39届国际物理竞赛试题IPhO(英文)理论第一题
2022年第-39-届全国物理竞赛预赛试题及解析
第39届全国中学生物理竞赛预赛试题（2022年9月3日9:00-12:00）考生必读1、 考生考试前务必认真阅读本须知。
2、 本试题共5页，总分为400分。
3、 需要阅卷老师评阅的内容一定要写在答题纸相应题号后面的空白处；阅卷老师只评阅答题纸上的内容；选择题和填空题也必须在答题纸上作答；写在试题纸和草稿纸上的解答一律无效。
一、 选择题（本题60分，含5小题，每小题12 分。
在每小题给出的4个选项中，有的小题只有一项符合题意，有的小题有多项符合题意。
将符合题意的选项前面的英文字母写在答题纸对应小题后面的括号内。
全部选对的得 12分，选对但不全的得6分，有选错或不答的得0分。
） 1.1964年10月16日，中国第一枚原子弹试爆成功。
该原子弹核反应物的主要成分是235U 。
天然235U 是不稳定的，它将通过若干次α衰变和β衰变最终成为稳定的元素Pb 。
235U 衰变形成的稳定的Pb 同位素为 A ．204PbB.205Pb C.206Pb D.207Pb2.将相同材料做成的两根内径不同、两端开口的圆柱形毛细管竖直插入某种与管壁浸润的液体内。
下列说法正确的是A 毛细管内液面都下降，且液面与管壁接触处比液面中间低B 毛细管内液面都上升，且液面与管壁接触处比液面中间高C 内径小的毛细管内液面上升的高度比内径大的高D 内径小的毛细管内液面上升的高度比内径大的低3. 三个质量皆为m 的小球a 、b 、c 由三段长度皆为l 的不可伸长的轻细线1L 、2L 、3L 相继连接，竖直悬挂，并处于静止状态，如图3a 所示。
在某一时刻，小球a 、b 受到水平方向的冲击，分别获得向右、向左的大小为v 的速度。
此时，中间那段细线2L 的张力大小为A ．2mgB. 242m mg l+v C.252m mg l+v D. 2112m mg l+v 4.如图4a 所示，由两块相距为0.50 mm 的薄金属板L M 、构成的平行板电容器，被屏蔽在一个金属盒K 内，连接M 端的导线同时与K 相连，金属盒上壁与L 板相距0.25 mm ，下壁与M 板相距亦为0.25 mm ，金属板面积为23040 mm ⨯。
39届物理竞赛决赛试题
39届物理竞赛决赛试题
第39届物理竞赛决赛试题
一、多项选择题
1. 关于电荷守恒定律，正确的表述是（）
A、在物理反应中，电荷总量不变
B、在物理反应中，电荷量可以增加，可以减少
C、在物理反应中，质子的电荷量会随时间而减少
D、在电解装置中，电荷量不变
2. 下列物质最有可能引发火药爆炸的是（）
A、汽油
B、醋酸
C、硫酸
D、尿素
3. 利用空气作为介质来传播声音，需要的基本物理要素有（）
A、压强
B、温度
C、湿度
D、光强
二、填空题
4. 沿着光束的方向，平行四边形波的响应强度 ___________ 随
着距离的增加而减弱。
5. 如果只考虑电磁波在空气中的消散，则垂直于电磁波传播方向
的功率会随距离的增加而减弱，逐渐衰减至 __0___。
三、判断题
6. 在空气中，声波、电磁波、重力波的速度为不同。
7. 任何物体都可以吸收声波。
A、对
B、错
四、问答题
8. 电荷守恒定律告诉我们什么？
电荷守恒定律告诉我们在任何物理反应过程中，电荷量（正负电荷量）总和不变，也就是说一定的电荷量在物理反应中会一直存在。
这个定
律可以用来解释许多物理现象，如电解溶液、带电粒子的电离等。
39届国际物理竞赛试题IPhO(英文) 理论第三题
CHANGE OF AIR TEMPERATURE WITH ALTITUDE,ATMOSPHERIC STABILITY AND AIR POLLUTIONVertical motion of air governs many atmospheric processes, such as the formation of clouds and precipitation and the dispersal of air pollutants. If the atmosphere is stable , vertical motion is restricted and air pollutants tend to be accumulated around the emission site rather than dispersed and diluted. Meanwhile, in an unstable atmosphere, vertical motion of air encourages the vertical dispersal of air pollutants. Therefore, the pollutants’ concentrations depend not only on the strength of emission sources but also on the stability of the atmosphere.We shall determine the atmospheric stability by using the concept of air parcel in meteorology and compare the temperature of the air parcel rising or sinking adiabatically in the atmosphere to that of the surrounding air. We will see that in many cases an air parcel containing air pollutants and rising from the ground will come to rest at a certain altitude, called a mixing height . The greater the mixing height, the lower the air pollutant concentration. We will evaluate the mixing height and the concentration of carbon monoxide emitted by motorbikes in the Hanoi metropolitan area for a morning rush hour scenario, in which the vertical mixing is restricted due to a temperature inversion (air temperature increases with altitude) at elevations above 119 m.Let us consider the air as an ideal diatomic gas, with molar mass μ = 29 g/mol.Quasi equilibrium adiabatic transformation obey the equation , whereconst pV γ=p Vc c γ=is the ratio between isobaric and isochoric heat capacities of the gas.The student may use the following data if necessary: The universal gas constant is R =8.31 J/(mol.K).The atmospheric pressure on ground is 0p =101.3 kPa The acceleration due to gravity is constant, g =9.81 m/s 2The molar isobaric heat capacity is 72p c =R for air. The molar isochoric heat capacity is 52V c =R for air.Mathematical hints a. ()()11ln d A Bx dx A Bx A Bx B A Bx B+==++∫∫+b. The solution of the differential equation =dxAx B dt+ (with A and B constant) is()()1B x t x t A =+ where ()1x t is the solution of the differential equation =0dxAx dt +.c. 11lim xx e x →∞⎛⎞+=⎜⎟⎝⎠1. Change of pressure with altitude.1.1. Assume that the temperature of the atmosphere is uniform and equal to .Write down the expression giving the atmospheric pressure as a function of the altitude .0T p z 1.2. Assume that the temperature of the atmosphere varies with the altitude according to the relation()()0T z T z =−Λwhere is a constant, called the temperature lapse rate of the atmosphere (the vertical gradient of temperature is -).ΛΛ 1.2.1. Write down the expression giving the atmospheric pressure as a function of the altitude .p z 1.2.2. A process called free convection occurs when the air density increases with altitude. At which values of does the free convection occur? Λ2. Change of the temperature of an air parcel in vertical motionConsider an air parcel moving upward and downward in the atmosphere. An air parcel is a body of air of sufficient dimension, several meters across, to be treated as an independent thermodynamical entity, yet small enough for its temperature to be considered uniform. The vertical motion of an air parcel can be treated as a quasi adiabatic process, i.e. the exchange of heat with the surrounding air is negligible. If the air parcel rises in the atmosphere, it expands and cools. Conversely, if it moves downward, the increasing outside pressure will compress the air inside the parcel and its temperature will increase.As the size of the parcel is not large, the atmospheric pressure at different points onthe parcel boundary can be considered to have the same value ()p z , with - the altitude of the parcel center. The temperature in the parcel is uniform and equals toz ()parcel T z , which is generally different from the temperature of the surrounding air ()T z . In parts 2.1 and 2.2, we do not make any assumption about the form of T (z ).2.1. The change of the parcel temperature parcel T with altitude is defined byparcel dT G dz=−. Derive the expression of (T , T G parcel ).2.2. Consider a special atmospheric condition in which at any altitude z the temperature of the atmosphere equals to that of the parcel T parcel T , ()()parcel T z T z =. We use Γ to denote the value of when G parcel T T =, that is parcel dT dzΓ=−(with parcel T T =). Γ is called dry adiabatic lapse rate . 2.2.1. Derive the expression of Γ2.2.2. Calculate the numerical value of Γ.2.2.3. Derive the expression of the atmospheric temperature ()T z as a functionof the altitude.2.3. Assume that the atmospheric temperature depends on altitude according to the relation ()()0T z T z =−Λ, where Λ is a constant. Find the dependence of the parcel temperature ()parcel T z on altitude . z2.4. Write down the approximate expression of ()parcel T z when ()0z T Λ<< andT (0) ≈ T parcel (0).3. The atmospheric stability.In this part, we assume that changes linearly with altitude.T 3.1. Consider an air parcel initially in equilibrium with its surrounding air at altitude0z , i.e. it has the same temperature ()0T z as that of the surrounding air. If the parcel ismoved slightly up and down (e.g. by atmospheric turbulence), one of the three following cases may occur:- The air parcel finds its way back to the original altitude , the equilibrium of the parcel is stable. The atmosphere is said to be stable.0z - The parcel keeps moving in the original direction, the equilibrium of the parcelis unstable. The atmosphere is unstable.- The air parcel remains at its new position, the equilibrium of the parcel is indifferent. The atmosphere is said to be neutral.What is the condition on for the atmosphere to be stable, unstable or neutral?Λ3.2. A parcel has its temperature on ground ()parcel 0T higher than the temperature()0T of the surrounding air. The buoyancy force will make the parcel rise. Derive theexpression for the maximal altitude the parcel can reach in the case of a stable atmosphere in terms of and Γ.