IPhO2014-第45届国际物理奥林匹克竞赛理论试题与解答
国际物理林匹克竞赛试题25th_IPhO_1994
THE EXAMINATIONXXV INTERNATIONAL PHYSICS OLYMPIADBEIJING, PERPLE’S REPUBLIC CHINATHEORETICAL COMPETITIONJuly 13, 1994Time available: 5 hoursREAD THIS FIRST!INSTRUCTIONS:1. Use only the ball pen provided.2. Your graphs should be drawn on the answer sheets attached to the problem.3. Your solutions should be written on the sheets of paper attached to the problems.4. Write at the top of the first page of each problem:●The total number of pages in your solution to the problem●Your name and code numberTheoretical Problem 1RELATIVISTIC PARTICLEIn the theory of special relativity the relation between energy E and momentum P or a free particle with rest mass m 0 is242022mc c m c p E =+=When such a particle is subject to a conservative force, the total energy of the particle, which is the sum of42022c m c p + and the potential energy, is conserved. Ifthe energy of the particle is very high, the rest energy of the particle can be ignored (such a particle is called an ultra relativistic particle).1) consider the one dimensional motion of a very high energy particle (in which rest energy can be neglected) subject to an attractive central force of constant magnitude f . Suppose the particle is located at the centre of force with initial momentum p 0 at time t =0. Describe the motion of the particle by separately plotting, for at least one period of the motion: x against time t , and momentum p against space coordinate x . Specify the coordinates of the “turning points ” in terms of given parameters p 0 and f . Indicate, with arrows, the direction of the progress of the mothon in the (p , x ) diagram. There may be short intervals of time during which the particle is not ultrarelativistic. However, these should be neglected.Use Answer Sheet 1.2) A meson is a particle made up of two quarks. The rest mass M of the meson is equal to the total energy of the two-quark system divided by c 2.Consider a one--dimensional model for a meson at rest, in which the two quarks are assumed to move along the x -axis and attract each other with a force of constant magnitude f It is assumed they can pass through each other freely. For analysis of the high energy motion of the quarks the rest mass of the quarks can be neglected. At time t=0 the two quarks are both at x =0. Show separately the motion of the two quarks graphically by a (x , t ) diagram and a (p , x ) diagram, specify the coordinates of the “turning points ” in terms of M and f , indicate the direction of the process in your (p , x ) diagram, and determine the maximum distance between the two quarks. Use Answer Sheet 2.3) The reference frame used in part 2 will be referred to as frame S , the Lab frame, referred to as S , moves in the negative x -direction with a constant velocity v =0.6c . the coordinates in the two reference frames are so chosen that the pointx =0 in S coincides with the point 0='x in S '' at time 0='=t t . Plot the motion of the two quarks graphically in a (x ', t ') diagram. Specify the coordinates of the turning points in terms of M , f and c , and determine the maximum distance between the two quarks observed in Lab frame S '. Use Answer Sheet 3.The coordinates of particle observed in reference frames S and S '' are related by the Lorentz transformation⎪⎩⎪⎨⎧+='+=')()(c x t t ct x x βγβγwhere c v /=β,21/1βγ-= and v is the velocity of frame S moving relative to the frame S ''.4) For a meson with rest energy Mc 2=140 MeV and velocity 0.60c relative to theLab frame S '', determine its energy E ' in the Lab Frame S ''.ANSWER SHEET 1 ANSWER SHEET 2 1)2)OOtOxptx 1, x 2Ox 1p 1 Ox 2p 2Quark1Quark2The maximum distance betweenthe two quarks is d =ANSWER SHEET 3 3)Theoretical Problem 1—Solution1) 1a. Taking the force center as the origin of the space coordinate x and the zero potential point, the potential energy of the particle is||)(x f x U =(1)The total energy is||42022x f c m c p W ++=.1b. Neglecting the rest energy, we get||||x f c p W +=,(2)Since W is conserved throughout the motion, so we havec p x f c p W 0||||=+=,(3)Let the x axis be in the direction of the initial momentum of the particle,cp fx pc c p fx pc c p fx pc cp fx pc 0000=--=-=+-=+⎪⎪⎭⎪⎪⎬⎫<<><<>>>.0,0;0,0;0,0;0,0p x p x p x p x (4)The maximum distance of the particle from the origin, let it be L , corresponds to p =0. It isf