英国2011年物理竞赛半决赛试题
2011高中应用物理竞赛试题答案
2011年全國高中應用物理知識競賽試題一、本題包括11小題，考生只需做10小題，每小題5分。
共50分．在每小題給出的四個選項中。
有的小題只有一個選項正確，有的小題有多個選項正確．請把符合題目要求的選項的序號填入題後的( )內．全選對的得5分，選不全的得3分，有選錯或不選的得O分．1.圖1為汽車發動機的冷卻風扇設計的一個自動控制電路. 要求發動機的點火開關閉合且溫度過高時(熱敏電阻R阻值變小), 繼電器J有電流通過, 就會使風扇自動開啟. 關於這個自動控制電路, 以下判斷正確的是( )A:虛線框內的門電路是“或”門B: 虛線框內的門電路是“與”門C:虛線框內的門電路是“非”門D:虛線框內的門電路是“與非”門2.我國的國土範圍東西方向大致分佈在東經70˚到東經137˚, 所以我國發射的通信衛星一般定點在赤道上空3.6×107m、東經100˚附近. 若某通信衛星計劃定點在赤道上空東經104˚的位置, 但測得該衛星剛入軌時位於赤道上空的3.6×107m,東經101˚處. 為了把它調整到計劃定點位置, 可以先分兩次短時間啟動衛星上的小型噴氣發動機調整衛星圓軌道的高度(稱為漂移軌道), 便其“漂移”到預定經度后再分兩次短時間啟動衛星上的小型噴氣發動機使它回到原來的圓軌道(稱為靜止軌道). 則衛星在漂移軌道中運動與在靜止軌道上運動相比較( ) A:高度較低, 速度較大B: 高度較高, 速度較大C:高度較低, 速度較小D: 高度較高, 速度較小3.鐵路上使用一種電磁裝置向控制中心傳輸信號以確定火車的位置和運動情況. 這種裝置的核心結構如圖2(甲)所示(俯視圖), 在列車首節車廂下面安有磁鐵, 它下方的磁場可看作勻強磁場, 當它經過安放在兩鐵軌間的線圈時, 便會產生一電信號並輸送到控制中心. 若某列火車通過線圈時, 控制中心接收到線圈兩端的電壓信號為圖2(乙)所示, 則說明火車在這段時間內的運動情況是( )A: 勻速直線運動B: 勻加速直線運動C: 勻減速直線運動D: 加速度逐漸加大4.燃燒汽油的汽車, 其發動機的汽缸中的汽油---空氣混合氣是靠火花塞產生電火花點燃的.但是汽車蓄電池兩端的電壓只有12V, 不能使火花塞中產生電火花, 因此要使用如圖3所示的點火裝置. 此裝置的核心是一個變壓器, 該變壓器的初級線圈通過點火開關連在蓄電池上, 次級線圈接到火花塞的兩端, 點火開關由機械自動控制其通斷狀態, 以保證在做功衝程開始時次級線圈中產生上萬伏的電壓, 使火花塞產生電火花. 對於這個點火裝置, 下列說法中正確的是( ) A:點火開關由斷開變為閉合和由閉合變為斷開, 一定能使火花塞產生相同的火花B: 點火開關若始終閉合, 次級線圈的兩端也會有高電壓C: 火花塞產生的火花是混合氣被電離時產生的弧光放電D: 點火裝置中變壓器的次級線圈匝數必須遠大於初級線圈的匝數5.一名中學生在進行籃球比賽時, 獲得到一次站在罰球線罰球的機會, 他要將籃球罰入籃框, 需要對籃球所做的功約為( )A.2JB.20JC.100JD. 200J6.一輛客車在平直的公路上行駛時, 固定於路邊的照相機在2S的時間內連續對車拍照, 得到多張清晰照片,其中第1和最后一張照片如圖4所示, 所有各張照片中懸挂在客車頂部的拉手均向后傾斜, 拉手懸繩與豎直方向夾角約為10˚且保持不變. 已知客車中間的三個大窗戶的寬度約為1.5m. 小剛根據照片的情景及上述已知條件作出如下的判斷, 其中正確的是( )A. 客車在做勻加速運動B.可估算出車在t=1s時的運動速度C.可估算出車在t=2s時的運動速度D.客車可能正在減速后退7.在研究由核子誘發的非彈性核反應過程中, 需要對“彈核”進行加速, 使其獲得足夠的動能. 現將質子, 氘核和氚核分別置於回旋加速器的中心點A并開始加速, 如圖5, 若它們的初速度大小可忽略不計, 則當它們由同一出口離開D形盒時( )A.質子的動能較大B.氚核的動能較大C.氘核的動能較大D.三种粒子的動能大小相等8.在一種速降娛樂項目中, 人乘坐在吊籃中, 吊籃通過滑輪沿一條傾斜的鋼索向下滑行.現有兩條彼此平行的鋼索, 它們的起、終點分別位於同一高度. 小紅和小明分別乘吊籃從速降的起點由靜止開始下滑, 在他們下滑過程中, 當吊籃與滑輪達到相對靜止狀態時, 分別拍下一張照片, 如圖6所示, 已知二人運動過程中, 空氣阻力的影響可以忽略, 則( )A. 小明到達終點用時較短B. 小紅到達終點用時較短C. 小明到達終點時的速度較大D.兩人的運動都一定是勻速運動9.一位同學為了表演“輕功”,他用打氣筒給4只相同的氣球充以相等質量的空氣, 然後將這4只氣球以相同的方式放在水平塑料板上, 在氣球的上方放置一輕質塑料板, 如圖7所示, 然後這位同學站到輕質塑料板上正中間位置, 對於他站上的過程, 下列說法正確的是( )A. 球內氣體體積變大Ｂ.球內氣體體積變小C. 球內氣體內能變大D.球內氣體內能不變10.十字路口的交通信號燈, 用紅燈指示停車等候, 綠燈指示可以通行.在一些道路施工的工地上, 為了提示人們注意路面的情況, 常在工地周圍挂起紅色的電燈, 這樣做除了紅色光容易引起人的視覺注意以外, 還有一個重要的原因, 是紅色光( )A.比其他可見光更容易發生衍射B.比其他可見光的光子能量大C. 比其他可見光更容易發生干涉D.比其他可見光更容易發生全反射11.右表給出了四种放射性同位素的輻射線的半衰期. 在醫療技術中有時需要利用放射線治療腫瘤. 所用的放射源必須具有以下兩個條件:(1)放出的放射線有較強的穿透能力,以便輻射到體內腫瘤所在處; (2)能在較長時間內提供比較穩定的輻射強度.為此, 在表中所列的四種同位素中, 應選用的是( )A.釙210B.鍶90C. 99D.鈷60二本題包括5小題。
2011年第28届物理竞赛决赛试卷及答案
第28届全国中学生物理竞赛决赛试卷一、（15分）在竖直平面内将一半圆形光滑导轨固定在A 、B 两点，导轨直径2A R =，AB 与竖直方向间的夹角为o 60，在导轨上套一质量为m 的光滑小圆环，一劲度系数为k 的轻而细的光滑弹性绳穿过圆环，其两端系于A 、B 两点，如图所示，当圆环位于A 点正下方C 点时，弹性绳刚好为原长。
现将圆环从C 点无初速度释放，圆环在时刻t 运动到'C 点，'C O 与半径OB 的夹角为θ。
重力加速度为g 。
试分别对下述良好总情形，求导轨对圆环的作用力大小； 1. o 90θ=； 2. o 30θ=。
二、（15分）如图，在水平地面上有一质量为M 、长度为L 的小车。
车内两端靠近底部处分别固定两个轻弹簧，两弹簧位于同一直线上，其原长分别为1l 和2l ，劲度系数分别为1k 和2k ；两弹簧的另一端前分别放着一个质量为1m 、2m 的小球，弹簧与小球都不相连。
开始时，小球1压缩弹簧1并保持整个系统处于静止状态，小球2被锁定在车底板上，小球2与小车右端的距离等于弹簧2的原长。
现无初速释放小球1，当弹簧1的长度等于其原长时，立即解除对小球2的锁定；小球1与小球2碰撞后合为一体，碰撞时间极短。
已知所有解除都是光滑的；从释放小球1至弹簧2达到最大压缩量时，小车移动了距离3l 。
试求开始时弹簧1的长度l 和后来弹簧2所达到最大压缩量2l ∆。
三、（20分）某空间A 绕地球作圆周运动，轨道半径为66.7310A r m =⨯。
一人造地球卫星B在同一轨道平面内作圆周运动，轨道半径为32B A r r =，A 和B 均沿逆时针方向运行，现从空间站上发射一飞船（对空间站无反冲）前去回收卫星。
为零节省燃料，除了短暂的加速或减速变轨过程外，飞船在往返过程中均采用同样形式的逆时针椭圆转移轨道，作无动力飞行，往返两过程的椭圆轨道均位于空间站和卫星的圆轨道平面内，且其近地点和远地点都分别位于空间站和卫星的圆轨道上，如图所示。
第二十一届（2011年）全国初中物理竞赛试题-完整版
2011年第二十—届全国初中应用物理竞赛试题一、本题共10小题，每小题2分，共20分。
以下各小题给出的四个选项中只有一个是正确的, 把正确选项前面的字母填在题后的括号内。
1.目前，很多汽车的驾驶室里都有一个叫做GPS （全球卫星定位系统）接收器的装置。
GPS 接收器 通过接收卫星发射的导航信号，实现对车辆的精确定位并导航。
卫星向GPS 接收器传送信息依靠 的是 （）A.红外线B.紫外线C.电磁波D.激光于这两种方法比较省力勺是（）A.图甲朗方法比较省力。
法比较省力。
列说法中2.图1为工人师傅用于推车运送纯净水的 图片，对法中哪种方的判断，下B.图乙的方C.图甲和图乙两种方法所用力的大小相等。
D.无法判断。
图33. 如图2所示，教室里的天花板下面装有多挡位吊扇， 当吊扇正常工作时，对于吊扇对天花板的拉力大小与其重力大小的判断，下列说法中正确的是（）A. 吊扇在高挡位工作时，拉力等于重力。
B. 吊扇在低挡位工作时，拉力大于重力。
C. 电扇在任意挡位工作时，拉力等于重力。
D. 电扇在任意挡位工作时，拉力小于重力。
4. 在学校的田径运动会上，小亮注意到发令台上有一块“板子”，如图3所示：关于这块板子的作用，他做了如下猜想，其中正确的是（）A.主要用来反射发令枪的枪声，所以必须正对着终点计时裁判的方向。
B.主要用来衬托发令枪冒出的白烟，所以应该是深色或黑色的。
C.主要用来减弱噪声，所以要把它放在看台和发令枪之间OD.主要用来遮阳，所以要把它放在裁判员的头的上方。
5.随着我国航天员在轨道舱内停留时间的增加，在轨道舱内进行体育锻炼必将成为航天员需要完成的一项重要工作；下列适合航天员在轨道舱中进行体育锻炼的运动项目是（）A.举哑铃B.跳绳C.踢毯子D.拉弹簧拉力器6.动车组列车进站前的减速过程分为两个阶段进行:第一阶段采用“再生刹车”技术，速度从250km/h减至90km/h,这期间停止动力供给，列车依靠惯性继续前行，并带动发电机发电：第二阶段采用机械刹车，速度从90km/h开始不断减小直至停止：关于列车进站过程中的能量转化,下列说法中正确的是（）A.第一阶段减小的动能全部转化为电能，即电能等于减小的动能。
英国2011年中学物理竞赛答案
2011 GCSE PHYSICS CHALLENGE PAPERONE HOUR PHYSICS COMPETITION PAPERFriday 11th MarchWe hope teachers will set and mark the enclosed paper for their GCSE Physics students, or equivalent students in Scotland.The solutions and marking scheme are contained herein.It is intended that the paper should be taken on Friday 11th March. However, if this is not possible, any date during the period 9th to 15th March will be acceptable. Scripts of the Gold Medallists, the entry form and the requests for certificates must be posted in sufficient time to arrive by first class post on Wednesday16th March 2011 at the Olympiad Office at the University of Oxford. Any scripts arriving after this date cannot be considered for an award.After the scripts have been marked please send to the Oxford office: ∙those scripts with marks of 38 and above (the scripts of the Gold Medal∙Certificate students) in order to be considered for the award of a book∙Prize (it is recommended that you keep a photocopy of the scripts)∙the entry form, which is on the following page∙the request form for certificates∙the completed teacher questionnaireWe will invite the five outstanding