v 2,p = v 3,p. The total energy at P is then mv 2 p = 6.68mv 2 p 4.49Gm2 d. (3) P 2 O 3 r o Gm = v2 p d2 P 3

Transcription

1 Nordic-Baltic hyic Olympiad 08 Solution GRAVITATIONAL RACING i) a) Since all three bodie move along the ame trajectory, they mut be T 3 away from each other at any moment of time Thu, it take T 3 to get from O to O b) From ymmetry, time taken to go from to O mut be T 4 Furthermore, it take T 3 to get from to and from O to O 3 Thi mean that it take T 3 + T 3 + T 4 = T to get from to O 3 or T T = T to get from O 3 to ii) Since there are no external force at play, the centre of ma of the three body ytem mut tay in place and, due to ymmetry, be located at O Thu, r + r + r 3 = 0, where r, r and r 3 are poition vector from O Differentiating, v + v + v 3 = 0 ) iii) The total angular momentum i conerved Thu, we can find the angular momentum at a moment of time that mot convenient for u, uch a the configuration when one of the bodie i at O Due to ymmetry, r = r 3 and v = v 3 The total angular momentum i then J = m r v + m r v + m r 3 v 3 = = m r v + r 3 v 3 ) = m r v r v ) = 0 iv) Uing equation?? and the fact that v = v 3, v + v = 0 mut hold Thu, v = v Note that thi can only happen when v, v and v 3 are parallel Since the velocitie of the bodie are tangent to the trajectory at their location, the trajectorie of the bodie mut have the ame lope at O, O and O 3 Thi give a convenient way of recontructing the poition of O and O 3 uing a ruler a hown in the figure in red v) We can ue ymmetry to deduce that r,x = r 3,x and r,y = r 3,y Additionally, r,x + r,x + r 3,x = 0, o r,x = r 3,x = r,x Thi give a imple way to find the x-coordinate of and 3 and recontruct their poition hown in blue in the figure Let apply the conervation of angular momentum at point Due to ymmetry, the velocity vector at and 3 mut interect at the x-axi For the total angular momentum to be 0, they mut interect at Therefore, and 3 have tangent that pa through and 3 can then be recontructed a hown in green in the figure O Q d vi) The general idea i to apply energy conervation in both configuration and ue the radiu of curvature at to find the value of Gm From the figure, = 47, r 0 = 45, d = 4 and = 6deg the unit for ditance are arbitrary, only the ratio carry any meaning) The total energy at O i E= mv,o + mv,o r o 3 O 3 + mv 3,o Gm Gm Gm r,o r 3,o r 3,o We have v,o = v,o = v 3,o = v o and r,o = r 3,o = r 3,o = r o Thu, E = 3mv o 4 Total Energy at i mv,p mv,p mv 3,p 5Gm r o ) v,p = v 3,p The total energy at i then E = mv p ) + in Gm d + ) = in = 668mv p 449Gm d 3) When a body i at, the gravitational force i equal to the centrifugal force Thi mean that E= + + Gm Gm Gm ii) Uing the trigonometric identity given in r,p r 3,p r 3,p the problem, we can expre the product of two wave imply a a um of wave Additionally, r,p = r 3,p = d and r 3,p = d in Furthermore, applying equation?? coπf t)coπf 0 t) = on the y-axi, v,p v,p in v 3,p in = 0 Thu, v,p = v p = v,p in = v 3,p in ince = co[πf + f 0 )t] + co[πf f 0 )t], mv p = Gm d co = 96Gm d, Gm = 050 v p d 4) Combining equation??,?? and?? give 3v o 4 = 668v p v p d ), r o d rearranging, v o 4 = d v p 3 SEED CAMERA )) = 8 r 0 d i) The Doppler hift formula ha to be applied twice Firt, the oberver on the approaching car ee both the incoming and reflected wave with frequency f = f 0 +v/c) Now, the oberver at the peed camera ee the reflected wave Doppler hifted to f = f = f +v/c) = f 0 +v/c) Finally, we can implify: f = f 0 + v/c) f 0 + v/c) where we can eaily identify two frequency component f high = f + f 0 and f low = f f 0 iii) We can expre f low = f f 0 = f 0 v/c and calculate the peed of car a v = f low f 0 c = 30m/ 3 WEATHER FORECAST i) The angle φ i equal to the latitude Thi mean that on the northern hemiphere the Corioli force vector i rotated 90 clockwie from the velocity vector if both are drawn on the map To maintain force balance, the Corioli force need to be directed oppoite to the preure gradient force, ie in the direction of increaing preure Thu the velocity need to be directed along the iobar The force hould be directed counter-clockwie around the preure minimum, ie to the north in A and to the outhwet in B The anwer may alo be accepted if the velocity ha a mall component toward the preure minimum, a long a thi i much maller than the component along the iobar ii) In point A the iobar are approximately traight, meaning that the velocity i contant and thu that all force um to zero A mall lab of air with area A and thickne ha the ma dm = A The force from the preure difference dp between oppoite ide in F p = Adp, uch that the force per ma i F p dm = Adp A = dp The preure gradient can be etimated by meauring the ditance between a few nearby iobar in the map, and kg m 3 Force balance give the equation vωinφ = dp = v = dp Ωinφ

