2-3)\cdot5+2\cdot3\cdot0}{5}=\dfrac{-5}{5}=-1\,\text{m/s}$. $v_2=\dfrac{2\cdot2\cdot5+(3-2)\cdot0}{5}=\dfrac{20}{5}=4\,\text{m/s}$. 819 | \end{solutionbox} 820 | \end{cheatproblem} 821 | \begin{cheatproblem} 822 | In a perfectly inelastic collision, $m_1=4\,\text{kg}$ at $u_1=6\,\text{m/s}$ collides with $m_2=2\,\text{kg}$ at rest. Find the final velocity. 823 | \begin{solutionbox} 824 | $v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}=\dfrac{4\cdot6+0}{6}=4\,\text{m/s}$. 825 | \end{solutionbox} 826 | \end{cheatproblem} 827 | \end{KnowledgePoint} 828 | 829 | \begin{KnowledgePoint}{Center of Mass and System Dynamics \DOne} 830 | \KPFormulas 831 | \begin{formulabox} 832 | \textbf{Conceptual explanation:} The center of mass (COM) is an average of the mass distribution of a system; external forces accelerate the COM as if all the mass were centered at the COM. 833 | 834 | \textbf{Core formulas:} 835 | \[ 836 | \left\{\begin{aligned} 837 | &\vec R=\frac{\sum m_i\vec r_i}{\sum m_i},\quad \vec V=\dot{\vec R}=\frac{\sum m_i\vec v_i}{M},\\ 838 | &M\,\ddot{\vec R}=\sum \vec F_{\text{ext}}.\ (\text{Internal forces cancel in pairs}) 839 | \end{aligned}\right. 840 | \] 841 | 842 | \textbf{Variable definitions:} $\vec r_i,\vec v_i$ positions/velocities; $M=\sum m_i$ total mass; $\vec R$ COM position. 843 | 844 | 845 | 846 | \textbf{Prerequisites \& scope:} Requires Newton's third law in internal pairs; for variable mass, include momentum flux. 847 | \end{formulabox} 848 | 849 | \KPHeuristics 850 | \begin{heuristicsbox} 851 | \begin{itemize}[leftmargin=*] 852 | \item Compute COM motion to track overall translation, then analyze internal relative motion separately. 853 | \item In explosions/fragmentation, the COM continues with pre-event velocity if external forces are negligible. 854 | \end{itemize} 855 | \vspace{0.4em} 856 | \begin{itemize}[leftmargin=*] 857 | \item Treating internal impulses as changing COM momentum. Fix: only external impulse changes total momentum. 858 | \end{itemize} 859 | \end{heuristicsbox} 860 | 861 | \KPProblems 862 | \begin{cheatproblem} 863 | Two masses $m_1=3\,\text{kg}$ at $x_1=0$ and $m_2=2\,\text{kg}$ at $x_2=5\,\text{m}$ lie on a line. Find the COM position. 864 | \begin{solutionbox} 865 | $X=\dfrac{m_1x_1+m_2x_2}{m_1+m_2}=\dfrac{0+10}{5}=2\,\text{m}$. 866 | \end{solutionbox} 867 | \end{cheatproblem} 868 | \begin{cheatproblem} 869 | A system has two equal masses $m$ moving at velocities $\vec v_1=(2,0)$ and $\vec v_2=(-1,3)$ m/s. Find the velocity of the COM. 870 | \begin{solutionbox} 871 | $\vec V=\dfrac{m\vec v_1+m\vec v_2}{2m}=\dfrac{(1,3)}{2}=(0.5,1.5)\,\text{m/s}$. 872 | \end{solutionbox} 873 | \end{cheatproblem} 874 | \end{KnowledgePoint} 875 | 876 | \begin{KnowledgePoint}{Variable Mass and Rockets \DTwo} 877 | \KPFormulas 878 | \begin{formulabox} 879 | \textbf{Concept explanation:} For mass-exchange systems (rockets), momentum conservation for rocket+exhaust yields logarithmic change in velocity. 880 | 881 | \textbf{Core formulas:} 882 | \[ 883 | \left\{\begin{aligned} 884 | &\text{Continuous: } m\,d\vec v=-\,\vec v_e\,dm,\\ 885 | &\text{with } dm<0\ (\text{mass loss}),\ \vec v_e\ \text{ exhaust speed relative to rocket},\\ 886 | &\text{Ideal Tsiolkovsky (1D): } \Delta v=v_e\ln\frac{m_i}{m_f}.\\ 887 | \end{aligned}\right. 888 | \] 889 | 890 | \textbf{Variable definitions:} $\vec v_e$ exhaust velocity relative to rocket (magnitude $v_e$); $m_i,m_f$ initial/final mass; $m$ instantaneous mass. 891 | 892 | 893 | 894 | \textbf{Prerequisites \& scope:} Neglect external forces (or include gravity drag separately); $v_e$ constant; exhaust ejected at steady relative speed. 895 | \end{formulabox} 896 | 897 | \KPHeuristics 898 | \begin{heuristicsbox} 899 | \begin{itemize}[leftmargin=*] 900 | \item Include gravity losses by subtracting $g\,\Delta t$ from $\Delta v$ when appropriate (vertical ascent approximation). 901 | \item Use staging by summing $v_e\ln(m_i/m_f)$ per stage. 902 | \end{itemize} 903 | \vspace{0.4em} 904 | \begin{itemize}[leftmargin=*] 905 | \item Using exhaust speed relative to Earth instead of rocket. Fix: $v_e$ is defined relative to the rocket. 906 | \end{itemize} 907 | \end{heuristicsbox} 908 | 909 | \KPProblems 910 | \begin{cheatproblem} 911 | A rocket has initial mass $m_i=1000\,\text{kg}$, final mass $m_f=400\,\text{kg}$, and exhaust speed $v_e=2000\,\text{m/s}$. Find $\Delta v$ in space (ignoring gravity). 912 | \begin{solutionbox} 913 | $\Delta v=v_e\ln\dfrac{m_i}{m_f}=2000\ln\dfrac{1000}{400}=2000\ln2.5\approx1833\,\text{m/s}$. 914 | \end{solutionbox} 915 | \end{cheatproblem} 916 | \begin{cheatproblem} 917 | A rocket ejects mass at rate $\dot m=10\,\text{kg/s}$ with effective exhaust speed $v_e=1500\,\text{m/s}$. Find the instantaneous thrust force. 918 | \begin{solutionbox} 919 | Thrust $=\dot m\,v_e=10\cdot1500=15000\,\text{N}$. 920 | \end{solutionbox} 921 | \end{cheatproblem} 922 | \end{KnowledgePoint} 923 | 924 | \Unit[Angular kinematics; torque ($|\boldsymbol{\tau}|=rF\sin\theta$); moment of inertia; $\tau_{\text{net}}=I\alpha$; rotational kinetic energy; angular momentum and conservation; rolling without slipping; inertia integrals]{Unit 5: Rotational Motion} 925 | 926 | \begin{KnowledgePoint}{Angular Kinematics and Dynamics \DOne} 927 | \KPFormulas 928 | \begin{formulabox} 929 | \textbf{Concept description:} Rotational motion is analog to linear motion: torque is the rotational analog to force, and moment of inertia of mass. 930 | 931 | \textbf{Core formulas:} 932 | \[ 933 | \left\{\begin{aligned} 934 | &\omega=\frac{d\theta}{dt},\quad \alpha=\frac{d\omega}{dt},\ \text{with constant-}\alpha\ \text{kinematics}.\\ 935 | &\boldsymbol{\tau}=\vec r\times\vec F,\quad |\boldsymbol{\tau}|=r_\perp F=rF\sin\theta,\quad \sum \tau=I\alpha,\\ 936 | &K_r=\tfrac12 I\omega^2,\quad P=\tau\,\omega. 937 | \end{aligned}\right. 938 | \] 939 | 940 | \textbf{Variable definitions:} $\theta,\omega,\alpha$ angular position/velocity/acceleration; $\tau$ torque; $I$ moment of inertia. 941 | 942 | 943 | 944 | \textbf{Prerequisites \& scope:} Rigid body about a fixed axis; $I$ constant in time. 945 | \end{formulabox} 946 | 947 | \KPHeuristics 948 | \begin{heuristicsbox} 949 | \begin{itemize}[leftmargin=*] 950 | \item Use the perpendicular lever arm $r_\perp$ for torques; sum about convenient pivots to kill unknown forces. 951 | \item Prefer energy when forces are complicated but conservative; otherwise use $\sum \tau=I\alpha$ about the COM or a fixed axis. 952 | \end{itemize} 953 | \vspace{0.4em} 954 | \begin{itemize}[leftmargin=*] 955 | \item Mixing signs of torques from different reference senses. Fix: choose a positive rotation sense and stick with it. 956 | \end{itemize} 957 | \end{heuristicsbox} 958 | 959 | \KPProblems 960 | \begin{cheatproblem} 961 | A disk of radius $r=0.5\,\text{m}$ and moment of inertia $I=2\,\text{kg·m}^2$ experiences a net torque $\tau=10\,\text{N·m}$. Find its angular acceleration. 962 | \begin{solutionbox} 963 | $\alpha=\dfrac{\tau}{I}=\dfrac{10}{2}=5\,\text{rad/s}^2$. 964 | \end{solutionbox} 965 | \end{cheatproblem} 966 | \begin{cheatproblem} 967 | A flywheel rotating at $\omega=20\,\text{rad/s}$ has $I=5\,\text{kg·m}^2$. Find its rotational kinetic energy. 968 | \begin{solutionbox} 969 | $K_r=\tfrac12 I\omega^2=\tfrac12\cdot5\cdot400=1000\,\text{J}$. 970 | \end{solutionbox} 971 | \end{cheatproblem} 972 | \end{KnowledgePoint} 973 | 974 | \begin{KnowledgePoint}{Angular Momentum and Conservation \DOne} 975 | \KPFormulas 976 | \begin{formulabox} 977 | \textbf{Explanation of concept:} Angular momentum is conserved when no external torque is applied. 978 | 979 | \textbf{Core formulas:} 980 | \[ 981 | \left\{\begin{aligned} 982 | &\vec L=I\,\vec\omega\ (\text{about fixed axis}),\ \ \sum \tau_{\text{ext}}=\frac{d\vec L}{dt}.\\ 983 | &\text{If }\sum \tau_{\text{ext}}=0,\ \vec L\ \text{conserved}. 984 | \end{aligned}\right. 985 | \] 986 | 987 | \textbf{Variable definitions:} $\vec L$ angular momentum; $I$ moment of inertia. 988 | 989 | \textbf{Prerequisites \& scope:} Axis and point of reference must be specified; rolling applies at instantaneous point of contact. 990 | \end{formulabox} 991 | 992 | \KPHeuristics 993 | \begin{heuristicsbox} 994 | \begin{itemize}[leftmargin=*] 995 | \item For isolated systems with negligible external torques, apply $\vec L$ conservation about a fixed axis. 996 | \item Choose the reference point wisely to eliminate unknown torques. 997 | \end{itemize} 998 | \vspace{0.4em} 999 | \begin{itemize}[leftmargin=*] 1000 | \item Forgetting that static friction can act either way in rolling. Fix: determine its direction from torque/acceleration requirements. 