IPhO 2026

Problem 1 Coastal mapping (MC problem) (5.0 pts.) With the help of lasers mounted on aircraft, coastal waters can be mapped. To do this, short laser pulses are emitted in various directions, which are diffusely reflected partly at the water surface and partly at the sea bed. As a result, part of the emitted light returns to the aircraft, where it is detected and analyzed. Consider an aircraft flying at a constant altitude along a stretch of coast. 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 displayed. Assume that the laser pulse strikes the water surface at an angle of incidence of $40^{\circ }$ and that the refractive index of the water is 1.33. Fig. 1. Illustration of the 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 under investigation? A 2.7 m B 3.1 m C 4.1 m D 6.1 m Answer section Calculations and explanations 56. IPhO 2026 - 2nd Round exam Code: Code Calculations and explanations (continued) Correct answer: 56. IPhO 2026 - 2nd Round exam Code: Code

Topic: Geometric Optics Metodi: Snell’s Law, Physical Modeling, Kinematic Equations Competenze: Physical Reasoning, Mathematical Modeling Objects: — Fonte: Testo (PDF) — p.2

Problema 1 Mapping costiero (problema MC) (5,0 p. d.) Con l’aiuto di laser montati su aerei, le acque costiere possono essere

• Mappa. Per fare questo, vengono emessi brevi impulsi laser Le direzioni di questo tipo sono diffuse, 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 analizzata. Considerate un aereo che vola ad altitudine costante lungo un tratto di costa. 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 è visualizzato un segnale intero riflesso. Supponiamo che il laser pulse colpisca la superficie dell’acqua ad un angolo di incidenza di $40^{\circ }$ e che l’indice di rifrazione dell’acqua è di 1,33. Fig. 1. Illocalizzazione del 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 al luogo in esame? A 2.7 m B 3.1 m C 4.1 m D 6.1 m Answer section Calcoli e spiegazioni

56. IPhO 2026 - 2° round Codice: Codice Calcoli e spiegazioni (continuato) Corretta risposta:
57. IPhO 2026 - 2° round Codice: Codice

Topic: Geometric Optics Metodi: Snell’s Law, Physical Modeling, Kinematic Equations Competenze: Physical Reasoning, Mathematical Modeling Objects: — Fonte: Testo (PDF) — p.2

Problem 1 Coastal mapping (MC problem) (5.0 p.p.) With the help of lasers mounted on aircraft, coastal waters can be mapped. To do this, short laser pulses are emitted into The water is reflected in different directions, which are diffusely reflected partly at the water surface and partly at the sea bed. As a result, part of the emitted light returns to the aircraft, where it is detected and analyzed. Consider an aircraft flying at a constant altitude along A stretch of coast. 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 displayed. Assume that the laser pulse strikes the water surface at an angle of incidence of $40^{\circ }$ And that the refractive index of the water is 1.33. Fig. 1. Illustration of the 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 under investigation? A 2.7 m B 3.1 m C 4.1 m D 6.1 m Answer section Calculations and explanations 56. IPhO 2026 - Second Round exam The code: Code Calculations and explanations (continued) Correct answer: 56. IPhO 2026 - Second Round exam The code: Code

Topic: Geometric Optics Metodi: Snell’s Law, Physical Modeling, Kinematic Equations Competenze: Physical Reasoning, Mathematical Modeling Objects: — Fonte: Testo (PDF) — p.2

Problem 2 Atwood machine (MC problem) (5.0 pts.) Professor Atwood has just returned from a conference, and he is already tinkering with his next experiment: He places 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 axis of the pulley fixed and measures the force $F$ with which he must pull it upward in order to hold the pulley with the bodies in place. Now he hangs, as shown in the figure, an additional body of mass $4m$ on the right-hand mass piece. Again he holds the axis fixed. What is the magnitude of the force $F^{\prime }$ with which, in the situation shown, he must pull on the axis in order to hold it in place? A $\frac{5}{3}F$ B $2F$ C $\frac{5}{2}F$ D $3F$ String Pulley $F^{\prime }$ $m$ $m$ $4m$ Answer section Calculations and explanations Correct answer: 56. IPhO 2026 - 2nd Round exam Code: Code

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

Il problema 2 della macchina Atwood (problema MC) (5,0 p. d.) Il professor Atwood è appena tornato da una conferenza, e lui è già Tinkering with his next experiment: Egli mette una corda su un pallone 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. Lui ora tiene l’asse del pallone fixed and measures the force $F$ with which he must pull it upward in order to Tenete il pallone con i corpi in posto. Now he hangs, as shown in the figure, an additional body of mass $4m$ sul pezzo di massa a destra. Ancora una volta, ha il suo asse

