# 交大 李威儀 物理考古題彙整
### 一 期中
* 請列出四種基本作⽤⼒並比較他們的強弱以及作⽤範圍。
:::spoiler ans
強弱：強力 > 電磁力 > 弱力 > 萬有引力
作用範圍：
* 電磁力、萬有引力：無限
* 強力：原子核內（約 $10^{-15}$ m）
* 弱力：$10^{-17}$ m
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* (a) 請盡可能詳細的說明托勒密的地⼼說如何解釋某些星體在天空上看起來忽進忽退。
(b) 請問托勒密的行星運動理論是不是科學？
:::spoiler ans
1. 托勒密為了解釋行星乎近乎退的現象，創出了「大輪、小輪」之說。小輪之心環繞地球以等速圓周運動，而行星則在小輪上繞小輪之心。
2. 是，經過科學的演繹與推理而得來的知識就是科學，儘管其結果在今日看來不盡正確，仍然是科學的一種。
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* Three blocks of mass $m_1$ = 3 kg, $m_2$ = 4 kg, $m_3$ = 6 kg - resting on a frictionless table and connected by strings with tensions T1 and T2 are being pulled to the right by a force of 6 N. [4.11]
(a) What is the acceleration of the blocks?
(b) What are the tensions in the strings?
:::spoiler ans
$a = \frac{F}{m_1+m_2+m_3} = 0.46\ m/s^2$
$T_1 = m_1a = 1.36\ N$
$T_2 = F - m_3a = 6 - 2.76 = 3.24\ N$
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* 請盡可能的說明
(a) 能量守恆定律
(b) 為何⼀個東⻄的能量是「此物體能給其他物體做功的能⼒」
:::spoiler ans
(a) 一個封閉系統內的能量總和不變
(b) IDK : (
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* 在 15 世紀科學家就發現了摩擦⼒和接觸⾯所受到的合⼒成正比，請解釋為何摩擦⼒和接觸⾯積無關
:::spoiler ans
接觸面積變大，單位面積所受的力就變小，使得微觀來看兩物體的接觸點變小，兩者間的電磁力變小，摩擦力就變小，但再乘上面積的效應，雖然單位面積的摩擦力小，但總面積大，所以總摩擦力相同
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* A bug sits on a phonograph record 0.18 m from the center. If the record turns at 30 rev/min, what is the radial acceleration of the bug? If it has a mass of 0.5 mg, what is the centripetal force acting on it? [7.12]
:::spoiler ans
$a_c = r\omega^2 = 0.18\times (\dfrac{30\cdot 2\pi}{60})^2 = 1.76$
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* Below is a picture about a spinning wheel which axis is hung by a rope. Please redraw the picture on your answer sheets, on which mark the direction of $\tau$, $L$and $\Delta L$.
:::spoiler ans
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* 請⽤理想氣體公式解釋為何在深海潛⽔的⼈突然往海⾯上游是⼀件很危險的事情？
:::spoiler ans
由 $PV = nRT$，體積正比於壓力，由於在水下壓力變化較大，假設一個人在50 公尺深向上游，則他初始的壓力為 6 atm，水面上為 1 atm 為水下的六分之一倍，根據理想氣體方程式，體積會增為六倍，使得人的肺部爆炸
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* 請畫出當溫度 T > 0K 時的⼀個封閉系統當中空氣粒⼦速率分佈圖，以及在同⼀張圖上分別標注「當溫度 T' > T 時的空氣粒⼦速率分佈圖」和「當溫度 T'' = 0K 時的分佈圖」

* The temperature of a room 7m x 5m x 3m is 27° C. [9.11]
(a) How much energy is contained in the air of that room?
(b) If that energy could be converted to electrical energy, for how long could a 100-W bulb be lit? Assume the air behaves as an ideal gas.
