奥林匹克物理第20届试卷及答案

2004年20届物理奥赛决赛卷 10.27一、5个质量相等的匀质球，其中4个半径均为a 的球，静止放在半径为R 的半球形碗内，它们的球心在同一水平面内．另1个半径为b 的球放在4球之上．设接触面都是光滑的，试求碗的半径R 的值满足什么条件时下面的球将相互分离．二、一人造地球卫星绕地球做椭圆运动，地心是椭圆的一个焦点，在直角坐标系中，椭圆的轨迹方程为 12222=+by a x a 、b 分别是椭圆的长半轴和短半轴，为已知常数．当该人造卫星在轨道的远地点时，突然以很大的能量沿卫星运行方向从卫星上发射出一个质量为m 的太空探测器，这探测器在地球引力作用下做双曲线运动，此双曲线的焦点位于地心，实半轴的长度正好等于原来椭圆远地点到地心的距离．试问在发射时，给探测器的能量为多大?设地球的质量m E 、万有引力常量G 为已知，不计地球以外星体的影响． 三、如图决所示，金属飞轮具有n 条辐条，每条辐条长l ，可绕转轴OO ′旋转．飞轮处在匀强磁场中，磁场方向与转轴平行，磁感强度为B ．转轴与飞轮边缘通过电刷与电阻R 、电感L 连成闭合回路．飞轮及转轴的电阻和转动过程的摩擦均不计．1．现用一恒定的外力矩M 作用于飞轮，使它由静止开始转动，求当飞轮转动达到稳定时，转动角速度Ω0和通过电阻R 的电流I 0．2．在飞轮转速达到稳定后，突然撤去外力矩M ，则飞轮的角速度Ω和通过电阻R 的电流，都将随时间变化．当R 取不同值时，角速度随时间变化的图线Ω(t)和电流随时间变化的图线，I(t)将不同．现给出了Ω(t)图线各6条(附本题后)，各图中都把刚撤去外力矩M 的时刻作为起始时刻，即t=0的时刻，此时刻的角速度Ω=Ω0，电流I=I 0．角速度Ω各图的纵坐标单位相同，电流 I 各图的纵坐标单位相同，各图的时间轴横坐标的单位均相同．试从这些图线中选出可能正确表示飞轮的角速度Ω随时间变化的图线Ω(t)，以及与所选图线对应的最接近正确的电流，随时间变化的图线I(t)．把你选出的图线Ω(t)与对应的I(t)图线在下面的图线符号之间用直的实线连接起来．注意：只有全部连接正确的才能得全分，连对但不全的可得部分分，有连错的得零分．四、如图决所示，y轴右边存在磁感强度为2B0的匀强磁场，y轴左边存在磁感强度为B0的匀强磁场，它们的方向皆垂直于纸面向里．在原点O处，一个带正电的电荷量为q、质量为m的粒子a，在t=0时以大小为2v0的初速度沿x轴方向运动．在粒子a开始运动后，另一质量和电荷量都与a相同的粒子b从原点0以大小为v0的初速度沿负x轴方向开始运动．要想使a和b能在运动过程中相遇，试分析和计算它们出发的时间差的最小值应为多大，并求出与此对应的相遇地点的坐标．设整个磁场区域都处于真空中，且不考虑重力及a、b两粒子之间相互作用力．五、一位近视眼朋友不戴眼镜时，能看清的物体都在距眼睛 a=20 cm以内．他发现，如果在眼前靠近眼睛处放一个有小圆孔的卡片，通过小圆孔不戴眼镜看远处的物体时也能看得清晰些．1．若小圆孔直径为D，试根据几何光学原理求出当近视眼直视远处的一个点物时，眼的视网膜上产生的光斑的直径．2．再考虑小圆孔的衍射效应，求小圆孔直径最恰当的大小．计算时可取可见光的平均波长为600 nm．[提示]1．人眼是一个结构比较复杂的光学系统，在本题中，可将人眼简化成一个焦距．厂可调的薄透镜和一个可成像的视网膜，透镜的边缘为瞳孔，两侧介质均为空气，视网膜与透镜的距离为b．2．小圆孔的存在对透镜成像的影响介绍如下:在几何光学中，从远处物点射向透镜的、平行于光轴的平行光束将会聚于透镜的焦点上，这就是像．如果在透镜前放一直径为D的小圆孔，则将发生光的衍射，在焦点处像屏上将出现如图左所示的衍射图样，其中央是一个明亮的圆斑，圆斑外周是一组亮度逐渐减弱的亮暗相间的同心圆环，由于这些圆环亮度比中央圆斑弱得多，观察时可以不予考虑．中央圆斑的半径对薄透镜中心的张角△θ的大小与D有关．理论计算得到△θ=1．22λ/D 式中λ是所用光的波长，这圆斑就是有小圆孔时观察到的物点的“像”，如图右所示．由上式可见，D越大，像斑就越小，点物的像就越接近一个点，物体的像越清晰；反之，D越小，点物的衍射像斑就越大，物体的像就越不清晰．如果观察屏不在焦点处而在焦点附近，屏上将出现类似的衍射图样，其中央亮斑对透镜中心的张角△θ可近似地用上式计算．六、设地球是一个半径为6370 km的球体．在赤道上空离地面1千多公里处和赤道共面的圆与赤道形成的环形区域内，地磁场可看作是均匀的，其磁感强度为B=3.20×10-6T．某种带电宇宙射线粒子，其静质量为m0=6.68×10-27kg，其电荷量为q=3.20×lO-19C，在地球赤道上空的均匀地磁场中围绕地心做半径为R=7370 km的圆周运动．已知在相对论中只要作用于粒子的力F的方向始终与粒子运动的速度v的方向垂直，则运动粒子的质量m和加速度a与力F的关系仍为F=ma，但式中的质量m为粒子的相对论质量．问：1．该粒子的动能为多大?2．该粒子在圆形轨道上运动时与一个不带电的静质量为 m2=4m0的静止粒子发生碰撞，并被其吸收形成一个复合粒子，试求复合粒子的静质量m1．。

2020年第二十届全国应用物理竞赛试题一、选择题（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。

1．2020年的诺贝尔物理学奖由三人分享，其中在光纤通信技术方所做的原创性工作而被称为“光纤之父”的美籍华人是（）A．钱永健B．朱棣文C．高锟D．丁肇中2．晴朗无风的早晨，当飞机从空中飞过，在蔚蓝的天空中会留下一条长长的“尾巴”，如图1所示，这种现象俗称为“飞机拉烟”。