• Non è stato fatto. What is the magnitude of the force $F^{\prime }$ with which, in the situation shown, he must Tirare sull’asse per tenerlo in posto? A $\frac{5}{3}F$ B $2F$ C $\frac{5}{2}F$ D $3F$ String Pulley $F^{\prime }$ $m$ $m$ $4m$ Answer section Calcoli e spiegazioni Corretta risposta:

56. IPhO 2026 - 2° round Codice: Codice

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

The following is the list of the types of products which are used in the manufacture of the product: (5.0 p.p.) Professor Atwood has just returned from a conference, and he’s already tinkering with his next experiment: He puts 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 axis of the pulley fixed and measures the force $F$ with which he must pull it upward in order to Hold the pulley with the bodies in place. Now he hangs, as shown in the figure, an additional body of mass $4m$ on the right-hand mass piece. Again he holds the axis I’m not going to lie. What is the magnitude of the force $F^{\prime }$ with which, in the situation shown, he must pull on the axis in order to hold it in place? A $\frac{5}{3}F$ B $2F$ C $\frac{5}{2}F$ D $3F$ String Pulley $F^{\prime }$ $m$ $m$ $4m$ Answer section Calculations and explanations Correct answer: 56. IPhO 2026 - Second Round exam The code: Code

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

Problem 3 Submerging cup (MC problem) (5.0 pts.) A cup of mass $m=200\text{g}$ with a filling volume $V_0=250\text{mL}$ 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 cup wall and base 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 Answer section Calculations and explanations Correct answer: 56. IPhO 2026 - 2nd Round exam Code: Code

Topic: Fluid Mechanics, Thermodynamics Metodi: Hydrostatic Equilibrium, Ideal Gas Law Competenze: Mathematical Modeling, Physical Reasoning Objects: Container, Gas Fonte: Testo (PDF) — p.5

Problema 3 Coppa di immersione (problema MC) (5,0 p. d.) Una coppa di massa $m=200\text{g}$ con un volume di riempimento $V_0=250\text{mL}$ e una altezza $H$ è si è trasformata di testa in giù e submersa in un grande serbatoio di acqua a 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 sia di di coppa, il muro e la base possono essere trascurati. Supponiamo inoltre che La coppa rimane sempre al contrario e non punta al lato. A quali profondità può fluttuare la coppa in equilibrio? Qui, profondità 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 Answer section Calcoli e spiegazioni Corretta risposta: 56. IPhO 2026 - 2° round Codice: Codice

Topic: Fluid Mechanics, Thermodynamics Metodi: Hydrostatic Equilibrium, Ideal Gas Law Competenze: Mathematical Modeling, Physical Reasoning Objects: Container, Gas Fonte: Testo (PDF) — p.5

Problem 3 Submerging cup (MC problem) (5.0 p.p.) A cup of mass $m=200\text{g}$ with a filling volume $V_0=250\text{mL}$ 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 cup wall and base 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 Answer section Calculations and explanations Correct answer: 56. IPhO 2026 - Second Round exam The code: Code

Topic: Fluid Mechanics, Thermodynamics Metodi: Hydrostatic Equilibrium, Ideal Gas Law Competenze: Mathematical Modeling, Physical Reasoning Objects: Container, Gas Fonte: Testo (PDF) — p.5

Problem 4 Rotated capacitor (MC problem) (5.0 pts.) Two capacitors with capacitances $C_1=2,0\text{nF}$ and $C_2=8,0\text{nF}$ are connected in series, as shown in the following figure, to a voltage source with voltage $U_0=10,0\text{V}$: $U_0$ $C_2$ $C_1$ The capacitors are then disconnected from the voltage source, and one of the capacitors is rotated by $180^{\circ }$. The capacitors are now short-circuited: $C_2$ $C_1$ $C_2$ $C_1$ What voltage appears across the rotated capacitor? A 0 V B 1.6 V C 3.2 V D 5.0 V Answer section Calculations and explanations 56. IPhO 2026 - 2nd Round exam Code: Code Calculations and explanations (continued) Correct answer: 56. IPhO 2026 - 2nd Round exam Code: Code

Topic: Electrostatics, Circuits Metodi: Kirchhoff’s Laws, Electric Potential Method Competenze: Physical Reasoning, Mathematical Modeling Objects: Capacitor, Battery Fonte: Testo (PDF) — p.6