:::spoiler ans
(a) $E_k = \dfrac{3}{2}NkT = \dfrac{3}{2}PV = \dfrac{3}{2}\cdot 101300\ (N/m) \cdot 7 \cdot 5 \cdot 3 = 1.59\times 10^7\ J$
(b) $\dfrac{1.59\times 10^7}{100} = 1.59\times 10^5\ (s) = 1.84\ (days)$
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* Derive that the magnitude of the gravity is inversely proportional to the square of the object’s distance from Newton’s second law of motion and Kepler’s third law of planetary motion
> hint: Kepler’s third law of planetary motion indicates $\frac{a^3}{T^2}$ that must be a constant.
:::spoiler ans
$F = ma\omega^2 = \dfrac{GMm}{a^n} \Rightarrow a(\dfrac{2\pi}{T})^2 = \dfrac{GM}{a^n} \Rightarrow \dfrac{a^{n+1}}{T^2} = \dfrac{GM}{2\pi^2}$
Also, from Kepler's 3rd law we can get $\dfrac{a^3}{T^2} = const.$ Thus $n+1 = 3 \Rightarrow n = 2$
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* Premised on elastic collision, explain why the small ball in the picutre below will rebound to about nine times the original height. The collision is called “slingshot collision”.
:::spoiler ans
也可直接用彈碰公式下去解：
大球地面彈回時，$\vec{v_1} = \vec{v}$，$\vec{v_2} = -\vec{v}$，且 $m_1 \gg m_2$
又 $\vec{v_{2,f}} = \dfrac{2m_1}{m_1+m_2}\ \vec{v_1} + \dfrac{m_2-m_1}{m_1+m_2}\ \vec{v_2} \Rightarrow \vec{v_{2, f}} = \dfrac{2m_1}{m_1}\vec{v} + \dfrac{-m_1}{m_1}(-\vec{v}) = 3\vec{v}$
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* Please give an example of friction force doing negative work and explain why it does negative work. Then give an example of a frictional force doing positive work and explain why it does positive work.
:::spoiler ans
兩木塊上下疊合，拉動下木塊，地表對下木塊的摩擦力做負功，下木塊對上木塊的摩擦力做正功。
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* What is “centrifugal force”? Is centrifugal force a real force from a physical point of view? If centrifugal force doesn’t exist, please explain why lots of people believe in centrifugal force.
:::spoiler ans
在進行圓周運動的物體，因有向心加速度，故對該物體而言有一個向外的假想力，即為離心力，但其不是一個真正的力，不過因為我們在那當下都有向心加速度，故對我們而言會感受到假想力。
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* Please tell a story about thermodynamics related to the pictures below. What theory or law is expounded in your story?
:::spoiler ans
左圖為帶有鐵鍊的染色冰塊，右圖的染色冰塊直接浮在水面上。冰塊在杯底融化時，因冰水密度較大，故繼續沉在杯底，不易產生對流。相反地，在水面上融化的冰塊，冰水會向下流動，導致杯中出現對流，使得整杯水都被染色
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* A 20-g bullet is shot into a ballistic pendulum with a velocity of 1000 m/sec (see Fig. 9-8). The mass of the wooden block is 2 kg. If the bullet remains embedded in the block and 80% of the energy lost in the collision is absorbed as heat by the bullet, what is the increase in the temperature of the bullet? The specific heat of the bullet is 0.1 cal/g℃. [9.16]

* 找出兩種作用力與反作用力，並解釋原因

* 等速度移動的物體，角動量是否一定為零？是否一定守恆
:::spoiler ans
1-1. 物體可能自行等速率旋轉並等速度向前進，此時物體的質心仍為等速度運動，但物體本身會有不為零的角動量。
1-2. 參考點隨便選，可能導致等速度直線移動的物體有角動量
2. 物體不受外力時，角動量維持守恆。
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* (1) A children's merry-go-round of radius 4 m and mass 100 kg has an 80-kg man standing at the rim. The merry-go-round coasts on a frictionless bearing at 0.2 rev/sec. The man walks inward 2 m toward the center. What is the new rotational speed of the merry-go-round? What is the source of this energy? (The moment of inertia of a solid disk is $I = \frac12mr^2$ ).