1001 | \end{itemize} 1002 | \end{heuristicsbox} 1003 | 1004 | \KPProblems 1005 | \begin{cheatproblem} 1006 | A disk of radius $R$ rolls without slipping at $v_{cm}=5\,\text{m/s}$. Find $\omega$. 1007 | \begin{solutionbox} 1008 | $v_{cm}=\omega R\Rightarrow\omega=\dfrac{v_{cm}}{R}=\dfrac{5}{R}\,\text{rad/s}$. 1009 | \end{solutionbox} 1010 | \end{cheatproblem} 1011 | \begin{cheatproblem} 1012 | A solid sphere ($I=\tfrac25 mR^2$) rolls at $\omega=10\,\text{rad/s}$ and $v_{cm}=2\,\text{m/s}$. Find its total kinetic energy if $m=3\,\text{kg}$. 1013 | \begin{solutionbox} 1014 | Using no-slip $v_{cm}=\omega R$, we have $I\omega^2=\tfrac25 mR^2\omega^2=\tfrac25 m v_{cm}^2$. Thus 1015 | $K=\tfrac12 m v_{cm}^2+\tfrac12 I\omega^2=\Big(\tfrac12+\tfrac15\Big)m v_{cm}^2=\tfrac{7}{10}\,m v_{cm}^2=\tfrac{7}{10}\cdot3\cdot4=8.4\,\text{J}$. 1016 | \end{solutionbox} 1017 | \end{cheatproblem} 1018 | \end{KnowledgePoint} 1019 | 1020 | \begin{KnowledgePoint}{Moments of Inertia \DTwo} 1021 | \KPFormulas 1022 | \begin{formulabox} 1023 | \textbf{Concept description:} Moment of inertia is a measure of resistance to angular acceleration; composition rules and standard shapes allow for rapid calculation. 1024 | 1025 | \textbf{Core formulas:} 1026 | \[ 1027 | \left\{\begin{aligned} 1028 | &I=\int r^2\,dm\ (\text{axis distance }r),\\ 1029 | &\text{Parallel axis: } I=I_{\text{cm}}+Md^2,\\ 1030 | &\text{Perpendicular axis (planar lamina): } I_z=I_x+I_y. 1031 | \end{aligned}\right. 1032 | \] 1033 | 1034 | \textbf{Variable definitions:} $I_{\text{cm}}$ about COM axis; $d$ offset; $M$ total mass. 1035 | 1036 | 1037 | \textbf{Prerequisites \& scope:} Perpendicular-axis requires lamina in the plane; parallel-axis requires fixed, parallel axes. 1038 | \end{formulabox} 1039 | 1040 | \KPHeuristics 1041 | \begin{heuristicsbox} 1042 | \begin{itemize}[leftmargin=*] 1043 | \item Decompose into standard shapes and sum moments about the same axis. 1044 | \item Use symmetry to eliminate products of inertia; choose axes through COM when possible. 1045 | \end{itemize} 1046 | \vspace{0.4em} 1047 | \begin{itemize}[leftmargin=*] 1048 | \item Applying perpendicular-axis to 3D bodies. Fix: valid only for planar laminae. 1049 | \end{itemize} 1050 | \end{heuristicsbox} 1051 | 1052 | \KPProblems 1053 | \begin{cheatproblem} 1054 | A thin rod of mass $m$ and length $L$ rotates about one end. Find its moment of inertia. 1055 | \begin{solutionbox} 1056 | $I=\int_0^L x^2\,dm=\int_0^L x^2\dfrac{m}{L}\,dx=\dfrac{m}{L}\dfrac{L^3}{3}=\dfrac{mL^2}{3}$. 1057 | \end{solutionbox} 1058 | \end{cheatproblem} 1059 | \begin{cheatproblem} 1060 | A disk of mass $M$ and radius $R$ has $I_{cm}=\tfrac12 MR^2$. Using the parallel-axis theorem, find $I$ about a point on its rim. 1061 | \begin{solutionbox} 1062 | $I=I_{cm}+MR^2=\tfrac12 MR^2+MR^2=\tfrac32 MR^2$. 1063 | \end{solutionbox} 1064 | \end{cheatproblem} 1065 | \end{KnowledgePoint} 1066 | 1067 | \begin{KnowledgePoint}{Rolling Without Slipping \DOne} 1068 | \KPFormulas 1069 | \begin{formulabox} 1070 | \textbf{Concept description:} Rolling is a mixture of translation of the center of mass and rotation; the no-slip condition ties them together. 1071 | 1072 | \textbf{Core formulas:} 1073 | \[ 1074 | \left\{\begin{aligned} 1075 | &v_{\text{cm}}=\omega R,\\ 1076 | &K=\tfrac12 M v_{\text{cm}}^2+\tfrac12 I\omega^2.\\ 1077 | \end{aligned}\right. 1078 | \] 1079 | 1080 | \textbf{Variable definitions:} $R$ radius; $M$ mass; $I$ moment of inertia; $v_{\text{cm}}$ center-of-mass speed. 1081 | \end{formulabox} 1082 | 1083 | \KPHeuristics 1084 | \begin{heuristicsbox} 1085 | \begin{itemize}[leftmargin=*] 1086 | \item Static friction can accelerate or decelerate rolling bodies but does no work on the body in pure rolling. 1087 | \item Use energy for ramp problems; use $\sum \tau=I\alpha$ when forces/accelerations are requested. 1088 | \end{itemize} 1089 | \end{heuristicsbox} 1090 | 1091 | \KPProblems 1092 | \begin{cheatproblem} 1093 | A solid cylinder of mass $m$ and radius $R$ is released from rest to roll without slipping down an incline of height $h$. Find its speed at the bottom. 1094 | \begin{solutionbox} 1095 | Energy: $mgh=\tfrac12 m v_{cm}^2+\tfrac12 I\omega^2=\tfrac12 m v_{cm}^2+\tfrac12(\tfrac12 mR^2)(v_{cm}^2/R^2)=\tfrac12 m v_{cm}^2+\tfrac14 m v_{cm}^2=\tfrac34 m v_{cm}^2$, so $v_{cm}=\sqrt{\tfrac{4}{3}gh}$. 1096 | \end{solutionbox} 1097 | \end{cheatproblem} 1098 | \begin{cheatproblem} 1099 | A rolling sphere moves to the right and speeds up. Determine the direction of static friction on the sphere. 1100 | \begin{solutionbox} 1101 | For a solid sphere with $a>0$, no slip implies friction acts up the incline or in the direction that provides a torque to increase $\omega$: here, friction acts forward (to the right) at the contact point to produce a counterclockwise torque consistent with $\dot\omega>0$. 1102 | \end{solutionbox} 1103 | \end{cheatproblem} 1104 | \end{KnowledgePoint} 1105 | 1106 | % --- New consolidated Practice Pointers for Part I --- 1107 | \subsection*{Part I: Mechanics Practice Pointers} 1108 | \addcontentsline{toc}{subsection}{Part I: Mechanics Practice Pointers} 1109 | \begin{itemize}[leftmargin=*] 1110 | \item Physics Bowl Kinematics \& Momentum Problem 1 Page: 2 1111 | \item Physics Bowl Circular Motion \& Energy Problem 2 Page: 3 1112 | \item Physics Bowl Simple Machines \& Statics Problem 5 Page: 6 1113 | \item Physics Bowl Kinematics Problem 10 Page: 11 1114 | \item Physics Bowl Kinematics Problem 13 Page: 14 1115 | \item Physics Bowl Work \& Energy with Friction Problem 14 Page: 14 1116 | \item Physics Bowl Rotational Kinematics Problem 15 Page: 16 1117 | \item Physics Bowl Rolling Dynamics Problem 16 Page: 17 1118 | \item Physics Bowl Gravitation Problem 17 Page: 18 1119 | \item Physics Bowl Rolling Energy Problem 20 Page: 21 1120 | \item Physics Bowl Statics Problem 23 Page: 24 1121 | \item Physics Bowl Relative Motion Problem 26 Page: 27 1122 | \item Physics Bowl Circular Motion Problem 27 Page: 28 1123 | \item Physics Bowl Work \& Energy with Friction Problem 28 Page: 28 1124 | \item Physics Bowl Center of Mass Problem 29 Page: 29 1125 | \item Physics Bowl Kinematics Problem 30 Page: 30 1126 | \item Physics Bowl Momentum \& Impulse Problem 33 Page: 32 1127 | \item Physics Bowl Kinematics Problem 35 Page: 33 1128 | \item Physics Bowl Dynamics \& Friction Problem 36 Page: 33 1129 | \item Physics Bowl Statics Problem 37 Page: 35 1130 | \item Physics Bowl Rotational Energy Problem 38 Page: 36 1131 | \item Physics Bowl Kinematics Problem 39 Page: 37 1132 | \end{itemize} 1133 | 1134 | % ===================== Part II: Oscillations, Waves, Thermo/Fluids ===================== 1135 | \clearpage 1136 | \parttheme{waves} 1137 | \section*{Part II: Oscillations, Waves, Thermodynamics \& Fluids} 1138 | \addcontentsline{toc}{section}{Part II: Oscillations, Waves, Thermodynamics and Fluids} 1139 | 1140 | \Unit[SHM; pendulum (small-angle); wave speed $v=f\lambda$; superposition/standing waves; sound and Doppler; SHM ODE]{Unit 6: Oscillations \& Waves} 1141 | 1142 | \begin{KnowledgePoint}{Simple Harmonic Motion \DOne} 1143 | \KPFormulas 1144 | \begin{formulabox} 1145 | \textbf{Concept explanation:} SHM occurs when acceleration is proportional to and opposite to displacement; solutions are sinusoidal with constant amplitude (no damping). 1146 | 1147 | \textbf{Core formulas:} 1148 | \[ 1149 | \left\{\begin{aligned} 1150 | &x=A\cos(\omega t+\phi),\quad v=-A\omega\sin(\omega t+\phi),\quad a=-\omega^2 x,\\ 1151 | &T=\frac{2\pi}{\omega},\quad \omega=\sqrt{\frac{k}{m}}\ (\text{mass--spring}),\quad T_{\text{pend}}\approx2\pi\sqrt{\frac{\ell}{g}}\ (\text{small angle}). 1152 | \end{aligned}\right. 1153 | \] 1154 | 1155 | \textbf{Variable definitions:} $A$ amplitude; $\omega$ angular frequency; $\phi$ phase; $k$ spring constant; $\ell$ pendulum length. 1156 | 1157 | 1158 | 1159 | \textbf{Prerequisites \& scope:} No damping/driving; small-angle approximation for pendulum. 1160 | \end{formulabox} 1161 | 1162 | \KPHeuristics 1163 | \begin{heuristicsbox} 1164 | \begin{itemize}[leftmargin=*] 1165 | \item Use energy partition $K+U=\tfrac12kA^2$ to find speeds at positions; use phase to compute time fractions. 1166 | \item For compound oscillators, reduce to effective $k_{\text{eff}}$ or $\ell_{\text{eff}}$ before applying SHM formulas. 1167 | \item Small-angle pendulum: check $\theta_{\max}\lesssim 10^\circ$ for $T\approx2\pi\sqrt{\ell/g}$ to be within a percent; otherwise expect longer $T$. First-order correction (radians; \DTwo result, memorize only): $\displaystyle T\approx 2\pi\sqrt{\ell/g}\,\big(1+\tfrac{\theta_0^2}{16}\big)$. 