IPhO 2026

Problem 1 Coastal mapping (multiple-choice problem) (5.0 pts.) (Problem group of the PhysicsOlympiad - Stefan Petersen) Using lasers mounted on aircraft, coastal waters can be mapped. To do this, short laser pulses are emitted in various directions, which are reflected diffusely partly at the water surface and partly at the seabed. As a result, part of the emitted light returns to the aircraft, where it is detected and evaluated. Consider an aircraft flying at constant altitude along a coastal strip. The following graph shows the signal strength of the reflected signal of a laser pulse as a function of time. The time origin is chosen so that the entire reflected signal is shown. Assume that the laser pulse strikes the water surface at an angle of incidence (measured from the normal) of $40^{\circ }$ and that the refractive index of the water is 1.33. Fig. 1. Illustration of mapping with a laser (image source GEUS). Time in ns Signal strength in arbitrary units Fig. 2. Signal strength as a function of time (the point $t=0$ is arbitrary). How deep is the water at the location investigated? A 2.7 m B 3.1 m C 4.1 m D 6.1 m Solution Calculations and explanations The time difference $\Delta t\approx 60$ ns $-29$ ns = 31 ns between the two maxima in the signal-strength curve corresponds to the time the light ray needs in the water to travel from the water surface to the seabed and back. The angle $\beta$ of the light ray in the water (measured from the normal) is, according to Snell’s law of refraction, given by (cf. Fig. 3) $\beta =\arcsin \left(\frac{1}{n}\sin \alpha \right)=28,9^{\circ },$ (1.1) where $\alpha =40,0^{\circ }$ is the angle of incidence of the light ray. $d$ $\alpha$ $\beta$ Water Bottom Fig. 3. Sketch of the ray path. The distance travelled by the light in the water during the time $\Delta t$ is $\frac{c\Delta t}{n}$, where $n=1,33$ denotes the refractive index of water. From this, the water depth $d$ follows by projection $d=\frac{c\Delta t}{2n\cos \beta }=\frac{c\Delta t}{2n\cos \left(\arcsin \left(\frac{1}{n}\sin \alpha \right)\right)}\approx 3,1\text{ m}.$ (1.2) Correct answer: B Remark: Answer option A results if the refraction at the water surface is not taken into account, answer option C without the lengthening of the optical path in the water, and answer option D if the factor 2 for the outward and return path of the light ray is not considered. Grading - Coastal mapping (multiple-choice problem) Points Recognising that the time difference of the maxima is decisive and reading off the value from the graph 1.0 Using the law of refraction and determining the refraction angle (1.1) 1.0 Taking into account the lengthening of the optical path 0.5 Taking into account the factor 2 for the light path 0.5 Setting up an expression for the water depth with projection (1.2) 1.0 Stating the correct solution 1.0 5.0

Topic: Geometric Optics Metodi: Snell’s Law, Ray Tracing, Experimental Data Analysis Competenze: Diagrammatic Reasoning, Physical Reasoning, Experimental Data Analysis Objects: — Fonte: Testo (PDF) — p.2

Problema 1 Mapping costiero (problema di scelta multipla) (5,0 p. d.) (Problema group of the PhysicsOlympiad - Stefan Petersen) Usando laser montati su aeromobili, le acque costiere possono essere

• Mappa. Per fare questo, vengono emessi brevi impulsi laser Le direzioni di questo fenomeno sono diverse, che si riflettono in modo diffuso, in parte sulla superficie dell’acqua e in parte sul fondo del mare. Di conseguenza, parte della luce emessa ritorna all’aeromobile, dove viene rilevata e valutata. Considerate un aereo che vola a costante altitudine lungo Una striscia costiera. Il grafico seguente mostra la forza del segnale del segnale riflesso di un impulso laser come funzione
• Non è tempo. Il tempo di origine è scelto in modo che il intero segnale riflesso è mostrato. Supponiamo che il pulso laser colpisca la superficie dell’acqua ad un angolo di incidenza (measure from the normal) di $40^{\circ }$ e che l’indice di rifrazione dell’acqua è di 1,33. Fig. 1. Illocalizzazione di mapping with a laser (Image source GEUS) Tempo in ns Signal strength in unità arbitrarie Fig. 2. Signal strength as a function of time (il punto $t=0$ è arbitrario). Quanto profondo è l’acqua alla località indagata? A 2.7 m B 3.1 m C 4.1 m D 6.1 m Soluzione Calcoli e spiegazioni La differenza di tempo $\Delta t\approx 60$ ns $-29$ ns = 31 ns due massime nella curva di forza del segnale corrisponde al tempo Il raggio di luce ha bisogno nell’acqua per viaggiare dalla superficie dell’acqua al fondo del mare e indietro. L’angolo $\beta$ del raggio luminoso nell’acqua (measure from the normal) è, according to La legge di refrazione di Snell, data da (cfr. Fig. 3) $\beta =\arcsin \left(\frac{1}{n}\sin \alpha \right)=28,9^{\circ },$ (1.1) dove $\alpha =40,0^{\circ }$ è l’angolo di incidenza del raggio di luce. $d$ $\alpha$ $\beta$ Acqua Sotto Fig. 3. Sketch del sentiero dei raggi. La distanza percorsa dalla luce nell’acqua durante il tempo $\Delta t$ è $\frac{c\Delta t}{n}$, dove $n=1,33$ indica la Indice di refraczione dell’acqua. From this, the water depth $d$ follows by projection $d=\frac{c\Delta t}{2n\cos \beta }=\frac{c\Delta t}{2n\cos \left(\arcsin \left(\frac{1}{n}\sin \alpha \right)\right)}\approx 3,1\text{ m}.$ (1.2) Risposta corretta: B Remark: Answer option A results if the refraction at the water surface is not “Tatto è il caso di un’opzione C senza il lungo percorso ottico in acqua”. e risposta opzione D se il fattore 2 per l’outward and return path del raggio di luce non è considerato. Classificazione - Mapping costiero (problema di scelta multipla) Punti Riconoscendo che la differenza di tempo del massimo è decisiva e leggendo fuori il Valore dal grafico 1.0 Usando la legge della refrazione e determinando l’angolo di refrazione (1.1) 1.0 Considerando il lungo percorso ottico 0.5 Considerando il fattore 2 per il percorso della luce 0.5 Setting up an expression for the water depth with projection (1.2) 1.0 Stating the correct solution 1.0 5.0

Topic: Geometric Optics Metodi: Snell’s Law, Ray Tracing, Experimental Data Analysis Competenze: Diagrammatic Reasoning, Physical Reasoning, Experimental Data Analysis Objects: — Fonte: Testo (PDF) — p.2

Problem 1 Coastal mapping (multiple choice problem) (5.0 p.p.) (Problem group of the PhysicsOlympic - Stefan Petersen) Using lasers mounted on aircraft, coastal waters can be mapped. To do this, short laser pulses are emitted into The water is reflected diffusely partly at the water surface and partly at the seabed. As a result, part of the emitted light returns to the aircraft, where it is detected and evaluated. Consider an aircraft flying at constant altitude along A coastal strip. The following graph shows the signal strength of the reflected signal of a laser pulse as a function of time. The time origin is chosen so that the entire reflected signal is shown. Assume that the laser pulse strikes the water surface at an angle of incidence (measured from the normal) of $40^{\circ }$ And that the refractive index of the water is 1.33. Fig. 1. Illustration of mapping with a laser (image source GEUS) Time in ns Signal strength in arbitrary units Fig. 2. Signal strength as a function of time (the point $t=0$ is arbitrary). How deep is the water at the location investigated? A 2.7 m B 3.1 m C 4.1 m D 6.1 m The solution Calculations and explanations The time difference $\Delta t\approx 60$ ns $-29$ ns = 31 ns between the Two maxima in the signal strength curve corresponds to the time The light ray needs to travel in the water from the water surface to the seabed and back. The angle $\beta$ of the light ray in the water (measured from the normal) is, according to Snell’s law of refraction, given by (cf. Fig. 3) $\beta =\arcsin \left(\frac{1}{n}\sin \alpha \right)=28,9^{\circ },$ (1.1) where $\alpha =40,0^{\circ }$ is the angle of incidence of the light ray. $d$ $\alpha$ $\beta$ Water Bottom Fig. 3. Sketch of the ray path. The distance travelled by the light in the water during the time $\Delta t$ is $\frac{c\Delta t}{n}$, where $n=1,33$ denotes the refractive index of water. From this, the water depth $d$ follows by projection $d=\frac{c\Delta t}{2n\cos \beta }=\frac{c\Delta t}{2n\cos \left(\arcsin \left(\frac{1}{n}\sin \alpha \right)\right)}\approx 3,1\text{ m}.$ (1.2) Correct answer: B Note: Answer option A results if the refraction at the water surface is not taken into account, answer option C without the lengthening of the optical path in the water, and answer option D if the factor 2 for the outward and return path of the light ray is not considered. The Commission shall adopt implementing acts in accordance with Article 21 of Regulation (EC) No 1272/2009. Points Recognizing that the time difference of the maxima is decisive and reading off the value from the graph 1.0 Using the law of refraction and determining the refraction angle (1.1) 1.0 Taking into account the lengthening of the optical path 0.5 Taking into account the factor 2 for the light path 0.5 Setting up an expression for the water depth with projection (1.2) 1.0 Stating the correct solution 1.0 5.0

Topic: Geometric Optics Metodi: Snell’s Law, Ray Tracing, Experimental Data Analysis Competenze: Diagrammatic Reasoning, Physical Reasoning, Experimental Data Analysis Objects: — Fonte: Testo (PDF) — p.2

Problem 2 Atwood machine (multiple-choice problem) (5.0 pts.) (Problem group of the PhysicsOlympiad - Titus Bornträger) Professor Atwood has just returned from a conference and is already tinkering with his next experiment: he lays a string over a pulley and attaches to each end of the string a body of mass $m$. Assume that the string and the pulley are massless and that the pulley can rotate freely about its central axis. He now holds the pulley’s axis fixed and measures the force $F$ with which he must pull it upward in order to hold the pulley with the bodies in its position. Now, as shown in the figure, he hangs an additional body of mass $4m$ on the right mass piece. He again holds the axis fixed. What is the force $F^{\prime }$ with which, in the situation shown, he must pull on the axis in order to hold it in its position? A $\frac{5}{3}F$ B $2F$ C $\frac{5}{2}F$ D $3F$ String Pulley $F^{\prime }$ $m$ $m$ $4m$ Solution Calculations and explanations In the first case the weights of the bodies on both sides of the pulley are identical. The weight of the body on one side is transmitted by the string tension to the body on the other side of the pulley, so that no resultant force acts on either body and they are therefore not accelerated. The force $F$ is therefore given as the sum of the weights of the two bodies, that is by $F=2mg.$ (2.1) In the second experiment sketched, a resultant force acts on the bodies, since the total mass on the two sides of the pulley is different. The total accelerating force on the bodies corresponds to the difference of the weights of the bodies on the two sides, that is $4mg$. All bodies are accelerated, however, so that the accelerated mass corresponds to the sum of the masses of all bodies, that is $6m$. Thus the magnitude of the acceleration is given by $a=\frac{4mg}{6m}=\frac{2}{3}g.$ (2.2) The force $F^{\prime }$ with which the pulley must be held at the axis can be determined by finding the tension in the string on the left or right side of the pulley and taking it times two${}^a$. Thus $F^{\prime }=2\cdot m(g+a)=\frac{10}{3}mg=\frac{5}{3}F$ or $F^{\prime }=2\cdot 5m(g-a)=\frac{10}{3}mg=\frac{5}{3}F.$ (2.3) Correct answer: A Remark: The solution can also be obtained by other means. For example, one can consider the time rate of change $\dot{p}$ of the total momentum of the three bodies, which is produced by the sum of the weights and the force $F^{\prime }$ on the pulley’s axis. One must pay attention here to the orientation of the forces and accelerations. It holds that $F^{\prime }-(2+4)mg=\dot{p}=ma-(1+4)ma.$ (2.4) With the acceleration (2.2) this likewise yields $F^{\prime }=m(6g-4a)=\left(6-\frac{8}{3}\right)mg=\frac{10}{3}mg=\frac{5}{3}F.$ (2.5) ${}^a$The factor 2 arises because the tension in the string must be identical on both sides of the pulley; otherwise the string would tear apart or no longer be taut. Grading - Atwood machine (multiple-choice problem) Points Determining the force $F$ in the first experiment (2.1) 0.5 Recognising that the accelerating mass and the accelerated mass are different 1.0 Determining the acceleration (2.2) 1.0 Determining the tension in the string ((2.3) without the factor 2) 1.0 Taking into account the factor 2 0.5 Stating the correct solution 1.0 5.0

Topic: Newtonian Mechanics Metodi: Free-Body Diagram, Kinematic Equations, Conservation of Momentum Competenze: Physical Reasoning, Mathematical Modeling Objects: String, Pulley Fonte: Testo (PDF) — p.4

Il problema 2 della macchina Atwood (problema di scelta multipla) (5,0 p. d.) (Problema group of the PhysicsOlympiad - Titus Bornträger) Il professor Atwood è appena tornato da una conferenza e è già Tincando con il suo prossimo esperimento: and attaches to each end of the string a body of mass $m$. Supponiamo che la corda e la polla siano senza massa e che la polla può girare liberamente intorno al suo asse centrale. # Ora tiene l’asse del pallino # fixed and measures the force $F$ with which he must pull it upward in order Per tenere il carrozzino con i corpi in posizione. Now, as shown in the figure, he hangs an additional body of mass $4m$ sul giusto pezzo di massa. # Ha ancora il suo asse #

• Non è stato fatto. What is the force $F^{\prime }$ with which, in the situation shown, he must pull on the
• Axis per tenerlo in posizione? A $\frac{5}{3}F$ B $2F$ C $\frac{5}{2}F$ D $3F$ String Pulley $F^{\prime }$ $m$ $m$ $4m$ Soluzione Calcoli e spiegazioni Nel primo caso, i pesi dei corpi su entrambi i lati della polla sono identici. Il peso del corpo su un lato è trasmesso dalla tensione della corda al corpo su l’altro lato del pollice, in modo che non si agisca forza risultante su entrambi i corpi e quindi non sono accelerati. La forza $F$ è quindi data come la somma dei pesi dei due corpi, cioè $F=2mg.$ (2.1) Nel secondo esperimento, una forza risultante agisce sui corpi, poiché la massa totale sui due lati del pollice è diversa. La forza totale di accelerazione on the bodies corrisponde alla differenza dei pesi dei corpi sui due lati, cioè $4mg$. Tutti i corpi sono accelerati, tuttavia, in modo che la massa accelerata corrisponde alla somma delle masse di tutti corpi, cioè $6m$. Quindi la grandezza dell’accelerazione è data da $a=\frac{4mg}{6m}=\frac{2}{3}g.$ (2.2) The force $F^{\prime }$ with which the pulley must be held at the axis can be determined by trovare la tensione nella corda sul lato sinistro o destro del pallone e prenderla times two${}^a$. Così $F^{\prime }=2\cdot m(g+a)=\frac{10}{3}mg=\frac{5}{3}F$ or $F^{\prime }=2\cdot 5m(g-a)=\frac{10}{3}mg=\frac{5}{3}F.$ (2.3) Risposta corretta: A Nota: La soluzione può essere ottenuta anche con altri mezzi. Per esempio, si può considerare il tempo rate of change $\dot{p}$ of the total momentum of the three bodies, which is produced by the sum of the weights and the force $F^{\prime }$ on the pulley’s axis. Uno deve prestare attenzione qui L’orientamento delle forze e delle accelerazioni. Tieni che $F^{\prime }-(2+4)mg=\dot{p}=ma-(1+4)ma.$ (2.4) Con l’accelerazione (2.2) questo rende $F^{\prime }=m(6g-4a)=\left(6-\frac{8}{3}\right)mg=\frac{10}{3}mg=\frac{5}{3}F.$ (2.5) ${}^a$Il fattore 2 si verifica perché la tensione nella corda deve essere identica su entrambi i lati del pulley; altrimenti La corda sarebbe strappata o non sarebbe più viva. Grading - Atwood machine (problema di scelta multipla) Punti Determinare la forza $F$ nel primo esperimento (2.1) 0.5 Riconoscendo che la massa accelerante e la massa accelerata sono diverse 1.0 Determinare l’accelerazione (2.2) 1.0 Determinare la tensione nella stringa ((2.3) without the factor 2) 1.0 Considerando il fattore 2 0.5 Stating the correct solution 1.0 5.0