Λ4. The mixing height4.1. Table 1 shows air temperatures recorded by a radio sounding balloon at 7: 00 am on a November day in Hanoi. The change of temperature with altitude can be approximately described by the formula ()()0T z T z =−Λ with different lapse rates Λ in the three layers 0 << 96 m, 96 m << 119 m and 119 m<< 215 m.z z z Consider an air parcel with temperature ()parcel 0T = 22o C ascending from ground. On the basis of the data given in Table 1 and using the above linear approximation, calculate the temperature of the parcel at the altitudes of 96 m and 119 m.4.2. Determine the maximal elevation the parcel can reach, and the temperatureH ()parcel T H of the parcel.H is called the mixing height. Air pollutants emitted from ground can mix with theair in the atmosphere (e.g. by wind, turbulence and dispersion) and become diluted within this layer.Table 1Data recorded by a radio sounding balloon at 7:00 am on a November day in Hanoi.Altitude, m Temperature, o C5 21.560 20.664 20.569 20.575 20.481 20.390 20.296 20.1102 20.1109 20.1113 20.1119 20.1128 20.2136 20.3145 20.4153 20.5159 20.6168 20.8178 21.0189 21.5202 21.8215 22.0225 22.1234 22.2246 22.3257 22.35. Estimation of carbon monoxide (CO) pollution during a morning motorbike rush hour in Hanoi.Hanoi metropolitan area can be approximated by a rectangle with base Ldimensions and W as shown in the figure, with one side taken along the south-west bank of the Red River.It is estimated that during the morning rush hour, from 7:00 am to 8:00 am, there are 8x105 motorbikes on the road, each running on average 5 km and emitting 12 g of CO per kilometer. The amount of CO pollutant is approximately considered as emitted uniformly in time, at a constant rate M during the rush hour. At the same time, the clean north-east wind blows perpendicularly to the Red River (i.e. perpendicularly to the sides L of the rectangle) with velocity u, passes the city with the same velocity, and carries a part of the CO-polluted air out of the city atmosphere.Also, we use the following rough approximate model:• The CO spreads quickly throughout the entire volume of the mixing layerC t of CO at time cant above the Hanoi metropolitan area, so that the concentration ()be assumed to be constant throughout that rectangular box of dimensions L, W and H.• The upwind air entering the box is clean and no pollution is assumed to be lost from the box through the sides parallel to the wind.• Before 7:00 am, the CO concentration in the atmosphere is negligible.5.1. Derive the differential equation determining the CO pollutant concentration ()C t as a function of time.C t.5.2. Write down the solution of that equation for ()C t at 8:00 a.m.5.3. Calculate the numerical value of the concentration ()L W uGiven = 15 km,= 8 km, = 1 m/s.。
国际物理林匹克竞赛试题ipho_BUL81The
XII International Physics OlympiadVarna, Bulgaria, July 1981The problems and the solutions are adapted by Miroslav Abrashev University of Sofia, Faculty of Physics, 5 James Bourchier Blvd., 1164 Sofia, BulgariaReference: O. F. Kabardin, V. A. Orlov, in “International Physics Olympiads for High School Students”, ed. V. G. Razumovski, Moscow, Nauka, 1985. (In Russian).Theoretical Problem 1A static container of mass M and cylindrical shape is placed in vacuum. One of its ends is closed. A fixed piston of mass m and negligible width separates the volume of the container into two equal parts. The closed part contains n moles of monoatomic perfect gas with molar mass M0 and temperature T. After releasing of the piston, it leaves the container without friction. After that the gas also leaves the container. What is the final velocity of the container?The gas constant is R. The momentum of the gas up to the leaving of the piston can be neglected. There is no heat exchange between the gas, container and the piston. The change of the temperature of the gas, when it leaves the container, can be neglected. Do not account for the gravitation of the Earth.Theoretical Problem 2An electric lamp of resistance R0= 2 Ωworking at nominal voltage U0= 4.5 V is connected to accumulator of electromotive force E = 6 V and negligible internal resistance.1. The nominal voltage of the lamp is ensured as the lamp is connected potentiometrically to the accumulator using a rheostat with resistance R. What should be the resistance R and what is the maximal electric current I max, flowing in the rheostat, if the efficiency of the system must not be smaller than η0 = 0.6?2. What is the maximal possible efficiency η of the system and how the lamp can be connected to the rheostat in this case?Theoretical Problem 3A detector of radiowaves in a radioastronomical observatory is placed on the sea beach at height h = 2 m above the sea level. After the rise of a star, radiating electromagnetic waves of wavelength λ= 21 cm, above the horizont the detector registers series of alternating maxima and minima. The registered signal is proportional to the intensity of thedetected waves. The detector registers waves with electric vector, vibrating in a direction parallel to the sea surface.1. Determine the angle between the star and the horizont in the moment when the detector registers maxima and minima (in general form).2. Does the signal decrease or increase just after the rise of the star?3. Determine the signal ratio of the first maximum to the next minimum. At reflection of the electromagnetic wave on the water surface, the ratio of the intensities of the electric field of the reflected (E r ) and incident (E i ) wave follows the low:ϕϕcos cos +-=n n E E ir ,where n is the refraction index and ϕ is the incident angle of the wave. For the surface “air -water” for λ = 21 cm, the refraction index n = 9.4. Does the ratio of the intensities of consecutive maxima and minima increase ordecrease with rising of the star?Assume that the sea surface is flat.Solution of the Theoretical Problem 1Up to the moment when the piston leaves the container, the system can be considered as a closed one. It follows from the laws of the conservation of the momentum and the energy:0)(10=-+mu v nM M (1)U mu v nM M ∆=++22)(2210, (2)where v 1 – velocity of the container when the piston leaves it, u – velocity of the piston in thesame moment, ∆U – the change of the internal energy of the gas. The gas is perfect and monoatomic, therefore)(2323f T T nR T nR U -=∆=∆;(3)T f - the temperature of the gas in the moment when the piston leaves the container. This temperature can be determined by the law of the adiabatic process:.const pV =γUsing the perfect gas equation nRT pV =, one obtains.1const TV=-γ,11--=γγff V T TVUsing the relation V V f 2=, and the fact that the adiabatic coefficient for one-atomic gas is352325===RR c c vp γ, the result for final temperature is:3232122)(--===T T V V T T ff γ (4)Solving the equations (1) – (4) we obtain))(()21(30132M nMm M nMmnRTv +++-=- (5)If the gas mass nM 0 is much smaller than the masses of the container M and the piston m , then the equation (5) is simplified to:)()21(3321M m M mnRT v +-=- (5’)When the piston leaves the container, the velocity of the container additionally increases to value v 2 due to the hits of the atoms in the bottom of the container. Each atom gives the container momentum: x A v m p ∆=2, where m A – mass of the atom; AA N Mm 0=, and x v can be obtained by the averaged quadraticvelocity of the atoms 2v as follows:2222vv v v zyx=++, and 222zyxv v v ==, therefore 32vv x =. It appears that due to the elasticimpact of one atom the container receives averaged momentum322vN Mp A=All calculations are done assuming that the thermal velocities of the atoms are much larger than the velocity of the container and that the movement is described using system connected with the container. Have in mind that only half of the atoms hit the bottom of the container, the total momentum received by the container is3212vnMp nNp At ==(6)and additional increase of the velocity of the container is3202vMM nMp v t ==. (7)Using the formula for the averaged quadratic velocity23M RT vf =as well eq. (4) for the temperature T f , the final result for v 2 isMRTM n v 03122-=.(8)Therefore the final velocity of the container is21v v v +==))(()21(3032M nMm M nMmnRT+++--+MRTM n 0312-≈≈)()21(332M m M mnRT +--+MRTM n 0312-.(9)Solution of the Theoretical Problem 21) The voltage U 0 of the lamp of resistance R 0 is adjusted using the rheostat ofresistance R . Using the Kirchhoff laws one obtains:xR R U R U I -+=00, (1)where x R R - is the resistance of the part of the rheostat, parallel connected to the lamp, R x is the resistance of the rest part, x IR E U -=0(2) The efficiency η of such a circuit isRIEU IE r U P P accum lamp 2020.===η. (3)From eq. (3) it is seen that the maximal current, flowing in the rheostat, is determined by theminimal value of the efficiency:2min2max ηηRE U RE U I ==.(4)The dependence of the resistance of the rheostat R on the efficiency η can determined replacing the value for the current I , obtained by the eq. (3), ηRE U I 2=, in the eqs. (1) and (2):x R R R RE U -+=1100η, (5)20)(U RE U E R x η-=.