1168 | \end{itemize} 1169 | \vspace{0.4em} 1170 | \begin{itemize}[leftmargin=*] 1171 | \item Using pendulum period formula at large angles. Fix: restrict to small angles or use elliptic corrections. 1172 | \end{itemize} 1173 | \end{heuristicsbox} 1174 | 1175 | \KPProblems 1176 | \begin{cheatproblem} 1177 | A mass on a spring follows SHM with amplitude $A$ and period $T$. What fraction of the period is spent with $|x|>\tfrac{A}{2}$? 1178 | \begin{solutionbox} 1179 | Let $x(t)=A\cos(\omega t)$. The condition $|x(t)|>A/2$ is equivalent to $|\cos(\omega t)|>1/2$. It is simpler to calculate the fraction of time for the complementary condition, $|x(t)|\le A/2$, which corresponds to $|\cos(\omega t)|\le 1/2$. 1180 | In one full cycle $\theta=\omega t \in [0, 2\pi)$, this holds for $\theta \in [\pi/3, 2\pi/3]$ and $\theta \in [4\pi/3, 5\pi/3]$. The total angular duration is $(\frac{2\pi}{3}-\frac{\pi}{3}) + (\frac{5\pi}{3}-\frac{4\pi}{3}) = \frac{\pi}{3}+\frac{\pi}{3}=\frac{2\pi}{3}$. 1181 | The fraction of the period for this condition is $\frac{2\pi/3}{2\pi}=\frac{1}{3}$. 1182 | Therefore, the fraction of the period spent with $|x|>A/2$ is $1 - 1/3 = 2/3$. 1183 | \end{solutionbox} 1184 | \end{cheatproblem} 1185 | \end{KnowledgePoint} 1186 | 1187 | \begin{KnowledgePoint}{Waves (Traveling) \DOne} 1188 | \KPFormulas 1189 | \begin{formulabox} 1190 | \textbf{Concept explanation:} Traveling waves follow $v=f\lambda$; boundary conditions set standing-wave modes; source/observer motion shifts frequency (Doppler). 1191 | 1192 | \textbf{Core formulas:} 1193 | \[ 1194 | \left\{\begin{aligned} 1195 | &v=f\lambda,\ \ y(x,t)=A\cos(kx-\omega t+\phi),\ k=\tfrac{2\pi}{\lambda},\ \omega=2\pi f.\\ 1196 | \end{aligned}\right. 1197 | \] 1198 | 1199 | \textbf{Variable definitions:} $v$ wave speed in medium; $L$ length; $v_o$ observer speed; $v_s$ source speed. 1200 | 1201 | \textbf{Prerequisites \& scope:} Linear superposition; small amplitudes; Doppler formula assumes $v_o,v_s\ll v$ (nonrelativistic). 1202 | \end{formulabox} 1203 | 1204 | \KPHeuristics 1205 | \begin{heuristicsbox} 1206 | \begin{itemize}[leftmargin=*] 1207 | \item Draw mode shapes to match node/antinode boundary conditions before writing $f_n$. 1208 | \item Use a sign diagram for Doppler to avoid sign errors; approaching increases frequency. For the formula given, this means observer towards source ($+v_o$, numerator) and source towards observer ($-v_s$, denominator). For receding, reverse these signs. 1209 | \end{itemize} 1210 | \vspace{0.4em} 1211 | \begin{itemize}[leftmargin=*] 1212 | \item Using $v$ of sound/light incorrectly across media. Fix: use the correct medium speed for $v=f\lambda$. 1213 | \end{itemize} 1214 | \end{heuristicsbox} 1215 | 1216 | \KPProblems 1217 | \begin{cheatproblem} 1218 | A wave has wavelength $\lambda=2\,\text{m}$ and frequency $f=50\,\text{Hz}$. Find its speed. 1219 | \begin{solutionbox} 1220 | $v=f\lambda=50\cdot2=100\,\text{m/s}$. 1221 | \end{solutionbox} 1222 | \end{cheatproblem} 1223 | \end{KnowledgePoint} 1224 | 1225 | \begin{KnowledgePoint}{Standing Waves (Strings/Pipes) \DOne} 1226 | \KPFormulas 1227 | \begin{formulabox} 1228 | \textbf{Concept explanation:} Boundaries reflect waves; interference of forward and backward waves creates nodes and antinodes with discrete mode frequencies set by geometry and boundary conditions. 1229 | \textbf{Core formulas:} 1230 | \[ 1231 | \left\{\begin{aligned} 1232 | &\text{String fixed ends: } f_n=\frac{nv}{2L},\ n=1,2,\dots\\ 1233 | &\text{Open/closed pipe: }\begin{cases} 1234 | f_n=\dfrac{nv}{2L}, & n=1,2,3,\dots\ (\text{both ends open}),\\ 1235 | f_n=\dfrac{(2n-1)v}{4L}, & n=1,2,3,\dots\ (\text{one end closed}). 1236 | \end{cases} 1237 | \end{aligned}\right. 1238 | \] 1239 | \end{formulabox} 1240 | 1241 | \KPProblems 1242 | \begin{cheatproblem} 1243 | A string of length $L=1.2\,\text{m}$ fixed at both ends has fundamental frequency $f_1=200\,\text{Hz}$. Find the wave speed. 1244 | \begin{solutionbox} 1245 | $f_1=\dfrac{v}{2L}\Rightarrow v=2Lf_1=2\cdot1.2\cdot200=480\,\text{m/s}$. 1246 | \end{solutionbox} 1247 | \end{cheatproblem} 1248 | \end{KnowledgePoint} 1249 | 1250 | \begin{KnowledgePoint}{Doppler Effect (Fixed Medium) \DOne} 1251 | \KPFormulas 1252 | \begin{formulabox} 1253 | \textbf{Concept explanation:} Relative motion between source and observer shifts the detected frequency: approaching raises $f'$ and receding lowers it; the medium is stationary. 1254 | \textbf{Core formula:} 1255 | \[ 1256 | f'=\frac{v\pm v_o}{v\mp v_s}\,f\ (\text{approach uses top signs}). 1257 | \] 1258 | \textbf{Variable definitions:} $v$ wave speed; $v_o$ observer speed; $v_s$ source speed. 1259 | \end{formulabox} 1260 | 1261 | \KPProblems 1262 | \begin{cheatproblem} 1263 | An acoustic source emits $f=440\,\text{Hz}$ in air with $v=340\,\text{m/s}$. The observer moves toward the source at $v_o=10\,\text{m/s}$ while the source moves toward the observer at $v_s=20\,\text{m/s}$. Find the observed frequency. 1264 | \begin{solutionbox} 1265 | $f'=\dfrac{v+v_o}{v-v_s}\,f=\dfrac{350}{320}\cdot 440\approx 481\,\text{Hz}$. 1266 | \end{solutionbox} 1267 | \end{cheatproblem} 1268 | \end{KnowledgePoint} 1269 | 1270 | \begin{KnowledgePoint}{SHM ODE and Driven Forms \DTwo} 1271 | \KPFormulas 1272 | \begin{formulabox} 1273 | \textbf{Concept explanation:} The homogeneous SHM ODE has sinusoidal solutions; with driving and damping, the steady-state response depends on drive frequency and damping ratio. 1274 | 1275 | \textbf{Core formulas:} 1276 | \[ 1277 | \left\{\begin{aligned} 1278 | &x''+\omega_0^2 x=0 \Rightarrow x(t)=C\cos\omega_0 t+D\sin\omega_0 t,\\ 1279 | &\text{Damped: } x''+2\zeta\omega_0 x'+\omega_0^2 x=0,\ \text{under/critical/over-damped by }\zeta,\\ 1280 | &\text{Driven: } x''+2\zeta\omega_0 x'+\omega_0^2 x=\frac{F_0}{m}\cos\omega t,\\ 1281 | &\Rightarrow\ \text{amplitude peaks near }\omega\approx\omega_0,\\ 1282 | &\text{for small damping }\omega_{\text{peak}}\approx\omega_0\sqrt{1-2\zeta^2}\ (\zeta\ll1). 1283 | \end{aligned}\right. 1284 | \] 1285 | 1286 | \textbf{Variable definitions:} $\omega_0$ natural frequency; $\zeta$ damping ratio; $F_0$ drive amplitude. 1287 | 1288 | \textbf{Prerequisites \& scope:} Linear oscillator model; small oscillations; steady-state assumes transients have decayed. 1289 | \end{formulabox} 1290 | 1291 | \KPHeuristics 1292 | \begin{heuristicsbox} 1293 | \begin{itemize}[leftmargin=*] 1294 | \item Identify regime via $\zeta$; near resonance, estimate amplification and phase shift. 1295 | \end{itemize} 1296 | \vspace{0.4em} 1297 | \begin{itemize}[leftmargin=*] 1298 | \item Confusing natural and driving frequencies. Fix: keep $\omega_0$ (system) distinct from $\omega$ (drive). 1299 | \end{itemize} 1300 | \end{heuristicsbox} 1301 | 1302 | \KPProblems 1303 | \begin{cheatproblem} 1304 | For $x''+\omega^2x=0$ with $x(0)=0$ and $\dot x(0)=v_0$, find $x(t)$ and the maximum speed. 1305 | \begin{solutionbox} 1306 | The general solution is $x(t)=C\cos(\omega t)+D\sin(\omega t)$. 1307 | $x(0)=0 \Rightarrow C=0$. 1308 | $\dot x(t) = D\omega\cos(\omega t)$, so $\dot x(0)=v_0 \Rightarrow D\omega=v_0 \Rightarrow D=v_0/\omega$. 1309 | Thus, $x(t)=\dfrac{v_0}{\omega}\sin(\omega t)$. The velocity is $v(t) = \dot x(t) = v_0\cos(\omega t)$. 1310 | The maximum speed is the amplitude of $v(t)$, which is $|v_0|$. 1311 | \end{solutionbox} 1312 | \end{cheatproblem} 1313 | \end{KnowledgePoint} 1314 | 1315 | \Unit[Hydrostatics (pressure/buoyancy); continuity; Bernoulli; ideal gas; First Law and engines; entropy]{Unit 7: Fluids \& Thermodynamics} 1316 | 1317 | \begin{KnowledgePoint}{Hydrostatics and Buoyancy \DOne} 1318 | \KPFormulas 1319 | \begin{formulabox} 1320 | \textbf{Concept explanation:} Static fluids exert pressure that increases with depth; the buoyant force equals the weight of displaced fluid (Archimedes). 1321 | 1322 | \textbf{Core formulas:} 1323 | \[ 1324 | \left\{\begin{aligned} 1325 | &P=P_0+\rho g h\ (\text{hydrostatic pressure}),\\ 1326 | &F_b=\rho g V_{\text{disp}}\ (\text{buoyancy}). 1327 | \end{aligned}\right. 1328 | \] 1329 | 1330 | \textbf{Variable definitions:} $P_0$ reference pressure (often atmospheric at $h=0$); $\rho$ fluid density; $h$ depth; $V_{\text{disp}}$ displaced volume. 1331 | 1332 | 1333 | 1334 | \textbf{Prerequisites \& scope:} Fluid at rest (no flow), constant $\rho$ with depth (or integrate if varying); neglect surface tension unless specified. 1335 | \end{formulabox} 1336 | 1337 | \KPHeuristics 1338 | \begin{heuristicsbox} 1339 | \begin{itemize}[leftmargin=*] 1340 | \item Draw free-body diagrams (FBD) of floating/sinking bodies: set $F_b$ vs weight vs any tension to solve equilibrium. 