ESTONIAN FINNISH PHYSICS OLYMPIAD PROBLEMS & SOLUTIONS (2003-2014) ESTONIAN FINNISH PHYSICS OLYMPIAD PROBLEMS & SOLUTIONS (2003) Tallinn 2003 Complete solution of each subquestion gives two points. You have five hours for solving. 1. Volleyball (8 points) 2) A wire is made of diﬀerent alloys, its heat resistivity ρ as a function of the coordinate along the wire is given in the attached graph. The crosssection area of the wire is S = 1 mm2 , its length l = 4 cm. Find the heat flux through the wire, if one end of the wire is kept at the temperature 100◦ C, and the other end — at 0◦ C. I(mA) the brick velocity change during a semi-period is negligible? Further we assume that this condition is satisfied. 10 2) The brick is kept in motion along x-axis by a force Fx in such a way that the mean brick velocity is v. Sketch graphically the dependance 5 Fx (v). ρ (Km/W) 3) The brick is kept in motion along (horizontal) y-axis by a force Fy in such a way that the mean U(V) 0,3 0,2 0,4 brick velocity is v. Find the dependance Fy (v). 0 0,1 0,2 4) Until now we have ignored the dependance tunnel diode Uin of the friction coeﬃcient on the sliding velocity. (input-signal) Further let us assume this dependance is given Uout 0,1 E R by the graph below. The brick is kept in motion (output-signal) along x-axis by a force Fx in such a way that the x(cm) 1) Find the current in the circuit, if Uin + E = mean brick velocity is v. Sketch graphically the 3 2 4 0 1 dependance Fx (v) taking u = 34 w0 . 0,08 V. µ 2) Find the output voltage Uout 0 if Uin = 0 V. µ 1 3) Find the output signal Uout − Uout 0 if Uin = 1 mV. 3. Gravitation (6 points) 4) The input signal is given in the graph below. µ0 1) Find the free falling acceleration g0 at the sur- Sketch the output signal as a function of time. I(mV) face of such a spherical planet, which has mass 1 M and material density ρ (in what follows, M w and ρ are assumed to be constant). _ _ Consider a simplified model of a volleyball: a thin spherical envelope filled with air. The envelope material is non-stretchable, easily foldable. The excess pressure inside the ball ∆p = 20 kPa, the ball radius R = 10 cm and mass m = 400 g (the air mass inside the ball is negligible). You can neglect the dependence of the excess pressure on the deformation of the ball. 1) The ball is pressed between two parallel rigid plates, the distance between of which is 2R − 2h (so that the height of the deformed segments is h = 1 cm). Find the force between the ball and a plate N . 2) Ball moves with velocity v0 = 2 m/s and hits a rigid wall. Find the maximal height of the deformed segment hm . . . 3) . . . and the collision time τ . 4) For small excess pressures, the ball can lose its spherical shape even in those points, which are not in touch with the wall. Which condition t (ms) between the quantities ∆p, R, m and h has to 2) Is it possible that at the surface of a non2 1 be satisfied in order to ensure that such a loss of spherical planet, there is a point with free falling acceleration g > g0 ? Motivate your answer. sphericity is negligible? 3) For which planet shape the maximum of the -1 2. Heat flux (4 points) free falling acceleration is achieved? Answer can Heat resistivity is equal to the ratio of the tem- be given in polar coordinates, the expression can 5. Vibration (10 points) perature diﬀerence between the end-points of a contain one unspecified constant. Consider a smooth horizontal surface, which is wire of unit cross-section and unit length, and moved periodically back and forth along the hothe heat flux (unit: W) through this wire. rizontal x-axis: during the first semi-period τ , the surface velocity is u, during the second semi1) Microprocessor of power P = 90 W has a 4. Tunnel diode (8 points) period — −u. A brick of mass m is put on that water-cooling system. The chip and flowing water are separated by a copper plate of thickness Tunnel diode is a semi-conductor device, similar surface; the friction coeﬃcient between the surd = 5 mm and cross-section area s = 100 mm2 . to the ordinary diode, the volt-amper characte- face and brick is µ, free falling acceleration is g. What is the temperature diﬀerence between the ristic of which is given in the attached graph. The 1) The brick has initial x-directional velocity v. processor and water? The copper heat resistivity circuit below describes a simple amplifier. The Which condition between the quantities g, µ, v, is ρ = 2,6 mm·K/W. resistance R = 10 Ω, battery voltage E = 0,25 V. and τ has to be satisfied in order to ensure that w0 w0 w0 5) The brick is put on the surface, there are no external forces. What is the brick’s terminal velocity v? Provide the answer as a function of u. case when the pulses are of variable polarity: for the 2n-th pulse, the electric field is +E, and for A particle of mass m and charge q is in a hothe 2n+1st pulse −E. Find the particles average mogeneous magnetic field with induction B (the velocity, (vectorially) averaged over the cyclotron vector is parallel to the z-axis). The characterisperiod. tic time of the system is the cyclotron period of the particle TB = 2πm/Bq. The system is si- 7. Telescope (12 points) tuated in between two parallel electrodes, which As it is well known, a telescope makes it possibcan be used to create an homogeneous, parallel le to see the stars in daylight. Let us study the to the x-axis electric field E. problem in more details. Consider a simplified 1) The particle is at rest. At the moment of time model of the eye: a single lens with focal length t = 0, the electric field E is switched on; after a f = 4 cm and diameter d = 3 mm creating an short time interval τ (τ TB ), it is switched oﬀ, image on screen (retina). The model of a telescoagain. What will be the trajectory of the particle? pe is similar: a lens of focal length F = 2 m and 2) Let px and py denote the x- and y compo- diameter D = 20 cm creating an image in focal nents of the momentum of the particle. Sketch plane (where eg. a film can be put). In your calcuthe trajectory of the particle in (px , py )-plane lations, the following quantities can be used: the and depict the vectors of the momentum for the density of the light energy radiated from a unit moments of time tn = nTB /4 (n = 1, 2, 3 and Solar surface in unit time w (the light power 0 4). surface density); the ratio of the star and Sun 3) Consider the situation when the on-oﬀ switc- distances q = 4 · 105 (we assume that the star hing of the electric field is done periodically, is identical to the Sun); Solar angular diameter starting with t = 0, after equal time intervals φ ≈ 9 mrad. Remark: If the answer contains w0 ∆t = TB /4. Sketch the particle trajectories in then numerical answer is not required. (px , py )- and (x, y)-planes. 1) Consider a sheet of paper, the normal of which 4) Let the period be short, ∆t TB (but still is directed towards the Sun. What is the surmuch longer than the duration of pulse, ∆t face density of the light power w1 arriving to the τ ). Show that after the n-th pulse (at the time sheet from the Sun? moment tn = n∆t), the momentum of the par- 2) Find the net power P2 of the light, which is ticle can be represented as the sum of n vectors focused by the telescope into the image of the pi , where all the component-vectors are equal star. in modulus (the modulus being independent of 3) Assume that blue sky is as bright as a sheet n), and the neighboring vectors ( pi and pi+1 , of gray paper illuminated by Sun. You may asi = 1, 2, . . .) have equal angles between them. sume that in the direction, perpendicular to the 5) Consider the limit case ∆t → 0, so that sheet, the ratio of the light power scattered by Eτ /∆t → Ek (Ek denotes the time-average of the paper into a 1-steradian space angle, to the the elctric field). Sketch the particles trajectory net light power arriving to the sheet, is α ≈ 0,1 in (px , py )-plane and express the particles mean (this corresponds to the dissipation of ca 70 % velocity (vectorially; averaged over the cyclotron light energy in the gray paper). What is the surperiod) via the quantities Ek and B. face density of the light power in the focal plane 6) Let us return to the non-zero (but still small, of the telescope w3 , due to the blue sky? ∆t TB ) time-intervals. Let us consider the 4) While studying the star image, let us ignore 6. Charged particle (12 points) all the eﬀects other than diﬀraction. Estimate the surface density of the light power in the center of the star image w2 (in the focal plane of the telescope), due to the light arriving from the star. 5) Provide an expression for the ratio of the surface densities of the light powers k in the middle of the star image, and in a point farther away from it. 6) Is it possible to see a star in daylight using a telescope? Plain eye? Motivate yourself. 8. Experiment (12 points) Determination of attraction force between iron plate and a permanent magnet as a function of distance. Tools: iron plate, wooden brick, ruler, dynamometer, paper stripes. Attention! the permanent magnets are very strong, keep them far from credit cards etc. Avoid also hitting them against each other and against the iron plate, because they are fragile and can be broken. 1) Determine the static and dynamic coeﬃcients of friction between the iron plate and a paper stripe. Draw the scheme of your set-up. 2) (4 points) Determine the attraction force between the iron plate and a magnet for those distances which allow direct usage of dynamometer. Draw the scheme of your set-up. 3) (4 points) Determine the attraction force between the iron plate and a magnet for smaller distances. For that purpose, you can use the wooden brick sliding down an inclined plate and hitting the magnet. You do not need to study the zero-distance (direct contact of magnet and iron plate) case. Draw the scheme of your set-up. Depict all the measurement result graphically. 4) Join two permanent magnets by a bridge made of a piece of iron (a) as shown in figure. Put a stripe of paper (b) on the iron plate (c) and put the system of magnets upon it. Determine the attraction force between the system of magnets and iron plate. N S S N a b c Solutions 1. Volleyball (8 pts) 1) F = ∆pS, where S = πr2 is the segment base surface. It is easy to see that r2 = (2R − h)h, hence F = ∆pπh(2R − h) ≈ 120 N. 2) During the collision the ball is deformed as shown in Figure: the envelope is not stretchable, hence it retains the spherical shape (except where in touch with the wall). Using the approximation h R we can neglect the term h2 in the expression for the force. Then, the force is proportional to h, ie. the ball behaves as a spring of stiﬀness k = 2πR∆p. According to the energy conservation law mv 2 = 2πR∆ph2 , hence h = v m/2πR∆p ≈ 11 mm. h(2R−h) < 4R2 . This condition is always satisfied, no additional constraint is needed. Notice that we considered only the worst case requiring the largest compensating force when the force of inertia is normal to the surface. Remark: The case of stretchable envelope is completely diﬀerent, sphericity disappears over all the surface (try to press a balloon against a glass!). 2. Heat flux (4 pts) 1) The heat flux P = ∆T s/ρd, hence ∆T = P ρd/s ≈ 12 K. 2) By a constant heat flux P , the temperature change along the wire ∆T = P ρ∆x/S, where ∆x is a displacement along the wire. Hence the temperature drop t1 − t2 = P S/S, where S is the surface under the graph. Thus, P = (t1 − t2 )S/S. Using the graph we find S ≈ 50 Kcm2 /W and P ≈ 20 mW. 3. 2) F = 0, when |v| < u; F = µmg, when 1) For voltages below 0.08V, the graph is almost a |v| > u. straight line corresponding to a constant resis- 3) The x-component of the frictional force out, the y-component is left: tance RD = 0.05 V/6.5 mA ≈ 7.7 Ω. Hence cancels in average √ 2 F = µmgv/ v + u2 . I = (Uin + E)/(R + RD ) ≈ 4.5 mA. 2) The output voltage can be found graphically: 4) F = [µ(v + u) + µ(v − u)]mg, if v > u and the diode voltage U (I) = E − IR, hence, the F = [µ(u + v) − µ(u − v)]mg, if v < u (F > 0 intersection point of the graph and the straight means that F and v are opposite to each other). line U = E − IR, gives us the diode current It is easy to see that by small values of v, the force 6 mA; then, the output voltage IR = 60 mV starts linearly decreasing [with F (v = 0) = 0] (see the graph). (F < 0 implies that force and velocity are in 4. Tunnel diode (8 pts) I(mA) 10 5 U(V) Gravitation (6 pts) 0 3) This is the half of the harmonic oscillations period, τ = π m/2πR∆p = πm/2R∆p ≈ 18 ms. 4) Let us use the ball’s system of reference. The envelope surface element dS is exerted by the force of inertia dFi = amdS/4πR2 , where a = ∆pπh(2R − h)/m. Thus, dFi = ∆ph(2R − h)dS/4R2 . In order to keep the spherical shape, this force has to be compensated by the force due to the excess pressure dFr = ∆pdS, hence the same direction). At u = v, the graph exerts a jump, F becomes positive, and starts decreasing. The attached graph presents a sketch of the eﬀective friction coeﬃcient; the construction has been based on the lengths µk1 = µ(w0 /2) − µ(w0 ), µk2 = µ(w0 /4) − µ(5w0 /4), and µk3 = µ(0) − µ(3w0 /2). 0,1 0,2 0,3 0,4 1) g0 = γM/R2 , where R can be found from the 3) One millivolt input shifts the line intersecrelationship 43 πR3 ρ = M . Hence, ting the graph a little-bit sideward, but the shift 4πρ 2/3 g0 = γM ( ) . is so small that the graph can be approximated 3M by a straight line. The cotangent of the slope of 2) Taking a piece of ground from a certain point that line gives us the diﬀerential resistance of the of the planet surface and carrying it into another diode, Rd = −16 Ω. Then, a small change in point, the free fall acceleration can be changed the input voltage ∆U will lead to a current chan(the sign of the change depends on the direction ge ∆I given by the relationship (R + Rd )∆I = of the transport). ∆U ; hence, ∆I = ∆U/(Rd + R). The output 3) Let use the polar coordinates with the origin voltage change ∆Uout = IR = R∆U/(Rd + at the point where the free fall acceleration is to R), and the amplification factor ∆Uout /∆U = be maximized. Let the axis φ = 0 be given by the R/(Rd + R) ≈ 1.7. Consequently, the output direction of the acceleration. Carrying a small voltage is 1.7 mV, and . . . piece of ground from a point (r1 , φ1 ) to anot- 4) the output graph is exactly the same as the inher point (r2 , φ2 ) must keep the modulus of the put graph, except that it is vertically stretched by acceleration vector g constant, i.e. the vector of a factor of -1.7. the small change must be perpendicular to the vector g. Consequently, cos φ1 /l12 = cos φ2 /l22 , 5. Vibration (10 pts) √ hence l = l0 cos φ. 1) µmgτ v. µ µ1 µk2 µ0 µk3 µk1 w µef ekt w0 _ w0 5) The rest position is unstable, if u < w0 : the particle obtains the (stable) velocity u. If u > w0 , the rest position is stable, and the particle velocity remains 0. 6. Charged particle (12 pts) 1) The particle acquires the velocity v = Eqτ /m and starts moving along a circle of radius R, with mv 2 /R = Bvq, hence R = Eτ /B. px 4 py 3 1 2 2) px py endpoint of the particles momentum lies on that circle. Thus, averaged over the moments of time 2n∆t, the average velocity is vx = −Eqτ /2m. For odd number of impulses, one has to add the lastly given momentum P = (Eqτ, 0); hence, a similar circle is formed, except that the center is shifted by P : the center coordinates are mvx = Eqτ /2, mvy = 0. Correspondingly, averaged over the moments of time 2n∆t, the average velocity is vx = +Eqτ /2m. Averaged over all the moments of time, the final result is vx = vy = 0. 3) P r’es 4) Let us consider the vectorial sum of the momenta given to the particle in diﬀerent moments of time. During the time interval ∆t, all the component-vectors are rotated by the angle 2π∆t/TB = τ Bq/m. Thus, with each impulse, a vector P with modulus P = Eqτ is added; the angle between the lastly added vector, and the previously added vector is α = ∆tBq/m. P1 P P2n re s P2 5) All these vectors, when added according The figure represents the net moment Pres after to the triangle rule, form a circle of radius 2n-th impulse, and also the net impulse Pres for R = P/ sin α → P/α = Ek m/B. another time moment 2n ∆t. For an odd numpx ber of impulses, the pattern is exactly the same, except that all the vectors have opposite direcpy tion (because the lastly added component, the vertical vector, has opposite direction). 7. Telescope (12 pts) Hence, the average velocity vy = −R/m = 1) The light flux density decreases inversely proEk /B, vx = 0. portionally to the square of the distance, the6) Two subsequent momenta along x-axes result refore w1 = w0 Rp2 /L2p , where Rp is the solar in net moment along y-axes Py = P α. The radius, and Lp — the solar distance. Due to sequence of such moment pairs form a (near- φ = 2Rp /Lp , we obtain w1 = w0 φ2 /4. ly) circle (actually, regular equilateral polygon), 2) The previous result can be applied to the composed of vectors (with modulus Py ), the star flux density, which is q −2 w1 ; hence P2 = angle between of which is 2α (see Fig.). The ra- 14 πD2 w1 q −2 = w0 π(φD/4q)2 . dius of the circle is Py /2α = P/2 = 12 Eqτ , 3) The paper surface area S radiates towards and its center coordinates are mvx = −Eqτ /2, the lens of the telescope the power P3 = mvy = 0. After an even number of impulses, the w1 αS( π4 D2 /L2 ), where L is the telescope dis- tance. The image of this piece of paper has size s = SF 2 /L2 ; thus, w3 = P3 /s = w1 α( π4 D2 /F 2 ) = w0 απ(φD/4F )2 . 4) The angular distance of the first diﬀraction minimum (using the single slit approximation — circle is actually not a slit) is λ/D. Hence, the bright circle radius can be estimated as δ = F λ/D. Consequently, w2 = P2 /πδ 2 = w0 (φD2 /4qF λ)2 . 5) k = (w2 +w3 )/w3 = 1+(απ)−1 (D/λq)2 ≈ 4 (assuming λ ≈ 500 nm). 6) k − 1 ∼ 1 (or k − 1 > 1) means that the star can be easily seen (as is the case for the telescope); k − 1 1 means that the star cannot be seen (for the eye, k − 1 ≈ 1 · 10−4 ). 8. Experiment (12 pts) 1) We incline the plate until sheet starts sliding: √ the static coeﬃcient is found as µstatic = h/ l2 − h2 , where h is height of the plate endpoint, and l — the plate length. Now we push the sheet laying on the plate slightly, and find the inclination angle, for which the sheet will slide down with a constant velocity; we use again √ the formula µkinetic = h/ l2 − h2 . The reasonable numerical values are µstatic ≈ 0.37 and µkinetic ≈ 0.29. 2) We put several paper stripes on the plate, and the magnet on the top of them. We make a loop of cord, put it around the magnet, and pull it using the dynamometer sideward (sliding the whole system of paper and magnet). The attraction force F ≈ N (where N is the reaction force) is found as the ratio of the reading of the dynamometer Fd and the appropriate friction coefficient (depends, which reading is taken: either the maximal one, or the one corresponding to sliding), F ≈ Fd /µ. The distance d is measured in the number of paper stripes (one stripe had a thickness of ≈ 0.2 mm). For large distances (approximately d > 4 mm), the weight of the paper Fp stripes and magnet is no longer negligib- le, the accuracy of the results can be enhanced by subtracting this weight from N : F = Fd /µ−Fp . 3) We use a similar set-up, except that smaller number of paper stripes is used (totaling up to around 2 mm), and a steep slope of the plate. We let the brick slide down the slope and hit on the magnet. We keep the falling height and plate slope constant, and measure the sliding path, which is covered by the papers and the magnet after having been hit by the brick. This path is inversely proportional to the attraction force N . If this path turns out to be too short for an accurate measurement (for very small distances between the magnet and the plate), several brick hits can be used. In that case, the single-hit path can be found as the measured path, divided by the number of hits. The constant of proportionality can be found by comparing the results of this and previous question, for those distances, which are covered by both measuring techniques. Reasonable measurement results are given in the attached graph. F /N/ 30 25 20 15 10 5 d /mm/ 0 0 1 2 3 4 5 6 4) The same technique as in the case of previous question is applied, except that a larger number of hits has to be used (≈ 10 − −20). Reasonable result for d = 0.2 mm (one paper stripe) is F ≈ 270 N. Note that the result is much larger than the double result in the case of a single magnet; this is due to closing the ferromagnetic loop of magnetic field lines. ESTONIAN FINNISH PHYSICS OLYMPIAD PROBLEMS & SOLUTIONS (2004) 1. Rubber fiber (12 pts) Fibers made of elastic rubber can be stretched to lengths l, much longer than the length in undeformed state l0 . For such rubbers, the net volume of the fiber remains constant. 1) Express the cross-sectional area S of such a fiber in a deformed state via its length l and initial dimensions l0 , S0 (1 pt). 2) For small deformations of an elastic material, the stretching force F and deformation x are related to each other by the Hooke’s law F = k0 x, where the stiffness k0 = E0 S0 /l0 and E0 is the Young’s modulus of the rubber. For non-small (possibly large, l ≫ l0 ) deformations of elastic rubber, however, the Hooke’s law is substituted by a non-linear law, F (l) = a + bl (breaking of this law at very large values of l will not be studied here). Express the constants a and b in terms of l0 , S0 , and E0 (2 pt). 3) Suppose such a fiber is stretched by some force up to the length l. A small change ∆F of the stretching force results in a small change in the length ∆l ≪ l. Express ∆F in terms of l, l0 , S0 , E0 , and ∆l (1 pt). 4) Suppose a small body is fixed to an one end of the fiber and the system is put into rotation around the other end of the fiber. In the case of a circular motion of the body, express the length of the fiber l via l0 , S0 , E0 , and the kinetic energy of the body K (the kinetic energy of the fiber and gravity can be neglected). (1.5 pts) 5) Let us analyse a slightly non-circular motion of the body. Let us describe the motion of the system by the length change of the fiber r(t) = l(t) − l(0), the radial vr (t) and tangential vt (t) velocities of the body (the components respectively parallel and perpendicular to the fiber). The initial values of these quantities are designated as L ≡ l(0), Vr ≡ vr (0), and Vt ≡ vt (0). The values L and Vt are chosen so that if the initial radial velocity were zero, the motion would be circular. Write down two independent equations relating r(t), vr (t), and vt (t) to each other (using also the mass of the body m, together with the parameters L, Vr , Vt , l0 , S0 , E0 ). (3.5 pts). 6) Find the relationship between r(t) and vr (t) (containing also the parameters m, L, Vr , Vt , l0 , S0 , E0 ) assuming that |r| ≪ L, and find the period T of small oscillations of r(t). Simplify the expression of T for L ≫ l0 (3 pts). 2. Planets (6 pts) the main optical axis and the straight line lay in the same (x, y) surface. Hint: use the coordinate system, where the origin coincides with the center of the lens and represent lines algebraically, e.g. y = ax + b. Make use of the formula of thin lens f −1 = x−1 − x′−1 (x > 0 and x′ are the x-coordinates of a point and its image, respectively) (2 pts). 2) In figure (a), draw the image of the given line and indicate, which parts of the image are virtual, and which are real (2 pts). Two planets move along circular orbits around a star of mass M = 2.0 · 1030 kg; gravitational constant G = 6.67 · 10−11 m3 /kg ·s2 . The dependence of the angular distance between a planet and the star on time, as seen from the other planet, is depicted in figure. 3) Photographer wants to take a photo of a field of flowers. In order to get image where all the flowers (both the close and far ones) are sharp, he has to use a lens with tilt-shift (TS) capabilities (either an ordinary camera with TS lens, or a large-format camera, where the entire lens compartment can be freely positioned). The field of flowers (which extends effectively to infinity) and the image of its distant edge, together with the image plane are depicted in figure (b). Reconstruct the position of the lens, the focal length of which is provided as a scale (2 pts). 1) What is the ratio of the radii of the planets k (2 pts)? 2) Determine the value of the unit on the vertical axis (or express it in terms of k, if you were unable to find it) (2 pts). 3) What are the orbital radii of the planets, if the unit on the horizontal axis equals to one year 4. Transparent film (6 pts) (2 pts)? A thick glass plate is coated by a thin transparent 3. Tilt-shift lens (6 pts) film. The transmission spectrum of the system is 1) Show that an image of a straight line created depicted in graph (light falls normal to the plate). by a thin lens is also a straight line. Consider The refractive index of the film n ≈ 1.3. What is two-dimensional geometry only, i.e. assume that the thickness of the film d? 5. 4th order ellipse (6 pts) 4 4 4th order ellipse is defined by equation xa4 + yb4 = 1, where a and b are the lengths of the half- axes. Consider an homogeneous cylinder, the crosssection of which is 4th order ellipse. The position of the cylinder is measured by the angle 0 ≤ ϕ ≤ π/2 between the vertical direction and a longer half-axes, see figure. 1) What are the equilibrium positions of the cylinder laying on an horizontal surface (3.5 pts)? 2) Sketch on graph the net torque of gravity and surface reaction forces with respect to the contact point of the cylinder and surface as a function of ϕ (0 ≤ ϕ ≤ π/2). For the axis of torque, you do not need to indicate any quantitative scale (1.3 pts). 3) Which equilibrium positions are stable and which are not? Motivate your answer (1.2 pts). 6. Magnets (6 pts) Passive air-cooling (9 pts) b Certain type of magnetic toys are made up of ferromagnetic spheres and permanent magnets of cylindrical shape. These building blocks can be used to build, for instance, a tetrahedron, see figure (letter “N ” marks the northern end of a magnet). Assume that all these permanent magnets are identical and each of them alone can create a magnetic flux Φ (assuming the both ends of the magnet are in contact with a Ushaped large piece of ferromagnetic material, so that a closed ferromagnetic contour is formed). Assume also that due to high magnetic permeability of the material of the building blocks, all the magnetic field lines are constrained inside of them (i.e. in the surrounding medium, the magnetic inductance B = 0). contain also the parameters defined above) (2 Consider a passive cooling system depicted in pts). figure. Cold air (at normal conditions: p0 = 5) What is the temperature T of the outflowing 105 Pa, T0 = 293 K) flows over the heat sink of a air? In your calculations, you may use approxichip of power dissipation P = 100 W, into a ver- mation T − T0 ≪ T0 (2 pts)? tical pipe of length L = 1 m and cross-sectional area S = 25 cm2 . After passing the pipe, air enters the ambient room. Assume that the air 8. Loop of wire (7 pts) inside the pipe becomes well mixed; neglect the Consider a rectangular loop of wire with dimenviscous and turbulent friction of air inside the pisions a = 0.03 m and b = 1.0 m, one side pe and heat sink. Air can be considered as an of which is parallel to another long straight wiideal gas with adiabatic exponent γ = 1.4 and re carrying current I0 = 1000 A, at distance molar mass µ = 29 g/mol. l = 0, 01 m, see figure. The magnetic inductance of such current is plotted as a function of the distance from the wire in attached graph. The Ohmic resistance of the loop is R = 1, 0 Ω, the in- 9. Experiment (15 pts) ductance is negligible. The black box contains a nonlinear element (active resistance) and a capacitor, connected sequentially. Find the capacitance C of the capacitor (5 pts) and the V − I characteristic of the nonlinear element (6 pts). Note that (a) the I0 electrolytic capacitor accepts only one polarity of charge (indicated by the colors of the output wires of the black box); (b) the V − I characteristic cannot be expected to be symmetric with respect to I = 0. However, you are requested l a to study the range I > 0 corresponding to the discharge of the capacitor. Tabulate your measu1) Express heat capacitance at constant pressure rement data and draw appropriate graphs (4 pts). cp via quantites γ and R (1 pt). Experimental equipment: batteries, wires, multi2) Find a relationship between the outflowing air meter, stopwatch, graphic paper. 1) Calculate the magnetic flux Φ through the density ρ and temperature T (the relationship loop (2 pts). may contain also the parameters defined above) 2) At a certain moment of time, the current in (2 pts). 