Problema 4 Capacitatore rotato (problema MC) (5,0 p. d.) Due condensatori con capacità $C_1=2,0\text{nF}$ e $C_2=8,0\text{nF}$ sono collegati in serie, come mostrato nel seguente: figure, to a voltage source with voltage $U_0=10,0\text{V}$: $U_0$ $C_2$ $C_1$ I condensatori sono quindi disconnessi dalla fonte di voltage, e uno dei condensatori è rotato da $180^{\circ }$. I condensatori sono ora short-circuited: $C_2$ $C_1$ $C_2$ $C_1$ Che voltage appare attraverso il condensatore rotato? A 0 V B 1.6 V C 3.2 V D 5.0 V Answer section Calcoli e spiegazioni 56. IPhO 2026 - 2° round Codice: Codice Calcoli e spiegazioni (continuato) Corretta risposta: 56. IPhO 2026 - 2° round Codice: Codice

Topic: Electrostatics, Circuits Metodi: Kirchhoff’s Laws, Electric Potential Method Competenze: Physical Reasoning, Mathematical Modeling Objects: Capacitor, Battery Fonte: Testo (PDF) — p.6

The following is the list of the types of electrical power generators: (5.0 p.p.) Two capacitors with capacitances $C_1=2,0\text{nF}$ and $C_2=8,0\text{nF}$ are connected in series, as shown in the following Figure, to a voltage source with voltage $U_0=10,0\text{V}$: $U_0$ $C_2$ $C_1$ The capacitors are then disconnected from the voltage source, and one of the capacitors is rotated by $180^{\circ }$. The capacitors are now short-circuited: $C_2$ $C_1$ $C_2$ $C_1$ What voltage appears across the rotated capacitor? A 0 V B 1.6 V C 3.2 V D 5.0 V Answer section Calculations and explanations 56. IPhO 2026 - Second Round exam The code: Code Calculations and explanations (continued) Correct answer: 56. IPhO 2026 - Second Round exam The code: Code

Topic: Electrostatics, Circuits Metodi: Kirchhoff’s Laws, Electric Potential Method Competenze: Physical Reasoning, Mathematical Modeling Objects: Capacitor, Battery Fonte: Testo (PDF) — p.6

Problem 5 Lissajous pendulum (MC problem) (5.0 pts.) A simple pendulum is suspended, as sketched alongside, from two strings that converge in a V-shape. The height of the V is equal to the length $\ell$ of the simple 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 pictures result. Which of the following pictures could have been produced with the pendulum above? $\ell$ $\ell$ Fig. 3. Sketch of the pendulum setup. A B C D Fig. 4. Candidates for traces of the pendulum bob in the horizontal plane. Answer section Calculations and explanations 56. IPhO 2026 - 2nd Round exam Code: Code Calculations and explanations (continued) Correct answer: 56. IPhO 2026 - 2nd Round exam Code: Code

Topic: Oscillations & Waves Metodi: Simple Harmonic Motion Analysis, Physical Modeling Competenze: Physical Reasoning, Diagrammatic Reasoning Objects: Pendulum, String Fonte: Testo (PDF) — p.8

Il problema 5 del pendolo di Lissajous (problema MC) (5,0 p. d.) Un semplice pendolo è sospeso, come disegnato insieme, da due stringhe che convergono in forma di V. La altezza di V è pari alla lunghezza $\ell$ del semplice pendolo 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, belle immagini

• il risultato. Quale delle seguenti immagini potrebbe essere stato prodotto con il pendolo sopra? $\ell$ $\ell$ Fig. 3. Sketch del pendolo di impostazione. A B C D Fig. 4. Candidati per tracce del pendolo bob in piano orizzontale. Answer section Calcoli e spiegazioni

56. IPhO 2026 - 2° round Codice: Codice Calcoli e spiegazioni (continuato) Corretta risposta:
57. IPhO 2026 - 2° round Codice: Codice

Topic: Oscillations & Waves Metodi: Simple Harmonic Motion Analysis, Physical Modeling Competenze: Physical Reasoning, Diagrammatic Reasoning Objects: Pendulum, String Fonte: Testo (PDF) — p.8

