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Hong Kong Physics Olympiad 2006 – Preliminary Contest二零零六年香港物理奥林匹克预选赛

Hong Kong Physics Olympiad 2006 – Preliminary Contest 二零零六年香港物理奧林匹克預選賽 Organized by Physics Club 物理學會主辦 March 10th, 2006 二零零六年三 二零零六年三月十日

Rules of the Contest

1. The competition comprises a 70-minute written test. 比賽以筆試形式進行,為時 70 分鐘。 2. The full mark of this paper is 25.6. 全卷滿分為 25.6 分。 3. There will be multiple-choice and open-ended questions in bilingual versions. Contestants must answer all questions. They can answer in either Chinese or English. 試題分爲選擇題及開放式問答題,題目中英對照,參賽學生可選擇以中文 或英文作答,但必須解答全卷所有題目。 4. Contestants should present the solutions clearly when solving open-ended questions. 作答開放式問答題時,參賽學生須清楚交代解題步驟。 5. All answers should be either exact or corrected to three or more significant figures unless otherwise specified. 除特別指明外,所有最終數值需完全準確或取三個或以上的有效數字。 6. Calculators that are HKEAA-approved can be used for the event. Measuring instruments like rulers, compasses, etc. can also be used. 比賽時,學生可使用香港考試及評核局認可之計算器。直尺、圓規及其它 量度工具亦可作輔助之用。 7. The following conditions will be applied unless otherwise specified: ? ? ? All objects are near Earth surface and the gravity is pointing downwards. Neglect air resistance. All speeds are much lower than the speed of light.

除特別指明外,本卷將使用下列條件: ? ? ? 所有物體都處於地球表面,重力向下。 忽略空氣阻力。 所有速度均遠低於光速。

Physical Constants and Formulae for Reference

? ? ? ? ? ? Acceleration due to gravity 地心加速度 g = 9.80665 m s-2. Atmospheric pressure 大氣壓強 p0 = 1.01325 × 105 Pa. ? Gas constant 氣體常數 R = 8.31447 J K-1 mol-1. Permittivity in free-space 真空介電常數 ε0 = 8.85419 × 10-12 C2 N-1 m-2. Speed of light in vacuum 真空光速 c = 2.99792 × 108 m s-1. Zero degrees Celsius 攝氏零度 0 °C = 273.15 K. ? ? Properties of water: 水的性質: ? Specific heat capacity 比熱容 cw = 4.18 × 103 J kg-1 K-1 Latent heat of vaporization 氣化潛熱 Lv,w = 2.26 × 106 J kg-1 Density of steam 水蒸氣密度 ρs = 8.03 × 10-1 kg m-3 Molar mass 摩爾質量 Mw = 1.80 ×10-2 kg mol-1


Ideal gas law 理想氣體方程 pV = nRT p: Pressure in gas 氣體中的壓強 V: Volume occupied by gas 氣體佔的體積 T: Temperature (in kelvins) of gas 氣體的絕對溫度 n: Number of moles of gas 氣體的摩爾數


Coulomb force between two charges 電荷間的庫倫力 電荷間的庫倫 1 q1 q 2 F= 4 ?ε0 r 2
F: Force experienced by the charges 電荷感受到的力 r: Distance between the charges 電荷間的距離 q1, q2: Electric charges 電荷強度


Hooke’s law of Springs 彈弓的拉力(虎克定律) 彈弓的拉力(虎克定律)
F = ?kx F: Force exerted by spring 彈弓的拉力 k: Spring coefficient 彈力系數 x: Extension of spring 彈弓延長了長度

Multiple Choice Questions 多項選擇題
(1) Which of the followings leads to Newton’s third law? 以下哪一條定律可導出牛頓第三定律? a) Conservation of momentum 動量守恆定律 b) Conservation of energy 能量守恆定律 c) Conservation of mass 質量守恆定律 d) Conservation of electric charge 電荷守恆定律 e) None of the above 以上皆不是

(0.5 marks per question) (每題 0.5 分)