3) Find a relationship between the air flow ve- the long wire is switched off. What is the net locity in the pipe v and outflowing air density ρ charge Q flowing through a fixed cross-section (the relationship may contain also the parame- of the wire of the loop (3 pts)? 7. b b 1) Let us designated the fluxes in each permanent magnet (magnets A–F in figure) by ΦA – ΦF . Write down equation relating ΦA , ΦB , and ΦC to each other (and possibly to Φ) (1 pt). 2) Write down equation relating ΦA , ΦB , and ΦF to each other (and possibly to Φ) (1 pt). 3) Find the ratio ΦF /ΦC (1 pt). 4) Find the magnetic fluxes in each permanent magnet (2 pts). 5) Which of the magnets is the most difficult one to remove? Motivate your answer (1 pts). ters defined above) (2 pts). 4) Express the power disspation P in terms of the air flow velocity v, the outflowing air temperature T , and density ρ (the relationship may 3) What is the net momentum p given to the loop during the switch-off of the current (express it in terms of Q and the given quantities, if you were unable to calculate Q) (2 pts)? and maximum, t1 ≈ 1.2, and between neighbouring minima t2 ≈ 4.6 (in graph units), respecti1) Volume conservation: Sl = S0 l0 , hence S = k vely. µ = tt12 = π−2 arcsin ≈ 0.261, hence 2π S0 l0 /l. 1 2) At the limit of small deformations, F (l) = a+ k = sin[( 2 − µ)π] ≈ 1.47 ≈ 1.5. b 2 2 l ≈ a − bx/l0 = k0 x, hence E0 S0 /l0 = −b/l0 , 2) For the maximal angular displacement ϕm , hence b = −E0 S0 l0 (1 pt). Besides, at l = l0 , sin ϕ = k = sin[( 1 − µ)π], hence ϕ = m m 2 F = 0 (0.5 pts), hence a + bl = 0 and a = E0 S0 ( 1 − µ)π = 0.75 rad ≈ 3.6 units. Therefore, the 2 (1 pt). unit is ϕm /( 21 − µ)π ≈ 4.8. l0 δl 3) ∆F ≈ dF dl dl = E0 S0 l2 . 4) Newton II law: E0 S0 (1 − ll0 ) = 2K/l, hence 3) If the angular velocities of the planets are ω1 and ω2 , the seeming angular velocity (as seen l = l0 + E2K . 0 SO 5) Conservation of angular momentum: lvt = from the system, where both star and the obserL LVt , hence vt = Vt L+r (1.5 pts). Conservation ver planet are at rest) is ω = ω1 − ω2 . From m 2 2 the Newton II law, GM ri−2 = ωi2 ri , where of energy: 2 (vt + vr ) + E0 S0 (r − l0 ln r+L L )= m 2 2 i = 1, 2 q and ri is the planet’s orbital radius. 2 (Vt + Vr ) (2 pts). √ 6) Substituting vt from the angular momen- So, ωi = GM ri−3 and ω = GM (r2−1.5 − √ tum conservation law into the energy equation r−1.5 ) = GM r−1.5 (1 − k −1.5 ). Finally, the 1 1 L 2 2 2 2 2 we obtain m 2 [Vt ( L+r ) + vr ] + E0 S0 (r − square of the observed period T = (2π/ω) = m 2 2 2 3 −1.5 2 2 l0 ln r+L ) = (V + V ) (0.4 pts). Furtt r 4π r1 /GM (1 − k ) and r1 = [T GM (1 − L 2 her we make use of condition |r| ≪ L and k −1.5 )2 /4π2 ]1/3 . Using T ≈ 4.6 years≈ 1.45 · L 2 substitute ( L+r ) ≈ 1 − 2 Lr + 3( Lr )2 (0.4 108 s, we arrive at r1 ≈ 2.5 · 1011 m; corresponr+L pts), ln L ≈ Lr − 12 ( Lr )2 (0.4 pts). Line- dingly, r2 = kr1 ≈ 3.7 · 1011 m. ar in r terms cancel out due to the condition E0 S0 (1 − lL0 ) = mVt2 /L (0.4 pts). So, we arrive 1 r 2 m 2 2 r 2 2 at m 2 [3Vt ( L ) + vr ] + 2 E0 S0 l0 ( L ) = 2 Vr (0.4 pts). This is the energy conservation law for a pendulum consisting of a spring with effective 3. Tilt-shift lens (6 pts) stiffness keff = (3mVt2 + E0 S0 l0 )L−2 and of a x′ f body with effective mass meff = m (0.5 pts). So, From f −1 = x−1 − x′−1 we obtain x = x′ +f . p √ T = 2πL/ 3Vt2 + E0 S0 l0 m−1 ≈ 2πL/ 3Vt Since the ray passing through the centre of a lens without refraction, from similar triangles we ob(0.3+0.2 pts). tain the relationship between the y-coordinates ′ 2. Planets (6 pts) f of the image: y = y ′ xx′ = xy′ +f . Substituting ′ 1) First method: determine the tangents to the into y = ax + b we result in y f = a x′ f + b, x′ +f x′ +f graph at the points where the curve crosses the x′ ′ ′ ′ hence y = ax + b( f + 1) = x (a + fb ) + b, horizontal axis, a1 ≈ −2.8 and a2 ≈ 16 (in which defines also a straight line. −k graph units), respectively. Then, a2 = ω 1+k k and a2 = ω 1−k . The graph units will cancel out 1) First we notice that the line, its image, and from the ratio of these to tangents, ε = − aa12 = lens plane intersect in one point, because the 1+k 1−ε image of that point of the line which lays at the 1−k ≈ 5.7, hence k = 1+ε ≈ 1.4. Second method (more precise): determine lens plane, coincides with itself. Now, it is easy to the distances between neighbouring minimum construct the image, see the figure. 1. Rubber fiber (12 pts) 2) First we notice that the distance of the image of the far end of the field (let us designate it by A) from the focal plane equals to the focal length f . So, the lens plane must touch the circle of radius f , drawn around A, see figure. Next we notice that there is one such ray connecting far end of the field and its image, which does not refract — the one passing through the enter of the lens, see figure. 4. Transparent film (6 pts) The short-wavelength oscillations on the graph are due to the diffraction on the film, therefore the local maximum condition is 2dn = λN = cN/ν. So, 2dnν = cN and 2dn(ν + δ)ν = c(N + 1), hence 2dnδν = c and d = c/2nδν. In order to measure the distance between two maxima more precisely, we take a longer frequency interval , e.g. ∆ν = 80 THz and count the number of maxima between them, m ≈ 34. Consequently, δν = ∆ν/m ≈ 2.35 THz, and d ≈ 50 µm 5. 4th order ellipse (6 pts) 1) There are trivial positions ϕ = 0 and ϕ = π/2. Besides, there is a position between these two. At the equilibrium, the vector from the origin to the touching point ~r = (x, y) has to be perpendicular to the tangent at that point. In order to find the tangent, let us differentiate the el3 3 lipse formula: 4 xa4 dx + 4 yb4 dy = 0, hence, with 3 4 dx = 1, dy = − xy3 ab 4 , a tangent vector is ~τ = [1, −( xy )3 ( ab )4 ]. The vectors are perpendicular, if the scalar product is zero: x − y( xy )3 ( ab )4 , i.e. y y b 2 b 2 x = ( a ) = ϕ = arctan x = ( a ) . 2) Around each zero ϕ changes sign. At ϕ = 0, small increase in ϕ will result in a torque trying to return to the initial position, i.e. the torque becomes negative. So, the graph looks like the one below. 3) If the derivative of the graph at equilibrium point is negative, the position is stable; otherwise it is unstable. ϕ = 0 and ϕ = π/2 are stable, ϕ = ( ab )2 is unstable. 6. Magnets (6 pts) 1) Each permanent magnet can be considered as a solenoidal molecular current at the surface of the magnets. Suppose that each magnet has net surface current I. Consider triangular contour going through the interiors of the magnets A, B, C. According to the circulation theorem for that contour, the circulation BA l + BB l + BC l is proportional to the overall molecular current through that contour: BA l + BB l + BC l = 3kI. Here, BA designates magnetic inductance inside the magnet A; BB and BC are defined analogously. For a single magnet attached to a massive U-shaped ferromagnetic, the circulation theorem yields B0 l = kI (where B0 is the magnetic inductance inside the magnet; the contribution to the circulation inside a massive Ushaped ferromagnetic can be neglected, because the magnetic field there is much smaller than inside the magnet). So, BA + BB + BC = 3B0 and ΦA + ΦB + ΦC = 3Φ. 2) There are no sources of magnetic field lines (and hence of the flux), so ΦA = ΦB + ΦF . 3) Due to symmetry, ΦF /ΦC = 1. 4) Upon using symmetry, ΦF = ΦC and ΦD = ΦB . From the circulation theorem for the triangle CDE, ΦC + ΦE − ΦB = Φ. From the noflux-source condition for the vertex with magnets E, C, B we obtain ΦE = ΦC − ΦB . Together with the equations form questions 1 and 2, ΦA = 32 Φ, ΦC = ΦF = Φ, ΦB = ΦD = ΦE = 12 Φ. 5) The larger the flux, the more difficult to remove a magnet, because the magnetic flux needs to go through the air gap which will be formed (enlarging the magnetic energy), when starting the removal. So, the answer is “A”. process is by constant pressure, otherwise there would be huge acceleration due to pressure drop). 3) Different air densities inside and outside the pipe give rise to small residual (as compared to the static pressure distribution inside the pipe) pressure difference between the open ends of the pipe, ∆p = −∆ρgL. This pressure difference is responsible for the acceleration of the air, from zero, up to the velocity of the air flow v. The momentum balance for small time interval τ yields S∆pτ = ρ(Svτ )v, hence (ρ0 − ρ)gL = ρv 2 . . Here, the cold air density ρ0 = p0 µ/RT0. Finalp0 µ ly, ( RT − ρ)gL = ρv 2 . 0 γ 4) Heat flux: P = γ−1 R(T − T0 )Svρ/µ. 5) From the result of question 2, we obtain ∆ρ = − ∆T From the result of question ρ T . µ0 I/2πl and B1 = µ0 I/2π(l + a), we end a 0I up with dp = Rb µ2π l(l+a) dΦ. Using the rea+l 0 sult of first question, dΦ = bµ 2π ln l dI, i.e. bµ0 2 a a+l dp = ( 2π ) Rl(l+a) ln l IdI. Finally, p = a(bµ0 I0 )2 8π 2 Rl(l+a) −6 ln a+l kg·m/s2 . l ≈ 2.08 · 10 The same result could have been obtained using the graph and approach used in the alternative solution of the question 1. 9. Experiment (15 pts) The idea: take readings of discharge current, as a function of time. The surface area under the graph is the outflown charge Q. Taking the readings of voltage U0 and U1 at the beginning and at the end of discharge, we obtain Q/C = U0 − U1 , i.e. C = Q/(U0 −U1 ). As for V-I characteristic, interrupt from time to time discharge, take ∆ρ v2 3, ρ = − gL . Substituting these values in- reading of discharge current I just before interto the equation obtained for question 4, P = ruption, measure voltage U , and continue discγ v3 harging. Collect enough data to draw V-I characγ−1 R gL T Sρ/µ. Using the gas equation, this γ gL P 3 simplifies to v = γ−1 S p0 . So, T = T0 [1 + teristic. γ gL P 2/3 ( γ−1 /gL] ≈ 322 K. S p0 ) 8. Loop of wire (7 pts) 1) At the distance r from the current I0 , the 0 I0 magnetic induction B = µ2πr . Then, the R l+a flux through the contour Φ = l Bbdx = R l+a bµ0 I0 bµ0 I0 a+l 2πx dx = 2π ln l . l Alternatively, we can find it using the graph by determining the area S under the curve, from r = r1 = 0, 01 m to r = r2 = 0, 04 m: S ≈ 0, 28 mT·m, further, Φ = Sb = 280 µWb. 2) After switching off the current, the flux through the tends to zero. From the Ohm’s law dΦ R dq dt = dt , hence Rdq = dΦ, i.e. RQ = ∆Φ = Φ. Finally, Q = Φ/R = 280 µC. 3) We calculate the force as difference between 7. Passive air-cooling (9 pts) the forces at the two loop segments parallel 1) Using γ = cp /cV and cp = cV + R we arrive to the straight line: F1 = biB1 and F2 = γ at cp = γ−1 R. biB2 , where i = R−1 dΦ dt . So, dp = (F1 − ρ 2) From the ideal gas equation, p0 = µ RT (the F2 )dt = bR−1 (B1 − B2 )dΦ. Using B1 = ESTONIAN FINNISH PHYSICS OLYMPIAD PROBLEMS & SOLUTIONS (2005) 1. Rock Climber (6 points) 2. Magnetic break (12 points) Foucault currents can be used to decelerate moving metal objects, e.g. a rotating disc. Consider the following simple model. For the sake of simplicity, the disc is substituted by a “circular railway”, see Figure. Plastic “can” of radius r = 15 cm, mass m = 100 g, and height h = 1 cm consists of a homogeneous disc (the bottom of the “can”), and of a much thinner cylindrical wall. Along the edges of the cylinder, there are two wire rings (“rails”), which are connected with a set of parallel wire bars (“sleepers”). Both the “sleepers” and “rails” are made of a copper wire of diameter δ = 0,2 mm; the distance between the “sleepers” L = αh, where α = 0,3. Frictionless rotation of the system is decelerated with an homogeneous magnetic field (B = 1 T) in the slit between the poles of a permanent magnet, see Figure. Assume that the homogeneous field fills a region of rectangular cross-section, equal to the area between three subsequent “sleepers” (i.e. of size h × 2αh). Outside of that region, the 1) Assume that the distance between the climber field is negligible. The specific resistance of the −8 and the last carabiner is L (see Figure). If the copper ρ = 1.724 · 10 Ωm. climber happens to fall, the distance between the highest carabiner and the climber will reach a maximal value l (afterwards, the elasticity of the rope starts lifting the climber). Which inequality S N αh should be satisfied for l ? (1.5 pts) h A rock climber of mass m = 80 kg ascends along a vertical rock. For self-protection, the climber uses the following method. One end of an elastic rope is anchored to the ground. The rope goes through smooth protection loops (carabiners), which are anchored to the rock. The height of the last carabiner is H = 20 m. The other end of the rope goes through a special braking clip which is tied to the harness of the climber. During the climb, this clip keeps rope tight, but enables the climber to lengthen protective part of the rope. (Assume that the rope between the clip and carabiners is always tight) When falling, the maximum acceleration must not exceed amax = 5g (to protect from injuries). You may assume that the rope is always vertical, the distance between the clip and the centre of mass of the climber is very small, and friction between the rope and carabiners is negligible. Relationship between the strain and stress of the rope is sketched on the graph below. 2α h 2) Find the maximal safe length L between the climber and the last carabiner (upon reaching of which he has to anchor a next carabiner; 4.5 pts). B σ(kN) L 6 2r A Suppose the system is motionless. H 4 2 10 20 30 ε(%) istance between A and B is approximately given by R0 = R[ α(α + 2) − α] (2 pts). Now suppose the system is rotating with an angular speed ω = 1 rad/s, and the “rails” are uncut. 3) Sketch an equivalent DC circuit, so that the currents through the resistors are equal to the currents in the respective “railway” elements: “sleepers” and “rail” segments (between subsequent “sleepers”; 2 pts). 4) Using the above obtained results, prove that the (Joule) dissipation power is given by formula P = kB 2 ω 2 /R, and express the constant k (3 pts). 5) Find the decelerating torque M (2 pts). 6) Prove that the angular speed will vanish as ω = ω0 e−t/τ , and determine the time constant τ (2 pts). 3. Ballistic rocket (8 p) A rocket is launched from a pole of the Earth with the first cosmic velocity (near-Earth orbital velocity) in such a way that it lands at the Equator. The radius of the Earth R = 6400 km. 1) Find the longer semi-axes a of the rocket’s orbit (1.5 pts). 2) What is the maximal height of the rocket’s orbit h (from the Earth’s surface; 3.5 pts)? 3) What is the rocket’s flight time τ (3 p)? Remark: The mechanical energy of a planet rotating around a star E = −GM m/2a, where G is the gravitation constant, M — the star mass, m — the planet mass, and a — the orbit’s longer semi-axes (zero potential energy corresponds to an infinite departure). The surface area of an ellipse S = πab, where b is the shorter semi-axes. 4. with water; there are holes of net cross-sectional area S2 along the perimeter of the tank, which are open for the operating regime of the pump. The height of the tank from the free water surface of the reservoir is h (the height of the tank itself is small). An electric engine keeps the vessel rotating at the angular velocity ω. The water density is ρ, the air pressure — p0 , and the saturated vapour pressure — pk . Assume the water flow to be laminar (neglect the energy of vortices) and neglect the friction. Inside the tank, there are metal blades, which make the water rotating together with the tank. Water pump (10 p) 1) Find the resistance R of a single “sleeper” Consider the following construction of a water (1 pt). pump. A vertical tube of cross-sectional area S1 2) Suppose the “rails” are cut near the end-points leads from an open water reservoir to a cylindric of a certain “sleeper” A and B. Prove that the res- rotating tank of radius r. All the vessels are filled 1) Calculate the pressure p2 at the perimeter of the tank, when all the holes are closed (2 p). 2) From now on, all the holes are open. Find the velocity v2 of the water jets with respect to the ground (2 p). 3) If the tank rotates too fast, the pump eﬃciency drops due to cavitation: the water starts “boiling” in some parts of the pump. Find the highest cavitation-free angular speed ωmax (3 pts). 4) If the power of the electric engine is P , what is the theoretical upper limit of the volume productivity of the pump µmax (volume of the pumped water per unit time)? 5. Anemometer (6 points) Anemometer is a device measuring flow rate of a gas or a fluid. Let us look the construction of a simple laser-anemometer. In a rectangular pipe with thin glass walls flows a fluid (refractive index n = 1,3), which contains light dissipating particles. Two coherent plane waves with wavelength λ = 515 nm and angle α = 4◦ between their wave vectors, are incident on a plate so that (a) angle bisector of the angle between wave vectors is normal to one wall of the pipe and (b) pipe is parallel to the plane defined by wave vectors. Behind the pipe is a photodetector, that measures the frequency of changes in dissipated light intensity. 6. Mechano-electrical oscillator (7 points) Mechanical and electrical processes are sometimes strongly coupled. Very important examples are systems containing piezoelectric materials, e.g. quartz resonator. Here we investigate a somewhat simpler situation. There are two metal plates with area S and mass m. One plate is situated atop of the other one. Plates are connected to each other with springs, whose total spring constant is k and what are made of insulator. The lower plate is mounted on a steady base. Equilibrium distance between the plates is X0 . x flow 1) How long is the (spatial) period ∆ of the interference pattern created along x-axis (see Figure; 1) Let us assume that there is a small vertical 2 pts)? shift x of the upper plate from its equilibrium po2) Let the oscillation frequency of the photo- sition. Derive acceleration ẍ of x in terms of sysmeter signal be ν = 50 kHz. How large is the tem parameters. What is the angular frequency fluid’s speed v? What can be said about the dir- ω0 of the small vertical oscillations of the upper ection of the fluid flow (2 pts)? plate (1 pts)? 3) Let us consider a situation, when the 2) Plates are now connected to a constant high wavelengths of the plane waves diﬀer by δλ = voltage source, so that they form a capacitor. 4,4 fm (1 fm= 10−15 m). What is the frequency Electrostatic force between the plates causes an of signal oscillations now (fluid’s speed is the additional shift of the upper plate. The equilibsame as in previous section)? Is it possible to de- rium distance between the plates is now X1 . Determine the flow direction with such a device (2 rive expressions of the electrical attractive force Fe and voltage applied to the plates U in terms pts)? of X0 , X1 , S, m and k (2 pts). photodetector 3) System is set to oscillate again, keeping voltage U constant. Let x stand still for the shift from the equilibrium position. Derive an expresd re w a ve 1 tte sion for the acceleration ẍ of x in terms of X0 , a sc ht g X 1 , S, m, k and shift x. What is the angular freli quency ω1 of upper plate’s small vertical oscillations (2 pts)? 2 wave 4) Let us modify the situation of the previous question and connect an inductor with inductance L in series to the capacitor and voltage source. We describe the situation in terms of plate shift x and capacitor’s charge q. Derive expressions for the accelerations ẍ and q̈ in terms of X0 , X1 , S, m, k, x and q. Which angular frequencies of harmonic oscillation are possible in the system? (2 pts) P(W) 300 200 100 7. Heat exchange (8 points) 1) Consider a simplified model of the air ventilation system of a house using a passive heat exchanger. The exchanger consists of a metal plate of length x and width y and thickness d dividing the air channel into two halves, one for incoming cold air, and another for outgoing warm air. Both channels have constant thickness h, air flow velocity is v see Figure. Thermal conductance of the metal is σ (the heat flux through a unit area of the plate per unit time, assuming that the temperature drops by one degree per unit thickness of the plate). Specific heat capacity of the air by constant pressure is cp , air density is ρ (neglect its temperature dependance). You may assume that the air is turbulently mixed in the channel, so that the incoming and outgoing air temperatures Tin and Tout depend only on the coordinate x (the x-axes is taken parallel to the flow velocity), i.e. Tin ≡ Tin (x) and Tout ≡ Tout (x). Assuming that the inside and outside temperatures are T0 and T1 , respectively, what is the temperature T2 of the incoming air at the entrance to the room (4 pts)? cold air metal plate 200 8. 400 o 600T( C) Balloon (8 points) Find the mass of the balloon (including the gas inside it). Equipment: balloon (floats in air), digital scales, rope, measuring tape, rope fasteners, dynamometer, paper sheets for folding and rough angle measurements, 100-g weight, thread. Remark: You may find it useful to know that if a rope is tied around a balloon so that the tension of the rope is T , the excess pressure inside the balloon is ∆p, the angle between the tangents of the balloon envelope near the rope is 2α (average over the perimeter; see Figure), and the radius of the circular loop formed by the rope is R, then ∆p = T tan α/R2 . The universal gas constant R = 8.31 J/K·mol, the molar mass of air µ = 29 g/mol. 2α warm air 2) Attached is a plot of the heat exchange rate P of the wire of an electric heater as a function of temperature (assuming the room temperature is T0 = 20◦ C). The operating temperature of the wire is T1 = 800◦ C. The heater is switched oﬀ; find the time after which the temperature of the wire will drop down to T2 = 100◦ C. The heat capacitance of the wire is C = 10 J/K (4 pts). 9. Mechanical black box (7 points) There is something small inside the cylindrical “black box”. Find the mass of it, as well as the friction coeﬃcient between it and the inner surface of the box. Equipment: black box, ruler, a wooden plank, timer, scales. 1. Rock Climber 1) In the case of falling, the acceleration should not exceed 5g , which means that σ(ε) m − g < 5g . Maximum strain is the solution of the following equation σ(ε) = 6gm = 6×9.8 sm2 ×80kg = 4.7kN . According to the graph, ε = 0.315; hence, l < 0.315(L + H) + L 2) In the case of falling, the climber reaches the lowest point, when its velocity become zero. This means that the energy absorbed by the rope becomes equal to the change of the potential energy: E = mg(2L + x), where x = l − L. Energy absorbed by the rope is given by E= σ(ε)dx = 0.31 σ(ε)dε ≈ 564.8N S(ε) = 4) Electromotive force in the sleeper is E = Bvh. Energy is dissipated E2 into heat P = Rcircuit . where Rcircuit = 12 (RR + 2αR + R), Rcircuit = Thus, (L + H)S(ε) = mg(2L + x) = mg(2L + ε(L + H)), hence H(mgε − S(ε)) ≈ 5.08m. S(ε) − mg(ε + 2) So, the new carabiner must be anchored within next L = 5.08 m. 2. Magnetic brake 1) Sleeper is a simple cylindrical conductor: ρh R = δ 2 ≈ 5.59mΩ. (2) π 2) Length of the railway element is αR, hence the resistance is R2 = αR. Main ideas: first - we can imagine that railway is infinite; second — the resistance (RR ) of this infinit array remain same even if we cut of one periodic element. Hence, R(2R2 + RR ) . RR = 2R2 + RR + R After solving the equation RR = −R2 ± R22 + 2R2 R = R22 + 2R2 R − R2 and noting that the negative solution of the equation has to be dropped (it does not have physical meaning), we arrive at α(α + 2) − α). 3) Important ideas: P = R( 2B 2 ω 2 r2 h2 α(2 + α) + α + 1) Eventually, 2r2 h2 k= ≈ 2.12 × 10−6 . α(2 + α) + α + 1 5) Since the power equls to M ω = P , the torqe can be found as M= P 2B 2 ωr2 h2 = ≈ 0.39 mNm. ω R( α(2 + α) + α + 1) 6) Disc has a momentum of inertia eual to I = 12 mr2 ; the angular dω acceleration ε = M I = dt . Consequently (using decelerating M), kB 2 ω dω =− . IR dt If we group the variables I and t into diﬀerent side of the equation, we obtain kB 2 dω dt = . IR ω Integrating the both sides of the equation yields • electromotive force is generated when conductors move in magnetic field; t 0 2 kB dt = − IR • There is always two sleepers moving between magnets (in magnetic field); ω = ω0 e − • Those sleepers act as a sources of electromotive force (like a battery); 3. Ballistic rocket • those sleepers also have internal resistance R. 1 R( α(2 + α) + α + 1). 2 Consequently L= RR = R( A R 0 O σ(ε)dε. We know that the maximal value is ε = 0.315, which makes it possible to calculate the integral numerically, as the area under the graph. C αR σ(ε)(L + H)dε = (L + H) B O’ Notice also that we can take account symmetry and connect points with equal potential; this allows us later to simplify cyclic railway to previously solved infinite (actually, very long) railway. We can also see that there is no current between the two sleepers residing in the magnetic field (there is no potential diﬀerence), hence we can disconnect them. So, we can obtain two indipendent (almost) infinite railways and both have their own source of elecromotive force. kB 2 IR ω ω0 t, and finally τ = 2 dω kB ω ⇒ t = −ln , ω IR ω0 IR kB 2 ≈ 2.9s. 1) The net energy depends only on the longer semi-axes. Hence, the longer semi-axes is the same as in the case of near-Earth orbit: a = R. 2) The ellipse has a property that the sum of lengths from each point on the orbit to the both foci of the orbit is constant (equals to 2a). Hence, the other focus (i.e. which is not the centre of Earth) is at the distance R from both the launching point and landing point, see Fig. So, the√height h = |CB| = |OB| − R; since |OB| = R + 12 |OO | = R(1 + 22 ), we finally obtain h = √R2 . 3) The ratio of the flight time to the period along the elliptic orbit equals to the ratio of two surface areas: the one painted dark grey in Fig, and the overall area of the ellipse. The rotation period is the same as in the case of near-Earth orbit (due to Kepler’s third law), T = 2πR/v = 2π R/g . The dark gray surface area is calculated as the sum of half of the ellipse area, and a triangle area. So, √ τ = T · ( π2 R · √R2 + R2 /2)/πR · √R2 = (π + 2) R/g. 4. Water pump 1) Let us consider the process in the system, rotating together with the tank. Then, there is a potential energy related to the centrifugal force: r Uc = 0 ω 2 rdr = 12 ω 2 r2 . So, the pressure p2 = p0 − ρgh + 12 ω 2 r2 . 2) From the Bernoulli formula, 12 ρu2 = p2 − p0 = 12 ω 2 r2 − ρgh, hence the squared velocity in the rotating reference system u2 = ω 2 r2 − 2gh. 2 2 2 2 2 The laboratory speed v2 = u + ω r = 2(ω r − gh), i.e. v2 = 2 2 2(ω r − gh). 3) The point of lowest pressure pm inside the pump is the upmost point of the tube. Using the Bernoulli formula, p0 = pm + ρgh + 12 ρv12 , where the velocity in the tube can be found from the continuity condition: S1 v1 = S2 u = S2 ω 2 r2 − 2gh. Therefore, pm = p0 − ρgh − 1 2 2 ρ(ω 2 r2 − 2gh)( S ) . Notice that the “boiling” starts when pm = pk . 2 S1 2 2 k So, ωm r = 2gh + ( p0 −p − gh)( SS21 )2 ; finally we obtain ρ ωm = r −1 2gh + p0 − pk − gh ρ S1 S2 2 . 4) The maximal productivity is apparently achieved for the highest efficiency. The eﬃciency is highest, when the residual velocity is lowest: u → 0, and ω → √ ωmin . According to the results of the second question, ωmin = r−1 2gh. So, the √ minimal residual velocity of the water streams is vmin = ωmin r = 2gh. The associated lost power is 2 1 µvmin = µgh. The useful power is associated with the potential ener2 gy increase (by gh),i.e. the total power P = 2µgh. Hence, µ = P/2gh. 5. Anemometer 1) First we need to find the angle after the refraction β : For small incidence angles we find approximately β = α/n. In the liquid, the wavelength is decreased n times: λ = λ/n. The requested wavelength can be found as the distance between the lines connecting the intersection points of the equal phase lines of the two beams. Alternatively (and in a simpler way), it is found as the diﬀerence of the two wavevectors: k = kβ , where k = 2π/λ = 2πn/λ is the wavevector of the incident beams. So, ∆ = 2π/k = λ/α ≈ 7,4 µm. 2) The scattered light fluctuates due to the motion of the scattering particles; the frequency is ν = v/∆ = vα/λ. There is no way to determine the direction of the flow, but the modulus is obtained easily: v = νλ/α ≈ 0.37 m/s. 3) The spatial structure of the interference pattern remains essentially unchanged (the wavelength diﬀerence is negligible). However, the pattern obtains temporal frequency δω = δ(c/λ) ≈ cδλ/λ2 . The velocity of the interference pattern u = ∆δω = αc δλ λ . If the fluid speed is v ≈ 0.37 m/s, then the relative speed of the pattern and the fluid is ν = αc δλ ± v , depending on the direction of the flow (in both cases, λ ν ≈ 740 kHz). So, the output frequency allows us to determine the flow direction as long as we can be sure that the interference pattern velocity is larger than the flow velocity. 