Gold Medallists, together with their teachers, to the Physics Challenge Presentation Ceremony at The Royal Society in London on Thursday28th April 2011. Prizes and certificates will be despatched to all remaining medallists, who are not amongst those invited to the Presentation in May. Teachers are requested to complete the certificates according to the medal scheme specified on the last page, and present them to their students.Oxford Office: Dr S. OwenBPhO OfficePhysics Challenge CompetitionDepartment of PhysicsClarendon LaboratoryParks Road,University of OxfordOxford, OX1 3PUPhysics Challenge 2011 - Mark schemePlease award marks as indicated below.Equivalent valid reasoning should gain equal credit to the solutions presented here.Error carried forward marks may be awarded where an incorrect answer is used as part of the data needed for a subsequent question, providing that the resulting answer is not plainly ridiculous.If incorrect units are used more than once then one mark should be deducted from the total.If an inappropriate number of significant figures are given more than once in final answers then one mark should be deducted from the total.Section 1 – Multiple Choice QuestionsSection 2 – Short Answer QuestionsMarks for these two questions should be awarded for a clear explanation of the underlying Physical principals using correct scientific terminology. Answers that are incomplete, contain errors in Physics or use terminology incorrectly cannot be awarded full credit.Award 0 marks: No valid attempt made to answer questionAward 1 mark: Valid point presented but other-wise incorrect or incomplete answer Award 2 marks: Partially correct answer but major error or omission in reasoningAward 3 marks: Mostly correct answer, only minor errors or omissions in reasoningAward 4 marks: Completely correct answer, no errors, omissions of reasoning or incorrectuse of terminologyQuestion 11. (4 max) ∙Water / ethanol evaporates∙Only highest energy particles have enough energy to escape / evaporate∙Therefore average energy of remaining particles decreases∙Temperature depends on average energy of particles∙Therefore temperature of water / ethanol goes down (and you feel cold)Question 12. (4 max) ∙Temperature increases and so resistance of thermistor decreases (2)∙Therefore current in circuit increases∙Therefore voltage across resistor increases(Give full marks for resistance of thermistor increases, current decreases, voltage decreases)Section 3 – Longer QuestionsQuestion 13(a)(i)∙Volume = 0.2 x 0.2 x 0.2 = 0.008 m3→ mass = 3.2kg ∙Weight = 32 N [1] [1](a)(ii)∙Mass = 8kg → Weight = 80N[1] (a)(iii)∙T = U - W∙T = 80 – 32 = 48 N [1] [1](b)∙Use of F = ma rearranged to give a = F/m ∙Giving a = 48 / 3.2 = 15 m/s2[1] [1](c)∙Velocity and time axes labelled∙Curve, with initial velocity = 0, reaching a steady terminal velocity [1] [1](d)(i)∙Acceleration axes labelled with an initial value of 15 m/s2∙Following a smooth curve to show acceleration becoming zero [1] [1](d)(ii)∙Initial acceleration shown as greater∙Labelled as 15.6 m/s2∙Following a smooth curve to show acceleration becoming zero (any curve tending to zero is acceptable as long as it is similar to graph A) [1] [1] [1](d)(iii)∙Acceleration axes labelled with initial value of 15m/s2∙Smooth curve to reach zero acceleration more quickly than graph A [1] [1]Question 14(a)∙Correct circuit shown including ammeter and voltmeter (-1 for any error) [2] (b)∙Repeat readings (or any other appropriate technique) [1](c)∙Appropriate scales used∙Points plotted accurately∙Best fit line – accept either straight line through first 6 points or through all 8 points. [1] [1] [1](d)∙Resistance increases as temperature increases ∙Points at 80o C and 90o C do not fit trend [1] [1](e)∙Any suitable method to give α = 0.1 Ω /o C∙Any suitable method to give R o = 18 Ω (allow ± 0.5 Ω) [1] [1](f)∙Units Ω /o C [1]Question 15(a)∙Volume ice = 0.2 x 10-3 m3∙Mass ice = 920 x 0.2 x 10-3 = 0.184 kg∙Energy required to turn ice to water = 0.184 x 334 x103 = 61456 J ∙Energy lost by water = V x 4200 x 72 = 302400 x V∙Thus V = 61456 / 302400 = 0.203 litres = 0.203 kg (accept 0.2 kg) [1] [1] [1] [1](b)∙Energy transferred to the surroundings [1]。
美国2011年物理竞赛决赛试题
2011F=ma Contest25QUESTIONS-75MINUTESINSTRUCTIONSDO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN•Use g=10N/kg throughout this contest.•You may write in this booklet of questions.However,you will not receive any credit for anything written in this booklet.•Your answer to each question must be marked on the optical mark answer sheet.•Select the single answer that provides the best response to each question.Please be sure to use a No.2pencil and completelyfill the box corresponding to your choice.If you change an answer,the previous mark must be completely erased.•Correct answers will be awarded one point;incorrect answers will result in a deduction of14point.There isno penalty for leaving an answer blank.•A hand-held calculator may be used.Its memory must be cleared of data and programs.You may use only the basic functions found on a simple scientific calculator.Calculators may not be shared.Cell phones may not be used during the exam or while the exam papers are present.You may not use any tables,books,or collections of formulas.•This test contains25multiple choice questions.Your answer to each question must be marked on the optical mark answer sheet that accompanies the test.Only the boxes preceded by numbers1through25are to be used on the answer sheet.