2 Uing Ω = , ϕ = 56, dp/ 08/50 a m = 0003 a m we get the etimation v = m Since the tudent are only aked for an etimation, a wide range of numerical anwer are accepted, a long a the method i correct iii) Now the iobar are curved, and from the map one can etimate the radiu of curvature r 06 km The difference of the preure gradient force and the Corioli force mut equal the centripetal force: r = dp vωinφ v Thi i a econd order equation in v with poitive olution v = rωinφ + rωinφ) + r dp With ϕ = 60, dp/ a m we get the etimation v = 4 m A a comparion, if we neglect the curvature of the iobar we get m 4 FRESNEL RISM i) In order to find the grating pitch, we et up a imple diffraction experiment: direct laer light through the grating to the creen; there will be a long erie of bright pot which correpond to a erie of main maxima; all angle are mall, o we can apply mall-angle approximation A compared with a pair of beam exiting the grating from two neighbouring lit perpendicularly, a pair of beam exiting at a mall angle φ obtain an additional optical path difference equal to d inφ dφ, ee figure Suppoe that angle φ 0 correpond to a main diffraction maximum of a certain order n o that the optical path difference between the two beam i equal to an integer number n of wavelength Then, for the n + j-th main maximum, oberved at angle j, the optical path difference between the neighbouring beam i n+ j)λ Hence, dφ j dφ 0 = jλ o that φ j φ 0 = jλ/d Angle difference φ j φ 0 reult in the ditance of bright pot at creen being equal to a j = φ j φ 0 )L, where L i the ditance from the grating to the creen So, we can meaure the ditance a j between uch a pair of bright pot on the creen which are eparated by j bright pot, and calculate the = grating contant a For y = and h = we obtain = d = jλl a j In order to obtain better accuracy, it i neceary to ue a large a poible value of j the larget uch value that the both dot remain on the creen) With L =, j =, and a 0, we obtain d = ii) There are two way of determining the prim angle Firt, one can ue laer light and creen to determine, to which ditance x i the brightet pot on the creen the zeroth main maximum) hifted when the Frenel prim i inerted into the path of the beam at ditance L from the creen It appear that the angle β by which the prim deflect the beam remain mall, o that we can till ue the mall angle approximation: β = x/l Simple geometrical optic calculation yield = β n = x Ln ) For L = and x = we obtain = An alternative approach i uing the cyan tripe on the heet We look through the prim o that we can ee tripe both through the prim, and bypaing the tream imultaneouly We find uch two neighbouring tripe and uch ditance h between the prim and the heet that one tripe een through the prim eem to be exactly at the ame poition a the other tripe een beyond the edge of the prim We meaure the ditance y between thee two tripe on the heet Then, the deflection angle of the prim i found a β = y/h, o that β n = y hn ) dn dλ = z Hλ m λ c Uing z = and H = we obtain dn dλ = 5 MAGNETIC BILLIARD i) After the firt colliion, let the velocitie of the firt and econd ball be v and v repectively Applying the conervation of energy give mv = mv + mv or v = v + v Conervation of momentum yield mv = mv + mv or v = v +v Combining the two equation give v = v +v v ) = v vv +v and v = 0; v Since the firt olution correpond to the cae when the colliion doen t happen, the peed of the econd ball mut be v = v ii) The ball experience Lorentz force due to the external magnetic field Since the Lorentz force i perpendicular to the line of motion and contant in magnitude, the ball move along a circular orbit Equating the Lorentz force with centrifugal force give mv R and ω = v R = m mv = qvb Thu, R = Thi mean that one of the charge move along the orbit clockwie and