6. Mechano-electrical oscillator k 1) From the Newton’s second law, mẍ = −kx, hence ẍ = − m x, hence ω = k/m. 2) From the Gauss’ law, the charge on the plate Q = Sε0 E = Sε0 U/X1 . The force acting on it Fe = k(X0 − X1 ) = Q E, where E is the average electric field (averaged over the charges). Let us look at the charge layer (at the surface of the plate) with a high magnification: the electric field there depends linearly on the net charge inwards (in the plate) from the current point. Therefore, the average field is just the arithmetic average of the fields on both sides of the layer: E = E/2. Finally, Fe = k(X0 − X1 ) = QE/2 (this result could have been obtained from energetic considerations, using infinitesimal virtual displacement of the plate andthe energy conservation law). So, Fe = S2 ε0 (U/X1 )2 , hence U = X1 2k(X0 − X1 )/Sε0 . 3) If the plates move by x, the change of the force due to electric field d S is δFe = x| dX ε0 (U/X1 )2 | = Xx1 Sε0 (U/X1 )2 ; bearing in mind 1 2 that S2 ε0 (U/X1 )2 = k(X0 − X1 ), we obtain δFe = 2 Xx1 k(X0 − X1 ). There is also force cahnge due to elasticity: δFk = −kx; the two forces have opposite sign (while approaching the discs, δFk tries to push back, 0 and δFe tries to pull disks even closer). So, δF = −kx[1 − 2( X X1 − X0 X0 k 1)] = −kx(3 − 2 X1 ). Finally, ẍ = δF/m = −x m (3 − 2 X1 ), and ω= k m (3 0 − 2X X1 ) 4) Now we have two oscillating variables, x and q . First, we write down d the equation due to Kirchoﬀ ’s laws: Lq̈ = − Cq − xQ dX C −1 . He1 re, the second term describes the voltage change on the capacitor due to the change of the capacitance (we approximate the real change by diﬀerential, valid for small shifts x). Note that C −1 = X1 /Sε0 and Q = Sε0 U/X1 ; hence −1 d dX1 C = 1/Sε0 , and x q −U . C X1 Here, the sign of the second term assumes that the x-axes is directed upwards (there is no current in the inductance and Lq̈ = 0, if the voltage on the capacitor keeps constant; for increasing charge q > 0, this assumes increasing capacitance, i.e. x < 0; in a full agreement with the signs of the above expression). The second equation describes the Newton second law. First we note that the expression for Fe can be rewritten as Fe = Q2 /2Sε0 . So, if the charge on the plate does not change (q = 0), neither does chand Q2 /2Sε0 = qQ/Sε0 . The infinitesimal force ge Fe . So, δFe = q dQ changes (δFk and δFe ) can be simply added: mẍ = −kx − qQ/Sε0 . Now, let us look for a sinusoidal solution of circular frequency ω . Then, ẍ = −ω 2 x and q̈ = −ω 2 q . Substituting this into the two above obtained equations, we find (Lω 2 − C −1 )q = xU/X1 . (ω 2 m − k)x = qQ/Sε0 This has a non-zero solution for x and q only if (Lω 2 − C −1 )(ω 2 m − k) = U Q/X1 Sε0 . Bearing in mind that U Q/X1 = 2k(X0 − X1 ) and C = ε0 S/X1 , we can rewrite the equation as (ε0 SLω 2 − X1 )(ω 2 m − k) = 2k(X0 − X1 ). Introducing ω02 = k/m and ω12 = X1 /ε0SL we canfurther rewrite as X0 ω 4 − ω 2 (ω12 + ω02 ) + ω02 ω12 3 − 2 = 0. X1 Therefore, 2ω 2 = ω12 + ω02 ± ω14 + ω04 + 2ω12 ω02 (X0 X1−1 − 5), 3 0 i.e. this system has two eigenfrequencies, if X X1 < 2 (and becomes unstable, otherwise). Lq̈ = − 7. Heat exchange 1) It is easy to see that the temperature profile along the plate is linear, and the temperature diﬀerence ∆T between the two plates is constant, ∆T ≡ T0 − T2 . Indeed, then the heat exchange rate q (per unit plate area) is also constant, which in its turn corresponds to a linear temperature profile. Let us use a reference frame moving together with the incoming air. Then, the temperature increase rate at a given point is Ṫ = v(T2 − T1 )/x. Then, the heat balance for a air element of volume V = s × h is written as ρshcp Ṫ = q = sσ∆T /d = sσ(T0 − T2 )/d. So, ρshcp v(T2 − T1 )/x = sσ(T0 − T2 )/d, hence xσT0 + ρhcp vdT1 T2 = . xσ + ρhcp vd 2) Rewriting the heat balance equation P = −C dT as dt −1 dt = −CP dT we conclude that time can be found via the area S under the graph, where P −1 is plotted versus the temperature as t = SC . The graph data: T (K) 100 200 300 400 500 600 700 800 P (W) 13 30 55 83 122 177 258 395 100P −1 7.7 3.3 1.8 1.2 .82 .57 .39 .25 Substituting the region with smooth boundaries with a superposition of trapezoids we find S ≈ 12 K/W. Consequently, t = 120 s. 8. Balloon We can measure the lift of the ball by attaching a weight M = 100 g to it and taking the reading of the scales F/g = m+M −ρV = 73.4 g, where ρ = µp0 /RT ≈ 1.2 g/l. Hence, ρV −M = 27.6 g. So, we need to determine the volume of the ball. To that end, we tighten the rope around the ball as tightly as the fasteners can hold (note that smaller tension values would results in a too small volume decrease, and hence, in a large uncertainty of the final answer). We weight the ball with rope (93.3 g) and subtract the mass of the rope (19.1 g) to find ρV − M = 28.4 g, where V is the ball volume, when the rope is tightened. Hence, the volume decrease ∆V = V − V = (28.4 − 27.6)/1.2 l≈ .67 l. Later, we determine the maximal tension in rope (which can be hold by fasteners) with dynamometer, T ≈ 30 N. We also estimate the average angle tan α ≈ 1.4. The radius of the loop is calculated from the measurement of the perimeter, R ≈ 15 cm. According to the given formula, these data correspond to ∆p ≈ 1900 Pa. Due to gas law, ∆p ≈ ∆V , p0 V ≈ 33 l. Therefore, M = ρV − 27.6 g ≈ 12 g. hence V ≈ p0 ∆V ∆p 9. Mechanical black box First we determine the coeﬃcient of friction as follows. We make sure that the object is at the bottom of the box. We put the box on the plank so that the axes of the cylinder is parallel to the axes of the plank. We start inclining the plank so that the bottom of the cylindrical box gets higher. We determine the angle of the plank α, when the object starts sliding: it hits the cover of the box. It is convenient to put the cover (and the rim of the cover) of the box hanging slightly over an end of the plank. In that case, object hiting the cover of the box results in box falling down from the plank. We measure the tangent of the plank at that moment: µ = tan α = 0.17 ± 0.2. Now we turn the axes of the cylinder perpendicular to the axes of the plank, but keep lying on its side. We start again inclining the plank and determine the angle β , at which the box starts rolling down. Assuming that the object is small (as compared to the radius of the box), the following relationship can be derived: M sin β = m(sin α − sin β), where m is the mass of the object and M — the mass of the empty box. 25±3 Using sin α = 60±5 350 and sin β = 350 we obtain M/m = 35/25 = 1.4 ± 0.2. From the measurement of the net weight M + m = 10.4 g we find m = 10.4 g/2.4 = (4.3 ± 0.4) g. Note that the actual mass was 4.5 g. The mass ratio can be, in principle, determined from the period of small oscillations, T ≈ 0.4 s. Then, if we estimate the moment of inertia of the system box+object as (M + m)r2 , where r is the radius of the m g cylinder, then I ϕ̈ = −mrgϕ, i.e. ω 2 = m+M r . Using r = 16 mm g 2 2 M+m we obtain m = r T /4π = 3.9. This result, however, is rather approximate, because it is diﬃcult to measure such a short oscillation period (at the high dissipation rate). Furthermore, the period is taken to the second power, this explains the unrealistic result (so,it does not make sense to try to improve the approach by taking account the geometric factors for the calculation of the moent of inertia of the box etc). ESTONIAN FINNISH PHYSICS OLYMPIAD PROBLEMS & SOLUTIONS (2006) 1. Drying (12 p) r > 100% According to the wide-spread belief, it is useful to keep window open wen drying laundry even if the relative humidity outside is 100%, because the temperature of the incoming air rises and thereby the relative humidity drops. Let us analyse, do these arguments hold, when heating is switched o. air ure air ure Suppose that inside a room, the volume of V1 = 20 m3 from inside at the temperatt1 = 25 C◦ is mixed with the volume of V2 = 10 m3 from outside at the temperatt2 = 1 C◦ . The specic heat of the air (by xed pressure) cp = 1005 J/(kg·K) can be as- sumed to be constant for the given temperature range; the heat exchange with the medium can 4) is formed)? through the LED, it emits light. If you happened to obtain r > 100% then the oversaturated vapour breaks down into a fog which contains tiny water droplets. In that case, what is the mass m of the condensed water (i.e. the total mass of the water droplets)? Air dens3 ity ρ0 = 1,189 Kg/m ; latent heat of vaporization for water 2. q = 2500 kJ/kg. If a capacitor of capacitance Photography (7 p) istor of resistance voltage will approach exponentially its asymp−t/RC totic value: U = E ± U0 e . camera lens used for this photo. You can as- sume that images created by this camera lens are identical to ones created by ideal thin lens of matching focal length and diameter. Let a large vessel be lled with an incompress- the vapour. ible dielectric liquid of density of mass density ρm Sucking (7 p) (the relative dielectric permeability ε ≈ 1). not change i.e. that the volume of the mixed air This liquid carries homogeneous volume charge V = V1 + V2 . ρe which is so small that the electric E0 created by it is negligible: E0 ρe ≪ gρm , where g is the free fall acceleration. Surface tenof density What is the temperature of the mixed air T? ρ (g/m3) eld sion can be also neglected. All the heights will be measured from the unperturbed height of the liquid surface. 2,0 −q A point charge sign is brought to the height H, of opposite due to which a kink is formed on the liquid surface. 1) 2) 1,5 Determine the height of the kink a. If the height of the charge is slowly decreased, at which height h the liquid will start owing to the point charge? 4. Electric experiment (12 p) Find 1,0 the capacitance of an unknown capa- citor and estimate the experimental uncertainty. Equipment: red light emitting diode (LED), three resistors one of resistance 0 0 5 10 15 20 t ( C) 25 3) The graph below shows the dependence of the saturated vapour density for water as a function of temperature. Before mixing, both the interior and exterior air had relative humidity of r r0 = 100%. What is the relative humidity of the mixed air (if it happens to increase then assume that an oversaturated vapour with E an electromotive force 3. 2) and res- end of the page, determine the diameter of the lect the possibility of (partial) condensation of Prove that the total volume of the air will C are connected in series to By taking measurements from the photo at the be neglected. For the time being, you may neg- 1) R of resistance R2 = 6.2kΩ, R1 = 1.5kΩ one and one of unknown resistance; a battery of unknown electromotive force (internal resistance is smaller than 500Ω), wires, timer, unknown capacitor. Remarks: below is provided a typical V −I curve of a LED; during this experiment, the V −I curve of the LED can be approximated with that of an ideal diode, cf. graph. value of the opening voltage not known. Uc The of the LED is When there is a non-zero current then the capacitors 5. Empty bag (12 p) 6. Car (7 p) between the entry and exit points is l (see gure) ing from rest, as shown in gure. able fabric, impermeable for air, which has sur- meter of its wheel face mass density σ; l. its perimeter less than its length L is much If this bag is lled with air, it resembles a sausage. The bag is laid on a horizontal smooth oor (coecient of friction µ = 0). The excess pressure inside the bag is the free fall acceleration g. p, The density of air d = 1 m, total, see gure. The dia- µ = 1. b; The of those ions which hit B, which diverges near the maximum of the func- ergy per impact parameter range riers which can driven over with such a car, as. axis (i.e. wheels rotate without friction), and the 1) 1) Find the width of the contact surface between driving wheels are the centre of the detector via the quantities 1) 2) 3) l, U 2) Prove that the tension in the bag's fabric T = αx + β , where x is the height of the given point P above the oor, and nd the coecient α. Remark: fabric's tension is (loosely speaking) the force per unit length. Actually, fabric's 2) rear axis. Suppose that the car has four-wheel-drive, and the road barrier is substituted with a wall. Let the highest point of the bag be at height from the oor. What is the tension T1 at this highest point? Express your answer in terms of (or α, if you were unable to answer the previous question). C 4) Hint: consider the force balance between two α Now let us assume that r is negligibly small, Due to dierence in energies, the within the range from 7. Mass-spectrometer(9 p) mass-spectrometer is given. It is a device for T P T x to M + δM . of the mass- 8. Optics experiment (10 p) The sub- Equipment: a cylindrical bottle lled with wa- stance under investigation is ionised by heat- ter and having a millimetre scale across half of measuring the masses of molecules. ing up to a temperature T on a hot lament The ions are accelerated using voltage U. At and l M − δM δM spectrometer? In the gure below, a simplied scheme of a Assuming that p ≫ σg , determine the quantb−a ity ε = b+a , where b is the width of the bag. c of those ions which can still hit the What is the resolving power rst, let us neglect the thermal energy of the b ∆φ eU ≫ kT ). halves of the bag. 4) b enter the detector; but we cannot neglect the thermal eects (still (molecules undergo a single-electron ionisation). a M + ∆M will range width ∆M . to detector can be hit by ions of dierent masses, gure). p Because of the nite size of the detector? Hint: consider the force balance for a small and r (r ≪ l). Under the assumptions of the previous ques- angles acting across piece of fabric (using the vertical cut shown in α(b).] tion, what is the width of the range of the exit an horizontal cut of the sausage. a, σ radius detector entrance, ions within a mass range from 3) describe the fabric's tension here with a single a tion The entrance to the detector is a circle with nd the ponent along the sausage's axis. Thus, we can 3) M e. by driving slowly against the wall? sible; here, however, we neglect the force com- T and M − ∆M sion, because dierent force directions are pos- number, the force per unit length Express the mass Would it be possible to rise the front of the car tension is not as simple concept as a rope's ten- falls onto the droplet with all pos- b < r, the light en∆b is 2I0 πb∆b; hence, the energy per exit angle interval ∆α −1 is ∆I/∆α = 2I0 πb∆b//∆α = 2I0 πb(dα/db) , Determine the maximal height of those road bar- front wheels; I0 sible impact parameters suming that there is no drag in the non-driving c. ions (eU ≫ kT , where e is elementary charge its perimeter (seen when looking through the bottle); measuring tape. 1) Determine, how long arc of the bottle's meas- uring scale could be seen simultaneously if looking at it through the bottle from a very distant A nar- point (much larger than the bottle's diameter), row beam of accelerated ions enters a region assuming that the observation point lies at the with magnetic eld. height of the millimetre scale. k the Boltzmann's constant). For the sake of simpli- city, let us assume that the region has a rect- 2) angular shape, and the magnetic eld is ho- determine the angular radius of a rainbow (the mogeneous inside it. Using the results of the previous experiment The magnetic elds de- angle between a ray coming to the observer's eye ects the ions and depending on their mass, they from a rainbow, and the axis of the cone formed may hit the detector. Let us assume that those by other such rays). ions which hit the centre of the detector enter has a max- the angular radius of the rainbow equals to tensity marks the position of the car's centre of mass. is negligible. the bag and the oor α the maximal exit angle [indeed, if light of in- sketch is drawn using correct proportions, point C The exit angle imum as a function of the impact parameter coecient of fric- tion between the wheels and ground Note that the internal reection is only partial, and not A car attempts driving over a road barrier, startA cylindrical bag is made from a freely deform- the droplet after second refraction. Remark: Rainbow is formed due to those and exit the region with magnetic eld perpen- rays which enter a spherical water droplet, re- dicularly with its boundary, and the distance ect once from its surface internally, and exit 1. Drying 1) Let the number of moles of cold and warm air be ν1 and ν2 ; letCV designate the molar heat capacitance at a fixed volume. Then the total change of internal energy is ∆U = CV [ν1 (T − T1 ) + ν2 (T − T2 ) = (CV p0 /R)(V − V1 − V2 ) (using the ideal gas law). Internal energy change must be equal to the work of the external pressure: (CV p0 /R)(V − V1 − V2 ) = p0 (V − V1 − V2 ), hence V − V1 − V2 (since CV /R 6= 1). 2) The molar amount of gas (p0 /R)(V1 /T1 + V2 /T2 ) = (p0 /R)(V1 + V2 )/T∗ , hence T∗ = (V1 + V2 )/(V1 T1−1 + V2 T2−1 ), i.e. t∗ ≈ 16,5 ◦ C. 3) The vapor mass ma = ρa (t1 )V1 + ρa (t2 )V2 , the mass of saturating vapor at the given temperature mak = ρa (t∗ )(V1 + V2 ). Relative humidity r = ma /mak , because at the fixed temperature, the pressure is proportional to the density. So, r = ρ̃a /ρa (t∗ ), where the weighted average of the vapor ρ̃a = [ρa (t1 )V1 + ρa (t2 )]/(V1 + V2 ) — this value can be found from the graph as the coordinate of the point C : we draw the line at + b, connecting points A and B , and take the reading for the point C lying on the line at∗∗ +b ≈ 1,68 g/m3 at t∗∗ = 17 ◦ C (this value divides the interval [t2 ; t1 ] in the proportions V1 : V2 ). The saturating vapor pressure at the given temperature is found as the coordinate of the point D: pa (t∗ ) ≈ 1,38 g/m3 . Finally we obtain r ≈ 1,22 = 122%. ρ (g/m3) B 2,0 C’ 1,5 C 2. Photographing Let us notice that at the lower part of the photo, there are few brighter spots of regular circular shape and clear edges — unlike all the rest at the smudged (out of focus) part of the image. This can be only due to the point sources in that far area. Let the distance of the linear from the lens be l, and the distance between the sensor and the focus — x. Then, according to the Newton formula, x(l −f ) = f 2 , where f is the = fx . Let the spot diamefocal distance; hence l−f f ter be δ . Then the lens diameter d = δ fx = δ l−f . Let f the size of the image of the linear be a, and the size l . From the of the linear itself — A. Then A = a x+f l−f l−f 1 lens formula, x+f = f l , hence A = a f . Comparing with the previous result we obtain d = δA/a, i.e. the lens diameter equals to the spot diameter, using the scale of the linear. From the figure, we find d = 17 mm. 3. Sucking 1) Let x be the horizontal axes, and y — the vertical axes. At the liquid surface, the potential energy of a unit volume is constant (so that the liquid will not flow towards the lower potential energy). So, the formula for the height χ(x) of the liquid surface is 1 given by Πvp = ρm gχ − 4πε ρe q/r = 0, where 0 E D 1,0 p A 0 4) In order to find the condensating mass, we write down heat balance: cp ρ0 ∆t = q[ρ̃a − ρa (t∗ + ∆t)], where ∆t is the temperature change due to the condensation. By designating t∗ + ∆t = τ we can rewrite the balance as ρa (τ ) = ρ̃a − cp ρ0 (τ − t∗ )/q . So, we need to find the intersection point E of the curve ρa (τ ) with the line ρ̃a − cp ρ0 (τ − t∗ )/q = ρ̃a − 0,478 g·m−3 K−1 · (τ − t∗ ) (line C 0 E in Fig.). Using the graph we find ∆ρ ≈ 0,25g/m3 — this is the length of the line with arrows. So, the condensating mass ∆m = ∆ρ(V1 + V2 ) ≈ 7,5g. Thus, when meteorologists tell us that at the meeting point of cold and hot air, there are heavy rains, the phenomenon can be explained by this problem. 5 10 15 20 t (0C) 25 r = x2 + (χ − H)2 is the distance of the given point from the charge. Let us designate χ0 ≡ χ(0). From the previous formula we obtain (bearing in mind that for x = 0 we have r = H − χ0 ) the result ρe q 1 χ0 (χ0 − H) + 4πε = 0. Using the designation 0 ρm g ρe q 1 4πε0 ρm g = A, the result can be written as p 1 χ0 = (H − H 2 − 4A). 2 2) It is clear that flowing starts at the point x = 0, where the fluid surface is the highest. When the flowing starts, this surface point [with coordinates (0, χ0 )] realizes the potential energy maximum, when moving along the y -axes towards the charge. 1 So, the function Π(y) = ρm gy − 4πε ρe q/(h − y) 0 has a maximum at y = H0 . This gives us two equations: 1 ρm gχ0 − 4πε ρe q/(h − χ0 ) = 0, 0 1 ρm g − 4πε0 ρe q/(h − χ0 )2 = 0. Comparing these, we find h = 2χ0 and χ20 = 1 4πε0 ρe q/ρm g , hence h= p ρe q/πε0 ρm g. 4. Electrical experiment We start with charging the capacitor (waiting long enough, to allow equalizing the voltages of the source and the capacitor, of the order of the discharge time below). The capacitor will be discharge on the diode and two resistances (the unknown one r is parallel to the diode), using the scheme in the figure. We perform two experiments using for the sequentially connected resistor R the both supplied resistors with known resistance, R = R1 and R = R2 . r R Initial voltage of the capacitor U0 = E ; the voltage drop on the diode is constant (while emitting light)— exactly as on a voltage source. Therefore, the voltage on the capacitor approaches that value exponentially: U − Uc = (E − Uc )e−t/RC . Diode stops burning, when all the current I = (U − Uc )/R goes through the unknown resistor, I = Uc /r. Thus, at the fading moment (t = τ ): r(E − Uc )e−τ /RC = RUc . Rewriting the latter equality for the both experiments, r(E − Uc )e−τ1 /R1 C = R1 Uc . r(E − Uc )e−τ2 /R2 C = R2 Uc . Dividing these and taking the logarithm results in τ2 τ1 R1 C=( − )/ ln . R2 R1 R2 Performing for both cases 3–5 measurements and finding the average (τ1 ≈ 37 s, τ2 ≈ 32,4 s), we find C ≈ 13 µF. 5. Empty sack 1) The pressure at the floor P = p + σg , hence p σLg = (p + σg)c, from which c = L/( σg + 1). 2) Here we provide a solution departing from the recommendations (finding the other solution is left for the reader). Let the tension of the material at some contact point with floor P0 be T0 . Consider the energy balance of a piece of material between the points P and P0 for a tiny virtual displacement δ , tangential everywhere to the material (thus, the shape of the material is preserved). The potential energy change (per unit length of the sack) is σδgx (because the piece of material of length δ will get from the floor to the height x); the work done equals to (T − T0 )δ . The energy balance yields T = σgx + T0 , hence α = σg . 3) The force balance between the left and right halves of the sack can be written as T1 + T0 = pa. Bearing in mind that T0 = T1 −σga, we find T1 = (p+σg) a2 . 4) The force balance between the lower and upper halves of the sack: 2T2 + L1 σg = pb, where T2 is the tension at the widest point, and L1 ≈ L/2 — is the length of the upper half. The tension T grows linearly with the height, and the widest point is approximately at the half height; hence 2T2 ≈ T1 +T0 = pa. Substituting it into the first equation, we obtain p(b − a) = Lσg/2. Taking into account that the sack is almost of a circular cross-section, we write π(b + a) ≈ 2L; hence, we finally obtain ε ≈ πgσ 4p . 6. Car 1) Let us consider the force balance projected to the horizontal axes. The only force, which could create a non-zero projection, is the resultant of the friction and reaction force, applied by the corner of the delimiter. Due to the balance, this must be also zero, i.e. this resultant force is directed vertically, hence √ H = d4 (2 − 2) ≈ 15 cm. 2) Consider the torque balance with respect to the point O — the intersection point of the lines of the resultant force applied to the rear wheel by ground, and of the gravity force (vertical line through C ). At the equilibrium, the line of the reaction force applied to the front wheel by the delimiter must go through the same point. Thus, the intersection point of the line OP with the wheel gives us the corner of the delimiter (P is the center of the front wheel). Using the scale of the figure yields H ≈ 10 cm. 3) Consider the torque balance with respect to the point Q — intersection point of the lines of the resultant forces applied to the touching points of the front- and rear wheels with the wall and floor, respectively. Only the gravity force can contribute to the net torque; since Q lies leftwards to the center of mass, this torque rotates car rising its front. So, the front will start rising. Q O case of very large distance (black line d). In the latter case, the ray (in Fig, a) is refracted at the entrance to the bottle by a certain angle ; when observing from smaller distances, one ray (b in Fig) is refracted by the same angle. These two rays coincide after rotation by an angle β around the center of the bottle. So, the part of the scale, given by the gray line in Fig, is longer than the black line at least by 2Rβ . We should perform the measurements with as large L as possible; the result of the measurement is to be adjusted by subtracting 2Rβ , where β = arcsin(l/2L). a β b C R L c d l P 7. Mass-spectrometer 1) The trajectory of a charged particle √ in the magnetic field is circle of radius R = l/ 2. Lorenz force is responsible for the acceleration, Bev = M v 2 /R, hence BeR = p. Substituting p2 = 2M U e = B 2 e2 R2 , we obtain M = B 2 l2 e/4U. 2) Now, the radius can be√R ± r. Approximate calculus yields ∆R/R = r 2/l ≈ ∆M/2M , hence √ ∆M ≈ M r2 2/l, i.e. √ ∆M = B 2 lre/ 2U. 3) Ion leaves the magnetic field at the distance r before (or after) performing a quarter of the circle. So, √ ∆ϕ ≈ r/R = r 2/l. 4) Certain initial energy kT implies that the terminal energy U e+kT = e(U +kT /e); this is equivalent to the change of the voltage by δU = kT /e. Using approximate calculus and the result of the first question, we obtain: δM = dM dU kT /e, i.e. δM = B 2 l2 kT /4U 2 . 8. Optical experiment 1) Looking at the bottle from a distance reveals that the central part of the scale is not reversed, unlike the image at the extreme edge of the bottle. The turning point corresponds to an one end of the visible part of the glued scale (the other end-point is symmetrically situated). Looking from smaller distances results in large visible part (Gray line c in Fig.), than in the β Alternatively, we can measure c by different values of L, and present the results on graph. It makes sense to use 1/L as the scale for the horizontal axes. Then, L = ∞ represents the origin, to which the curve can be easily extrapolated). The measurements yield d ≈ 22 mm (by R = 31 mm). 2) Comparing the ray geometry for the previous problem (in connection to the piece scale c), and the ray geometry in the rainbow, it turns out that the geometry is actually identical, with d2 = R α2 , see Fig. So, α = d/R. Using the data from the previous part, α ≈ 41◦ . α/2 α/2 d ESTONIAN FINNISH PHYSICS OLYMPIAD PROBLEMS & SOLUTIONS (2007) 1. Wire (7 pts) A conducting wire is formed of a cylindrical copper core with a diameter of a = 2,5 mm. The core is wrapped in a concentric, cylindrical aluminium coating, the total diameter of the wire being b = 4 mm. A current of I = 2,4 A flows through the wire. The specific resistivity of copper is ρc = 0,0168 · 10−6 Ω·m and of aluminium, ρa = 0,028 · 10−6 Ω·m. 1) What are the current densities j in different parts of the wire (current density is defined as the current per cross-section area)? 2) What is the magnetic inductance B1 at the distance c = 1 cm from the axis of the wire? 3) What is the magnetic inductance B2 at the surface between the copper and aluminium? R ~ = ~ dl Remark It may be useful to know the circulation theorem: B· µ0 I , where the integral is taken along a closed trajectory (loop) and I is the net current flowing though that loop; µ0 = 4π · 10−7 H·m−1 . This formula is completely analogousRto the formula for the work done by a ~. force along a trajectory: F~ , A = F~ · dl 2. Pendulum (7 pts) Consider an elastic rod, the mass and the compressibility of which (i.e. the length change) can be neglected in this problem. It can be assumed that if one end of the rod is firmly fixed, and a force F is applied to the other end of the rod, perpendicularly to the rod at the point of application, then the rod takes a form of a circle segment. The radius of that circle is inversely proportional to the force, R = k/F , where the factor k is a characteristic of the rod. F R 1) Let the rod be fixed vertically, at its bottom end, and a ball of mass m be attached to its upper end. Knowing the factor k, the length of the rod l, and the free fall acceleration g , find the period of small oscillations of the ball. In this question, you may assume that gml k. 2) What is the maximal mass M of the ball, which can be stably held on such a vertical rod? Remark: you may use approximate expressions sin x ≈ x − x3 /6 and cos x ≈ 1 − x2 /2 (for x 1). 