c p L /0=.1c. From Eq. 3 and Newton ’s law⎩⎨⎧<>-==;0,;0,x f x f F dt dp(5)we can get the speed of the particle asc dtdp f c dt dx ==, (6)tx 1′, x 2′OThe maximum distance between the two quarks observed in S ′frame is d ′=when when when wheni.e. the particle with very high energy always moves with the speed of light except that it is in the region extremely close to the points L x ±=. The time for the particle to move from origin to the point L x =, let it be denoted by τ, isf p c L //0==τ.So the particle moves to and for between L x = and L x -= with speed c and periodf p /440=τ. The relation between x and t is⎪⎪⎭⎪⎪⎬⎫≤≤-=≤≤-=≤≤-=≤≤=,43,4,32,2,2,20,τττττττt L ct x t ct L x t ct L x t ct x(7)The required answer is thus as given in Fig. 1 and Fig. 2.Fig. 12) The total energy of the two-quark system can be expressed as||||||21212x x f c p c p Mc -++=,(8)where 1x ,2x are the position coordinates and 1p , 2p are the momenta of quark 1 and quark 2 respectively. For the rest meson, the total momentum of the two quarks is zero and the two quarks move symmetrically in opposite directions, we have021=+=p p p , 21p p -=, 21x x -=.(9)Let p 0 denote the momentum of the quark 1 when it is at x =0, then we havec p Mc 022=or2/0Mc p =(10)From Eq. 8, 9 and 10, the half of the total energy can be expressed in terms of 1p and1x of quark 1:||||110x f c p c p +=,(11)just as though it is a one particle problem as in part 1 (Eq. 3) with initial momentumL= p 0/fc/f2/0Mc p =. From the answer in part 1 we get the (x , t ) diagram and (p , x ) diagram of the motion of quark 1 as shown in Figs. 3 and 4. For quark 2 the situation is similar except that the signs are reversed for both x and p ; its (x , t ) and (p , x ) diagrams are shown in Figs. 3 and 4.The maximum distance between the two quarks as seen from Fig. 3 isf Mc f c p L d //2220===. (12)Fig. 3Fig. 4a Quark1 Fig. 4bQuark23) The reference frame S moves with a constant velocity V =0.6c relative to the Lab frame S '' in the x ' axis direction, and the origins of the two frames are coincident at the beginning (0='=t t ). The Lorentz transformation between these two frames is given by:),/(),(c x t t ct x x βγβγ+='+='(13)where c V /=β, and 21/1βγ-=. With c V 6.0=, we have 5/3=β, and4/5=γ. Since the Lorenta transformation is linear, a straight line in the (x , t ) diagram2L=Mc 2/2fx 1: solid linetransforms into a straight line the (x ', t ') diagram, thus we need only to calculate the coordinates of the turning points in the frame S '.For quark 1, the coordinates of the turning points in the frames S and S ' are as follows:Frame SFrame S ' 1x 1t)(111ct x x βγ+=' )/(111e x t t βγ+=' 114345ct x +=c x t /434511+= 0 0 0L τL L 2)1(=+βγ ττβγ2)1(=+τ2L L 232=γβτγτ252=L - τ3 L L =-)13(βγττβγ3)3(=-0 τ4L L 34=γβτγτ54=where f Mc f c p L 2//20==, f Mc f p 2//0==τ.For quark 2, we have Frame SFrame S ' 2x 2t)(222ct x x βγ+=' )/(222c x t t βγ+=' 224345ct x +=c x t /434522+= 0 0 0L - τL L 21)1(-=--βγττβγ21)1(=-τ2 L L 232=γβτγτ252=L τ3 L L 27)13(=+βγττβγ29)3(=+τ4L L 34=γβτγτ54=With the above results, the (x ', t ') diagrams of the two quarks are shown in Fig. 5.The equations of the straight lines OA and OB are:t c t x '='')(1; ττβγ2)1(0=+≤'≤t ; (14a) t c t x '-='')(2;ττβγ21)1(0=-≤'≤t(14b)The distance between the two quarks attains its maximum d ' when τ21='t , thus wehave maximum distancefMc L c d 2)1(2)1(22=-=-='βγτβγ. (15)Fig. 54) It is given the meson moves with velocity V =0.6 crelative to the Lab frame, its energy measured in the Lab frame is1751408.01122=⨯=-='βMc E MeV . Grading SchemePart 1 2 points, distributed as follows:0.4 point for the shape of x (t ) in Fig. 1; 0.3 point for 4 equal intervals in Fig. 1; (0.3 for correct derivation of the formula only)0.1 each for the coordinates of the turning points A and C , 0.4 point in total; 0.4 point for the shape of p (x ) in fig. 2; (0.2 for correct derivation only)0.1 each for specification of 0p , f c p L /0=, 0p -, L - and arrows, 0.5 pointin total.(0.05 each for correct calculations of coordinate of turning points only).Part 2 4 points, distributed as follows:0.6 each for the shape of )(1t x and )(2t x , 1.2 points in total;0.1 each for the coordinates of the turning points A, B, D and E in Fig. 3, 0.8 pointin total;0.3 each for the shape of )(11x p and )(22x p , 0.6 point in total;0.1 each for 2/0Mc p =, f Mc L 2/2=, 0p -, L - and arrows in Fig. 4a andFig. 