产生这一现象的原因之一是飞机在飞行过程中排出的暖湿气体遇冷所致。

在这一过程中，暖湿气体发生的物态变化是（）A．熔化B．液化C．蒸发D．升华3.2020年1月2日起，我国北方大部分地区遭大范围降雪天气袭击。

大雪严重影响了民航、铁路和高速工路等交通，如图2所示。

在遇到这种天气时，为了尽快清除积雪，常用的办法是撒“融雪盐”，这是因为（）A.“融雪盐”与少量水发生化学反应，产生的热量使周围的冰雪熔化B.“融雪盐”产生“保暖层”，使冰雪吸收足够的“地热”而熔化C.使雪形成“含融雪盐的雪”，“含融雪盐的雪”熔点低于当地温度，使雪熔化D.“融雪盐”有利于冰雪对阳光的吸收，从而加快冰雪的熔化4.小明发现户外地面以上的冬季供热管道每隔一段距离总呈现型，如图3所示。

其主要原因是（）A.为了避开行人和建筑物B.为了美观C.为了避免管道因热胀冷缩导致的损坏D.为了方便工人师傅安装和检修5.体操、投掷、攀岩等体育活动都不能缺少的“镁粉”，它的学名是碳酸镁。

体操运动员在杠前都要在手上涂擦“镁粉”，其目的是A.仅仅是为了利用“镁粉”吸汗的作用，增加手和器械表面的摩擦而防止打滑B.仅仅是为了利用手握着器械并急剧转动时“镁粉”能起到衬垫作用，相当于在中间添加了一层“小球”做“滚动摩擦”C.仅仅是为了利用“镁粉”填平手掌的褶皱和纹路，使手掌与器械的接触面积增大，将握力变得更加实在和均匀D.上述各种功能都有6.小新同学家中的墙壁上竖直悬挂着一指针式电子钟，当其因电池电能不足而停止时。

第二十届全国中学生物理竞赛预赛题参考答案、评分标准一、参考解答（1） 右 f 实 倒 1 。

（2） 左 2f 实 倒 1 。

评分标准：本题20分，每空2分。

二、参考解答波长λ与频率ν的关系为 cνλ=， （1）光子的能量为 E h νν=， （2） 由式（1）、（2）可求得产生波长74.8610λ-=⨯m 谱线的光子的能量194.0910E ν-=⨯J （3）氢原子的能级能量为负值并与量子数n 的平方成反比： 21n E kn=-，n =1，2，3，… （4） 式中k 为正的比例常数。

氢原子基态的量子数n =1，基态能量1E 已知，由式（4）可得出1k E =- （5）把式（5）代入式（4），便可求得氢原子的n =2，3，4，5，… 各能级的能量，它们是19221 5.45102E k -=-=-⨯J ， 193212.42103E k -=-=-⨯J ，194211.36104E k -=-=-⨯J ，205218.72105E k -=-=-⨯J 。

比较以上数据，发现1942 4.0910E E E ν-=-=⨯J 。

（6）所以，这条谱线是电子从4n =的能级跃迁到2n =的能级时发出的。

评分标准：本题20分。

式（3）4分，式（4）4分，式（5）4分，式（6）及结论共8分。

三、参考解答1. 操作方案：将保温瓶中90.0t =℃的热水分若干次倒出来。

第一次先倒出一部分，与温度为010.0t =℃的构件充分接触，并达到热平衡，构件温度已升高到1t ，将这部分温度为1t 的水倒掉。

再从保温瓶倒出一部分热水，再次与温度为t 的构件充分接触，并达到热平衡，此时构件温度已升高到2t ，再将这些温度为2t 的水倒掉。

然后再从保温瓶中倒出一部分热水来使温度为2t 的构件升温……直到最后一次，将剩余的热水全部倒出来与构件接触，达到热平衡。

只要每部分水的质量足够小，最终就可使构件的温度达到所要求的值。

2. 验证计算：例如，将1.200kg 热水分5次倒出来，每次倒出0m ＝0.240kg ，在第一次使热水与构件达到热平衡的过程中，水放热为1001()Q c m t t =- （1）构件吸热为110()Q cm t t '=- （2） 由11Q Q '=及题给的数据，可得1t ＝27.1℃ （3）同理，第二次倒出0.240kg 热水后，可使构件升温到2t ＝40.6℃ （4）依次计算出1t ～5t 的数值，分别列在下表中。