3. Temporal focusing (10 pts) Suppose that at the point A, there is a source of thermal electrons (of negligible thermal energy), which are accelerated initially by the voltage U0 = 36 V in horizontal direction (see figure). At the path of the electrons, there are two voltage gaps B and C of negligible size, at the distance a from each other. These gaps receive a voltage signal UC (t) = −UB (t) ≡ U (t) from a waveform generator. We can assume that |U (t)| U0 . The electrons starting at different moments of time are to be gathered together (focused) at the probe D, which is at distance b from the gap C . In order to analyze this setup, answer the following questions. A e U(t) B D C a b 1) Assuming that U (t) ≡ 0, what is the time needed for the electrons to travel from the gap B to the probe D? 2) Find the same travel time assuming that the voltage U (t) ≡ U U0 is constant (your approximating expression should be a linear function of U ). 3) What functional equation should be satisfied for the waveform U (t) in order to ensure the focusing of all the electrons at the probe D. Solve this equation by assuming a b and |U (t)| U0 . 4) The waveform generator yields a periodic signal of period T in such a way that the profile U (t) is followed up to achieving some maximal value Um ; after that, the voltage drops immediately to 0 and the process starts repeating. What is the fraction of electrons missing the time focus at the probe D? 4. Coefficient of friction (12 pts) Equipment: a wooden brick, a spherical ball, board and ruler (the mass ration of the brick and ball is provided). 1) Determine the static coefficient of friction between the board and the brick. 2) Determine the static coefficient of friction between the ball and the brick. 5. Rotating disk (7 pts) A lamp is attached to the edge of a disk, which moves (slides) rotating on ice. The lamp emits light pulses: the duration of each pulse is negligible, the interval between two pulses is τ = 100 ms. The first pulse is of orange light, the next one is blue, followed by red, green, yellow, and again orange (the process starts repeating periodically). The motion of the disk is photographed using so long exposure time that exactly four pulses are recorded on the photo (see figure). Due to the shortness of the pulses and small size of the lamp, each pulse corresponds to a colored dot on the photo. The colors of the dots are provided with lettering: O— orange, S — blue, P —red, R — green, and K — yellow). The friction forces acting on the disk can be neglected. 1) Mark on the figure by numbers (1–4) the order of the pulses (dots). Motivate your answer. What can be said about the value of the exposure time? 2) Using the provided figure, find the radius of the disk R, the velocity of the center of the disk v and the angular velocity ω (it is known that ω < 60 rad/s). The scale of the figure is provided by the image of a line of length l = 10 cm; 6. Truck (7 pts) 1) A rope is put over a pole so that the plane of the rope is perpendicular to the axes of the pole, and the length of that segment of the rope, which touches the pole is l, much shorter than the radius of the pole R, see figure (a). To the one end of the rope, a force T is applied; the sliding of the rope can be prevented by applying a force T1 to the other end of the rope. Express the ratio T1 /T via l, R, and µ, where µ is the coefficient of friction between the rope and the pole. 2) Answer the first question, if l is not small (i.e. without the assumption l R). Remark: you may use equality lim (1 + nx)1/x = en . x→0 3) The rope makes exactly n = 2 winds around the pole. One end of the rope is attached to a truck standing on a slope (slanting angle φ = 10◦ ); the mass of the truck m = 20 t, see figure (b). Find the force F , needed to apply to the other end of the rope, in order to keep the truck at rest. Use the numerical value µ = 0,3. All the other friction forces acting upon the truck can be neglected. 4) How does the answer change, if the cross-section of the pole is not circular, but instead, egg-shaped? Motivate your answer. (a) T1 (b) T 7. To the Mars (10 pts) In this problem, we study a project for flying to Mars. At the first stage, the space ship switches on the rocket engines and obtains an initial velocity v0 . You may assume that during the first stage, the height of the space ship (from the surface of the Earth) remains much smaller than the radius of the Earth R0 = 6400 km. At the second stage, the space ship performs a ballistic motion in the gravity field of the Earth: upon achieving a height, which is much larger than the radius of the Earth R0 , but much smaller than the orbital radius of the Earth, Re = 1.5·108 km. 1) Find the relationship between the residual velocity v1 (with respect to the Earth) at the end of the second stage, and the quantities v0 , R0 , and the free fall acceleration at the surface of the Earth g (for the subsequent questions you may use the numeric value g ≈ 9.8 m/s2 ). At the third stage, the space ship performs a ballistic motion in the gravity field of the Sun, up to reaching an immediate neighborhood of the Mars. The trajectory is chosen by minimizing the residual launch velocity v1 (required for achieving the Mars). 2) Sketch the trajectory. 3) Find the flight time T . You may use the following numeric data: the orbital velocity of the Earth ve = 30 km/s, orbital radius of the Mars Rm = 2.3 · 108 km. 4) Find the previously considered launch velocity v0 and the terminal (i.e. at the end of the third stage) velocity vt of the space ship with respect to the Mars. 5) The required mass of the fuel M can be found from the formula v = u ln[(M + m)/m], where u is the speed of the gas at the outlet of the engine (with respect to the space ship), and m is the useful mass of the ship (when all the fuel is exhausted). You may assume m M and use the numerical value u = 1 km/s. Find, how much more fuel is needed for the Mars flight, as compared to simply escaping the Earth gravity field, if the useful space ship mass is equal in both cases. 8. Laser (12 pts) Apparatus: Laser (wavelength λ = 650 nm), ruler, stand, a strip of reflecting material, a sheet of paper with a circular hole (in your pack of paper sheets), pencil. Note that the reflecting material provided to you is coated with a layer of densely packed tiny glass spheres of equal diameter. 1) Describe the position and geometry of the diffraction pattern, which can be observed, when the laser beam falls onto the strip; use different incidence angles. 2) Provide a qualitative (approximate) explanation for the observed phenomenon. 3) Estimate the diameter of these glass spheres. 1. Wire (7 pts) 1) Using the Ohm’s law in differential form, E = ρj , and noting that the electric field must be the same both in the core and in the coating, we conclude jc ρc = ja ρa ⇒ jc = æa ρa /ρc . On the other hand, the net current I = π4 [a2 jc + (b2 − a2 )ja ] = π4 ja (b2 − a2 + a2 ρa /ρc ), hence I 4 ≈ 0.15A/mm2 ; jc ≈ 0.25A/mm2 . ja = π b2 + a2 (ρa − ρc )/ρc 2) Writing down the circulation theorem for a circular concentric loop of radius c around the straight wire, 2πcB1 = µ0 I , we obtain B1 = µ0 I/2πc = 2 Ic · 10−7 H·m−1 = 4.8 · 10−5 T . 3) Using the technique as before, but noting that the current flowing through the smaller loop of radius a/2 is Ic = jc π4 a2 , we obtain B2 = µ0 Ic /πa = µ0 jc a/4 = πjc a · 10−7 H·m−1 = 2.0 · 10−4 T . 3) For the changing voltage, we can use the result of the previous question, but the voltage value should be taken at the moment of electron passing the gap. Also, the terminal velocity is changed, v1 − v0 is related to the change of the kinetic energy e[U (t + ta ) − U (t)]. The travel time of the later electrons passing B at a certain moment of time t > 0 should as much shorter as it was delayed, i.e. r m t1 − t2 = {a[U (0) − U (t)] + b[U (t + ta ) − U (t)]} = t, 8U03 e where the flight time ta ≈ a/v0 [since U U0 ]. Using the suggested approximation, we obtain r m b[U (t + a/v0 ) − U (t)]} = t. 8U03 e If we seek a quadratic solution U (t) = At2 + Bt, we get 2. Pendulum (7 pts) 1) The smallness of the oscillations means that the angle of the arc formed by the rod is small, α = l/R 1. Therefore, the force returning the ball is almost horizontal, F = k/R = kα/l, and the horizontal displacement of the ball is x = R(1 − cos α) ≈ Rα2 /2 = lα/2, hence α = 2x/l. For horizontal equation of motion, we can neglect the gravity force, which is of the order of mgα αk/l; the right-hand side here hap2 pens p to be the elastic force. So, mẍ = −kα/l p = −2kx/l , hence ω = 2k/m/l; the period T = 2π/ω = πl 2m/k . 2) Elastic energy of the rod depends apparently only on its shape, i.e. on the curvature radius R, or, equivalently, on the binding R x angle α. First, we derive the expression for that energy: Πr = 0 F dx = R (2kx/l2 )dx = kx2 /l2 = kα2 /4. This is to be compared with the change of the gravitational potential energy Πg = −mg(l − R sin α) = −mgl(1 − sin α/α) ≈ −mglα2 /6. So, the vertical position is stable, if kα2 /4 − mglα2 /6 > 0, i.e. 3k > 2mgl. Remark: a simple-minded force balance of the ball to the direction, perpendicular to the rod leads to another result k > mgl. This is not correct, because if we bind the rod with a force applied to its tip, small displacements of the tip are not perpendicular to the rod (as one might think). In particular, this means that if there are both perpendicular force F and tangential tension T in the rod, the curvature radius R will depend also on the tension T . In our solution, we avoided such kind of complications by noting that the elastic energy depends only on the shape of the rod (if we bind the rod with some force F while keeping T = 0, we don’t need to bother about this effect). 3. Temporal focusing (10 pts) p 1) Apparently, t0 = (a + b)/v0 = (a + b) m/2U0 e. 2) Apparently, t = a/va + b/vb ≈ t0 − [a(va − v0 ) + b(vb − v0 )]/v02 . On the other hand, for constant voltage U (t), vb = v0 . For the interval between B and C , we have ∆(mv 2 /2) ≈ mv0 (va − v0 ) = −U e. So, r U a m t = t0 + U ea/mv03 = t0 + . U0 2 2U0 e t 2Ata (t + ta /2) + Bta = b So, r r r 8U03 e . m p 8U03 e 8U03 e 2eU0 /m 4U02 e /2ta b = = , m m ab mab and B = −Ata . Finally we obtain p 4U02 e t t − a m/2U0 e . U (t) = mab 4) Lost are those electrons, which are in the interval between B and C , when the voltage drops to 0. The duration of the time interval, when these werepemitted, is τ = a/v0 . So, the asked fraction is given by τ /T = Ta m/2U0 e. A= 4. Coefficient of friction (12 pts) 1) This is a straightforward question: we measure the tangent of the slope of the board, when the brick starts sliding, µ1 = tan α1 , for the given setup, µ1 ≈ 0.24. 2) We put the ball and brick together on the slope, the ball touching both the brick and board and being upwards on the slope. So, the ball pushes the brick down and the sliding is expected to take place for somewhat smaller sloping angles. For the given setup, that critical angle turns out to give tan β ≈ 0.12. Now we need to express µ knowing µ2 ≈ 0.24 and β = arctan 0.12. First we write the torque balance for the ball with respect to the touching pint with the board: mgR sin β = µNR + NR, where N is the pressure force between the ball and the brick. Hence, mg N = µ+1 sin β . Now, we use the normal (to the board) force balance for the brick, to find the pressure force between the board and the brick: N2 = M g cos β + µmg µ+1 sin β . Finally, we have the tangential force balance for the brick: mg µm sin β + M g sin β = µ2 g M cos β + sin β . µ+1 µ+1 This equation can be simplified to h i m m + µ + 1 = µ2 (µ + 1) cot β + µ , M M from where µ= m M + 1 − µ2 cot β . m µ2 (cot β + M )−1 Using the measured values we get µ ≈ 0.2. 5. Rotating disk (7 pts) 1) We notice that there is no image of the orange pulse, hence it must have taken place immediately before the shutter release. So the blue pulse is first, red — the second etc. The exposure time must have been triple and quadruple flash interval, 300 ms < t < 500 ms. 2) The displacement of the lamp between two subsequent pulses can be represented as the sum of two components: ~ri = ~v τ +2R sin(ωτ /2)~ei , where each next unit vector ~ei+1 is rotated with respect to the previous one (~ei ) by angle ωτ . So, if the starting points of the displacement vectors ~ri coincide, then the end-points must be on a circle, at equal angular distances ωτ from each other, see figure. In our case we redraw the displacement vectors 1, 2 and 3 as vectors with common origin, P~A, P~B , and P~C . Since the starting points of the vectors 2R sin(ωτ /2)~ei are brought together to the point O, their endpoints lay on the circle, the center of which can be found as the center of the circle drawn around the triangle ABC . B 2 3 2 O P 1 1 A L 3 C The velocity of center of the disk is found as the ratio of the length P O and the interval τ : v ≈ 65 cm/s. The angular velocity is found as the ratio of the angle 6 AOB = 6 BOC and the interval τ : ω ≈ 23 rad/s. Radius of the disk is found from the length |OA| = 2R sin(ωτ /2) = 2R sin 6 BOC ≈ 1.5R; using the scale of the figure, 1.5R ≈ 8 cm and R ≈ 5 cm. 6. Truck (7 pts) 1) First, since l is small, T1 ≈ T . From the radial force balance, N ≈ T α = T l/R, where α is the angle, by which the direction of the rope is changed. From the tangential balance, T1 = T − µN = T (1 − µα). 2) If the angle is not small, we divide the touching segment into M small segments and use the previous result: Ti+1 = Ti (1 − µα/M ). So, TM = T (1 − µα/M )M → T e−µα (as M → ∞). So, T 0 = T e−µl/R . 3) Using the previous formula and the provided numerical values (α = 2π ), we obtain T = M g sin φe−µα ≈ 800 N. 4) If we re-examine the solution to the second question, we notice that we haven’t used the assumption of cylindrical shape. What matters, is just the rotation angle of the tangent to the rope. So, the answer does not change for egg-like cross-section. 7. To the Mars (10 pts) 1) Apparently v12 = v02 − 2gR0 . 2) The trajectory is ellipse touching Earth orbit at its perihelion and Mars orbit at its apohelion.. 3) According to the Kepler’s third law, the time T = 21 T0 (a/Re )3/2 , where T0 = 1 year is the period of Earth and a = (Re + Rm )/2 is the longer semi-axes of the trajectory. Numerically, T = 0, 70 years. 4) The full energy (kinetic plus potential) of the space ship is E = −Gm/2a = −Gm/(Re + Rm ) = −Gm/Re + m (v1 + ve )2 . So, 2 2 Rm 2 Rm 1 2 (v1 + ve ) = G Re (Re +Rm ) = ve Re +Rm and v1 = ve r ! 2Rm −1 Re + Rm ≈ 0.095ve ≈ 2.86 km/s. √ Hence, v0 = 11.22 + 2.862 = 11.6 km/s. The relative speed near the Mars can be written using the appropriate change of indices in the expression for v1 (alternatively, it can be found analogously to v1 ): we first find the speed with respect to Sun from the expression of the total energy: v3 = v3 = vm r ! 2Re −1 Re + Rm ≈ −0.106ve r Re ≈ −2.59 km/s. Rm 5) Apparently Mi = mevi /u , so that Ma /Mb = e(va −vb )/u = e0.4 ≈ 1.5. 8. Laser (12 pts) 1) These are concentric circles reflected back towards the laser. They can be seen, if laser light is directed through the hole in the paper and the paper is used as the screen. The position and size of the circles is independent of the incidence angle (for large incidence angles, the circles will be somewhat elongated along the direction of slanting). 2) The exact calculation of the diffraction pattern on such microspheres is very difficult task, evidently beyond the possibilities of an olympiad problem. So, we have to work as a detective, drawing conclusions from the observed data. First, the diffraction pattern is not where the reflection form the stripe would be; so, it doesn’t work as a reflecting diffraction grating. If the packing were regular, and the optical path difference (required for the diffraction) were gathered due to scattering on neighboring spheres, there would be a pattern, characteristic to diffraction on crystal lattices (regularly positioned spots). If the optical path difference were gathered due to scattering on neighboring spheres, with irregular packing, the path difference would be random, and hence, the diffraction pattern would be also random. This would not explain the regular circular pattern. On the other hand, if the optical path difference were gathered on a single sphere, the pattern would be axially symmetric, as is the sphere itself. So, this fits well with all the experimental observations (except for slight elongation of the circles for large incidence angles, which may, perhaps, be explained by the refraction in the coating layer). On a single sphere, the light is reflected partially back on the front surface, and partially refracts into the sphere, reflects internally and refracts back into the air (in our rough analysis, we neglect multiple partial reflections). For beams falling close to the axes of the sphere, the outgoing beams are reflected almost in the opposite direction. So, in its central part, the sphere acts almost like a reflecting disk. In our very rough model, we substitute the spheres by disks of approximately the same diameter as the spheres, oriented perpendicularly to the incident beam. Then, all the microspheres provide maxima and minima in the same directions. 3) We measure the diameter of the first dark ring d ≈ 1.1 cm, and the distance between the laser and the strip l ≈ 80 cm. Then, the diameter of the spheres can be estimated as the diameter of the effective lightreflecting disk, D ≈ λl/d ≈ 50 µm. ESTONIAN FINNISH PHYSICS OLYMPIAD PROBLEMS & SOLUTIONS (2008) Eestonian-Finnish Olympiad - 2008 1. Dumbbell (a) (6 points) Two perfectly elastic identic balls of mass m are connected by a spring of stiness k so that a dumbbell-like system is formed. That dumbbell lays at rest on a slippery horizontal surface (all the friction forces can be neglected). Third ball (identic to the ones making up the dumbbell) approaches coaxially the dumbbell from the left side with velocity v (see Figure). Fourth ball (identic Ti nano-disk heater thermal to the other ones) lays coaxially rightwards to the dumbbell. temperature sensor bridge 1) Find the velocity of the centre of mass of the dumbbell after 1) Find the thermal resistance R between the microcalorimeter being hit by the ball approaching from the left. and the wafer (i.e. the ratio of the temperature dierence and 2) For which distances L between the dumbbell and the right- heat ux). most ball, the nal velocity of the latter will be exactly the same, For questions (ii) and (iii), use quantity R, without substias the initial velocity v of the leftmost ball? tuting it via the answer of question (i). v L 2) Write down the heat balance equation for the microcalorimeter and nd the temperature of the microcalorimeter as a function of time T (t) [you may seek it in the form T = T + ∆T sin(ωt + ϕ)]. 3) In order to study the thermal properties of the Ti-nanodisks, the amplitude of the sinusoidal oscillations of T (t) should change by as large as possible value, as a response to a small change of 2. Microcalorimeter C (which is caused by the Ti-disks). Find the optimal circular frequency ω . A microcalorimeter is a thin circular silicon nitride membrane, 4) We have assumed that the temperature prole along the thermally isolated from the surroundings, except that it is bridges is linear, i.e. their heat capacity can be neglected. For thermally connected to the wafer by four thin and narrow thermal high frequencies ω & ω , this is not the case. Estimate the critbridges (see Figure). The microcalorimeter is equipped with a ical frequency ω in terms of κ, l, specic heat c and density ρ of small heater in the middle of the membrane and a similar struc- the bridge material. ture on the edge of the membrane working as a thermometer. This micro calorimeter is used to study the thermal properties of nanoscale Ti disks (light tiny dots in Fig). The thermal power of the heater depends sinusoidally on time, P = P cos(ωt) (negative power implies a withdrawal of heat). The circular frequency ω 3. Tractor is suciently low, so that for any moment of time t, the temperature of the microcalorimeter T (t) can be considered constant Provided sketches (a) and (b) are made on the basis of satelacross its entire surface, and the temperature prole along the lite images, preserving proportions. They represent tractors, tothermal bridges can be considered linear. The wafer, to which gether their smoke trails. The tractors were moving along the the bridges are connected, is large and thick enough, so that its roads in the direction indicated by the arrows. The velocity of temperature T can be considered to be constant all the time. the tractors was v = 30 km/h. For sketch (a), the direction Each of the four bridges have length L and cross sectional area of all S ; the thermal conductance of them is κ. Thermal conductance of wind is indicated by another arrow. is dened as the heat ux (measured in Watts) per surface area, 1) Using the provided sketch, nd the wind speed for case (a). assuming that the temperature drop is 1 C per 1 m. The heat capacity of the microcalorimeter (with Ti-disks) is C . 2) Using the provided sketch, nd the wind speed for case (b). 0 (9 points) 0 c c 0 (6 points) 0 0 ◦ (b) 4. Magnetic eld (6 points) Magnetic eld with inductance B (parallel to the z-axis) lls the region x + y < R . Let us consider an electron of velocity v = RBe/m (where e is its charge and m its mass). 1) Sketch the trajectory of the particle, if initially, it moves along the line y = 0 towards the region lled with magnetic eld. 2) How long time does the electron spend inside the magnetic eld? 3) Now, let us consider the situation, when initially the electron moves along the line y = a (a < R). Find the angle α, by which the electron is inclined after passing through magnetic eld. 2 5. Ball 2 (9 points) 2 ruler, a glass ball, sheet of paper, marker. Find the coecient of friction between the glass ball and the ruler. Estimate the uncertainty of you result. Equipment: 6. Rectier (8 points) A voltage rectier is made according to the circuit depicted in Figure. The load R = 10 kΩ is fed with DC, equal to I = 2 mA. In what follows we approximate the U-I characteristic of the diode with the curve depicted in Figure. The relative variation of the current at the load has to satisfy the condition ∆I/I < 1%. 9. Asteroid (7 points) It is believed that the impacts of the Earth with asteroids have played an important role in the history of Earth. In this problem, you are required to study such an impact. As an example, let us use the orbital data of the Apollo asteroid. Its perihelion is 0.65 AU, i.e. its closest distance from the Sun is r = βR, where β = 0.65 and R denotes the radius of the Earth's orbit; its aphelion is 2.3 AU, i.e. its farthest distance from the Sun is r = αR, where α = 2.3. In your calculations, you may use the orbital velocity of Earth, v = 30 km/h, Earth radius R = 6400 km, and the free fall acceleration at the Earth's surface, g = 9.8 m/s . You may also use the formula E = −GmM/2a, expressing the total energy of a body of mass m, which moves along an elliptic orbit of longer semiaxis a, in the gravity eld of a much heavier body of mass M . You may assume that the orbital planes of the Earth and of the asteroid coincide, and that they rotate in the same direction around the Sun. 1) Find the velocity v of the asteroid in the vicinity of the Earth, in the Sun's system of reference, neglecting the eect of Earth's attraction. 2) Find the radial and tangential components v and v of that velocity (i.e. the components, respectively parallel and perpendicular to the vector, drawn from the centre of the Sun to current position of the asteroid). 3) Find the same velocity components u and u in the Earth's system of reference. 4) Find the velocity of the asteroid w, immediately before entering the Earth's atmosphere (at the height h = 100 km from the Earth's surface). 1 2 0 0 2 Find the average power dissipation at the diode at the working regime of such a circuit. 2) Determine the amplitude of the AC voltage (with frequency ν = 50 Hz), which has to be applied at the input of the circuit. 3) Find the required capacitance C . 4) Find the average power dissipation at the diode during the rst period (of AC input voltage) immediately following the application of AC voltage to the input of the circuit. 1) 7. Fire (6 points) r t There is wet wood burning in a replace on the ground. Seven meters above ground, the smoke is at a temperature of t = 40 C. Disregard the exchange of heat with the surrounding air and assume that the atmospheric pressure at the ground is constant in time and equal to p = 1000 hPa; the air temperature t = 20 C is independent of height . Assume that the smoke represents an ideal gas of a molar mass µ = 29 g/mol (i.e. equal to the molar 10. Glass plate Laser (λ = 650 nm), thin glass plate, lens, ruler. mass of the air), and of a molar specic heat at constant volume C = 2.5R; universal gas constant R = 8.31 J/kg · K. How high NB! glass plate is xed to a stand; avoid touching the glass itself (because its edges are sharp, and because it can break easily). will the smoke column rise? Find the thickness of the plate and estimate the uncertainty of the result. Draw the scheme of your experimental setup. 8. Electron Electron rests at the origin. At the moment of time t = 0, an electric eld is switched in: its modulus is constant and equal to E , but its direction rotates with a constant circular velocity ω in the x − y plane. At the moment t = 0, it is directed along the x-axes. 1) Find the average velocity of the electron over a long time interval for t > 0. 2) Sketch the trajectory of the electron and calculate the geometrical characteristics of it. 7 0 1 0 ◦ ◦ r t (10 points) Equipment: V (5 points) 0 1 Actually, during day time, this is not the case: air temperature decreases with height. However, during evening and night, due to heat radiation, the lower layers of air cool more rapidly than upper layers, and it may easily happen that the temperature is roughly independent of height. The heat contained in the bridge must be comparable with the 2) Circular part of the trajectory is a quarter of the full circle, heat, which ows through it during one half-period (if it is much so t = πR/2v. 1) During the rst collision, we can neglect the eect of the the stationary linear prole will develop very soon). So, 3) Let O be the centre of the circular orbit of the electron and B spring, because during the collision time, the balls almost don't smaller, the intersection point of the trajectory with the region boundmove, hence the spring doesn't deform. Two absolutely elastic AcρSL ≈ AκS/(Lω ); hence, ary. The polygon COBO is rhomb, because all the sides are ω ≈ κ/cρL . identic balls exchange velocity during a central collision. So, the equal to R. So, the line BO is vertical (because O C is vertical rst ball will remain at rest, and the second one will obtain the and BO is parallel to it). Hence, the inclination angle of the velocity v. So, the velocity of the centre of mass of the dumbbell 3. Tractor electron is is v/2. π a α = ∠CO B = ∠AOB + ∠AOC = − arcsin . 2) After the impact, the dumbbell will oscillate in the sys2 R tem √of reference of its centre of mass with circular frequency 5. Ball ω = 2k/m (balls oscillate so that the middlepoint of the spring is at rest; twice shorter spring has a twice larger stiness). Due to the energy conservation law, the only way for the fourth ball to acquire the velocity v is such that all the other balls remain at a complete rest after the interaction. Therefore, before the impact of the third and fourth balls, the third ball must have velocity v (and the second ball must be at rest). This is the opposite phase of the moment, when the dumbbell started its motion. Hence, the travel time t of the dumbbell must be a halfinteger multiple of the period T = 2π√m/2k. For that phase of oscillation, the spring is, again, undeformed, i.e. the travel We lay one of the rulers horizontally on the table. Then, we put distance of the centre of mass is also L. So, 2L/v = T (n + ), the ball on that ruler, and the other ruler laying on the ball. With hence ( ) nger, we keep one end of the second ruler in contact with the 1 √ L = πv n + m/2k. rst ruler and nd the closest stable position of the ball (resulting 2 in the largest inclination angle of the second ruler), see Figure. 