(2) A non-deformable container of volume 4189 cm3 is kept at 0 °C and then filled with air at that temperature. Later the sphere is heated. If the container can withstand a pressure difference of 50,000 Pa, find the minimum temperature at which the container explodes. 一個體積為 4189 cm3 的不變形球形容器在 0 °C 的氣溫下充滿了空氣,然後將其 加熱。如果容器能承受 50,000 Pa 的壓強差,試找出令它爆開的最低溫度。 a) b) c) d) 14 °C 135 °C 408 °C The container will not explode 容器不會爆開

e) None of the above 以上皆不是

(3) Theoretically, photons consist of no mass but finite momentum. If it is known that the energy and frequency of a photon are in direct proportion (E = hf), guess an expression for the momentum of photon (λ is photon’s wavelength). 理論上,光子沒有質量但有有限的動量。已知光子的能量和頻率成正比 (E = hf) , 試估算光子動量的公式(λ 是光子的波長) 。 a) b) c) d) p = h/λ p = hλ p = h/f p = hf

e) p = hfλ

(4) We know that a transverse wave will cause particles to vibrate perpendicular to the direction of the wave, and a longitudinal wave will cause them to vibrate along it. If a transverse wave and a longitudinal wave move together in the same direction, what will be the possible motion of the particles? 我們知道一橫波 橫波會令周遭粒子沿波的垂直 橫波 a) b) c) 方向振動,而縱波 縱波則會令它們沿波的方向 縱波 振動。如果一橫波和縱波同時沿相同方向傳播,那麼粒子可怎樣運動? d)

a) Perform circular motion on a plane parallel to the direction of waves. 在與波方向平行的平面上作圓周運動。 b) Perform circular motion on a plane perpendicular to the waves. 在與波方向垂直的平面上作圓周運動。 c) Oscillate along the direction of the waves. 沿波的方向振動。 d) Oscillate along the perpendicular of the wave 沿波的垂直方向振動。 e) None of the above. 以上皆不是。

(5) As shown in the figure, a light ray enters a media, which the refractive index n is related to the depth x via the relation n = x + 1. Find θ. 如右圖所示,一光線進入了一介質,其折射率 n 與深度 x 的關係為 n = x + 1。求 θ 的值。
63.0° x



a) 0.505° b) 26.5° c) 61.9° d) 63.0° e) The ray will not emerge from the bottom 光線不會從底射出

(6) Although in many physical models we ignore the air friction, it does present.
1 Empirical result shows the air friction acting on an object is f = C d ρAv 2 , 2

where Cd is the drag coefficient, ρ is the density of air, A is the area of object facing motion and v is the speed. Which of the following correctly shows the velocity-time graph of a free-falling object of mass m? 在很多物理模形中,我們都忽略空氣阻力,可是,它確是存在的,而根據實驗數
1 據,空氣阻力為 f = C d ρAv 2 ,其中 Cd 是「阻力常數」 ρ 是空氣密度、A 是物 、 2 體面向運動的表面積、v 是速率。以下哪幅圖表正確表示質量為 m 的物體自由落 體時速度與時間的關係? v v
2 mg C d ρA

2mg Cdρ2A3


t v


t v
2 mg C d ρA



2mg Cdρ2A3





In the following figures, is sound source (e.g. a loudspeaker) emitting a note of 512 Hz continuously, and ? is a sound receiver (e.g. a human being). The arrows denote which way they are moving to. Then which of them ? will receive the sound of highest pitch?


是一個長期發出 512 Hz 聲調的聲源(如擴音器) ,而 ? 是一個

聲音接收者(如一個人) 。而箭號則代表他們行進的方向。那哪幅圖中 ? 會接 收到最高的聲調?
a) b) c) d) e)

Two particles of charge +5 × 10-4 C are attached to both ends of a spring of original length 5 m and spring constant 7 N m-1. Find the length of the spring

in equilibrium. 一條長 5 m、彈性係數為 7 N m-1 的彈弓兩端各連上了電荷為 +5 × 10-4 C 的粒 子。求彈弓延長後的長度。 a) 1.39 m b) 3.98 m c) 2.83 × 103 m d) 6.39 m e) 8.98 m (9) As shown in the figure, 2007 metal spheres are arranged horizontally. The first sphere is induced with a positive charge, and the last is grounded. What +


kind of charge does the last sphere carry in electrostatic equilibrium? 如右圖所示,2007 個金屬球順序地平排。第一個球加入了一些正電荷,而最後 一個則接了地。達到靜電平衡時,最後一球帶何種電荷? a) b) c) d) e) Positive charge 正電荷 Negative charge 負電荷 Neutral 不帶電 Same as that of the 2006th 與第 2006 個球相同 Cannot be decided 無法判斷