(6)Thenηηη20201)1(1U E U E UE R R --+=. (7)To answer the questions, the dependence )(ηR must be investigated. By this reason we find the first derivative ηR ':'⎪⎪⎪⎪⎭⎫⎝⎛--+∝'ηηηη21)1(U EU E R ∝ +--+∝)1)(1(2100ηηU EU E =⎥⎦⎤⎢⎣⎡-+002)1(U E U E ηη1)1)(2(00+--U E U E ηη. η < 1, therefore the above obtained derivative is positive and the function R(η) is increasing. It means that the efficiency will be minimal when the rheostat resistance is minimal. ThenΩ≈--+=≥53.81)1(10020200min ηηηU E U E UE R R R .The maximal current I max can be calculated using eq. (4). The result is: I max ≈660 mA.2) As the function R(η) is increasing one, max ηη→, when ∞→R . In this case thetotal current I will be minimal and equal to RU 0. Therefore the maximal efficiency is75.00max ==E U ηThis case can be realized connecting the rheostat in the circuit using only two of itsthree plugs. The used part of the rheostat is R 1:Ω≈-=-=67.000001R U U E I U E R .Solution of the Theoretical Problem 31) The signal, registered by the detector A, is result of the interference of two rays: the ray 1, incident directly from the star and the ray 2, reflected from the sea surface (see the figure).The phase of the second ray is shifted by π due to the reflection by a medium of larger refractive index. Therefore, the phase difference between the two rays is: =⎪⎭⎫ ⎝⎛-+=-+=∆)2cos(sin 2sin 2ααλαλh h AB ACαλααλsin 22)]2cos(1[sin 2h h +=-+=(1)The condition for an interference maximum is: λαλk h =+max sin 22, orhk hk 4)12(2)21(sin max λλα-=-=, (2)where k = 1,2,3,…,19. (the difference of the optical paths cannot exceed 2h , therefore kcannot exceed 19). The condition for an interference minimum is: 2)12(sin 22max λαλ+=+k h , orhk 2sin min λα=(3)where k = 1,2,3, (19)2) Just after the rise of the star the angular height α is zero, therefore the condition foran interference minimum is satisfied. By this reason just after the rise of the star, the signal will increase. 3) If the condition for an interference maximum is satisfied, the intensity of the electric field is a sum of the intensities of the direct ray E i and the reflected ray E r , respectively: r i E E E +=max .Because ϕϕcos cos +-=n n E E ir , then ⎪⎪⎭⎫⎝⎛+-+=max maxmax cos cos 1ϕϕn n E E i . From the figure it is seen that maxmax 2απϕ-=, we obtain)2sin(2sin sin 1max maxmaxmaxααα+=⎪⎪⎭⎫ ⎝⎛+-+=n n E n n E E i i .(4)At the interference minimum, the resulting intensity is:minmin min sin sin 2αα+=-=n E E E E ir i . (5)The intensity I of the signal is proportional to the square of the intensity of the electricfield E , therefore the ratio of the intensities of the consecutive maxima and minima is:2max 2min min 222minmax minmax )sin ()sin (sin ααα++=⎪⎪⎭⎫ ⎝⎛=n n n E E I I . (6)Using the eqs. (2) and (3), the eq. (6) can be transformed into the following form:22222minmax 4)12(24⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡-++=h k n h k n k h n I I λλλ. (7)Using this general formula, we can determine the ratio for the first maximum (k =1) and thenext minimum:2222min max 424⎪⎪⎪⎪⎭⎫ ⎝⎛++=h n h n h n I I λλλ= 3.104 4) Using that hn 2λ>>, from the eq. (7) follows :2222minmax 4λk h n I I ≈.So, with the rising of the star the ratio of the intensities of the consecutive maxima and minima decreases.。
第39届全国中学生物理竞赛预赛试题解答
第39届全国中学生物理竞赛预赛试题解答（2022年9月3日9:00-12:00）一、 选择题（本题60分，含5小题，每小题12 分。
在每小题给出的4个选项中，有的小题只有一项符合题意，有的小题有多项符合题意。
将符合题意的选项前面的英文字母写在答题纸对应小题后面的括号内。
全部选对的得 12分，选对但不全的得6分，有选错或不答的得0分。
）1. D2. BC3. D4. C5. B二、填空题（本题100 分，每小题20分，每空10分。
请把答案填在答题纸对应题号后面的横线上。
只需给出结果，不需写出求得结果的过程。
）bmg cF b c ] 7. [0020(4)3RT P V V V ，043RT ] 8. 1588， 9. [380052:2043, 380052:2043]或[42251:227, 42251:227] 或[186.0:1, 186.0:1]或[186.1:1, 186.1:1]]三、计算题（本题 240分，共6小题，每小题40 分。
计算题的解答应写出必要的文字说明、方程式和重要的演算步骤, 只写出最后结果的不能得分。
有数值计算的, 答案中必须明确写出数值，有单位的必须写出单位。
） 11.（1）一套平凸-平板透镜相切式干涉装置如解题图11a 所示，在平行于系统光轴OA 的光线傍轴垂直（垂直于平板）入射的条件下，即e R 时，在直角三角形OAB 中，运用勾股定理得 222()R e r R ①式中，r 是入射光线到系统光轴OA 的距离，e 是该光线在平板上反射光线所走过的距离。
略去2e 项（二阶小量），由①式得 22r e R②后文的讨论都基于此傍轴垂直入射时的公式。
对于题干图11a 所示的平凸-平凹内切结构的干涉装置。
在平行于系统光轴的光线（与光轴相距r ）傍轴入射的条件下，利用公式②得2112r e R ③解题图11a2222r e R ④这里，12e e 、是入射光线的延长线分别与两球面的交点到该光线与过两球面内切点的切平面的交点之间的距离，如解题图11b 所示。
39届国际物理竞赛试题IPhO（英文）理论第一题
39届国际物理竞赛试题IPhO（英⽂）理论第⼀题WATER-POWERED RICE-POUNDING MORTARA. IntroductionRice is the main staple food of most people in Vietnam. To make white rice from paddy rice, one needs separate of the husk (a process called "hulling") and separate the bran layer ("milling"). The hilly parts of northern Vietnam are abundant with water streams, and people living there use water-powered rice-pounding mortar for bran layer separation. Figure 1 shows one of such mortars., Figure 2 shows how it works.B. Design and operation1. Design .The rice-pounding mortar shown in Figure 1 has the following parts: The mortar , basically a wooden container for rice.The lever , which is a tree trunk with one larger end and one smaller end. It can rotate around a horizontal axis. A pestle is attached perpendicularly to the lever at the smaller end. The length of the pestle is such that it touches the rice in the mortar when the lever lies horizontally. The larger end of the lever is carved hollow to form a bucket. The shape of the bucket is crucial for the mortar's operation.2. Modes of operation The mortar has two modes.Working mode . In this mode, the mortar goes through an operation cycle illustrated in Figure 2.The rice-pounding function comes from the work that is transferred from the pestle to the rice during stage f) of Figure 2. If, for some reason, the pestle never touches the rice, we say that the mortar is not working.Rest mode with the lever lifted up . During stage c) of the operation cycle (Figure 2), as the tilt angle α increases, the amount of water in the bucket decreases. At one particular moment in time, the amount of water is just enough to counterbalance the weight of the lever. Denote the tilting angle at this instant by β. If the lever is kept at angle β and the initial angular velocity is zero, then the lever will remain at this position forever. This is the rest mode with the lever lifted up. The stability of this position depends on the flow rate of water into the bucket, Φ. If exceeds some value Φ2Φ, then this rest mode is stable, and the mortar cannot be in the working mode. In other words, is the minimal flow rate for the mortar not to work. 2ΦFigure 1A water-powered rice-pounding mortarOPERATION CYCLE OF A WATER-POWERED RICE-POUNDING MORTARFigure 2a)b)a) At the beginning there is no water in the bucket, the pestle rests on the mortar. Water flows into the bucket with a small rate, but for some time the lever remains in the horizontal position.b) At some moment the amount of water is enough to lift the lever up. Due to thetilt, water rushes to the farther side of the bucket, tilting the lever more quickly.Water starts to flow out at 1αα=.c) As the angle α increases, water starts to flow out. At some particular tilt angle, αβ=, the total torque is zero.d) α continues increasing, watercontinues to flow out until no water remains in the bucket.e) α keeps increasing because of inertia. Due to the shape of the bucket, water falls into the bucket but immediately flows out. The inertial motion of the lever continues until α reaches the maximal value 0α.f) With no water in the bucket, the weight of the lever pulls it back to the initial horizontal position. The pestle gives the mortar (with rice inside) a pound and a new cycle begins.C. The problemConsider a water-powered rice-pounding mortar with the following parameters (Figure 3)The mass of the lever (including the pestle but without water) is M =30 kg,The center of mass of the lever is G .. The lever rotates around the axis T (projected onto the point T on the figure).The moment of inertia of the lever around T is I =12 kg ?m 2.When there is water in the bucket, the mass of water is denoted as , the center of mass of the water body is denoted as N.m The tilt angle of the lever with respect to the horizontal axis is α.The main length measurements of the mortar and the bucket are as in Figure 3. Neglect friction at the rotation axis and the force due to water falling onto the bucket. In this problem, we make an approximation that the water surface is always horizontal.Figure 3 Design and dimensions of the rice-pounding mortar1. The structure of the mortarAt the beginning, the bucket is empty, and the lever lies horizontally. Then water flows into the bucket until the lever starts rotating. The amount of water in the bucket at this moment is 1.0 kg.m =1.1. Determine the distance from the center of mass G of the lever to the rotation axis T. It is known that GT is horizontal when the bucket is empty.1.2. Water starts flowing out of the bucket when the angle between the lever and the horizontal axis reaches 1α. The bucket is completely empty when this angle is 2α. Determine 1αand 2α.1.3. Let ()µα be the total torque (relative to the axis T) which comes from theweight of the lever and the water in the bucket. ()µα is zero when αβ=. Determineβ and the mass of water in the bucket at this instant.1m2. Parameters of the working modeLet water flow into the bucket with a flow rate Φ which is constant and small. The amount of water flowing into the bucket when the lever is in motion is negligible. In this part, neglect the change of the moment of inertia during the working cycle.2.1. Sketch a graph of the torque µ as a function of the angle α, ()µα, duringone operation cycle. Write down explicitly the values of ()µα at angle α1, α2, andα = 0.2.2. From the graph found in section 2.1., discuss and give the geometric interpretation of the value of the total energy produced bytotal W ()µαand the workthat is transferred from the pestle to the rice.pounding W2.3. From the graph representing µ versus α, estimate 0α and (assumethe kinetic energy of water flowing into the bucket and out of the bucket is negligible.) You may replace curve lines by zigzag lines, if it simplifies the calculation. pounding W3. The rest modeLet water flow into the bucket with a constant rate Φ, but one cannot neglect the amount of water flowing into the bucket during the motion of the lever. 