Topic: Newtonian Mechanics Metodi: Free-Body Diagram, Kinematic Equations, Conservation of Momentum Competenze: Physical Reasoning, Mathematical Modeling Objects: String, Pulley Fonte: Testo (PDF) — p.4

The following is the list of the following problems: (5.0 p.p.) (Problem group of the PhysicsOlympic - Titus Bornträger) Professor Atwood has just returned from a conference and is already tinkering with his next experiment: he lays a string over a pulley and attaches to each end of the string a body of mass $m$. Assume that the string and the pulley are massless and that the pulley can rotate freely around its central axis. He now holds the pulley’s axis fixed and measures the force $F$ with which he must pull it upward in order To hold the pulley with the bodies in its position. Now, as shown in the figure, he hangs an additional body of mass $4m$ on the right mass piece. He ‘s holding the axis again . I’m not going to lie. What is the force $F^{\prime }$ with which, in the situation shown, he must pull on the Axis in order to hold it in its position? A $\frac{5}{3}F$ B $2F$ C $\frac{5}{2}F$ D $3F$ String Pulley $F^{\prime }$ $m$ $m$ $4m$ The solution Calculations and explanations In the first case the weights of the bodies on both sides of the pulley are identical. The weight of the body on one side is transmitted by the string tension to the body on The other side of the pulley, so that no resultant force acts on either body And they’re not accelerated. The force $F$ is therefore given as the sum of the weights of the two bodies, that is by $F=2mg.$ (2.1) In the second experiment sketched, a resultant force acts on the bodies, since the total mass on the two sides of the pulley is different. The total accelerating force on the bodies corresponds to the difference of the weights of the bodies on the two sides, that is $4mg$. All bodies are accelerated, however, so that the accelerated mass corresponds to the sum of the masses of all bodies, that is $6m$. Thus the magnitude of the acceleration is given by $a=\frac{4mg}{6m}=\frac{2}{3}g.$ (2.2) The force $F^{\prime }$ with which the pulley must be held at the axis can be determined by finding the tension in the string on the left or right side of the pulley and taking it times two${}^a$. Thus $F^{\prime }=2\cdot m(g+a)=\frac{10}{3}mg=\frac{5}{3}F$ or $F^{\prime }=2\cdot 5m(g-a)=\frac{10}{3}mg=\frac{5}{3}F.$ (2.3) Correct answer: A Note: The solution can also be obtained by other means. For example, one can consider the time rate of change $\dot{p}$ of the total momentum of the three bodies, which is produced by the sum of the weights and the force $F^{\prime }$ on the pulley’s axis. One must pay attention here The first is the orientation of forces and accelerations. It holds that $F^{\prime }-(2+4)mg=\dot{p}=ma-(1+4)ma.$ (2.4) With the acceleration (2.2) this also yields $F^{\prime }=m(6g-4a)=\left(6-\frac{8}{3}\right)mg=\frac{10}{3}mg=\frac{5}{3}F.$ (2.5) ${}^a$The factor 2 arises because the tension in the string must be identical on both sides of the pulley; otherwise The string would tear apart or no longer be taut. Grading - Atwood machine (multiple choice problem) Points Determining the force $F$ in the first experiment (2.1) 0.5 Recognizing that the accelerating mass and the accelerated mass are different 1.0 Determining the acceleration (2.2) 1.0 Determining the tension in the string ((2.3) without the factor 2) 1.0 Taking into account the factor 2 0.5 Stating the correct solution 1.0 5.0

Topic: Newtonian Mechanics Metodi: Free-Body Diagram, Kinematic Equations, Conservation of Momentum Competenze: Physical Reasoning, Mathematical Modeling Objects: String, Pulley Fonte: Testo (PDF) — p.4

Problem 3 Diving cup (multiple-choice problem) (5.0 pts.) (Problem group of the PhysicsOlympiad - Eugen Dizer & Arne Wolf) A cup of mass $m=200$ g with a filling volume $V_0=250$ mL and a height $H$ is turned upside down and submerged in a large water tank at constant water temperature. As a result, a layer of air is trapped and pushes the cup upward. The density of water is $\rho =1,0\cdot 10^3{\text{ kg/m}}^3$. Assume that the cup has a cylindrical cross section and that the thickness of the wall and of the cup bottom can be neglected. Assume furthermore that the cup always remains upside down and does not tip to the side. At which depths can the cup float in equilibrium? Here “depth” means the difference in the heights of the water levels in the cup and in the tank. A $\frac{1}{5}H$ and 2.6 m B $\frac{1}{5}H$ and 12.9 m C $\frac{4}{5}H$ and 2.6 m D $\frac{4}{5}H$ and 12.9 m Solution Calculations and explanations The cup floats in equilibrium when the mass of the displaced water equals the mass of the cup. The first equilibrium case occurs upon immersion. The cup floats when it is immersed exactly far enough that it displaces $200$ g $=\rho \cdot 200$ mL $=\frac{4}{5}\rho V_0$ of water, that is at the depth $h=\frac{4}{5}H.$ (3.1) Here it is assumed that the water does not enter the cup for small immersion depths. To push the cup deeper into the water, work must be done against the buoyant force. The layer of air in the cup is thereby compressed by the water pressure. The second equilibrium case occurs at greater depth, when exactly 200 mL of air remain in the cup. Using the hydrostatic pressure $p=\rho gh$ at depth $h$ and the ideal gas law $pV=nRT$, it follows that $p_0{V}_0=(p_0+\rho gh)V=\frac{4}{5}(p_0+\rho gh)V_0.$ (3.2) Here $p_0\approx 1,01\cdot 10^5{\text{ N/m}}^2$ denotes the atmospheric pressure. Rearranging gives $h=\frac{p_0}{4\rho g}\approx 2,6\text{ m}$ (3.3) Correct answer: C Remark: If one uses $\rho gh$ instead of $\rho gh+p_0$ for the pressure, one obtains instead $h\approx 12,9$ m. Grading - Diving cup (multiple-choice problem) Points Considering the buoyant force 1.0 Deriving the first equilibrium case 1.0 Using the hydrostatic pressure $p=\rho gh$ 0.5 Using the gas equation $pV=$ const. 0.5 Deriving the second equilibrium case 1.0 Correct answer 1.0 5.0

Topic: Fluid Mechanics, Thermodynamics Metodi: Hydrostatic Equilibrium, Ideal Gas Law, Conservation Laws Competenze: Physical Reasoning, Mathematical Modeling Objects: Container, Gas Fonte: Testo (PDF) — p.6

Problema 3 Diving cup (problema di scelta multipla) (5,0 p. d.) (Problem group of the PhysicsOlympiad - Eugen Dizer & Arne Wolf) Una coppa di massa $m=200$ g con un volume di riempimento $V_0=250$ mL e una altezza $H$ è si è trasformata di testa in giù e è stata immersa in un grande serbatoio ad una temperatura di acqua costante. Come risultato, Un livello di aria è intrappolato e spinge la tazza verso l’alto. La densità dell’acqua è $\rho =1,0\cdot 10^3{\text{ kg/m}}^3$. Supponiamo che la coppa abbia una sezione incrociata cilindrica e che lo spessore di fronte al muro e al fondo della coppa può essere trascurato. Supponiamo inoltre che La coppa rimane sempre al contrario e non punta al lato. A quali profondità può fluttuare la coppa in equilibrio? Qui “depth” significa la differenza tra le altezze dei livelli di acqua nella tazza e nel serbatoio. A $\frac{1}{5}H$ e 2,6 m B $\frac{1}{5}H$ e 12,9 m C $\frac{4}{5}H$ e 2,6 m D $\frac{4}{5}H$ e 12,9 m Soluzione Calcoli e spiegazioni La coppa flotta in equilibrio quando la massa dell’acqua spostata è uguale alla massa della coppa. Il primo caso di equilibrio si verifica dopo l’immersione. La coppa flotta quando è immersa esattamente lontano sufficiente a dislocare $200$ g $=\rho \cdot 200$ mL $=\frac{4}{5}\rho V_0$ di acqua, che è alla profondità $h=\frac{4}{5}H.$ (3.1) Qui si presume che l’acqua non entri nella coppa per piccole profondità di immersione. Per spingere la coppa più in profondità nell’acqua, il lavoro deve essere fatto contro la forza buoyant. Il livello di aria presente nella tazza è quindi compresso dalla pressione dell’acqua. Il secondo equilibrio si verifica a maggiore profondità, quando esattamente 200 ml di aria rimangono nella coppa. Usando the hydrostatic pressure $p=\rho gh$ at depth $h$ and the ideal gas law $pV=nRT$, it follows that $p_0{V}_0=(p_0+\rho gh)V=\frac{4}{5}(p_0+\rho gh)V_0.$ (3.2) Qui $p_0\approx 1,01\cdot 10^5{\text{ N/m}}^2$ indica la pressione atmosferica. Riorganizzazione dà $h=\frac{p_0}{4\rho g}\approx 2,6\text{ m}$ (3.3) Corretta risposta: C Nota: se uno utilizza $\rho gh$ invece di $\rho gh+p_0$ per la pressione, uno ottiene invece $h\approx 12,9$ m. Classificazione - Coppa di immersione (problema di scelta multipla) Punti Considerando la forza buoyant 1.0 Deriving the first equilibrium case 1.0 Using the hydrostatic pressure $p=\rho gh$ 0.5 Usando l’equazione gas $pV=$ const. 0.5 Deriving the second equilibrium case 1.0 Corretta risposta 1.0 5.0

Topic: Fluid Mechanics, Thermodynamics Metodi: Hydrostatic Equilibrium, Ideal Gas Law, Conservation Laws Competenze: Physical Reasoning, Mathematical Modeling Objects: Container, Gas Fonte: Testo (PDF) — p.6

Problem 3 Diving cup (multiple choice problem) (5.0 p.p.) (Problem group of the PhysicsOlympiad - Eugen Dizer & Arne Wolf) A cup of mass $m=200$ g with a filling volume $V_0=250$ mL and a height $H$ is turned upside down and submerged in a large water tank at constant water temperature. As a result, A layer of air is trapped and pushes the cup upward. The density of water is $\rho =1,0\cdot 10^3{\text{ kg/m}}^3$. Assume that the cup has a cylindrical cross section and that the thickness of the wall and of the cup bottom can be neglected. Furthermore, assume that The cup always stays upside down and doesn’t tip to the side. At what depths can the cup float in equilibrium? Here “depth” means the difference in the heights of the water levels in the cup and in the tank. A $\frac{1}{5}H$ and 2.6 m B $\frac{1}{5}H$ and 12.9 m C $\frac{4}{5}H$ and 2.6 m D $\frac{4}{5}H$ and 12.9 m The solution Calculations and explanations The cup floats in equilibrium when the mass of the displaced water equals the mass of the cup. The first equilibrium case occurs upon immersion. The cup floats when it is immersed exactly far enough that it displaces $200$ g $=\rho \cdot 200$ mL $=\frac{4}{5}\rho V_0$ of water, that is at the depth $h=\frac{4}{5}H.$ (3.1) Here it is assumed that the water does not enter the cup for small immersion depths. To push the cup deeper into the water, work must be done against the buoyant force. The layer of air in the cup is thus compressed by the water pressure. The second equilibrium case occurs at greater depth, when exactly 200 ml of air remain in the cup. Using The hydrostatic pressure $p=\rho gh$ at depth $h$ and the ideal gas law $pV=nRT$, it follows that $p_0{V}_0=(p_0+\rho gh)V=\frac{4}{5}(p_0+\rho gh)V_0.$ (3.2) Here $p_0\approx 1,01\cdot 10^5{\text{ N/m}}^2$ denotes the atmospheric pressure. Rearranging gives $h=\frac{p_0}{4\rho g}\approx 2,6\text{ m}$ (3.3) Correct answer: C Note: If one uses $\rho gh$ instead of $\rho gh+p_0$ for the pressure, one obtains instead $h\approx 12,9$ m. Grading - Diving cup (multiple choice problem) Points Considering the booyant force 1.0 Deriving the first equilibrium case 1.0 Using the hydrostatic pressure $p=\rho gh$ 0.5 Using the gas equation $pV=$ const. 0.5 Deriving the second equilibrium case 1.0 Correct answer 1.0 5.0

Topic: Fluid Mechanics, Thermodynamics Metodi: Hydrostatic Equilibrium, Ideal Gas Law, Conservation Laws Competenze: Physical Reasoning, Mathematical Modeling Objects: Container, Gas Fonte: Testo (PDF) — p.6

Problem 4 Flipped capacitor (multiple-choice problem) (5.0 pts.) (Problem group of the PhysicsOlympiad - Fabian Bühler) Two capacitors with capacitances $C_1=2,0$ nF and $C_2=8,0$ nF are, as shown in the following figure, connected in series to a voltage source with voltage $U_0=10,0$ V: $U_0$ $C_2$ $C_1$ The capacitors are then disconnected from the voltage source, and one of the capacitors is rotated by $180^{\circ }$. Now the capacitors are short-circuited: $C_2$ $C_1$ $C_2$ $C_1$ What voltage results across the flipped capacitor? A 0 V B 1.6 V C 3.2 V D 5.0 V Solution Calculations and explanations The total capacitance of the capacitors connected in series is $C_{\text{Reihe}}=\frac{1}{\frac{1}{C_1}+\frac{1}{C_2}}=1,6\text{ nF}.$ (4.1) For the charge of the capacitors it holds that $Q_1=Q_2=C_{\text{Reihe}}\cdot U_0=16\text{ nC}.$ (4.2) The short-circuited capacitors are equivalent to a parallel connection of the two capacitors with capacitance $C_{\text{Parallel}}=C_1+C_2=10\text{ nF}.$ (4.3) Because one of the capacitors was rotated by $180^{\circ }$, the total charge of the capacitors equals the sum of the individual charges $Q_{\text{Parallel}}=Q_1+Q_2=32\text{ nC}.$ (4.4) The voltage across both capacitors is the same, namely $U=\frac{Q_{\text{Parallel}}}{C_{\text{Parallel}}}=\frac{2C_{\text{Reihe}}{U}_0}{C_{\text{Parallel}}}=2U_0\frac{C_1{C}_2}{(C_1+C_2{)}^2}=\frac{32}{10}\text{ V}=3,2\text{ V}.$ (4.5) Correct answer: C Grading - Flipped capacitor (multiple-choice problem) Points Calculating the total capacitance of the capacitors in series 1.0 Calculating the original charge of the capacitors 1.0 Recognising that the flipping is equivalent to a parallel connection 1.0 Calculating the total capacitance of the capacitors in parallel 1.0 Stating the correct solution 1.0 5.0