the other anticlockwie After each colliion, one of the ball move at peed v while the other one i at ret The moving ball travel a part of the full cyclotron period either clockwie or anticlockwie, depending on the charge) before making a head- iii) Finally, we ue that part of the heet where there are neighbouring cyan and magenta tripe We ue a cloely poitioned pair of uch tripe, and look at it through the prim Depending on the orientation of the prim the pair of tripe i either brought cloe to each other, or, vice vera, moved apart We ue uch orientation for which the tripe are brought cloer to on colliion with the ball at ret each other, and find uch a ditance H between the prim and the heet for which the two tripe overlap exactly reulting in a eemingly yellowih tripe) We alo meaure the ditance z between the tripe Uing our expreion for the deflection angle β = n, we obtain an expreion for the change of the deflection angle -q δβ = δn, where δn denote the difference of v r the refraction index for the cyan and magenta Therefore, δn = δβ/ We can find the change of the deflection angle from our meaurement data a δβ = z/h So, δn = z/h), and The momentum i given over to the firt ball and the previouly moving ball tay at ret and the motion tart once again During the ubequent colliion, the ball tart drifting in one direction a can be een in the figure r R iii) The average velocity of the ball i equal to the average peed of the colliion point From the figure, it can be een that the direction of the average velocity i π clockwie from the initial direction of the incoming ball, where = arctan r R The colliion point move by +q d = r co between two ubequent colliion

3 In between the two colliion, one of the ball move π along a cyclotron orbit The time taken i then t = π = m ) ω π arctan r R and the average velocity i v avg = d t = rωco π ) = vrr R 4r + R π ) = v = ) 4 + R r π arctan r R iv) Let the velocitie of the two ball at any moment of time be v and v The velocity of the centre of ma i then v CM = v + v The equation of motion of the ytem i q v B F + q v B + F = m v + m v, where F i the force between the two ball, either the elatic force during a colliion or the electrotatic force Then q v + v ) B = m d dt v + v ), q v CM B d = m dt v CM Thi mean that the centre of ma of the ytem undergoe cyclotronic motion with a radiu of R = mv Becaue every colliion point can only be located where the center of ma i, the colliion point mut alo be limited to the ame circle Thu, the maximum ditance between any two colliion i R = mv 6 CUBE The cube get puhed by the light reflecting againt it urface Since there i no partial reflection, light can only reflect inide the cube via total internal reflection Let the cube face be aligned to x-y-z axi and let the light enter from the face which i perpendicular to the z-axi Before entering the cube, let the unit vector directed along the motion of the light be t = t x, t y, t z ), after entering the cube, r = r x, r y, r z ), before leaving the cube, r = r x, r y, r z ) and after leaving the cube, t = t x, t y, t z ) Every time the light bounce againt one of the ide of the cube, the repective component of r get flipped i) The laer beam i limited to propagate in a two-dimenional plane Take t y = 0, r y = 0, r y = 0 and t y = 0 In time dt, the laer pointer generate light with total energy dt carrying momentum c dt In that time, the ame amount of light enter the cube and exit it, only with different direction Applying Newton III law, the cube mut attain a momentum of d p = c dt t t ) and thu experience a force of F d p = dt = c t t ) = c tx t x) + t z t z) Thi mean that we wih to maximize the quantity t x t x ) + t z t z ) Snell law can be written a nr x = t x and nr x = t x ince t x and r x are the ine of angle of incidence and departure repectively The laer beam can only reflect againt the ide that i perpendicular to the x-axi, beam path with internal reflection i hown inthe figure Thu, r z = r z and t z = t z Let invetigate the reflection againt the x-face The angle of incidence i co = r x The condition for total internal reflection i inn Rearranging the term yield co < n or r x < Thi mean that t x < n n The force i maximal when the laer beam bounce againt the cube odd number of time Then r x = r x and t x t x < n Thu, F = c tx t x) + t