3.1. Assuming the bucket is always overflown with water,3.1.1. Sketch a graph of the torque µ as a function of the angle α in the vicinity of αβ=. To which kind of equilibrium does the position αβ= of the lever belong?3.1.2. Find the analytic form of the torque ()µα as a function of αΔ whenαβ=+Δα, and αΔ is small.3.1.3. Write down the equation of motion of the lever, which moves with zero initial velocity from the position αβα=+Δ (αΔ is small). Show that the motion is, with good accuracy, harmonic oscillation. Compute the period τ.3.2. At a given , the bucket is overflown with water at all times only if the lever moves sufficiently slowly. There is an upper limit on the amplitude of harmonic oscillation, which depends on . Determine the minimal value ΦΦ1Φof (in kg/s) so that the lever can make a harmonic oscillator motion with amplitude 1Φo .3.3. Assume that is sufficiently large so that during the free motion of the lever when the tilting angle decreases from Φ2α to 1α the bucket is always overflown with water. However, if is too large the mortar cannot operate. Assuming that the motion of the lever is that of a harmonic oscillator, estimate the minimal flow rate for the rice-pounding mortar to not work.Φ2Φ。
39届国际物理竞赛试题IPhO(英文) 理论第二题
CHERENKOV LIGHT AND RING IMAGING COUNTERLight propagates in vacuum with the speed . There is no particle which moves with a speed higher than . However, it is possible that in a transparent medium a particle moves with a speed higher than the speed of the light in the same medium c c v c n , where is the refraction index of the medium. Experiment (Cherenkov, 1934) and theory (Tamm and Frank, 1937) showed that a charged particle, moving with a speed in a transparent medium with refractive indexsuch that n v n c n>v , radiates light, called Cherenkov light1arccos nθβ= (1) where cβ=v . 1. To establish this fact, consider a particle moving at constant velocity c n >v on a straight line. It passes A at time 0 and B at time . As the problem is symmetric with respect to rotations around AB, it is sufficient to consider light rays in a plane containing AB.1t At any point C between A and B, the particle emits a spherical light wave, which propagates with velocity c n. We define the wave front at a given time as the envelope of all these spheres at this time.t 1.1. Determine the wave front at time and draw its intersection with a plane containing the trajectory of the particle.1t 1.2. Express the angle ϕ between this intersection and the trajectory of the particle in terms of and n β.2. Let us consider a beam of particles moving with velocity c n>v , such that the angle θ is small, along a straight line IS. The beam crosses a concave spherical mirror of focal length f and center C, at point S. SC makes with SI a small angle α (see the figure in the Answer Sheet). The particle beam creates a ring image in the focal plane of the mirror.Explain why with the help of a sketch illustrating this fact. Give the position of the center O and the radius of the ring image.r This set up is used in ring imaging Cherenkov counters (RICH) and the medium which the particle traverses is called the radiator .Note: in all questions of the present problem, terms of second order and higher in α and θ will be neglected.3. A beam of particles of known momentum 100GeV/.p c = consists of three types of particles: protons, kaons and pions, with rest mass 2p 094GeV ./M c =, 2κ050GeV ./M c = and 2π014GeV ./M c =, respectively. Remember that and pc 2Mc have the dimension of an energy, and 1 eV is the energy acquired by an electron after being accelerated by a voltage 1 V , and 1 GeV = 109 eV , 1 MeV = 106 eV .The particle beam traverses an air medium (the radiator) under the pressure . The refraction index of air depends on the air pressure according to the relation where a = 2.7×10P P 1n a =+P -4 atm -13.1. Calculate for each of the three particle types the minimal value of the air pressure such that they emit Cherenkov light.min P 3.2. Calculate the pressure 12P such that the ring image of kaons has a radius equalto one half of that corresponding to pions. Calculate the values of κθ and πθ in this case.Is it possible to observe the ring image of protons under this pressure?4. Assume now that the beam is not perfectly monochromatic: the particles momenta are distributed over an interval centered at 10 having a half width at half height . This makes the ring image broaden, correspondingly GeV /c p Δθ distribution has a half width at half height θΔ. The pressure of the radiator is 12P determined in 3.2.4.1. Calculateκp θΔΔ and πp θΔΔ, the values taken by p θΔΔin the pions and kaons cases.4.2. When the separation between the two ring images, πκθθ−, is greater than 10πtimes the half-width sum κθθΔ=Δ+Δθ, that is πκ10θθθ−>Δ, it is possible to distinguish well the two ring images. Calculate the maximal value of such that the two ring images can still be well distinguished.p Δ5. Cherenkov first discovered the effect bearing his name when he was observing a bottle of water located near a radioactive source. He saw that the water in the bottle emitted light.5.1. Find out the minimal kinetic energy of a particle with a rest mass min T M moving in water, such that it emits Cherenkov light. The index of refraction of water is n = 1.33.5.2. The radioactive source used by Cherenkov emits either α particles (i.e. helium nuclei) having a rest mass 2α38GeV ./M c = or β particles (i.e. electrons) having a rest mass 2e 051MeV ./M c =. Calculate the numerical values of for α particles and β particles.min T Knowing that the kinetic energy of particles emitted by radioactive sources never exceeds a few MeV , find out which particles give rise to the radiation observed by Cherenkov.6. In the previous sections of the problem, the dependence of the Cherenkov effect on wavelength λ has been ignored. We now take into account the fact that the Cherenkov radiation of a particle has a broad continuous spectrum including the visible range (wavelengths from 0.4 µm to 0.8 µm). We know also that the index of refraction of the radiator decreases linearly by 2% of n 1n − when λ increases over this range.6.1. Consider a beam of pions with definite momentum of moving in air at pressure 6 atm. Find out the angular difference 100GeV .c /δθ associated with the two ends of the visible range.6.2. On this basis, study qualitatively the effect of the dispersion on the ring image of pions with momentum distributed over an interval centered at and having a half width at half height 10GeV /p c =03GeV ./p c Δ=.6.2.1. Calculate the broadening due to dispersion (varying refraction index) and that due to achromaticity of the beam (varying momentum).6.2.2. Describe how the color of the ring changes when going from its inner to outer edges by checking the appropriate boxes in the Answer Sheet.。
第39届物理竞赛月考试卷