(10) For the same setup from question 9, but all sphere from the second to the last are all grounded. What kind of charge does the last sphere carry now? 承上題,但第二至最後的球均先接地。此時最後一球又帶何種電荷? a) Positive charge 正電荷 b) Negative charge 負電荷 c) Neutral 不帶電 d) Same as that of the 2006th 與第 2006 個球相同 e) Cannot be decided 無法判斷

Open-Ended Questions 開放題

(11) As shown in the figure, two identical charged metal spheres of charge q and mass m are attached to the end of inelastic strings of length l, and the two strings are attached to the same point on the ceiling. Due to the Coulomb force between the two charges, the strings are separated with an angle of 2θ. 如右圖,兩個帶有電荷 q、質量 m 的金屬球繫在兩條長度 q m 為 l 不具彈性的繩子,而兩條繩子都連結於天花版的同一 點上。因為電荷間庫倫力的關係,兩條繩分隔了 2θ 的角度。 (a) Assume θ is small, express θ in terms of q, m, l and other necessary physical constants. (Hint: for small θ in radian, tan θ ≈ sin θ ≈ θ and cos θ ≈ 1.) 假設 θ 很小,試以 q、m、l 和其他需要的物理常數表達 θ。 (提示:對於小的 θ,如使用弧度,tan θ ≈ sin θ ≈ θ 和 cos θ ≈ 1。) (b) If a stream of alpha particles passes through the system just between the two spheres, what will the value of θ be after achieving equilibrium? (Note: air is present.) 如果一束 α 粒子流在兩個球中間穿過,θ 會變成多少? (注意:空氣存在。) α2+ (1 marks) (1 分) (3 marks) (3 分) m q 2θ l l

(c) If the system is free-falling to Earth’s surface, what will the value of θ be? 如果整個系統在地球表面自由落體,θ 會變成多少? (1 marks) (1 分)

(12) Historically, electric current in conductors was thought to be carried by positive charge carriers. However, it was later discovered that currents are actually carried by negative charge carriers called “electrons” with the development of electrochemistry. Here, we introduce an experiment to prove the above result using the Hall Effect. 在發現電流時,它被認為是因帶正電荷的粒子移動造成的。然而,電化學的發展 讓我們知道電流其實是帶負電荷的「電子」產生的。現在,我們利用「庫爾效應」 來設計一實驗證明上述結果。 (a) In the figure below, a positively-charged proton and a negatively-charged electron are shot horizontally to the right. A uniform magnetic field pointing inside the page is applied. Sketch the trajectory of the two particles. 如下圖,一粒帶正電荷的質子和帶負電荷的電子向右發射,然後進入一指向頁內 的均勻磁場。試完成二粒子的彈道路徑。 (1.4 marks) (1.4 分)

p e?

(b) Using the following apparatus, design an experiment to prove that the charge carrier of electric current in a conductor is negatively-charged: Strong (3.6 marks) magnets, copper plate, conducting wires, batteries, resistor, and voltmeter. 試設計一實驗,利用以下儀器來證明電流的電荷載子是帶負電的:強力磁石、銅 板、電線、電池、電阻和伏特計。 (3.6 分)

In this question, give your answers to 2 significant figures.

本題答案只需準至兩個有效數字 兩個有效數字。 兩個有效數字

(a) What happens at thermal equilibrium when 1 kg of water at 0 °C is mixed with 1 kg of steam at 100 °C in standard atmospheric pressure? (Assume no heat is lost to the environment.) 在標準大氣壓強下,1 kg 在 0 °C 的水和 1 kg 在 100 °C 的水蒸氣混合並到達熱 平衡後會怎樣?(假設熱沒有散失到外界。) (b) Now, the system is put inside a rigid insulating vacuum container that provides the same pressure as (a) before any heat exchange takes place. Describe briefly how the mixture will become after thermal equilibrium in reached here compared to (a) in terms of (i) the final temperature and (ii) the mass of the product(s). (2.2 marks) 現在,整個系統放進了一個整固絕熱的真空容器內,使得容器內的壓強在任何熱 交換開始前與 (a) 相同。試簡略描述達熱平衡後混合物在(甲)溫度和(乙) 質量上與 (a) 的分別。 (c) It is known that the relationship between the surrounding pressure p (in Pa) and the boiling point of water T (in °C) is: 已知氣壓 p(在 Pa 時)和水的沸點 T(在 °C 時)有如下關係:

(2 marks) (2 分)

(2.2 分)

T = 237.611 ?