1341 | \item Choose a definite reference level for $h$ and keep $P_0$ the same when comparing points. 1342 | \item Submerged fraction for floating: $\dfrac{V_{sub}}{V}=\dfrac{\rho_b}{\rho}$ (with $0<\rho_b\le\rho$). 1343 | \item If $\rho_b>\rho$ and the object is released freely, the initial net force is downward: $mg-F_b>0$. 1344 | \item If later supported (bottom contact or tension), static equilibrium requires $T+F_b=mg$. 1345 | \item Gauge vs absolute pressure: $\Delta P=\rho g h$ is a \emph{gauge} difference; absolute pressure is $P= P_{\text{atm}}+\rho g h$ when the surface is open to atmosphere. 1346 | \end{itemize} 1347 | \vspace{0.4em} 1348 | \begin{itemize}[leftmargin=*] 1349 | \item Using object's volume instead of displaced volume for $F_b$. Fix: use actual displaced fluid volume (submerged part only). 1350 | \end{itemize} 1351 | \end{heuristicsbox} 1352 | 1353 | \KPProblems 1354 | \begin{cheatproblem} 1355 | A block of volume $V$ and density $\rho_b$ floats in a liquid of density $\rho$. What fraction of its volume is submerged? 1356 | \begin{solutionbox} 1357 | At equilibrium $\rho g V_{sub}=\rho_b g V$, so $V_{sub}/V=\rho_b/\rho$. 1358 | \end{solutionbox} 1359 | \end{cheatproblem} 1360 | % (duplicate problem removed) 1361 | \end{KnowledgePoint} 1362 | 1363 | \begin{KnowledgePoint}{Continuity and Bernoulli \DOne} 1364 | \KPFormulas 1365 | \begin{formulabox} 1366 | \textbf{Concept explanation:} In steady incompressible flow, mass conservation gives $Av=\text{const}$; along a streamline with negligible viscosity, mechanical energy per volume is constant (Bernoulli). 1367 | 1368 | \textbf{Core formulas:} 1369 | \[ 1370 | \left\{\begin{aligned} 1371 | &\text{Continuity: } A_1 v_1=A_2 v_2\ (\rho \ \text{constant}).\\ 1372 | &\text{Bernoulli: } P+\tfrac12\rho v^2+\rho g y=\text{const}\ (\text{along a streamline}). 1373 | \end{aligned}\right. 1374 | \] 1375 | 1376 | \textbf{Variable definitions:} $A$ cross-sectional area; $v$ speed; $P$ pressure; $y$ elevation; $\rho$ density. 1377 | 1378 | 1379 | 1380 | \textbf{Prerequisites \& scope:} Steady, incompressible, non-viscous flow; apply Bernoulli along a streamline, not across shocks or with pumps/turbines unaccounted. 1381 | \end{formulabox} 1382 | 1383 | \KPHeuristics 1384 | \begin{heuristicsbox} 1385 | \begin{itemize}[leftmargin=*] 1386 | \item Check assumptions (steady/incompressible/irrotational) before using Bernoulli; otherwise use energy loss terms. 1387 | \item Combine continuity with Bernoulli to eliminate speeds or pressures efficiently. 1388 | \item Use stagnation points: where $v=0$, the total (stagnation) pressure is $P_0=P+\tfrac12\rho v^2$ upstream along a streamline. 1389 | \end{itemize} 1390 | \vspace{0.4em} 1391 | \begin{itemize}[leftmargin=*] 1392 | \item Using Bernoulli across different streamlines where viscous losses or pumps exist. Fix: apply along a single streamline and include head gains/losses when needed. 1393 | \end{itemize} 1394 | \end{heuristicsbox} 1395 | 1396 | \KPProblems 1397 | \begin{cheatproblem} 1398 | Water flows through a horizontal pipe from diameter $D_1=0.1\,\text{m}$ to $D_2=0.05\,\text{m}$. If $v_1=2\,\text{m/s}$, find $v_2$ using continuity. 1399 | \begin{solutionbox} 1400 | $A_1v_1=A_2v_2\Rightarrow\pi(D_1/2)^2v_1=\pi(D_2/2)^2v_2\Rightarrow v_2=v_1\dfrac{D_1^2}{D_2^2}=2\dfrac{0.01}{0.0025}=8\,\text{m/s}$. 1401 | \end{solutionbox} 1402 | \end{cheatproblem} 1403 | \begin{cheatproblem} 1404 | At point 1 in a pipe, $P_1=1.0\times10^5\,\text{Pa}$, $v_1=2\,\text{m/s}$, $y_1=0$. At point 2, $v_2=5\,\text{m/s}$, $y_2=3\,\text{m}$. Find $P_2$ for water ($\rho=1000\,\text{kg/m}^3$). 1405 | \begin{solutionbox} 1406 | Bernoulli: $P_1+\tfrac12\rho v_1^2+\rho gy_1=P_2+\tfrac12\rho v_2^2+\rho gy_2$. 1407 | Compute in SI with scientific notation: 1408 | $\tfrac12\rho(v_1^2-v_2^2)=0.5\times10^3\,(4-25)=-1.05\times10^4\,\text{Pa}$ and 1409 | $\rho g(y_2-y_1)=10^3\times9.8\times3=2.94\times10^4\,\text{Pa}$. 1410 | Hence $P_2=1.00\times10^5-1.05\times10^4-2.94\times10^4=6.01\times10^4\,\text{Pa}$. 1411 | \end{solutionbox} 1412 | \end{cheatproblem} 1413 | \end{KnowledgePoint} 1414 | 1415 | \begin{KnowledgePoint}{Ideal Gas and First Law \DOne} 1416 | \KPFormulas 1417 | \begin{formulabox} 1418 | \textbf{Concept explanation:} $PV=nRT$ holds for ideal gases; the First Law links changes in internal energy to heat and work with clear sign conventions. 1419 | 1420 | \textbf{Core formulas:} 1421 | \[ 1422 | \left\{\begin{aligned} 1423 | &PV=nRT,\quad U=\tfrac{f}{2}nRT,\\ 1424 | &\text{where } f\ \text{is dof (e.g., }f=3\ \text{monatomic},\ 5\ \text{diatomic at room T)},\\ 1425 | &\Delta U=Q+W_{\text{on}}\ \ (W_{\text{on}}=\text{work done on the system}),\\ 1426 | &W_{\text{on}}=-\int P\,dV\quad (\text{so }W_{\text{by}}\equiv-\,W_{\text{on}}=\int P\,dV). 1427 | \end{aligned}\right. 1428 | \] 1429 | 1430 | \textbf{Variable definitions:} $P,V,T$ pressure/volume/temperature; $n$ moles; $R$ gas constant; $Q$ heat into system; $W_{\text{on}}$ work on system; $W_{\text{by}}$ work done by the gas; $U$ internal energy. 1431 | 1432 | 1433 | 1434 | \textbf{Prerequisites \& scope:} Ideal gas approximation; $U$ depends only on $T$ for ideal gases; sign convention must be consistent. 1435 | \end{formulabox} 1436 | 1437 | \KPHeuristics 1438 | \begin{heuristicsbox} 1439 | \begin{itemize}[leftmargin=*] 1440 | \item Identify process (isochoric/isobaric/isothermal/adiabatic) to pick $W,\,Q,\,\Delta U$ quickly. 1441 | \item Draw $P$–$V$ diagrams: areas give work; direction indicates sign. 1442 | \end{itemize} 1443 | \vspace{0.4em} 1444 | \begin{itemize}[leftmargin=*] 1445 | \item Mixing sign conventions for work. Fix: adopt $\Delta U=Q+W_{\text{on}}$ (work on system positive) consistently; then $W_{\text{by}}=-W_{\text{on}}$. 1446 | \end{itemize} 1447 | \end{heuristicsbox} 1448 | 1449 | \KPProblems 1450 | \begin{cheatproblem} 1451 | An ideal gas undergoes an isothermal expansion from volume $V_1$ to $V_2$ at temperature $T$. Find the work done by the gas. 1452 | \begin{solutionbox} 1453 | $W=\int_{V_1}^{V_2} \frac{nRT}{V}\,dV=nRT\ln\!\frac{V_2}{V_1}$. 1454 | \end{solutionbox} 1455 | \end{cheatproblem} 1456 | \begin{cheatproblem} 1457 | An ideal gas at $P=2\times10^5\,\text{Pa}$ and $V=0.01\,\text{m}^3$ has $n=1\,\text{mol}$. Find the temperature $T$ using $PV=nRT$ with $R=8.314\,\text{J/(mol·K)}$. 1458 | \begin{solutionbox} 1459 | $T=\dfrac{PV}{nR}=\dfrac{2\times10^5\cdot0.01}{1\cdot8.314}\approx240\,\text{K}$. 1460 | \end{solutionbox} 1461 | \end{cheatproblem} 1462 | \end{KnowledgePoint} 1463 | 1464 | \begin{KnowledgePoint}{Entropy and Carnot \DOne} 1465 | \KPFormulas 1466 | \begin{formulabox} 1467 | \textbf{Concept explanation:} Entropy quantifies thermal disorder and increases in irreversible processes; Carnot gives the maximum efficiency of heat engines between two reservoirs. 1468 | 1469 | \textbf{Core formulas:} 1470 | \[ 1471 | \left\{\begin{aligned} 1472 | &\Delta S=\int_{\text{rev}} \frac{\delta Q}{T},\\ 1473 | &\text{Carnot bound: } \eta_{\max}=1-\frac{T_c}{T_h}. 1474 | \end{aligned}\right. 1475 | \] 1476 | 1477 | \textbf{Variable definitions:} $S$ entropy; $T_h,T_c$ hot/cold absolute temperatures; $\delta Q$ infinitesimal heat (reversible path). 1478 | 1479 | 1480 | 1481 | \textbf{Prerequisites \& scope:} Absolute temperatures (Kelvin); reversible paths for definition; real engines achieve less than Carnot due to irreversibilities. 1482 | \end{formulabox} 1483 | 1484 | \KPHeuristics 1485 | \begin{heuristicsbox} 1486 | \begin{itemize}[leftmargin=*] 1487 | \item Compute $\Delta S$ along a convenient reversible path (e.g., isothermal + isochoric steps). 1488 | \item For engine limits, compare cycle temperatures to $T_h,T_c$ to bound $\eta$ quickly. 1489 | \end{itemize} 1490 | \vspace{0.4em} 1491 | \begin{itemize}[leftmargin=*] 1492 | \item Using Celsius in $\eta=1-T_c/T_h$. Fix: convert to Kelvin. 1493 | \end{itemize} 1494 | \end{heuristicsbox} 1495 | 1496 | \KPProblems 1497 | \begin{cheatproblem} 1498 | A Carnot engine operates between $T_h=500\,\text{K}$ and $T_c=300\,\text{K}$. Find its maximum efficiency. 1499 | \begin{solutionbox} 1500 | $\eta_{\max}=1-\dfrac{T_c}{T_h}=1-\dfrac{300}{500}=0.4=40\%$. 1501 | \end{solutionbox} 1502 | \end{cheatproblem} 1503 | \begin{cheatproblem} 1504 | An ideal gas expands reversibly at constant $T=400\,\text{K}$ from $V_1=1\,\text{m}^3$ to $V_2=2\,\text{m}^3$. Find the entropy change if $n=1\,\text{mol}$. 1505 | \begin{solutionbox} 1506 | $\Delta S=\dfrac{Q_{rev}}{T}=\dfrac{nRT\ln(V_2/V_1)}{T}=nR\ln2=8.314\ln2\approx5.76\,\text{J/K}$. 