4b, 1 point in total; 0.4 point for f Mc d /2=Part 3 3 point, distributed as follows:0.8 each for the shape of )(1t x '' and )(2t x '', 1.6 points in total; 0.1 each for the coordinates of the turning points A, B, D and E in Fig. 5, 0.8 pointin total; (0.05 each for correct calculations of coordinate of turning points only).0.6 point for f Mc d 2/2='.Part 4 1 point (0.5 point for the calculation formula; 0.5 point for the numerical valueand units)Theoretical Problem 2SUPERCONDUCTING MAGNETSuper conducting magnets are widely used in laboratories. The most common form of super conducting magnets is a solenoid made of super conducting wire. The wonderful thing about a superconducting magnet is that it produces high magnetic fields without any energy dissipation due to Joule heating, since the electrical resistance of the superconducting wire becomes zero when the magnet is immersed in liquid helium at a temperature of 4.2 K. Usually, the magnet is provided with a specially designed superconducting switch, as shown in Fig. 1. The resistance r of the switch can be controlled: either r =0 in the superconducting state, or n r r = in the normal state. When the persistent mode, with a current circulating through the magnet and superconducting switch indefinitely. The persistent mode allows a steady magnetic field to be maintained for long periods with the external source cut off.The details of the superconducting switch are not given in Fig. 1. It is usually a small length of superconducting wire wrapped with a heater wire and suitably thermally insulated from the liquid helium bath. On being heated, the temperature of the superconducting wire increases and it reverts to the resistive normal state. The typical value of n r is a few ohms. Here we assume it to be 5Ω. The inductance of a superconducting magnet depends on its size; assume it be 10 H for the magnet in Fig. 1. The total current I can be changed by adjusting the resistance R . This problem will be graded by the plots only!The arrows denote the positive direction of I , I 1 and I 2.Fig. 11) If the total current I and the resistance r of the superconducting switch are controlledto vary with time in the way shown in Figs, 2a and 2b respectively, and assuming the currents I 1 and I 2 flowing through the magnet and the switch respectively are equal at the beginning (Fig. 2c and Fig. 2d), how do they vary with time from t 1 to t 4? Plot your answer in Fig. 2c and Fig. 2d2) Suppose the power switch K is turned on at time t =0 when r =0, I 1=0 and R =7.5Ω, and the total current I is 0.5A. With K kept closed, the resistance r of the superconducting switch is varied in he way shown in Fig. 3b. Plot the corresponding time dependences of I , I 1 and I 2in Figs. 3a, 3c and 3d respectively.3) Only small currents, less than 0.5A, are allowed to flow through theFig.2a2b 2c 2dFig. 3a3b 3c 3dsuperconducting switch when it is in the normal state, with larger currents the switch will be burnt out. Suppose the superconducting magnet is operated in a persistent mode, i. e. I =0, and I 1=i 1(e. g. 20A), I 2=-i 1, as shown in Fig. 4, from t =0 to t =3min. If the experiment is to be stopped by reducting the current through the magnet to zero, how would you do it? This has to be done in several operation steps. Plot the corresponding changes of I , r , I 1 and I 2in Fig. 44) Suppose the magnet is operated in a persistent mode with a persistent current of 20A (t =0 to t =3min. See Fig. 5). How would you change it to a persistent mode with acurrent of 30a? plot your answer in Fig. 5.Fig. 4a4b 4c 4dFig. 5a5b 5cTheoretical Problem 2—Solution 1) For t =t 1 to t 3Since 0=r , the voltage across the magnet dt LdI V M /1==0, therefore,011121)(I t I I ==; 01221I I I I I -=-=.For t =t 3 to t 4Since I 2=0 at t =t 3, and I is kept at 021I after3t t =, 02==n M r I V , therefore, 1I and 2I will not change.0121I I =; 02=IThese results are shown in Fig. 6.5dFig. 6a6b 6c 6d2) For 0=t to 1=t min: Since 0=r , 0/1==dt LdI V M0)0(11==I I5.012=-=I I I A.At 1=t min, r suddenly jumps from O to n r , I will drop from R E / to)/(n r R E + instantaneously, because 1I can not change abruptly due to the inductance of the magnet coil. For R E /=0.5A, Ω=5.7R and Ω=5n R . I will drop to 0.3A. For 1=t min to 2 min:I , 1I and 2I gradually approach their steady