The following is the list of the problems: (5.0 p.p.) A simple pendulum is suspended, as sketched alongside, from two Strings that converge in a V-shape. The height of the V is equal to the length $\ell$ of the simple 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, beautiful pictures The result. Which of the following pictures could have been produced with the pendulum above? $\ell$ $\ell$ Fig. 3. Sketch of the pendulum setup. A B C D Fig. 4. Candidates for traces of the pendulum bob in the horizontal plane. Answer section Calculations and explanations 56. IPhO 2026 - Second Round exam The code: Code Calculations and explanations (continued) Correct answer: 56. IPhO 2026 - Second Round exam The code: Code

Topic: Oscillations & Waves Metodi: Simple Harmonic Motion Analysis, Physical Modeling Competenze: Physical Reasoning, Diagrammatic Reasoning Objects: Pendulum, String Fonte: Testo (PDF) — p.8

Problem 6 Freezing water (MC problem) (5.0 pts.) Using a heat pump with an electrical power of 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 $334{\text{kJ kg}}^{-1}$. What minimum time is in any case required for the freezing? A about 6 min B about 11 min C about 15 min D about 20 min Answer section Calculations and explanations Correct answer: 56. IPhO 2026 - 2nd Round exam Code: Code

Topic: Thermodynamics Metodi: Thermodynamic Cycle Analysis, First Law of Thermodynamics Competenze: Mathematical Modeling, Physical Reasoning Objects: Heat Engine, Container Fonte: Testo (PDF) — p.10

Problema 6 Acqua di congelatore (problema MC) (5,0 p. d.) Usare una pompa di calore con una potenza elettrica di 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 $334{\text{kJ kg}}^{-1}$. 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 Answer section Calcoli e spiegazioni Corretta risposta: 56. IPhO 2026 - 2° round Codice: Codice

Topic: Thermodynamics Metodi: Thermodynamic Cycle Analysis, First Law of Thermodynamics Competenze: Mathematical Modeling, Physical Reasoning Objects: Heat Engine, Container Fonte: Testo (PDF) — p.10

The following is the list of the main problems: (5.0 p.p.) Using a heat pump with an electrical power of 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 $334{\text{kJ kg}}^{-1}$. What minimum time is required for the freezing? A about 6 min B about 11 min C about 15 minutes D about 20 minutes Answer section Calculations and explanations Correct answer: 56. IPhO 2026 - Second Round exam The code: Code

Topic: Thermodynamics Metodi: Thermodynamic Cycle Analysis, First Law of Thermodynamics Competenze: Mathematical Modeling, Physical Reasoning Objects: Heat Engine, Container Fonte: Testo (PDF) — p.10

Problem 7 Kaon decay (MC problem) (5 pts.) A kaon moving with a speed of 0.80 c, that is, 80 % of the speed of light, in the laboratory frame decays into two pions, which afterward move along and opposite to the original direction of motion of the kaon, respectively. No further particles are produced by the decay. For the rest energies $E_{0,K}$ and $E_{0,\pi }$ of the kaon and the pion, respectively, the relation holds $\kappa =\frac{E_{0,K}}{E_{0,\pi }}=3,68.$ With what speeds do the two pions move after the decay in the laboratory frame? A $-0,039$ c and 0.99 c B $-0,12$ c and 0.98 c C $-0,19$ c and 0.92 c D $-0,21$ c and 0.89 c Answer section Calculations and explanations Correct answer: 56. IPhO 2026 - 2nd Round exam Code: Code Long-answer problems Work on the following two problems also in the boxes provided for them. Unlike the multiple-choice problems, no answer options are given. Describe your solution path so that it is easy to follow but not unnecessarily long. So if, for example, you use the law of conservation of energy, write this down briefly.

Topic: Special Relativity, Nuclear & Particle Physics Metodi: Relativistic Energy-Momentum, Conservation Laws Competenze: Mathematical Modeling, Physical Reasoning Objects: — Fonte: Testo (PDF) — p.11

Problema 7 decadimento del caon (problema MC) (cfr. Un kaon che si muove a una velocità di 0,80 c, cioè l’80% della velocità della luce, nel laboratorio decade in due pioni, che successivamente si muovono lungo e opposto alla direzione originale di movimento del kaon, rispettivamente. Non ci sono ulteriori particelle prodotte dal decadimento. Per il rest energies $E_{0,K}$ and $E_{0,\pi }$ of the kaon and the pion, respectively, the relation holds $\kappa =\frac{E_{0,K}}{E_{0,\pi }}=3,68.$ Con che velocità i due pioni si muovono dopo il decadimento nel laboratorio? A $-0,039$ c e 0,99 c B $-0,12$ c e 0,98 c C $-0,19$ c and 0,92 c D $-0,21$ c e 0,89 c Answer section Calcoli e spiegazioni Corretta risposta: 56. IPhO 2026 - 2° round Codice: Codice Problemi di risposta lunga La Commissione ha inoltre presentato una serie di proposte di risoluzione sulle misure di sicurezza e di sicurezza. A differenza dei problemi di scelta multipla, non sono state indicate le opzioni di risposta. Descrivere il tuo percorso di soluzione in questo modo che è facile da seguire ma non troppo lungo. Quindi se, per esempio, si usa la legge della conservazione dell’energia, scrivete brevemente.