z t z) < c n ) Note that t x + t z = o t x < Thi mean that the force can t be larger than c The maximal force i then { F = c n, if n < c, otherwie ii) We proceed in a imilar way a in the previou part, the main difference being that the y-component doen t have to be 0 The act of entering the cube keep the light moving in the ame direction in the x-y plane t Thu, x t y = r x r y Snell law can be written a t x + t y = n r x + r y, ince t x + t y and r x + r y are the ine of the angle of incidence and departure repectively Combining thee equation, we get r x = t x n, r y = t y n Similarly, t x = nr x and t y = nr y The act of reflecting againt the ide of the cube doen t change the magnitude of r x and r y Thu, t z = t z Thi mean that the quantity t x t x ) + t y t y ) need to be maximized and thi happen when r y = r y and r x = r x o F = c t x + t y = n c r x + r y Uing the ame argumentation a in the previou ubtak, the condition for a reflection to happen againt the x-face i t x < n Similarly, t y < n mut hold for the y-face Thi mean that r x +r y < ) On the n other hand, t x +t y +t z = o t x +t y < and r ) x + r y < Thu, r n x + r y < min ), n n = n min n ), ) The maximum force the cube can experience i then { F = c n, if n < 3/ c, otherwie 7 LCR-CIRCUIT i) Let u conider firt the upper branch of the circuit coniting of the capacitor C and reitor R There i the ame current I through the both element o that the complex voltage amplitude are I /iωc) and I R, repectively Diviion by imaginary unit rotate a vector in complex plane clock-wie by π/, hence the voltage vector on reitor i rotated with repect to the voltage on the capacitor counter-clock-wie by π/ Similar analyi lead u to the concluion that the voltage on the inductor L i rotated with repect to the voltage on the reitor R 0 counter-clock-wie by π/, and that the voltage on the reitor R i rotated with repect to the voltage on the inductor L 0 clock-wie by π/ The reulting phaor diagram i hown below ii) From Thale theorem we can conclude that the point F, D, and E in the figure above lay on the circle drawn around the egment AB a a diameter Hence, the voltage V AB which we want to know equal by modulu to the diameter AB of the circumcircle of the triangle FDE for which we know the ide length By making ue of the two formula for the urface area of a triangle, the Heron formula A = pp a)p b)p c), with p = a + b + c), and A = abc 4R with R denoting the radiu of the circumcircle, we conclude that the diameter of the circumcircle R = abc pp a)p b)p c)

4 With a = 7V, b = 5V and c = 0V we obtain p = V and V AB = R = 5V 8 AIR IN A SUBMARINE i) We are uppoed to calculate the volume rate in m3 ) at which the water flow in We know A = 0cm We apply Bernoulli equation, where the initial point i in the ea and the final point i in the hole: from which we get: v f = i + v i = f + }{{} v f 5) =0 gh = 767m/ 6) Here v f i the peed at which the water flow in Thi we can inert into the equation for the volume rate: Q = Av f = 053 m3 50 litre 7) ii) Atmopheric air conit mainly of diatomic nitrogen and oxygen ga At the temperature involved thee molecule have f = 5 degree of freedom: 3 tranlational and rotational The adiabatic contant γ i γ = f + )/f = 7/5 One can alo obtain thi reult from γ = c V + R)/c V For adiabatic compreion we have p i V γ i = p f V γ f 8) The final preure i the preure from the ea, which i approximately p 0 + gh = )a 3Ma Thi give ) 5 pi 7 V f = V i 09m 3 p f 9) Note: The final temperature i only about 6 time the initial temperature, uch that the vibrational degree of freedom of the molecule doe not have to be conidered iii) The work W done on the ytem coniting of the whole ubmarine) by the urrounding water i W = c V, where c i the contant preure of the urrounding water There i no heat exchange, o thi work mut be equal to the change in internal energy of the ytem: W = U ga + U water = c V n T + K turb 0) where K turb i the quantity that we are after and get: K turb = c V c V n T ) We need the value of n NB you can alo figure it out from the ideal ga law): n = m M ) where m = 3 kg 0m 3 = 3kg and M = m kg mol lugging in the value we get n 44mol The final temperature can be calculated from the initial temperature by uing that p γ T γ i conerved By plugging in all the other value we get: K turb 0 7 J 