values:5.0==REI A, 5.01==I I A 02=I .The time constantnn Rr r R L )(+=τ. When 10=L H, Ω=5.7R and Ω=5n r , 3=τsec. For 2=t min to 3 min:Since 0=r , 1I and 2I will not change, that is5.01=I A and 02=IFig. 7a7b3) The operation steps are: First stepTurn on power switch K , and increase the total current I to 20 A, i. e. equal to 1I .Since the superconducting switch is in the state 0=r , so that L V M = 0/1=dt dI , that is, 1I can not change and 2I increases by 20A, in other words, 2I changes from 20- A to zero. Second stepSwitch r from 0 to n r . Third stepGradually reduce I to zero while keeping 5.02<I A: since n M r V I /2= anddt dI L V m /1=, when 10=L H, Ω=5n r , the requirement 5.02<I A corresponds to 25.0/1<dt dI A/sec, that is, a drop of <15A in 1 min. In Fig. 8 dt dI /～0.1A/sec and dt dI /1 is around this value too, so the requirement has been fulfilled. Final stepSwitch r to zero when 0=M V and turn off the power switch K . These resultsare shown in Fig. 8.7c 7dFig. 8a8b 8c4) First step and second step are the same as that in part 3, resulting in 02=I .Third step Increase I by 10 A to 30 A with a rate subject to the requirement5.02<I A. Fourth step Switch r to zero when 0=M V .Fifth step Reduce I to zero, 301=I A will not change because M V is zero.12I I I -= will change to 30- A. The current flowing through the magnet is thus closed by the superconducting switch. Final step Turn off the power switch K . The magnet is operating in the persistent mode.These results are shown in Fig. 9.Grading SchemePart 1,2 points:0.5 point for each of 1I , 2I from 1t t = to 3t and 1I , 2I from 3t t = to 4t .Part 2,3 points:0.3 point for each of 1I , 2I from 0=t to 1 min, I , 1I , 2I at 1=t min,8dFig. 9a9b 9c 9dand 0I , 1I , 2I from 1=t to 2 min;0.2 point for each of I , 1I , and 2I from 2=t to 3 min. Part 3, 2 points: 0.25 point for each section in Fig. 8 from 3=t to 9 min, 8 sections in total. Part 4,3 points:0.25 point for each section in Fig. 9 from 3=t to 12 min, 12 sections in total.Theoretical Problem 3COLLISION OF DISCS WITH SURFACE FRICTIONA homogeneous disc A of mass m and radius R A moves translationally on a smoothhorizontal x -y plane in the x direction with a velocity V (see the figure on the next page). The center of the disk is at a distance b from the x-axis. It collides with a stationary homogeneous disc B whose center is initially located at the origin of the coordinate system. The disc B has the same mass and the same thickness as A, but its radius is R B . It is assumed that the velocities of the discs at their point of contact, in the direction perpendicular to the line joining their centers, are equal after the collision. It is also assumed that the magnitudes of the relative velocities of the discs along the line joining their centers are the same before and after the collision.1) For such a collision determine the X and Y components of the velocities of the twodiscs after the collision, i. e. AXV ', AY V ', BX V ' and BY V ' in terms of m , A R , B R , V and b .2) Determine the kinetic energies A E ' for disc A and B E ' for disc B after the collision in terms of m , A R , B R , V and b .Theoretical Problem 3—Solution1) When disc A collides with disc B, let n be the unit vector along the normal to the surfaces at the point of contact and t be the tangential unit vector as shown in the figure. Let ϕ be the angle between n and the x axis. Then we haveϕsin )(B A R R b +=The momentum components of A and B along n and t before collision are:0,cos ==Bn An mV mV mV ϕ,0,sin ==Bt At mV mV mV ϕ.Denote the corresponding momentum components of A and B after collision byAnV m ', Bn V m ', At V m ', and Bt V m '. Let A ω and B ω be the angular velocities of A and B about the axes through their centers after collision, and A I and B I be their corresponding moments of intertia. Then,221A A mR I =,221B B mR I =The conservation of momentum givesBn AnV m V m mV '+'=ϕcos , (1) tn AtV m V m mV '+'=ϕsin ,(2)The conservation of angular momentum about the axis through O givesB B A A B A AtI I R R V m mVb ωω+++'=)( (3)The impulse of the friction force exerted on B during collision will cause amomentum change of AtV m ' along t and produces an angular momentum B B I ω simultaneously. They are related by.B B b BtI R V m ω=' (4)During the collision at the point of contact A and B acquires the same tangential