第20届北京市高一物理竞赛第20届北京市高一物理（力学）竞赛决赛试题（北京二中杯）（全卷满分100分）2007年5月27日9：00~11：00题号一二总分7 8 9 10 11分数阅卷人复查人一、填空题（共30分，每小题5分）1．如图1所示，从倾角 =30°的斜坡上部水平抛出一小球，小球初速度为v，不计空气阻力，经过时间t，小球未落到斜坡上，此时小球的切向加速度为，法向加速度为．2．如图2所示，质量为m的物体A放在粗糙水平地面上，需要水平力3N才能推动．现有一质量也为m的物体B，将A、B之间用水平弹簧连接起来，B与水平地面间的摩擦忽略．用水平恒力F=2N推B，A能否被推动？答：．理由是得第 2 页共 16 页第 3 页 共 16 页．3．如图3所示，飞行员驾驶教练机在竖直平面沿半径R =1200m 的半圆周上飞行，飞行员的质量m =54kg ，从最低点A 到最高点C 飞机速率均匀减小，飞行员在A 、C 处时对座椅的压力F A =1680N ，F C =350N ，则飞行员在B 处时对座椅的压力B F = N ．4．如图4所示，A 、B 滑块的质量顺次为m 和2m ，它们之间用长为l 的细绳连结，开始时细绳水平，从静止释放，不计摩擦，则A 下降l2时B的速度B v = ．5．如图5所示，A 、B 两车的质量均为m ，A 车地板上放置一质量为m 的货箱，在光滑水平地面上，A 车向右运动的速度为0v ，B 车向左运动的速度为02v ，两车迎面相碰后不再分开，货箱在A 车地板上滑动l 相对静止，由此可知货箱与地板间的动摩擦系数为 ，从两车相碰到货箱相对车静止的这段时间内B 车位移的大小和方向图图F图CO 图υ第 4 页 共 16 页为 ．6．如图6所示．质量为m 的飞船关闭发动机后速度为0v ，飞船要降落到远处的一行星上，行星质量为M ，半径为R ．飞船要正好降落在行星表面，飞船速度0v 与瞄准距离L 之间应满足的关系是 ． 二、计算题（共70分）7.（12分）．如图7所示，一根长为a 的轻质直杆（质量不计），上端固连着一质量为m 的小球A ．小球可看作质点，直立在粗糙的跳台上，直杆的下端B 位于跳台的边缘，直杆受扰动就绕B 点翻倒．试求：（1）直杆翻倒后，离开跳台下落时直杆转过的锐角 为多少？ （2）此时直杆的角速度大小为多少？得图图图8．（12分）光滑的水平面上有一个固定的钢性粗糙挡板AB，得一个圆柱放在水平面上，挡板BC与AB在B端用固定铰链连接，如图8所示．现在C端施加一个垂直于BC的作用力F，AB和BC与柱的摩擦力相同且柱无转动．求两挡板摩擦系数满足什么条件，无论力F 多大，柱都不会被挤出．图第 5 页共 16 页第 6 页 共 16 页9．（12分）一仓库高25m 、宽40m ，如图9所示．今在仓库前L m 、高5m 的A 处抛一石块，使石块抛过屋顶，问距离L多大时，初速度0v 最小．（取g =10m/s 2）得图第 7 页 共 16 页10．（16分）如图10所示，质量分别为m 和()M M 2m =的重物由绕过两个定滑轮的绳子相连，当重物M 沿四角块上的竖槽下降时，m 和四角块移动，设四角块的质量为m ，略去一切摩擦及绳子和定滑轮的质量．求，当M 由静止释放并下降h （h 小于竖槽的深度）时，四角块的速度大小及相对原位置的位移．得图第 8 页 共 16 页11．（18分）在跳台边缘固定一个半径为R 的半圆形光滑柱面，一根不可伸长的细绳两端分别系物体A 和B ，且B A m 2m =，如图11所示．现从图示位置开始，由静止释放B 物体， （1）试证明A 物体在到达圆柱面的最高点前，就离开圆柱面． （2）A 物体离开圆柱面时，转过的角度θ变量与其他已知量之间的关系？（3）若A 物体刚好能到达最高点，求A m 和B m 的比值应满足的关系．得BAR图第 9 页第20届北京市高一物理竞赛（力学）决赛试题参考答案（北京二中杯）2007年5月27日一、填空题（每小题5分，共30分） 1．解：()220cos v gt α=+0第 10 页 共 16 页()2t 220g t a g sin v gt α==+ ；()0n 220gv a g cos v gt α==+2．解答：能．理由是：设A 不动，F 将使弹簧压缩x ，21Fx=kx 2，弹力kx=2F=4N ，大于3N ． 3．BF =1015N ．解：2AA v F mg mR-=)(212C A BB F F R mF +==ν2CC v F mg mR-=2222A B B Cv v v v -=- )(21222C A Bννν+=4．Bv=5gl ．解：22A B L 11mgmv 2mv 222=+A B v cos v sin θθ=2B 2glv 2tg θ=+5、203v 4gl；向左、l 3 ．解：01mv 2mv =1v v 2=012mv 2mv 3mv -=2v 0=222010113mgl=mv 2mv mv 224μ+=12a 2a =6、L =202GM R1+v R．解：0mv L mvR=22011GmMmv mv 22R=-2200v L 11GmMmv m 22R R⎛⎫=- ⎪⎝⎭二、计算题（共70分） 7．（12分）解： 2ma mg cos Nωθ=-（1）()21m a mga cos mga 2ωθ+= （2）平台为势能零点 联立求解：2221ma mga cos mga 22g(1cos )aωθθω+=-=（1）离开跳台：N 0=2g(1cos )N mamg cos 0aθθ-=-+=解出：2cos 3θ= 2arccos 3θ⎛⎫= ⎪⎝⎭（2） 222g(1)2g3a 3aω-==ag32=ω8．（12分）解：y 方向： f f cos N sin αα+= x方向：N cos f sin N'αα+=其中 f N μ= f 'N'μ=sin 'tg 1cos 2ααμμα===+9．（12分）解：0v 最小，且能抛过屋顶，在B点的速度方向应与BC 成045角，设在B 点速度为v2v sin 2BC gα=2BCgv 400sin 2α==mv 20s=22011mv mv mg(255)22=+-0mv 202s=设0v 与水平面夹角为θ0v cos v cos 45θ=1cos 2θ= 060θ= 00v sin 45v sin gtθ=-()t 231=- ()001L v cos60t 20223114.64m 2=⨯=⨯⨯-=10．（16分）设m 相对四角滑块的速度大小为v ，四角滑块相对地的速度为大小为v' 水平方向动量守恒：νννννννν'='='+'='-︒8221)60cos (M m m机械能守恒： ()()2200222011111m v sin60m v cos60v'Mv Mv'mv'Mgh mgh sin6022222+-+++=-M 2m=()2222213111113m 4v'm 4v'v'2m 4v'2mv'mv'mgh 222222222⎛⎫⎛⎫⎛⎫⨯+⨯-+++=- ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭)34(47121-='gh ν设四角块移动距离为S ：()()2122v'2a Svcos60v'2a hcos60S =-=-102a v'a v cos60v'=-00v'Sv cos60v'S hcos60=--+8h S =11．（18分）（1）若A 能到达最高点，则：22B A B B A A111m g R m gR m v m v 222π-=+B A v v v==解出：()22v gR 13π=-A在最高点时的向心力：()()2A A A 2gR 1v 32m m m g 13R Rππ-==-A在最高点实际能提供的最大向心力为：A m g()AA 2gm 1m g3π->所以A 不能到达最高点，应在到达最高点之前就离开圆柱面.（2）设A 在某一位置θ处离开柱面，其动能： 2A12m g R mgR sin 3mv 2θθ-=⨯()21v 4g R 2gR sin 3θθ=-离开时不受柱面作用力，只有重力的径向分力提供向心力：()2AA A m v m g sin m 4g 2g sin R 3θθθ==-5sin 4θθ=（3）若使A 到达最高点：()2BAAB1m g R m gR m m v 22π-=+ 且：A 恰能通过最高点，需满足：v gR= 代入上式：A B m 1m 3π-∴=即：A B m 1m 3π-< A 在到达最高点之前脱离BARθA B m 1m 3π-> A 可以顺利通过最高点.。