•All questions are equally weighted,but are not necessarily the same level of difficulty.•In order to maintain exam security,do not communicate any information about the questions (or their answers or solutions)on this contest until after February20,2011.•The question booklet and answer sheet will be collected at the end of this exam.You may not use scratch paper.DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN1.A cyclist travels at a constant speed of 22.0km/hr except for a 20minute stop.The cyclist’s average speed was 17.5km/hr.How far did the cyclist travel?(A)28.5km (B)30.3km (C)31.2km (D)36.5km(E)38.9kmQuestions 2to 4refer to the three graphs below which show velocity of three objects as a function of time.Each object is moving only in one dimension.246810−20+2+4v e l o c i t y (m /s )time (s)246810−20+2+4v e l o c i t y (m /s )time (s)246810−20+2+4v e l o c i t y (m /s )time (s)Object I Object II Object III2.Rank the magnitudes of the average acceleration during the ten second interval.(A)I >II >III (B)II >I >III (C)III >II >I (D)I >II =III (E)I =II =III3.Rank the magnitudes of the maximum velocity achieved during the ten second interval.(A)I >II >III (B)II >I >III (C)III >II >I (D)I >II =III (E)I =II =III4.Rank the magnitudes of the distance traveled during the ten second interval.(A)I >II >III (B)II >I >III (C)III >II >I (D)I =II >III (E)I =II =III5.A crude approximation is that the Earth travels in a circular orbit about the Sun at constant speed,at a distanceof150,000,000km from the Sun.Which of the following is the closest for the acceleration of the Earth in this orbit?(A)exactly0m/s2(B)0.006m/s2(C)0.6m/s2(D)6m/s2(E)10m/s26.A child is sliding out of control with velocity v c across a frozen lake.He runs head-on into another child,initiallyat rest,with3times the mass of thefirst child,who holds on so that the two now slide together.What is the velocity of the couple after the collision?(A)2v c(B)v c(C)v c/2(D)v c/3(E)v c/47.An ice skater can rotate about a vertical axis with an angular velocityω0by holding her arms straight out.Shecan then pull in her arms close to her body so that her angular velocity changes to2ω0,without the application of any external torque.What is the ratio of herfinal rotational kinetic energy to her initial rotational kinetic energy?(A)√2(B)2(C)2√2(D)4(E)88.When a block of wood with a weight of30N is completely submerged under water the buoyant force on the blockof wood from the water is50N.When the block is released itfloats at the surface.What fraction of the block will then be visible above the surface of the water when the block isfloating?(A)1/15(B)1/5(C)1/3(D)2/5(E)3/59.A spring has an equilibrium length of2.0meters and a spring constant of10newtons/meter.Alice is pulling onone end of the spring with a force of3.0newtons.Bob is pulling on the opposite end of the spring with a force of3.0newtons,in the opposite direction.What is the resulting length of the spring?(A)1.7m(B)2.0m(C)2.3m(D)2.6m(E)8.0m10.Which of the following changes will result in an increase in the period of a simple pendulum?(A)Decrease the length of the pendulum (B)Increase the mass of the pendulum(C)Increase the amplitude of the pendulum swing(D)Operate the pendulum in an elevator that is accelerating upward(E)Operate the pendulum in an elevator that is moving downward at constant speed.11.A large metal cylindrical cup floats in a rectangular tub half-filled with water.The tap is placed over the cup andturned on,releasing water at a constant rate.Eventually the cup sinks to the bottom and is completely submerged.Which of the following five graphs could represent the water level in the sink as a function of time?w a t e r l e v e ltime w a t e r l e v e ltime w a t e r l e v e ltime (A)(B)(C)w a t e r l e v e ltime w a t e r l e v e ltime (D)(E)12.You are given a large collection of identical heavy balls and lightweight rods.When two balls are placed at the endsof one rod and interact through their mutual gravitational attraction (as is shown on the left),the compressive force in the rod is F .Next,three balls and three rods are placed at the vertexes and edges of an equilateral triangle (as is shown on the right).What is the compressive force in each rod in the latter case?(A)1√3F (B)√32F(C)F(D)√3F (E)2F13.The apparatus in the diagram consists of a solid cylinder of radius 1cm attached at the center to two disks ofradius 2cm.It is placed on a surface where it can roll,but will not slip.A thread is wound around the central cylinder.When the thread is pulled at the angle θ=90◦to the horizontal (directly up),the apparatus rolls to the right.Which below is the largest value of θfor which it will not roll to the right when pulling on the thread?(A)θ=15◦(B)θ=30◦(C)θ=45◦(D)θ=60◦(E)None,the apparatus will always roll to the right14.You have5different strings with weights tied at various point,all hanging from the ceiling,and reaching down tothefloor.The string is released at the top,allowing the weights to fall.Which one will create a regular,uniform beating sound as the weights hit thefloor?(A)(B)(C)(D)(E)15.A vertical mass-spring oscillator is displaced2.0cm from equilibrium.The100g mass passes through the equilib-rium point with a speed of0.75m/s.What is the spring constant of the spring?