Topic: Electrostatics, Circuits Metodi: Equivalent Circuit Reduction, Electric Potential Method, Conservation Laws Competenze: Physical Reasoning, Mathematical Modeling Objects: Capacitor, Battery Fonte: Testo (PDF) — p.7

Il problema 4 è il problema del condensatore a pieghe (problema di scelta multipla) (5,0 p. d.) (Problema group of the PhysicsOlympiad - Fabian Bühler) Due condensatori con capacità $C_1=2,0$ nF e $C_2=8,0$ nF sono, come mostrato nel seguente figure, connected in series to a voltage source with voltage $U_0=10,0$ V: $U_0$ $C_2$ $C_1$ I condensatori sono quindi disconnessi dalla fonte di voltage, e uno dei condensatori è rotato da $180^{\circ }$. Ora i condensatori sono a corto circuito: $C_2$ $C_1$ $C_2$ $C_1$ Che voltage si ottiene attraverso il condensatore inverso? A 0 V B 1.6 V C 3.2 V D 5.0 V Soluzione Calcoli e spiegazioni La capacità totale dei condensatori collegati in serie è $C_{\text{Reihe}}=\frac{1}{\frac{1}{C_1}+\frac{1}{C_2}}=1,6\text{ nF}.$ (4.1) Per la carica dei condensatori che contiene $Q_1=Q_2=C_{\text{Reihe}}\cdot U_0=16\text{ nC}.$ (4.2) I condensatori a corto circuito sono equivalenti a una connessione parallela dei due condensatori con capacità $C_{\text{Parallel}}=C_1+C_2=10\text{ nF}.$ (4.3) Poiché uno dei condensatori è stato rotato da $180^{\circ }$, la carica totale dei condensatori pari alla somma dei singoli carichi $Q_{\text{Parallel}}=Q_1+Q_2=32\text{ nC}.$ (4.4) La tensione attraverso entrambi i condensatori è la stessa, cioè $U=\frac{Q_{\text{Parallel}}}{C_{\text{Parallel}}}=\frac{2C_{\text{Reihe}}{U}_0}{C_{\text{Parallel}}}=2U_0\frac{C_1{C}_2}{(C_1+C_2{)}^2}=\frac{32}{10}\text{ V}=3,2\text{ V}.$ (4.5) Corretta risposta: C Capacitatore invertito (problema di scelta multipla) Punti Calcolare la capacità totale dei condensatori in serie 1.0 Calcolo della carica originale dei condensatori 1.0 Riconoscendo che il flipping è equivalente a una connessione parallela 1.0 Calcolare in parallelo la capacità totale dei condensatori 1.0 Stating the correct solution 1.0 5.0

Topic: Electrostatics, Circuits Metodi: Equivalent Circuit Reduction, Electric Potential Method, Conservation Laws Competenze: Physical Reasoning, Mathematical Modeling Objects: Capacitor, Battery Fonte: Testo (PDF) — p.7

The problem is that the power supply is not a single unit. (5.0 p.p.) (Problem group of the PhysicsOlympiad - Fabian Bühler) Two capacitors with capacitances $C_1=2,0$ nF and $C_2=8,0$ nF are, as shown in the following Figure, connected in series to a voltage source with voltage $U_0=10,0$ V: $U_0$ $C_2$ $C_1$ The capacitors are then disconnected from the voltage source, and one of the capacitors is rotated by $180^{\circ }$. Now the capacitors are short-circuited: $C_2$ $C_1$ $C_2$ $C_1$ What voltage results across the flipped capacitor? A 0 V B 1.6 V C 3.2 V D 5.0 V The solution Calculations and explanations The total capacitance of the capacitors connected in series is $C_{\text{Reihe}}=\frac{1}{\frac{1}{C_1}+\frac{1}{C_2}}=1,6\text{ nF}.$ (4.1) For the charge of the capacitors it holds that $Q_1=Q_2=C_{\text{Reihe}}\cdot U_0=16\text{ nC}.$ (4.2) The short-circuited capacitors are equivalent to a parallel connection of the two capacitors with capacitance $C_{\text{Parallel}}=C_1+C_2=10\text{ nF}.$ (4.3) Because one of the capacitors was rotated by $180^{\circ }$, the total charge of the capacitors equals the sum of the individual charges $Q_{\text{Parallel}}=Q_1+Q_2=32\text{ nC}.$ (4.4) The voltage across both capacitors is the same, namely $U=\frac{Q_{\text{Parallel}}}{C_{\text{Parallel}}}=\frac{2C_{\text{Reihe}}{U}_0}{C_{\text{Parallel}}}=2U_0\frac{C_1{C}_2}{(C_1+C_2{)}^2}=\frac{32}{10}\text{ V}=3,2\text{ V}.$ (4.5) Correct answer: C Grading - Flipped capacitor (multiple choice problem) Points Calculating the total capacitance of the capacitors in series 1.0 Calculating the original charge of the capacitors 1.0 Recognizing that the flipping is equivalent to a parallel connection 1.0 Calculating the total capacitance of the capacitors in parallel 1.0 Stating the correct solution 1.0 5.0

Topic: Electrostatics, Circuits Metodi: Equivalent Circuit Reduction, Electric Potential Method, Conservation Laws Competenze: Physical Reasoning, Mathematical Modeling Objects: Capacitor, Battery Fonte: Testo (PDF) — p.7

Problem 5 Lissajous pendulum (multiple-choice problem) (5.0 pts.) (Problem group of the PhysicsOlympiad - Johannes Rothe) A string pendulum is, as sketched alongside, suspended from two strings that converge in a V shape. The height of the V is equal to the length $l$ of the string pendulum below it. The size of the pendulum bob and the masses of the strings are negligible. The pendulum bob is now slightly displaced from its rest position and released. If one records the motion of the pendulum bob in the horizontal plane, beautiful patterns result. Which of the following patterns could have been produced with the pendulum above? $l$ $l$ Fig. 4. Sketch of the pendulum setup. A B C D Fig. 5. Candidates for traces of the pendulum bob in the horizontal plane. Solution Calculations and explanations The V is rigid in the plane of the drawing but can be deflected perpendicular to it. Therefore, in these two planes pendulums of different pendulum lengths result - $l$ in the plane of the drawing and $2l$ perpendicular to it. The harmonic oscillations resulting at small deflections therefore have different periods. For a string pendulum at small deflections, $T$ is proportional to the square root of the length of the string, and thus the periods of the pendulum are in the ratio $1:\sqrt{2}$. Since this ratio is irrational, the (sections of the) trajectories seen in the patterns are never closed. The answer options A and D are thereby ruled out, since they show closed trajectories. From the turning points in the patterns it can also be concluded that the patterns shown are oriented such that the plane of the drawing in the sketch in Fig. 4 runs either vertically (in B and D) or horizontally (in A and C). To distinguish the remaining options B and C, one can count half oscillation periods. For this one chooses a starting point with maximal deflection in one of the two directions, that is with a horizontal or vertical tangent in the respective figure. If one now counts three of the faster half-oscillations, that is in the figure vertical oscillations in B or horizontal ones in C, then in case B one sees slightly more than two half-oscillations in the other direction, but in case C slightly fewer. Denote by $T_1$ the period of the pendulum motion in the plane of the drawing of the sketch and by $T_2>T_1$ that of the pendulum motion perpendicular to it. Then, with the observation above, it holds for the ratio of the respective periods ${\frac{T_2}{T_1}|}_B<\frac{3}{2}<{\frac{T_2}{T_1}|}_C.$ (5.1) Since $\frac{T_2}{T_1}=\sqrt{2}\approx 1,41<1,5$, only option B remains as a possible answer. Correct answer: B Remark: Answer option A results for the ratio $T_1:T_2=1:2$, answer option C for the ratio $T_1:T_2=1:\sqrt{3}$, and answer option D for the ratio $T_1:T_2=2:3$. The orientation of the figure in the horizontal plane is not identical across the patterns. Grading - Lissajous pendulum (multiple-choice problem) Points Recognising harmonic oscillations of different frequencies 1.0 Using $T\sim \sqrt{L}$ and stating the correct period ratio $1:\sqrt{2}$ 1.0 Ruling out the closed trajectories in A and D 1.0 Counting oscillations to distinguish B and C 1.0 Stating the correct solution 1.0 5.0

Topic: Oscillations & Waves, Newtonian Mechanics Metodi: Simple Harmonic Motion Analysis, Small-Angle Approximation, Superposition Principle Competenze: Physical Reasoning, Mathematical Modeling, Diagrammatic Reasoning Objects: Pendulum, String Fonte: Testo (PDF) — p.9

Il problema 5 del pendolo di Lissajous (problema di scelta multipla) (5,0 p. d.) (Problema group of the PhysicsOlympiad - Johannes Rothe) Un pendolo a corda è, come disegnato accanto, sospeso da due stringhe che convergono in forma di V. La altezza di V è pari alla lunghezza $l$ del pendolo della stringa sotto di esso. Bob e le masse delle corde Le condizioni di lavoro sono trascurabili. Il pendolo Bob è ora leggermente spostato dalla sua posizione di riposo e rilasciato. Se si registra il movimento del pendolo bob in piano orizzontale,

• E’ un risultato di belle schemi. Quale dei seguenti modelli potrebbe essere stato prodotto con il pendolo sopra? $l$ $l$ Fig. 4. Sketch del pendolo di impostazione. A B C D Fig. 5. Candidati per tracce del pendolo bob in piano orizzontale. Soluzione Calcoli e spiegazioni Il V è rigido nel piano del disegno ma può essere deflezionato perpendicolare a esso. Pertanto, in questi due piani Pendulums of different pendulum lengths result - $l$ in the plane of the drawing and $2l$ perpendicolare a essa. Le oscillazioni armoniche risultanti da piccole deflezioni hanno quindi periodi diversi. Per un pendolo a string a small deflections, $T$ è proporzionale al la radice quadrata della lunghezza della corda, e quindi i periodi del pendolo sono in rapporto $1:\sqrt{2}$. Poiché questo rapporto è irrazionale, le (sezioni del) traiettorie viste nei modelli non sono mai chiuse. Le opzioni A e D sono quindi escluse, poiché
• Show closed trajectories. Dal punto di svolta dei modelli si può anche concludere che i modelli mostrati sono orientati così che il piano del disegno nello schizzo in Fig. 4 run verticalmente (in B e D) o orizzontalmente (in A e C). Per distinguere le rimanenti opzioni B e C, si possono contare i periodi di oscillazione a metà. Per questo uno sceglie un punto di partenza con massima deflessione in una delle due direzioni, cioè con un tangente orizzontale o verticale nella rispettiva figura. Se uno ora conta tre delle oscillazioni a metà più veloci, cioè in B le oscillazioni verticali o orizzontali in C, allora nel caso B uno vede leggermente più di due oscillazioni a metà l’altra direzione, ma in caso C leggermente meno. Denote by $T_1$ the period of the pendulum motion in the plane of the drawing of the sketch and by $T_2>T_1$ che del movimento del pendolo perpendicolare a esso. Poi, con l’osservazione sopra, si ritiene che ratio dei rispettivi periodi ${\frac{T_2}{T_1}|}_B<\frac{3}{2}<{\frac{T_2}{T_1}|}_C.$ (5.1) Dal momento che $\frac{T_2}{T_1}=\sqrt{2}\approx 1,41<1,5$, only option B remains as a possible answer. Risposta corretta: B Remark: Answer option A results for the ratio $T_1:T_2=1:2$, answer option C per il rapporto $T_1:T_2=1:\sqrt{3}$, e risposta opzione D per il rapporto $T_1:T_2=2:3$. L’orientamento della figura nel piano orizzontale non è identico attraverso i modelli. Grading - Pendale di Lissajous (problema di scelta multipla) Punti Riconoscere oscillazioni armoniche di diverse frequenze 1.0 Usando $T\sim \sqrt{L}$ e dichiarando il corretto rapporto di periodo $1:\sqrt{2}$ 1.0 Ruling out the closed trajectories in A and D 1.0 Counting oscillations to distinguish B and C 1.0 Stating the correct solution 1.0 5.0

Topic: Oscillations & Waves, Newtonian Mechanics Metodi: Simple Harmonic Motion Analysis, Small-Angle Approximation, Superposition Principle Competenze: Physical Reasoning, Mathematical Modeling, Diagrammatic Reasoning Objects: Pendulum, String Fonte: Testo (PDF) — p.9

The following is the list of the problems to be solved: (5.0 p.p.) (Problem group of the PhysicsOlympiad by Johannes Rothe) A string pendulum is, as sketched alongside, suspended from two Strings that converge in a V shape. The height of the V is equal to the length $l$ of the string pendulum below it. The size of the pendulum bob and the masses of the strings The Commission’s proposal is not yet available. The pendulum bob is now slightly displaced from its rest position and released. If one records the motion of the pendulum bob in the horizontal plane, The result is beautiful patterns. Which of the following patterns could have been produced with the pendulum above? $l$ $l$ Fig. 4. Sketch of the pendulum setup. A B C D Fig. 5. Candidates for traces of the pendulum bob in the horizontal plane. The solution Calculations and explanations The V is rigid in the plane of the drawing but can be deflected perpendicular to it. Therefore, in these two planes pendulums of different pendulum lengths result - $l$ in the plane of the drawing and $2l$ perpendicular to it. The harmonic oscillations resulting at small deflections therefore have different periods. For a string pendulum at small deflections, $T$ is proportional to the square root of the length of the string, and thus the periods of the pendulum are in the ratio $1:\sqrt{2}$. Since this ratio is irrational, the trajectories seen in the patterns are never closed. The answer options A and D are thus ruled out, since they are Show closed trajectories. From the turning points in the patterns it can also be concluded that the patterns shown are oriented So that’s the plane of the drawing in the sketch in Fig. 4 runs either vertically (in B and D) or horizontally (in A and C). To distinguish the remaining options B and C, one can count half oscillation periods. For this one chooses a starting point with maximum deflection in one of the two directions, that is with a horizontal or vertical tangent in the respective figure. If one now counts three of the faster half-oscillations, that is in the figure vertical oscillations in B or horizontal ones in C, then in case B one sees slightly more than two half-oscillations in The other direction, but in case C slightly less. Denote by $T_1$ the period of the pendulum motion in the plane of the drawing of the sketch and by $T_2>T_1$ that of the pendulum motion perpendicular to it. Then, with the observation above, it holds for the ratio of the respective periods ${\frac{T_2}{T_1}|}_B<\frac{3}{2}<{\frac{T_2}{T_1}|}_C.$ (5.1) Since $\frac{T_2}{T_1}=\sqrt{2}\approx 1,41<1,5$, only option B remains as a possible answer. Correct answer: B Note: Answer option A results for the ratio $T_1:T_2=1:2$, answer option C for the ratio $T_1:T_2=1:\sqrt{3}$, and answer option D for the ratio $T_1:T_2=2:3$. The orientation of the figure in the horizontal plane is not identical across the patterns. Grading - Lissajous pendulum (multiple choice problem) Points Recognising harmonic oscillations of different frequencies 1.0 Using $T\sim \sqrt{L}$ and stating the correct period ratio $1:\sqrt{2}$ 1.0 Ruling out the closed trajectories in A and D 1.0 Counting oscillations to distinguish B and C 1.0 Stating the correct solution 1.0 5.0