奥林匹克物理竞赛试题及答案国际奥林匹克物理竞赛是国际中学生的物理大赛，高中同学可以用来提升物理解题能力。

下面店铺给大家带来奥林匹克物理竞赛试题，希望对你有帮助。

奥林匹克物理竞赛试题国际物理奥林匹克竞赛简介竞赛设立由参赛成员国组成的国际物理奥林匹克委员会。

竞赛章程规定：目的是为增进中学物理教学的国际交流，通过竞赛促进开展物理学科的课外活动，以加强不同国家青年之间的友好关系和人民间的相互了解合作。

同时帮助参赛者发展物理方面的创造力，把从学校学到的知识用于解决实际问题的能力。

国际物理奥林匹克竞赛每年举办一次。

由各会员国轮流主办，并由各代表团团长和一名主办国指定的主席组成国际委员会。

国际委员会的任务是公平合理地评卷，监督章程规定的执行情况，决定竞赛结果。

每一会员国可选派5名高中学生或技术学校学生参加竞赛。

参加者的年龄到竞赛开始的那一天不能超过20岁。

参赛代表队要有2名团长，2名团长是国际委员会的成员，条件是能胜任解答赛题，能参加竞赛试卷的讨论和评分工作，并能通晓一种国际物理奥林匹克的工作语言。

国际物理奥林匹克的工作语言是英文、法文、德文和俄文。

代表团到达主办国时，团长要将参加学生及团长的情况告诉主办国家组织人员。

竞赛于每年6月底举行。

竞赛分两天进行。

第一天进行3道理论计算题竞赛，另一天的竞赛内容是1—2道实验题。

中间有一天的休息。

参赛者可使用计算尺、不带程序编制的计算器和对数表、物理常数表和制图工具，但不能使用数学和物理公式一览表。

竞赛题由参加国提供题目，主办国命题。

在竞赛前，赛题要保密。

竞赛题内容包括中学物理的4个部分(力学、热力学和分子物理学、光学及原子和核物理学、电磁学) ，解题要求用标准的中等数学而不要用高等数学。

主办国提出评卷标准并指定评卷人。

每题满分为10分。

各代表团团长同时对自己团员竞赛卷的复制品进行评定，最后协商决定成绩。

评奖标准是以参赛者前三名的平均分数计为100%，参赛者达90% 以上者为一等奖，78—90%者为二等奖，65—78%者为三等奖，同时发给证书。

选择题：1. 光的速度在以下哪个介质中最大？a) 真空b) 水c) 空气d) 金属2. 根据相对论，当物体以接近光速运动时，时间会_________。

a) 加速b) 减速c) 停止d) 没有改变3. 当一个物体在自由下落过程中，重力做功，而空气阻力不做功。

这是因为___________。

a) 重力大于空气阻力b) 重力等于空气阻力c) 重力小于空气阻力d) 空气阻力可以抵消重力4. 在电路中，电流由正极流向负极，这是因为___________。

a) 正电荷从正极流向负极b) 负电荷从正极流向负极c) 正电荷从负极流向正极d) 负电荷从负极流向正极5. 根据热力学第一定律，一个系统的内能变化等于传入系统的热量减去系统做功的大小。

这个定律也被称为___________。

a) 能量守恒定律b) 熵增定律c) 理想气体定律d) 等温定律填空题：1. 根据普朗克公式E = hf，其中E 是能量，h 是普朗克常数，f 是___________。

答案: 光的频率2. 波的衍射现象可以解释为光的_________ 。

答案: 波动性3. 根据库仑定律，两个电荷之间的电力相互作用与它们的__________ 乘积成反比。

答案: 电荷量4. 波的速度等于频率乘以波长，即v = λf。

当波长减小时，频率_________。

答案: 增大5. 根据能量守恒定律，能量不可以从一个热量温度低的物体传递到一个热量温度高的物体，这被称为_________。

答案: 热传导定律。

2020年第二十届全国初中物理竞赛试题2020年第二十届全国初中应用物理竞赛试题注意事项：1.请在密封线内填写所在地区、学校、姓名和考号。

2.用蓝色或黑色钢笔或圆珠笔书写。

3.本试卷共有六个大题，满分100分。

4.答卷时间：2020年3月28日（星期日）上午9:30-11:00一、选择题（每小题2分，共20分）1.2020年的诺贝尔物理学奖由三人分享，其中因在光纤通信技术方面所做的原创性工作而被称为“光纤之父”的美籍华人是(。

)A.XXXB.XXXC.XXXD.XXX2.晴朗无风的早晨，当飞机从空中飞过，在蔚蓝的天空中会留下一条长长的“尾巴”，飞机如图1所示，这种现象俗称为“飞机拉烟”。

产生这一现象的原因之一是飞机在飞行过程中排出的暖湿气体遇冷所致。

在这一过程中，暖湿气体发生的物态变化是(。

)A.熔化B.液化C.蒸发D.升华3.2020年1月2日起，我国北方大部地区遭遇大范围降雪天气袭击。

大雪严重影响了民航、铁路和高速公路等交通，如图2所示。

在遇到这种天气时，为了尽快清除积雪，常用的办法是使用“融雪盐”，这是因为(。

)A。

“融雪盐”与少量水发生化学反应，产生的热量使周围的冰雪熔化B。

“融雪盐”产生“保暖层”，使冰雪吸收足够的“地热”而熔化C。

使雪形成“含融雪盐的雪”，“含融雪盐的雪”熔点低于当地温度，使雪熔化D。

“融雪盐”有利于冰雪对阳光的吸收，从而加快冰雪的熔化4.XXX发现户外地面以上的冬季供热管道每隔一段距离总呈现波浪型，如图3所示。

其主要原因是(。

)A。

为了避开行人和建筑物B。

为了美观C。

为了避免管道因热胀冷缩导致的损坏D。

为了方便工人师傅安装和检修5.体操、投掷、攀岩等体育运动都不能缺少的“镁粉”，它的学名是碳酸镁。

体操运动员在上杠前都要在手上涂擦“镁粉”，其目的是(。

)A。

仅仅是为了利用“镁粉”，吸汗的作用，增加手和器械表面的摩擦而防止打滑B。

仅仅是为了利用手握着器械并急剧转动时“镁粉”，能起到衬垫作用，相当于在中间添加了一层“小球”做“滚动摩擦”C。

2020年河北省中考物理奥赛试题试卷学校：__________ 姓名：__________ 班级：__________ 考号：__________注意事项：1．答题前填写好自己的姓名、班级、考号等信息2．请将答案正确填写在答题卡上一、单选题1．对下列物理量估测符合实际的是............................................................................... （）A．正常人的体温是32℃B．你的脉搏每跳动一次的时间约1minC．正在考试用的一张物理答题卡的质量约100 gD．一张书桌的高度约0.8m2．如图所示的杠杆中，使用时费力的是（）3．我们知道，惯性是造成许多交通事故的原因。

下列各项规则中，不是为了防止惯性早晨交通事故的是......................................................................................................................... （）A．转向时，机动车应减速慢行B．车辆快速行驶时前后要保持车距C．车辆要靠道路右侧行驶D．小型客车的驾驶员和前排乘客必须系上安全带4．关于原子和原子核，以下叙述正确的是:（）A ．原子核位于原子的中央，带负电;B ．原子核外的电子带负电，并固定在某一位置;C ．原子核带正电，电子带负电;D ．原子核内有带正电的质子和带负电的中子.5．一只家用电能表标有“1500R/kWh”的字样，若只让一台录音机工作测得表盘转过一周所用的时间恰好为100秒，则主台录音机的电功率为 ....................................................... （）A．15W B．24W C．41.7W D．60.7W6．如右图所示，刚从酒精中拿出来的温度计示数会变小，这是因为温度计玻璃泡上的酒精 ............................................................................................................................................ （）A．熔化吸热B．汽化吸热C．升华吸热D．液化吸热7．下列关于声的说法中，错误的是（）A．声音是由于物体振动产生的B．人潜入水中，仍能听到岸上人的讲话声C．在空气中，声音传播的速度小于光传播的速度D．城市里可用噪声监测设备来消除噪声污染8．下列家用电器中应用电磁波工作的是：（）A．微波炉；B．电熨斗； C、空调；D．洗衣机。