(A)90N/m(B)100N/m(C)110N/m(D)140N/m(E)160N/mQuestions16and17refer to the information and diagram below. Jonathan is using a rope to lift a box with Beckyin it;the box is hanging offthe side of a bridge,Jonathan is on top.A rope is hooked up fromthe box and passes afixed railing;Jonathan holdsthe box up by pressing the rope against the rail-ing with a massless,frictionless physics textbook.The static friction coefficient between the rope andrailing isµs;the kinetic friction coefficient betweenthe rope and railing isµk<µs;the mass of the box and Becky combined is M;and the initial height of the bottom of the box above the ground is h. Assume a massless rope.BeckyLoose ropeFloorJonathan, pushesropefixed hardrailingon book against16.What magnitude force does Jonathan need to exert on the physics book to keep the rope from slipping?(A)Mg(B)µk Mg(C)µk Mg/µs(D)(µs+µk)Mg(E)Mg/µs17.Jonathan applies a normal force that is just enough to keep the rope from slipping.Becky makes a small jump,barely leaving contact with thefloor of the box.Upon landing on the box,the force of the impact causes the rope to start slipping from Jonathan’s hand.At what speed does the box smash into the ground?Assume Jonathan’s normal force does not change.(A)√2gH(µk/µs)(B)√2gH(1−µk/µs)(C)√2gHk s(D)√2gHk s(E)√2gH(µs−µk)18.A block of mass m=3.0kg slides down one ramp,and then up a second ramp.The coefficient of kinetic frictionbetween the block and each ramp isµk=0.40.The block begins at a height h1=1.0m above the horizontal.Both ramps are at a30◦incline above the horizontal.To what height above the horizontal does the block rise on the second ramp?(A)0.18m(B)0.52m(C)0.59m(D)0.69m(E)0.71mQuestions19and20refer to the following informationA particle of mass2.00kg moves under a force given byF=−(8.00N/m)(xˆi+yˆj)whereˆi andˆj are unit vectors in the x and y directions.The particle is placed at the origin with an initial velocity v=(3.00m/s)ˆi+(4.00m/s)ˆj.19.After how much time will the particlefirst return to the origin?(A)0.785s(B)1.26s(C)1.57s(D)2.00s(E)3.14s20.What is the maximum distance between the particle and the origin?(A)2.00m(B)2.50m(C)3.50m(D)5.00m(E)7.00m21.An engineer is given afixed volume V m of metal with which to construct a spherical pressure vessel.Interestingly,assuming the vessel has thin walls and is always pressurized to near its bursting point,the amount of gas the vessel can contain,n(measured in moles),does not depend on the radius r of the vessel;instead it depends only on V m (measured in m3),the temperature T(measured in K),the ideal gas constant R(measured in J/(K·mol)),and the tensile strength of the metalσ(measured in N/m2).Which of the following gives n in terms of these parameters?(A)n=23V mσRT(B)n=233√V mσRT(C)n=233√V mσ2 RT(D)n=233√V m2σRT(E)n=233V mσ2RT22.This graph depicts the torque output of a hypothetical gasoline engine as a function of rotation frequency.Theengine is incapable of running outside of the graphed range.IIIIIIEngine Revolutions per Minute0102030O u t p u t T o r q u e (N m )1,0002,000At what engine RPM (revolutions per minute)does the engine produce maximum power?(A)I(B)At some point between I and II (C)II(D)At some point between II and III (E)III23.A particle is launched from the surface of a uniform,stationary spherical planet at an angle to the vertical.Theparticle travels in the absence of air resistance and eventually falls back onto the planet.Spaceman Fred describes the path of the particle as a parabola using the laws of projectile motion.Spacewoman Kate recalls from Kepler’s laws that every bound orbit around a point mass is an ellipse (or circle),and that the gravitation due to a uniform sphere is identical to that of a point mass.Which of the following best explains the discrepancy?(A)Because the experiment takes place very close to the surface of the sphere,it is no longer valid to replacethe sphere with a point mass.(B)Because the particle strikes the ground,it is not in orbit of the planet and therefore can follow a non-elliptical path.(C)Kate disregarded the fact that motions around a point mass may also be parabolas or hyperbolas.(D)Kepler’s laws only hold in the limit of large orbits.(E)The path is an ellipse,but is very close to a parabola due to the short length of the flight relative to thedistance from the center of the planet.24.A turntable is supported on a Teflon ring of inner radius R and outer radius R+δ(δ R),as shown in the diagram.To rotate the turntable at a constant rate,power must be supplied to overcome friction.The manufacturer of the turntable wishes to reduce the power required without changing the rotation rate,the weight of the turntable,or the coefficient of friction of the Teflon surface.Engineers propose two solutions:increasing the width of the bearing (increasingδ),or increasing the radius(increasing R).What are the effects of these proposed changes?(A)Increasingδhas no significant effect on the required power;increasing R increases the required power.(B)Increasingδhas no significant effect on the required power;increasing R decreases the required power.(C)Increasingδincreases the required power;increasing R has no significant effect on the required power.(D)Increasingδdecreases the required power;increasing R has no significant effect on the required power.(E)Neither change has a significant effect on the required power.25.A hollow cylinder with a very thin wall(like a toilet paper tube)and a block are placed at rest at the top of aplane with inclinationθabove the horizontal.The cylinder rolls down the plane without slipping and the block slides down the plane;it is found that both objects reach the bottom of the plane simultaneously.What is the coefficient of kinetic friction between the block and the plane?(A)0tanθ(B)13tanθ(C)12(D)2tanθ3(E)tanθ。