Topic: Oscillations & Waves, Newtonian Mechanics Metodi: Simple Harmonic Motion Analysis, Small-Angle Approximation, Superposition Principle Competenze: Physical Reasoning, Mathematical Modeling, Diagrammatic Reasoning Objects: Pendulum, String Fonte: Testo (PDF) — p.9

Problem 6 Freezing water (multiple-choice problem) (5.0 pts.) (Problem group of the PhysicsOlympiad - Stefan Petersen) With a heat pump of electrical power 50 W, 2.0 kg of water at a temperature of $0{}^{\circ }\text{C}$ is to be frozen in a thermally perfectly insulated vessel. The outside temperature is $25{}^{\circ }\text{C}$. The enthalpy of fusion of water is $334\text{ kJ/kg}$. What minimum time is needed for the freezing in any case? A about 6 min B about 11 min C about 15 min D about 20 min Solution Calculations and explanations In the freezing of the water, the heat $Q=334\text{ kJ/kg}\cdot 2,0\text{ kg}=6,7\cdot 10^5\text{ J}$ (6.1) is released. This heat must be removed by the heat pump. The heat pump operates between two reservoirs at the temperatures $T_k=0{}^{\circ }\text{C}=273$ K and $T_w=25{}^{\circ }\text{C}=298$ K. The efficiency $\epsilon$ of a refrigerator operating between these two reservoirs is thermodynamically limited. Denote by $W$ the work done by the heat pump, which at constant power equals the product of the electrical power $P$ and the time $t$ for which the pump runs. Then for the efficiency it holds that $\epsilon =\frac{Q}{W}=\frac{Q}{Q_{\text{ab}}-Q},$ (6.2) where $Q_{\text{ab}}$ denotes the heat given off to the surroundings at temperature $T_w$. If the refrigerator operates reversibly, that is theoretically optimally, the entropy is conserved and it holds for the heat absorbed and released that $\frac{Q}{T_k}=\frac{Q_{\text{ab}}}{T_w}.$ (6.3) Substituted into (6.2), this gives the maximum possible efficiency${}^a$ as ${\epsilon }_{\text{max}}=\frac{Q}{Q\left(\frac{T_w}{T_k}-1\right)}=\frac{T_k}{T_w-T_k}.$ (6.4) From this the minimum time $t_{\text{min}}$ for the freezing can be estimated as $t_{\text{min}}=\frac{Q}{P}\frac{1}{{\epsilon }_{\text{max}}}=\frac{Q}{P}\frac{T_w-T_k}{T_k}\approx \frac{6,7\cdot 10^5\text{ J}}{50\text{ W}}\frac{25}{273}\approx 1,2\cdot 10^3\text{ s}\approx 20\text{ min}.$ (6.5) Correct answer: D ${}^a$This can also be expressed through the Carnot efficiency ${\eta }_C=1-T_k/T_w$ at the two temperatures as ${\epsilon }_{\text{max}}=1/{\eta }_C-1$. Grading - Freezing water (multiple-choice problem) Points Determining the heat (6.1) 1.0 Recognising that the efficiency is limited 0.5 Using the conservation of entropy or the Carnot efficiency 0.5 Stating the maximum efficiency (6.4) 1.0 Using that work is the product of power and time 0.5 Deriving a formula for the time (6.5) 0.5 Correct answer 1.0 5.0

Topic: Thermodynamics Metodi: Thermodynamic Cycle Analysis, First Law of Thermodynamics, Physical Modeling Competenze: Mathematical Modeling, Physical Reasoning Objects: Heat Engine, Container Fonte: Testo (PDF) — p.11

Il problema 6 è quello dell’acqua congelata (problema di scelta multipla) (5,0 p. d.) (Problema group of the PhysicsOlympiad - Stefan Petersen) Con una pompa di calore di potenza elettrica 50 W, 2,0 kg di acqua a temperatura di $0{}^{\circ }\text{C}$ è da congelare in un recipiente termicamente perfettamente isolato. La temperatura esterna è $25{}^{\circ }\text{C}$. L’enthalpy of fusion of water è $334\text{ kJ/kg}$. Qual è il minimo di tempo necessario per il congelamento? A circa 6 minuti B circa 11 minuti C circa 15 minuti D circa 20 minuti Soluzione Calcoli e spiegazioni Nel gelo dell’acqua, il caldo $Q=334\text{ kJ/kg}\cdot 2,0\text{ kg}=6,7\cdot 10^5\text{ J}$ (6.1) è rilasciato. Questo calore deve essere rimosso dalla pompa di calore. La pompa di calore operato tra due serbatoi a temperature $T_k=0{}^{\circ }\text{C}=273$ K e $T_w=25{}^{\circ }\text{C}=298$ K. L’efficienza $\epsilon$ di un frigorifero che opera tra questi due serbatoi è Termodinamica limitata. Denote by $W$ the work done by the heat pump, which at La potenza costante è pari al prodotto della potenza elettrica $P$ e al tempo $t$ per il quale la potenza elettrica $P$ è

• La pompa corre. Quindi per l’efficienza che contiene $\epsilon =\frac{Q}{W}=\frac{Q}{Q_{\text{ab}}-Q},$ (6.2) dove $Q_{\text{ab}}$ denota il calore rilasciato agli ambienti circostanti a temperatura $T_w$. Se il il frigorifero opera in modo reversibile, cioè teoricamente ottimale, l’entropia è conservata e tiene per il calore assorbito e rilasciato che $\frac{Q}{T_k}=\frac{Q_{\text{ab}}}{T_w}.$ (6.3) Substituito in (6.2), this gives the maximum possible efficiency${}^a$ as ${\epsilon }_{\text{max}}=\frac{Q}{Q\left(\frac{T_w}{T_k}-1\right)}=\frac{T_k}{T_w-T_k}.$ (6.4) From this the minimum time $t_{\text{min}}$ for the freezing can be estimated as $t_{\text{min}}=\frac{Q}{P}\frac{1}{{\epsilon }_{\text{max}}}=\frac{Q}{P}\frac{T_w-T_k}{T_k}\approx \frac{6,7\cdot 10^5\text{ J}}{50\text{ W}}\frac{25}{273}\approx 1,2\cdot 10^3\text{ s}\approx 20\text{ min}.$ (6.5) Risposta corretta: D ${}^a$This can also be expressed through the Carnot efficiency ${\eta }_C=1-T_k/T_w$ at the two temperatures as ${\epsilon }_{\text{max}}=1/{\eta }_C-1$. Grading - Freezing water (problema di scelta multipla) Punti Determinare il calore (6.1) 1.0 Riconoscendo che l’efficienza è limitata 0.5 Usando la conservazione dell’entropia o l’efficienza di Carnot 0.5 Stating the maximum efficiency (6.4) 1.0 Usando quel lavoro è il prodotto del potere e del tempo 0.5 Deriving a formula for the time (6.5) 0.5 Corretta risposta 1.0 5.0

Topic: Thermodynamics Metodi: Thermodynamic Cycle Analysis, First Law of Thermodynamics, Physical Modeling Competenze: Mathematical Modeling, Physical Reasoning Objects: Heat Engine, Container Fonte: Testo (PDF) — p.11

The following is the list of the types of water freezing (multiple choice problem) (5.0 p.p.) (Problem group of the PhysicsOlympic - Stefan Petersen) With a heat pump of electrical power 50 W, 2.0 kg of water at a temperature of $0{}^{\circ }\text{C}$ is to be frozen in a thermally perfectly insulated vessel. The outside temperature is $25{}^{\circ }\text{C}$. The enthalpy of fusion of water is $334\text{ kJ/kg}$. What minimum time is needed for the freezing in any case? A about 6 min B about 11 min C about 15 minutes D about 20 minutes The solution Calculations and explanations In the freezing of the water, the heat $Q=334\text{ kJ/kg}\cdot 2,0\text{ kg}=6,7\cdot 10^5\text{ J}$ (6.1) is released. This heat must be removed by the heat pump. The heat pump operates between two reservoirs at the temperatures $T_k=0{}^{\circ }\text{C}=273$ K and $T_w=25{}^{\circ }\text{C}=298$ K. The efficiency $\epsilon$ of a refrigerator operating between these two reservoirs is The thermodynamically limited. Denote by $W$ the work done by the heat pump, which at The constant power equals the product of the electrical power $P$ and the time $t$ for which the pump runs. Then for the efficiency it holds that $\epsilon =\frac{Q}{W}=\frac{Q}{Q_{\text{ab}}-Q},$ (6.2) where $Q_{\text{ab}}$ denotes the heat given off to the surroundings at temperature $T_w$. If the refrigerator operates reversibly, that is theoretically optimally, the entropy is conserved and it holds for the heat absorbed and released that $\frac{Q}{T_k}=\frac{Q_{\text{ab}}}{T_w}.$ (6.3) Substituted into (6.2), this gives the maximum possible efficiency${}^a$ as ${\epsilon }_{\text{max}}=\frac{Q}{Q\left(\frac{T_w}{T_k}-1\right)}=\frac{T_k}{T_w-T_k}.$ (6.4) From this the minimum time $t_{\text{min}}$ for the freezing can be estimated as $t_{\text{min}}=\frac{Q}{P}\frac{1}{{\epsilon }_{\text{max}}}=\frac{Q}{P}\frac{T_w-T_k}{T_k}\approx \frac{6,7\cdot 10^5\text{ J}}{50\text{ W}}\frac{25}{273}\approx 1,2\cdot 10^3\text{ s}\approx 20\text{ min}.$ (6.5) Correct answer: D ${}^a$This can also be expressed through the Carnot efficiency ${\eta }_C=1-T_k/T_w$ at the two temperatures as ${\epsilon }_{\text{max}}=1/{\eta }_C-1$. Grading - freezing water (multiple choice problem) Points Determining the heat (6.1) 1.0 Recognizing that efficiency is limited 0.5 Using the conservation of entropy or the Carnot efficiency 0.5 Stating the maximum efficiency (6.4) 1.0 Using that work is the product of power and time 0.5 Deriving a formula for the time (6.5) 0.5 Correct answer 1.0 5.0

Topic: Thermodynamics Metodi: Thermodynamic Cycle Analysis, First Law of Thermodynamics, Physical Modeling Competenze: Mathematical Modeling, Physical Reasoning Objects: Heat Engine, Container Fonte: Testo (PDF) — p.11

Problem 7 Kaon decay (multiple-choice problem) (5 pts.) (Problem group of the PhysicsOlympiad - Thomas Hellerl) A kaon moving with velocity $0,80c$, that is 80 % of the speed of light, in the laboratory frame decays into two pions, which afterwards move along and against, respectively, the kaon’s original direction of motion. No further particles are produced in the decay. For the rest energies $E_{0,K}$ and $E_{0,\pi }$ of the kaon and the pion, respectively, the relation $\kappa =\frac{E_{0,K}}{E_{0,\pi }}=3,68.$ holds. At what velocities do the two pions move after the decay in the laboratory frame? A $-0,039c$ and $0,99c$ B $-0,12c$ and $0,98c$ C $-0,19c$ and $0,92c$ D $-0,21c$ and $0,89c$ Solution Calculations and explanations In the rest frame of the kaon the momentum is zero. After the decay the pions must therefore, due to conservation of momentum, have opposite but equal-magnitude momenta. Hence their energies are also identical, so that in this frame the kaon’s initial rest energy $E_{0,K}$ is split symmetrically between the two pions. It therefore holds that $E_{0,K}=2E_{\pi }=2\gamma E_{0,\pi }=\frac{2}{\sqrt{1-\frac{u^2}{c^2}}}{E}_{0,\pi },$ (7.1) where $\gamma ={\left(1-\frac{u^2}{c^2}\right)}^{-1/2}$ denotes the Lorentz factor and $u$ the velocity of the pions in that frame. With the given value of $E_{0,K}/E_{0,\pi }=\kappa$, the velocity $u$ can be determined from this. It is $\kappa =\frac{2}{\sqrt{1-\frac{u^2}{c^2}}}\text{and thus}u=\sqrt{1-\frac{4}{{\kappa }^2}}c\approx 0,839c,$ (7.2) The velocities $v_{1,2}$ of the two pions in the laboratory frame are obtained by relativistic velocity addition. Let $w=0,80c$ be the velocity of the kaon in the laboratory frame. Then $v_{1,2}=\frac{w\pm u}{1\pm \frac{wu}{c^2}}.$ (7.3) The forward-moving pion thus has the velocity: $v_1=\frac{0,80+0,839}{1+0,80\cdot 0,839}c\approx 0,98c.$ (7.4) The pion moving in the opposite direction has the velocity $v_2=\frac{0,80-0,839}{1-0,80\cdot 0,839}c\approx -0,12c.$ (7.5) Correct answer: B Grading - Kaon decay (multiple-choice problem) Points Working in a suitable inertial frame (e.g. the rest frame of the kaon) 1.0 Using conservation of momentum and energy (7.1) 1.0 Determining the pion velocities in that frame (7.2) 1.0 Using relativistic velocity addition (7.3) 1.0 Stating the correct solution 1.0 5.0 Long problems Work the following two problems likewise in the boxes provided for them. Unlike the multiple-choice problems, no answer options are given. Describe your solution so that it is easy to follow but not unnecessarily long. So if you use, for example, the law of conservation of energy, write this down briefly.