1507 | \end{solutionbox} 1508 | \end{cheatproblem} 1509 | \end{KnowledgePoint} 1510 | 1511 | \begin{KnowledgePoint}{Heat Engines and Efficiency \DOne} 1512 | \KPFormulas 1513 | \begin{formulabox} 1514 | \textbf{Core formulas:} 1515 | \[ 1516 | \left\{\begin{aligned} 1517 | &W_{\text{by}}=\int P\,dV,\quad W_{\text{on}}=-W_{\text{by}},\quad \eta=\frac{W_{\text{by}}}{Q_h},\\ 1518 | &\eta_{\max,\,\text{Carnot}}=1-\frac{T_c}{T_h}. 1519 | \end{aligned}\right. 1520 | \] 1521 | \textbf{Variable definitions:} $Q_h$ heat absorbed from hot reservoir; $T_h,T_c$ absolute temperatures of hot/cold reservoirs; $W_{\text{by}}$ work done by the gas. 1522 | \end{formulabox} 1523 | \end{KnowledgePoint} 1524 | 1525 | % --- New consolidated Practice Pointers for Part II --- 1526 | \subsection*{Part II: Oscillations, Waves, Thermodynamics and Fluids Practice Pointers} 1527 | \addcontentsline{toc}{subsection}{Part II: Oscillations, Waves, Thermodynamics and Fluids Practice Pointers} 1528 | \begin{itemize}[leftmargin=*] 1529 | \item Physics Bowl Waves \& Sound Problem 3 Page: 4 1530 | \item Physics Bowl Thermodynamics \& Phase Change Problem 4 Page: 5 1531 | \item Physics Bowl Thermodynamics \& Phase Equilibrium Problem 6 Page: 7 1532 | \item Physics Bowl Fluid Mechanics Problem 11 Page: 12 1533 | \item Physics Bowl Fluid Mechanics \& Projectile Motion Problem 18 Page: 19 1534 | \item Physics Bowl Thermodynamics \& Engines Problem 19 Page: 20 1535 | \item Physics Bowl Oscillations Problem 24 Page: 25 1536 | \item Physics Bowl Oscillations Problem 31 Page: 31 1537 | \end{itemize} 1538 | 1539 | % ===================== Part III: Electricity & Magnetism ===================== 1540 | \clearpage 1541 | \parttheme{em} 1542 | \section*{Part III: Electricity \texorpdfstring{\&}{&} Magnetism} 1543 | \addcontentsline{toc}{section}{Part III: Electricity \& Magnetism} 1544 | 1545 | \Unit[Coulomb force; electric field/potential; capacitors and energy; Gauss's law]{Unit 8: Electrostatics} 1546 | 1547 | \begin{KnowledgePoint}{Coulomb, Field and Potential \DOne} 1548 | \KPFormulas 1549 | \begin{formulabox} 1550 | \textbf{Concept explanation:} Point charges interact by an inverse-square law; electric field and potential describe force per unit charge and energy per unit charge. 1551 | 1552 | \textbf{Core formulas:} 1553 | \[ 1554 | \left\{\begin{aligned} 1555 | &|\vec F|=k\,\frac{|q_1 q_2|}{r^2}\ (\text{along the line of centers}),\\ 1556 | &\vec E=\frac{\vec F}{q},\quad V=\frac{U}{q},\quad \Delta U=-q\int \vec E\cdot d\vec r,\\ 1557 | &\text{Point charge: } E=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2},\ V=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r}. 1558 | \end{aligned}\right. 1559 | \] 1560 | 1561 | \textbf{Variable definitions:} $q,Q$ charges; $r$ separation; $k=1/(4\pi\varepsilon_0)$; $U$ potential energy. 1562 | 1563 | \textbf{Prerequisites \& scope:} Electrostatics (charges at rest); superposition holds; signs determine directions. 1564 | \end{formulabox} 1565 | 1566 | \KPHeuristics 1567 | \begin{heuristicsbox} 1568 | \begin{itemize}[leftmargin=*] 1569 | \item Apply symmetry (dipoles, rings, infinite sheets) to cancel parts before integrating. 1570 | \item Apply potential for conservative additions first, then differentiate to get fields. 1571 | \end{itemize} 1572 | \vspace{0.4em} 1573 | \begin{itemize}[leftmargin=*] 1574 | \item Not remembering vector directions for $\vec E$ and $\vec F$. Fix: graph direction first, calculate magnitude second. 1575 | \end{itemize} 1576 | \end{heuristicsbox} 1577 | 1578 | \KPProblems 1579 | \begin{cheatproblem} 1580 | Two point charges $+Q$ are at $(\pm a,0)$. Find the electric field on the $y$-axis at $(0,y)$. 1581 | \begin{solutionbox} 1582 | Horizontal components cancel; vertical add: $E_y=2\,kQ\,\dfrac{y}{(a^2+y^2)^{3/2}}$. 1583 | \end{solutionbox} 1584 | \end{cheatproblem} 1585 | \end{KnowledgePoint} 1586 | 1587 | % Constants and units quick reference (placed near EM section for convenience) 1588 | \begin{insightbox} 1589 | \textbf{Constants/units:} $\varepsilon_0=8.85\times10^{-12}\,\mathrm{F/m}$, $\mu_0=4\pi\times10^{-7}\,\mathrm{H/m}$, $c=3.00\times10^8\,\mathrm{m/s}$, $e=1.60\times10^{-19}\,\mathrm{C}$. Use SI unless specified. 1590 | \end{insightbox} 1591 | 1592 | % (Removed redundant KnowledgePoint: Point Charge Field; content covered in Unit 8 Electrostatics) 1593 | 1594 | \begin{KnowledgePoint}{Capacitors and Energy \DOne} 1595 | \KPFormulas 1596 | \begin{formulabox} 1597 | \textbf{Concept explanation:} A capacitor is a device that holds equal and opposite charge on two conductors with a gap or dielectric between them. The stored charge at a given potential difference will be proportional to the geometry and material. Networks simplify by simple series/parallel formulas and energy can be traced by $U=\tfrac12 C V^2$. 1598 | 1599 | \textbf{Core formulas:} 1600 | \[ 1601 | \left\{\begin{aligned} 1602 | &C=\varepsilon_0\,\frac{A}{d}\ \text{(parallel plates in vacuum)},\quad C=\varepsilon_r\varepsilon_0\,\frac{A}{d}\ \text{(uniform dielectric)},\\ 1603 | &\text{Series: } \frac{1}{C_s}=\sum \frac{1}{C_i},\quad \text{Parallel: } C_p=\sum C_i,\\ 1604 | &U=\tfrac12 C V^2=\tfrac12 QV=\frac{Q^2}{2C}. 1605 | \end{aligned}\right. 1606 | \] 1607 | 1608 | \textbf{Variable definitions:} $A$ plate area; $d$ separation; $\varepsilon_0$ vacuum permittivity; $\varepsilon_r$ relative permittivity; $Q$ charge; $V$ voltage. 1609 | 1610 | \textbf{Prerequisites \& scope:} Edge effects neglected; linear dielectrics; use equivalent capacitance to reduce networks. 1611 | \end{formulabox} 1612 | 1613 | \KPHeuristics 1614 | \begin{heuristicsbox} 1615 | \begin{itemize}[leftmargin=*] 1616 | \item Parallel by series and symmetry simplify before writing node/loop equations. 1617 | \item Use $U=\tfrac12 CV^2$ to compare energy storage or redistribution after reconfiguration. 1618 | \end{itemize} 1619 | \vspace{0.4em} 1620 | \begin{itemize}[leftmargin=*] 1621 | \item Assuming charge conservation on each plate when switches change connectivity. Fix: conserve charge on isolated conductors only. 1622 | \end{itemize} 1623 | \end{heuristicsbox} 1624 | 1625 | \KPProblems 1626 | \begin{cheatproblem} 1627 | Two capacitors $C_1=2\,\mu\text{F}$ and $C_2=3\,\mu\text{F}$ are in series. Find the equivalent capacitance. 1628 | \begin{solutionbox} 1629 | $\dfrac{1}{C_s}=\dfrac{1}{C_1}+\dfrac{1}{C_2}=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\Rightarrow C_s=\dfrac{6}{5}=1.2\,\mu\text{F}$. 1630 | \end{solutionbox} 1631 | \end{cheatproblem} 1632 | \begin{cheatproblem} 1633 | A capacitor with $C=10\,\mu\text{F}$ is charged to $V=50\,\text{V}$. Find the energy stored. 1634 | \begin{solutionbox} 1635 | $U=\tfrac12 CV^2=\tfrac12\cdot10\times10^{-6}\cdot2500=12.5\times10^{-3}=12.5\,\text{mJ}$. 1636 | \end{solutionbox} 1637 | \end{cheatproblem} 1638 | \end{KnowledgePoint} 1639 | 1640 | \begin{KnowledgePoint}{Gauss's Law \DOne} 1641 | \KPFormulas 1642 | \begin{formulabox} 1643 | \textbf{Concept explanation:} The flux of $\vec E$ through a closed surface equals enclosed charge over $\varepsilon_0$; symmetry lets you get fields without integration. 1644 | 1645 | \textbf{Core formula:} 1646 | \[ 1647 | \oint \vec E\cdot d\vec A=\frac{Q_{enc}}{\varepsilon_0}. 1648 | \] 1649 | 1650 | \textbf{Variable definitions:} $Q_{enc}$ charge enclosed; $d\vec A$ outward area element; choose Gaussian surface aligned to symmetry. 1651 | 1652 | 1653 | 1654 | \textbf{Prerequisites \& scope:} Use for infinite planes/cylinders/spheres; for conductors, $E=0$ inside and charges reside on surfaces. Within uniform dielectrics/insulators with embedded charge, fields may exist inside the material (i.e., $E$ need not vanish). \textit{Under \DOne: memorize the integral statement; derivations are not required.} 1655 | \end{formulabox} 1656 | 1657 | \KPHeuristics 1658 | \begin{heuristicsbox} 1659 | \begin{itemize}[leftmargin=*] 1660 | \item Pick surfaces where $E$ is constant and parallel to $d\vec A$ over large patches (sphere/cylinder/plane). 1661 | \item For conductors in electrostatics, set $E_{\text{inside}}=0$ and use boundary conditions for surface charge; for dielectrics, prefer symmetry and superposition without advanced $\vec D$ formalism. 1662 | \end{itemize} 1663 | \vspace{0.4em} 1664 | \begin{itemize}[leftmargin=*] 1665 | \item Choosing a Gaussian surface that doesn't match symmetry, forcing difficult integrals. Fix: reselect surface to exploit symmetry. 1666 | \end{itemize} 1667 | \end{heuristicsbox} 1668 | 1669 | \KPProblems 1670 | \begin{cheatproblem} 1671 | Using Gauss's law, find $E(r)$ outside a uniformly charged sphere of radius $R$ and total charge $Q$. 