(2) 為何角速度會改變？

* (a) 慣性座標的定義？
(b) 慣性座標為什麼重要？
(c\) 在非慣性座標系統，有哪些物理量不會改變大小？
:::spoiler ans
(a) 靜止或等速度運動的坐標系
(b) 在慣性座標中，牛頓第一定律恆成立，且無假想力
(c\) 質量、體積、溫度......
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* 絕對溫度代表何種物理量？潛熱為何不會影響溫度？
:::spoiler ans
組成粒子平均動能的大小，潛熱用來進行相變，使得粒子間的位能增減，不影響其動能，因此溫度不變。
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* The position of a particle moving in a straight line is given by $x = 5+2t+4t^2-t^3$ , where $x$ is in meters.
(a) Find an expression for the instantaneous velocity as a function of time.
(b) Find the position of the particle at t = 0, 1, 0.1, and 0.01 sec.
(c\) What is the average velocity between t = 0 sec and t = 1 sec, between t = 0 sec and t = 0.1 sec, and between t = 0 sec and t = 0.01 sec?
(d) What is the instantaneous velocity at t = 0 sec?
(e) What conclusion do you draw from the answers in (c\) and (d)?

* A 2 kg block rests on the ground. The coefficient of friction between the block and the ground is 0.4. A man fires a 0.01 kg bullet parallel to the ground. It lodges in the block, and the block and bullet are observed to slide 2 m before coming to rest. What was the velocity of the bullet? [6.21]
:::spoiler ans
假設子彈原本的速度為 $v_0$，則射入木塊後，兩者的速度 $v = \frac{0.01}{2.01}v_0$
其動能被轉換成摩擦力所做的功 $\Rightarrow \frac{1}{2}mv^2 = mg\mu S \Rightarrow \frac{1}{2}(\frac{0.01}{2.01}v_0)^2 = 9.8 \times 0.4 \times 2 \Rightarrow v_0 = 796\ m/s$
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* The torque of the motor is turned at right angles to the wheels by the differential gear. Assume that in low gear the angular velocity of the rear wheels is 0.1 that of the motor. If the motor has 50 hp and is turning over at a rate of 500 rev/min, how much torque is delivered to the rear wheels? [8.12]
:::spoiler ans
$P = \tau\omega \Rightarrow \tau = \dfrac{P}{\omega}$
又 $\omega_{wheel} = \frac{1}{10}\omega_{motor}$
所以 $\tau_{wheel} = \dfrac{50\times 746}{\frac{500\times 2\pi \times 0.1}{60}} = 7.13\times 10^3$
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* A mole of an ideal gas is taken from state A to state C along the path ABC. [9.20]
(a) If 1000 cal of heat flow into the gas and the gas does 2100 J of work, what is the change in the internal energy of the gas?
(b) When the gas is returned from C to A along the path CDA, 700 cal of heat flow out of the gas. How much work is done on the gas?
:::spoiler ans
$1000\times 4.18 - 2100 = 2080\ J$
$700\times 4.18 - 2080 = 846\ J$
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* 請以牛頓定律評估，在不危及生命安全的前提下，一個 100 公斤的人從靜止加速到光速，至少要多少時間，並請述明假設理由。
:::spoiler ans
人類可承受的最大加速度大約為 3g ~ 5g，折衷用 4g 算
$4\times 9.8 = \frac{3\times 10^8}{t} \Rightarrow t \approx 7653061\ s \approx 55.58\ days$
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* 請問，一個原點固定於地球表面上的參考座標是不是一個慣性座標？如果不是，請問為什麼不是？
:::spoiler ans
不是，因地球在自轉，所以有向心加速度，因此非慣性座標。最顯而易見的例子即是颱風受到科氏力（一種假想力）的影響，北半球逆時針轉，南半球順時針轉
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* If the radius of a carbon atom is about $2.5 \times 10^{-8}$ cm
(a) How many could fit in a row 1 cm long ?