质心GPhO 标准试通知:2022全国中学生物理竞赛五一集训营已上线,上课时间：5月1日—5月8日复赛刷题班:70%使用2022年新出原创题，30%使用近几年出过的原创题题1摇滚圆桌如图一个半径为R 的质量为M 的均匀圆盘,水平放置,下方用三个固定轴承支撑.每个轴承半径为r ,保持恒定的旋转角速度ω0=v 0/r ,使得三个轴承和圆盘的接触点的速度都指向原点.三个轴承ABC 恰好形成一个正三角形,到达原点的距离都是l <R .圆盘和轴承面之间的摩擦系数都是µ,重力加速度为g .俯视图侧视图已知积分:∫d x x √a 2x 2+1=√1+a 2x 2+ln √1+a 2x 2−1ax+C (1)当圆盘偏离中心的平衡位置一个小距离(x ,y )时,求三个轴承上的支持力N A ,N B ,N C .(2)当圆盘位有一个小位移(x ,y ),且有一个小速度为(˙x,˙y )时,求出水平方向上圆盘受到的合力.(3)初态圆盘偏离了原点很小的距离,并具有很小的运动速度,请给出接下来圆盘能够尽快回到稳态的条件.(4)初态圆盘质心保持在原点,而获得一个竖直方向的角速度Ω0,注意Ω0不是小量,求之后圆盘角速度随时间变化.题2动力履带如图所示的动力传送装置由左边绕固定轴转动的主动轮与右边绕固定轴转动的从动轮组成,两轮皆为半径为R ,质量为M 的质量均匀分布的圆盘.主动轮与从动轮之间通过绷紧的履带传动,履带线密度为λ,总长度为L ,本题中重力可忽略不计.已知轮与履带间的摩擦系数为µ,当两轮角速度打到ω时,为了保证不打滑,从动轮可以获得的最大角速度为β,求此时履带上张力最小处的张力F 0.题3磁镜聚焦如图所示的磁镜由半径为R的N匝线圈组成,流过电流为I.这一磁镜可以令水平速度v z恒定的傍轴运动电子(质量为m,电荷量为q)汇聚到一点.由于焦距f远远大于磁场运动区域,可以将这一磁镜看作一薄透镜.已知积分∫+∞−∞1(1+x2)3d x=3π8(1)通过磁场性质,以线圈轴线为z轴建立柱坐标系,求出轴线附近r处径向磁场B r(z,r).(2)通过电子运动求出焦距f.(3)在做某次观测的时候,理想情况物体在物距u处,使用焦距f可以表达像平面位置.当物体的实际物距偏离u一小段距离,例如u±∆的时候,像平面上将获得一个弥散圆,当这个圆的半径超过δ的时候,即认为成像失去意义,称临界∆c为此成像系统景深.本题中磁镜系统可视为有效半径为R c的理想薄透镜,请利用u,R c,δ表达上述磁镜系统的景深∆c.题4家庭用电功率因数是指交流电路中元件有功功率对视在功率的比值,有功功率是元件实际平均功率,视在功率是元件的有效电压乘以有效电流.现在用一个220V,50Hz的市电电源给一个功率因数为0.6的感性负载供电,负载吸收的实际功率为550W.(1)欲使功率因数提高到1,需要串联多大的电容?(2)串联电容后电源输出的功率是多少?(3)如果导线上有小电阻r=0.1Ω,求出两种情况下导线上的功率之比.题5电热水壶现有总体热容为C的热水壶处于温度为T0的媒质中,若用上题的用电器以P0的功率加热,它所能达到的最高温度为T1.设系统的散热遵从牛顿冷却定律,牛顿冷却定律可以表示为d Q=−a(T−T0)d t其中T0为环境温度.T为系统温度,a为未知常数.(T1+T0)时所需的时间,请注意a为未知常数.试帮质心姐姐计算一下,加热电路切断后,物体温度从T1降为12题6重温牛二已知在相对论中,力可以定义为F=d p,请解决以下两个问题.d t(1)如图所示,一个“哑铃”形状的结构,两端是静止质量为m质点,中间用轻杆连接.“哑铃”绕着中心,在平面内发生旋转.地面系中某时刻两个端点速度大小为v,,旋转周期为T,存在一个外力F如图作用在杆的中心,使杆中心具有加速度a,本题中时间尺度关系为T≪v.求哑铃一周运动中,需要的平均作用外力¯F.a(2)一个光子具有能量E.在x轴上发生一维的来回运动.它被束缚在一个质量忽略不计的盒子当中,光子来回运动周期为T.存在外力使盒子的加速度为a,本题中时间尺度关系为T≪c.求光子一个来回运动中,盒子上a需要的平均作用外力¯F.(提示：考虑光子由盒子壁反射造成的频率变化)题7电容话筒贝尔实验室在1916年发明了电容话筒.迄今为止,因为电容话筒低频处理较好,以及噪声影响小这两个技术上的优势,而被广泛应用.但是它有个缺点,就是贵.在本题当中,模型如下：理想电压源E=9.00V.电阻R=2.00Ω.平板电容器电容为C0=10.0pF.两板之间间距为d0.三个元件串联.电容的一个板很轻,可以根据声波的振动而振动.可以振动的这个板通过装置悬挂,其回复力可以忽略.本题中两个电容板之间的静电吸引力可以忽略.(1)因为声波的影响,两板之间间距随时间变化为d(t)=d0[1+αsin(ωt)].这里ω是角频率,以及α≪1.这样电阻元件的电压U R(t)=U R0sin(ωt+φ).请求出这里的U R0和φ.(2)请给出U R(t)的表达式,有效声音频率范围50Hz∼20kHz(人耳一般听力范围附近).(3)已知声音的波速c,标准大气压p0时空气密度为ρ.如果探测的声音是稳定的单频率声波u(x,t)= u0sin(kx−ωt),这里u0是声波振幅.而声波也可描述为空气的压缩和膨胀p(x,t)=p0+p1sin(kx−ωt+ψ),其中p1远小于p0,请求出这里的p1,结果用c,ρ,ω,u0表示.(4)接下来我们分析声波的能量.假设模型中的浮动的板质量为m=0.500g.面积为S=10.0cm2.我们用I表示单位时间单位面积声波传递的能量,这样角频率为ω,振幅为u0的声音就有I=12ρcω2u2.声音强度的常用单位是[dB]（分贝）,它和国际单位[Wm2]的换算是：I[dB]=10log10(II0).这里I0=1×10−12Wm2可以理解为“听力极限”也就是小于这个功率就听不到了.如果强度I=70dB.频率f=1.00kHz.空气密度ρ=1.20kgm3,音速c=340ms .真空介电常数ϵ0=8.85×10−12Fm.请计算,通过我们的这个电容话筒测量,电阻两端的最大电压U0能达到多少？(5)计算两板之间间距d0.(6)如果收音板是铝材料制成,密度为ρc=5.00×103kgm3.请求出收音板的厚度h.题8继续摇滚在一个半径为R的圆柱形弹性固定空腔中,有一个半径为r质量为m的均匀小弹球,在垂直于圆柱的轴线平面中运动.弹球和空腔壁发生碰撞的时候,沿着圆柱半径方向的法向速度会反向,而摩擦力的作用会改变角速度和切向速度,但是保持能量守恒.（1）初态弹球位于边缘处,角速度ω0=0,速度大小为v0,方向和半径夹角为θ0.求第n次反弹后,小球的速度大小v n和方向θn,以及之后运动中小球距离轴线的最近距离.（2）小球初态和上一问相同,但是圆柱筒壁以恒定的角速度Ω转动,小球反弹的属性和上一问相同,但是筒壁可以对小球做功,这个过程中没有能量耗散.求之后运动中小球距离轴线的最近距离.（3）接上一问,要求在筒壁的旋转参照系中,写出小球轨迹应当满足的方程,你应当能重复出第二问的结果.。
历届IPho试题-历届IPho试题-O_13
Problems of the 13th International Physics Olympiad(Malente, 1982)Gunter Lind & Gunnar Friege 1Leibniz-Institute for Science Education (IPN) at the University of Kiel, GermanyAbstractThe 13th International Physics Olympiad took place in 1982 in the Federal Republic of Germany. This article contains the competition problems, their solutions and a grading scheme.IntroductionIn 1982 the Federal Republic of Germany was the first host of the Physics Olympiad outside the so-called Eastern bloc. The 13th International Physics Olympiad took place in Malente, Schleswig-Holstein. The competition was funded by the German Federal Ministry of Science and Education. The organisational guidelines were laid down by the work group “Olympiads for pupils” of the conference of ministers of education of the German federal states. The Institute for Science Education (IPN) at the University of Kiel was responsible for the realisation of the event. A commission of professors, whose chairman was appointed by the German Physical Society, were concerned with the formulation of the competition problems. All other members of the commission came from physics department of the university of Kiel or from the college of education at Kiel.The problems as usual covered different fields of classical physics. In 1982 the pupils had to deal with three theoretical and two experimental problems, whereas at the previous Olympiads only one experimental task was given. However, it seemed to be reasonable to put more stress on experimental work. The degree of difficulty was well balanced. One of the theoretical problems could be considered as quite simple (problem 3: “hot-air balloon”). Another theoretical problem (problem 1: “fluorescent lamp”) had a mean degree of difficulty and the distribution of the points was a normal distribution with only a few1 Contact:Leibniz-Institute for Science Education (IPN) at the University of KielOlshausenstrasse 62, 24098 Kiel, Germanyipho@ipn.uni-kiel.deexcellent and only a few unsatisfying solutions. The third problem (problem 2: “oscillation coat hanger”) turned out to be the most difficult problem. This problem was generally considered as quite interesting because different ways of solving were possible. About one third of the pupils did not find an adequate start to the problem, but nearly one third of the pupils was able to solve the substantial part of the problem. That means, this problem polarized between the pupils. The two experimental tasks were quite different in respect of the input for the experimental setup and the time required for dealing with the problems, whereas they were quite similar in the degree of difficulty. Both required demandingly theoretical considerations and experimental skills. Both experimental problems turned out to be rather difficult. The tasks were composed in a way that on the one hand almost every pupil had the possibility to come to certain partial results and that there were some difficulties on the other hand which could only be solved by very few pupils. The difficulty in the second experimental problem (problem5: “motion of a rolling cylinder”) was the explanation of the experimental results, which were initially quite surprising. The difficulty in the other task (problem 4: “lens experiment") was the revealing of an observation method with a high accuracy (parallax). The five hours provided for solving the two experimental problems were slightly too short. According to that, in both experiments only a few pupils came up with excellent solutions. In problem 5 nobody got the full points.The problems presented here are based on the original German and English versions of the competition problems. The solutions are complete but in some parts condensed to the essentials. Almost all of the original hand-made figures are published here.Theoretical ProblemsProblem 1: Fluorescent lampAn alternating voltage of 50 Hz frequency is applied to the fluorescent lamp shown in the accompanying circuit diagram.The following quantities are measured:overall voltage (main voltage)U = 228.5 V electric currentI = 0.6 A partial voltage across the fluorescent lampU’= 84 V ohmic resistance of the series reactor d R 26.3=ΩThe fluorescent lamp itself may be considered as an ohmic resistor in the calculations.a) What is the inductance L of the series reactor?b) What is the phase shift ϕ between voltage and current?c) What is the active power P w transformed by the apparatus?d) Apart from limiting the current the series reactor has another important function. Name and explain this function!Hint: The starterincludes a contact which closes shortly afterswitching on the lamp, opens up again and stays open.e) In a diagram with a quantitative time scale sketch the time sequence of the luminous flux emitted by the lamp.f) Why has the lamp to be ignited only once although the applied alternating voltage goes through zero in regular intervals?g) According to the statement of the manufacturer, for a fluorescent lamp of the described type a capacitor of about 4.7 µF can be switched in series with the series reactor. How does this affect the operation of the lamp and to what intent is this possibility provided for?h) Examine both halves of the displayed demonstrator lamp with the added spectroscope.Explain the differences between the two spectra. You may walk up to the lamp and you may keep the spectroscope as a souvenir.Solution of problem 1:a)The total resistance of the apparatus is 228.5V Z 380.80.6A==Ω ,the ohmic resistance of the tube isR 84V R 1400.6A ==Ω.Hence the total ohmic resistance isR 14026.3166.3=Ω+Ω=Ω.Therefore the inductance of the series reactor is:L 342.6ω⋅==Ω.This yields 1342.6L 1.09H 100s−Ω==π.b)The impedance angle is obtained fromL 342.6tan 2.06R 166.3ω⋅Ωϕ===Ω.Thus o 64.1ϕ=.c)The active power can be calculated in different ways:1)o w P U I cos 228.5V 0.6A cos 64.159.88W=⋅⋅ϕ=⋅⋅=2)22w P R I 166.3(0.6A)59.87W=⋅=Ω⋅=d)By opening the contact in the starter a high induction voltage is produced across the series reactor (provided the contact does not open exactly the same moment, when the current goes through zero). This voltage is sufficient to ignite the lamp. The main voltage itself, however, is smaller than the ignition voltage of the fluorescent tube.e)f) The recombination time of the ions and electrons in the gaseous discharge is sufficiently large.g) The capacitive resistance of a capacitor of 4.7 µF is611(100 4.710)677.3C −−=⋅π⋅⋅Ω=Ωω⋅.The two reactances subtract and there remains a reactance of 334.7 Ω acting as a capacitor.The total resistance of the arrangement is nowZ'373.7=Ω=Ω,which is very close to the total resistance without capacitor, if you assume the capacitor to be loss-free (cf. a) ). Thus the lamp has the same operating qualities, ignites the same way, and a