3) Alternative olution: Alternatively, one can look at the ga and the water inide the ubmarine) a eparate ubytem The work done on the ga i equal to the change in internal energy of the ga: g dv = c V n T 4) The work done on the water inide the ubmarine by the water outide the ubmarine i c V The water inide the ubmarine alo doe work on the ga given by g dv The change in internal energy of the water in the ubmarine i then K turb = c V g dv = c V c V n T 5) where the lat equality follow from eqn??) From here one proceed a already written above 9 BLACK BOX By meauring current with with poitive lead of multimeter connected to blue and negative lead connected to black we get I 0 95mA From thi meaurement alone, ince we made a circuit that continuouly conducted current, we can deduce that the circuit inide the black box ha to be one of following: A white black blue B white black blue Note that the order of the inductor and diode in erie doe not change anything The actual meaured current value varie a bit from one black box to another and alo change very lightly due battery voltage dropping and inductor heating up lightly i) Meauring voltage between blue and black we determine the electromotive force of the battery U = 95V ii) We get the internal reitance of the inductor from R l = U/I 0 00ohm We can alo get ome hint to the magnitude of the inductance a we aw no exponential ramp up of current when meauring, meaning L/R t iii) Determining which of the two poible circuit i inide the black box i trickier One way to do it, i to notice that when we diconnect blue and black we can get a mall park, or feel a mall pule of current if we happen to touch the wire at that point That i becaue L I t = U l the current through the inductor can t change intantaneouly and the voltage will generated by the inductor enough for park or high voltage pule Meanwhile if we have capacitor in parallel with inductor while diconnecting the circuit we wont get the effect By teting with white lead parallel with black or with blue we can determine that no park happen in latter cae and the circuit in the black box i circuit B from the figure iv) By connecting voltmeter between white and black or white and blue we can ee the voltage decaying exponentially That mean indeed, that the capacitor i connected to the white wire and depending if the other lead i connected to black or blue we are charging the capacitor to negative battery voltage through voltmeter or dicharging it through voltmeter and inductor and diode We have to be careful not to touch both wire at the ame time, ince the reitance of good kin conductance i much maller then the reitance of the voltmeter We can meaure the capacitance by meauring the exponent: for example by taking two voltage reading and meauring the time interval between the reading C = ln U U τr m v) uf Firt we connect all three wire together That mean we have current running through inductor and the capacitor i charged to negative of battery voltage We connect voltmeter between white and black, o that we meaure the total of the capacitor and battery voltage The reading i zero at the tart ince all black box lead are connected Now we diconnect the

5 inductor battery current loop by diconnecting black from other box lead and the multimeter reading will jump to U c +U ap prox3v and tart to decay exponentially a before We can do thi many time to get a maximum reading After diconnecting the black, the current goe through LCR circuit formed by inductor and capacitor, but intead of ocillating it top due to the diode when current through inductor ha reached zero We can get the upper and lower bound for the inductance value by conidering two different cae Upper bound we can get when we neglect the reitive loe In that cae all the energy at the end i in the capacitor When we write down the energy balance we get: U C + I L = U c C L 30mH Lower bound we can get when we aume that mot of the energy went to reitive loe, in that cae the inductor current decay exponentially and we can write expreion for the down the total charge: U c +U) C = I R l L L 336mH The correct value for the inductor L 00mH i between thoe bound and cloer to the upper bound a we may gue ince U c > U It i poible to get more accurate value by looking at it without the aumption - a a damped harmonic ocillation