1) Let us draw from an arbitrary point B on the road a line par2. Microcalorimeter we consider the torque balance (for the ball) with respect to allel to the direction of the wind, and let it intersect the smoke Now, the ball ruler touching point B. Gravity force has no torque, 1) Every bridge has thermal resistance L/κS ; so, the overall restrail at point C . Then, the smoke emitted by the tractor at B because and it is to the centre of the ball O. So, the resultant istance is R = L/4κS. has travelled the distance |BC| = ut, where u is the wind speed. vector of the applied friction reaction forces at C must have also zero 2) The power dissipation P results in an heat ux through the Tractor itself has travelled the distance |AB| = v t. So, we can torque, i.e. it has to goandthrough B . At the threshold of sliding, the bridges, Φ = ∆T /R, and in the change of the heat contained in measure the microcalorimeter, Q̇ = C Ṫ = C∆Ṫ (here, dot denotes the late the distances |AB| and |BC from the gure and calcu- angle between this vector and the surface normal CO is arctan µ. So, µ = tan ∠BCO = tan ∠OAB = |OB|/|AB| = R/|AB|. The time derivative). So, |BC| 18 mm u=v = 30 km/h ≈ 13 km/h. radius of the ball R ≈ 40 mm can be measured by rolling the ball P cos(ωt) = C∆Ṫ + ∆T /R. |AC| 42 mm Now, we can search the√ solution as ∆T = A cos(ωt + ϕ), and 2) If the second tractor (at the right-hand-side) had started some- on the ruler by angle 2π. The distance |AB| can be measured denote ψ = arcsin(Cω/ √C ω + R ). Then, what earlier, the two tractors had been at the crossroad simul- directly using the ruler. Several measurements are needed, to taneously. Now, the tractors would be at the same distance from nd the critical position of the ball more accurately. P cos(ωt) = A C ω + R √cos(ωt + ϕ − ψ). the crossroad, i.e. for the current position of the second tractor We have used one ruler as the basis, because if the surface has So, we must have ϕ = ψ( and A = P / C√ω + R , i.e.) C , |OC| = |AO| = v t (this is how we nd the point C ). Its smaller coecient of friction than the ruler, the sliding starts at P cos ωt + arcsin(Cω/ C ω + R ) √ T =T + . smoke trail can be found as a line, parallel to its smoke trail at the point B, hence we are not able to obtain the required result. C ω +R its actual position B. Such a meeting of the tractors would have 6. Rectier ; it been resulted 3) The amplitude of the oscillations is A = √ crossing of the smoke trails. which would be 1) Since none of the DC current through the load can come from must be as sensitive with respect to the small changes of C , i.e. now in positionin Dthe, with OD = ut. So, we nd dA/dC must be maximal by modulus. dA/dC = P (C ω + the capacitor, all must come through the diode. Hence, the aver|OD| 27 mm = 30 km/h ≈ 21 km/h. u=v R ) Cω ; if we denote x = (Cω) , we need to minimize the age current through the diode is also I = 2 mA, and the average |AO| 39 mm following function of x : power dissipation is obtained by multiplying it with the diode [ ] 4. Magnetic eld ln (dC/dA) = 3 ln(x + R ) − ln x + ln C. voltage u = 1 V: P = 2 mW. Upon taking derivative and putting it equal to 0, we obtain 1) Since the radius of the cyclotron orbit is equal to the radius 2) If the diode is open, Uload(t) = U cos(ωt) − u. If the di3x = x + R , from where x = R of the region R, the trajectory is given by the curve DABE in ode is closed [i.e. U cos(ωt) < Uload(t) + u], the capacitor dis√ /2, i.e. the Figure (AB is a circle fragment). ω = 1/ 2CR. charges through the load. However, the relative change of the 1. Dumbbell 4) (6 points) ′ c c ′ 2 ′ (6 points) ′ (9 points) 1 2 (9 points) 0 0 0 2 2 0 −2 2 −2 2 2 0 2 0 0 2 −2 2 2 −2 0 −2 2 P0 (8 points) C 2 ω 2 +R−2 −2 −3/2 0 2 2 −2 2 −2 −2 2 2 0 (6 points) 0 0 voltage of the capacitor has to be small (otherwise ∆I/I would We can integrate these equations over time (bearing in mind that not be small). The respective load voltage as a function of time initial velocity is zero): is sketched in the Figure. So, we can use the above written mẋ = −eE ω sin ωt, Kircho's law with Uload(t) ≈ IR, hence U = IR + u = 21 V. mẏ = eE ω (1 − cos ωt). Now, we can integrate once more, bearing in mind that the initial coordinates are zero: 0 0 −1 eE0 (cos ωt − 1), mω 2 eE0 eE0 y= cos ωt + t. mω 2 ωm x= The change of the voltage of the capacitor during the discharge cycle can be estimated as ∆U = ∆Q/C , where the capacitor's charge drop ∆Q = It, and t is the discharge time. Since the discharge cycle occupies almost all the period (see Figure), we can use t ≈ 1/ν . Further, ∆I/I = ∆U/U = ∆Q/CU = ∆Q/CIR = 1/CRν . Hence, C ≥ 100/Rν = 200 µF. 4) Initially, the capacitor is empty, so that the charge owing through the capacitor during the rst cycle is Q = CIR. Hence, the average power P = Quν = CIRuν = 200 mW. 1 7. Fire (6 points) The smoke will rise until its density becomes equal to the density of the air at the same height. Since the molar masses and pressures of the smoke and air are equal, this implies also equal temperatures (pressures are equal, because otherwise, there would be no mechanical equilibrium). Temperature of the smoke will drop with increasing height due to adiabatic expansion. If we combine the law of the adiabatic process pV = Const with the ideal gas law (pV /T ) = Const, we obtain p /T = Const. Taking a logarithm and dierential from this equation, we obtain = 0, hence we can use approximate expression (γ − 1) − γ for the require temperature change So, the electron performs circular motion in the system of reference, moving with velocity (parallel to the y-axis) u = . The radius of the orbit is R = . In the laboratory system, this is a cycloid (the curve drawn by a point on the edge of a rolling disk); the distance between the neighbouring loops is ∆ = 2πu/ω = 2πR. dp p γ−1 γ dT T γ − 1 ∆p 20 K = ∆T = T . γ p ∆p = ρgh ρ = pµ/RT ≈ 1.2 kg/m3 γ = cp /cV = (cV + R)/cV We can use , where density. Also, we can substitute we obtain R hence µgh ∆T = , cV + R R ( cV ) ∆T R h= 1+ ≈ 2040 m. R µg (7 points) The longer semiaxis is a = (r + r ) = (α + β)R. So, the full energy of the asteroid at the Earth's location is 1) 1 2 1 Bearing in mind that , we can rewrite this as v02 = v02 − α+β So, we obtain v = v0 1 2 v . 2 Tangential component can be found from the angular momentum conservation law: v = v β, where the velocity at the perihelion can√be found analogously to v: 2) r ( vp = v0 2 1 1 − β α+β p √ ) = v0 2α ≈ 37.5 km/h. α+β is the air So, √ . So, 2α ≈ 24.4 km/h. v =v β α+β Radial component √ t 0 vr = v 2 − vt2 ≈ 24.4 km/h. The required components can be found by subtracting the Earth's orbital velocity. Apparently u = v , and 3) r ut = vt − v0 ≈ −5.6 km/h. (5 points) Let us write the Newton's II law for x- and y-components of the 4) When the asteroid approaches Earth's surface along the parabolic orbit, the energy due to the Earth gravity force gR is added electrons coordinates: to its kinetic energy√in the Earth's system of reference: mẍ = −eE ω cos ωt, 0 0 mÿ = eE0 sin ωt. w= There are two possible setups. First, we consider the interference of the beams, reected from the upper and lower surfaces of the glass plate, see Figure, upper drawing. Second, we direct the beam on the edge of the plate. As a result, on the screen, there will be almost the same diraction pattern, as from a single slit (lower drawing in the Figure). In the rst case, we need to calculate the optical path difference, see Figure. ∆l = 2(n|CD| − |AB|) = 2(nd/ cos β − d sin β sin α) = 2d(n/ cos β − sin α/n); we keep in mind that sin β = sin α/n. We need to nd such a change in α, which gives rise to the change of ∆l by λ (this corresponds to a transition from one diraction minimum to another one): ∆α · = λ. Then, we can relate the measured quantity, the distance between the minima on the screen a = L∆α (where L is the path length |AB| + |BC|) to the plate thickness. = 2d(sin α/ cos β − sin 2α/n) = 2d sin α(cos β − 2 cos α/n). So, Lλ = 2ad sin α(cos β − 2 cos α/n); hence, d = Lλ/2a sin α(cos β − 2 cos α/n). We can easily measure α and calculate β ; for n, we can use typical value n ≈ 1.4, or use the Brewster angle α measurement to nd n = tan α . For the precise measurement of a, we count several, e.g. 10, inter-minima intervals, and divide the distance between the farthest minima by 10. In the second case, the angular distance between the minima is given by ∆α = 2λ/d, so that a = 2Lλ/d and d = 2Lλ/a. Numerically, the thickness was d ≈ 0.20 mm. 2 d(∆l) dα √ 2[1 − (α + β)−1 ] ≈ 34.5 km/h. r 8. Electron 1 2 2 γM 1 γM E =− = v2 − . m 2a 2 R v02 = γM R γ γ eE0 ωm eE0 ω2 m 9. Asteroid (10 points) −1 0 3) 10. Glass plate u2t + u2r + 2gR0 ≈ 27.4 km/h d(∆l) dα −2 2 −2 −2 B B ESTONIAN FINNISH PHYSICS OLYMPIAD PROBLEMS & SOLUTIONS (2009) Estonian-Finnish Olympiad - 2009 1. Boat (9 points) Consider a boat of effective mass m, which moves along the (horizontal) x-axes, and for which the friction force of the water is given by Ff = −αv , where v is the velocity of the boat. (Effective mass is somewhat larger than the real mass, because any change of the boat’s velocity incurs a velocity change of those regions of water, which are close to the boat; hence, the boat’s inertia is complemented by the inertia of the surrounding water.) i) Prove that during this motion, v + kx = Const, and find the factor k . ii) Now, a boy of mass M moves inside the boat, back and forth, with velocity u(t) (also, along the x-axis). What term must be added to the conservation law v + kx = Const, to describe this situation? iii) Now, let the boat rest initially close to the coast and ready to depart. The boy jumps from the coast into the boat, turns around and jumps back to the coast. The horizontal component of the boy’s velocity during the first jump was u1 , and during the second jump −u2 . What is the maximum distance s travelled by the boat? Neglect the effect of a vertical motion on the horizontal friction force of the water. 2. Nanoclock (10 points) Nanotechnology allows controllable fabrication of very small structures. Let us consider a tiny homogeneously charged thin ring, having a radius R and carrying a positive net charge Q. i) Find the electric potential ϕ at a point P , which is on the axes of the ring, at a distance z from the centre of the ring. ii) Find the electric field E at the point P . iii) Show that the force acting on an electron moving along the symmetry axes in the vicinity of the centre of the ring (|z| ≪ R) is harmonic (i.e. depends linearly on z ). iv) Define the frequency of oscillations of such an electron. Use numerical values R = 1µm and Q = 1.0 × 10−13 C. v) Now, let us assume that the electron can perform also off-axis movements. Is the position at the symmetry centre of the ring (onaxis, z = 0) stable or unstable? Motivate your answer. Hint: You may use the approximate formula (1 + x)α ≈ 1 + αx + 12 α(α − 1)x2 . 3. Ball (8 points) A homogeneous ball of radius R and mass m is thrown horizontally onto a table at the moment of time t = 0. Its initial velocity before contact is purely horizontal and equal to v ; it is non-rotating. Coefficient of kinetic friction between the table and the ball is µ. i) Find the moment of time t, when the ball stops sliding, i.e. starts rolling without sliding. ii) Calculate the angular velocity of the ball ω∗ and its total mechanical energy E∗ at the moment when it stops sliding. In the case of a hollow sphere, would the energy E∗ be larger or smaller than in the case of a homogeneous ball? iii) Now, assume that the horizontal surface is treated so that the coefficient of kinetic friction depends on the horizontal coordinate x as µ = a + b cos x (with a > b). Find the expression for the terminal mechanical energy E∗′ in this new case. iv) Now, let us return to the case of constant kinetic friction coefficient µ. However, let us assume that the surface is not perfectly rigid (e.g. covered by a felt cloth). This gives rise to a second friction force — the rolling friction force Fr = µr mg . Unlike the kinetic friction force, it is not tangential to the touching point of the ball and the surface. Instead, it can be interpreted as the horizontal component of the surface reaction force (the entire reaction force is, of course, normal to the surface), see Fig. Find the expression for the terminal mechanical energy E∗′′ in this case. What is the most important (qualitative) difference between the expressions for E∗ and E∗′′ ? Hint: The moment of inertia of a ball is I = 25 M R2 . 4. Black box (9 points) Equipment: electrical black box with three outlets, battery, voltmeter. It is known that inside the black box, there are three resistors (connected with wires in an unknown manner), the smallest of which is R1 = 100 Ω. Find the value of the largest resistor R3 . What can be said about the middle-valued resistor R2 ? Estimate the uncertainties of your results. i) Find the current through the resistors R1 and R2 at the moment of time t1 = 5 ms. ii) Find the current through the resistors R1 and R2 at the moment of time t2 = 15 ms. iii) What is the net charge passing through the resistor R2 ? 9. Stratostat (5 points) i) Show that the pressure of an isothermal gas of molar mass µ follows the law p = p0 e−αz , where p0 is the pressure at the origin, and z is the height. Find the constant α. The temperature is T , the free fall acceleration is g . ii) Consider a stratostat, the envelope of which (a freely deformable non-elastic sack) is filled at the Earth’s surface by helium to the volume fraction of β = 10%. At which height h does the helium expand so that it fills the entire volume of the stratostat? The molar masses of the air and helium are µa = 29 g/mol and µHe = 4 g/mol, respectively. You may neglect the temperature variations of the atmosphere, and use the value T = 250 K. 10. Wedge (5 points) A wedge of mass M is kept at rest on an horizontal surface, and a block of mass m is kept on the wedge at the height h from the surface. The angle of the wedge is α, see Fig. There is no friction neither between the block and wedge nor Hint: you may use a model, according to which the already between the surface and the wedge. The system is released into broken part of the soap film gathers into a single front and moves a free motion. Find the time t needed for the block to reach the all together towards the still preserved part of the film. surface. 8. Magnetic pulse (7 points) Consider an electric circuit consisting of a coil of negligibly small inductance, consisting of N = 10 turns and with the surface area of a single loop S = 10 cm2 ), resistors R1 = R2 = 3 Ω, capacitor C = 0.2 F, and an inductance L = 1H , connected as shown in Fig. At the moment of time t = 0, a magnetic field, parallel to the axis of the coil is switched on. The inductance of the magnetic field starts growing linearly, starting from B = 0 until the maximal value B = 1 T is achieved at t = 10 ms. Further, the inductance of the magnetic field remains constant (and equal to 1T). 5. Pencil (6 points) Equipment: pencil, paper, ruler. Determine the coefficient of friction of the pencil’s graphite core against the paper. Estimate the uncertainty. 6. Spring (7 points) Equipment: helical spring of known mass m = 19 ± 0.5 g, measuring tape, a load of unknown mass. Determine the mass of the load. Estimate the uncertainty. 7. Soap film (6 points) Lord Rayleigh had in 1891 a lecture about taking photos of physical processes. Among others, he showed a photo of a soap film, which is falling apart (see Fig.). Instead of a flash, he used an electric spark (well, nowadays the flashes are also based on electric sparks). Estimate, how precise must have been the timing, i.e. estimate the time for a soap film to fall apart. Let the thickness of the soap film be h = 1 µm, the ring diameter D = 10 cm and the surface tension σ = 0.025 N/m. Estonian-Finnish Olympiad - 2009 1. Boat (9 points) i) From the Newton II law, m dv + α dx = 0. Multiplying this dt dt α dx = 0. Integrating (i.e. equation by dt, we obtain dv + m summing over all the small increments dx and dv ) this equation leads us to α v + x = Const, m α i.e. k = m . ii) We proceed in the same way as before, but we need to add the + α dx = interaction force between the boat and the boy: m dv dt dt du M dt . [Note that since the right-hand-side of this equation is the interaction force, it goes to zero, if the boy leaves the boat. Therefore, if we want to keep this equation correct even after the boy leaves the boat at the moment of time t = t∗ , we must assume u(t) ≡ u(t∗ ) for t > t∗ .] Similarly to the previous section, we obtain α M u(t) + x = Const. v+ m m iii) We use the conservation law of the previous section, and compare the value of the left-hand-side immediately before the boy lands into the boat with its value after a very long time. Bearing in mind that we need to substitute u(t → ∞) = −u2 (see above), we obtain M M α u1 + 0 = 0 − u2 + s. 0+ m m m So, M s= (u1 + u2 ), α i.e. the result is independent of how long time did the boy spend in the boat. 2. Nanoclock (10 points) i) All the√charges of the ring are at the same distance from the point P , l = R2 + z 2 . So, according to the superposition principle, the potential is the sum of potential of all the charges, √ ϕ = kQ/ R2 + z 2 . ii) E = − dϕ = kQz/(R2 + z 2 )3/2 . dz iii) For |z| ≪ R we may approximate (R2 + z 2 )3/2 ≈ R3 , so that E =≈ kQz/R3 . 3 2 iv) From the Newton II law, q mz̈ = −ekQz/R , i.e. ω = ekQ 1 ekQ/mR3 and f = 2πR ≈ 5.6 × 1012 Hz. mR v) At the origin, there are no charges; hence, the electric field lines cannot neither start nor end there. Consider a tiny coaxial cylinder embracing the origin. The field lines exit the cylinder through its bottom and top surfaces (because there is an electric field E =≈ kQz/R3 ). Hence there must be field lines entering the cylinder through its side surface. This implies a radial repelling force for an electron situated at that surface, i.e. instability. Remark: In the same way, one can prove a theorem, electrostatic equlibria are always unstable. 3. Ball (8 points) A straightforward way to solve parts i), ii), and iii) is to use conservation of angular momentum with respect to any axis laying on the surface (there is no torque with respect to these axes). Angular momentum at the beginning: L0 = mvr . Rolling ball is always rotating around the touching point with the ground (although the location of that point is constantly changing). From Steiner’s theorem we know that for a rotation axis that is located at the distance r form the center of mass, the moment of inertia is I ′ = I + mr 2 . Therefore the angular momentum for rotating ball is L′ = I ′ ω and clearly L0 = L′ , mvr therefore mvr = (I + mr 2 )ω, ω = I+mr 2. i) The torque with respect to center of mass: M = mgµr . Since M ∆t = I∆ω , and the sliding stops when angular speed has Iω reached value ω , we obtain the corresponding time: t = mgµr = Iv . gµ(I+mr 2 ) mvr and E∗ = 12 (mv∗2 + I+mr 2 2 2 2 m v r = 2(I+mr 2 ) . If I increases then E∗ Iω∗2 ) = 1 (mr 2 ω∗2 + Iω∗2 ) decreases. 2 iii) Since we didn’t make any assumptions about the functional form of µ while deriving w∗ , the results w∗ and E∗ of ii) are still valid. iv) The net force F = Fµ +Fr causes the decrease of translational velocity: m∆v = F ∆t, v ′ (t) = v − (µ + µr )gt. The sliding stops when ω(t) = v ′ (t)/r . Note that the torque with respect to center of mass is only caused by kinetic frictional force, therefore ω(t) = mgµr t. We get an equation for the time of the termination I mgµr 2 Iv of sliding: I t = v − (µ + µr )gt, t = gµ(I+mr 2 )+gµ I . r mvr ′′ The corresponding angular speed ω∗ = I+mr2 + µr I , and energy ii) Clearly, ω∗ = µ E∗ ′′ = 12 (I + mr 2 )ω∗2 . Clearly, E∗ ′′ < E∗ . 4. Black box (9 points) Independetly of whether there is a triangular or a star connection, one can measure the ratio of the resistances a pair of resistors by connecting the battery to two outlets — let these be the outlets 1 and 3, and measuring the voltages between the outlets 1 and 2, and between the outlets 2 and 3. For a star connection, R3 V23 = , R1 V12 where Ri denotes the resistor closest to the i-th outlet; for a triangular connection, V12 R3 = , R1 V23 where Ri denotes the resistor farthest away from the i-th outlet; for a triangular connection, In such a way we find R3 R2 R3 ≈ 4,7, ≈ 3,3, and ≈ 1,4, R1 R2 R1 implying R1 = 100 Ω, R2 = 140 Ω and R3 = 470 Ω; or R1 R2 R1 ≈ 4,7, ≈ 3,3, and ≈ 1,4, R3 R3 R2 R3 = 100 Ω, R2 = 330 Ω and R1 = 470 Ω. So, we can conclude that the largest resistance is 470 Ω, and the middlevalued resistance is either 330 Ω or 140 Ω. 5. Pencil (6 points) We put the pencil with its graphite end against a horizontal sheet of paper, and push the other end with a fingertip (without holding between the fingers and thereby possibly giving a torque to it). Then, there will be a purely longitudinal stress inside the pencil (there is no bending of the pencil). If we push strongly enough, we may neglect the weight of the pancil in the balance of torque with respect to the fingertip. So, equilibrium implies that the resultant force of the friction and reaction forces at the graphite end are along the axis of the pencil. This is possible, if tan α ≤ µ, where α is the angle between the pencil and a vertical line. So, we increase α wile pushing the pencil, and determine the position α0 , when the√pencil starts sliding; tan α0 can be calculated as tan α0 = x/ l2 − x2 , where l is the length of the pencil, and x is the length of the projection of the pencil onto a horizontal plane. 6. Spring (7 points) First, we measure the lengthening of the spring under its own weight. The relative lengthening of each loop is propotional to the order number of the loop. So, the average value of those legth increments is half of the largest increment (for the topmost loop). The topmost loop is deformed by the weight of the whole spring, hence the average deformation corresponds to the half-weight of the spring, kx1 = mg/2. Now we add a load to the lower end of the spring and measure the new deformation x2 . Since the additional weight of the load increases the stress of all the loops by the same value M g , each loop is deformed additionally by the same length increment. The sum of those additional increments is x2 − x1 ; according to the Hook’s law, k(x2 − x1 ) = M g . So, xx12 − 1 = 2M , and m m x2 − x1 M= ≈ 6 g. 2 x1 7. Soap film (6 points) Suppose the area of a broken part of the soap film is S . The corresponding surface energy is ES = 2σS . The mass of the soap water that was previously located at the broken part is m = Shρ, where ρ = 103 kg/m3 . Assuming that the surface energy is transformed to the kinetic energy of the moving front, we get an equation for the speed of the q front: 4σ 2σS = 12 mv 2 , 4σS = Shρv 2 . Therefore v = = hρ q 4 · 0.025 m/s = 10m/s and we can estimate the time of 10−6 · 103 1 breakup of the soap film to be t = D/v = 100 s. 8. Magnetic pulse (7 points) Since we can neglect the inductance of the coil, it performes as a voltage source, which outputs U = N SB/τ = 1 V during the time period between t = 0 ms and 10 ms, and 0 V otherwise. The characteristic time scales of the RC and LC cirquits are τ1 = R1 C = 0.6 s and τ2 = L/R2 ≈ 0.3 s. So, for both cirquits, the processes are very fast, i.e. the capacitor is effectively short-circuited, and almost all the voltage falls on the inductance. i) According to the considerations given above, I1 = U/R1 ≈ 0.33 A. As for I2 , it starts growing from 0 A at t = 0 at a rate, given by L dI = U , i.e. I2 = U t1 /L = 5 mA. dt ii) When the voltage U is switched off (at t = τ = 10 ms), the capacitor will (almost completely, because t2 − τ ≪ R1 C ) retain the charge it has accumulated, Q = I1 τ . All the voltage of the capacitor (Q/C ) will fall on the resistor R1 , so that I1′ = Q/R1 C = U τ /R12 C ≈ 5.6 mA. As for the inductance, it will retain (almost completely, because t2 − τ ≪ L/R2 ) the current it has acquired during the first 10 ms, I2′ = U τ /L = 10 mA. iii) Since the current in R2 will decay very slowly, as compared to its growth during the first 10 ms, we can neglect the charge passing thorugh it during t < τ . Then we can write the Kirhoff ’s law in the form L dI +R2 dq = 0, from where LdI +R2 dq = 0, dt dt and L∆I = −R2 ∆q . Since ∆I = −I2′ , we obtain ∆q = LI2′ /R2 = 3.3 mC. 9. Stratostat (5 points) i) Consider the pressure difference at heights z + dz and z : dp = −ρgdz (the difference is simply due to the weight of the layer dz ). The density can be found using the state equation of ideal µp µg . Therefore p1 dp . gas: pV = m RT, ρ = m = RT = − RT µ V dz Note that we have a derivative of a logarithm: p1 dp = d lndzp(z) . dz µg Therefore ln(p/p0 ) = −αz, p = p0 e−αz , where α = RT . Alternatively, we can derive this law from the Boltzmann distribution for the particle density n = n0 e−U/kT , where U is the potential energy of a molecule. Bearing in mind that for a constant temperature, the pressure is propotional to the density, we obtain p = p0 e−U/kT . Substituting U = mgz = NµA gz and R = kNA , we obtain the same result as above. ii) Clearly, the pressures inside and outside of the stratostat are equal and depending on the height as p = p0 e−αz , where ag α = µRT . For helium inside the sack pV = const. Let the volume of the stratostat be VS , therefore p0 βVS = pVS , where p is the pressure at the height, where helium has filled the entire ag volume. Since p = βp0 , we obtain µRT h = − ln β, h = RT 1 ln β , h = 17 km. µa g 10. Wedge (5 points) The center of mass of the system doesn’t move, therefore M u = mvh , where u and vh are the horisontal components of the velocities of the wedge and block in the lab frame at some instant of time. In the wedge’s frame, the block has m ). Since the horisontal velocity v ′ = vh + u = vh (1 + M block is sliding down the wedge, the vertical component of the block has to be v↓ = v ′ tan α. The total kinetic energy in the lab frame EK = 12 (M u2 + mvh2 + mv↓2 ). Substituting the relevant quantities and simplifying, we obtain m M Ek = v↓2 β, where β ≡ cot2 α + 1. 2 M +m The kinetic energy equals to the change in potential energy: m 2 ′ βv ↓ = ∆EP = mg(h − h ). By differentiating and noting 2 that dh = dt · v↓ , we get βdv↓ = gdt, t = βg v↓max . From the q above written energy conservation law we get v↓max = 2gh ; so β r q M we finally have t = 2βh = 2h cot2 α + 1 . g g M+m ESTONIAN FINNISH PHYSICS OLYMPIAD PROBLEMS & SOLUTIONS (2010) Estonian-Finnish Olympiad - 2010 1. Charges in E (8 points) Two particles (the blue and the red) of mass m are connected with a spring, the stress-free length of which is L and stiffness k ; the blue carries charge q (q > 0) and the red is chargeless. In the region x > 0, there is an homogeneous electric field E , antiparallel to the x-axis; In the region x < 0, there is no electric field. Initially, the “dumbbell” of charges moves in region x < 0 with velocity v , parallel to the x-axis; the dumbbell’s axis is also parallel to the x-axis and the spring is stress-free. It is known that after a while, the dumbbell moves in the region x < 0 with velocity −v , and that the red particle never enters the region x > 0. Also, the spring’s length achieves minimum only once. i) (2.5 pt) How long time τ does the blue particle spend in the region x > 0? The process takes place exactly as described, if one equality and one inequality are satisfied for the quantities m, v , k , q , E and L. ii) (3 pt) Which equality must be satisfied? iii) (2.5 pt) Which inequality must be satisfied? 2. Thermos bottle (6 points) In order to study the thermal properties of a thermos bottle, let us model it as two concentric spherical vessels, with radii R1 = 7 cm and R2 = 10 cm. The gap between the walls of the vessels contains vacuum (hence, the heat conductivity can be neglected). i) (3,5 pt) Find the radiative heat flux (i.e. transmitted heat per unit time) between the walls of the bottle, assuming that the ambient temperature is T2 = 293 K and the inner sphere is filled with liquid nitrogen at the boiling temperature T1 = 77 K. The emissivities of all the surfaces are equal to that of stainless steel: ε = 0.1. Remark: The emitted heat flux per unit area is given Stefan-Boltzmann’s law P = εσT 4 , where σ = 5.67 × 10−8 W/m2 K4 (assuming that ε is independent of the wavelength. ii) (2,5 pt) Estimate, how long time does it take for a full evaporization of the liquid nitrogen (the vapor escapes through an over pressure valve). For the liquid nitrogen, density ρ = 810 g/l and latent heat for vaporization λ = 5.580 kJ/mol). NB! If you were unable to find P (for question i), express the evaporization time symbolically (i.e. using the symbol P ). 