difference is found only in the impedance angle ϕ’, which is opposite to the angle ϕ calculated in b):()1L C 334.7tan ' 2.01R 166.3−ω⋅−ω⋅ϕ==−=−o '63.6ϕ=−.Such additional capacitors are used for compensation of reactive currents in buildings with a high number of fluorescent lamps, frequently they are prescribed by theelectricity supply companies. That is, a high portion of reactive current is unwelcome,because the power generators have to be layed out much bigger than would be really necessary and transport losses also have to be added which are not payed for by the customer, if pure active current meters are used.h) The uncoated part of the demonstrator lamp reveals the line spectrum of mercury, the coated part shows the same line spectrum over a continuous background. Thecontinuous spectrum results from the ultraviolet part of the mercury light, which is absorbed by the fluorescence and re-emitted with smaller frequency (energy loss of the photons) or larger wavelength respectively.Problem 2: Oscillating coat hangerA (suitably made) wire coat hanger can perform small amplitude oscillations in the plane of the figure around the equilibrium positions shown. In positions a) and b) the long side ishorizontal. The other two sides have equal length. The period of oscillation is the same in all cases.What is the location of the center of mass, and how long is the period?a) b)c)The figure does not contain any information beyond the dimensions given. Nothing is known, e.g., concerning the detailed distribution of mass.Solution of problem 2First method:The motions of a rigid body in a plane correspond to the motion of two equal point masses connected by a rigid massless rod. The moment of inertia then determines their distance. Because of the equilibrium position a) the center of massis on the perpendicular bipartition of the long side of thecoat hanger. If one imagines the equivalent masses andthe supporting point P being arranged in a straight line ineach case, only two positions of P yield the same periodof oscillation (see sketch). One can understand this byconsidering the limiting cases: 1. both supporting pointsin the upper mass and 2. one point in the center of mass and the other infinitely high above. Between these extremes the period of oscillation grows continuously. The supporting point placed in the corner of the long side c) has the largest distance from the center of mass, and therefore this point lies outside the two point masses. The two other supporting points a), b) then have to be placed symmetrically to the center of mass between the two point masses, i.e., the center of mass bisects the perpendicular bipartition. One knows of the reversible pendulum that for every supporting point of the physical pendulum it generally has a second supporting point of the pendulum rotated by 180o, with the same period of oscillation but at a different distance from the center of mass. Thesection between the two supporting points equals the length of the corresponding mathematical pendulum. Therefore the period of oscillation is obtained through thecorresponding length of the pendulum s b + s c , where s b = 5 cm and c s =be T = 1.03 s .Second method:Let s denote the distance between the supporting point and the center of mass, m the mass itself and θ the moment of inertia referring to the supporting point. Then we have the period of oscillation T :T 2= ,(1)where g is the acceleration of gravity, g = 9.81 m/s 2. Here θ can be obtained from the moment of inertia θo related to the center of mass:20m s θ=θ+⋅(2)Because of the symmetrical position incase a) the center of mass is to be found on the perpendicular bisection above thelong side. Now (1) and (2) yield220T m s m g s for 2⎛⎞θ+⋅=⋅⋅⋅⎜⎟⋅π⎝⎠s = s a , s b and s c . (3)because all periods of oscillation are the same. This quadratic equation has only twodifferent solutions at most. Therefore at least two of the three distances are equal. Because of s c > 21 cm > s a + s b , only s a and s b can equal each other. Thus we haves a = 5 cm(4)The moment of inertia θ0 is eliminated through (3):()222c a c a T m (s s )m g s s 2⎛⎞⋅−=⋅⋅⋅−⎜⎟⋅π⎝⎠and we have T 2=⋅(5)with the numerical value T 1.03s =,which has been rounded off after two decimals because of the accuracy of g.Third method:This solution is identical to the previous one up to equation (2).From (1) and (2) we generally have for equal periods of oscillation T 1 = T 2:2011m s m g s θ+⋅⋅⋅ = 222m s m g s θ+⋅⋅⋅and therefore ()()22201102s m s s m s ⋅θ+⋅=⋅θ+⋅or ()()21012s s m s s 0−⋅θ−⋅=(6)The solution of (6) includes two possibilities:1212s s or s s m θ=⋅=Let 2⋅a be the length of the long side and b the height of the coat hanger. Because of T b = T c we then have 0b c b c either s s or s s mθ=⋅=, where c s =which excludes the first possibility. Thus 0b c s s m θ⋅=.(7)For T a = T b the case s a ⋅ s b =0m θ is excluded because of eq. (7), for we have a b s s ⋅<0c b s s m θ⋅=.Hence a b s s ==and T 22=⋅=The numerical calculation yields the valueT 1.03s =.Problem 3: Hot-air-balloonConsider a hot-air balloon with fixed volume V B = 1.1 m 3. The mass of the balloon-envelope, whose volume is to be neglected in comparison to V B , is m H = 0.187 kg.The balloon shall be started, where the external air temperature is 1ϑ = 20 o C and the normal external air pressure is p o = 1.013 ⋅ 105 Pa. Under these conditions the density of air is ρ1 = 1.2 kg/m 3.a)What temperature 2ϑ must the warmed air inside the balloon have to make the balloon just float?b)First the balloon is held fast to the ground and the internal air is heated to a steady-state temperature of 3ϑ = 110 o C. The balloon is fastened with a rope.Calculate the force on the rope.c)Consider the balloon being tied up at the bottom (the density of the internal air stays constant). With a steady-state temperature 3ϑ = 110 o C of the internal air the balloon rises in an isothermal atmosphere of 20 o C and a ground pressure ofp 0 = 1.013 ⋅ 105 Pa. Which height h can be gained by the balloon under theseconditions?d)At the height h the balloon (question c)) is pulled out of its equilibrium position by 10 m and then is released again.Find out by qualitative reasoning what kind of motion it is going to perform!Solution of problem 3:a) Floating condition:The total mass of the balloon, consisting of the mass of the envelope m H and the mass of the air quantity of temperature 2ϑ must equal the mass of the displaced air quantity with temperature 1ϑ = 20 o C.V B ⋅ ρ2 + m H = V B ⋅ ρ1H 21Bm V ρ=ρ− (1)Then the temperature may by obtained from1221T T ρ=ρ,1212T T ρ=⋅ρ= 341.53 K = 68.38 °C (2)b) The force F B acting on the rope is the difference between the buoyant force F A and the weight force F G :F B = V B ⋅ ρ1 ⋅ g - (V B ⋅ ρ3 + m H ) ⋅ g(3)It follows with ρ3 ⋅ T 3 = ρ1 ⋅ T 1F B = V B ⋅ ρ1 ⋅ g ⋅ 13T 1T ⎛⎞−⎜⎟⎝⎠ - m H ⋅ g = 1,21 N (4)c) The balloon rises to the height h , where the density of the external air ρh has the same value as the effective density ρeff , which is evaluated from the mass of the air of temperature ϑ3 = 110 o C (inside the balloon) and the mass of the envelope m H :10g h 3B H 2eff h 1B BV m m e V V ρ⋅⋅−ρρ⋅+ρ===ρ=ρ⋅ (5)Resolving eq. (5) for h gives:o 11eff p h 1n 843m g ρ=⋅=ρ⋅ρ(6).d) For small height differences (10 m in comparison to 843 m) the exponential pressure drop (or density drop respectively) with height can be approximated by a linearfunction of height. Therefore the driving force is proportional to the elongation out of the equilibrium position.This is the condition in which harmonic oscillations result, which of course are damped by the air resistance.Experimental Problems Problem 4: Lens experimentThe apparatus consists of a symmetric biconvex lens, a plane mirror, water, a meter stick,an optical object (pencil), a supporting base and a right angle clamp. Only these parts may be used in the experiment.a)Determine the focal length of the lens with a maximum error of ± 1 %.b)Determine the index of refraction of the glass from which the lens is made.The index of refraction of water is n w = 1.33. The focal length of a thin lens is given by()12111n 1f r r ⎛⎞=−⋅−⎜⎟⎝⎠,where n is the index of refraction of the lens material and r 1 and r 2 are the curvature radii of the refracting surfaces. For a symmetric biconvex lens we have r 1 = - r 2 = r, for a symmetric biconcave lens r 1 = - r 2 = - r .Solution of problem 4:a) For the determination of f L , place the lens on the mirror and with the clamp fix the pencil to the supporting base.Lens and mirror are then moved around until the vertically downward looking eye sees the pencil and its image side by side.In order to have object and image in focus at the