velocities, so we haveB B Bt A A AtR V R V ωω-'=-' (5)It is given that the magnitudes of the relative velocities along the normal directionof the two discs before and after collision are equal, i. e.An BnV V V '-'=ϕcos . (6)From Eqs. 1 and 6 we get0='AnV , ϕcos V V Bn='. From Eqs. 2 to 5, we get ϕsin 65V V At=', ϕsin 61V V Bt=',A A R V 3sin ϕω=,BB R V 3sin ϕω=.The x and y components of the velocities after collision are:,)(65sin cos 22B A At An Ax R R Vb V V V +='+'='ϕϕ (7)222)(6)(5cos sin B A B A At An AyR R b R R Vb V V V +-+='+'-='ϕϕ, (8)⎥⎦⎤⎢⎣⎡+-='+'='22)(651sin cos B A Bt Bn Bx R R b V V V ϕϕ, (9)222)(6)(5cos sin B A B A Bt Bn By R R b R R Vb V V V +-+-='+'-='ϕϕ,(10)2) After the collision, the kinetic energy of disc A is222222)(8321)(21B A A A Ay Ax A R R b mV I V V m E +=+'+'='ω(11)while the kinetic energy of disc B is⎥⎦⎤⎢⎣⎡+-=+'+'='222222)(121112121)(21B A B B By Bx BR R b mV I V V m E ω (12)Grading Scheme1. After the collision, the velocity components of discs A and B are shown in Eq. 7, 8, 9 and 10 of the solution respectively. The total points of this part is 8. 0. If the result in which all four velocity components are correct has not been obtained, the point is marked according to the following rules.0.8 point for each correct velocity component;0.8 point for the correct description of that the magnitudes of the relative velocities of the discs along the line joining their centers are the same before and after the collision.0.8 point for the correct description of the conservation for angular momentum;0.8 point for the correct description of the equal tangential velocity at the touching point;0.8 point for the correct description of the relation between the impulse and the moment of the impulse.2. After the collision, the kinetic energies of disc A and disc B are shown in Eqs. 11 and 12 of the solution respectively.1.0 point for the correct kinetic energies of disc A;1.0 point for the correct kinetic energies of disc B;The total points of this part is 2.0XXV INTERNATIONAL PHYSICS OLYMPIADBEIJING, PEOPLE’S REPUBLIC OF CHINAPRACTICAL COMPETITIONJuly 15, 1994Time available: 2.5 hoursREAD THIS FIRST!INSTRUCTIONS:1. Use only the ball pen provided.2. Your graphs should be drawn on the answer sheets attached to the problem.3. Write your solution on the marked side of the paper only.4. The draft papers are provided for doing numerical calculations and draft drawings.5. Write at the top of every page:●The number of the problem●The number of the page of your report in each problem●The total number of pages in your report to the problem●Your name and code numberEXPERIMENTAL PROBLEM 1Determination of light reflectivity of a transparent dielectric surface.Experimental Apparatus1. He-Ne Laser(～1.5mW).The light from this laser is not linearly polarized.2. Two polarizers (P1, P2) with degree scale disk (Fig. 1), one (P1) has been mounted in front of the laser output window as a polarizer, and another one can be fixed in a proper place of the drawing board by push-pins when it is necessary.3. Two light intensity detectors (D1, D2) which consisted of a photocell and a microammeter (Fig. 2).4. Glass beam splitter(B).5. Transparent dielectric plate, whose reflectivity and refractive index are to be determined.6. Sample table mounted on a semicircular degree scale plate with a coaxial swivel arm(Fig. 3).7. Several push-.pins for fixing the sample table on the drawing board and as its rotation axis.8. Slit aperture and viewing screen for adjusting the laser beam in the horizontal direction and for alignment of optical elements.9. Lute for adhere of optical elements in a fixed place.10. Wooden drawing board.11. Plotting papersExperiment Requirement1. Determine the reflectivity of the p-component as a function of the incident angle(the electric field component, parallel to the plane of incidence is called the p-component).(a) Specify the transmission axis of the polarizer (A) by the position of the markedline on the degree scale disk in the p-componet measurement(the transmission axis is the direction of vibration of the electric field vector of the transmitted light).(b) Choose any one of the light intensity detector and set its micro-ammeter at therange of "×5". Verify the linear relation ship between the light intensity and the micro-ammeter reading. Draw the optical schematic diagram. Show your measured data and calculated results(including the calculation formula)in the farm of a table. Plot the linear relationship curve.