Topic: Special Relativity, Nuclear & Particle Physics Metodi: Relativistic Energy-Momentum, Conservation of Momentum, Conservation of Energy Competenze: Mathematical Modeling, Physical Reasoning Objects: — Fonte: Testo (PDF) — p.12

Il problema 7 del decadimento dei caoni (problema di scelta multipla) (cfr. (Problema group of the PhysicsOlympiad - Thomas Hellerl) Un kaon che si muove con velocità $0,80c$, cioè l’80% della velocità della luce, nel laboratorio decade in due pioni, che successivamente si muovono lungo e contro, rispettivamente, la direzione originale del movimento del kaon. Non si producono ulteriori particelle nel decadimento. Per il rest energies $E_{0,K}$ and $E_{0,\pi }$ of the kaon and the pion, respectively, the relation $\kappa =\frac{E_{0,K}}{E_{0,\pi }}=3,68.$

• Tieni. A che velocità si muovono i due pioni dopo il declino nel laboratorio? A $-0,039c$ e $0,99c$ B $-0,12c$ e $0,98c$ C $-0,19c$ e $0,92c$ D $-0,21c$ e $0,89c$ Soluzione Calcoli e spiegazioni Nel resto del kaon, la velocità è zero. Dopo il decadimento i pioni devono quindi, a causa di Conservation of momentum, have opposite but equal magnitude moments. Per questo Le loro energie sono identiche, quindi in questo frame l’energia restante iniziale del kaon è $E_{0,K}$ è diviso simmetricamente tra i due pioni. Il rapporto di cui sopra $E_{0,K}=2E_{\pi }=2\gamma E_{0,\pi }=\frac{2}{\sqrt{1-\frac{u^2}{c^2}}}{E}_{0,\pi },$ (7.1) dove $\gamma ={\left(1-\frac{u^2}{c^2}\right)}^{-1/2}$ denota il fattore di Lorentz e $u$ la velocità dei pioni in quel frame. Con il valore dato di $E_{0,K}/E_{0,\pi }=\kappa$, la velocità $u$ può essere determinata da questo. It is $\kappa =\frac{2}{\sqrt{1-\frac{u^2}{c^2}}}\text{and thus}u=\sqrt{1-\frac{4}{{\kappa }^2}}c\approx 0,839c,$ (7.2) Le velocità $v_{1,2}$ dei due pioni nel laboratorio sono ottenute da relativistici
• Insomma, velocità di somma. Let $w=0,80c$ be the velocity of the kaon in the laboratory frame. Allora $v_{1,2}=\frac{w\pm u}{1\pm \frac{wu}{c^2}}.$ (7.3) Il pion in movimento in avanti ha quindi la velocità: $v_1=\frac{0,80+0,839}{1+0,80\cdot 0,839}c\approx 0,98c.$ (7.4) Il pion che si muove nella direzione opposta ha la velocità $v_2=\frac{0,80-0,839}{1-0,80\cdot 0,839}c\approx -0,12c.$ (7.5) Risposta corretta: B La classificazione - decadimento del caon (problema di scelta multipla) Punti Working in a suitable inertial frame (es. il resto del kaon) 1.0 Usando la conservazione del momento e dell’energia (7.1) 1.0 Determinare le velocità di pion in quel frame (7.2) 1.0 Usando relativistic velocity addition (7.3) 1.0 Stating the correct solution 1.0 5.0 Long problemi La Commissione ha inoltre presentato una serie di proposte di risoluzione. A differenza dei problemi di scelta multipla, non sono state indicate le opzioni di risposta. Descrivere la soluzione Quindi è facile da seguire, ma non è troppo lungo. Quindi se usi, per esempio, la legge della conservazione dell’energia, scrivi questo brevemente.

Topic: Special Relativity, Nuclear & Particle Physics Metodi: Relativistic Energy-Momentum, Conservation of Momentum, Conservation of Energy Competenze: Mathematical Modeling, Physical Reasoning Objects: — Fonte: Testo (PDF) — p.12

The problem is that the number of cells in the cell is less than the number of cells in the cell. (five points) (Problem group of the PhysicsOlympic - Thomas Hellerl) A kaon moving with velocity $0,80c$, that is 80 percent of the speed of light, in the laboratory frame decays into two pions, which afterwards move along and against, respectively, the kaon’s original direction of motion. No further particles are produced in the decay. For the rest energies $E_{0,K}$ and $E_{0,\pi }$ of the kaon and the pion, respectively, the relation $\kappa =\frac{E_{0,K}}{E_{0,\pi }}=3,68.$ Hold on a second. At what speeds do the two pions move after the decay in the laboratory frame? A $-0,039c$ and $0,99c$ B $-0,12c$ and $0,98c$ C $-0,19c$ and $0,92c$ D $-0,21c$ and $0,89c$ The solution Calculations and explanations In the rest frame of the kaon the momentum is zero. After the decay the pions must therefore, due to The conservation of momentum, have opposite but equal magnitude moments. Hence Their energies are also identical, so in this frame the kaon’s initial rest energy $E_{0,K}$ is divided symmetrically between the two pions. It therefore holds that $E_{0,K}=2E_{\pi }=2\gamma E_{0,\pi }=\frac{2}{\sqrt{1-\frac{u^2}{c^2}}}{E}_{0,\pi },$ (7.1) where $\gamma ={\left(1-\frac{u^2}{c^2}\right)}^{-1/2}$ denotes the Lorentz factor and $u$ the velocity of the pions in that frame. With the given value of $E_{0,K}/E_{0,\pi }=\kappa$, the velocity $u$ can be determined from this. It is $\kappa =\frac{2}{\sqrt{1-\frac{u^2}{c^2}}}\text{and thus}u=\sqrt{1-\frac{4}{{\kappa }^2}}c\approx 0,839c,$ (7.2) The velocities $v_{1,2}$ of the two pions in the laboratory frame are obtained by relativistic The velocity addition. Let $w=0,80c$ be the velocity of the kaon in the laboratory frame. Then $v_{1,2}=\frac{w\pm u}{1\pm \frac{wu}{c^2}}.$ (7.3) The forward-moving pion thus has the velocity: $v_1=\frac{0,80+0,839}{1+0,80\cdot 0,839}c\approx 0,98c.$ (7.4) The pion moving in the opposite direction has the velocity $v_2=\frac{0,80-0,839}{1-0,80\cdot 0,839}c\approx -0,12c.$ (7.5) Correct answer: B Grading - Kaon decay (multiple choice problem) Points Working in a suitable inertial frame (e.g. the rest frame of the kaon) 1.0 Using conservation of momentum and energy (7.1) 1.0 Determining the pion velocities in that frame (7.2) 1.0 Using relativistic velocity addition (7.3) 1.0 Stating the correct solution 1.0 5.0 Long problems Work the following two problems equally in the boxes provided for them. Unlike the Multiple-choice problems, no answer options are given. Describe your solution So that it’s easy to follow but not unnecessarily long. So if you use, for example, the law of conservation of energy, write this down briefly.

Topic: Special Relativity, Nuclear & Particle Physics Metodi: Relativistic Energy-Momentum, Conservation of Momentum, Conservation of Energy Competenze: Mathematical Modeling, Physical Reasoning Objects: — Fonte: Testo (PDF) — p.12

Problem 8 Pencil lead (17.0 pts.) (Problem group of the PhysicsOlympiad - Joachim Brucherseifer & Pascal Reeck) A pencil lead slides without friction in a vertical homogeneous magnetic field of magnetic flux density $B$ down two parallel, ideally conducting metal rails inclined at an angle $\alpha =30^{\circ }$ to the horizontal. The spacing of the rails is $L=10$ cm and the overhang of the lead beyond the metal rails can be neglected. An ideal voltage source of voltage $U_0=10$ mV with a switch is connected to the rails. Lead, rails, switch and voltage source together form an electric circuit. The setup is sketched in Figure 6. Fig. 6. Sketch of the sliding pencil lead When the switch is closed, the pencil lead stays at rest. When the switch is opened, the lead slides on. Use for the resistivity of the lead the value ${\rho }_{\text{el}}=5,0\cdot 10^{-6}\Omega \text{m}$ and for the density $\rho =2,3\cdot 10^3{\text{ kg/m}}^3$. 8.a) State whether the front rail or the rear rail in the figure is connected to the positive terminal of the DC voltage source, and justify this physically. (2.0 pts.) 8.b) Determine the magnitude $B$ of the magnetic flux density and check the correctness of the units of your result with a units check. (8.0 pts.) 8.c) Determine the velocity $v_H$ that becomes established, with which the pencil lead slides down the slope, when the flux density of the magnetic field is halved. Carry out a units check for the result. (7.0 pts.) Solution 8.a) Calculations and explanations When the lead comes to rest with the switch closed, only the external voltage source acts; its current produces a Lorentz force that is equal in magnitude and opposite to the down-slope component of the weight. According to the right-hand rule, this is only possible if the positive terminal is connected to the front rail. Fig. 7. Sketch explaining the current direction and the polarity of the voltage source. 8.b) Calculations and explanations On the lead at rest on the rails act the vertical weight ${\vec{F}}_G$ and the horizontal Lorentz force ${\vec{F}}_L$. The sum of these forces must be perpendicular to the rails, so that the lead does not slide along the rails. The figure alongside illustrates the addition of the forces. For the magnitudes of the forces it accordingly holds that: $\frac{F_L}{F_G}=\tan 30^{\circ }=\frac{1}{\sqrt{3}}.$ (8.1) The Lorentz force is given by ${\vec{F}}_L=L\cdot \vec{I}\times \vec{B}.$ (8.2) Since current and flux density are perpendicular to each other, the scalar form also holds $F_L=LIB.$ (8.3) Fig. 8. Sketch of the force decomposition. With the lead at rest, the current is determined exclusively by the voltage source and the resistance of the lead. The current $I$ follows from Ohm’s law and the resistance $R$ from the resistance law as $I=\frac{U_0}{R}=\frac{U_0}{{\rho }_{\text{el}}\frac{L}{A}},$ (8.4) where $A$ denotes the cross-sectional area of the pencil lead. Substituting into equation (8.3) for the Lorentz force gives $F_L=\frac{L{U}_0}{R}B=\frac{U_{0}A}{{\rho }_{\text{el}}}B.$ (8.5) For the weight, the cylindrical geometry of the lead gives $F_G=\rho LAg.$ (8.6) Substituting (8.5) and (8.6) into the force ratio (8.1) gives $\frac{\frac{U_{0}A}{{\rho }_{\text{el}}}B}{\rho LAg}=\frac{1}{\sqrt{3}}$ and thus $B=\frac{\rho Lg{\rho }_{\text{el}}}{\sqrt{3}{U}_0}.$ (8.7) The flux density sought is thus independent of the conductor cross section and has the value $B=\frac{2,3\cdot 10^3{\text{ kg/m}}^3\cdot 0,10\text{ m}\cdot 9,81{\text{ m/s}}^2\cdot 5,0\cdot 10^{-6}\Omega \text{m}}{\sqrt{3}\cdot 0,01\text{ V}}\approx 0,65\text{ T}.$ (8.8) The following check shows that the units are also correct: $[B]=\frac{{\text{kg/m}}^3\cdot \text{m}\cdot {\text{m/s}}^2\cdot \text{V/A}\text{m}}{\text{V}}.$ (8.9) 8.c) Calculations and explanations At half the flux density the Lorentz force is no longer sufficient to compensate the down-slope component of the weight. Now a conductor moves across the magnetic field, whereby a voltage is induced in it that, according to Lenz’s rule, opposes the cause of the induction, that is the motion. Accordingly, a current must be induced that reinforces the current produced by the voltage source. The lead reaches a limiting velocity at which the forces are once again in equilibrium. The following figure sketches the motion of the electrons in the downward-sliding lead, seen from above Fig. 9. Sketch of the induction in the lead, illustrating the direction by the left-hand rule. The charge separation thus caused yields an induced voltage $U_i$ on the rails, which is connected in series with the external voltage $U_0$. The total voltage across the lead is thus $U=U_0+L\cos (\alpha )v_H\frac{B}{2}.$ (8.10) The resulting induced current reinforces the existing field current. For the Lorentz force, in this situation, it now holds with (8.5) $F_L^{\prime }=\frac{UA}{{\rho }_{\text{el}}}\frac{B}{2}=\left(U_0+L\cos (\alpha )v_H\frac{B}{2}\right)\frac{AB}{2{\rho }_{\text{el}}}.$ (8.11) The contribution of the first term corresponds, because of $B/2$ instead of $B$, exactly to half of the force necessary to hold the lead according to (8.5). In the case in which the lead no longer experiences any acceleration along the rails, the contribution of the second term must therefore supply the other half of the force necessary for force equilibrium and thus be exactly as large. From $U_0=L\cos (\alpha )v_H\frac{B}{2}$ it follows for the slope velocity of the lead that becomes established $v_H=\frac{2U_0}{L\cos (\alpha )B}=\frac{2\cdot 0,01\text{ V}}{0,10\text{ m}\cos (30^{\circ })0,65\text{ T}}\approx 0,36\text{ m/s}.$ (8.12) The following check shows that the units are also correct: $[v_H]=\frac{\text{V}}{{\text{m kg A}}^{-1}{\text{s}}^{-2}}=\frac{{\text{V A s}}^2}{\text{kg m}}=\frac{{\text{kg m}}^2{\text{s}}^{-1}}{\text{kg m}}={\text{m s}}^{-1}.$ (8.13) Here it was used that $1\text{ V A s}=1\text{ J}=1{\text{ kg m}}^2{\text{s}}^{-2}$. Grading - Pencil lead Points 8.a) Stating the correct polarity 1.0 Justifying the polarity with the direction of the Lorentz force 1.0 8.b) Stating the relevant forces with direction (also implicitly) 1.0 Recognising the force equilibrium and stating the force ratio (8.1) 1.0 Expressing the Lorentz force through given quantities and $B$ (8.5) 2.0 Expressing the weight through given quantities (8.6) 1.0 Deriving an expression for the flux density $B$ (8.7) 1.0 Calculating the value of the flux density (8.8) 1.0 Carrying out a units check (8.9) 1.0 8.c) Recognising that a voltage is induced that opposes the cause 1.0 Recognising that a terminal velocity becomes established 1.0 Stating an expression for the induced voltage as in (8.10) 1.0 Recognising that the induced voltage must equal $U_0$ 1.0 Deriving an expression for the velocity $v_H$ (8.12) 1.0 Calculating the value of the velocity in (8.12) 1.0 Carrying out a units check (8.13) 1.0 17.0

Topic: Electromagnetic Induction, Magnetism, Newtonian Mechanics Metodi: Lorentz Force Analysis, Faraday’s Law of Induction, Free-Body Diagram Competenze: Mathematical Modeling, Diagrammatic Reasoning, Physical Reasoning Objects: Rod, Battery, Switch, Wire Fonte: Testo (PDF) — p.14

Problema 8 Pennale lead (P. 17,0 p. (Problem group of the PhysicsOlympiad - Joachim Brucherseifer & Pascal Reeck) Un matita lead scorre senza attrito in un campo magnetico verticale omogeneo di densità di flusso magnetico $B$ down two parallel, ideally conducting metal rails inclined at an angle $\alpha =30^{\circ }$ to the horizontal. Il spaziamento dei binari è $L=10$ cm e l’overhang di Lead Beyond the Metal Rails può essere trascurato. Per il meglio voltage source of voltage $U_0=10$ mV with a switch is connected to the rails. Lead, rails, Scommutazione e fonte di tensione insieme formano un circuito elettrico. La struttura è schizzata in figura 6. Fig. 6. Sketch of the sliding pencil lead Quando il switch è chiuso, il lead della matita resta a riposo. Quando il switch è aperto, il Lead slides on. Use for the resistivity of the lead the value ${\rho }_{\text{el}}=5,0\cdot 10^{-6}\Omega \text{m}$ and for the density $\rho =2,3\cdot 10^3{\text{ kg/m}}^3$. 8. (a) Indicare se la linea frontale o la linea posteriore nella figura è collegata al terminale positivo della fonte di tensione DC, e giustificare questo fisicamente. (punto 2.0) 8.b) Determina la magnitude $B$ della densità del flusso magnetico e verifica la correttezza del flusso magnetico. unità del tuo risultato con un controllo delle unità. (8,0 pts.) 8.c) Determine la velocità $v_H$ che diventa established, with which the pencil lead slides Down the slope, quando la densità di flusso del campo magnetico è dimezzata. Carry out a units check for il risultato. (7,0 p.s.) Soluzione 8.a) Calcoli e spiegazioni Quando il lead viene a riposo con il switch chiuso, solo la fonte di tensione esterna agisce; la sua corrente produce una forza di Lorentz che è uguale in magnitudo e opposta alla componente di slope del peso. Secondo la regola di destra, questo è possibile solo se il Il terminal positivo è collegato al primo binario. Fig. 7. Sketch che spiega la direzione corrente e la polarità della fonte di tensione. 8.b) Calcoli e spiegazioni On the lead at rest on the rails act the vertical weight ${\vec{F}}_G$ and the La forza di Lorentz orizzontale ${\vec{F}}_L$. La somma di queste forze deve essere perpendicolare ai binari, Così il lead non scivola lungo le binarie. La figura che si trova al fianco illustra il