3. Tyrannosaur (T. Rex) (6 points) Paleontologists have discovered tracks of a tyrannosaur where the footprints of the same leg are A = 4.0 meters apart. They have also recovered a piece of a tyrannosaur leg bone that has bone cross-sectionalo area N = 10000 times that of a chicken (which the tyrannosaur is related to). i) (3 pt) Knowing that the chicken leg is approximately l = 15 cm tall, estimate the length of a tyrannosaur leg L. You may assume that the length of a leg scales as the length of the whole animal, and that the bone stress (force per area) is the same for both animals. Is your result consistent with the step length A? ii) (3 pt) Estimate the natural walking speed of the tyrannosaur by approximating the walking motion of a leg with a freely oscillating pendulum motion. State clearly all the assumptions you make. 4. Ball (6 points) Massive spherical ball has a mass M = 100 kg; an attempt is made to roll the ball upwards, along a vertical wall, by applying a force F to some point P on the ball. The coefficient of friction between the wall and the ball is µ = 0.7. i) (5 pt) What is the minimal force Fmin required to achieve this goal? ii) (1 pt) On a side view of the ball and the wall, construct geometrically the point P , where the force has to be applied to, together with the direction of the applied force. 5. Elastic thread (10 points) Equipment: ruler, tape, an elastic thread, a wooden rod, marker, a known weight. The purpose of this problem is to study the elastic properties of an elastic thread for large relative deformations ε = (l − l0 )/l0 , where l0 and l are the lengths in initial and stretched states, respectively. If the Hook’s law were valid, the ratio F/ε of the elastic force F and ε would be constant: F/ε = SE , where S is the cross-section area of the thread and E — the Young modulus of the thread material. i) Collect the data needed to plot the ratio F/ε as a function ε, up to ε ≈ 4. Plot the appropriate graph, and indicate the uncertainties. ii) By making assumption that the Young modulus E = F/Sε remains constant, study, how does the volume of the thread depend on ε. Plot the appropriate graph. 6. Charges in B (5 points) There is an homogeneous, parallel to the z -axis magnetic field of inductance B in region x > 0. In region x < 0, there is no field. There are two particles of mass m and charge q . Initially, the particles have coordinates y = z = 0, and respectively x = −L0 and x = −2L0 (with L0 > 0). Initial velocity of both particles is v , along the x-axis, towards the magnetic field. Neglect the electrostatic repulsion force of the two charges. i) (1,5 pt) Sketch the trajectory of the first particle, and the dependance of its y -coordinate on time. ii) (3,5 pt) Sketch the distance L between the particles as a function of time t, assuming that πmv/Bq > L. What is the minimal distance Lmin ? 7. Satellite (5 points) i) (3 pt) A large ball of mass m1 is kept at the height h from the floor (so that the center of the ball is at the height h + d/2, where d is its diameter). A small ball of mass m2 is placed upon the large one, and the system is released (so that it starts falling). To which height (from the floor) will the small ball rise, assuming that the collision between the lower ball and ground, and the collision between the balls are absolutely elastic, and m1 ≫ m2 ? ii) (2 pt) Consider the following satellite launching project. There are N absolutely elastic balls of masses m1 ≫ m2 ≫ ... ≫ mN : the first ball (the heaviest) is the lowest; the second ball is placed on top of the first; the third — on top of the second etc. The upmost ball is supposed to become a satellite, i.e. to obtain the velocity vN = 7.8 km/s). The lowest ball is at the height h = 1 m from the floor, and the system is released. What should be the number of balls N ? What should be the mass of the lowest ball, if mi /mi+1 = 10, and the mass of the satellite MN = 1 kg? 8. Sprinkler (3 points) A sprinkler has a shape of hemisphere, which has small holes drilled into the spherical part of its surface. From these small holes, water flows out with velocity v = 10 m/s. Near the sprinkler, the water flow is distributed evenly over all the directions of the upper half-space. The sprinkler is installed at the ground level so that its axis is vertical. In what follows, the air resistance can be neglected, and the dimensions of the sprinkler can be assumed to be very small. i) (1.5 pt) Find the surface area of the ground watered by the sprinkler. i) (2 pt) What is the coefficient of friction, assuming a four-wheel ii) (1.5 pt) At which distance from the sprinkler is the watering drive? Because of a manual gear change, there is time period of intensity (mm/h) the highest? τ1 = 0.5 s, during which there is no driving force (so that the 9. Power supply (6 points) car decelerates due to air friction). Except for that period, the acceleration follows the law given by the graph. As a result, the terminal velocity vt = 40 m/s is reached τ2 = 1.0 s later than it would have been reached, if there were no delay caused by the gear change. Upon reaching the terminal velocity, the car continues moving at constant speed. In your calculations, you can assume i) (2 pt) Consider the cirquit given in Fig (a), where the diode that the air friction was constant during the gear change period. can be assumed to be ideal (i.e. having zero resistance for forward ii) (2,5 pt) At which speed the gear was changed? current and infinite resistance for reverse current. The key is iii) (2,5 pt) How many meters shorter distance will be covered switched on for a time τc and then switched off, again. The input during the first 100 seconds, as compared to ideal acceleration (i.e. and output voltages are during the whole process constant and without the delay due to the gear change)? equal to Ui and Uo , respectively (2Ui < Uo ). Plot the graphs 11. Black box (10 points) Equipment: a black box, multimeter, battery, timer (on the screen). of input and output currents as functions of time. Determine the electrical scheme inside the black box, and the ii) (2 pt) Now, the key is switched on and off periodically; each values of all the resistors inside it. Estimate the characteristics of time, the key is kept closed for time interval τc and open — also other electrical components. It is known that apart from the wires, for τc . Find the average output current. the total number of components is three. iii) (2 pt) Now, cirquit (a) is substituted by cirquit (b); the switch is switched on and off as in part ii. What will be the voltage on the load R, when a stationary working regime has been reached? You may assume that τc ≪ RC , i.e. the voltage variation on the load (and capacitor) is negligible during the whole period (i.e. the charge on the capacitor has no time to change significantly). 10. Ice-rally (7 points) The car accelerates on a slippery ground so that the wheels are always at the limit of slipping (e.g. via using an electronic traction control). Such an acceleration would result in the velocity vs time graph as given in the Figure. ESTONIAN FINNISH PHYSICS OLYMPIAD PROBLEMS & SOLUTIONS (2011) Estonian-Finnish Olympiad - 2011 1. Spool (12 points) A spool with inner radius r and outer radius R lies on a horizontal table; the axis of the spool is horizontal. A weightless rope is wound around the inner part as shown in the picture. The loose end of the rope makes an angle α with the horizont (the angle α can be also negative). The moment of inertia of the spool is J and mass — M . In what follows you may assume that the spool rolls on the table without slipping. i) (2 pts ) We pull the loose end of the rope with velocity u (parallel to the loose part of the rope; that loose part can be thought to be very long). What is the velocity of the spool? ii) (3 pts) Suppose now that the spool is at rest, and we apply force F to the loose end of the rope (parallel to the loose part of the rope). What is the acceleration of axis of the spool? iii) (2 pts) How large does the coefficient of friction µ need to be (as a function of α) to ensure that there is no slipping between the spool and the table? iv) (1.5 pts) Now the spool rolls, again, with velocity u; however there is no rope. The spool hits a threshold of height H (see Figure); the impact is perfectly inelastic. What is the speed v of the spool immediately after the impact? v) (1.5 pts) What is the speed w of the spool after rolling over the threshold? Assume that u is such that the spool will roll over the threshold without losing contact with its edge. vi) (2 pts) If the speed u is too large, u > u0 , the spool will jump up and lose contact with the edge of the threshold. Determine u0 . 2. Capacitor (6 points) An ideal plate capacitor has plates with area A and separation d and is charged so that the electric field between the plates equals to E . i) (2 pts) Find the energy density of the electric field inside the capacitor and the total energy of the field. ii) (1.5 pts) What is the force required to keep the plates separated? iii) (2.5 pts) Now, this capacitor is submerged into distilled water of dielectric permittivity ε = 80; the electric field between the plates equals still to E . What is the hydrostatic pressure between the plates if the atmospheric pressure is p0 and the pressure of the water column can be neglected? 3. Charged cylinder (8 points) Dielectric cylinder of radius r carries a charge of surface density σ on its cylindrical surface and rotates with angular velocity ω . i) (3 pts) Determine the magnetic induction B inside the cylinder. Remark: if you wish, you can use the expression for the inductance L = µ0 N 2 S/l of a solenoidal coil of radius r , length l ≫ r , area of cross-section S and number of loops N . ii) (3 pts) A radial conducting wire connects the axis of the cylinder with the cylindrical surface (it rotates together with the cylinder). Find the electromotive force (voltage) E between the ends of the wire. iii) (2 pts) Suppose that the wire connecting the axis of the cylinder with the cylindrical surface is not radial and has an arbitary shape (still, there are no segments protruding outside the cylinder). Show that E does not depend on the shape of the wire. 4. Black box (10 points) Equipment: a black box with three terminals, voltmeter, timer. Inside the black box, there are two capacitors and a battery, connected as shown in Figure. The capacitance C1 = (3400 ± 400)µF; you are asked to determine the capacitance C2 and estimate the ucertainty. Remark: the terminal “+” is a wire, long enough to be conected to either terminal “A” or terminal “B”. 5. Plutonium decay (3 points) Plutonium is an unstable element, a Pu239 atom decays with a half-life of τ1/2 = 24 000 years by creating smaller nuclei, including an α-particle. Find the αparticle flux density (i.e the number of passing nuclei per unit time and per unit cross-sectional area) near the surface of a plate of Pu239 . The plate has thickness d = 1 mm; its width and length are much larger than that. The density of plutonium ρ = 19 800 kg/m3 . Remark: half-life is the period of time it takes for a substance undergoing decay to decrease in size (in the number of particles) by half. The mass of an atom of Pu239 is m0 = 3.84 × 10−25 kg. 6. Violin string (9 points) The motion of a bow puts a violin string into a periodic motion. Let us make a simplified model of this process. The string has elasticity and inertia, so we substitute it by a block of mass m, fixed via a spring of stiffness k to a motionless wall and laying on a frictionless horizontal surface. The bow is substituted by a horizontal plate, which is pressed with constant force N downwards, and which moves with a constant velocity u, parallel to the axis of the spring, see Figure. The static coefficient of friction between the plate and the block is µ1 , and the kinetic coefficient of friction is µ2 < µ1 . So, as long as the plate does not slide with respect to the block, the coefficient of friction equals to µ1 ; as soon as there is some slip, it decreases down to µ2 . i) (2 pts) For questions (i) and (ii), let us assume that the speed of the plate u is very small as compared to the maximal velocity of the block. What is the maximal velocity of the block vmax (maximized over time)? ii) (2 pts) Sketch qualitatively the graph of the displacement of the block as a function of time and indicate on the graph the durations of the prominent stages of the block motion (graph segments). iii) (1,5 pts) Now, let us abandon the assumption about the smallness of u. Sketch qualitatively the graph of the velocity of the block as a function of time. iv) (2,5 pts) Determine the amplitude A of the block’s oscillations. v) (1 pt) Which condition (strong inequality, ≫ or ≪) must be satisfied for u in order to ensure that the oscillations will be almost harmonic? 7. Vacuum bulb (8 points) Let us study how a vacuum can be created inside a bulb by pumping. Let the volume of the bulb be V , and the pump consist in a piston moving inside a cylinder of volume αV , where α ≪ 1. The pumping cycles starts with piston being pulled up; when the pressure inside the cylinder becomes smaller than inside the bulb, a valve VA (connecting the cylinder and the bulb) opens and remains open as long as the piston moves up. When piston is released, it starts moving down, at that moment, the valve VA closes. As long as the valve VA is open, the pressures of the bulb and the cylinder can be considered as equal to each other. When the piston moves down, the pressure in the cylinder increases adiabatically until becoming equal to the outside pressure p0 = 105 Pa; at that moment, another valve VB opens leting the gas out of the cylinder. When the piston reaches the bottommost position, there is no residual air left inside the cylinder. Now, the piston is ready for being lift up: the valve VB closes and VA opens, marking the beginning of the next pumping cycle. The air inside the bulb can be considered isothermal, with the temperature being equal to the surrondin temperature T0 . The adiabatic exponent of air γ = cp /cV = 1.4. i) (2 pts) How many pumping cycles N needs to be done to reduce the pressure inside bulb from p = p0 down to p = βp0 , where β ≪ 1? ii) (2 pts) What is the net mechanical work done during such a pumping (covering all the N cycles)? iii) (2 pts) What is the temperature of the air released from the cylinder to the surroundings at the end of the pumping process (when the pressure inside the bulb has become equal to βp0 )? iv) (2 pts) According to the above described pumping scheme, there is a considerable loss of mechanical work during the period when the piston is released and moves down. Such a loss can be avoided if there is another pump, which moves in an opposite phase: the force due to outside air pressure pushing the piston down can be transmitted to the other pump for lifting the piston up. What is the net mechanical work done when such a pumping scheme is used? 8. Heat sink (6 points) Consider a heat sink in the form of a copper plate of a constant thickness (much smaller than the diameter d of the plate). An electronic component is fixed to the plate, and a temperature sensor is fixed to the plate at some distance from that compnent. You may assume that the heat flux (i.e. power per unit area) from the plate to the surrounding air is proportional to the difference of the plate temperature at the given point (the coefficient of proportionality is constant over the entire plate, including the site of the electronic component). i) (2 pts) The electronic component has been dissipating energy with a constant power of P = 35 W for a long time, and the average plate temperature has stabilized at the value T0 = 49 ◦ C. Now, the component is switched off, and the average plate temperature starts dropping; it takes τ = 10 s to reach the value T1 = 48 ◦ C. Determine the heat capacity C (units J/◦ C) of the plate. The capacities of the electronic component and the temperature sensor are negligible. ii) (4 pts) Now, the electronic component has been switched off for a long time; at the moment t = 0, a certain amount of heat Q is dissipated at it during a very short time. In the Figure and Table, the temperature is given as a function of time, as recorded by the sensor. Determine the dissipated heat amount Q. t (s) T (◦ C) t (s) T (◦ C) 0 20.0 400 39.9 20 20.0 600 33.4 30 20.4 800 27.9 100 32.9 1000 24.4 200 41.6 1200 22.3 300 42.2 1400 21.2 9. Coefficient of refraction (10 points) Equipment: A thick glass plate having the shape of a half-cylinder, a glass prism, a container with an unknown liquid, a laser pointer, graph paper, ruler. i) (5 pts) Determine the coefficient of refraction of the halfcylindrical glass plate and estimate the uncertainty of the result. ii) (5 pts) Determine the coefficient of refraction of the liquid and estimate the uncertainty of the result. Estonian-Finnish Olympiad - 2011: solutions Using the result of the previous task, we can use this equation directly to obtain an expression for the minimal allowed value of the coefficient of friction: J r − MR 2 cos α R µmin = Mg J 1 + MR − sin α 2 F 1. Spool (12 points) i) First solution. The momentary rotation centre of the spool is the contact point P with the floor (since this point is at rest). So, the velocity of the spool is u′ = Rω , where ω is the angular velocity. Consider triangle P OA, where A is defined as the point where the loose end of the rope meets the inner part of the spool at the current moment of time, but which is iv) The angular moment of the spool with respect to the edge of actually a point of the spool, i.e. it rolls together with the spool); the threshold conserves during the impact (since the impact force O is the centre of the spool. Let us denote 6 P AO = β ; it is easy has zero arm): v u to see that 6 AOP = π − α. The velocity ~vA of the point A ⇒ M u(R − H) + J = J + M R2 R R is perpendicular to P A and, hence, forms angle β with the loose ! H/R end of the rope. Its projection to the rope equals to u, therefore v =u 1− J vA = u/ cos β . Further, ω = vA /l,√where l = |AP | can be 1 + MR 2 2 2 found from the cosine theorem: l = R + r + 2Rr cos α. v) From the energy conservation law we obtain immediately 2 2 The angle β can be found using the sine theorem for the triangle 2 v 2 w α (J + M R ) = (J + M R ) + 2M gH ⇒ AOP : sin β = R sin . Combining everything together we end l R2 s R2 up with 2gH uR uR w = v2 − J . u′ = √ 2 . = 1 + MR 2 |R cos α + r| R cos2 α + r 2 + 2Rr cos α vi) The spool is the most prone to jumping immediately after the Second solution. Let us decompose the velocity ~vA into two impact; the gravity force needs to be large enough to bind the components: the tangential component (parallel to the rope) centre of mass to the rotational motion around the edge of the equals (by modulus) to u; let us denote the radial component as threshold: ur . Since the distance between O and A is constant, the projecM v2 R−H g tion of the velocities of O and A to the line OA are equal: ≤g ⇒ v2 ≤ (R − H) ⇒ R R M ur = v sin α ⇒ v = ur / sin α ⇒ ω = ur /R sin α. r J 1 + MR g 2 The vertical component of the velocity of the point A remains un(R − H) u0 = J H . M 1 + − changed if we switch the laboratory system of reference with the MR2 R 2. Capacitor (6 points) system associated with point O ; hence, i) The energy is W = CU 2 /2 = 12 ε0 Ad E 2 d2 = 12 ε0 AdE 2 ; u sin α − ur cos α = ωr sin α = ur r/R ⇒ hence, the energy density w = W/Ad = 12 ε0 E 2 . ur uR v= = . ii) There are two ways to calculate the force. First, we notice that sin α R cos α + r the innermost charges q at the capacitor plates are affected by the ii) (2 pts) The easiest way to solve this part is to use the energy electric field E, therefore there is a force qE acting upon these. The balance for infinitesimal displacement of the cylinder and apply outermost charges, however, have no electric field around them the answer to the previous question: (because outside the inter-plate space, there is no electric field). J J M 2 F u · dt = d v 1+ = M vdv 1 + ⇒Due to the Gauss law, the electric field decreases linearly with the 2 M R2 M R2 net charge left below the level of the current point (i.e. towards dv Fu F cos α + Rr the inter-plate space). Therefore, the electric field averaged over = a= = · . J J the charges is just half of the maximal value E : hEi = 12 E , and dt M M v 1 + MR 1 + 2 MR2 net force acting on the plate is F = Q hEi = CEd hEi = iii) Let us write the force balance projection to the horizontal axis the 1 2 ε assuming that the spool is at the edge of slipping, i.e. the friction 2 0 AE . The second way includes writing the energy balance for a small force Ff = µmin N , where N = mg − F sin α is the normal 2 displacement of a plate: F · δd = δ(Q2 /2C) = 2εQ0 A δd = force: M a = F cos α + µmin N = F cos α + µmin (M g − F sin α). 12 C 2 E 2 d · δd ⇒ F = 12 ε0 AE 2 . iii) Let us push away part of the water from the inter-plate space so that there will be a small region of plate area dA, where there is no water between the plates (here, ∆p is the pressure difference between the inter-plate space and the outside regions). By doing so, we perform work d · δA · ∆p, and increase the capacitor’s energy: Q2 d 1 1 2 δW = δ(Q /2C) = − . 2ε0 εA ε(A − δA) + δA So, Q2 d(ε − 1) · δA 1 δW = = ε0 E 2 d(ε − 1) · δA; 2 2 2ε0 ε A 2 comparing this with the pressure work d · δA · ∆p we conclude that 1 1 ∆p = ε0 E 2 (ε − 1) ⇒ p = p0 + ε0 E 2 (ε − 1). 2 2 3. Charged cylinder (8 points) i) Moving surface charge creates a solenoidal surface current with the surface density j = σv = σωr . From the circulation theorem for a rectangular loop embracing a segment of surface cur= jl, where l is the length of the surface current rent we obtain Bl µ0 segment (so that jl gives the current flowing through the loop). Hence, B = µ0 j = µ0 σωr . , where S is the area covered ii) Using formula E = dΦ = B dS dt dt by the wire, we obtain E = Bωr 2 /2. Indeed, during a small time interval dt, the wire covers a equilateral triangle of side lengths r , r , and rωdt; its area is apparently r 2 ωdt/2. By using the earlier obtained expression for B we end up with E = µ0 σω 2 r 3 /2. iii) We need to show that from the previous task, dS is independdt ent of the wire shape. First we note that due to rotational symmetry, dS , it cannot depend on the rotation angle, i.e. dS ≡ Ṡ = dt dt Const. Further we note that regardless of the wire shape, during the entire rotation period 2π/ω , the whole circle area is covered; Ṡ · 2π/ω = πr 2 ⇒ Ṡ = r 2 ω/2. 4. Black box (10 points) There are several ways to perform this task. First one can notice that if two capacitors discharge at the same resistor, starting with equal voltages and ending also with equal voltages, the ratio of the discharge times equals to the ratio of the capacitances (because for each given voltage, the discharge currents are the same, but larger capacitor has more charge — proportionally to the capacitance). Therefore we can first charge the known capacitor (using the battery), and let it discharge on the voltmeter (which has some finite resistance), measuring the time t1 required for it to reach a pre-defined final voltage. Then we need to repeat the procedure with the other capacitor and measure the time t2 and calculate C2 = C1 t2 /t1; the uncertainty is estimated ∆t1 2 1 as ∆C1 = C1 t1 + ∆t + ∆C . t2 C1 It is recommended to check the negligibility of the leak current across the plates of the capacitor. To this end, one can charge a capacitor, measure the voltage, remove the voltmeter and wait for some time (of the order t1 and t2 ), and check again the voltage. Another way is to discharge completely one capacitor by short-circuiting its terminals and charge the other capacitor up to the voltage of the battery. Further, we connect the terminals A and B so that the capacitors re-distribute the charge Q = EC1 and take the same voltage: Q1 /C1 = (Q − Q1 )/C2 ⇒ Q1 = QC1 /(C1 + C2 ) = EC12 /(C1 + C2 ). Consequently, the new voltage (which we measure) is U = Q1 /C1 = EC1 /(C1 +C2 ), from where C2 = ( UE − 1)C1 . 5. Plutonium decay (3 points) Let the number of Pu239 -atoms be reduced during time interval t = 1 s by a factor of 1 − λ (with λ ≪ 1). Then, during the time period of τ1/2 , it is reduced by a factor of (1 − λ)τ1/2 /t ≈ e−λτ1/2 /t = 12 ⇒ λ = t ln 2/τ1/2 . Therefore, the number of atom decay events is Nd = N t ln 2/τ1/2 , where N = ρdS/m0 is the number of atoms, i.e. the α-particle flux is Φ = Nd /2St (where the factor 2 accounts for the fact that the particles are emitted towards the both sides of the plate). Upon bringing all the expressions together, we obtain ρd ln 2 Φ= ≈ 2.36 × 1013 m−2 · s−1 . 2τ1/2 m0 6. Violin string (9 points) i) When the plate slides, there is a constant friction force µ2 N acting upon the block, which means that the equilibrium deformation of the spring is x0 = µ2 N/k ; the net force acting upon the block (due to spring and friction) is given by F = −kξ , where we have defined ξ = x − x0 . Therefore, while sliding, the block oscillates harmonically around the point ξ = 0. Slipping starts when the static friction will be unable to keep equilibrium, i.e. at kx = µ1 N , which corresponds to ξ0 = (µ1 − µ2 )N/k . If the plate moves slowly, the block is released with essentially missing kinetic energy, and the energy conservation law yields p 1 1 2 2 kξ = mv ⇒ v = ξ k/m . max 0 0 max 2 2 ii) As mentioned, when the plate slides, the motion of the block is harmonic, i.e. the graph of x(t) is a segment of a sinusoid; when there is no sliding, the block moves together with the plate, i.e. the graph of x(t) is a straight line. At the moment when slipping starts or stops, the oscillatory speed is equal to the speed of plate, i.e. the straight line is tangent to the sinusoid. The length of a straight segment can be calculated as T1 = 2ξ0 /u = 2(µ1 − µ2 )N/ku; the sinusoidal segment p corresponds to a half-period and therefore has a length of T2 = π m/k . iii) The speed v(t) = dx is the derivative of x(t); therefore, the dt sinusoidal segment of x(t) will correspond to a sinusoidal segment of v(t), and a straight segment of x(t) — to a horizontal segment of v(t). The resulting graph is depicted below. iv) Let the amplitude of the oscillations be A, i.e. the sinusoidal p segments follow the law ξ(t) = A cos(ωt), where ω = k/m. Correspondingly, v(t) = Aω sin(ωt) ⇒ A sin(ωt) = v(t)/ω ; hence, for any point at a sinusoidal segment, ξ 2 + v 2 /ω 2 = A2 . At a point, where a sinusoid and a straight line meet, the straight line and sinusoid have equal values for ξ = ξ0 = (µ1 − µ2 )N/k and v = u. Consequently, (µ1 − µ2 )2 N 2 /k2 + u2 /ω 2 = A2 ⇒ 1q A= (µ1 − µ2 )2 N 2 + u2 mk. k v) The oscillations will be almost harmonic when the straight segments are very short, √i.e. when u/ω ≫ (µ1 − µ2 )N/k ⇒ u ≫ (µ1 − µ2 )N/ mk. 7. Vacuum bulb (8 points) i) Each pumping cycle reduces the number of molecules inside the bulb by a factor of (1 − α); therefore, after N cycles, the number of molecules (and hence, the pressure) by a factor of β = (1 − α)N ≈ e−N α ⇒ ln β N =− . α ii) Majority of the pumping cycles are done when the pressure inside the bulb is negligible as compared to the outside pressure. During such a cycle, a work equal to p0 V α is done. Therefore, A ≈ N p0 V α = p0 V | ln β|. iii) Due to adiabatic law, pV γ = Const; when combined with the gas law pV ∝ T we obtain pγ−1 ∝ T γ . During the last downwards motion of the piston, the pressure inside the cylinder 1 is increased by a factor of 1/β ; thus, T = T0 β γ −1 . iv) According to the modified pumping scheme, the work/energy loss is only due to the release of the hot air. Note that if we had a cylinder of volume V , we could be able to create vacuum inside there using only one pumping motion, i.e. by performing work A = p0 V and without any energy loss. Now, we perform an excess work, which is converted into internal energy of the released hot air, which needs to be calculated. Let ξ = pp0 be an intermediate rarefaction factor; then, we can apply the previous result to calculate the internal energy of released air, if its quantity is dν 1 moles: dU = T0 (ξ γ −1 − 1)cV dν . Let us note that the nump0 V 0 ξV ber of moles inside the bulb is ν = pRT ⇒ dν = RT dξ . So, 0 0 R 1 1 −1 cV c V U = p0 V R 0 (ξ γ − 1)dξ = (γ − 1)p0 V R . Now, recall 1 that γ = cp /cV = 1 + cRV , hence cRV = γ−1 and U = p0 V . This gives us the energy loss due to heating the released air; another p0 V is required for loss-free creation of the vacuum. Hence, the total required work is A = 2p0 V . 