(c) Determine the reflectivity of the p -component as a function of the incident angle. Draw the optical schematic diagram. Show your measured data and calculated reflectivity(including the calculation formula)in the form of a table. Plot the reflectivity as a function of the incident angle.2. Determine the refractive index of the sample as accurate as possible. Explanation and Suggestion1. Laser radiation avoid direct eye exposure.2. Since the output power of the laser beam may fluctuate from time to time, the fluctuation of light output has to be monitored during the performance of the experiment and a correction of the experimental results has to be made.3. The laser should be lighting all the time, even when you finish your experiment and leave the examination hall, the laser should be keeping in work.4. The reflected light is totally plane polarized at an incident angle B θ while tg B θn = (refractive index).Fig. 1 polarizers with degree scale diskFig. 2 Light intensity detector(1) Insert the plug of photocell into the “INPUT” socket of microammeter (2) Switching on the microammeter.(3) Blocd off the light entrance hole in front of the photocell and adjust the scale reading of micro ammeter to “0”.(4) Set the “Multiple” knob to a proper range.Fig.3 Sample table mounted on a semicircular degree scale plateExperimental Problem 1——Solution1. (a) Determine the transmission axis of the polarizer and the Brewster angle B θ ofthe sample by using the fact that the rerlectivity of the p -component 0=p R at the Brewster angle.Change the orientation of the transmission axis of 1P , specified by the position of the marked line on the degree scale disk (ψ) and the incident angle (i θ) successively until the related intensity 0=r I .Now the incident light consists of p -component only and the incident angle is B θ, thecorresponding values 1ψ and B θ are shown below:︒±︒=5.05.1401ψ or 322.3°±0.1°The Brewster angle B θ is 56.3°±0.1° 1. (b) Verification of the linear relationship between the light intensity and themicroammenter reading.The intensity the transmitted light passing through two polarized 1P and 2Pobeys Malus ’ lawθθ20cos )(I I =where 0I is the intensity of the light polarized by 1p and incident, I is the intensity of the transmitted light, and θ is the angle between the transmission axes of1P and 2p . Thus we can obtain light with various intensities for the verification by using two polarizers. The experimental arrangement is shown in the figure.The light intensity detector 1D serves to monitor the intensity fluctuation of theincident beam (the ratio of 1I to 2I remain unchanged), and 2D measures 2I . Let)(1θi and )(2θi be the readings of 1D and 2D respectively, and )(2θψ be the reading of the marked line position. 02=i when 90=θ°, the corresponding 2ψ is 2ψ(90°), and the value of θ corresponding to 2ψ is|90)90(|22︒±︒-=ψψθData and results;︒=︒4)90(2ψFrom the above data we can obtain the values of )(/)(2θθI I from the formula)0()0()()()(21120i i i i I I ⋅=θθθ and compare them with θ2cos for examining the linear relationship. The results obtained are:1. (c) Reflectivity measurementThe experimental arrangement shown below is used to determine the ratio of 0Ito 1I which is proportional to the ratio of the reading )(20i of 2D to the corresponding reading )(10i of 1D .Then used the experimental arrangement shown below to measure the relativityp R of the sample at various incident angle )(θ while the incident light consists of p -component only. Let )(1θi and )(2θi be the readings of 1D and 2Drespectively.Then the reflectivity is2010120)()()()(i i i i I I R p ⋅==θθθθ Data and results:Ai A i μμψ3.1358.195.14010201=⨯=︒=。
2013-2014年物理奥林匹克竞赛国家集训队 热学练习题答案解析汇总
2013-2014年物理奥林匹克竞赛国家集训队热学练习题姓名：所在中学：成绩：注意：必须写出完整步骤，否则得不到步骤分。
答卷请勿涂改，无法看清的地方一律不给分。
1、（12分）一端开口，横截面积处处相等的长管中充有压强p的空气。
先对管子加热，使它形成从开口端温度1000K均匀变为闭端200K的温度分布，然后把管子开口端密封，再使整体温度降为100K，试问管中最后的压强是多大？2、（12分）一容积为1 升的容器，盛有温度为300 K，压强为30⨯10Pa的氩气，氩的摩尔质量为0.040 kg。
若器壁上有一面积为1.0×10㎝的小孔，氩气将通过小孔从容器内逸出，经过多长时间容器里的原子数减少为原有原子数的 1/e？3、（12分）若认为地球的大气是等温的, 则把所有大气分子压缩为一层环绕地球表面的、压强为一个大气压的厚度为H的均匀气体球壳，试证：这层球壳厚度H 就是大气标高。
4、（12分）标准状态下氦气的粘度为η1，氩气的粘度为η2，他们的摩尔质量分别为M1和M2.。
试问：（1）氦原子和氦原子碰撞的碰撞截面σ1和氩原子与氩原子的碰撞截面σ2之比等于多少？（2）氦的导热系数κ1与氩的导热系数κ2之比等于多少？（3）氦的扩散系数D1与氩的扩散系数D2之比等于多少？（4）此时测得氦气的粘度-3 24η1=1.87⨯10-3N⋅s⋅m-2和氩气的粘度η2=2.11⨯10-3N⋅s⋅m-2。