(b) How many could fit in a layer one atom deep and area 1 $cm^2$ ?
:::spoiler ans
(a) $\dfrac{1}{2\times 2.5\times 10^{-8}} = 2\times 10^7$個
(b) $2\times 10^7\times 2\times 10^7 = 4\times 10^{14}$個
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### 一 期末
* 在 $t=0$，脈衝波的波函數為 $y(x)=\dfrac{A}{B-x^3}$，波以 $5\ m/s$ 的速度往 $+x$ 方向前進，$t$ 秒的時候波函式 $y(x,t)=$?
:::spoiler ans
$x = x - 5t$ 代入 $\Rightarrow y(x,t)=\dfrac{A}{B-(x-5t)^3}$
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* The equation of a transverse wave traveling along a very long string is given by $y=6\sin(0.02\pi x+4\pi t)$ where $x$ and $y$ are expressed in centimeters and $t$ in seconds. Find: [11.8]
(a) the amplitude
(b) the wavelength
(c\) the frequency
(d) the speed of propagation
(e) the direction of propagation of the wave
(f) the maximum transverse speed of a particle in the string
:::spoiler ans
(a) 6 cm
(b) $k=0.02\pi=\dfrac{2\pi}{\lambda}\Rightarrow \lambda = 100$ cm
(c\) $\omega=4\pi=2\pi f \Rightarrow f=2$ Hz
(d) $v=f\lambda = 200$ cm/s
(e) $-x$
(f) $v_{max} = A\omega = 24\pi$
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* 人耳能接收最清楚的頻率為 $3000\ Hz$，試算出人的耳道大約長度（聲速$=340\ m/s$）
:::spoiler ans
耳道視為一開一閉的管子
$\lambda = \dfrac{v}{f} = \dfrac{340}{3000} \approx 0.1133$
$d = \dfrac{\lambda}{4} \approx 2.83$ cm
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* Sodium chloride (NaCl) has a crystal structure similar to that of silver bromide (AgBr). The atomic weight of NaCl is $58.44\ g/mole$ and its density is $2.16\ g/cm^3$. [12.14]
(a) Calculate the spacing between the atoms in a NaCl crystal
(b) If X rays of wavelength $1.5Å$ are incident on a NaCl crystal, at what angle $\theta$ will the first order diffraction maximum be observed?
:::spoiler ans
(a) 
(b) $2d\sin\theta = n\lambda\ (n = 1)\Rightarrow \sin\theta = 0.266$
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* An electron is projected with an initial velocity $v_i=3\times10^6\ m/s$ in the x direction in the region between two oppositely charged plates. By the time the electron leaves the region between the plates, it has undergone a vertical deflection of $2\ cm$. Assume that the electric field between the plates is uniform and perpendicular to the plates and that the electric field outside the region of the plates is zero. [14.9]
(a) What is the strength of the electric field between the plates?
(b) At what point $y_f$ on a screen 1m away from the plates where the electron land?