历届国际物理奥林匹克竞赛试题与解答第１届（1967年于波兰的华沙）【题１】质量Ｍ＝0.2kg 的小球静置于垂直柱上，柱高ｈ＝5m 。

一粒质量ｍ＝0.01kg 、以速度0＝500m/s 飞行的子弹水平地穿过球心。

球落在距离柱s ＝20m 的地面上。

问子弹落在地面何处？子弹动能中有多少转换为热能？解：在所有碰撞情况下，系统的总动量均保持不变：MV mv mv +=0其中v 和V 分别是碰撞后子弹的速度和小球的速 度. 两者的飞行时间都是01.12==ght s 球在这段时间沿水平方向走过20m 的距离，故它在水平方向的速度为：8.1901.120==V （m/s ） 由方程0.01×500＝0.01v ＋0.2×19.8 可求出子弹在碰撞后的速度为：v ＝104m/s子弹也在1.01s 后落地，故它落在与柱的水平距离为S ＝vt ＝104×1.01＝105m 的地面上。

碰撞前子弹的初始动能为=2021mv 1250 J 球在刚碰撞后的动能为=221MV 39.2 J 子弹在刚碰撞后的动能为=221mv 54 J与初始动能相比，两者之差为1250 J －93.2 J ＝1156.8 J这表明原来动能的92.5%被系统吸收而变为热能。

这种碰撞不是完全非弹性碰撞。

在完全弹性碰撞的情形下，动能是守恒的。

而如果是完全非弹性碰撞，子弹将留在球内。

【题2】右图（甲）为无限的电阻网络，其中每个电阻均为r ，求Ａ、Ｂ两点间的总电阻。

解：如图（乙）所示Ａ、Ｂ两点间的总电阻应等于Ｃ、Ｄ两点间的总电阻与电阻ｒ的并联，再与ｒ串联 图（甲） 后的等效电阻。

如果网络是无限的，则Ａ、Ｂ两点间的总电阻应等于Ｃ、Ｄ 两点间的总电阻，设为Ｒx 。

根据它们的串并联关系有：m M h SsυABr r r r r r r rA B r r r r r r r r ＣＤrR rR r R x xx ++= 图（乙） 解上式可得： r R x 251+=【题3】给定两个同样的球，其一放在水平面上，另一个以细线悬挂。