1672 | \begin{solutionbox} 1673 | Gaussian sphere: $E\cdot4\pi r^2=Q/\varepsilon_0$, so $E(r)=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r^2}$ for $r\ge R$. 1674 | \end{solutionbox} 1675 | \end{cheatproblem} 1676 | \end{KnowledgePoint} 1677 | 1678 | % (Removed outdated Practice Pointers for Part III Unit 8) 1679 | 1680 | \Unit[Ohm's law and power; series/parallel reductions; Kirchhoff (KCL/KVL); RC qualitative; RC exact]{Unit 9: DC Circuits} 1681 | 1682 | \begin{KnowledgePoint}{Ohm's Law and Reductions \DOne} 1683 | \KPFormulas 1684 | \begin{formulabox} 1685 | \textbf{Concept explanation:} Ohm's law relates voltage, current, and resistance; power forms help rank dissipation; series/parallel laws simplify networks. 1686 | 1687 | \textbf{Core formulas:} 1688 | \[ 1689 | \left\{\begin{aligned} 1690 | &V=IR,\quad P=IV=I^2R=\frac{V^2}{R},\\ 1691 | &R_s=\sum R_i,\quad \frac{1}{R_p}=\sum \frac{1}{R_i}. 1692 | \end{aligned}\right. 1693 | \] 1694 | 1695 | \textbf{Variable definitions:} $V$ voltage; $I$ current; $R$ resistance; $P$ power. 1696 | 1697 | \textbf{Prerequisites \& scope:} Ohmic elements only; temperature dependence ignored unless specified. 1698 | \end{formulabox} 1699 | 1700 | \KPHeuristics 1701 | \begin{heuristicsbox} 1702 | \begin{itemize}[leftmargin=*] 1703 | \item Reduce networks with series/parallel and symmetry; then solve KCL/KVL for the rest of unknowns alone. 1704 | \item Equate power by using $I^2 R$ or $V^2/R$ based on conditions of fixed current/voltage. 1705 | \item In bridge-type circuits, look for equal potentials in a branch (through symmetry or KCL) to remove it. 1706 | \end{itemize} 1707 | \vspace{0.4em} 1708 | \begin{itemize}[leftmargin=*] 1709 | \item Mixing fixed-voltage and fixed-current contexts when comparing brightness. Fix: pick the power form consistent with constraints. 1710 | \end{itemize} 1711 | \end{heuristicsbox} 1712 | 1713 | \KPProblems 1714 | \begin{cheatproblem} 1715 | Three resistors of $R$ are in parallel; find the equivalent resistance. 1716 | \begin{solutionbox} 1717 | $\dfrac{1}{R_{eq}}=\dfrac{1}{R}+\dfrac{1}{R}+\dfrac{1}{R}$ so $R_{eq}=R/3$. 1718 | \end{solutionbox} 1719 | \end{cheatproblem} 1720 | \end{KnowledgePoint} 1721 | 1722 | \begin{KnowledgePoint}{Kirchhoff Laws (KCL/KVL) \DOne} 1723 | \KPFormulas 1724 | \begin{formulabox} 1725 | \textbf{Concept explanation:} Kirchhoff's current and voltage laws enforce charge and energy conservation; first-order RC circuits charge and discharge exponentially with time constant $\tau$. 1726 | 1727 | \textbf{Core formulas:} 1728 | \[ 1729 | \left\{\begin{aligned} 1730 | &\text{KCL: } \sum I_{\text{in}}=\sum I_{\text{out}}\ (\text{at a node}),\\ 1731 | &\text{KVL: } \sum \Delta V=0\ (\text{around a loop}),\\ 1732 | &\text{Charging (step from }V_0\to V):\\ 1733 | &\quad V_C(t)=V+\big(V_0-V\big)e^{-t/RC},\\ 1734 | &\quad I(t)=\frac{V-V_0}{R}\,e^{-t/RC},\\ 1735 | &\text{Discharging (to 0): }\\ 1736 | &\quad V_C(t)=V_0\,e^{-t/RC},\\ 1737 | &\quad I(t)=-\frac{V_0}{R}e^{-t/RC},\\ 1738 | &\text{General: }\ V_C(t)=V_\infty+\big(V_0-V_\infty\big)e^{-t/\tau}. 1739 | \end{aligned}\right. 1740 | \] 1741 | 1742 | \textbf{Variable definitions:} $\tau=RC$ time constant; $V$ source voltage; $V_C$ capacitor voltage; $I$ branch current. 1743 | 1744 | \textbf{Prerequisites \& scope:} Linear time-invariant components; piecewise-constant sources for standard transients. 1745 | \end{formulabox} 1746 | 1747 | \KPHeuristics 1748 | \begin{heuristicsbox} 1749 | \begin{itemize}[leftmargin=*] 1750 | \item Use series/parallel reductions or source transformations conceptually when helpful, but solve RC timing with baseline KCL/KVL forms. 1751 | \item Check for limiting values at $t=0^+$ and $t=\infty$ to test expressions; impose continuity of $V_C$ at switching times. 1752 | \item Time constant: $\tau=RC$ for the basic first-order RC considered here. 1753 | \end{itemize} 1754 | \vspace{0.4em} 1755 | \begin{itemize}[leftmargin=*] 1756 | \item Letting capacitor voltage jump at $t=0$. Fix: enforce continuity of $V_C$ and initial condition from prior steady state. 1757 | \end{itemize} 1758 | \end{heuristicsbox} 1759 | 1760 | \KPProblems 1761 | \begin{cheatproblem} 1762 | An $RC$ circuit with $V$ applied at $t=0$ has $R=2\,\Omega$, $C=1\,\text{F}$. Find $V_C(t)$. 1763 | \begin{solutionbox} 1764 | $V_C(t)=V(1-e^{-t/RC})=V(1-e^{-t/2})$. 1765 | \end{solutionbox} 1766 | \end{cheatproblem} 1767 | \begin{cheatproblem} 1768 | Quick check for limits: For any first-order $RC$ with step to $V_\infty$, verify $V_C(0^+)=V_C(0^-)$ (no jump) and $V_C(\infty)=V_\infty$. With $V_C(0^-)=V_0$, the standard form $V_C(t)=V_\infty+(V_0-V_\infty)e^{-t/\tau}$ satisfies both. 1769 | \begin{solutionbox} 1770 | Evaluate: $V_C(0^+)=V_\infty+(V_0-V_\infty)=V_0$ (continuous). As $t\to\infty$, $e^{-t/\tau}\to0$, so $V_C\to V_\infty$. 1771 | \end{solutionbox} 1772 | \end{cheatproblem} 1773 | \end{KnowledgePoint} 1774 | 1775 | \begin{KnowledgePoint}{RC Transients (First Order) \DOne} 1776 | \KPFormulas 1777 | \begin{formulabox} 1778 | \textbf{Core results:} 1779 | \[ 1780 | \left\{\begin{aligned} 1781 | &\tau=RC,\\ 1782 | &\text{Charge to }V:\ V_C(t)=V+\big(V_0-V\big)e^{-t/RC},\\ 1783 | &\text{Discharge to 0: } V_C(t)=V_0 e^{-t/RC}. 1784 | \end{aligned}\right. 1785 | \] 1786 | \end{formulabox} 1787 | 1788 | \KPProblems 1789 | \begin{cheatproblem} 1790 | An $RC$ circuit with $V$ applied at $t=0$ has $R=2\,\Omega$, $C=1\,\text{F}$. Find $V_C(t)$. 1791 | \begin{solutionbox} 1792 | $V_C(t)=V(1-e^{-t/RC})=V(1-e^{-t/2}).$ 1793 | \end{solutionbox} 1794 | \end{cheatproblem} 1795 | \end{KnowledgePoint} 1796 | 1797 | % (Removed outdated Practice Pointers for Part III Unit 9) 1798 | 1799 | \Unit[Lorentz force (charges/wires); Faraday-Lenz induction; EM spectrum; Ampere law]{Unit 10: Magnetism \& Induction} 1800 | 1801 | \begin{KnowledgePoint}{Lorentz Force \DOne} 1802 | \KPFormulas 1803 | \begin{formulabox} 1804 | \textbf{Concept explanation:} A moving charge feels $q\vec E$ and $q\,\vec v\times\vec B$; the magnetic force stays perpendicular to velocity, so it deflects direction without changing speed. 1805 | 1806 | \textbf{Core formulas:} 1807 | \[ 1808 | \left\{\begin{aligned} 1809 | &\vec F=q\,(\vec E+\vec v\times\vec B) 1810 | \end{aligned}\right. 1811 | \] 1812 | 1813 | \textbf{Variable definitions:} $q$ charge; $\vec v$ particle velocity; $\vec E,\vec B$ fields. 1814 | 1815 | \textbf{Prerequisites \& scope:} Nonrelativistic; right-hand rule for cross products. 1816 | \end{formulabox} 1817 | 1818 | \KPHeuristics 1819 | \begin{heuristicsbox} 1820 | \begin{itemize}[leftmargin=*] 1821 | \item Use right-hand rule consistently; reverse direction for negative charges. 1822 | \item Magnetic force does no work (always perpendicular to $\vec v$), so magnetic fields alone cannot change particle speed. 1823 | \end{itemize} 1824 | \vspace{0.4em} 1825 | \begin{itemize}[leftmargin=*] 1826 | \item Using $q$'s sign incorrectly in $q\,\vec v\times\vec B$. Fix: compute direction for positive charge, then flip if $q<0$. 1827 | \end{itemize} 1828 | \end{heuristicsbox} 1829 | 1830 | \KPProblems 1831 | \begin{cheatproblem} 1832 | A particle of charge $q$ enters a uniform magnetic field $\vec B$ perpendicular to its velocity with speed $v$. Find the radius and period of its circular motion (neglect $\vec E$). 1833 | \begin{solutionbox} 1834 | Magnetic force provides centripetal: $qvB=\dfrac{mv^2}{r}\Rightarrow r=\dfrac{mv}{qB}$. The period is $T=\dfrac{2\pi r}{v}=\dfrac{2\pi m}{qB}$. 1835 | \end{solutionbox} 1836 | \end{cheatproblem} 1837 | \end{KnowledgePoint} 1838 | 1839 | \begin{KnowledgePoint}{Magnetic Force on Wires \DOne} 1840 | \KPFormulas 1841 | \begin{formulabox} 1842 | \textbf{Core formulas:} 1843 | \[ 1844 | \vec F=I\,\vec L\times\vec B,\quad |\vec F|=ILB\sin\theta. 1845 | \] 1846 | \textbf{Variable definitions:} $I$ current; $\vec L$ directed along the current segment with magnitude $L$; $\vec B$ magnetic field; $\theta$ angle between $\vec L$ and $\vec B$. 1847 | \end{formulabox} 1848 | 1849 | \KPHeuristics 1850 | \begin{heuristicsbox} 1851 | \begin{itemize}[leftmargin=*] 1852 | \item For loops, integrate $d\vec F=I\,d\vec l\times\vec B$ and exploit symmetry. 1853 | \end{itemize} 1854 | \end{heuristicsbox} 1855 | 1856 | \KPProblems 1857 | \begin{cheatproblem} 1858 | A wire of length $L$ carries current $I$ in a uniform field $\vec B$ perpendicular to the wire. Find the magnitude of magnetic force. 1859 | \begin{solutionbox} 1860 | $F=ILB$. 