2010-2011年度高一物理竞赛试题参考答案
2010-2011年度高一物理竞赛试题参考答案一．选择题（共13小题，每小题4分，共52分．选对的得4分，选错或不答的得0分．）1.A2.B3.B4.D5.C6.C7.C8.B9.A 10.B 11.C 12.A 13.B二．实验题（共1小题，共16分）14．（15分）（1）① O 点位置（2分）， 两弹簧秤的读数 （2分），两细绳套的方向 （2分） ② F ′（2分）③ B （2分）（2）① 0.21m/s （2分） ② 2 （2分） ③ 9.5m/s 2（2分）三．论述计算题（共2小题，共32分。
）15．（16分）参考解答：解：（1）因物体向左运动，摩擦力方向向右，设物体加速度为a,则 ma F mg =+-)(μ--------------（2分） 解得：2/3)(s m mF mg a -=+-=μ--------------（2分） 负号表示方向向右--------------（2分）（2）物体向左作匀减速直线运动，设经过时间t 速度为V t =0,由运动学公式有at v v t +=0--------------（2分） 解得时间：s s a v v t t 431200=--=-=--------------（2分） 故4秒末物体速度为0，此后由于F=12N<N mg 24=μ,物体静止，根据二力平衡可得物体受到的静摩擦力为：N F F f 12== 方向水平向左--------（2分）（3）5s 内物体的位移实际等于4s 内物体的位移：)(244)3(2141221220m at t v s =⨯-+⨯=+=--------------（4分） 16．（16分）参考解答：解：（1）设儿童下滑的加速度大小为a ，由牛顿第二定律有：mgsin37°-μmgcos37°=ma 1 ①-------------（2分）解得：a 1=2 m/s 2． ② -------------（2分）（2）因为H=2 m ，圆弧CD 的半径R=0．2 m ，所以AB 的长度m 337sin R -H L 01== ③ -------（2分） 设儿童滑到B 点的速率为v B ，由运动学公式有：v B 2=2a 1L 1 ④-----------------------（2分）由②③④解得：/B v s = ⑤--------------（2分） （3）设儿童在C 点恰做平抛运动滑出时的速率为v C ，则：2C v mg m R= ⑥-------------------------（2分） 物体由B 到C 的加速度设为a 2,由牛顿第二定律有： 2ma mg =-μ ⑦----------------- (1分)L a v v B c 2222=- ⑧--------------------(1分) 由⑤⑥⑦⑧解得：L=1 m ．-------------------(1分)所以水平滑槽BC 的长度L 应大于或等于1m------(1分)。
2011年第28届物理竞赛复赛试卷+答案
第28届全国中学生物理竞赛复赛试卷一、（20分）如图所示，哈雷彗星绕太阳S 沿椭圆轨道逆时针方向运动，其周期T 为76.1年，1986年它过近日点P 0时与太阳S 的距离r 0=0.590AU ，AU 是天文单位，它等于地球与太阳的平均距离，经过一段时间，彗星到达轨道上的P 点，SP 与SP 0的夹角θP =72.0°。
已知：1AU=1.50×1011m ，引力常量G=6.67×10－11Nm 2/kg 2，太阳质量m S =1.99×1030kg ，试求P 到太阳S 的距离r P 及彗星过P 点时速度的大小及方向（用速度方向与SP 0的夹角表示）。
二、（20分）质量均匀分布的刚性杆AB 、CD 如图放置，A 点与水平地面接触，与地面间的静摩擦系数为μA ，B 、D 两点与光滑竖直墙面接触，杆AB 和CD 接触处的静摩擦系数为μC ，两杆的质量均为m ，长度均为l 。
1、已知系统平衡时AB 杆与墙面夹角为θ，求CD 杆与墙面夹角α应该满足的条件（用α及已知量满足的方程式表示）。
2、若μA =1.00，μC =0.866，θ=60.0°。
求系统平衡时α的取值范围（用数值计算求出）。
三、（25分）在人造卫星绕星球运行的过程中，为了保持其对称转轴稳定在规定指向，一种最简单的办法就是让卫星在其运行过程中同时绕自身的对称轴转，但有时为了改变卫星的指向，又要求减慢或者消除卫星的旋转，减慢或者消除卫星旋转的一种方法就是所谓消旋法，其原理如图所示。
一半径为R ，质量为M 的薄壁圆筒，，其横截面如图所示，图中O 是圆筒的对称轴，两条足够长的不可伸长的结实的长度相等的轻绳的一端分别固定在圆筒表面上的Q 、Q ′（位于圆筒直径两端）处，另一端各拴有一个质量为2m 的小球，正常情况下，绳绕在圆筒外表面上，两小球用插销分别锁定在圆筒表面上的P 0、P 0′处，与卫星形成一体，绕卫星的对称轴旋转，卫星自转的角速度为ω0。
2011全国物理竞赛复赛试题及答案
得分 评卷人
图1
1．当风速为 6.0m/s 时，风力发电机的输出功率将变为 50W，在这种情况下，将蓄电池的电量由 20%充 至 90%需多长时间？ 2．如果当地垂直于太阳光的平面得到的太阳辐射最大强度约为 240W/m2，要想太阳能电池的最大功率 达到 36W，太阳能电池板的面积至少要多大？ 3．若遇到阴天无风的天气，仅靠蓄电池供电，最多可供灯具正常发光多长时间？
1．图 6 为安装示意图，安装时要将 L 形挂钩用膨胀螺钉固定在承重墙上，上表是热水器的部分数据， 请估算 L 形挂钩受到的最大向下拉力 F 及 对应此时膨胀 F 承重墙 螺钉与承重墙体间的摩擦力 f。 悬挂孔 L 形挂钩 2．这台热水器共有三挡加热功率，销 售人员曾介绍 E 悬挂架 说：热水器内部有两个加热管。请你根据 销售人员的介 D 绍，猜测并画出内部加热电路的主要结构 （简化的等效 安全阀 出水管 B 进水管 电路图） 。然后依据你的猜想，计算出每个 加热管的电阻 C A 值。 图6 3．产品说明书上标明：当达到设定 温度，停止加 热后，如果水温降到低于设定温度 4℃时，热水器将自动以设置功率将桶内的水加热至设定温度，如此反复 直至手动停止加热。小芳想到了国家提倡的峰谷电价制度的有关规定：2011 年起，本地的分时电价为：峰 时段 (08： 00～ 22： 00)0.6 元 /度，谷时段 (22： 00～次日 08： 00)0.3 元 /度。她想，如果每天下午 6 点洗浴 时需要把 20℃的冷水加热至 50℃，可以下午直接加热，也可以早晨 8 点前加热好，让热水器自动保温至 下午 6 点，哪种方式更省钱呢？为此她利用热水器提供的测温功能测量了水温从 50℃降至 46℃需要 190 分钟，已知 c 水 =4.2× 103J/(kg•℃)，请帮她通过计算做出哪种方式更省钱的判断。
英国2011年中学物理竞赛
Name TotalMarkSchool2011 Physics ChallengeTime allowed: 1 hourAttempt all questionsWrite your answers on this question paperYou may use a calculatorAssume the gravitational field strength has a value of 10 N / kgSection A:Ten multiple choice questions worth 1 mark each (worth 10 marks in total)Allow about 10 minutes for this sectionSection B:Two short answer questions (worth 8 marks in total)Questions require a clear explanation of the underlying physical principlesAllow about 10 minutes for this sectionSection C:Longer answer questions requiring calculations (worth 32 marks in total)Questions may be set on unfamiliar topics. Additional informationnecessary to answer the question will be given in the questionAllow 40 minutes for this sectionTotal 50 marks; mark allocations for each sub-section are shown in bracketsSection A: Multiple Choice AnswersWrite the correct letter in the grid. The first column has been done as an example if the answer to question zero were CQuestion 0 1 2 3 4 5 6 7 8 9 101.In imperial units, length can be measured in inches where 1.00 inch =2.54 cm.A block of metal has a volume of 2.00 cubic inches. The volume in metric units is:A. 3.28 x 10-5 m3B.32.77 x 10-4 m3C. 5.08 x 10-2 m3D.12.90 x 10-2 m3E. 3.28 x 10-1 m32. A group of explorers near the equator leave base camp and travel 7km North and then 2km East andthen finally 5km South. They then realise that they are late for dinner! In what direction should they travel to return directly to base camp?A.WestB.South WestC.SouthD.South EastE.North East3. A 50kg brick starts from rest and slides down a slope converting gravitational potential energy tokinetic energy. In the process it has to do work against a constant force due to friction of 65N. Air resistance may be ignored. The slope is 40m long and the top of the slope is 6.0m vertically above the bottom of the slope. What will be the speed of the block at the bottom of the slope?A.16 m/sB.11 m/sC. 4 m/sD.0 m/sE.Cannot be determined form the information given4. A solar panel is used to heat water. Each minute 20 litres of water pass through the panel. The waterentering the panel