8. Heat sink (6 points) i) When the average temperature is stable at T0 , all the power dissipated at the electronic component is eventually given to the air: the air is being heated with power P . As the heat flux depends linearly on the temperature difference between a point on the plate and the air, the average heat flux and therefore the net power dissipated into the air depends linearly on the average temperature of the plate. The average temperature determines the radiated power. Now consider the situation after the heating has ended. The average temperature is initially the same, so the radiated heat power is initially still P . By the definition of heat capacity, an infinitesimal heat amount given to the surroundings is dQ = −C dTavg with the minus sign encoding the direction of the heat flow. Thus, at the first moment, P = dQ = −C dTdtavg . Assumdt ing that during τ the average temperature depends approximately linearly on time (because T0 − T1 = 1 ◦ C is much less than the dT τ 0 usual ambient temperature), dtavg ≈ T1 −T and C ≈ T0P−T = τ 1 ◦ 350 J/ C. Actually the graph of Tavg (t) is slightly curved downwards (as it is an exponential eventually stabilizing at the ambient temperature) and initially somewhat steeper, so C is a bit smaller. ii) The average temperature of the heat sink falls off exponentially, therefore, if the “tail” of the given graph turns out to be exponential, we can presume the “tail” depicts the situation where the sensor is sensing the average temperature and the initial “bump” in the temperature distribution has evened out. Extrapolating the exponential to t = 0 we get the initial average temperature Tavg,0 (immediately after the Q has been dissipated into the sink) and, by Q = C(Tavg,0 − Tamb ), the heat Q. The ambient temperature Tamb can be read off from the beginning of the given graph where the sensor’s surroundings have not yet heated up. This is furthermore a check for the assumption T0 − T1 ≪ Tamb made in the first part of the solution. From the table, Tamb = 20.0 ◦ C. Let us analyse the (yet hypothetical) exponential Tavg −Tamb ought to obey, so that eventually we expect T ∼ Tavg = t Tamb + Tc e− tc where Tc and tc are, respectively, a characteristic temperature and a characteristic time. (The “∼” means “is asymptotical to” or “approaches”.) We plot ln(T − Tamb ) using the data from the table. Then approximate the “tail” linearly (valuing the end of it most) to get ln[(T − Tamb )/◦ C] ∼ 4.89 − 300t s . Therefore Tc ≈ e4.89 ◦ C ≈ 133 ◦ C. On the other hand, plugging t = 0 into our exponential function shows that Tavg,0 − Tamb = Tc and, finally, Q = CTc ≈ 46 700 J. Actually, quite a good result can be obtained without replotting anything, by just considering the last three datapoints of the table. Denote ∆Ti ≡ Ti − Tamb . If the times t3 − t2 = t2 − t1 , then with an exponential we should observe that ∆T3 /∆T2 = ∆T2 /∆T1 . The last three timepoints are good indeed, so we check ∆T1 = 4.4 ◦ C, ∆T2 = 2.3 ◦ C and ∆T3 = 1.2 ◦ C. Their ratios are ∆T3 /∆T2 ≈ 0.522 and ∆T2 /∆T1 ≈ 0.523, a splendid match. This confirms the exponential “tail”. As in every equal time interval the ∆T is multiplied by the same number (that is the essence of exponentials), 3 t t−t 3 2 2 Tc = ∆Tavg,0 = ∆T3 × ∆T ≈ 114 ◦ C. From this, ∆T3 Q ≈ 39 900 J. This is discrepant from our previous calculation, but not too much: Tc is exponentially sensitive to the T -intercept of the straight line fitted to the “tail” (its crossing point with the T -axis) on the logarithmic plot. The bump has still not yet disappeared completely enough. 9. Coefficient of refraction (10 points) i) We direct the laser beam radially into the semi-cylinder: perpendicularly through its cylindrical surface. The beam enters the plate without refraction and reaches the opposing flat face at the axis of the cylinder. Depending on the angle between that face and the beam, there may or may not be a refracting beam, but there is always a reflecting (from the flat face) beam . We rotate the semicylinder around its axis to find the position, when the refracting beam appears/disappears; the angle α between the flat face and the incident beam correspond to the angle of complete internal reflection, i.e. n = 1/ cos α. We can measure cos α using the graph paper: we draw the beam as a segment AO and the flat face of the semi-cylinder as a line BO so that 6 ABO = π/2; then, n = |AO|/|BO|. The uncertainty can be found using the formula ∆n = n( ∆|AO| + ∆|BO| ) and by estimating the uncer|AO| |BO| tainties of the direct length measurements ∆|AO| and ∆|BO|. ii) We drop the liquid on the prism and press it against the flat face of the semi-cylindrical plate. Further we study the complete internal reflection at the boundary between the semi-cylinder and prism (which is filled with the liquid) by repeating the above described experiment. Thereby we measure new lengths A′ O and B ′ O ; the condition of complete internal reflection is now n/nl = |A′ O|/|B ′ O| ⇒ nl = n|B ′ O|/|A′ O|, where nl stands for the coefficient of refraction of the liquid. The uncer′ ′ O| O| tainty is now calculated as ∆nl = nl ( ∆|A + ∆|B + ∆n ). |A′ O| |B ′ O| n ESTONIAN FINNISH PHYSICS OLYMPIAD PROBLEMS & SOLUTIONS (2013) Estonian-Finnish Olympiad 2013 The focal length of its camera f = 4.3 mm been bent into the shape of a square with 1. PRISM (8 points) i) (4 points) A right prism that has an equilateral triangular base with length a is placed in a horizontal slit between two tables, so that one of the side faces is vertical. How small can the width d of the slit be made before the prism falls out of the slit? There is no friction between the prism and the tables and the prism is made of a homogeneous material. The edges of the slit are parallel. and the diameter of the lens D = 1.8 mm. The sensor is w = 4.6 mm wide corresponding to N = 3264 pixels. 3. MISSION TO MARS (7 points) A crew of astronauts is going to be sent to explore the polar region of Mars and search for buried water ice. Their spaceship will travel from Earth to Mars along an elliptic transfer orbit tangential to the orbits of both planets. Despite its shortcomings, this orbit is commonly used in space travel due to its relatively good fuel economy. Future manned missions to ii) (4 points) Now the prism is placed in the Mars are very likely based on this kind of slit so that one of its side faces is horizontal. transfer. In this problem you will examine How small can the width l of the slit be made some aspects of this orbit. before that position becomes unstable? The mean orbital radius of Mars is R a = 1 . 52 AU. The mean orbital radius of Earth a i) ii) is R g = 1 AU = 149 600 000 km. Mars has a mean radius of r a = 3397 km and surface a gravity g a = 3.71 m/s2 . Earth has a mean radius of r g = 6371 km. side length a. At some point on the wire is a small ideal current source that keeps current I flowing in the circuit in all situations. The magnetic moment m of a planar circuit is given by the relation m = I A , with vector ~ pointing in the normal direction of the cirm cuit according to the right hand rule ( A is the area bounded by the circuit). i) (3 points) The dipole is placed inside a ~ , so that the homogeneous magnetic field B ~ ~ and B is θ . Find the angles angle between m θs and θu that correspond to stable and unstable equilibria, respectively. Calculate the amount of work (w) needed to rotate the dipole from θs to θu . Give your answer in terms of m and B. We can use this model to calculate the magnetic properties of materials containing unpaired electrons that have negligibly weak interactions with one another. Let us consider a sample of material with n such uni) (1 point) Find the orbital period T a of paired electrons per unit volume, placed in~ l d Mars, i.e. find the length of Martian “year” side a homogeneous magnetic field B. Due to spin, each unpaired electron acts as a in Earth years. small magnetic dipole. However, owing to the 2. CELLPHONE CAMERA (6 points) A photographer focussed his camera to distance L ii) (1.5 points) How long ( t t ) does a one-way quantum nature the electron, the projection ~ can only be of its magnetic moment along B and took a photo. On the photo, all farther trip to Mars take? objects (up to infinity) turned also out to be iii) (1.5 points) The spaceship is put into this µB or −µB (µB is called the Bohr magneton). sharp. Additionally, all closer objects down orbit by using a powerful rocket. It is more ii) (4 points) Calculate M , the magnetic moto distance s were sharp. efficient to burn fuel as a short burst when ment per unit volume of the sample, if the i) (4 points) What is the minimum pos- the spaceship is still near Earth. How much temperature of the material is T and the exadditional speed (∆v1 ) does the booster have ternal magnetic field is B. sible L? to be able to give to the spaceship to enter the ii) (2 points) Find the corresponding s. transfer orbit, starting from the north pole? 5. FRICTION OF A STRING (8 points) Measure the dynamic coefficient of friction Background. We consider the image of Neglect the air resistance. µ1 between the ballpoint pen and the string. a pointlike object to be sharp if its image is iv) (1.5 points) Estimate the ∆v needed to 2 Estimate the uncertainty. It might help that smaller than one pixel on the sensor. Oth- enter a circular orbit close to Mars. the dynamic coefficient of friction between erwise the image is blurry. The lens of the the pencil and the same string was measured v) (1.5 points) What is the minimal duration camera may be viewed as a convex lens. The beforehand and µ2 = 0.20 ± 0.01 was obtained. of the trip to Mars and back? camera is focussed by changing the distance Equipment: dynamometer, string, ballbetween the sensor and the lens. 4. MAGNETIC DIPOLES (7 points) Let us point pen, pencil and weight. consider the following model for a magnetic Parameters. Calculate the answer for a cellphone made by a well-known company. dipole. Some wire with no resistance has Estonian-Finnish Olympiad 2013 8. ZENER DIODE (7 points) An inductance come to a halt. 6. SPHERE AND CYLINDER (7 points) A sphere and a cylinder are lying on an inclined surface with inclination angle α. Both have mass m and radius r . The bodies are released from equal initial heights H . The moments of inertia of the sphere and the cylinder are, respectively, I sph = 52 mr 2 and I cyl = 21 mr 2 . The coefficient of friction between the surface and the bodies is µ. L and a capacitor C are connected in series with a switch. Initially the switch is open and the capacitor is given a charge q 0 . Now the switch is closed. iv) (2 points) Find the decrease ∆ q in the maximum positive value of the capacitor’s charge q after one full oscillation. How long does it take before oscillation halts? 10. RESISTIVE HEATING (8 points) Measure the resistor. You are not asked to estimate the uncertainty. Equipment: resistor, voltage source (batteries), ammeter, calorimeter, thermometer, i) (1 point) What are the charge q on the 9. GLASS CYLINDER (7 points) In the fol- stopwatch. capacitor and the current I in the circuit as lowing figure, there is a half-cylinder, made The calorimeter has V = 0.80 dl of water functions of time? Draw the phase diagram of glass and put on a paper with stripes (the and m = 27 g of aluminum, specific heat caa of the system — the evolution of the system inter-stripe distance is everywhere the same). pacity of water c = 4.2 J/(K · g) and of aluw on a I − q graph — and note the curve’s para- Find the coefficient of refraction of the glass. minum c = 0.90 J/(K · g). Internal resistance a meters. Note the direction of the system’s of the batteries will vary. Your set of batteries i) (2 points) Which of the bodies comes down evolution with arrow(s). may become depleted, spares are available. faster? What was the relative lag of the A Zener diode is a non-linear circuit eleslower body γ = ( t 2 − t 1 )/ t 1 ? The times t 1 and ment that acts as a bi-directional diode: it t 2 , respectively, denote the traveling times of allows the current to flow in the positive dirthe faster or the slower body. Assume that ection when a forward voltage on it exceeds the rolling occurs without slipping. a certain threshold value, but it also allows ii) (2.5 points) Find the minimal angle of a current to flow in the opposite direction inclination α0 for which the cylinder starts when exposed to sufficiently large negative to slide in addition to rolling. voltage. Normally the two voltage scales are iii) (2.5 points) If α → 90◦ , the bodies obvi- quite different, but for our purposes we will ously lose contact with the surface and fall take a Zener diode with the following voltdown in free fall with equal times. What ampere characteristics: for forward currents, is the minimal angle of inclination αm , for the voltage on the diode is Vd , for reverse which both the sphere and the cylinder come currents, the voltage on the diode is −Vd , for zero current the voltage on the diode is down with equal times? −Vd < V < Vd . 7. BURNING WITH A LENS (7 points) SunNow we connect the inductance L, the carays are focused with a lens of diameter d = 10 cm and focal length of f = 7 cm to a pacitor C all in series with a switch and a black thin plate. Behind the plate is a mirror. Zener diode. The switch is initially open. The Angular diameter of the Sun is α = 320 and capacitor is again given the charge q 0 > CVd its intensity on the surface of the Earth is and the switch is then closed. I = 1000 W/m2 , Stefan-Boltzmann constant ii) (2 points) Make a drawing of the phase σ = 5.670 × 10−8 W/(m2 K4 ). diagram for the system. Note the direction of i) (4 points) Find the temperature of the heated point of the plate. the system’s evolution with arrow(s). iii) (2 points) Does the evolution of the sysii) (3 points) Using thermodynamic argu- tem only necessarily stop for q = 0? Find the ments, estimate the maximal diameter of the range of values of q on the capacitor for which the evolution of the system will necessarily lens for which this model can be used. Estonian-Finnish Olympiad 2013 ii) (4 points) Let the corners be denoted by A and B, and the tip of the prism (at Solutions 1. PRISM (8 points) i) (4 points) The prism is acted on by three ~ l from the left-hand forces: reaction force R table, directed perpendicularly to the prism’s ~ r from the right-hand face; reaction force R table, directed horizontally (with its point of action to be determined yet); and gravitational force mg, directed vertically and applied at the triangle’s centre. (Considering a planar triangular cross-section of the prism is enough.) ~r On the verge of falling out, the force R is applied at the lower corner of the triangle. If a body in equilibrium is acted on by three forces, then their lines of action must intersect at one point. This is because otherwise the torque of one of those forces would not be zero with respect to the intersection of the lines of action of the two other forces. its equilibrium position) by C . Let us consider a small rotation of the prism (assuming it remains in contact with the corners). The trajectory of the tip is a circle ascribed around the triangle ABC (it follows from the property of the inscribed angles because the 6 ACB remains equal to 60◦ ). The radius of p that circle r = l / 3; its centre will be denoted by O . Once the prism rotates by angle α, so that the new position of the tip will be D , the central angle 6 COD = 2α. Hence, the tip is raised by r − r cos(2α) ≈ 2 r α2 . The height of the centre of mass P of the prism is raised because the tip is raised, and lowered because the vertical projection of the segment CP is reduced p by |CP |(1 − cos α) ≈ |CP |α2 /2. Here, |CP | = a/ 3. So, the original position is stable if pa α2 /2 < 2 pl α2 , hence l > 14 a. 3 a 3 the values of a and b gives f2 ³ f+ L 1 + η/D η ´2 f2 = =f ≈ f 1+ f+ s 1 − η/D 1 − η/D D µ ¶ 2η ≈ f 1+ . D Let’s consider that the camera is focused to distance L and the image is formed exactly on the sensor’s plane. The object’s distance L and its image’s distance a (corresponds to p0 on the figure) are related by the lens formula 1 1 1 L + a = f , thus µ ¶ Lf Lf f f2 Lf = ≈ 1+ = f+ , a= L− f L(1 − f /L) L L L where the approximation (1 + x)−1 ≈ 1 − x (for small x) was used. Image’s distance exceeds the focal length by ∆a = a − f = f 2 /L. Finally, f 2 / s = 2 f η/D , or s = 12 D f /η = L/2 ≈ 2.75 m. 3. MISSION TO MARS (7 points) i) (1 point) We can find the orbital period of Mars from Kepler’s third law R a 3 /R g 3 = T a 2 /T g 2 , giving t t ≈ 1.87 yr. ii) (1.5 points) Again, we can use Kepler’s third law to calculate half of the orbital period. tt = T g (R a + R g )3/2 2 (2R g )3/2 ≈ 0.707 yr. iii) (1.5 points) Background. ∆v is important, because the sum of all ∆v determines As the distance between the triangle’s i) (4 points) The light coming from an in- how much fuel is needed for a given mission. p finitely far away object will pass the focal The fuel needed is exponential of total ∆v and P P0 centre and its side is 63 a, the distance point F and form a cone which is cut by the is described by Tsiolkovsky rocket equation. ~ ~ between the points of action of R l and Rpr is p α sensor’s plane. The diameter d of the cut on 3 3 1 1 ◦ ◦ Kinetic energy per unit mass of such a 6 a cos 30 = 4 a. Thus, d = 4 a cos 30 = 8 a. α the sensor’s plane can be found from similar transfer orbit where it intersects the Earth’s triangles d /D = ∆a/ f , thus d = D f /L. Taking GM GM orbit is − R g +Rs a + R g s . Using the orbital into account the sharpness condition d ≤ η, 2α O A B of Earth we can substitute where η = w/ N is the size of a single element angular speed 4π2 R 3g of the sensor, we find that the limiting value GM s = T . The speed at the beginning of g p of L is L = D f /η = D f N /w ≈ 5.5 m. the transfer orbit becomes 3 DC 6 a s ii) (2 points) We’ll now find the shortest disµ ¶ a l 1 1 tance s satisfying the sharpness condition. v t0 = 2GM s − ≈ 32.7 km/s R g R g + Ra m~ g ~ Object at distance s will have an image at Rl 2. CELLPHONE CAMERA (6 points) The distance b = f + f 2 /s and the light passing the distance L is often called hyperfocal distance lens will converge behind the sensor’s plane . 0 The speed in Earth’s inertial frame is 1 a in photography and it was calculated more forming a cone. The diameter d 2 of the cone’s v t0 = v t0 − v g ≈ 2.94 km/h. To achive that, we 4 than one hundred years ago by Louis Derr cut with the sensor’s plane can be calculated first need to escape Earth’s gravity, so v Ã (the figure is taken from his book Photo- from similar triangles: d 2 /D = ( b − a)/ b. Ac! u ~r R u v0t 2 GM g graphy for students of physics and chemistry, counting for sharpness condition d 2 = η, we 0 t ∆ v1 = 2 + . published in 1906). 2 rg d can express b = a/(1 − η/D ), and substituting —1— Using the surface gravity of Earth we can GM substitute r g g = g g r g so ∆v1 ≈ 11.2 km/s. iv) (1.5 points) We can calculate the speed of the transfer orbit where it intersects the orbit of Mars from Kepler’s second law v t1 = v t0 /1.52 ≈ 21.5 km/s. The speed of the spacecraft relative to Mars is v0t1 ≈ 3.25 km/s. The speed of the spacecraft once near Mars surface is v Ã ! u u v0t 2 1 t vt m = 2 + r a g a ≈ 5.98 km/s 2 Since the speed of low Mars orbit is v e a = p r a g a ≈ 3.55 km/s, we need to brake for ∆v2 ≈ 2.43 km/s. v) (1.5 points) The Earth–Sun–Mars angle α at the launch of the mission needs to be α = π − wa t t ≈ 0.77 for the spacecraft to reach Mars. Likewise for the return trip β = π − w g t t ≈ −1.301. If we go to the corotating frame of referense with earth, we can see that the minimal time between those two 2π−α+β angles is wa −w g ≈ 1.96 yr. The minimal duration of the trip is therefore longer by t t , giving 2.67 yr. 4. MAGNETIC DIPOLES (7 points) i) (3 points) There is no torque on the square if θ = 0 or θ = π, so one of them is stable and the other unstable. If we start from θ = 0 and turn the square to some θ , but keep two sides of the square perpendicular ~ , Lorentz forces on these two sides give a to B torque τ = −BIa · a sin θ = −Bm sin θ towards decreasing θ . By symmetry, we get the same result if we keep the other keep two sides ~ . It is posof the square perpendicular to B sible to conclude that the torque depends only on θ (at least near θ = 0), not on the exact orientation of the square. Since torque acts to restore θ = 0, we find that θs = 0 and θu = π. To find the work to get from θs to θu , we can again keep two sides perpendicu~ - the answer cannot depend on the lar to B path, so we choose the simplest one. Integrating τ = Bm sin θ from θ = 0 to π gives us w = 2Bm. ii) (4 points) Let us denote the number of electrons (per unit volume) with magnetic moment projection +µB as n + and the ones with −µB as n − . Their sum is always the same, n + + n − = n. Also, in thermal equilibrium, their ratio is given by nn−+ = ³ ´ 2µ B exp − k BT , where k B is Boltzmann’s conB stant. Solving the equations, we can find n + and n − . The total magnetic moment per ~ ) is given by unit volume (in the direction of B M = µB ( n + − n − ). After substituting, ³ ´ 2µ B ¶ µ 1 − exp k BT µB B B ³ ´ M = µB N . = µB n tanh 2µB B kB T 1 + exp kB T Additional comments. We see B and M ~ is always have the same sign, therefore M ~ parallel with B. This makes sense, as we ~ ) orientation ~ parallel to B saw that θ = 0 ( m had lowest energy. The graph of M vs B goes to µB n for very large B or to −µB n for very ~ ). At B = 0, small B (all spins aligned with B M = 0 as well, since both spin orientations have the same energy. Around zero, the curve is linear, as tanh x ≈ x for small x gives us M≈ 5. µ2B nB kB T . FRICTION OF A STRING (8 points) Let’s first calculate the difference of tension force T between two ends of a sliding string arced over a cylinder by an angle α. Furthermore, let’s look at a short piece of the arc that subtends an angle d α. On one hand, dT = µ dR is the friction force acting on the piece, where dR is the reaction force. On the other hand, dR ≈ T d α, because both ends of the piece are pulled by a force with a ra- dial component of T d2α (where α is measured stant acceleration a ∥ directed parallel to the in radians). Therefore, we get a differen- surface. Let’s express a ∥ from the equation tial equation: dT = µT d α or d ln T = µ d α, v2 = 2a ∥ x: whence T = T0 eµα . a ∥ = g sin α/(1 + k). As a solution to the problem, we can measThe times are now easy to calculate as t = ure the change of the tension force for differv/a ∥ , giving π 3π 5π ent angles α (for example, 2 , π, 2 , 2π, 2 µ ¶1 etc. for several turns; however, keeping the p 2 2H t = 1+k . strings vertical offers better precision) and 2 g sin α plot ln T with respect to α. The slope of the Replacing k s = 25 for sphere and k c = 12 for graph is the µ to be measured. Extra solution (not as exact). Those who cylinder, we find that the sphere is faster by cannot derive the necessary formula can still a relativesfactor r do the experiment by doing the same meas1 + kc 15 −1 = − 1 ≈ 0.035. γ= urements and noting from the plot that the 1 + ks 14 relationship between α and T looks exponential. Thus, we can make an ansatz that ii) (2.5 points) As found in previous subpart, T = T0 X µα : as α = 0 must correspond to the acceleration’s parallel component to the T = T0 , we cannot reasonably write the µ slope a ∥ is smaller than the contribution by anywhere else without over-complicating the gravity g sin α. The difference is contributed formula. Now, we can re-measure the given by the friction force F f = mg sin α− ma ∥ . Slidpencil (it may be reasonably enough approx- ing starts, if the necessary friction reaches imated with a cylinder here; more exact ap- the maximal value Fmax = µ N = µ mg cos α. proaches exist) and conclude that X ≈ 2.7. Equating the two expressions gives From there on, the calculation is the same. mg sin α − mg sin α/(1 + k) = µ mg cos α, 6. SPHERE AND CYLINDER (7 points) 1+k tan α = µ . i) (2 points) Since no energy is lost due to k friction on sliding, the change in potential For the cylinder the limiting angle is α = 0 energy ∆E p = mgH is transformed to kinetic arctan(3µ). energy consisting of both translational and rotational motion. Taking into account the iii) (2.5 points) When the maximal friction force is reached, the motion goes into rolling rolling condition v = ω r , we have and sliding mode, where the total force com1 1 ponent along the surface is given by the dif∆E p = E k = mv2 + I ω2 2 2 ference of gravity and friction: 1 1 1 = mv2 + kmv2 = (1 + k) mv2 , F∥ = mg sin α − Fmax = mg sin α − µ mg cos α. 2 2 2 where general expression I = kmr 2 for mo- We note that the acceleration in this mode ment of inertia is used. Therefore, v2 = does not depend on the moment of inertia any 2 gH /(1 + k). more. On the other hand, the bodies travel disCalculating the limiting angle of slipping tance x = H / sin α along the slope with a con- mode also for the sphere αsph = arctan( 72 µ) > —2— α0 shows that for all angles larger than αm = αsph both bodies are in the slipping mode and thus have equal accelerations and arrival times. 7. BURNING WITH A LENS (7 points) The solar energy flux which is focused by the lens to the image of the Sun can be calculated as P = π4 d 2 I ; the image of the Sun radiates according to the Stefan-Boltzmann law with the total power P = π4 (α f )2 σT 4 . From the heat balance we obtain π4 d 2 I = π4 (α f )2 σT 4 , hence v s u u d I t T= ≈ 4500 K. αf σ Due to the second law of thermodynamics, it is impossible to direct heat energy from a lower temperature body to a higher temperature body. Hence, the image temperature cannot exceed the temperature of the Sun. Now we can use the known temperature of Sun T0 = 6000 K, but it is better to use the Stefan-Boltzmann law for solar radiation flux density: near the Sun’s surface, I 0 = σT04 , with the total flux of P t = 4πR 2s I 0 . Near the Earth, the total flux P t = 4πL2 I ; here, R s is the Sun’s radius, and L — the orbital radius of the Earth. From here we obtain I = I 0 R 2s /L2 = σT04 R 2s /L2 ; using the previous result, s d Rs T = T0 . αf L the frequency ω = p 1 and we can immedi- summarize the equations as follows: LC ately write q( t) = q 0 cos ω t, while I ( t) = q̇( t) = q if q̇ < 0 L q̈ + = Vd −ω q 0 sin ω t. C q Note that L q̈ + = −Vd if q̇ > 0 C 1 q2 + 2 I 2 = q20 (sin2 ω t + cos2 ω t) = q20 , Let us introduce the new variables q 1,2 ω such that q 1 = q − CVd and q 2 = q + CVd . and therefore the phase diagram of the sys- Then we can rewrite the two equations above tem is an ellipse centred at the origin, with in a more familiar form: semi-axes q 0 and ω q 0 . Alternatively, this req1 lation comes directly from the conservation L q̈ 1 + =0 if q̇ < 0 C of energy: q2 L q̈ 2 + =0 if q̇ > 0 C q20 LI 2 q2 + = E0 = . Thus the introduction of the diode only serves 2 2C 2C to shift the equilibrium points for the othBy looking at q and I a quarter-period erwise simple harmonic orbits. For q̇ > 0, later from t = 0, say, it’s not hard to see that the equilibrium point is q 2 = 0 or q = −CVd , the system must evolve in a clockwise sense while for q̇ < 0 it is q = CVd . So the orbit will on the phase diagram. Note that in this in- consist of half-ellipses in the upper and the stance, only q = 0 is an equilibrium point: lower parts of the I − q diagram, centred at for all non-zero q there will be never-ending q = −CVd for the upper half and at q = CVd for the lower half. As the evolution is continuoscillations in the circuit. ous, these half-ellipses will join up at I = 0. I I Note that we have the right to talk about half- and full periods because the oscillations still happen at the immutable frequency ω = p 1 . Therefore the time between the LC two maxima is just a full period of oscillation, T = 2ωπ . Once q( t) has a zero derivative inside the region bounded by ±CVd , it will remain at that particular value forever. For a large initial ¯ ¯q 0 , we expect there to be approximately ¯ q0 ¯ | q0 | total oscillations. ¯ ∆ q ¯ = 4CV d GLASS CYLINDER (7 points) The axis of the half-cylinder is where the stripe and its image coincide (form a straight line). The front edge of the half-cylinder is at the 28th line, counting from the axis, hence the radius of the cylinder R ≈ 28. Let us consider the refracting ray s which is very close to a total internal reflection. One can see the images of 20 lines (ca 20.2, to be more precise), when counting from the central line upwards; the upper edge of the half-cylinder coincides with 9. −CVd CVd q which means that d ≤ 2 f . 8. ZENER DIODE (7 points) i) (1 point) Kirchoff’s 2nd law gives L İ + ii) (2 points) Now the sign of the voltage on 1 q/C = 0 or q̈ + LC q = 0. This is the equa- the diode depends on the direction of the cur- iii) (2 points) We can see on the diagram q tion of a simple harmonic oscillator with rent, giving either of L q̈ + C ± Vd = 0. We can that there is a “dead zone” between ±CVd —3— iv) (2 points) Let’s use the phase diagram to figure this out. Suppose the capacitor initially has the charge q 0 À CVd . Then the charge will first swing to the other way of CVd and will become q T /2 = CVd − ( q 0 − CVd ) = 2CVd − q 0 . Then it will perform the other half-oscillation around −CVd and the charge at the end of that is q T = −CVd + (−CVd − (2CVd − q 0 )) = q 0 − 4CVd , and therefore ∆ q = −4CVd . More exactly, the distance from the “dead zone” is initially | q 0 |− CVd and decreases during each half-oscillation by 2CVd .jThe total k | q 0 |−CVd number of half-oscillations is N = 2CVd p π and the total time t = N T2 = N ω = N π LC . q Let us note that αL = 2R s , hence s d T = T0 ≤ T0 , 2f (for I = 0). If a trajectory reaches any of the points in that segment, it will stay there forever. The extent of that region is 2CVd . the 42nd line at the background. the perpendicular of the paper surface. The Measure the temperature at the end, after ◦ So, the ray s arrives at the camera at incidence angle of the ray s is α + β ≈ 45.9 , waiting a bit or stirring the calorimeter. We want to get maximum temperature difference the angle α = arcsin(28/42) ≈ 41.8◦ with re- hence n = 1/ sin(α + β) ≈ 1.39. 