1861 | \end{solutionbox} 1862 | \end{cheatproblem} 1863 | \end{KnowledgePoint} 1864 | 1865 | \begin{KnowledgePoint}{Faraday-Lenz \DOne} 1866 | \KPFormulas 1867 | \begin{formulabox} 1868 | \textbf{Concept explanation:} Changing magnetic flux induces an emf that opposes the change (Lenz); steady currents set magnetic fields constrained by Amp\`ere's law. 1869 | 1870 | \textbf{Core formulas:} 1871 | \[ 1872 | \left\{\begin{aligned} 1873 | &\mathcal E=\oint_{\partial S}\vec E\cdot d\vec l=-\frac{d\Phi_B}{dt},\ \ \Phi_B=\iint_S \vec B\cdot d\vec A,\ \ \Phi_E=\iint_S \vec E\cdot d\vec A,\\ 1874 | &\text{(see Amp\`ere/Maxwell--Amp\`ere for magnetic circulation)} 1875 | \end{aligned}\right. 1876 | \] 1877 | 1878 | \textbf{Variable definitions:} $\mathcal E$ induced emf (scalar, $\mathcal E=\oint_{\partial S}\vec E\cdot d\vec l$); $\Phi_B=\iint_S \vec B\cdot d\vec A$ magnetic flux; $\Phi_E=\iint_S \vec E\cdot d\vec A$ electric flux; $I_{enc}$ enclosed current. 1879 | 1880 | \textbf{Prerequisites \& scope:} Under \DOne memorize the integral form (no derivation). Generally assume quasi-static fields; for time-varying fields use Maxwell--Amp\`ere with displacement current. 1881 | \end{formulabox} 1882 | 1883 | \begin{insightbox} 1884 | \textbf{Terminology note}: \emph{Electric circulation} refers to the line integral of the electric field that defines emf, $\mathcal E=\oint_{\partial S}\vec E\cdot d\vec l$. \emph{Magnetic circulation} refers to $\oint\vec B\cdot d\vec l$ as used in Amp\`ere/Maxwell--Amp\`ere. These are distinct: $\mathcal E$ is a scalar (emf), while $\vec E$ and $\vec B$ are fields. 1885 | \end{insightbox} 1886 | 1887 | \KPHeuristics 1888 | \begin{heuristicsbox} 1889 | \begin{itemize}[leftmargin=*] 1890 | \item Sketch the loop and determine the positive normal; apply Lenz's rule to deduce the direction of the induced current. 1891 | \item Apply circular/rectangular Amperian loops along symmetry for infinite wires/solenoids. 1892 | \end{itemize} 1893 | \vspace{0.4em} 1894 | \begin{itemize}[leftmargin=*] 1895 | \item Missing displacement current for charging capacitors. Fix: include $\varepsilon_0\,d\Phi_E/dt$ in Maxwell–Amp\`ere when fields vary. 1896 | \end{itemize} 1897 | \end{heuristicsbox} 1898 | 1899 | \KPProblems 1900 | \begin{cheatproblem} 1901 | In a loop of area $A$, the magnetic field increases as $B(t)=B_0+kt$. Find the induced emf. 1902 | \begin{solutionbox} 1903 | $\mathcal E=\left|\dfrac{d\Phi}{dt}\right|=\left|\dfrac{d(BA)}{dt}\right|=kA$. 1904 | \end{solutionbox} 1905 | \end{cheatproblem} 1906 | \end{KnowledgePoint} 1907 | 1908 | \begin{KnowledgePoint}{Ampere and Maxwell--Ampere \DTwo} 1909 | \KPFormulas 1910 | \begin{formulabox} 1911 | \textbf{Core formulas:} 1912 | \[ 1913 | \left\{\begin{aligned} 1914 | &\oint \vec B\cdot d\vec l=\mu_0 I_{enc}\ (\text{steady currents}),\\ 1915 | &\text{Maxwell--Ampere (general): }\oint \vec B\cdot d\vec l=\mu_0 I_{enc}+\mu_0\varepsilon_0\,\frac{d\Phi_E}{dt}. 1916 | \end{aligned}\right. 1917 | \] 1918 | \end{formulabox} 1919 | \end{KnowledgePoint} 1920 | 1921 | \begin{KnowledgePoint}{EM Spectrum and Maxwell (Concept) \DOne} 1922 | \KPFormulas 1923 | \begin{formulabox} 1924 | \textbf{Concept explanation:} Electromagnetic waves range from radio to gamma; Maxwell's equations couple $\vec E$ and $\vec B$ and give wave speed $c$ in vacuum. 1925 | 1926 | \textbf{Formulas \& Concepts:} 1927 | \[ 1928 | \left\{\begin{aligned} 1929 | &c=\frac{1}{\sqrt{\mu_0\varepsilon_0}},\\ 1930 | &\text{Spectrum ordering by frequency: radio }\to\ \text{microwave }\to\ \text{IR }\to\ \text{visible},\\ 1931 | &\text{then }\ \text{UV }\to\ \text{X }\to\ \text{gamma}. 1932 | \end{aligned}\right. 1933 | \] 1934 | 1935 | \textbf{Prerequisites \& scope:} Vacuum relations shown; material dispersion alters speed and wavelength. 1936 | \end{formulabox} 1937 | 1938 | \KPHeuristics 1939 | \begin{heuristicsbox} 1940 | \begin{itemize}[leftmargin=*] 1941 | \item Recall typical sources: antennas (radio), thermal (IR), electronic transitions (visible/UV), inner-shell transitions (X/gamma). 1942 | \item Use $c=f\lambda$ with medium refractive index $n$ via $v=c/n$. 1943 | \end{itemize} 1944 | \end{heuristicsbox} 1945 | 1946 | \KPProblems 1947 | \begin{cheatproblem} 1948 | Light in vacuum has wavelength $\lambda=600\,\text{nm}$ and speed $c=3\times10^8\,\text{m/s}$. Find its frequency. 1949 | \begin{solutionbox} 1950 | $f=\dfrac{c}{\lambda}=\dfrac{3\times10^8}{600\times10^{-9}}=5\times10^{14}\,\text{Hz}$. 1951 | \end{solutionbox} 1952 | \end{cheatproblem} 1953 | \begin{cheatproblem} 1954 | Rank the following by increasing photon energy: radio, visible, X-ray. 1955 | \begin{solutionbox} 1956 | Higher frequency means higher photon energy $E=hf$. Ordering: radio $<$ visible $<$ X-ray. 1957 | \end{solutionbox} 1958 | \end{cheatproblem} 1959 | \end{KnowledgePoint} 1960 | 1961 | % --- New consolidated Practice Pointers for Part III --- 1962 | \subsection*{Part III: Electricity \texorpdfstring{\&}{&} Magnetism Practice Pointers} 1963 | \addcontentsline{toc}{subsection}{Part III: Electricity \& Magnetism Practice Pointers} 1964 | \begin{itemize}[leftmargin=*] 1965 | \item Physics Bowl DC Circuits Problem 7 Page: 8 1966 | \item Physics Bowl Electricity \& Magnetism Problem 8 Page: 9 1967 | \item Physics Bowl DC Circuits Problem 9 Page: 10 1968 | \item Physics Bowl RC Circuits Problem 12 Page: 13 1969 | \item Physics Bowl DC Circuits Problem 22 Page: 23 1970 | \end{itemize} 1971 | 1972 | % ===================== Part IV: Optics & Modern Physics ===================== 1973 | \clearpage 1974 | \parttheme{optics} 1975 | \section*{Part IV: Optics \texorpdfstring{\&}{&} Modern Physics} 1976 | \addcontentsline{toc}{section}{Part IV: Optics and Modern Physics} 1977 | 1978 | \Unit[Reflection; refraction (Snell); thin lens and magnification; interference/diffraction]{Unit 11: Optics} 1979 | 1980 | \begin{KnowledgePoint}{Reflection and Refraction \DOne} 1981 | \KPFormulas 1982 | \begin{formulabox} 1983 | \textbf{Concept explanation:} Light reflects with equal incident and reflected angles; refraction across media obeys Snell's law. 1984 | 1985 | \textbf{Core formulas:} 1986 | \[ 1987 | \left\{\begin{aligned} 1988 | &\theta_i=\theta_r,\\ 1989 | &n_1\sin\theta_1=n_2\sin\theta_2,\quad \text{TIR when }\theta_1>\theta_c=\arcsin(n_2/n_1)\ (n_1>n_2). 1990 | \end{aligned}\right. 1991 | \] 1992 | 1993 | \textbf{Variable definitions:} $n$ refractive index; $\theta$ angles measured to the normal. 1994 | 1995 | \textbf{Prerequisites \& scope:} Geometric optics regime; isotropic media; polarization effects ignored here. 1996 | \end{formulabox} 1997 | 1998 | \KPHeuristics 1999 | \begin{heuristicsbox} 2000 | \begin{itemize}[leftmargin=*] 2001 | \item Draw the normal and principal rays first; search for total internal reflection when going to a lower-$n$ medium. 2002 | \item Use reversibility of light to validate constructions. 2003 | \end{itemize} 2004 | \end{heuristicsbox} 2005 | 2006 | \KPProblems 2007 | \begin{cheatproblem} 2008 | Light travels from air ($n_1=1$) into water ($n_2=1.33$) at incidence angle $\theta_1=40^\circ$. Find the refraction angle $\theta_2$. 2009 | \begin{solutionbox} 2010 | Snell: $n_1\sin\theta_1=n_2\sin\theta_2\Rightarrow\sin\theta_2=\dfrac{\sin40^\circ}{1.33}\Rightarrow\theta_2\approx28.9^\circ$. 2011 | \end{solutionbox} 2012 | \end{cheatproblem} 2013 | \begin{cheatproblem} 2014 | Light moves from glass ($n=1.5$) to air ($n=1$). Find the critical angle for total internal reflection. 2015 | \begin{solutionbox} 2016 | $\theta_c=\arcsin\dfrac{n_2}{n_1}=\arcsin\dfrac{1}{1.5}\approx41.8^\circ$. 2017 | \end{solutionbox} 2018 | \end{cheatproblem} 2019 | \end{KnowledgePoint} 2020 | 2021 | \begin{KnowledgePoint}{Thin Lenses and Sign Conventions \DOne} 2022 | \KPFormulas 2023 | \begin{formulabox} 2024 | \textbf{Concept explanation:} Thin lens imaging follows the lens equation with sign conventions; magnification uses image and object sizes/orientations. 2025 | 2026 | \textbf{Core formulas:} 2027 | \[ 2028 | \left\{\begin{aligned} 2029 | &\frac{1}{f}=\frac{1}{s}+\frac{1}{s'},\quad m=-\frac{s'}{s}=\frac{h'}{h}. 2030 | \end{aligned}\right. 2031 | \] 2032 | 2033 | \textbf{Variable definitions:} $f$ focal length; $s$ object distance; $s'$ image distance; $m$ magnification; $h',h$ image/object heights. 2034 | 2035 | \textbf{Prerequisites \& scope:} Use consistent sign convention (e.g., real is positive); paraxial approximation. 2036 | \end{formulabox} 2037 | 2038 | \KPHeuristics 2039 | \begin{heuristicsbox} 2040 | \begin{itemize}[leftmargin=*] 2041 | \item Combine equation + ray diagram: draw two principal rays to confirm the algebraic image location. 2042 | \item Remember that negative $m$ indicates inversion; $|m|>1$ indicates magnification. 