is at a temperature of 20o C and the water leaving the panel is at a temperature of 26o C. Water requires 4200J to raise the temperature of 1kg by 1o C and the mass of 1litre of water is 1kg. Assuming that the water does not lose any energy and that the solar panel is 100% efficient, how much radiant energy falls on the solar panel each second?A.500 kJB.84 kJC.25 kJD.8.4 kJE.0.42 kJ5. A student uses a radioactivity detector to measures the background count in the laboratory, with noradioactive sources present, to be 25 counts per minute. A radioactive isotope has an initial activity of 400 counts per minute and a half-life of 5 minutes. The student uses the same radioactivity detector to measure the radioactivity of the isotope. How long does it take for the detected count rate to reduce to50 counts per minute?A.10 minutesB.15 minutesC.20 minutesD.25 minutesE.30 minutes6. A heater is connected to a 12v battery and has a power output of 36W. The same heater is nowconnected to an 8v battery. Assume that the resistance of the heater remains constant. What is the power output of the heater?A.36WB.24WC.16WD.12WE.8W7.The best estimate for the wavelength of red light is:A.0.7mm (7 x 10-4 m)B.70µm (7 x 10-5 m)C.7µm (7 x 10-6 m)D.700nm (7 x 10-7 m)E.70nm (7 x 10-8 m)8. A large rock is dropped off a cliff and hits the ground below at 20 m/s. If a rock with three times themass was dropped off a cliff of four times the height, what speed would it hit the ground at (ignore air resistance)?A.20 m/sB.40 m/sC.60 m/sD.80 m/sE.240 m/s9.When a gas in a sealed container (which cannot expand) is heated the pressure increases.This is because:i.The particles of the gas hit the container walls more oftenii.The particles of the gas hit the container walls harderiii.The particles of the gas have more potential energyA.(i) onlyB.(ii) onlyC.(iii) onlyD.(i) and (ii) onlyE.(ii) and (iii) only10.In a simple d.c. electric motor:i.The split ring commutator changes the direction of the current in the armature coils every half turnii.The force acting on the coils of the armature only depends on the strength of the magnetic field iii.The armature is made of plastic. This is because it must not be magnetic and should not conduct electricityWhich of the above statements are TRUE?A.(i) onlyB.(ii) onlyC.(iii) onlyD.(i) and (ii) onlyE.(i) and (iii) onlySection B: Short Answer Questions11.Even on a warm day, when the air is warm, you still feel cold when coming out of the sea or out of aswimming pool. A similar effect can be observed in the laboratory by pouring a small amount ofethanol on to the back of the hand. Even though the ethanol is warm, your hand still feels cold.Explain why, in either of these examples, you feel cold even though the air around you is warm.[4 marks]…......................................................................................................................................…......................................................................................................................................…......................................................................................................................................…......................................................................................................................................…......................................................................................................................................…......................................................................................................................................12.Consider the circuit below. Explain what happens to the reading on the voltmeter as the temperatureincreases (i.e. state what happens and explain why).[4 marks]…......................................................................................................................................…......................................................................................................................................…......................................................................................................................................…......................................................................................................................................…......................................................................................................................................…......................................................................................................................................Section C: Longer Questions13.Archimedes principle and BuoyancyArchimedes principle is used to explain buoyancy and why things float. Archimedes principle states that any objects (such as a brick) immersed in a fluid (such as water) experiences an upthrust (U) equal to the weight of fluid displaced.Upthust = Weight of fluid displaceda)Consider a float, in the form of a cube, secured to the bottom of a deep pool of fresh water by a rope.Length of each side = 20 cmDensity of float material = 400 kg/m3Density of fresh water = 1000 kg/m3i.Calculate the weight of the float[2 marks]…............................................................................................................................