Problems of the 20th International Physics Olympiad 1(Warsaw, 1989)Waldemar GorzkowskiInstitute of Physics, Polish Academy of Sciences, Warsaw, Poland 2AbstractThe article contains problems given at the 20th International Physics Olympiad (1989) and their solutions. The 20th IPhO was the third IPhO organized in Warsaw, Poland.LogoThe emblem of the XX InternationalPhysics Olympiad contains a picture that is ahistorical record of the first hypernuclear eventobserved and interpreted in Warsaw by M.Danysz and J. Pniewski3. The collision of ahigh-energy particle with a heavy nucleus wasregistered in nuclear emulsion. Tracks of thesecondary particles emitted in the event, seen inthe picture (upper star), consist of tracks due tofast pions (“thin tracks”) and to much slowerfragments of the target nucleus (“black tracks”).The “black track” connecting the upper star (greater) with the lower star (smaller) in the figure is due to a hypernuclear fragment, in this case due to a part of the primary nucleus containing an unstable hyperon Λinstead of a nucleon. Hyperfragments (hypernuclei) are a new kind of matter in which the nuclei contain not only protons and neutrons but also some other heavy particles.In the event observed above the hyperon Λ, bound with nucleon, decays like a free particle through a week (slow) process only. This fact strongly suggested the existence of a new quantum number that could explain suppression of the decay, even in presence of nucleons. Indeed, this was one of the observations that, 30 months later, led to the concept of strangeness.1This article has been sent for publication in Physics Competitions in October 20032e-mail: gorzk@.pl3M. Danysz and J. Pniewski, Bull. Acad. Polon. Sci., 3(1) 42 (1952) and Phil. Mag., 44, 348 (1953). Later the same physicists, Danysz and Pniewski, discovered the first case of a nucleus with two hyperons (double hyperfragment).IntroductionTheoretical problems (including solutions and marking schemes) were prepared especially for the 20th IPhO by Waldemar Gorzkowski. The experimental problem (including the solution and marking scheme) was prepared especially for this Olympiad by Andrzej Kotlicki. The problems were refereed independently (and many times) by at least two persons after any change was made in the text to avoid unexpected difficulties at the competition. This work was done by:First Problem:Andrzej Szadkowski, Andrzej Szymacha, Włodzimierz UngierSecond Problem:Andrzej Szadkowski, Andrzej Szymacha, Włodzimierz Ungier, Stanisław WoronowiczThird Problem:Andrzej Rajca, Andrzej Szymacha, Włodzimierz UngierExperimental Problem:Krzysztof Korona, Anna Lipniacka, Jerzy Łusakowski, Bruno SikoraSeveral English versions of the texts of the problems were given to the English-speaking students. As far as I know it happened for the first time (at present it is typical). The original English version was accepted (as a version for the students) by the leaders of the Australian delegation only. The other English-speaking delegations translated the English originals into English used in their countries. The net result was that there were at least four English versions. Of course, physics contained in them was exactly the same, while wording and spelling were somewhat different (the difference, however, were not too great).This article is based on the materials quoted at the end of the article and on personal notes of the author.THEORETICAL PROBLEMSProblem 1Consider two liquids A and B insoluble in each other. The pressures p i (i = A orB) of their saturated vapors obey, to a good approximation, the formula:i i o i Tp p βα+=)/ln(; i = A or B,where p o denotes the normal atmospheric pressure, T – the absolute temperature of the vapor, and i α and i β (i = A or B) – certain constants depending on the liquid. (The symbol ln denotes the natural logarithm, i.e. logarithm with base e =2.7182818…)The values of the ratio p i /p 0 for the liquids A and B at the temperature 40︒C and 90︒C are given in Tab. 1.1.Table 1.1The errors of these values are negligible.A. Determine the boiling temperatures of the liquids A and B under the pressure p 0.B. The liquids A and B were poured into a vessel in which the layersshown in Fig. 1.1 were formed. The surface of the liquid B has been coveredwith a thin layer of a non-volatile liquid C, which is insoluble in the liquids A and B and vice versa, thereby preventing any free evaporation from the uppersurface of the liquid B, The ratio of molecular masses of the liquids A and B (in the gaseous phase) is:.8/==BA μμγp 0 p 0Fig. 1.1Fig. 1.2The masses of the liquids A and B were initially the same, each equal to m = 100g. The heights of the layers of the liquids in the vessel and the densities of the liquids are small enough to make the assumption that the pressure in anypoint in the vessel is practically equal to the normal atmospheric pressure p0.The system of liquids in the vessel is slowly, but continuously and uniformly, heated. It was established that the temperature t of the liquids changed with timeτ as shown schematically in the Fig. 1.2.Determine the temperatures t1 and t2 corresponding to the horizontal parts of the diagram and the masses of the liquids A and B at the time τ1. Thetemperatures should be rounded to the nearest degree (in ︒C) and the masses of the liquids should be determined to one-tenth of gram.REMARK: Assume that the vapors of the liquids, to a good approximation,(1)obey the Dalton law stating that the pressure of a mixture of gases isequal to the sum of the partial pressures of the gases forming the mixture and(2)can be treated as perfect gases up to the pressures corresponding tothe saturated vapors.SolutionPART AThe liquid boils when the pressure of its saturated vapor is equal to the external pressure. Thus, in order to find the boiling temperature of the liquid i (i - A or B), one should determine such a temperature T bi(or t bi) for which p i/p0 = 1.Hence,Thus, the boiling temperature of the liquid A is equal toT= 3748.49K/10.711 ≈349.95 K.bAIn the Celsius scale the boiling temperature of the liquid A ist(349.95 – 273.15)︒C = 76.80︒C ≈77︒C.=bAFor the liquid B, in the same way, we obtain:α-5121.64 K,≈Bβ13.735,≈BT372-89 K,≈bBt99.74°C ≈100°C.≈bBPART BAs the liquids are in thermal contact with each other, their temperatures increase in time in the same way.At the beginning of the heating, what corresponds to the left sloped part of the diagram, no evaporation can occur. The free evaporation from the uppersurface of the liquid B cannot occur - it is impossible due to the layer of the non-volatile liquid C. The evaporation from the inside of the system is considered below.Table 1.2t67︒ C (with required accuracy).Therefore, ≈1Now we calculate the pressures of the saturated vapors of the liquids A and B at t67°C, i.e. the pressures of the saturated vapors of the liquids A the temperature ≈1and B in each bubble formed on the surface separating the liquids. From the equations (1) and (2), we get:≈A p 0.7340p ,≈B p 0.2670p ,)001.1(00p p p p B A ≈=+.Marking Scheme１．physical condition for boiling 1 point2．boiling temperature of the liquid A (numerical value) 1 point ３．boiling temperature of the liquid B (numerical value) 1 point ４．analysis of the phenomena at the temperaturet 3 points1５．numerical value oft11 point６．numerical value of the mass ratio of the saturated vapors in the bubble 1 pointτ 1 point ７．masses of the liquids at the time1８．determination of the temperaturet 1 point2REMARK: As the sum of the logarithms is not equal to the logarithm of the sum, the formula given in the text of the problem should not be applied to the mixture of the saturated vapors in the bubbles formed on the surface separating the liquids. However, the numerical data have been chosen in such a way that even such incorrect solution of the problem gives the correct value of the temperaturet(within required1accuracy). The purpose of that was to allow the pupils to solve the part B of the problem even if they determined the temperaturet in a wrong way. Of course, one1cannot receive any points for an incorrect determination of the temperaturet even if1its numerical value is correct.Typical mistakes in the pupils' solutionsNobody has received the maximum possible number of points for this problem, although several solutions came close. Only two participants tried to analyze proportion of pressures of the vapors during the upward movement of the bubble trough the liquid B. Part of the students confused Celsius degrees with Kelvins. Many participants did not take into account the boiling on the surface separating the liquids A and B, although this effect was the essence of theproblem. Part of the students, who did notice this effect, assumed a priori that the liquid with lower boiling temperature "must" be the first to evaporate. In general, this need not be true: if γwere, for example, 1/8 instead 8, then liquid A ratherthan B would remain in the vessel. As regards the boiling temperatures,practically nobody had any essential difficulties.Problem 2Three non-collinear points P 1, P 2 and P 3, with known