Topic: Special Relativity, Nuclear & Particle Physics Metodi: Relativistic Energy-Momentum, Conservation Laws Competenze: Mathematical Modeling, Physical Reasoning Objects: — Fonte: Testo (PDF) — p.11

The problem is that the average rate of decay is not the same as the average rate of decay. (five points) A kaon moving at a speed of 0.80 c, that is, 80 percent of the speed of light, in the laboratory frame decays into two pions, which afterwards move along and opposite to the original direction of motion of the kaon, respectively. No further particles are produced by the decay. For the rest energies $E_{0,K}$ and $E_{0,\pi }$ of the kaon and the pion, respectively, the relation holds $\kappa =\frac{E_{0,K}}{E_{0,\pi }}=3,68.$ With what speeds do the two pions move after the decay in the lab frame? A $-0,039$ c and 0.99 c B $-0,12$ c and 0.98 c C $-0,19$ c and 0.92 c D $-0,21$ c and 0.89 c Answer section Calculations and explanations Correct answer: 56. IPhO 2026 - Second Round exam The code: Code Long-response problems Work on the following two problems also in the boxes provided for them. Unlike the Multiple-choice problems, no answer options are given. Describe your solution path like this That it’s easy to follow but not unnecessarily long. So if, for example, you use the law of conservation of energy, write this down briefly.

Topic: Special Relativity, Nuclear & Particle Physics Metodi: Relativistic Energy-Momentum, Conservation Laws Competenze: Mathematical Modeling, Physical Reasoning Objects: — Fonte: Testo (PDF) — p.11

Problem 8 Pencil lead (17.0 pts.) A pencil lead slides, in a vertical homogeneous magnetic field of magnetic flux density $B$, frictionlessly down two parallel, ideally conducting metal rails inclined at an angle $\alpha =30^{\circ }$ to the horizontal. The spacing of the rails is $L=10\text{cm}$ and the overhang of the lead beyond the metal rails can be neglected. An ideal voltage source of voltage $U_0=10\text{mV}$ with a switch is connected to the rails. The lead, rails, switch, and voltage source together form a circuit. The setup is sketched in Figure 5. Fig. 5. Sketch of the sliding pencil lead When the switch is closed, the pencil lead remains at rest. When the switch is opened, the lead continues to slide. Use for the resistivity of the lead the value ${\rho }_{el}=5,0\cdot 10^{-6}\Omega \text{m}$ and for the density $\rho =2,3\cdot 10^3{\text{kg m}}^{-3}$. 8.a) State whether the front or the rear rail in the figure is connected to the positive pole 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 unit check. (8.0 pts.) 8.c) Determine the steady-state speed $v_H$ with which the pencil lead slides down the incline when the flux density of the magnetic field is halved. Carry out a unit check for the result. (7.0 pts.) 56. IPhO 2026 - 2nd Round exam Code: Code Answer section 8.a) Calculations and explanations 56. IPhO 2026 - 2nd Round exam Code: Code 8.b) Calculations and explanations Expression and value for the magnetic flux density with unit check: 56. IPhO 2026 - 2nd Round exam Code: Code 8.c) Calculations and explanations Expression and value for the steady-state speed with unit check: 56. IPhO 2026 - 2nd Round exam Code: Code

Topic: Electromagnetic Induction, Newtonian Mechanics Metodi: Faraday’s Law of Induction, Lorentz Force Analysis, Free-Body Diagram Competenze: Mathematical Modeling, Physical Reasoning Objects: Rod, Wire, Battery, Switch Fonte: Testo (PDF) — p.12