2 spect to the plane of the paper. The pro- 10. RESISTIVE HEATING (8 points) After for precise measurement. Since P = R I , jection of the refraction point to the paper noting the temperature of the calorimeter, Q X ≈ ∆ t( I n /2 + I n+1 /2)2 , surface lies at the distance a = 28 sin α = connect the batteries, resistor (in the calorR 2 n 28 /42 ≈ 18.7 lines from the axis. There- imeter) and ammeter in series. Choose a ( c a m a + c w m w ) ∆T fore, before refraction, ray s forms an angle convenient time interval ∆ t and note the amR= P . 2 ◦ n ∆ t( I n /2 + I n+1 /2) β = arcsin[(20.2 − 18.7)/(28 cos α)] ≈ 4.1 with meter reading until batteries are depleted. —4— The resistance used was R = 0.47 Ω ± 5%. In the described circuit the batteries were depleted in 10 to 15 minutes and the temperature of the calorimeter rose by 7 to 10 degrees. ESTONIAN FINNISH PHYSICS OLYMPIAD PROBLEMS & SOLUTIONS (2014) Estonian-Finnish Olympiad 2014 iv) (2 points) Assuming that Vmax À V0 , Sun? 1. DC-DC CONVERTER (8 points) In order to obtain high voltage supply using a battery, the following circuit is used. An electromagnetic switch K 1 connects a battery of electromotive force E to an inductor of inductance L: it is closed if there is no current in the inductor (a spring keeps it closed), but if the inductor current reaches a critical value I 0 , magnetic field created by the inductor pulls it open. Due to inertia, once the key is open, it takes a certain time τK to close again even if the current falls to zero. what is the average power dissipation on the diode? iii) (1.5 points) What is the speed of the spaceship in the Earth’s frame of reference v) (2 points) Now, let the key K 2 be closed, when the distance from the Earth is much and let us assume simplifyingly that V0 = 0; larger than the Earth’s radius, but still much p also, RC À τL and τK > π LC . Suppose that shorter than the distance to the Sun? the circuit has been operated for a very long iv) (2.5 points) Answer the previous question time. Find the average voltage on the res- without making the approximation α ≈ 0°. istor. v) (1.5 points) What is the speed of the spacevi) (1 point) Find the amplitude of voltage ship in the Earth’s frame of reference when variations on the resistor. the distance from the Earth is much smaller 2. WASTE PROJECT (8 points) In 2114, than the Earth’s radius? Europarliament decided that all radioact- 3. MAGNETS (6 points) To explore the force ive wastes need to be sent to the Sun, so between two small magnets, the following as to avoid contamination of Earth and or- experiment is performed. One of the magbital space. In what follows, you can use the nets is hanged from a thread with length following numerical data: duration of one l = 1 m. Other magnet is moved slowly closer year T = 365.25 days, orbital speed on Earth while keeping the axes of the magnets alv0 = 29.8 km/s, angular diameter of Sun as ways on the same horizontal line. At the seen from the Earth α = 0.5°, radius of the moment when the distance between the magEarth R = 6400 km, free fall acceleration at nets is d = 4 cm and the hanged magnet has 1 the Earth’s surface g = 9.81 m/s2 . moved x = 1 cm from initial position, bal- According to the project, the waste is sent to the Sun using ballistic spaceships: the enFor the diode D you may assume that gine operates only during a short period of its current is zero for any reverse voltage time during which the displacement of the (VD < 0), and also for any forward voltage spaceship remains much shorter than the smaller than the opening voltage V0 (i.e. for radius of Earth. In the Earth’s frame of refer0 < VD < V0 ). For any non-zero forward curence, the spaceship obtains a velocity opposrent, the diode voltage VD remains equal to ite to the orbital velocity of Earth in the Sun’s V0 . frame of reference. Further, the ship moves You may express your answers in terms along a ballistic trajectory until it hits the of L, E , I 0 , V0 , R , and the capacitance C (see Sun. The trajectory is such as to minimize figure). the consumption of fuel. i) (1 point) At first, let the key K 2 be open. If i) (1 point) Sketch the trajectory of the spacethe initial inductor current is zero, how long ship. time τL will it take to open the key K 1 ? As a first approximation, calculations can ii) (1 point) Assuming (here and in what be done when neglecting the angular size of follows) that L/R ¿ τK ¿ τL , plot the in- the Sun (i.e. by putting α ≈ 0°); you can use ductor current as a function of time t (for this approach for the next two questions. 0 ≤ t < 3τL ). ii) (1.5 points) How long will it take for the iii) (1 point) What is the maximal voltage spaceship to travel from the Earth to the Vmax on the resistor R ? 1 ance is lost and the magnets pull together. By making the assumption that the pulling force between the magnets Fm depends on the distance d according to the relation Fm ∝ d −n , find the value of the exponent n. 4. SUPERBALLS (5 points) n + 1 elastic balls are dropped so that they are exactly above each other, with a very small gap between each. Bottom ball has a mass of m 0 , the one above has a mass of f m 0 , next f 2 m 0 and so on, until the topmost ball with mass f n m 0 , where f < 1. At the moment when the bottom ball touches the ground, all the balls are moving with the speed v. i) (1 point) After the collision between the two bottommost balls, what is the speed v1 of the second ball from the bottom? ii) (3 points) What is the speed of the topmost ball vn after all collisions? iii) (1 point) How many times higher would that ball fly compared to the initial drop height h? Take f = 0.5 and n = 10. It maybe useful that that sequence a 0 = 1, a k+1 = a k α + β has a general term a n = αn + n β αα−−11 , where α and β are constants. 5. PLANCK’S CONSTANT (8 points) In a simplistic model, light emitting diodes can be considered to only pass current when lit, and then they have a constant voltage drop Vt = E e across them. E = h f is the energy of the light quanta emitted and e = 1.60 × 10−19 C is the elementary charge. Speed of light in vacuum c = 3.00 × 108 m/s. You have a assorted light emitting diodes numbered 1–6, each with a R = 680 Ω series resistor. From the datasheets it is known for the peak wavelengths of the diodes to be 940 nm, 620 nm, 590 nm, 525 nm, 470 nm, and 450 nm. i) (2 points) Find out which wavelength corresponds to which diode. ii) (4 points) Measure the Planck’s constant h that corresponds to our simplistic model. This does not have to correspond to real Planck’s constant. iii) (2 points) Estimate the uncertainty. Equipment: voltage source (batteries) with an unknown voltage, ammeter, assorted light emitting diodes with series resistor. Take care not to short the battery with the ammeter. You may ignore the internal resistances of the batteries and the ammeter. 6. RUNNING ON ICE (4 points) A boy is running on a large field of ice with velocity v = 5 m/s toward the north. The coefficient of friction between his feet and the ice is µ = 0.1. Assume as a simplification that the reaction force between the boy and the ice stays constant (in reality it varies with every push, but the assumption is justified by the fact that the value averaged over one step stays constant). i) (2 points) What is the minimum time necessary for him to change his moving direction to point towards the east so that the final speed is also v = 5 m/s? placed at a distance l = N λ (where N is a large integer) from the point source, and the interference pattern is observed on a screen which is placed to a distance L À l from the point source (see figure). 3R Cv is sparse, so that the mean free path of the molecules is much larger than a. Assume that v2 ¿ c s where c s is the speed sound in the atmosphere surrounding the cube. 2R 10. R ✠ ✡ ✝ ✆☎ ☎✄ ✂✁ 0 100 200 T(K) 300 i) (1 point) What is the total heat energy of such a cube at the initial temperature T0 ? Now, the cube has 5 ii) (3 points) faces painted in white (reflects all relevant wavelengths) and one face painted in black ii) (2 points) What is the shape of the optimal (absorbs all these waves). The cube is surtrajectory called? rounded by vacuum at a very low temperat7. SPIN SYSTEM (8 points) Let us consider ure (near absolute zero); there is no gravity In what follows we use the x, y, and z coa system of N independent magnetic dipoles field. Initially, the cube is at rest; as it cools (spins) in a magnetic field B and temperature ordinates as defined in the figure. The screen down due to heat radiation, it starts slowly T . Our goal is to determine some properties is parallel to the mirror and lies in the y − z- moving. Estimate its terminal speed v . 1 of this system by using statistical physics. It plane. is known that the energy of a single spin is i) (2 points) At which values of the y- iii) (2 points) At very low temperatures, the heat capacitance of aluminium is proE = ² m, where m = ± 21 and ² = αB. coordinate (for z = 0) are the interference portional to T 3 , where T is its temperati) (2 points) What is the probability p ↑ for maxima observed on the screen? You may ure. Which functional dependance f ( t) dea spin to be in exited state, i.e. have positive assume that y ¿ L. scribes the temperature as a function of time energy? ii) (1 point) Sketch the shape of few smallest- [T = A · f (Bt), where A and B are constants] ii) (2 points) What is the average value of sized interference maxima on the screen (in for such very low temperatures under the assumptions of the previous question? the total energy E s of the spin system as a y − z-plane). function of B and T ? iii) (2 points) Now the flat screen is replaced iv) (3 points) Now, the cube has 5 faces iii) (2 points) Using high temperature ap- with a spherical screen of radius L, centred covered with a thermal insulation layer (you around the point source. How many interfer- may neglect heat transfer through these proximation T À αBm k , simplify the expresence maxima can be observed? faces). One face is left uncovered. The cube sion of E . ✟ ✞ s iv) (2 points) Using high temperature ap- 9. THERMAL ACCELERATION (9 points) Consider a cube of side length a = 1 cm, made proximation T À αBm k , find the heat capacity of aluminium (density ρ = 2.7 g/cm3 , molar C of the spin system. mass M A = 23 g/mol). The heat capacitance 8. MIRROR INTERFERENCE (5 points) of one mole of aluminium is given as a funcA point source S emits coherent light of tion of temperature in the graph below. The wavelength λ isotropically in all directions; speed of light c = 3 × 108 m/s, universal gas thus, the wavefronts are concentric spheres. constant R = 8.31 J/(kg · K). The initial temThe waves reflect from a dielectric surface perature of the cube is T = 300 K. 0 is surrounded by hydrogen atmosphere at a very low temperature (molar mass of hydrogen molecules M H = 2 g/mol). The cube starts cooling down due to heat transfer to the surrounding gas; you may neglect the heat radiation. Initially, the cube is at rest; as it cools down, it starts slowly moving. Estimate the order of magnitude of its terminal speed v2 . Assume that the surrounding gas YOUNG’S MODULUS OF RUBBER (12 points) The linear Hooke’s law for a rope made from an elastic material is supposed to held for small relative deformations ε = x/L (which is also called “strain”), where L is the undeformed length of the rope, and x is the deformation. Once ε becomes too large, the force-deformation relationship F = kx is no longer linear; what is “too large” depends on the material. For very elastic materials which can reach relative deformations considerably large than one, it may happen that the linear Hooke’s law with a constant stiffness k fails, but if we take into account the change of the cross-sectional area S of the rope with k = ES /L, where E is the Young’s modulus of the elastic material, such a nonlinear Hooke’s law remains valid. In that case we can say that there is still a linear stress-strain relationship σ = E · ε, where the stress σ = F /S . i) (7 points) Measure the relationship between the stress σ and strain ε in a rubber string and plot it. ii) (5 points) From your plot determine the Young’s modulus E with its uncertainty, and the maximum strain εm until which it applies. Note: the diameter of the thread is to be measured using the diffraction of laser light. Equipment: rubber thread, stand, measuring tape, 15 hex nuts with a known mass, a plastic bag for hanging a set of nuts to the thread, a green laser (λ = 532 nm), a screen. WARNING: AVOID LOOKING INTO A LASER BEAM, THIS MAY DAMAGE YOUR EYES! 1. interval when the diode is open, we can also neg- DC-DC converter 1) (1 pt) From the Kircho 's voltage law for L and E , E = L dI , hence dt = Et/L. From I0 = EτL /L we obtain the loop consisting of I lect the presence of the diode since Hence, the current i V0 Vmax . LC -loop changes in from i = I0 and end- in the time sinusoidally, starting hence its trajectory is an ellips with longer semiaxis equal to proportionality coecient. Then RE /2. According to the Kepler's 3/2 III law, the period on such an orbit is 2 times shorter than the Earth's orbital period T. 0 ∆Fm = Fm (d)∆d = because ∆d = −∆x. kn ∆x, dn+1 Therefore τL = LI0 /E. 2) (1 pts) Once the current I0 is reached, the key is opened; the current trough L cannot change i = 0 (then the diode will close disconnecting the LC -loop). During that process, the 2 1 magnetic energy of the inductor 2 LI0 is conver- 3) ted into the electric eld energy of the capacitor, needs to be zero; hence, in the Earth's frame instantaneously and therefore is forced to ow is described by this problem) which is later released as heat on the resistor. of reference, it is opposite to the orbital ve- have two equations with two unknowns (n and When the stationary regime is achieved, the en- locity of the Earth and by modulus equal to k): ergy lost by the capacitor during one period (of v0 = 29.8 km/s. 4) The speed vS R. through the resistor Since the characteristic time of this current loop (consisting of is very short (L/R τK ), L and R the current decays ing when τL ) as the heat dissipation on the res2 Q = Vav τL /R must be equal to the energy duration very fast and becomes essentially equal to zero istor while the key is still open. Now there is no cur- received from the inductor; so, rent through the inductor, so that the key will r close again and the process will start repeating from the beginning. As a result we'll have a periodic graph as shown in gure. I I0 2 Vav τL 1 = LI02 ⇒ Vav = I0 R 2 6) LR = 2τL r EI0 R . 2 found as qC = τL Vav /R (owing to RC τLC , the relative change of the capacitor's voltage is small). 3) Hence, the voltage drop is found as ∆V = qC /C = τL Vav /(RC). maximal values, so that the amplitude maximal when the current is maximal, which happens immediately after the switch is opened; I0 so that Vmax = RI0 . Vmax V0 , we can neglect U0 = the maximal current is 4) (2 pts) Due to the eect of the diode; so we have the Kircho 's dq dI voltage law L dt = RI = R dt (here we expressed the current via the charge q which ows through the resistor). Integration over a single cycle (when the inductor current drops from down to 0) yields LI0 = R∆q , I0 hence the charge own through the resistor (and through the diode) ∆q = I0 L/R. During that cycle, the diode V0 , A = V0 ∆q had a constant voltage so the electric eld performed work which was released as heat in the diode. So, the average power dissipation (2 pts) Now, since the characteristic time of RC -loop is very large, the capacitor main- 2. 1) Waste project the Earth's orbit needs to be as large as posit to the elliptical orbit), hence the full orbital GM m energy of the ship E = − 2a needs to be as small as possible. Here, M is the mass of the Sun, m the mass of the space ship, and the longer semiaxis. So, a a needs to be as large as possible, which means that the perihelion needs 2a = RE + rS , where the Earth and rS RE is the orbital radius of the radius of the Sun. The =− 2a RE + rS r vS = =− Re + 2 Part of the initial kinetic energy in the Sun 2) If we neglect the radius of the Sun, the space RC TLC ), we can neg- ship needs to fall directly to the Sun, which lect the presence of the resistor. During the time means that its initial orbital speed must be zero, 2 vE + 2gR ≈ 29.2 km/s. Magnets net: the downwards directed gravity force T~ , is almost equal to mg , horizontal projection is expressed as l kn mg = . dn+1 l 4. Superballs 1) During the bottom-most collision ball will −(x/l)mg , x = 1 cm is the length of the thread and the its oor, the speed and v0 = v . Let the k-th ball move up with a vk ; we'll consider the collision between this and (k + 1)-st ball, which moves down with the velocity v . In the frame of reference of the vk −f v centre of mass, u = ; hence, after the 1+f collision the upwards velocity equals to vk+1 = −f v 2 = 1+f vk + 1−f v . With v0 = v , we v + 2 vk1+f 1+f 3−f 4 can conclude that v1 = 1+f v = 1+f − 1 v . 2) speed velocity One can see that if we apply the recurrent formula repetitively, the result at the n-th step will n 2 vn = 2 1+f − 1 v. 3) Now we need to relate the speeds to the jump2 2 ing heights via v = 2gh0 and vn = 2ghn ; hence, hn /h0 = ~m . F so that its with retain change the direction of the velocity; its upwards m~g , Since the thread's angle is very small, the mod- T~ kn mg − = 0; dn+1 l xmg k = , dn l which is directed along the thread, and the horizontal magnetic force ulus of Thus we this can be rewritten as be There three forces acting on the hanging magthe tension force k xmg − = 0, dn l ∆F = 0. d/n = x, hence n = d/x = 4. of the potential energy due to the gravitational GME m = gmR hence pull of the Earth, ∆Π = R 2 2 vE u gR + 2 = 2 . Here, ME is the Earth's mass and u is the launching speed. So, 3. ∆x. equations we obtain √ 2rS α = v0 2 sin ≈ v0 α. R E + rS 2 Numerically this yields vS ≈ 2.8 km/s; the speed in the Earth's frame of reference vE = v0 − vS ≈ 27.0 km/s. vS = v0 p If we divide the corresponding sides of the two GM 2rS . R E R E + rS r u= kn mg − dn+1 l At the limit case of the loss of stability (which , This expressing can be rewritten by using equal2 GM as ity v0 = R rE ∆F = in the Sun's frame of reference hence where resulting trajectory is depicted below. LC -loop is and as the current to the resistor can be neglected (since − sible (we need to decelerate the ship to bring rbit TLC = 2π LC In order to minimize the fuel consumption, the speed near opens, the diode will open, and the capacitor formed. That √ loop admits oscillations of period In the Sun's frame of reference, the speed is found from the expression for the total energy, GM m GM m GM m mvS2 5) The trajectory is a very elliptical ellipse, peri- helion of which is within the Sun. of time when the diode is closed. Once the key is connected to the inductor so that a of the period, so that Earth's frame of reference goes to the change tains its charge (and voltage) during that period K1 I0 . 2RE Earth’s o the τL Vav I0 L ∆V = = 2 2RC 2C r to lie at the surface of the Sun, in which case A V0 I0 L V0 E P = = = . τL RτL R 5) The amplitude is half of the dierence between the minimal and (1 pt) The voltage through the resistor is t=2 (1 pt) The charge which ows away from the capacitor when the diode is closed can be t t is half T ≈ 64.6 days. The travel time −5/2 For f = 0.5 p vn /v0 = 2 2 1+f n − 1. n = 10 we obtain that the nal ca 1200 larger than the initial one. and height will be 5. s Planck's constant (the displacement from the initial position). The 1) net horizontal force battery, we can observe the light of the emit- equilibrium ted light; the mapping is as follows: is stable if F = Fm − (x/l)mg . At the point F = 0. The equilibrium point a small (virtual) displacement ∆x gives rise to a returning force ∆x ∆F = ∆Fm − mg l which needs to push towards the equilibrium point. Let F = kd−n , where k is an unknown m When we connect each of the diodes to the 940 nm 620 nm red, 590 nm orgreen; 470 nm blue; 450 nm invisible (infrared), ange, 525 nm violet. 2) We can measure the current I through the diode (which is also the current through the resistor R), so that the voltage on the diode would be Vd = E − IR, but we don't know the battery voltage. However, we do know that the diode's voltage equals approximately to the photon's energy Ep hc/(λe). if we plot Ep = IR = E − Vd , expressed in electron volts, Since we expect that IR versus 1/λ, we should obtain a straight line IR = E − 3) h = which allows us to calculate The major source of the uncertainty is not the instrument uncertainties, but the departure of the real diode data from the simplistic model. Therefore we can try to t the data points with dierent straight lines making the slope A as steep as possible (while still keeping a reasonable t with the data points, and also as at as possible; the uncertainty of ∆A = 21 (Amax − Amin ), and ∆h 6. 1) For small values of the argument of the hy- over all the photons, the perpendicular to the those molecules which collide with the cube at perbolic tangent, the last expression can be ap2 proximated as E ≈ −N /4kT . surface normal components cancel out (photons low temperatures to the overall momentum re- go to all the directions). mains still small. 4) the parallel component can be estimated just as pk ∼ Ec . According to the denition of the heat capa2 2 dE city, C = dT = N /4kT . 8. Mirror interference 1) 1 hc . λ e We can measure the slope of the straight line A = hc/e, eA/c. 3) A is found = h∆A/A. as For a position α = y/L angle the arriving rays form an imation; the angle is in radians). Then, the optical path dierence between the reected and 2 direct rays is ∆ = 2l cos α ≈ 2N λ − N λα . Since there is an additional phase shift for the reected rays at the reection from optically denser dielectric material, the total phase shift 2 is ϕ = 2π∆/λ = 4πN − π(2N α − 1). At the maxima, this equals to is an integer. 2π(2N − n), Let us vx − vy point A with n r r 2) n + 0.5 ⇒ yn = L N n = 0, 1, . . . N . n + 0.5 , N x-axis, n the max- ima form on the screen concentric circles; the plane: we need to move from the pitch between the neighbouring circles becomes (v, 0) (0, v) to a point B with coordinates while having a constant speed. Indeed, the velocity of a point in the vx − vy -plane is the acceleration of the body, which has here a constant modulus µg . Obviously,√the fastest path √ is a straight line of length v 2, so that smaller as the order number radii form a sequence √ 3) number of maxima stant, the trajectory is the same as for a body A 9. According to Boltzmann's distribution, p↑ = · e−m , where the constant A can be found from the condition that the probability of having either up or down orientation is one: e−/2 + A · e/2 = 1, hence A· 1 1 A = −/2kT = . 2 cosh(/2kT ) e + e/2kT Thus, 2) e−/2kT p↑ = −/2kT . e + e/2kT The average energy is the weighted average of up- and down-state energies for a single spin, multiplied by the number of spins: −/2kT /2kT N e −e N E= =− tanh(/2kT ). 2 e−/2kT + e/2kT 2 ϕmax = 4πN + π and ϕmin = π . The 1) hence S σ is the Stefan-Boltzmann con- the radiating area. This simpli- es to 4) T = 0, Using the graph we nd this as the area under the curve, q ≈ R·560 J/K. 3 The number of moles ν = a ρ/MA ≈ 0.117 mol, Q = qν ≈ 546 J. 2) Each photon of frequency ν radiated by the cube carries away heat energy equal to E = hν , and carries momentum p = h/λ = hν/c = E/c. If the photon departs at the angle α with rehence the total heat energy The setup is as follows. The rubber thread is xed to the stand, and the plastic bag is xed to the free end of the thread. The hex nuts are added, one-by-one, into the bag, starting with zero and ending with 15. The laser light is directed to the thread and the diraction pattern is observed on the screen (which is a vertically xed sheet of graphic paper on a support). The diraction pattern from a wire is the same as from a single slit (the superposition of the Huygens sources from those two cases gives a full set and opposite wave amplitudes and equal intensities). So, if we measure on the screen the dis- a between such maxima which are separ- ated by n (e.g. n = 10) periods of the diraction tance pattern then using small-angle-approximation, nλ/d = a/L, where In the case of a hydrogen atmosphere, the that the molecules colliding with the coated they came, but uncoated face gives away heat energy, and the molecules leave at higher temIf we assume that the departing a part of the energy is transferred), then the where vT = p RT /MH 1/vT , is the thermal speed of the molecules after the collision with the cube for the motion along the surface normal. So we 3 estimate a ρv ∼ Q/vT , hence v∼ Q ρa3 r d the thread dia- The strain is calculated by making markings on the thread and measuring the distance MH ≈ 180 m/s. RT b between these in a stretched state, b − b0 , b0 ε= faces bounce back with the same speed as the momentum-to-heat ratio is estimated as is the distance from the d = nLλ/a. momentum is given to the cube due to the fact perature. L thread to the screen and meter. So, dT = −Bt ⇒ T = A · e−Bt . T when particles of dierent masses collide, only For the heat energy of one mole of material no heat energy by is a constant, stant, and 1) Young's modulus of rubber elds from those two cases must provide equal cube (which serves us only as an estimate Thermal acceleration dq = Cv dT .R There is T hence q = Cv dT . 0 A 10. of sources on a at wave front, hence the electric Q v ∼ 3 ≈ 0.67 mm/s. ρa c molecules have the same temperature as the m = (ϕmax − ϕmin )/2π = 2N. in the Earth's eld of gravity a parabola. 1) 3 ≈ 1.73, only within a hemisphere, the phase shift varies 2) Spin system 1 = 1, Since the reected rays can reach the screen between 7. grows (using the etc). t = v 2/µg ≈ 7.2 s. Since the direction of the acceleration is con- n length unit dened by the √ smallest√radius, the 5 ≈ 2.23, momentum equals to The heat balance at very low temperatures 3 4 can be written as AT dT = −σST dt, where Since the rays of a given order number form a xed angle with the 1/c, the overall Q/c. Thus, a3 ρv ∼ Q/c, same for all the photons, equal to 3) consider the this graphically using the coordinates Since the momentum-energy ratio is the 1 If we apply the exact factor 3 (obtained above via integration), we end up with v ≈ 0.22 mm/s. maxima is written as where During the process, the velocity vector needs where If we want to obtain an exact result, we need to integrate over all the directions while keeping in mind that the light intensity is proportional to cos α. So, the momentum R 2 averaged over all the 1 directions p̄k = Ec 2π cos αdΩ, where the solid angle dierential dΩ = 2π sin αdα. Therefore, p̄k = R R π/2 2 π/2 E E . sin α cos2 αdα = E cos αd cos α = 3c c c 0 0 Therefore, the condition for the α= Running on ice to change its direction by 90 degrees. y, (we use the small-angle approx- The average value of where b0 is the length in a non-strained state. The stress is calculated as σ= where and m N 4N mg , πd2 is the number of hex nuts in the bag the mass of a single nut. The data are plotted in a graph; linear relationship corresponds to a straight line. The uncertainties are calculated using the rule of relative uncertainties, either using Pythagorean or simple addition, e.g. ∆ε = ε∆b where ∆b ≈ 0.5 mm 2 1 + b − b0 b0 , is the length measurement uncertainty. Similarly, ∆σ = 2σ It should be noted that in fact, one should have ∆d . d been careful with such an estimate, because the These uncertainties are marked in the graph as spect to the surface normal then the componE ent parallel to the surface normal pk = c cos α. thermal speed is at the denominator. This will error bars. increase the relative contribution of the heat ra- 2) The total momentum given by the photons to diated at low temperatures. value of the cube equals by modulus to the total mo- maining heat at low temperatures is propor4 tional to T , and therefore the contribution of mentum carried by the photons; when averaged However, the re- Using the plot, we need to determine such a ε = ε∗ that for ε1 > ε ∗ , it is impossible to draw a straight line intersecting the error bars of all the data points with ε1 < ε ∗ .