• Addition of the forces. - Addition of the forces. Per le magnitudini delle forze, si ritiene che: $\frac{F_L}{F_G}=\tan 30^{\circ }=\frac{1}{\sqrt{3}}.$ (8.1) La forza di Lorentz è data da ${\vec{F}}_L=L\cdot \vec{I}\times \vec{B}.$ (8.2) Poiché la densità di corrente e di flusso sono perpendicolari a ciascuna In altre parole, la forma scalare è $F_L=LIB.$ (8.3) Fig. 8. Sketch della decomposizione della forza. Con il lead a riposo, la corrente è determinata esclusivamente dalla fonte di voltage e dalla resistenza del lead. Il corrente $I$ segue dalla legge di Ohm e la resistenza $R$ from the resistance law as $I=\frac{U_0}{R}=\frac{U_0}{{\rho }_{\text{el}}\frac{L}{A}},$ (8.4) dove $A$ indica l’area cross-sectional del piombo della matita. Substituzione in equazione (8.3) per il Lorentz force dà $F_L=\frac{L{U}_0}{R}B=\frac{U_{0}A}{{\rho }_{\text{el}}}B.$ (8.5) Per il peso, la geometria cilindrica del lead dà $F_G=\rho LAg.$ (8.6) Substituendo (8.5) e (8.6) into the force ratio (8.1) $\frac{\frac{U_{0}A}{{\rho }_{\text{el}}}B}{\rho LAg}=\frac{1}{\sqrt{3}}$ e così $B=\frac{\rho Lg{\rho }_{\text{el}}}{\sqrt{3}{U}_0}.$ (8.7) Il flusso di densità ricercato è quindi indipendente dal cross-section del conduttore e ha il valore $B=\frac{2,3\cdot 10^3{\text{ kg/m}}^3\cdot 0,10\text{ m}\cdot 9,81{\text{ m/s}}^2\cdot 5,0\cdot 10^{-6}\Omega \text{m}}{\sqrt{3}\cdot 0,01\text{ V}}\approx 0,65\text{ T}.$ (8.8) I seguenti controlli mostrano che le unità sono corrette: $[B]=\frac{{\text{kg/m}}^3\cdot \text{m}\cdot {\text{m/s}}^2\cdot \text{V/A}\text{m}}{\text{V}}.$ (8.9) 8.c) Calcoli e spiegazioni A metà della densità del flusso la forza di Lorentz non è più sufficiente a compensare il componente di declino del peso. Ora un conduttore si muove attraverso il campo magnetico, in cui un voltage è indotto in esso che, secondo la regola di Lenz, opponente alla causa dell’induzione, cioè il
• Il movimento. In questo modo, un corrente deve essere indotta che rafforzi il corrente prodotta dal fonte di tensione. Il lead raggiunge una velocità limitante a cui le forze sono ancora una volta in equilibrio. La figura seguente descrive il movimento degli elettroni nel sliding verso il basso lead, seen from above Fig. 9. Sketch dell’induzione nella guida, illustrando la direzione dalla regola di sinistra. Il carico separato causò così rendimenti di voltage indotto $U_i$ sui binari, che è collegato in serie con la tensione esterna $U_0$. La tensione totale attraverso il Lead è così $U=U_0+L\cos (\alpha )v_H\frac{B}{2}.$ (8.10) Il risultante corrente indotta rafforza il corrente di campo esistente. Per il Lorentz forza, in questa situazione, ora si tiene con (8.5) $F_L^{\prime }=\frac{UA}{{\rho }_{\text{el}}}\frac{B}{2}=\left(U_0+L\cos (\alpha )v_H\frac{B}{2}\right)\frac{AB}{2{\rho }_{\text{el}}}.$ (8.11) Il contributo del primo termine corrisponde, per $B/2$ invece di $B$, esattamente a metà del La forza necessaria per mantenere il lead secondo (8.5). Nel caso in cui il lead non sperimenti più alcuna accelerazione lungo i binari, il contributo del secondo termine Il sistema di controllo deve quindi fornire l’altra metà della forza necessaria per l’equilibrio di forza e Quindi, essere esattamente grande. Da $U_0=L\cos (\alpha )v_H\frac{B}{2}$ segue per la velocità di slope del lead che diventa stabilito $v_H=\frac{2U_0}{L\cos (\alpha )B}=\frac{2\cdot 0,01\text{ V}}{0,10\text{ m}\cos (30^{\circ })0,65\text{ T}}\approx 0,36\text{ m/s}.$ (8.12) I seguenti controlli mostrano che le unità sono corrette: $[v_H]=\frac{\text{V}}{{\text{m kg A}}^{-1}{\text{s}}^{-2}}=\frac{{\text{V A s}}^2}{\text{kg m}}=\frac{{\text{kg m}}^2{\text{s}}^{-1}}{\text{kg m}}={\text{m s}}^{-1}.$ (8.13) Here it was used that $1\text{ V A s}=1\text{ J}=1{\text{ kg m}}^2{\text{s}}^{-2}$. Classificazione - Pennale di piombo Punti 8.a) Stating the correct polarity 1.0 Justificando la polarità con la direzione della forza di Lorentz 1.0 8.b) Stating the relevant forces with direction (also implicitly) 1.0 Recognising the force equilibrium and stating the force ratio (8.1) 1.0 Esprimendo la forza di Lorentz attraverso dati quantitativi e $B$ (8.5) 2.0 Esprimendo il peso attraverso quantitativi dati (8.6) 1.0 Deriving an expression for the flux density $B$ (8.7) 1.0 Calcolare il valore della densità del flusso (8.8) 1.0 Carrying out a units check (8.9) 1.0 8.c) Riconoscendo che una tensione è indotta che opponga la causa 1.0 Riconoscendo che una velocità terminale diventa established 1.0 Stating an expression for the induced voltage as in (8.10) 1.0 Recognising that the induced voltage must equal $U_0$ 1.0 Deriving an expression for the velocity $v_H$ (8.12) 1.0 Calcolare il valore della velocità in (8.12) 1.0 Carrying out a units check (8.13) 1.0 17.0

Topic: Electromagnetic Induction, Magnetism, Newtonian Mechanics Metodi: Lorentz Force Analysis, Faraday’s Law of Induction, Free-Body Diagram Competenze: Mathematical Modeling, Diagrammatic Reasoning, Physical Reasoning Objects: Rod, Battery, Switch, Wire Fonte: Testo (PDF) — p.14

Problem 8 Pencil lead The Commission has also adopted a proposal for a directive on the protection of workers’ rights. (Problem group of the PhysicsOlympiad - Joachim Brucherseifer and Pascal Reeck) A pencil lead slides without friction in a vertical homogeneous magnetic field of magnetic flux density $B$ down two parallel, ideally conducting metal rails inclined at an angle $\alpha =30^{\circ }$ to the horizontal. The spacing of the rails is $L=10$ cm and the overhang Of the lead beyond the metal rails can be neglected. The ideal voltage source of voltage $U_0=10$ mV with a switch is connected to the rails. Lead, rails, switch and voltage source together form an electrical circuit. The setup is sketched in Figure 6. Fig. 6. Sketch of the sliding pencil lead When the switch is closed, the pencil lead stays at rest. When the switch is opened, the Lead slides on. Use for the resistivity of the lead the value ${\rho }_{\text{el}}=5,0\cdot 10^{-6}\Omega \text{m}$ and for the density $\rho =2,3\cdot 10^3{\text{ kg/m}}^3$. 8. (a) State whether the front rail or the rear rail in the figure is connected to the positive terminal of the DC voltage source, and justify this physically. (b) the number of persons who are not members of the 8.b) Determine the magnitude $B$ of the magnetic flux density and check the correctness of the units of your result with a unit check. (8.0 pts.) 8.c) Determine the velocity $v_H$ that becomes established, with which the pencil lead slides down the slope, when the flux density of the magnetic field is halved. Carry out a units check for The result. (7.0 pts) The solution 8.a) Calculations and explanations When the lead comes to rest with the switch closed, only the external voltage source acts; its current produces a Lorentz force that is equal in magnitude and opposite to the down-slope component of the weight. According to the right-hand rule, this is only possible if the The positive terminal is connected to the front rail. Fig. 7. Sketch explaining the current direction and the polarity of the voltage source. 8.b) Calculations and explanations On the lead at rest on the rails act the vertical weight ${\vec{F}}_G$ and the horizontal Lorentz force ${\vec{F}}_L$. The sum of these forces must be perpendicular to the rails, So that lead doesn’t slide along the rails. The figure alongside illustrates the Addition of the forces. For the magnitudes of the forces it accordingly holds that: $\frac{F_L}{F_G}=\tan 30^{\circ }=\frac{1}{\sqrt{3}}.$ (8.1) The Lorentz force is given by ${\vec{F}}_L=L\cdot \vec{I}\times \vec{B}.$ (8.2) Since current and flux density are perpendicular to each other Other, the scalar form also holds $F_L=LIB.$ (8.3) Fig. 8. Sketch of the force decomposition. With the lead at rest, the current is determined exclusively by the voltage source and the resistance of the lead. The current $I$ follows from Ohm’s law and the resistance $R$ from the resistance law as $I=\frac{U_0}{R}=\frac{U_0}{{\rho }_{\text{el}}\frac{L}{A}},$ (8.4) where $A$ denotes the cross-sectional area of the pencil lead. Substituting into equation (8.3) For the Lorentz force gives $F_L=\frac{L{U}_0}{R}B=\frac{U_{0}A}{{\rho }_{\text{el}}}B.$ (8.5) For the weight, the cylindrical geometry of the lead gives $F_G=\rho LAg.$ (8.6) Substituting (8.5) and (8.6) into the force ratio (8.1) gives $\frac{\frac{U_{0}A}{{\rho }_{\text{el}}}B}{\rho LAg}=\frac{1}{\sqrt{3}}$ and thus $B=\frac{\rho Lg{\rho }_{\text{el}}}{\sqrt{3}{U}_0}.$ (8.7) The flux density sought is thus independent of the conductor cross section and has the value $B=\frac{2,3\cdot 10^3{\text{ kg/m}}^3\cdot 0,10\text{ m}\cdot 9,81{\text{ m/s}}^2\cdot 5,0\cdot 10^{-6}\Omega \text{m}}{\sqrt{3}\cdot 0,01\text{ V}}\approx 0,65\text{ T}.$ (8.8) The following check shows that the units are correct: $[B]=\frac{{\text{kg/m}}^3\cdot \text{m}\cdot {\text{m/s}}^2\cdot \text{V/A}\text{m}}{\text{V}}.$ (8.9) 8.c) Calculations and explanations At half the flux density the Lorentz force is no longer sufficient to compensate the down-slope component of the weight. Now a conductor moves across the magnetic field, where a voltage is induced in it that, according to Lenz’s rule, opposes the cause of the induction, that is the The motion. Accordingly, a current must be induced that reinforces the current produced by the The voltage source. The lead reaches a limiting velocity at which the forces are once again in equilibrium. The following figure sketches the motion of the electrons in the downward sliding Lead, seen from above Fig. 9. Sketch of the induction in the lead, illustrating the direction by the left-hand rule. The charge separation thus caused yields of induced voltage $U_i$ on the rails, which is connected in series with the external voltage $U_0$. The total voltage across the Lead is thus $U=U_0+L\cos (\alpha )v_H\frac{B}{2}.$ (8.10) The resulting induced current reinforces the existing field current. For the Lorentz force, in this situation, it now holds with (8.5) $F_L^{\prime }=\frac{UA}{{\rho }_{\text{el}}}\frac{B}{2}=\left(U_0+L\cos (\alpha )v_H\frac{B}{2}\right)\frac{AB}{2{\rho }_{\text{el}}}.$ (8.11) The contribution of the first term corresponds, because of $B/2$ instead of $B$, exactly to half of the force necessary to hold the lead according to (8.5). In the case where the lead no longer experiences any acceleration along the rails, the contribution of the second term The second part of the force needed for force balance and So be exactly as large. From $U_0=L\cos (\alpha )v_H\frac{B}{2}$ it follows for the slope velocity of the lead that becomes established $v_H=\frac{2U_0}{L\cos (\alpha )B}=\frac{2\cdot 0,01\text{ V}}{0,10\text{ m}\cos (30^{\circ })0,65\text{ T}}\approx 0,36\text{ m/s}.$ (8.12) The following check shows that the units are correct: $[v_H]=\frac{\text{V}}{{\text{m kg A}}^{-1}{\text{s}}^{-2}}=\frac{{\text{V A s}}^2}{\text{kg m}}=\frac{{\text{kg m}}^2{\text{s}}^{-1}}{\text{kg m}}={\text{m s}}^{-1}.$ (8.13) Here it was used that $1\text{ V A s}=1\text{ J}=1{\text{ kg m}}^2{\text{s}}^{-2}$. Grading - Pencil lead Points 8.a) Stating the correct polarity 1.0 Justifying the polarity with the direction of the Lorentz force 1.0 8.b) Stating the relevant forces with direction (also implicitly) 1.0 Recognising the force equilibrium and stating the force ratio (8.1) 1.0 Expressing the Lorentz force through given quantities and $B$ (8.5) 2.0 Expressing the weight through given quantities (8.6) 1.0 Deriving an expression for the flux density $B$ (8.7) 1.0 Calculating the value of the flux density (8.8) 1.0 Carrying out a unit check (8.9) 1.0 8.c) Recognizing that a voltage is induced that opposes the cause 1.0 Recognizing that a terminal velocity becomes established 1.0 Stating an expression for the induced voltage as in (8.10) 1.0 Recognising that the induced voltage must equal $U_0$ 1.0 Deriving an expression for the velocity $v_H$ (8.12) 1.0 Calculating the value of the velocity in (8.12) 1.0 Carrying out a unit check (8.13) 1.0 17.0

Topic: Electromagnetic Induction, Magnetism, Newtonian Mechanics Metodi: Lorentz Force Analysis, Faraday’s Law of Induction, Free-Body Diagram Competenze: Mathematical Modeling, Diagrammatic Reasoning, Physical Reasoning Objects: Rod, Battery, Switch, Wire Fonte: Testo (PDF) — p.14