Problema 8 Pennale lead (P. 17,0 p. Un matita lead slides, in un campo magnetico verticale omogeneo di densità di flusso magnetico $B$, frictionlessly down two parallel, idealmente con un’angolazione di $\alpha =30^{\circ }$ verso l’orizzontale. Il spaziamento dei binari è $L=10\text{cm}$ e l’overhang di Lead Beyond the Metal Rails può essere trascurato. Per il meglio voltage source of voltage $U_0=10\text{mV}$ with a switch is connected to the rails. Il lead, i rails, Smutare, e la fonte di tensione insieme formare un circuito. La struttura è schizzata in Figura 5. Fig. 5. Sketch of the sliding pencil lead Quando il cambio è chiuso, il lead della matita resta a riposo. Quando il switch è aperto, il Lead continua a slide. Use for the resistivity of the lead the value ${\rho }_{el}=5,0\cdot 10^{-6}\Omega \text{m}$ and for the density $\rho =2,3\cdot 10^3{\text{kg m}}^{-3}$. 8. (a) Indicare se la prima o la seconda linea della figura è collegata al polo 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 assegno unitario. (8,0 pts.) 8.c) Determina la velocità di stato staady $v_H$ con cui il matitale conduce Slide down the incline quando la densità di flusso del campo magnetico è dimezzata. Conduire un controllo unitario per il risultato. (7,0 p.s.) 56. IPhO 2026 - 2° round Codice: Codice Answer section 8.a) Calcoli e spiegazioni 56. IPhO 2026 - 2° round Codice: Codice 8.b) Calcoli e spiegazioni Espressione e valore per la densità del flusso magnetico con controllo unitario: 56. IPhO 2026 - 2° round Codice: Codice 8.c) Calcoli e spiegazioni Espressione e valore per la velocità di stato stabile con controllo unitario: 56. IPhO 2026 - 2° round Codice: Codice

Topic: Electromagnetic Induction, Newtonian Mechanics Metodi: Faraday’s Law of Induction, Lorentz Force Analysis, Free-Body Diagram Competenze: Mathematical Modeling, Physical Reasoning Objects: Rod, Wire, Battery, Switch Fonte: Testo (PDF) — p.12

Problem 8 Pencil lead The Commission has also adopted a proposal for a directive on the protection of workers’ rights. A pencil lead slides, in a vertical homogeneous magnetic field of magnetic flux density $B$, frictionlessly down two parallel, ideally conducting metal rails inclined at an angle $\alpha =30^{\circ }$ to the horizontal. The spacing of the rails is $L=10\text{cm}$ and the overhang Of the lead beyond the metal rails can be neglected. The ideal voltage source of voltage $U_0=10\text{mV}$ with a switch is connected to the rails. The lead, the rails, switch, and voltage source together form a circuit. The setup is sketched in Figure 5. Fig. 5. Sketch of the sliding pencil lead When the switch is closed, the pencil lead remains at rest. When the switch is opened, the Lead continues to slide. Use for the resistivity of the lead the value ${\rho }_{el}=5,0\cdot 10^{-6}\Omega \text{m}$ and for the density $\rho =2,3\cdot 10^3{\text{kg m}}^{-3}$. 8. (a) State whether the front or rear rail in the figure is connected to the positive pole 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 steady-state speed $v_H$ with which the pencil lead slides down the slope when the flux density of the magnetic field is halved. Carry out a unit check for The result. (7.0 pts) 56. IPhO 2026 - Second Round exam The code: Code Answer section 8.a) Calculations and explanations 56. IPhO 2026 - Second Round exam The code: Code 8.b) Calculations and explanations Expression and value for the magnetic flux density with unit check: 56. IPhO 2026 - Second Round exam The code: Code 8.c) Calculations and explanations Expression and value for the steady-state speed with unit check: 56. IPhO 2026 - Second Round exam The code: Code

Topic: Electromagnetic Induction, Newtonian Mechanics Metodi: Faraday’s Law of Induction, Lorentz Force Analysis, Free-Body Diagram Competenze: Mathematical Modeling, Physical Reasoning Objects: Rod, Wire, Battery, Switch Fonte: Testo (PDF) — p.12