masses m 1, m 2 andm 3, interact with one another through their mutual gravitational forces only; they are isolated in free space and do not interact with any other bodies. Let σ denote the axis going through the center-of-mass of the three masses, and perpendicular to the triangle P 1P 2P 3. What conditions should the angular velocities ω of the system (around the axis σ) and the distances:P 1P 2 = a 12, P 2P 3 = a 23, P 1P 3 = a 13,fulfill to allow the shape and size of the triangle P 1P 2P 3 unchanged during the motion of the system, i.e. under what conditions does the system rotate around the axis σ as a rigid body?SolutionAs the system is isolated, its total energy, i.e. the sum of the kinetic and potential energies, is conserved. The total potential energy of the points P 1, P 2 and P 3 with the masses 1m , 2m and 3m in the inertial system (i.e. when there are no inertial forces) is equal to the sum of the gravitational potential energies of all the pairs of points (P 1,P 2), (P 2,P 3) and (P 1,P 3). It depends only on the distances 12a , 23a and 23a which are constant in time. Thus, the total potential energy of the system is constant. As a consequence the kinetic energy of the system is constant too. The moment of inertia of the system with respect to the axis σ depends only on the distances from the points P 1, P 2 and P 3 to the axis σ which, for fixed 12a , 23a and 23a do not depend on time. This means that the moment of inertia I is constant. Therefore, the angular velocity of the system must also be constant:=ωconst. (1)This is the first condition we had to find. The other conditions will be determined by using three methods described below. However, prior to performingGM a =32ω (9)where321m m m M ++= (10)denotes the total mass of the system.In the same way, for the points P 2 and P 3, one gets the relations:a) the point P 2:1223a a =; GM a =32ωb) the point P 3:2313a a =; GM a =32ωSummarizing, the system can rotate as a rigid body if all the distances between the masses are equal:a a a a ===132312, (11)the angular velocity ω is constant and the relation (9) holds.SECOND METHODAt the beginning we find the moment of inertia I of the system with respect to the axis σ. Using the relation (2), we can write:222)(02323313121212323222221212332211r r r r r r r r r r r r m m m m m m m m m m m m +++++=++=Of course,22i i r =r i = 1, 2, 3and,011312313=-a a hence,The method described here does not differ essentially from the first method. In fact they are slight modifications of each other. However, it is interesting to notice how application of a proper mathematical language, e.g. the vector product, simplifies the calculations.The relation (9) can be called a “generalized Kepler’s law” as, in fact, it is very similar to the Kepler’s law but with respect to the many -body system. As far as I know this generalized Kepler’s law was presented for the first time right at the 20th IPhO.Marking scheme1. the proof that =ωconst 1 point2. the conditions at the equilibrium (conditions for the forcesand their moments or extremum of the total potential energy) 3 points3. the proof of the relation a a ij = 4 points4. the proof of the relation GM a =32ω 2 pointsRemarks and typical mistakes in the pupils' solutionsNo type of error was observed as predominant in the pupils' solutions.Practically all the mistakes can be put down to the students' scant experience in calculations and general lack of skill. Several students misunderstood the text of the problem and attempted to prove that the three masses should be equal. Of course, this was impossible. Moreover, it was pointless, since the masses were given. Almost all the participants tried to solve the problem by analyzingequilibrium of forces and/or their moments. Only one student tried to solve the problem by looking for a minimum of the total potential energy (unfortunately, his solution was not fully correct). Several participants solved the problem using a convenient reference system: one mass in the origin and one mass on the x -axis. One of them received a special prize.Problem 3The problem concerns investigation of transforming the electronmicroscope with magnetic guiding of the electron beam (which is accelerated with the potential difference U = 511 kV) into a proton microscope (in which the proton beam is accelerated with the potential difference –U ). For this purpose, solve the following two problems:A. An electron after leaving a device, which accelerated it with thepotential difference U , falls into a region with an inhomogeneous field Bgenerated with a system of stationary coils L 1, L 2, … , L n . The known currents in the coils are i 1, i 2, … , i n , respectively.What should the currents i 1’, i 2’, … , i n ’ in the coils L 1, L 2, … , L n be, inorder to guide the proton (initially accelerated with the potential difference –U ) along the same trajectory (and in the same direction) as that of the electron?HINT: The problem can be solved by finding a condition under which theequation describing the trajectory is the same in both cases. It may be helpful to use the relation:p dt d p = dt d 21p 2 = dtd 21p 2.B. How many times would the resolving power of the above microscopeincrease or decrease if the electron beam were replaced with the proton beam? Assume that the resolving power of the microscope (i.e. the smallest distance between two point objects whose circular images can be just separated) depends only on the wave properties of the particles.Assume that the velocities of the electrons and protons before theiracceleration are zero, and that there is no interaction between own magneticmoment of either electrons or protons and the magnetic field. Assume also that the electromagnetic radiation emitted by the moving particles can be neglected.NOTE: Very often physicists use 1 electron-volt (1 eV), and its derivativessuch as 1 keV or 1 MeV, as a unit of energy. 1 electron-volt is the energy gained by the electron that passed the potential difference equal to 1 V.Perform the calculations assuming the following data:Rest energy of electron: E e = m e c 2 = 511 keVRest energy of proton: E p = m p c 2 = 938 MeVSolutionPART AAt the beginning one should notice that the kinetic energy of the electron accelerated with the potential difference U = 511 kV equals to its rest energy 0E . Therefore, at least in the case of the electron, the laws of the classical physics cannot be applied. It is necessary to use relativistic laws.The relativistic equation of motion of a particle with the charge e in the magnetic field B has the following form:L dtd F p =It means that the value of the particle momentum (and the value of the velocity) is constant during the motion:==γv m p 0 const; v = const.The same result can be obtained without any formulae in the following way:The Lorentz force L F is perpendicular to the velocity v (and to theis the same in both cases.where c denotes the velocity of light.It is worthwhile to notice that our protons are 'almost classical', because their kinetic energy )(e k E E = is small compared to the proton rest energy p E . Thus, one can expect that the momentum of the proton can be determined, with a good accuracy, from the classical considerations. We have:.212121'22p e e p e e p e e p e p eE E c E E c E E E c E E E E E c E p =≈+=⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛+=In accordance with our expectations, we have obtained the same result as above.that λis inversely proportional to the momentum of the particle. Therefore, after replacing the electron beam with the proton beam the resolving power will be changed by the factor p/p' ≈1/35. It means that our proton microscope would allow observation of the objects about 35 times smaller than the electron microscope.Marking scheme1. the relativistic equation of motion 1 point2. independence of p and v of the time 1 point3. identity of e B/p in both cases 2 points4. scaling of the fields and the currents with the same factor 1 point5. determination of the momenta (relativistically) 1 point6. the ratio of the momenta (numerically) 1 point7. proportionality of the resolving power toλ 1 point8. inverse proportionality of λto p 1 point9. scaling of the resolving power 1 pointRemarks and typical mistakes in the pupils' solutionsSome of the participants tried to solve the problem by using laws of classicalmechanics only. Of course, this approach was entirely wrong. Some studentstried to find the required condition by equating "accelerations" of particles inboth cases. They understood the "acceleration" of the particle as a ratio of theforce acting on the particle to the "relativistic" mass of the particle. Thisapproach is incorrect. First, in relativistic physics the relationship between force and acceleration is more complicated. It deals with not one "relativistic" mass, but with two "relativistic" masses: transverse and longitudinal. Secondly, identity of trajectories need not require equality of accelerations.The actual condition, i.e. the identity of e B/p in both cases, can be obtained from the following two requirements:1° in any given