用这些数据近似的估算碰撞截面σ1,σ2。
5、（12分）在热水瓶里灌进质量为m=1.00 kg的水，热水瓶胆的内表面S=700 cm,瓶胆内外容器的间隙d=5.00 mm,间隙内气体压强p=1.00 Pa,假设热水瓶内的热量只是通过间隙内的气体的热传导而散失。
试确定需要多少时间容器内的水温从90℃降为80℃，取环境温度为20℃。
6、 (12分)加热室A（1000C）中蒸发出来的铍原子（相对原子质量为9）经小孔逸出，再经狭缝准直器B而形成原子束，最后进入另一真空室D中，（1）原子束将与真空室背景分子进行碰撞，若进行1m后其原子束强度（单位时间内通过的原子数）减少为1/e。
奥林匹克物理竞赛试题及答案
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。
国际物理奥林匹克竞赛试题.doc
国际物理奥林匹克竞赛试题理论试题题1A 某蹦迪运动员系在一根长弹性绳子的一端，绳的另一端固定在一座高桥上，他自静止高桥向下面的河流下落，末与水面相触，他的质量为m，绳子的自然长度为L，绳子的力常数（使绳子伸长lm所需的力）为k，重力场强度为g。
求出下面各量的表达式。
（a）运动员在第一次达到瞬时静止前所落下的距离y。
（b）他在下落过程中所达到的最大速率v。
（c）他在第一次达到瞬时静止前的下落过程所经历的时间t。
设运动员可以视为系于绳子一端的质点，与m相比绳子的质量可忽略不计，当绳子在伸长时服从胡克定律，在整个下落过程中空气的阻力可忽略不计。
B 一热机工作于两个相同材料的物体之间，两物体的温度分别为T A和T B（T A＞T B），每个物体的质量均为m，比热恒定，均为s。
设两个物体的压强保持不变，且不发生相变。
（a）假定热机能从系统获得理论上允许的最大机械能，求出两物体A和B最终达到的温度T?的表达式，给出解题全部过程。
（b）由此得出允许获得的最大功的表达式。
（c）假定热机工作于两箱水之间，每箱水的体积为2.50m3，一箱水的温度为350K，另一箱水的温度为300K。
计算可获得的最大机械能。
已知水的比热容= 4.19×103kg-1K-1，水的密度=1.00 x 103kgm.-3C 假定地球形成时同位素238U和235U已经存在，但不存在它们的衰变产物。
238U和235U的衰变被用来确定地球的年龄T。
（a）同位素238U以4.50×109年为半衰期衰变，衰变过程中其余放射性衰变产物的半衰期比这都短得多，作为一级近似，可忽略这些衰变产物的存在，衰变过程终止于铅的同位素206Ph。
用238U的半衰期、现在238U的数目238N表示出由放射衰变产生的206Pb原子的数目206n。
（运算中以109年为单位为宜）（b）类似地，235U在通过一系列较短半衰期产物后，以0.710×109年为半衰期衰变，终止于稳定的同位素207Pb。
第届国际物理奥林匹克竞赛试题与解答(一)
第届国际物理奥林匹克竞赛试题与解答(一)第50届国际物理奥林匹克竞赛是2021年7月17日至7月24日在立陶宛首都维尔纽斯举办的一场国际性的高中物理竞赛。
比赛采用英语作为交流语言，共有来自78个国家和地区的340名高中生参加。
本次比赛的试题难度较大，接下来将展示文本及其解答。
试题一：水球侦探有一个可预知的机器人，它能在20秒中将一个水球从一个瓶子上落下并射入一个篮子内。
现在，在较短的时间内，你需要发现该机器人和瓶子的相对位置，因为您希望以类似的方式射击水球。
您可以使用网球球拍（无障碍物的情况下）将水球从篮子向上弹起，并发现您需要瓶子的相对位置的任何信息；您还可以在所需的任何时间内以水平或垂直方向挡板的形式放置障碍物；在一次试验中，您可以在网球球拍打水球的起始位置上放几个记号来记录启动点。
您可以使用水球、网球球拍、篮子、木板和绳索等物品，还可以在最多五米范围内的工作桌上使用软件和硬件，但不能直接使用摄像机或激光仪来获取机器人方位信息和 distances.1 并且您需要在3次尝试中从一个新的启动点开始，每次都在20秒内打入水球。
问题1: 说明确定机器人的位置最少需要多少次尝试? 您将如何确定第一次尝试的起点和方向？如何优化您的尝试策略？解答1：最小值为3次尝试。
机器人与起始位置之间的最小距离为y，射中目标的距离为 x 在不知道任何具体数据的情况下，我们可以进行如下操作。
第一次尝试可以进行在中心处的最高点，平移到一个$y+\sqrt{2}x$的位置，这种位置需要花费一定的时间来判断，但如果我们成功射入目标，我们就可以找到100％精确的$y_0$值，后续垂直位移的知识问题。
而对于水平位置的了解，我们可以首先考虑将水球从一个中心点射出，使其成为一个匀加速运动，然后用一块水平板从左到右或从右到左预定一个固定距离，进一步将初始位置确定为距板最近的位置，然后调整初始角度。
这个角度可以通过打二角中心的两个球与板的交点来确定。
IPhO2014-第45届国际物理奥林匹克竞赛理论试题与解答
N
10.0 μm and is placed in vacuum. The soap film has the surface tension = 4.00 ∙ 10−2 m and the density
Consider the forces acting on the puck and the cylinder and
depicted in the figure on the right. The puck is subject to the
gravity force and the reaction force from the cylinder . The
instantly shorted. Find the maximum current max through the switch thereafter.
Theoretical competition. Tuesday, 15 July 2014
1/4
Problem 1
Solution
Part A
the plane as shown in the figure on the left. Find the interaction force
between the puck and the cylinder at the moment when the puck passes
the lowest point of its trajectory. Assume that the friction between the
历届IPho试题-历届IPho试题-IPhO-37ThCovermand
第37届国际奥林匹克物理竞赛
理论考试
（2006年7月10日，星期一，新加坡）
考试须知:
1.本场考试时间为5小时。
共有3题，每题10分。
2.只能使用考场所提供的笔。
3.只能使用考场所提供的纸张的正面。
4.不同的题目不要在同一张演算纸(Writing Sheets)上解答。
5.请在答卷纸(Answer Script)上答题。
另外还提供演算纸(Writing Sheets)。
根据
题中给出的数据，数值答案请保留正确的有效位数。
不要忘记写物理单位。
6.在演算纸上，写下解题所必需的步骤、内容，并标明希望考官批改的部分。
书
写时，希望用尽可能少的文字，只需提供公式、数值、符号和图表等。
7.请务必在所使用的每张答卷纸和演算纸上指定的方框内，填上国家号和学生
号；在所使用的每张演算纸上，填上题号、演算纸页码号和每题所使用的演算纸总数，以便考官批改。
此外，你可将题号和小题号写在空白纸张的顶部。
如果你不希望某张演算纸上的草稿被批改，请在不要的纸上画一个大叉，且不要把它算在演算纸的总页数内，亦不要编写页码。
8.考试结束时，请将所有答卷纸和演算纸按顺序整理，即每题答卷纸在前，希望
批改的演算纸随后，不希望批改的演算纸其次，未使用的纸张和印刷的考题放在最后。
将所有纸张按题号顺序放置，用所提供的夹子把它们夹好，留在桌面上。
不得把所提供的任何纸张携带出考场。
高中物理奥赛综合训练题(参考解答或答案)
高中物理奥赛综合训练题1、长方形风筝如图1所示，其宽度a = 40cm ，长度b = 50cm ，质量M = 200g（其中包括以轻绳吊挂的纸球“尾巴”的质量M′= 20g ，纸球可当作质点）。
AO 、BO 、CO 为三根绑绳，AO=BO，C为底边中点；绑绳及放风筝的牵绳均不可伸缩，质量不计。
放风筝时，设风速为零，牵绳保持水平拉紧状态。
当放风筝者以速度v持牵绳奔跑时，风筝单位面积所受的空气作用力垂直于风筝表面，量值为P = Kvsinα，K = 8N s/m3，α为风筝表面与水平面的夹角。
风筝表面为光滑平面，各处所受空气作用力近似相等，g取10m/s2。
试求：（1）放风筝者至少应以多大速度持牵绳奔跑，风筝才能做水平飞行?（2）这时风筝面与水平面的夹角应为何值?假设通过调整绑绳长度可使风筝面与水平面成任意角度α。
2、如图2是一个直径为D的圆柱体，其侧面刻有螺距为h的螺旋形凹槽，槽内有一小球，为使小球能自由落下，必须要以多大的加速度来拉缠在圆柱体侧面的绳子?3、（前苏联奥林匹克竞赛题）快艇系在湖边，湖岸是直线，系绳突然松脱，风吹着快艇以恒定速度v0 = 2.5km/h沿与湖岸成15°角的方向飘去，一人能在岸上以v1 = 4km/h行走或在水中以v2 = 2km/h游泳。
试问：（1）他能否赶上快艇；（2）当快艇速度多大时，他总可以赶上快艇。
4、（北京市高中物理竞赛题）一辆汽车沿水平公路以速度v无滑动地运动，如果车轮半径为R ，试求车轮抛出的水滴上升的最大高度和抛出点的位置。
5、（全国中学生物理竞赛题）图3中，AOB是一内表面光滑的楔形槽，固定在水平桌面(图中纸面)上，夹角α = 15°，现将一质点在BOA面内从C处以速度v = 3m/s射出，其方向与AO间的夹角为β = 30°，OC= 1m ，设质点与桌面的摩擦可忽略不计，质点与OB 面及OA面的碰撞都是弹性碰撞，且每次碰撞时间极短，可忽略不计。
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Theoretical competition. Tuesday, 15 July 2014 Problem 1