2043 | \item Sign convention (real-is-positive): take $s>0$ for real objects and $s'>0$ for real images on the opposite side of the lens from the object; $s'<0$ indicates a virtual image on the object side (then $m>0$ and the image is upright). 2044 | \end{itemize} 2045 | \end{heuristicsbox} 2046 | 2047 | \KPProblems 2048 | % Simple principal-ray diagram (converging lens, object inside f -> virtual image) 2049 | \begin{center} 2050 | \begin{tikzpicture}[x=0.06cm,y=0.06cm,>=Latex] 2051 | % Optical axis 2052 | \draw[-] (-60,0) -- (80,0); 2053 | % Lens at x=0 2054 | \draw[thick] (0,-25) -- (0,25); 2055 | % Focal points at ±60 cm (f=60 cm) 2056 | \fill (-60,0) circle (1.2pt) node[below=2pt] {$F$}; 2057 | \fill (60,0) circle (1.2pt) node[below=2pt] {$F'$}; 2058 | % Object (arrow) at s=30 cm left of lens (inside focal length) 2059 | \draw[thick,->] (-30,0) -- (-30,18) node[left] {object}; 2060 | % Principal ray 1: parallel to axis, refracted as if through F' 2061 | \draw[->] (-30,18) -- (0,18); % incident 2062 | \draw[->] (0,18) -- (60,0); % refracted 2063 | % Principal ray 2: through center (undeviated) 2064 | \draw[->] (-30,18) -- (0,0); 2065 | \draw[->] (0,0) -- (30,-18); 2066 | % Back-projections of refracted rays to locate virtual image at s'=-60 cm 2067 | \draw[dashed] (0,18) -- (-60,36); 2068 | \draw[dashed] (0,0) -- (-60,36); 2069 | % Image arrow (virtual, upright, magnified) 2070 | \draw[thick,->,blue] (-60,0) -- (-60,36) node[left] {virtual image}; 2071 | \end{tikzpicture} 2072 | \end{center} 2073 | \begin{cheatproblem} 2074 | An object at $s=30\,\text{cm}$ forms an image at $s'=-60\,\text{cm}$ using a thin lens. Find the focal length $f$ and magnification $m$. 2075 | \begin{solutionbox} 2076 | Lens equation: $\tfrac{1}{f}=\tfrac{1}{s}+\tfrac{1}{s'}=\tfrac{1}{30}+\tfrac{1}{-60}=\tfrac{1}{60}$, so $f=60\,\text{cm}$. 2077 | Magnification: $m=-s'/s = -(-60)/30=2$. 2078 | Since $f>0$, it is a converging lens. Since $s'<0$, the image is virtual. Since $m>0$, the image is upright. 2079 | \end{solutionbox} 2080 | \end{cheatproblem} 2081 | \end{KnowledgePoint} 2082 | 2083 | \begin{KnowledgePoint}{Interference and Diffraction \DOne} 2084 | \KPFormulas 2085 | \begin{formulabox} 2086 | \textbf{Concept explanation:} Coherent sources produce interference patterns; finite apertures make the light diffract, establishing angular scales by wavelength/aperture. 2087 | 2088 | \textbf{Core formulas:} 2089 | \[ 2090 | \left\{\begin{aligned} 2091 | &\text{Double-slit maxima: } d\sin\theta=m\lambda,\\ 2092 | &\text{Single-slit minima: } a\sin\theta=m\lambda,\ m=\pm1,\pm2,\dots 2093 | \end{aligned}\right. 2094 | \] 2095 | 2096 | \textbf{Variable definitions:} $d$ slit separation; $a$ slit width; $\lambda$ wavelength; $\theta$ diffraction angle. 2097 | 2098 | \textbf{Prerequisites \& scope:} Small-angle approximations $\sin\theta\approx\theta$ valid near axis; coherence required for stable fringes. 2099 | \end{formulabox} 2100 | 2101 | \KPHeuristics 2102 | \begin{heuristicsbox} 2103 | \begin{itemize}[leftmargin=*] 2104 | \item Map angles to screen positions with $y\approx L\tan\theta\approx L\theta$ for small $\theta$. 2105 | \item To resolve features, compare $\lambda$ to $a$ and $d$ to predict fringe spacing/envelope width. 2106 | \end{itemize} 2107 | \end{heuristicsbox} 2108 | 2109 | \KPProblems 2110 | \begin{cheatproblem} 2111 | For double-slit with spacing $d$ and wavelength $\lambda$, what is the angle of the $m$-th bright fringe? 2112 | \begin{solutionbox} 2113 | $d\sin\theta=m\lambda \Rightarrow \theta=\arcsin(m\lambda/d)$ (small-angle: $\theta\approx m\lambda/d$). 2114 | \end{solutionbox} 2115 | \end{cheatproblem} 2116 | \end{KnowledgePoint} 2117 | 2118 | % (Removed outdated Practice Pointers for Part IV Unit 11) 2119 | 2120 | \Unit[Special relativity ($\gamma$, time dilation, length contraction, $E=mc^2$); photoelectric effect; atomic spectra; nuclear decay/half-life]{Unit 12: Modern Physics} 2121 | 2122 | \begin{KnowledgePoint}{Special Relativity \DOne} 2123 | \KPFormulas 2124 | \begin{formulabox} 2125 | \textbf{Concept explanation:} At high speeds, time dilates and lengths contract; energy–mass equivalence relates rest mass to rest energy. 2126 | 2127 | \textbf{Core formulas (with proper vs observed):} 2128 | \[ 2129 | \left\{\begin{aligned} 2130 | &\gamma=\frac{1}{\sqrt{1-v^2/c^2}},\\ 2131 | &\text{Time dilation: } \Delta t=\gamma\,\Delta \tau\ \ (\Delta \tau\text{ proper time in moving clock's frame}),\\ 2132 | &\text{Length contraction: } L=\frac{L_0}{\gamma}\ \ (L_0\text{ proper length measured at rest with the rod}),\\ 2133 | &\text{Relativistic energy: } E=\gamma mc^2\ (E_0=mc^2),\quad p=\gamma mv,\quad E^2=(pc)^2+(mc^2)^2. 2134 | \end{aligned}\right. 2135 | \] 2136 | 2137 | \textbf{Variable definitions:} $\gamma$ Lorentz factor; $v$ relative speed; $c$ speed of light; $\Delta \tau$ proper time (clock's rest frame); $L_0$ proper length (object's rest frame); $E$ total energy; $E_0$ rest energy; $p$ relativistic momentum. 2138 | 2139 | \textbf{Prerequisites \& scope:} Inertial frames; $v$ along one axis for simple forms; proper quantities measured in an object's rest frame. 2140 | \end{formulabox} 2141 | 2142 | \KPHeuristics 2143 | \begin{heuristicsbox} 2144 | \begin{itemize}[leftmargin=*] 2145 | \item Label frames (S, S') and identify proper time/length before applying formulas. 2146 | \item [\DTwo] Approximation for $v\ll c$: $\gamma\approx1+\tfrac12(v/c)^2$ (derivation and series methods belong to \DTwo). 2147 | \end{itemize} 2148 | \end{heuristicsbox} 2149 | 2150 | \KPProblems 2151 | \begin{cheatproblem} 2152 | A spaceship moves at $0.8c$ relative to Earth. What factor relates proper time to dilated time? 2153 | \begin{solutionbox} 2154 | $\gamma=1/\sqrt{1-0.8^2}=\tfrac{5}{3}$. 2155 | \end{solutionbox} 2156 | \end{cheatproblem} 2157 | \end{KnowledgePoint} 2158 | 2159 | \begin{KnowledgePoint}{Photoelectric Effect \DOne} 2160 | \KPFormulas 2161 | \begin{formulabox} 2162 | \textbf{Concept explanation:} Electrons emit when photon energy exceeds the work function; the threshold frequency is $f_{th}=\phi/h$. 2163 | 2164 | \textbf{Core formulas:} 2165 | \[ 2166 | \left\{\begin{aligned} 2167 | &K_{\max}=hf-\phi,\quad f_{\text{th}}=\phi/h,\\ 2168 | \end{aligned}\right. 2169 | \] 2170 | 2171 | \textbf{Variable definitions:} $h$ Planck constant; $\phi$ work function. 2172 | 2173 | \textbf{Prerequisites \& scope:} Idealized models; surface effects and detector thresholds may alter observed $K_{\max}$. 2174 | \end{formulabox} 2175 | 2176 | \KPHeuristics 2177 | \begin{heuristicsbox} 2178 | \begin{itemize}[leftmargin=*] 2179 | \item Stopping potential depends on frequency (threshold via $f_{\text{th}}=\phi/h$), not intensity. 2180 | \item In a $K_{\max}$--$f$ plot, slope $=h$, vertical intercept $=-\phi$. 2181 | \item Increasing intensity raises saturation current but does not change stopping potential. 2182 | \end{itemize} 2183 | \end{heuristicsbox} 2184 | 2185 | \KPProblems 2186 | \begin{cheatproblem} 2187 | Light of frequency $f$ hits a metal with work function $\phi$. Write the maximum kinetic energy of ejected electrons. 2188 | \begin{solutionbox} 2189 | $K_{\max}=hf-\phi$. 2190 | \end{solutionbox} 2191 | \end{cheatproblem} 2192 | \end{KnowledgePoint} 2193 | 2194 | \begin{KnowledgePoint}{Nuclear Decay Basics \DOne} 2195 | \KPFormulas 2196 | \begin{formulabox} 2197 | \textbf{Core formulas:} 2198 | \[ 2199 | N(t)=N_0\,2^{-t/T_{1/2}}=N_0 e^{-\lambda t},\ \lambda=\frac{\ln 2}{T_{1/2}}. 2200 | \] 2201 | \end{formulabox} 2202 | \KPHeuristics 2203 | \begin{heuristicsbox} 2204 | \begin{itemize}[leftmargin=*] 2205 | \item For decay chains, use activity $A=\lambda N$; independent branches superpose exponentials. 2206 | \item Plot $\ln N$ vs $t$ to extract $\lambda$ from the slope. 2207 | \end{itemize} 2208 | \end{heuristicsbox} 2209 | \end{KnowledgePoint} 2210 | 2211 | % --- New consolidated Practice Pointers for Part IV --- 2212 | \subsection*{Part IV: Optics \texorpdfstring{\&}{&} Modern Physics Practice Pointers} 2213 | \addcontentsline{toc}{subsection}{Part IV: Optics \& Modern Physics Practice Pointers} 2214 | \begin{itemize}[leftmargin=*] 2215 | \item Physics Bowl Geometric Optics Problem 21 Page: 22 2216 | \item Physics Bowl Geometric Optics Problem 25 Page: 26 2217 | \item Physics Bowl Modern Physics Problem 32 Page: 31 2218 | \item Physics Bowl Geometric Optics Problem 34 Page: 32 2219 | \end{itemize} 2220 | 2221 | % (Removed redundant KnowledgePoint: Point Charge Field; content covered in Unit 8 Electrostatics) 2222 | 2223 | \end{document} 2224 | 2225 | 2226 | 2227 | 2228 | 2229 | 2230 | --------------------------------------------------------------------------------