:::spoiler ans
(a) Note: $m_e = 9.1\times10^{-31}$
(b) 10 cm
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* In the earth's atmosphere positive charges move toward the earth and negative charges move away from it. The total current is approximately $1800\ A$. The average value of the electric field responsible for this current near the surface of the earth is $100\ N/C$. What is the resistivity (RR) of the air at the surface of the earth? The radius of the earth is $6.37\times10^6\ m$. [15.6]
:::spoiler ans
$\rho = \dfrac{E}{J} = \dfrac{100}{\frac{1800}{4\pi\times(6.37\times 10^6)^2}} = 2.83\times 10^{13}\ \Omega\cdot m$
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* $R_x$ in terms of three known resistors $R_1$, $R_2$, and $R_s$. The value of $R_s$ is adjusted until no current flows through the galvanometer G. (The arrow over the resistor symbol of $R_s$ indicates that $R_s$ is a variable resistor.) Let $R_1 = 100\ Q$ and $R_2=100\ Q$. If no current flows through G when $R_s=470\ Q$, what is the value of $R_x$? [15.15]
:::spoiler ans
$R_sR_2 = R_xR_1 \Rightarrow R_x = 470\ \Omega$
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* (a) 請解釋為何電場必垂直物體表面
(b) 請解釋為何電力線必垂直等位線
:::spoiler ans
(a) 假設電場不垂直於物體表面，則可以將此電場分為垂直和平行於表面的兩個分量，對分布在表面的電荷而言，其會因平行表面電場的關係而產生移動，如此便不是一個靜電平衡的狀態。
(b) 同樣假設電力線不垂直於等位線，則代表電場可以分為垂直和平行於等位線的兩個分量，既然存在平行等位線的電場，代表電荷在等位線上會受力，既然會受力代表等位線電位不相同，矛盾。
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* 請比較萬有引力與靜電力的異同
:::spoiler ans
同：與距離平方成反比、不論距離皆會作用、皆為四大基本作用力之一
異：萬有引力只有吸引力而靜電力有排斥力、靜電力大小遠大於萬有引力（常數大）、
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* 請解釋為何磁鐵會吸引鐵而不吸引銅鋁等物質
:::spoiler ans
鐵為鐵磁性材料，在受到外加磁場的情況下，會使內部磁場的排列趨向外部磁場，並因此產生的磁場叫外加磁場強很多，使得磁場被加大，進而導致兩者相吸。而對於反磁性材料（銅）或順磁性材料（鋁），其內部磁場受外加磁場的影響較小，且所產生的感應磁場也不會遠大於外加磁場，因此較不容易相吸。
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* 請解釋為何會產生 VH
:::spoiler ans
電子由左向右移動，但電荷為負，根據 $\vec{F} = q\vec{v}\times \vec{B}$，電子會受力向上堆積，產生如平行版電容的效果，即為 VH。
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* The wire loop of carries a current of $2\ A$. It is placed in a region where there is a magnetic field $B=0.5\ T$ parallel to the plane of the loop. [15.5]
(a) Calculate the force on each side of the wire loop
(b) What is the torque on the wire loop?
:::spoiler ans
(a)平行磁場的兩邊不受力，垂直磁場的左右邊分別受到向上 0.6 N 和向下 0.6 N 的力
(b) $\tau = 2\times 0.6\times 0.15 = 0.18\ N-m$
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* 請寫出自然界中原子產生磁場的兩大主要原因
:::spoiler ans
電子自旋、電子繞原子核轉動
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* 帶電量 $3.2\times10^{-19}\ C$ 的粒子通過 3 V 的電壓，所需要的能量為何？
:::spoiler ans
$E = qV = 3.2\times 10^{-19}\times 3 = 9.6\times 10^{-19}\ J$
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* A copper wire whose current density is $10^6\ A/m^2$ and $V=4$ has $8\times10^{28}$ mobile electrons. What is the drift velocity of these electrons? [類 15.2]
:::spoiler ans
$i = qNAv_d = JA \Rightarrow v_d = \dfrac{J}{qN} = \dfrac{10^6}{1.6\times 10^{-19}\times \frac{8\times 10^{28}}{4}} = 3.125\times 10^{-4}\ m/s$
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* A mass is oscillating with amplitude A. When the displacement is x = 0.5A. What fraction of the energy is potential and what fraction is kinetic? [10.15]
:::spoiler ans
$E_p = \dfrac{1}{4}E$
$E_k = \dfrac{3}{4}E$
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### 二 期中
### 二 期末