772.079 log ( 4.05574 p )

Find the temperature mentioned in (b). You may express the answer in an (5.7 marks) equation in terms of T. (5.7 分) 求 (b) 部中的溫度。你可以以含 T 的方程式作答。
(Note: the heat capacitance per mole of any gas in constant volume is 1.5R.) (注意:任何氣體在不變體積時每摩爾的熱容是 1.5R。)

(d) The heat (energy) involved in boiling in used to overcome the intermolecular attractions, which in turn is an electrostatic force between the excess charges of the molecules. What happens to the answers in (a) when the quantity ε0 can be decreased? (0.7 mark) (0.7 分)

在沸騰中的熱(能)是用來克服分子之間的吸引力,而此力其實是分子多餘的電 荷引起的靜電力。如果 ε0 能夠減少,在 (a) 中的答案會怎樣改變?

~ End of Paper 全卷完 ~

Answer sheet 答題紙
Name 姓名: ____________ Class 班別: _______

1 2 3 4 5 6 7 8 9 10






(12) (a)

p e?

For the other open-ended questions, write your answers on the blank papers given. 其他開放題請於另外的提供的紙張上作答。

Answers for Reference
(1) A.

(2) B. As pV/T is a constant, we have
p0V p1V = T0 T1

where p1 is the ultimate pressure the container can withstand, T0 = 0 °C and T1 is the temperature during the explosion. Rewriting the equation gives 50, 000 Pa = p1 ? p0
?T ? = p0 ? 1 ? 1 ? ? T0 ? T1 ? ? = ( 1.01325 × 10 5 Pa ) ? ? 1? ? 273.15 K ? T1 = 407.9 K = 134.8° C

(3) A.
The unit of energy is [J] = [kg m2 s-2], that of frequency is [Hz] = [s-1], that of wavelength is [m] and that of momentum is [kg m s-1]. Hence, the unit of h is [J Hz-1] = [kg m2 s-1]. Obviously, by dimensional analysis, the momentum can be obtained by dividing h by a length-like quantity, which our only option is λ.

(4) A.
This can easily be pictured. In a quantitative treatment, we have

? y ′ = A1 cos ( k1 x ? ω2 t + φ1 ) ? ? ? x′ = A2 cos ( k2 x ? ω2 t + φ2 ) . ?
In the case that A1 = A2, k1 = k2, ω1 = ω2, φ1 = 0° and φ2 = 90°, we will have a circular motion (x’2 + y’2 = A2) along the x’y’ plane, i.e., the plane parallel to the direction of waves.

(5) D. Because the media where the ray from and to are the same, and the boundaries are all parallel to each other, the incident angle is equal to the final refractive angle, i.e., θ = 63.0°.

(6) B. In the presence of air friction, the rate of velocity of the falling object decreases gradually with time, so the increase of velocity will become slower and slower until reaching a constant point (terminal velocity) where the weight and air friction exactly cancels each other. This velocity can be calculated by simply

1 mg = f = C d ρAv 2 2
which means (b) is the answer.

? v=

2mg C d ρA

(7) B. By Doppler Effect (or visualizing the picture with moving wave-fronts), the frequency (thus pitch) received will be the highest when the source and receivers are approaching each other.

(8) E. In equilibrium, we have q2 k (L ? l) = 4 ?ε0 L2 ∴ L2 ( L ? ( 5 m ) ) = 4 ? ( 8.85419 × 10 ?12 C 2 N ?1 m ?2 )( 7 N m ?1 )

( +5 × 10




L2 ( L ? ( 5 m ) ) = 320.98 m 3 .

Substituting each choice into the equation above gives L = 8.98 m.

(9) B. For any ungrounded spheres, the left will be negatively charged because electrons are attracted by the positive charge of the previous sphere. This exposes the positive charges on the right, thus affecting the next sphere. Now,

the last sphere is grounded, which means there are infinite supply of electrons from the ground. Those electrons will rush up and fill the sphere, making it negatively charged.