…............................................................................................................................ii.Calculate the weight of the water displaced[1 mark]…............................................................................................................................…............................................................................................................................iii.Hence calculate the tension in the rope[2 marks]…............................................................................................................................…............................................................................................................................b)The rope now breaks. Calculate the initial acceleration of the float[2 marks]…............................................................................................................................…............................................................................................................................c)Sketch the velocity time graph for the float on the axes below. No scale is required on the velocity ortime axes. Assume the pool is very deep and the float reaches terminal velocity before reaching the surface.[2 marks]d)Consider the following acceleration-time graphs.In each case add a scale to the acceleration axis and continue the graph until such time as the float has reached its terminal velocity. No scale is required on the time axis.i.Sketch an acceleration-time graph for the float. Label this graph A[2 marks]ii.Add a second line to show the acceleration-time graph for the same float immersed in salt water having a density of 1024 kg/m3. Label this graph B[3 marks]iii.Add a third line to show the acceleration-time graph, in fresh water, for a float made of the same material but having twice the dimensions i.e. a cube with sides of length 40cm. Label thisgraph C[2 marks]ing Graphs – change of resistance with temperatureA group of students are investigating how the resistance of a particular material changes withtemperature. Their teacher suggests that the relationship is given byR = R0 + αT R = Resistance (Ω)R0 = Resistance at 0o C (Ω) - a constantT = temperature (o C)α = a constanta)Given an ammeter, voltmeter, variable power supply and wires etc. as necessary, draw a suitablecircuit that would enable the students to measure the resistance of the wire.[2 marks]b)The students take readings of resistance and temperature. Suggest how they could make their resultsas reliable as possible[1 mark]…............................................................................................................................…............................................................................................................................c)Use the results given in the table to plot a suitable graph of resistance and temperature.Add a line of best fit.[3 marks]Temperature (o C) Resistance (Ω)20 20.030 21.140 22.050 23.060 24.170 25.280 26.5d)To what extent do the results of the experiment support the relationship suggested by the teacher?[2 marks]…............................................................................................................................…............................................................................................................................…............................................................................................................................e)Use the graph, or the data in the table, to determine the best estimate for values for R0 and α[2 marks]…............................................................................................................................…............................................................................................................................…............................................................................................................................…............................................................................................................................f)State suitable units for α[1 mark]…............................................................................................................................15.Melting SnowA motorist decides to melt the snow off a car windscreen using hot waterThey find the following information:• 1 litre of snow can be compressed to 1/5th litre of ice• 1 litre = 1000 cm3•water has a density of 1000 kg/m3•ice has a density of 920 kg/m3•the energy required to turn 1kg ice at 0o C into 1kg of water at 0o C is 334 kJ•the energy required to change the temperature of 1kg of water by 1o C is 4200 J They use hot water with a temperature of 72o Ca)Calculate the minimum volume of hot water required to melt 1 litre of snow[4 marks]…............................................................................................................................…............................................................................................................................…............................................................................................................................…............................................................................................................................…............................................................................................................................…............................................................................................................................