same time, they must be placed at an equal distance to the eye.In this case object distance and image distance are the same and the magnification factor is 1 .It may be proved quite accurately, whether magnification 1 has in fact been obtained, if one concentrates on parallatical shifts between object and image when moving the eye:only when the distances are equal do the pencil-tips point at each other all the time.The light rays pass the lens twice because they are reflected by the mirror. Therefore the optical mapping under consideration corresponds to a mapping with two lenses placed directly one after another:L L111111,where g b f f f f +==+i.e. the effective focal length f is just half the focal length of the lens. Thus we find for magnification 1:L L22g b andi.e.f g.g f ===A different derivation of f L = g = b: For a mapping of magnification 1 the light rays emerging from a point on the optical axis are reflected into themselves. Therefore these rays have to hit the mirror at right angle and so the object distance g equals the focal length f L of the lens in this case.The distance between pencil point and mirror has to be determined with an accuracy,which enables one to state f L with a maximum error of ± 1 % . This is accomplished either by averaging several measurements or by stating an uncertainty interval, which is found through the appearance of parallaxe.Half the thickness of the lens has to be subtracted from the distance between pencil-point and mirror.'L L 1f f d,d 3.00.5mm 2=−=±The nominal value of the focal length of the lens is L f = 30 cm. However, the actual focal length of the single lenses spread considerably. Each lens was measured separately, so the individual result of the student can be compared with the exact value.b)The refractive index n of the lens material can be evaluated from the equation()L 12n 1f r=−⋅if the focal length f L and the curvature radius r of the symmetric biconvex lens are known. f Lwas determined in part a) of this problem.The still unknown curvature radius r of the symmetric biconvex lens is found in the following way: If one pours some water onto the mirror and places the lens in the water,one gets a plane-concave water lens, whichhas one curvature radius equalling the glass lens’ radius and the other radius is ∞ .Because the refractive index of water is known in this case, one can evaluate the curvature radius through the formula above, where r 1 = -r and r 2 = ∞ :()w w 11n 1.f r−=−⋅Only the focal length f ' of the combination of lenses is directly measured, for which we haveL w111f 'f f =+.This focal length is to be determined by a mapping of magnification 1 as above.Then the focal length of the water lens is'w L111f f f =−and one has the curvature radius ()w w r n 1f =−−⋅.Now the refractive index of the lens is determined by Lrn 12f =+⋅with the known values of f L and r, or, if one wants to express n explicitly through the measured quantities:()()w L f 'n 1n 2f 'f ⋅−=⋅−+ 1.The nominal values are: f ' = 43.9 cm, f w = -94.5 cm, r = 31.2 cm, n = 1.52.Problem 5: Motion of a rolling cylinderThe rolling motion of a cylinder may be decomposed into rotation about its axis and horizontal translation of the center of gravity. In the present experiment only thetranslatory acceleration and the forces causing it are determined directly.Given a cylinder of mass Μ, radius R, which is placed on a horizontal plane board. At a distance r i (i = 1 … 6) from the cylinder axis a force acts on it (see sketch). After letting the cylinder go, it rolls with constant acceleration.a)Determine the linear accelerations a i (i = 1 … 6) of the cylinder axis experimentallyfor several distances r i (i = 1 … 6).b)From the accelerations a i and given quantities, compute the forces F i which act inhorizontal direction between cylinder and plane board.c)Plot the experimental values F i as functions of r i. Discuss the results.Before starting the measurements, adjust the plane board horizontally. For present purposes it suffices to realize the horizontal position with an uncertainty of ± 1 mm of height difference on 1 m of length; this corresponds to the distance between adjacent markings on the level. What would be the result of a not horizontal position of the plane board?Describe the determination of auxiliary quantities and possible further adjustments; indicate the extent to which misadjustments would influence the results.The following quantities are given:R= 5 cm r1=0.75cmM= 3.275 kg r2= 1.50cmm= 2 x 50 g r3= 2.25cmD= 1.50 cm r4= 3.00cmd=0.1 mm r5= 3.75cmr6= 4.50cmMass and friction of the pulleys c may be neglected in the evaluation of the data.By means of knots, the strings are put into slots at the cylinder. They should be inserted as deeply as possible. You may use the attached paper clip to help in this job.The stop watch should be connected, as shown in the sketch, with electrical contacts at A and B via an electronic circuit box. The stop watch starts running as soon as the contact at A is opened, and it stops when the contact at B is closed.The purpose of the transistor circuit is to keep the relay position after closing of the contact at B, even if this contact is opened afterwards for a few milliseconds by a jump or chatter of the cylinder.Solution of problem 5:Theoretical considerations:a)The acceleration of the center of mass of the cylinder is22s a t ⋅=(1)b) Let a m be the acceleration of the masses m and T the sum of the tensions in the two strings, thenT = m ⋅ g - m ⋅ a m(2)The acceleration a of the center of mass of the cylinder is determined by the resultant force of the string-tension T and the force of interaction F between cylinder and the horizontal plane.M ⋅a = T - F(3)If the cylinder rotates through an angle θ the mass m moves a distance x m .It holdsx m = (R + r) ⋅ θ()m a a R r R=+⋅(4)From (2), (3) and (4) follows r F mg M m 1a R ⎡⎤⎛⎞=−+⋅+⋅⎜⎟⎢⎥⎝⎠⎣⎦.(5)c) From the experimental data we see that for small r i the forces M ⋅ a and T are in opposite direction and that they are in the same direction for large r i .For small values of r the torque produced by the string-tensions is not large enough to provide the angular acceleration required to prevent slipping. The interaction force between cylinder and plane acts into the direction opposite to the motion of the center of mass and thereby delivers an additional torque.For large values of r the torque produced by string-tension is too large and the interaction force has such a direction that an opposed torque is produced.From the rotary-impulse theorem we finda T r F R R⋅+⋅=Ι⋅θ=Ι⋅ ,where Ι is the moment of inertia of the cylinder.With (3) and (5) you may eliminate T and a from this equation. If the moment of inertia of the cylinder is taken as 21R 2Ι=⋅Μ⋅(neglecting the step-up cones) we find after some arithmetical transformations2r 12RF mg m r 321M R −⋅=⋅⎛⎞+⋅⋅+⎜⎟⎝⎠.For r = 0m gF m 32M ⋅→=+⋅> 0.For r = R m gF m 38M−⋅⇒=+⋅< 0.Becausem 1Mit is approximately12r F m g 33R=⋅−⋅.That means: the dependence of F from r is approximately linear. F will be zero if r m g R 2⋅=.Experimental results:s = L − (2 ⋅ R ⋅D + D 2)12− (2 ⋅ R ⋅d − d 2)12s = L − 4.5 cm = 39.2 cm − 4.5 cm = 34.7 cmGrading schemesTheoretical problemsProblem 1: Fluorescent lamp pts. Part a2 Part b1 Part c1 Part d1 Part e1 Part f1 Part g2 Part h110 Problem 2: Oscillating coat hanger pts. equation (1)1,5 equation (2)1,5 equation (4)3 equation (5)2 numerical value for T110 Problem 3: Hot-air-balloon pts. Part a3 Part b2 Part c3 Part d210Experimental problemsProblem 4: Lens experiment pts. correct description of experimental prodedure1 selection of magnification one0.5 parallaxe for verifying his magnification1 f L = g = b with derivation1 several measurements with suitable averaging or otherdetermination of error interval1 taking into account the lens thickness and computing f L, including the error0.5 idea of water lens0.5 theory of lens combination1 measurements of f ′0.5 calculation of n and correct result18Problem 5: Motion of a rolling cylinderpts.Adjustment mentioned of strings a) horizontally and b) in direction of motion0.5Indication that angle offset of strings enters the formula for the acting force only quadratically, i.e. by its cosine 0.5Explanation that with non-horizontal position, the force m ⋅g is to be replaced by m ⋅g ± M ⋅g ⋅ sin α1.0Determination of the running length according for formula ()()1/21/222s L 2R D D 2R d d =−⋅⋅+−⋅⋅+including correct numerical result 1.0Reliable data for rolling time1.0accompanied by reasonable error estimate 0.5Numerical evaluation of the F i 0.5Correct plot of F i (v i )0.5Qualitative interpretation of the result by intuitive consideration of the limiting cases r = 0 and r = R 1.0Indication of a quantitative, theoretical interpretation using the concept of moment of inertia1.0Knowledge and application of the formula a = 2 s / t 20.5Force equation for small mass and tension of the string m ⋅(g - a m ) = T1.0Connection of tension, acceleration of cylinder and reaction force T – F = M ⋅a1.0Connection between rotary and translatory motion()m x R r =+⋅θ0.5()m a 1r /R a=+⋅0.5Final formula for the reaction force ()()F m g M m 1r /R a=⋅−+⋅+⋅ 1.0If final formulae are given correctly, the knowledge for preceding equations must be assumed and is graded accordingly.12。