point of the trajectory the curvature should be the same in both cases;2° in the vicinity of any given point the plane containing a small arc of the trajectory should be oriented in space in both cases in the same way.Most of the students followed the approach described just above. Unfortunately, many forgot about the second requirement (they neglected the vector character of the quantity e B/p).EXPERIMENTAL PROBLEM 4The following equipment is provided:1. Two piezoelectric discs of thickness 10 mm with evaporated electrodes (Fig. 4.1) fixed in holders on the jaws of the calipers;Fig. 4.12. The calibrated sine wave oscillator with a photograph of the control panel, explaining the functions of the switches and regulators;3. A double channel oscilloscope with a photograph of the control panel, explaining the functions of the switches and regulators;4. Two closed plastic bags containing liquids;5. A beaker with glycerin (for wetting the discs surfaces to allow better mechanical coupling);6. Cables and a three way connector;7. A stand for support the bags with the liquids;8. Support and calipers.A piezoelectric material changes its linear dimensions under the influence of an electric field and vice-versa, the distortion of a piezoelectric material induces an electrical field. Therefore, it is possible to excite the mechanical vibrations in a piezoelectric material by applying an alternating electric field, and also to induce an alternating electric field by mechanical vibrations.A. Knowing that the velocity of longitudinal ultrasonic waves in the material of the disc is about 3104 m/s, estimate roughly the resonant frequency of the mechanical vibrations parallel to the disc axis. Assume that the disc holders do no4 The Organizing Committee planned to give another experimental problem: a problem on high T c superconductivity. Unfortunately, the samples of superconductors, prepared that time by a factory, were of very poor quality. Moreover, they were provided after a long delay. Because of that the organizers decided to use this problem, which was also prepared, but considered as a second choice.10 mm Electrodesrestrict the vibrations. (Note that other types of resonant vibrations with lower or higher frequencies may occur in the discs.)Using your estimation, determine experimentally the frequency for which the piezoelectric discs work best as a transmitter-receiver set for ultrasound in the liquid. Wetting surfaces of the discs before putting them against the bags improves penetration of the liquid in the bag by ultrasound.B.Determine the velocity of ultrasound for both liquids without opening the bags and estimate the error.C. Determine the ratio of the ultrasound velocities for both liquids and its error.Complete carefully the synopsis sheet. Your report should, apart from the synopsis sheet, contain the descriptions of:-method of resonant frequency estimation;-methods of measurements;-methods of estimating errors of the measured quantities and of final results.Remember to define all the used quantities and to explain the symbols.5In the real Synopsis Sheet the students had more space for filling.Solution (draft)6A. As the holders do not affect vibrations of the disc we may expectantinodes on the flat surfaces of the discs (Fig. 4.2; geometric proportions not conserved). One of the frequencies is expected forfv l 221==λ,where v denotes the velocity of longitudinal ultrasonic wave (its value is given in the text of the problem), f - the frequency and l - the thickness of the disc. Thus:lv f 2=.Numerically 5102⋅=f Hz = 200 kHz.6 This draft solution is based on the camera-ready text of the more detailed solution prepared by Dr. Andrzej Kotlicki and published in the proceedings [3]l = λ/2Fig. 4.2One should stress out that different modes of vibrations can be excited in the disc with height comparable to its diameter. We confine our considerations to the modes related to longitudinal waves moving along the axis of the disc as the sound waves in liquids are longitudinal. We neglect coupling between different modes and require antinodes exactly at the flat parts of the disc. We assume also that the piezoelectric effect does not affect velocity of ultrasound. For these reasons the frequency just determined should be treated as only a rough approximation. However, it indicates that one should look for the resonance in vicinity of 200 kHz.The experimental set-up is shown in Fig. 4.3. The oscillator (generator) is connected to one of the discs that works as a transmitter and to one channel of the oscilloscope. The second disc is connected to the second channel of the oscilloscope and works as a receiver. Both discs are placed against one of the bags with liquid (Fig. 4.4). The distance d can be varied.Fig. 4.3Fig. 4.4One searches for the resonance by slowly changing the frequency of theoscillator in the range 100 – 1000 kHz and watching the signal on theoscilloscope. In this way the students could find a strong resonance at frequency 220≈f kHz. Other resonance peaks could be found at about 110 kHz and 670 kHz. They should have been neglected as they are substantially weaker. (They correspond to some other modes of vibrations.) Accuracy of these measurements was 10 kHz (due to the width of the resonance and the accuracy of the scale on the generator).B. The ultrasonic waves pass through the liquid and generate anelectric signal in the receiver. Using the same set-up (Fig. 4.3 and 4.4) we can measure dependence of the phase shift between the signals at Y 1 and Y 2 vs.distance between the piezoelectric discs d at the constant frequency found in point A . This phase shift is 0/2ϕπϕ+=∆l v df , where l v denotes velocity of ultrasound in the liquid. 0ϕ denotes a constant phase shift occurring whenultrasound passes trough the bag walls (possibly zero). The graph representing dependence )(ϕ∆d should be a straight line. Its slope allows to determine l v and its error. In general, the measurements of ϕ∆ are difficult for manyreflections in the bag, which perturb the signal. One of the best ways is tomeasure d only for πϕn =∆ (n - integer) as such points can be found rather easy. Many technical details concerning measurements can be found in [3] (pp. 37 and 38).The liquids given to the students were water and glycerin. In the standardsolution the author of the problem received the following values:v water = 310)10.050.1(⋅± m/s; v glycerin = 310)10.096.1(⋅± m/s.The ratio of these values was 1531.1 ..0The ultrasonic waves are partly reflected or scattered by the walls of the bag. This effect somewhat affects measurements of the phase shift. To minimize its role one can measure the phase shift (for a given distance) or distance (at the same phase shift) several times, each time changing the shape of the bag. Asregards errors in determination of velocities it is worth to mention that the most important factor affecting them was the error in determination of the frequency.This error, however, practically does not affect the ratio of velocities.Marking SchemeFrequency estimation1.Formula 1 point2.Result (with units) 1 point3.Method of experimental determining the resonance frequency 1 point4.Result (if within 5% of standard value) 2 points5.Error 1 pointMeasurements of velocities1.Explanation of the method 2 points2.Proper number of measurements in each series 3 points3.Result for velocity in the first liquid (if within 5% of standard value) 2 points4.Error of the above 1 point5.Result for velocity in the second liquid (if within 5% of standard value) 2 points6.Error of the above 1 pointRatio of velocities1.Result (if within 3% of standard value) 2 points2.Error of the above 1 pointTypical mistakesThe results of this problem were very good (more than a half of competitors obtained more than 15 points). Nevertheless, many studentsencountered some difficulties in estimation of the frequency. Some of themassumed presence of nodes at the flat surfaces of the discs (this assumption is not adequate to the situation, but accidentally gives proper formula). In part B some students tried to find distances between nodes and antinodes for ultrasonicstanding wave in the liquid. This approach gave false results as the pattern ofstanding waves in the bag is very complicated and changes when the shape of the bag is changed.AcknowledgementI would like to thank very warmly to Prof. Jan Mostowski and Dr. Andrzej Wysmołek for reading the text of this article and for valuable critical remarks. I express special thanks to Dr. Andrzej Kotlicki for critical reviewing the experimental part of the article and for a number of very important improvements.Literature[1] Waldemar Gorzkowski and Andrzej Kotlicki, XX Międzynarodowa Olimpiada Fizyczna - cz. I, Fizyka w Szkole nr1/90, pp. 34 - 39[2] Waldemar Gorzkowski,XX Międzynarodowa Olimpiada Fizyc zna - cz. II, Fizyka w Szkole nr2/3-90, pp. 23 - 32[3] XX International Physics Olympiad - Proceedings of the XX International Physics Olympiad, Warsaw (Poland), July 16 - 24, 1989, ed. by W. Gorzkowski, World Scientific Publishing Company, Singapore 1990 [ISBN 981-02-0084-6]。