(10) A. As all except the first spheres are grounded, the second sphere will become negatively charges, and the third will be positive (electrons can move to the ground too), and so on. Therefore the 2007th sphere is positively charged.

(11) (a) The free body diagram of any of the sphere is T θ

q2 4 ?ε0 ( 2 x )

mg By Newton’s first law, the net force should be zero. By either investigating the components or Lami’s theorem, we found that q2 mg tan θ = . 16 ?ε0 x 2 But x = l sin θ ≈ lθ and tan θ ≈ θ, which simplifies the equation to mgθ = ∴ (b) When the alpha particles pass through the system, the air is continuously ionized and therefore charges on the metal spheres will be lost to the surroundings. Eventually, the spheres will become neutral and the Coulomb force separating the spheres apart will disappear. Thus θ = 0°. (c) θ= q2 16 ?ε0 ( lθ )
3 2

q2 16 ?ε0 l 2

When the system is free-falling, the effect of gravity will be canceled. The Coulomb will then make the spheres away from each other at the greatest extent. Hence θ = 90°.

(12) (a)

p e?

(b) Construct a circuit like this: Magnet (N)


Copper plate

Magnet (S)

Due to the magnetic field, the charge carriers will be pushed to side B of the copper plate, no matter it is positive or negative. Now, connect the voltmeter with side A and B of the plate. If it reads that the voltage at B is lower than that of A, the charge carrier is negative. Otherwise, the carrier is positive.

(13) (a) If all steams are condensed and all water are heated up to a temperature T0, we would have mcT0 = mc(100 – T0) + mLv, which gives T0 = 320 °C, and it is impossible. Neither do that all water are boiled (in which T0 = ?220 °C). So, what’s wrong? It turned out that part of steam is still unaffected, and the rest and all the water become water at 100 °C. Suppose finally there are mass ? of steam become water. Hence, mc (100 °C) = ?Lv, which gives ? = 0.185 kg. Thus in thermal equilibrium there is 1.2 kg of water and 0.82 kg of steam, both at 100 °C. (Alternative answer: No reaction will take place afterwards.) (b) In the container, the volume is fixed. But the number of steam molecules decreases as they are condensing, causing a decrease in pressure. This will lower the boiling point of water. This will bring more steam to water and lower the temperature at equilibrium (which, according to (a), is the boiling point of water). Hence the temperature will become lower and there is less water and more steam formed. (c) We have the following six equations:
3 ? mcT = 2 nR ( 100 ? T ) + ?L ( Conservation of energy ) ? ( Ideal gas law ) ? p ′V = n′RT ? 772.079 ? T = 237.611 ? ( Given ) log ( 4.05574 p ′ ) ? ? m ? n= ( Definition of moles ) ? M ? m?? ? n′ = ( Definition of moles ) M ? ? m V= , ( Definition of density ) ? ρ ? where m is the initial mass of water and steam (i.e., 1 kg), n is the number of

moles in 1 kg of water (and equally steam), c is cw, L is Lv,w, M is Mw, ρ is ρw (the subscripts are dropped for sake of simplicity), p’ is the final pressure, n’ is the final number of moles of steam, ? is the amount of steam converted to

water and T is the final temperature. Substituting the last three into the first three gives ? ? mcT = ? ? mp ′ = ? ? ρ ? ? p′ = ? 3mR ( 100 ? T ) + ?L 2M R (T + 273.15 ) ( m ? ? ) M 10772.079 ( 237.611?T ) . 4.05574

Substituting back all numeric data gives ? ? 4180T = 69287.3 ? 692.873T + ( 2.26 × 10 6 ) ? ? ? ? 1.24533 p ′ = 461.915 (T + 273.15 ) ( 1 ? ? ) ? 10772.079 ( 237.611?T ) ? p′ = . ? 4.05574 ? Solving the first equation gives ? = (?3.06581 × 10?2) + (2.15614 × 10?3) T. Substituting this and also the third equation into the second gives

10772.079 ( 237.611?T ) = 3.24358 ( 478.011 ? T )( 273.15 + T ) .
(d) If ε0 is decreased, the electrostatic (Coulomb) force will become stronger, thus increasing the strength of intermolecular force, and requiring more energy to overcome this. Thus it would be easier for steam to condensate and the boiling point is higher too, and there should be more water than (a).