…............................................................................................................................…............................................................................................................................b)Suggest why, in reality, the motorist is likely to need a greater volume of hot water than the amountcalculated in (a)[1 mark]…............................................................................................................................…............................................................................................................................…............................................................................................................................…............................................................................................................................。
bpho英国物理竞赛题目
bpho英国物理竞赛题目
以下是BPhO英国物理竞赛的部分题目：
1. 题目：一个钢球以30ms-1的速度从的高度扔到坚硬的表面上。
它在每次反弹时保留了70%的能量。
计算: (i) 物体第一次落地时的速度; (ii) 物体在反弹4次后达到的最大高度。
2. 题目：一个长途自行车手使用的自行车电脑是双动力的:它有一个内部电池和一个太阳能电池板。
3. 题目：圆柱体内装着n种不同体积的液体，这些液体形成水平层，每层的高度为h。
(ii)在围绕第二液体的表面A上的力F1是多少? (iii)给出对圆柱体底部Fn液体表面A第n小块的液体，底部的压力F1的表达式?
4. 题目：(iii)求出火箭加速度与时间的函数t的表达式，给定T，g，mt，m0和推力作用的总时间t0。
(iv)计算发射后宇航员的体重翻倍的时间。
希望以上信息对您有所帮助，如果您还有其他问题，欢迎告诉我。
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Question 1 A network of resistors, each of resistance R, is shown in Figure 1 below.
B A
X
C
Y
Figure 1.
(a) The overall resistance, measured between X and Y, is 33 kΩ. What is the value of R? [4]
mg Figure 3. Ladder leaning against a wall.
3
(11 marks)
Question 4 A typical value for therth is 101 kPa. The surface area of
(b) The power developed in resistor A, due to the current flowing from X to Y, is 1.8 mW. Calculate the power developed in the resistor
(i) B
(ii) C [4]
5
(g) As a man-made satellite orbits the Earth, there is always a point on the Earth directly below it. This point follows the path of a satellite’s orbit and is plotted on a map of the Earth, as shown below in Figure 4. Describe or sketch the satellite’s orbit i.e. how it is oriented about the Earth, and its shape. [3]
(e) The closest distance of approach to the Sun for the comet is 0.585 AU, when it is visible to the naked eye. Calculate the furthest distance of the comet from the Sun. [1]
(4 marks)
2
Question 3
(a) A broom used to sweep up dust is pushed steadily across the floor of a room, as shown in Figure 2 below. Sketch a free body diagram for the head of the broom, showing the push, the weight of the broom head, the reaction of the floor and the frictional force due to the floor.
(d) Halley’s Comet also orbits the Sun and so the value of k is the same as in (b). Its period of orbit is 75.3 years. Determine the value of R for its orbit about the Sun. [1]
(8 marks)
Question 2
When a beam of protons from a particle accelerator hits a target, particles are produced which have a very short lifetime. For a particle to be formed, one side of the particle must “be in touch with” the far side of the particle, so that a signal from one side of the particle must reach the other side before the particle decays.
(c) If the wall was made frictionless, the ladder would still remain upright. Draw a free body diagram for the ladder in this situation. [1]
(d) If the floor was made frictionless, whilst the wall is rough and produces friction, decide and explain whether the ladder would fall or remain upright. Use a free body diagram for the ladder to help explain your answer. [4]
Motion of broom
head of broom
Figure 2. Motion of broom to the right. [4]
(b) In Figure 3 below, a ladder is shown leaning against a wall. The ladder remains in place because there is both friction of the ladder on the wall and friction of the ladder on the floor. Draw a free body diagram for the ladder showing all of the forces acting upon it. [2]
the Earth A, is related to its mean radius R by the expression A = 4πR2, where R has the value 6400 km. (a) How is the weight of the atmosphere related to the pressure of the atmosphere
The diameter of the particle can be taken as the range of the Strong Nuclear Force which is about 1 x 10-15 m. If the signal propagates at the speed of light, what would be the shortest lifetime of such a particle?
(f) The height of the atmosphere calculated in (iii) is less than the height at which aircraft often fly. Explain why our calculation gives a low result for the height. [2]
BRITISH PHYSICS OLYMPIAD
British Physics Olympiad 2011 Paper 1
September /October 2010 Answer all questions
Allow 1 hour Total 50 marks g = 9.8 ms‐2 or N kg‐1 c = 3.0 x 108 ms‐1 NA = 6.02 x 1023 mol‐1
(f) The comet’s speed is 70.6 km s-1 at closest approach to the Sun. Is the speed greater or smaller than this at the comet’s furthest distance from the Sun? Give a reason for your answer. [3]
(c) The average distance of the Earth from the Sun is defined as 1 Astronomical Unit (1 AU). Determine the value of k for part (a), including units. (The period T can be measured in years). [1]
acting on the Earth’s surface? [1]
Calculate: (b) The weight of the atmosphere.
[2] (c) The mass of the atmosphere, assuming that g does not vary with height between the
bottom and the top of the atmosphere. [2]
(d) The number of molecules in the atmosphere, assuming that the mass of a mole of air is 30 g. [2]
(e) The height of the atmosphere if the density ρ = 1.2 kg m-3. [4]