国际物理奥赛试题及答案
国际物理奥赛试题及答案一、选择题（每题3分，共30分）1. 光在真空中的速度是多少？A. 299,792,458 m/sB. 299,792,458 km/sC. 299,792,458 cm/sD. 299,792,458 mm/s答案：A2. 根据量子力学，一个粒子的波函数代表什么？A. 粒子的位置B. 粒子的速度C. 粒子的概率密度D. 粒子的能量答案：C3. 以下哪项不是电磁波的特性？A. 波长B. 频率C. 质量D. 能量答案：C4. 根据牛顿第三定律，作用力和反作用力的关系是什么？A. 相等且相反B. 相等且相同C. 不相等且相反D. 不相等且相同答案：A5. 热力学第一定律表述了什么？A. 能量守恒定律B. 熵增定律C. 热力学第二定律D. 热力学第三定律答案：A6. 一个物体在自由落体运动中，其加速度是多少？A. 9.8 m/s²B. 10 m/s²C. 11 m/s²D. 12 m/s²答案：A7. 以下哪种力不属于基本力？A. 重力B. 电磁力C. 强核力D. 弱核力答案：A8. 一个完全弹性碰撞中，动量守恒和能量守恒是否都成立？A. 是B. 否答案：A9. 光的双缝干涉实验中，干涉条纹的间距与什么有关？A. 光的波长B. 双缝间的距离C. 观察屏的距离D. 所有以上因素答案：D10. 根据相对论，时间膨胀和长度收缩的效应是否只发生在高速运动的物体上？A. 是B. 否答案：A二、填空题（每题2分，共20分）1. 光年是______单位，表示光在一年内通过的距离。
答案：长度2. 绝对零度是温度的下限，其数值为______开尔文。
答案：03. 一个物体的动能与其速度的平方成正比，其比例系数是______。
答案：1/24. 根据库仑定律，两个点电荷之间的力与它们电荷量的乘积成正比，与它们距离的平方成______。
答案：反比5. 根据麦克斯韦方程组，变化的磁场可以产生______。
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WATER-POWERED RICE-POUNDING MORTARA. IntroductionRice is the main staple food of most people in Vietnam. To make white rice from paddy rice, one needs separate of the husk (a process called "hulling") and separate the bran layer ("milling"). The hilly parts of northern Vietnam are abundant with water streams, and people living there use water-powered rice-pounding mortar for bran layer separation. Figure 1 shows one of such mortars., Figure 2 shows how it works.B. Design and operation1. Design .The rice-pounding mortar shown in Figure 1 has the following parts: The mortar , basically a wooden container for rice.The lever , which is a tree trunk with one larger end and one smaller end. It can rotate around a horizontal axis. A pestle is attached perpendicularly to the lever at the smaller end. The length of the pestle is such that it touches the rice in the mortar when the lever lies horizontally. The larger end of the lever is carved hollow to form a bucket. The shape of the bucket is crucial for the mortar's operation.2. Modes of operation The mortar has two modes.Working mode . In this mode, the mortar goes through an operation cycle illustrated in Figure 2.The rice-pounding function comes from the work that is transferred from the pestle to the rice during stage f) of Figure 2. If, for some reason, the pestle never touches the rice, we say that the mortar is not working.Rest mode with the lever lifted up . During stage c) of the operation cycle (Figure 2), as the tilt angle α increases, the amount of water in the bucket decreases. At one particular moment in time, the amount of water is just enough to counterbalance the weight of the lever. Denote the tilting angle at this instant by β. If the lever is kept at angle β and the initial angular velocity is zero, then the lever will remain at this position forever. This is the rest mode with the lever lifted up. The stability of this position depends on the flow rate of water into the bucket, Φ. If exceeds some value Φ2Φ, then this rest mode is stable, and the mortar cannot be in the working mode. In other words, is the minimal flow rate for the mortar not to work. 2ΦFigure 1A water-powered rice-pounding mortarOPERATION CYCLE OF A WATER-POWERED RICE-POUNDING MORTARFigure 2a)b)a) At the beginning there is no water in the bucket, the pestle rests on the mortar. Water flows into the bucket with a small rate, but for some time the lever remains in the horizontal position.b) At some moment the amount of water is enough to lift the lever up. Due to thetilt, water rushes to the farther side of the bucket, tilting the lever more quickly.Water starts to flow out at 1αα=.c) As the angle α increases, water starts to flow out. At some particular tilt angle, αβ=, the total torque is zero.d) α continues increasing, watercontinues to flow out until no water remains in the bucket.e) α keeps increasing because of inertia. Due to the shape of the bucket, water falls into the bucket but immediately flows out. The inertial motion of the lever continues until α reaches the maximal value 0α.f) With no water in the bucket, the weight of the lever pulls it back to the initial horizontal position. The pestle gives the mortar (with rice inside) a pound and a new cycle begins.C. The problemConsider a water-powered rice-pounding mortar with the following parameters (Figure 3)The mass of the lever (including the pestle but without water) is M =30 kg,The center of mass of the lever is G .. The lever rotates around the axis T (projected onto the point T on the figure).The moment of inertia of the lever around T is I =12 kg ⋅m 2.When there is water in the bucket, the mass of water is denoted as , the center of mass of the water body is denoted as N.m The tilt angle of the lever with respect to the horizontal axis is α.The main length measurements of the mortar and the bucket are as in Figure 3. Neglect friction at the rotation axis and the force due to water falling onto the bucket. In this problem, we make an approximation that the water surface is always horizontal.Figure 3 Design and dimensions of the rice-pounding mortar1. The structure of the mortarAt the beginning, the bucket is empty, and the lever lies horizontally. Then water flows into the bucket until the lever starts rotating. The amount of water in the bucket at this moment is 1.0 kg.m =1.1. Determine the distance from the center of mass G of the lever to the rotation axis T. It is known that GT is horizontal when the bucket is empty.1.2. Water starts flowing out of the bucket when the angle between the lever and the horizontal axis reaches 1α. The bucket is completely empty when this angle is 2α. Determine 1αand 2α.1.3. Let ()μα be the total torque (relative to the axis T) which comes from theweight of the lever and the water in the bucket. ()μα is zero when αβ=. Determineβ and the mass of water in the bucket at this instant.1m2. Parameters of the working modeLet water flow into the bucket with a flow rate Φ which is constant and small. The amount of water flowing into the bucket when the lever is in motion is negligible. In this part, neglect the change of the moment of inertia during the working cycle.2.1. Sketch a graph of the torque μ as a function of the angle α, ()μα, duringone operation cycle. Write down explicitly the values of ()μα at angle α1, α2, andα = 0.2.2. From the graph found in section 2.1., discuss and give the geometric interpretation of the value of the total energy produced by total W ()μαand the workthat is transferred from the pestle to the rice.pounding W2.3. From the graph representing μ versus α, estimate 0α and (assumethe kinetic energy of water flowing into the bucket and out of the bucket is negligible.) You may replace curve lines by zigzag lines, if it simplifies the calculation. pounding W3. The rest modeLet water flow into the bucket with a constant rate Φ, but one cannot neglect the amount of water flowing into the bucket during the motion of the lever. 3.1. Assuming the bucket is always overflown with water,3.1.1. Sketch a graph of the torque μ as a function of the angle α in the vicinity of αβ=. To which kind of equilibrium does the position αβ= of the lever belong?3.1.2. Find the analytic form of the torque ()μα as a function of αΔ whenαβ=+Δα, and αΔ is small.3.1.3. Write down the equation of motion of the lever, which moves with zero initial velocity from the position αβα=+Δ (αΔ is small). Show that the motion is, with good accuracy, harmonic oscillation. Compute the period τ.3.2. At a given , the bucket is overflown with water at all times only if the lever moves sufficiently slowly. There is an upper limit on the amplitude of harmonic oscillation, which depends on . Determine the minimal value ΦΦ1Φof (in kg/s) so that the lever can make a harmonic oscillator motion with amplitude 1Φo .3.3. Assume that is sufficiently large so that during the free motion of the lever when the tilting angle decreases from Φ2α to 1α the bucket is always overflown with water. However, if is too large the mortar cannot operate. Assuming that the motion of the lever is that of a harmonic oscillator, estimate the minimal flow rate for the rice-pounding mortar to not work.Φ2Φ。