Solution Part A
1/4
Consider the forces acting on the puck and the cylinder and depicted in the figure on the right. The puck is subject to the gravity force ������������ and the reaction force from the cylinder ������. The cylinder is subject to the gravity force ������������, the reaction force from the plane ������1 , the friction force ������������������ and the pressure force from the puck ������ ′ = −������. The idea is to write the horizontal projections of the equations of motion. It is written for the puck as follows ������������������ = ������ sin ������, (A.1) where ������������ is the horizontal projection of the puck acceleration. For the cylinder the equation of motion with the acceleration ������ is found as ������������ = ������ sin ������ − ������������������ . (A.2) Since the cylinder moves along the plane without sliding its angular acceleration is obtained as ������ = ������ /������ (A.3) Then the equation of rotational motion around the center of mass of the cylinder takes the form ������������ = ������������������ ������, (A.4) where the inertia moment of the hollow cylinder is given by ������ = ������������ 2 . (A.5) Solving (A.2)-(A.5) yields 2������������ = ������ sin ������ . (A.6) From equations (A.1) and (A.6) it is easily concluded that ������������������ = 2������������ . (A.7) Since the initial velocities of the puck and of the cylinder are both equal to zero, then, it follows from (A.7) after integrating that ������������ = 2������������ . (A.8) It is obvious that the conservation law for the system is written as ������������������ = 2 + 2 + 2 , (A.9) where the angular velocity of the cylinder is found to be ������ ������ = ������ , (A.10) since it does not slide over the plane. Solving (A.8)-(A.10) results in velocities at the lowest point of the puck trajectory written as ������ = 2 ������ = ������
������ ������ 2 ������ 2
Theoretical competition. Tuesday, 15 July 2014
Problem 1 (9 points)
This problem consists of three independent parts.
1/1
Part A (3 points)
A small puck of mass ������ is carefully placed onto the inner surface of the thin hollow thin cylinder of mass ������ and of radius ������ . Initially, the cylinder rests on the horizontal plane and the puck is located at the height ������ above the plane as shown in the figure on the left. Find the interaction force ������ between the puck and the cylinder at the moment when the puck passes the lowest point of its trajectory. Assume that the friction between the puck and the inner surface of the cylinder is absent, and the cylinder moves on the plane without slipping. The free fall acceleration is ������.
g
Part C (3 points)
Initially, a switch ������ is unshorted in the circuit shown in the figure on the right, a capacitor of capacitance 2������ carries the electric chargபைடு நூலகம் ������0 , a capacitor of capacitance ������ is uncharged, and there are no electric currents in both coils of inductance ������ and 2������, respectively. The capacitor starts to discharge and at the moment when the current in the coils reaches its maximum value, the switch ������ is instantly shorted. Find the maximum current ������max through the switch ������ thereafter.
Part B (3 points)
A bubble of radius ������ = 5.00 cm, containing a diatomic ideal gas, has the soap film of thickness ℎ = N 10.0 μm and is placed in vacuum. The soap film has the surface tension ������ = 4.00 ∙ 10−2 m and the density ������ = 1.10 3 . 1) Find formula for the molar heat capacity of the gas in the bubble for such a process when cm the gas is heated so slowly that the bubble remains in a mechanical equilibrium and evaluate it; 2) Find formula for the frequency ������ of the small radial oscillations of the bubble and evaluate it under the assumption that the heat capacity of the soap film is much greater than the heat capacity of the gas in the bubble. Assume that the thermal equilibrium inside the bubble is reached much faster than the period of oscillations. Hint: Laplace showed that there is pressure difference between inside and outside of a curved 2������ surface, caused by surface tension of the interface between liquid and gas, so that ∆������ = ������ .
������ ������������������ ������������������ (2������ +������ ) (2������ +������ ) ������ ������ 2 ������������ 2 ������������ 2