Problem 9 Swing-by maneuver and Pioneer anomaly (18.0 pts.) The space probe Pioneer 10 was launched in 1972 to explore the outer solar system and was intended 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 gravitational field of the Earth with a speed of $v_0=9,4{\text{km s}}^{-1}$ relative to the Earth and tangential to its orbit. From there the probe flew toward Jupiter in order to perform a swing-by maneuver that was to enable it to leave the solar system. 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 can 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 $r_J=5,2\text{AE}$ Mass of the Sun $M_S=1,99\cdot 10^{30}\text{kg}$ Solar constant (power of the solar radiation at the Earth’s orbit) $E_0=1367{\text{W m}}^{-2}$ 9.a) Determine the orbital speed $v_2$ with which the probe arrived at the orbit of the planet Jupiter. Assume 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 maneuver through Jupiter’s gravitational field, the space probe was able to increase its speed relative to the Sun considerably. Assume that this swing-by maneuver was carried out in such a way that the probe can leave the solar system with the highest possible speed. You may also assume that the time during which Pioneer 10 interacts gravitationally with Jupiter is small compared with the orbital period and that during the maneuver the gravitational force between the probe and Jupiter is dominant. 9.b) Determine the radial speed $v_r(r)$ and in particular the limiting speed $v_r(\infty )$ of the probe after the swing-by maneuver as a function of the distance $r$ to the Sun. (6.0 pts.) After the probe in 1980, at 20 AE, had moved far enough away from the Sun to be able 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 can be assumed to be 241 kg, is, after gravitation, caused by the radiation pressure from 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 what acceleration results from the radiation pressure of the Sun on the probe at a distance $r\gg 1\text{AE}$ from the Sun. State how large the contribution was in 1990 for $r=50\text{AE}$ and compare it with the acceleration due to gravitation in the gravitational field of the Sun. (6.0 pts.) 56. IPhO 2026 - 2nd Round exam Code: Code The observed, unexplained acceleration was indeed of a similar order of magnitude, but pointed toward the Sun. Therefore further effects such as the solar wind, the influence of other celestial bodies, the recoil from the radio module, and above all non-isotropic thermal emission were also investigated, which in the end, apart from deviations within the uncertainties, were able to provide an explanation for the Pioneer anomaly. Answer section 9.a) Calculations and explanations Result for the orbital speed and its components: 56. IPhO 2026 - 2nd Round exam Code: Code 9.b) Calculations and explanations Expression for the radial speed and result for the limiting speed: 56. IPhO 2026 - 2nd Round exam Code: Code 9.c) Calculations and explanations Result for the acceleration due to radiation pressure and comparison with the gravitational acceleration: 56. IPhO 2026 - 2nd Round exam Code: Code Additional worksheet 56. IPhO 2026 - 2nd Round exam Code: Code Additional worksheet 56. IPhO 2026 - 2nd Round exam Code: Code Additional worksheet Graph paper

Topic: Gravitation, Astrophysics Metodi: Conservation Laws, Kepler’s Laws, Newton’s Law of Gravitation, Energy Conservation Method Competenze: Mathematical Modeling, Estimation & Approximation Objects: Satellite, Planet, Star Fonte: Testo (PDF) — p.16

Problematica 9 Swing-by manovra e anomalia di pioniere (8,0 pts.) 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 lasciò il campo gravitazionale della Terra con una velocità di $v_0=9,4{\text{km s}}^{-1}$ relativa alla Terra e tangenziale alla sua orbita. Da lì il test volava verso Giove per eseguire una manovra di swing-by che ha permettendole di 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 $r_J=5,2\text{AE}$ Mass of the Sun $M_S=1,99\cdot 10^{30}\text{kg}$ Costante solare (potenza della radiazione solare in orbita terrestre) $E_0=1367{\text{W m}}^{-2}$ 9.a) Determine the orbital speed $v_2$ 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 la sua velocità rispetto al sole in modo considerevole. Supponiamo che questo swing-by manovra cosa La sonda può lasciare il sistema solare con la massima velocità possibile. Potresti anche supporre che il tempo durante il quale Pioneer 10 interagisce gravitationally with Jupiter is small compared with the orbital period e che durante la manovra 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 $v_r(\infty )$ di questa sonda dopo la manovra di swing-by a funzione della distanza $r$ al Sole. (6,0 p.p.) Dopo la sonda nel 1980, a 20 AE, si era spostata abbastanza lontano dal Sole per essere in grado di prevedere l’influenza di varie forze con sufficiente precisione, in modo inspiegabile, 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 presunto essere di 241 kg, è, dopo gravità, causata dalla pressione delle radiazioni da parte della radiazione solare. L’antenna parabolica, che ha un diametro di $d=2,77\text{m}$ e a più grandi distanze punti circa verso il Il sole, assorbe il 20% della luce solare e riflette il resto verso il sole. 9.c) Determina approssimativamente quali sono i risultati dell’accelerazione derivanti dalla pressione di radiazione del Sole on the probe at a distance $r\gg 1\text{AE}$ from the Sun. Stat how large the contribution In particolare, il tasso di crescita è stato aumentato nel 1990 per $r=50\text{AE}$ e comparato con l’accelerazione dovuta a gravità nel campo gravitazionale del Sole. (6,0 p.p.) 56. IPhO 2026 - 2° round Codice: Codice L’accelerazione osservata, inesplicata, era indeed of a similar order of magnitude, but pointed verso il sole. Pertanto, altri 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 sono state anche investigate, che alla fine, a parte le deviazioni all’interno delle incertezze, sono state in grado di fornire un’efficacia di spiegazione dell’anomalia del pioniere. Answer section 9.a) Calcoli e spiegazioni Result for the orbital speed and its components: 56. IPhO 2026 - 2° round Codice: Codice 9.b) Calcoli e spiegazioni Espressione per la velocità radial e risultato per la velocità limitante: 56. IPhO 2026 - 2° round Codice: Codice 9.c) Calcoli e spiegazioni Risultato per l’accelerazione dovuta alla pressione di radiazione e confronto con l’accelerazione gravitazionale: 56. IPhO 2026 - 2° round Codice: Codice Ulteriori fogli di lavoro 56. IPhO 2026 - 2° round Codice: Codice Ulteriori fogli di lavoro 56. IPhO 2026 - 2° round Codice: Codice Ulteriori fogli di lavoro Carta grafica