Problem 9 Swing-by manoeuvre and the Pioneer anomaly (18.0 pts.) (Idea: Bastian Hacker) The space probe Pioneer 10 was launched in 1972 to explore the outer solar system and was to be one of the first spacecraft to leave the solar system for good. By now it is located about 140 astronomical units from the Sun. After launch, Pioneer 10 left the Earth’s gravitational field with a velocity of $v_0=9,4\text{ km/s}$ relative to the Earth and tangential to its orbit. From there the probe flew toward Jupiter to carry out a swing-by manoeuvre that was to make leaving the solar system possible. In the following, consider all processes within the ecliptic and assume that the planets orbit the Sun on circular orbits in the same sense of rotation. In addition, the following values may be used for the work: Radius of the Earth’s orbit (astronomical unit) $1\text{ AE}=149,6\cdot 10^6\text{ km}$ Radius of Jupiter’s orbit $r_J=5,2\text{ AE}$ Solar mass $M_S=1,99\cdot 10^{30}\text{ kg}$ Solar constant (power of the solar radiation at Earth’s orbit) $E_0=1367{\text{ W/m}}^2$ 9.a) Determine the orbital velocity $v_2$ with which the probe arrived at the orbit of the planet Jupiter. Assume here that only the gravitational force of the Sun acts on the probe. Determine the velocity components $v_{2,\varphi }$ tangential and $v_{2,r}$ perpendicular to Jupiter’s orbit. (6.0 pts.) With the help of a flight manoeuvre through Jupiter’s gravitational field, the space probe was able to increase its velocity relative to the Sun considerably. Assume that this swing-by manoeuvre was carried out so that the probe can leave the solar system with the highest possible velocity. You may furthermore assume that the time during which Pioneer 10 interacts gravitationally with Jupiter is small compared with the orbital period and that during the manoeuvre the gravitational force between probe and Jupiter is dominant. 9.b) Determine the radial velocity $v_r(r)$ and in particular the limiting velocity $v_r(\infty )$ of the probe after the swing-by manoeuvre as a function of the distance $r$ from the Sun. (6.0 pts.) After the probe had, in 1980 at 20 AE, moved far enough away from the Sun to predict the influence of various forces accurately enough, an inexplicable, tiny component was discovered in its acceleration. This so-called Pioneer anomaly has given rise to numerous speculations about possible modifications of the laws of nature. The strongest force on the probe, whose mass may be taken to be 241 kg, is, after gravity, caused by the radiation pressure of the solar radiation. The parabolic antenna, which has a diameter of $d=2,77\text{ m}$ and at larger distances points approximately toward the Sun, absorbs 20 % of the sunlight and reflects the rest back toward the Sun. 9.c) Determine approximately the acceleration that results from the radiation pressure of the Sun for the probe at a distance $r\gg 1$ AE from the Sun. State how large the contribution was in 1990 for $r=50$ AE, and compare it with the acceleration due to gravity in the Sun’s gravitational field. (6.0 pts.) The observed, unexplained acceleration was indeed of a similar order of magnitude, but pointed toward the Sun. Therefore further effects were also investigated, such as the solar wind, the influence of other celestial bodies, the recoil from the radio module, and above all non-isotropic thermal emission, which in the end, apart from deviations within the uncertainties, were able to provide an explanation for the Pioneer anomaly. Solution 9.a) Calculations and explanations The gravitational potential in the Sun’s gravitational field at a distance $r$ from the Sun is $\varphi (r)=-G\frac{M_S}{r},$ (9.1) where $G$ denotes the gravitational constant and $M_S$ the solar mass. For circular planetary orbits, the gravitational force equals the centripetal force, i.e. it holds that $F_G=m\frac{G{M}_S}{r^2}=m\frac{v^2}{r}=F_Z,$ or $v^2=\frac{G{M}_S}{r}=-\varphi (r).$ (9.2) This corresponds to the virial theorem, according to which for this case $E_{\text{kin}}=-\frac{1}{2}{E}_{\text{pot}}$ holds. For the orbital velocity $v_E$ of the Earth it therefore holds, with $r=r_E=1$ AE: $v_E^2=-\varphi (r_E)=\frac{G{M}_S}{1\text{ AE}}\approx {\left(29,8\text{ km/s}\right)}^2.$ (9.3) The velocity $v_0$ given in the problem statement denotes the velocity that the probe has relative to the Earth when it is still geometrically close to the Earth, but the influence of Earth’s gravity is already negligible. The probe therefore leaves the Earth’s orbit with the velocity${}^a$ $v_1=v_0+v_E\approx 9,4\text{ km/s}+29,8\text{ km/s}=39,2\text{ km/s}$ (9.4) and arrives at the Jupiter orbit (orbital radius $r_J=5,2r_E$) with the velocity $v_2$. For this it follows via conservation of energy $E_{\text{kin}}+E_{\text{pot}}=\frac{1}{2}m{v}^2(r)+m\varphi (r)=\text{const.}$ or $\frac{1}{2}{v}_1^2-\frac{G{M}_S}{r_E}=\frac{1}{2}{v}_2^2-\frac{G{M}_S}{r_J}.$ (9.5) Thus $v_2=\sqrt{v_1^2+2\varphi (r_E)\cdot (1-1/5,2)}\approx 10,1\text{ km/s}.$ (9.6) For the investigation of the swing-by manoeuvre, the radial component $v_{2,r}$ and the tangential component $v_{2,\varphi }$ of this velocity relative to the Sun are needed. The conservation of angular momentum $v_{\varphi }r=\text{const.}$ gives $v_{2,\varphi }=v_1/5,2\approx 7,5\text{ km/s}$ and hence $v_{2,r}=\sqrt{v_2^2-v_{2,\varphi }^2}\approx 6,7\text{ km/s}.$ (9.7) ${}^a$If the probe had been launched against the orbital velocity of the Earth, the difference would have to be considered here. But since the probe would then not have reached Jupiter, this case is not relevant. 9.b) Calculations and explanations Jupiter itself moves with an orbital velocity $v_J$ in the tangential direction, for which, with equation (9.2), it holds that: $v_J^2=-\varphi (r_J)\approx (13,1\text{ km/s}{)}^2.$ (9.8) Consider the swing-by manoeuvre in the rest frame of Jupiter. There the potential is static, so that the magnitude of the probe’s velocity is identical before and after the manoeuvre due to conservation of energy. It therefore holds that $\sqrt{v_{2,r}^2+(v_{2,\varphi }-v_J{)}^2}=\sqrt{v_{3,r}^2+(v_{3,\varphi }-v_J{)}^2},$ (9.9) where $v_3$ denotes the velocity in the Sun-fixed frame after the swing-by manoeuvre, and it was used that Jupiter has the same sense of orbital motion around the Sun as the Earth. The left-hand side is already known and the magnitude of $v_3$ is to be maximised. But the magnitude of a sum of two vectors of given length is maximal exactly when the vectors are aligned parallel. Therefore the exit velocity $v_3$ is, in the ideal case, again tangential to the orbit and $v_{3,r}=0$. Equation (9.9) then gives: $v_3=v_{3,\varphi }=v_J+\sqrt{v_{2,r}^2+(v_{2,\varphi }-v_J{)}^2}\approx 21,8\text{ km/s}.$ (9.10) The subsequent orbital velocity follows once again from conservation of energy: $v(r)=\sqrt{v_3^2+2(\varphi (r_J)-\varphi (r))}=\sqrt{v_3^2+2\varphi (r_J)(1-r_J/r)}.$ (9.11) The radial velocity can again be derived from conservation of angular momentum and is $v_r(r)=\sqrt{v_3^2+2\varphi (r_J)(1-r_J/r)}\sqrt{1-(r_J/r{)}^2}\approx \sqrt{(21,8\text{ km/s}{)}^2-2(13,1\text{ km/s}{)}^2(1-5,2\text{ AE}/r)}\sqrt{1-(5,2\text{ AE}/r{)}^2}.$ (9.12) In particular, the limiting velocity of the probe follows from equation (9.12) as $v_r(\infty )=\sqrt{v_3^2+2\varphi (r_J)}\approx \sqrt{(21,8\text{ km/s}{)}^2-2(13,1\text{ km/s}{)}^2}\approx 11,5\text{ km/s}.$ (9.13) With this velocity Pioneer 10 leaves the solar system. 9.c) Calculations and explanations For the photons of the sunlight, the relativistic energy-momentum relation holds $E=cp.$ (9.14) At a distance $r$ from the Sun, the probe receives a radiation power per area of $\frac{P}{A}=E_0{\left(\frac{1\text{ AE}}{r}\right)}^2,$ (9.15) where $A=\pi d^2/4$ gives the cross-sectional area of the probe. This results in a force on the probe of $F=\dot{p}=P/c$. Here the reflected part of the radiation transfers twice the momentum, so that for the acceleration $a_{\text{Strahlung}}(r)=\frac{F}{m}=\frac{(2\cdot 0,8+0,2)P}{mc}=\frac{1,8\pi d^2{E}_0}{4mc}{\left(\frac{1\text{ AE}}{r}\right)}^2$ (9.16) results. Thus the acceleration, expressed through the acceleration at $r=50$ AE, is $a_{\text{Strahlung}}(r)=8,2\cdot 10^{-11}{\text{ m/s}}^2\cdot {\left(\frac{50\text{ AE}}{r}\right)}^2.$ (9.17) The gravitational acceleration due to the Sun, on the other hand, expressed through the acceleration at $r=50$ AE, is $a_G(r)=-\frac{G{M}_S}{(50\text{ AE}{)}^2}$

Topic: Gravitation, Astrophysics, Modern-Quantum Physics Metodi: Newton’s Law of Gravitation, Conservation of Energy, Conservation of Momentum Competenze: Mathematical Modeling, Physical Reasoning, Estimation & Approximation Objects: Satellite, Planet, Star Fonte: Testo (PDF) — p.18

Problema 9 Manovra di svingamento e anomalia del pioniere (8,0 pts.) (idea: Bastian Hacker) La sonda spaziale Pioneer 10 è stata lanciata nel 1972 per esplorare il sistema solare esterno e che sarebbe stata una delle prime navi spaziali a lasciare il sistema solare per sempre. Ora è situato Circa 140 unità astronomiche dal Sole. Dopo il lancio, Pioneer 10 ha lasciato il campo gravitazionale terrestre con una velocità di $v_0=9,4\text{ km/s}$ relativa alla Terra e tangenziale alla sua orbita. Da lì il test volava verso Giove per effettuare una manovra di swing-by che ha avuto a rendere possibile il lasciare il sistema solare. In seguito, consideriamo tutti i processi all’interno dell’ecliptica e supponiamo che i pianeti orbitino intorno all’Ecliptica. Il Sole in orbita circolare nello stesso senso di rotazione. Inoltre, i seguenti valori possono essere: Usato per il lavoro: Radius of the Earth’s orbit (unit astronomical) $1\text{ AE}=149,6\cdot 10^6\text{ km}$ Radius di orbita di Giove $r_J=5,2\text{ AE}$ Massa solare $M_S=1,99\cdot 10^{30}\text{ kg}$ Costante solare (potenza della radiazione solare in orbita terrestre) $E_0=1367{\text{ W/m}}^2$ 9.a) Determine the orbital velocity $v_2$ with which the probe arrived at the orbit of the planet Giove. Supponiamo che solo la forza gravitazionale del Sun è in grado di intervenire. Determine i componenti di velocità $v_{2,\varphi }$ tangentiali e $v_{2,r}$ perpendicolare a L’orbita di Giove. (6,0 p.p.) Con l’aiuto di una manovra di volo attraverso il campo gravitazionale di Giove, la sonda spaziale è stata in grado di aumentare considerevolmente la sua velocità rispetto al Sole. Supponiamo che questo swing-by manovra cosa La sonda può lasciare il sistema solare con la massima velocità possibile. Potete inoltre supporre che il tempo durante il quale Pioneer 10 interagisce gravitationally con Giove La durata del manovra è limitata rispetto al periodo orbitale e il La forza gravitazionale tra la sonda e Giove è dominante. 9.b) Determina la velocità radial $v_r(r)$ e in particolare la velocità limitante $v_r(\infty )$ of the probe after the swing-by manoeuvre as a function of the distance $r$ from the Sun. (6,0 p.p.) Dopo che la sonda aveva, nel 1980 a 20 AE, si è spostata abbastanza lontano dal Sole per predict the influence of various forces accurately enough, inexplicable, inexplicable, inexplicable, inexplicable, inexplicable, inexplicable, inexplicable, inexplicable, inexplicable, inexplicable, inexplicable, inexplicable, inexplicable, inexplicable, inexplicable, inexplicable, inexplicable, inexplicable, inexplicable, inexplicable, inexplicable, inexplicable, inexplicable, inexplicable, inexplicable, inexplicable, inexplicable, Un piccolo componente che è stato scoperto nella sua accelerazione. Questa cosiddetta anomalia pionieristica ha Le sue origini sono state evidenti e hanno dato luogo a numerose speculazioni su possibili modifiche delle leggi della natura. La forza più forte sulla sonda, il cui peso può essere preso a essere di 241 kg, è, dopo gravità, causata dalla pressione delle radiazioni solari. L’antenna parabolica, che ha un diametro di $d=2,77\text{ m}$ e a più grandi distanze punti circa verso Il sole, assorbe il 20 per cento della luce solare e riflette il resto verso il sole. 9.c) Determinare approssimativamente l’accelerazione che si verifica dalla pressione di radiazione del Sole per la prova a distanza $r\gg 1$ AE dal Sole. Stat how large the contribution La crescita delle acque di carbonio è stata aumentata nel 1990 per $r=50$ AE, and compare it with the acceleration due to gravità nel campo gravitazionale del Sole. (6,0 p.p.) L’accelerazione osservata, inesplicata, era indeed of a similar order of magnitude, but pointed verso il sole. Pertanto, sono stati anche studiati ulteriori effetti, come il vento solare, l’influenza di altri celestial bodies, the recoil from the radio module, and above all non isotropic thermal Le emissioni che, a fine fine, a parte le deviazioni all’interno delle incertezze, sono state in grado di fornire un’efficacia di spiegazione dell’anomalia del pioniere. Soluzione 9.a) Calcoli e spiegazioni Il potenziale gravitazionale nel campo gravitazionale del Sole a una distanza $r$ dal Sole is $\varphi (r)=-G\frac{M_S}{r},$ (9.1) dove $G$ indica la costante gravitazionale e $M_S$ la massa solare. Per circolare Le orbite planetarie, la forza gravitazionale è uguale alla forza centripetal, cioè che $F_G=m\frac{G{M}_S}{r^2}=m\frac{v^2}{r}=F_Z,$ or $v^2=\frac{G{M}_S}{r}=-\varphi (r).$ (9.2) Questo corrisponde al teorema virile, secondo il quale per questo caso $E_{\text{kin}}=-\frac{1}{2}{E}_{\text{pot}}$ si mantiene. Per la velocità orbitale $v_E$ della Terra, quindi, con $r=r_E=1$ AE: $v_E^2=-\varphi (r_E)=\frac{G{M}_S}{1\text{ AE}}\approx {\left(29,8\text{ km/s}\right)}^2.$ (9.3) La velocità $v_0$ data nella dichiarazione del problema indica la velocità che la sonda ha relativa alla Terra quando è ancora geometricamente vicino alla Terra, ma l’influenza La gravità della Terra è già trascurabile. La sonda lascia quindi l’orbita della Terra con la velocità${}^a$ $v_1=v_0+v_E\approx 9,4\text{ km/s}+29,8\text{ km/s}=39,2\text{ km/s}$ (9.4) e arriva all’orbita di Giove (radius orbitale $r_J=5,2r_E$) con la velocità $v_2$. Per Questo è seguito attraverso la conservazione dell’energia $E_{\text{kin}}+E_{\text{pot}}=\frac{1}{2}m{v}^2(r)+m\varphi (r)=\text{const.}$ or $\frac{1}{2}{v}_1^2-\frac{G{M}_S}{r_E}=\frac{1}{2}{v}_2^2-\frac{G{M}_S}{r_J}.$ (9.5) Così $v_2=\sqrt{v_1^2+2\varphi (r_E)\cdot (1-1/5,2)}\approx 10,1\text{ km/s}.$ (9.6) Per l’investigazione della manovra swing-by, il componente radial $v_{2,r}$ e il tangenziale sono necessari i componenti $v_{2,\varphi }$ di questa velocità relativa al Sole. Il conservation of angular momentum $v_{\varphi }r=\text{const.}$ dà $v_{2,\varphi }=v_1/5,2\approx 7,5\text{ km/s}$ e quindi $v_{2,r}=\sqrt{v_2^2-v_{2,\varphi }^2}\approx 6,7\text{ km/s}.$ (9.7) ${}^a$If the probe had been launched against the orbital velocity of the Earth, the difference would have to be considerato qui. Ma poiché la sonda non avrebbe raggiunto Giove, questo caso non è rilevante. 9.b) Calcoli e spiegazioni Giove stesso si muove con una velocità orbitale $v_J$ nella direzione tangenziale, per il quale, con l’equazione (9.2), si ritiene che: $v_J^2=-\varphi (r_J)\approx (13,1\text{ km/s}{)}^2.$ (9.8) Considerate la manovra di swing-by nel resto del quadro di Giove. Il potenziale è statico. Quindi la grandezza della velocità della sonda è identica prima e dopo. La manovra dovuta alla conservazione dell’energia. Il rapporto di cui sopra $\sqrt{v_{2,r}^2+(v_{2,\varphi }-v_J{)}^2}=\sqrt{v_{3,r}^2+(v_{3,\varphi }-v_J{)}^2},$ (9.9) dove $v_3$ indica la velocità nel quadro fisso del Sole dopo la manovra di swing-by, e si è usato che Giove ha lo stesso senso di movimento orbitale intorno al Sole come la Terra. Il lato sinistro è già noto e la magnitudo di $v_3$ deve essere massimizzata. Ma il magnitude of a sum of two vectors of given length is maximal exactly quando i vettori sono allineati in parallelo. Pertanto la velocità di uscita $v_3$ è, in Il caso ideale, ancora tangenziale all’orbita e $v_{3,r}=0$. Equation (9.9) then gives: $v_3=v_{3,\varphi }=v_J+\sqrt{v_{2,r}^2+(v_{2,\varphi }-v_J{)}^2}\approx 21,8\text{ km/s}.$ (9.10) La velocità orbitale successiva segue ancora una volta dalla conservazione di energia: $v(r)=\sqrt{v_3^2+2(\varphi (r_J)-\varphi (r))}=\sqrt{v_3^2+2\varphi (r_J)(1-r_J/r)}.$ (9.11) La velocità radial può essere derivata da conservazione di momento angolare e è $v_r(r)=\sqrt{v_3^2+2\varphi (r_J)(1-r_J/r)}\sqrt{1-(r_J/r{)}^2}\approx \sqrt{(21,8\text{ km/s}{)}^2-2(13,1\text{ km/s}{)}^2(1-5,2\text{ AE}/r)}\sqrt{1-(5,2\text{ AE}/r{)}^2}.$ (9.12) In particolare, la velocità di limitazione della sonda segue da equation (9.12) come $v_r(\infty )=\sqrt{v_3^2+2\varphi (r_J)}\approx \sqrt{(21,8\text{ km/s}{)}^2-2(13,1\text{ km/s}{)}^2}\approx 11,5\text{ km/s}.$ (9.13) Con questa velocità Pioneer 10 lascia il sistema solare. 9.c) Calcoli e spiegazioni Per i fotoni della luce solare, la relazione relativistica del momento energetico $E=cp.$ (9.14) A una distanza $r$ dal Sole, la sonda riceve una potenza di radiazione per area di $\frac{P}{A}=E_0{\left(\frac{1\text{ AE}}{r}\right)}^2,$ (9.15) dove $A=\pi d^2/4$ dà l’area cross-sectional del campione. Questo comporta una forza sulla prova di $F=\dot{p}=P/c$. Here the reflected part of the radiation transfers Due volte l’impulso, quindi per l’accelerazione $a_{\text{Strahlung}}(r)=\frac{F}{m}=\frac{(2\cdot 0,8+0,2)P}{mc}=\frac{1,8\pi d^2{E}_0}{4mc}{\left(\frac{1\text{ AE}}{r}\right)}^2$ (9.16)