Topic: Gravitation, Astrophysics Metodi: Conservation Laws, Kepler’s Laws, Newton’s Law of Gravitation, Energy Conservation Method Competenze: Mathematical Modeling, Estimation & Approximation Objects: Satellite, Planet, Star Fonte: Testo (PDF) — p.16

Problem 9 Swing-by maneuver and pioneer anomaly The Commission has also adopted a proposal for a directive on the protection of workers’ rights. The space probe Pioneer 10 was launched in 1972 to explore the outer solar system and the Earth. intended 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 the gravitational field of the Earth with a speed of $v_0=9,4{\text{km s}}^{-1}$ relative to the Earth and tangential to its orbit. From there the probe flew Towards Jupiter in order to perform a swing-by maneuver that was to enable it to leave the solar system. 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 can 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 $r_J=5,2\text{AE}$ Mass of the Sun $M_S=1,99\cdot 10^{30}\text{kg}$ Solar constant (power of solar radiation at Earth’s orbit) $E_0=1367{\text{W m}}^{-2}$ 9. (a) Determine the orbital speed $v_2$ with which the probe arrived at the orbit of the planet Jupiter is here. Assume 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 speed relative to the Sun considerably. Assume that this swing-by maneuver was The probe can leave the solar system at the highest speed possible. You may also assume that the time during which Pioneer 10 interacts gravitationally with Jupiter is small compared to the orbital period and that during the maneuver the The gravitational force between the probe and Jupiter is dominant. 9.b) Determine the radial speed $v_r(r)$ and in particular the limiting speed $v_r(\infty )$ of the probe after the swing-by maneuver as a function of the distance $r$ to the Sun. (6.0 pts) After the probe in 1980, at 20 AE, had moved far enough away from the Sun to be able to predict 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 can be assumed to be 241 kg, is, after gravity, caused by the radiation pressure from 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 what acceleration results from the radiation pressure of the Sun on the probe at a distance $r\gg 1\text{AE}$ from the Sun. State how large the contribution was in 1990 for $r=50\text{AE}$ and compare it with the acceleration due to gravity in the gravitational field of the Sun. (6.0 pts) 56. IPhO 2026 - Second Round exam The code: Code The observed, unexplained acceleration was indeed of a similar order of magnitude, but pointed towards the sun. Therefore further effects 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 investigated emissions which, apart from deviations within the uncertainties, were able to provide an explanation for the pioneer anomaly. Answer section 9.a) Calculations and explanations Result for the orbital speed and its components: 56. IPhO 2026 - Second Round exam The code: Code 9.b) Calculations and explanations Expression for the radial speed and result for the limiting speed: 56. IPhO 2026 - Second Round exam The code: Code 9.c) Calculations and explanations Result for the acceleration due to radiation pressure and comparison with the gravitational acceleration: 56. IPhO 2026 - Second Round exam The code: Code Additional worksheet 56. IPhO 2026 - Second Round exam The code: Code Additional worksheet 56. IPhO 2026 - Second Round exam The code: Code Additional worksheet Graph paper

Topic: Gravitation, Astrophysics Metodi: Conservation Laws, Kepler’s Laws, Newton’s Law of Gravitation, Energy Conservation Method Competenze: Mathematical Modeling, Estimation & Approximation Objects: Satellite, Planet, Star Fonte: Testo (PDF) — p.16