• I risultati. Quindi l’accelerazione, espressa attraverso l’accelerazione a $r=50$ AE, è $a_{\text{Strahlung}}(r)=8,2\cdot 10^{-11}{\text{ m/s}}^2\cdot {\left(\frac{50\text{ AE}}{r}\right)}^2.$ (9.17) L’accelerazione gravitazionale dovuta al Sole, d’altra parte, espressa attraverso il Accelerazione a $r=50$ AE, è $a_G(r)=-\frac{G{M}_S}{(50\text{ AE}{)}^2}$

Topic: Gravitation, Astrophysics, Modern-Quantum Physics Metodi: Newton’s Law of Gravitation, Conservation of Energy, Conservation of Momentum Competenze: Mathematical Modeling, Physical Reasoning, Estimation & Approximation Objects: Satellite, Planet, Star Fonte: Testo (PDF) — p.18

Problem 9 Swing-by maneuver and the Pioneer anomaly The Commission has also adopted a proposal for a directive on the protection of workers’ rights. (Idea: Bastian Hacker) The space probe Pioneer 10 was launched in 1972 to explore the outer solar system and the Earth. What was to be one of the first spacecraft to leave the solar system for good. By now it ‘s located about 140 astronomical units from the Sun. After launch, Pioneer 10 left Earth’s gravitational field with a velocity of $v_0=9,4\text{ km/s}$ relative to the Earth and tangential to its orbit. From there the probe flew Towards Jupiter to carry out a swing-by maneuver that was to Make leaving the solar system possible. In the following, consider all processes within the ecliptic and assume that the planets orbit the Sun on circular orbits in the same sense of rotation. In addition, the following values may be used for the work: Radius of the Earth’s orbit (astronomical unit) $1\text{ AE}=149,6\cdot 10^6\text{ km}$ Radius of Jupiter’s orbit $r_J=5,2\text{ AE}$ Solar mass $M_S=1,99\cdot 10^{30}\text{ kg}$ Solar constant (power of solar radiation at Earth’s orbit) $E_0=1367{\text{ W/m}}^2$ 9. (a) Determine the orbital velocity $v_2$ with which the probe arrived at the orbit of the planet Jupiter is here. Assume here that only the gravitational force of the Sun acts on the probe. Determine the velocity components $v_{2,\varphi }$ tangential and $v_{2,r}$ perpendicular to Jupiter’s orbit is in motion. (6.0 pts) With the help of a flight maneuver through Jupiter’s gravitational field, the space probe was able to increase its velocity relative to the Sun considerably. Assume that this swing-by maneuver was The probe can leave the solar system with the highest possible velocity. You may further assume that the time during which Pioneer 10 interacts gravitationally with Jupiter The average time of the manoeuvre is small compared to the orbital period and that during the manoeuvre the The gravitational force between probe and Jupiter is dominant. 9.b) Determine the radial velocity $v_r(r)$ and in particular the limiting velocity $v_r(\infty )$ of the probe after the swing-by manoeuvre as a function of the distance $r$ from the Sun. (6.0 pts) After the probe had, in 1980 at 20 AE, moved far enough away from the Sun to predicts the influence of various forces accurately enough, to inexplicable, tiny component discovered in its acceleration. This so-called pioneer anomaly has The Commission has already taken a number of steps to improve the quality of life of the people of the Member States. The strongest force on the probe, whose mass may be taken to be 241 kg, is, after gravity, caused by the radiation pressure of the solar radiation. The parabolic antenna, which has a diameter of $d=2,77\text{ m}$ and at larger distances points approximately towards The Sun, absorbs 20 percent of the sunlight and reflects the rest back toward the Sun. 9.c) Determine approximately the acceleration resulting from the radiation pressure of the Sun for the probe at a distance $r\gg 1$ AE from the Sun. State how large the contribution The Commission’s proposal for a regulation on the use of the energy efficiency of the Community’s energy sector in 1990 for $r=50$ AE, and compare it with the acceleration due to gravity in the gravitational field of the Sun. (6.0 pts) The observed, unexplained acceleration was indeed of a similar order of magnitude, but pointed towards the sun. Therefore further effects were also investigated, such as the solar wind, the influence of other Celestial bodies, the recoil from the radio module, and above all non-isotropic thermal The Commission has also proposed that the Commission should be able to take into account the explanation for the pioneer anomaly. The solution 9.a) Calculations and explanations The gravitational potential in the Sun’s gravitational field at a distance $r$ from the Sun is $\varphi (r)=-G\frac{M_S}{r},$ (9.1) where $G$ denotes the gravitational constant and $M_S$ the solar mass. For circular planetary orbits, the gravitational force is equal to the centripetal force, i.e. It holds that $F_G=m\frac{G{M}_S}{r^2}=m\frac{v^2}{r}=F_Z,$ or $v^2=\frac{G{M}_S}{r}=-\varphi (r).$ (9.2) This corresponds to the virial theorem, according to which for this case $E_{\text{kin}}=-\frac{1}{2}{E}_{\text{pot}}$ holds. For the orbital velocity $v_E$ of the Earth it therefore holds, with $r=r_E=1$ AE: $v_E^2=-\varphi (r_E)=\frac{G{M}_S}{1\text{ AE}}\approx {\left(29,8\text{ km/s}\right)}^2.$ (9.3) The velocity $v_0$ given in the problem statement denotes the velocity that The probe has relative to the Earth when it is still geometrically close to the Earth, but the influence The gravity of the Earth is already negligible. The probe therefore leaves Earth’s orbit with the velocity${}^a$ $v_1=v_0+v_E\approx 9,4\text{ km/s}+29,8\text{ km/s}=39,2\text{ km/s}$ (9.4) and arrives at the Jupiter orbit (orbital radius $r_J=5,2r_E$) with the velocity $v_2$. For This is followed by conservation of energy $E_{\text{kin}}+E_{\text{pot}}=\frac{1}{2}m{v}^2(r)+m\varphi (r)=\text{const.}$ or $\frac{1}{2}{v}_1^2-\frac{G{M}_S}{r_E}=\frac{1}{2}{v}_2^2-\frac{G{M}_S}{r_J}.$ (9.5) Thus $v_2=\sqrt{v_1^2+2\varphi (r_E)\cdot (1-1/5,2)}\approx 10,1\text{ km/s}.$ (9.6) For the investigation of the swing-by manoeuvre, the radial component $v_{2,r}$ and the tangential components $v_{2,\varphi }$ of this velocity relative to the Sun are needed. The conservation of angular momentum $v_{\varphi }r=\text{const.}$ gives $v_{2,\varphi }=v_1/5,2\approx 7,5\text{ km/s}$ and hence $v_{2,r}=\sqrt{v_2^2-v_{2,\varphi }^2}\approx 6,7\text{ km/s}.$ (9.7) ${}^a$If the probe had been launched against the orbital velocity of the Earth, the difference would have to be considered here. But since the probe would not have reached Jupiter then, this case is not relevant. 9.b) Calculations and explanations Jupiter itself moves with an orbital velocity $v_J$ in the tangential direction, for which, with equation (9.2), it holds that: $v_J^2=-\varphi (r_J)\approx (13,1\text{ km/s}{)}^2.$ (9.8) Consider the swing-by maneuver in the rest frame of Jupiter. There the potential is static, So the magnitude of the probe’s velocity is identical before and after The manoeuvre due to energy conservation. It therefore holds that $\sqrt{v_{2,r}^2+(v_{2,\varphi }-v_J{)}^2}=\sqrt{v_{3,r}^2+(v_{3,\varphi }-v_J{)}^2},$ (9.9) where $v_3$ denotes the velocity in the Sun-fixed frame after the swing-by maneuver, and it was used that Jupiter has the same sense of orbital motion around the Sun as the Earth. The left hand side is already known and the magnitude of $v_3$ is to be maximized. But the magnitude of a sum of two vectors of given length is maximal exactly When the vectors are aligned parallel. Therefore the exit velocity $v_3$ is, in the The ideal case, again tangential to the orbit and $v_{3,r}=0$. Equation (9.9) then gives: $v_3=v_{3,\varphi }=v_J+\sqrt{v_{2,r}^2+(v_{2,\varphi }-v_J{)}^2}\approx 21,8\text{ km/s}.$ (9.10) The subsequent orbital velocity follows once again from conservation of energy: $v(r)=\sqrt{v_3^2+2(\varphi (r_J)-\varphi (r))}=\sqrt{v_3^2+2\varphi (r_J)(1-r_J/r)}.$ (9.11) The radial velocity can again be derived from conservation of angular momentum and is $v_r(r)=\sqrt{v_3^2+2\varphi (r_J)(1-r_J/r)}\sqrt{1-(r_J/r{)}^2}\approx \sqrt{(21,8\text{ km/s}{)}^2-2(13,1\text{ km/s}{)}^2(1-5,2\text{ AE}/r)}\sqrt{1-(5,2\text{ AE}/r{)}^2}.$ (9.12) In particular, the limiting velocity of the probe follows from equation (9.12) as $v_r(\infty )=\sqrt{v_3^2+2\varphi (r_J)}\approx \sqrt{(21,8\text{ km/s}{)}^2-2(13,1\text{ km/s}{)}^2}\approx 11,5\text{ km/s}.$ (9.13) With this velocity, Pioneer 10 leaves the solar system. 9.c) Calculations and explanations For the photons of the sunlight, the relativistic energy-momentum relation holds $E=cp.$ (9.14) At a distance $r$ from the Sun, the probe receives a radiation power per area of $\frac{P}{A}=E_0{\left(\frac{1\text{ AE}}{r}\right)}^2,$ (9.15) where $A=\pi d^2/4$ gives the cross-sectional area of the probe. This results in a force on the probe of $F=\dot{p}=P/c$. Here the reflected part of the radiation transfers So for the acceleration $a_{\text{Strahlung}}(r)=\frac{F}{m}=\frac{(2\cdot 0,8+0,2)P}{mc}=\frac{1,8\pi d^2{E}_0}{4mc}{\left(\frac{1\text{ AE}}{r}\right)}^2$ (9.16) The results. Thus the acceleration, expressed through the acceleration at $r=50$ AE, is $a_{\text{Strahlung}}(r)=8,2\cdot 10^{-11}{\text{ m/s}}^2\cdot {\left(\frac{50\text{ AE}}{r}\right)}^2.$ (9.17) The gravitational acceleration due to the Sun, on the other hand, expressed through the acceleration at $r=50$ AE, is $a_G(r)=-\frac{G{M}_S}{(50\text{ AE}{)}^2}$

Topic: Gravitation, Astrophysics, Modern-Quantum Physics Metodi: Newton’s Law of Gravitation, Conservation of Energy, Conservation of Momentum Competenze: Mathematical Modeling, Physical Reasoning, Estimation & Approximation Objects: Satellite, Planet, Star Fonte: Testo (PDF) — p.18