IPhO 2023

Problem 1 Sinking Body (MC problem) (5 pts.) (Idea: Problem group of the PhysicsOlympiad - Stefan Petersen) When a solid body with density 1.80 g ${\text{cm}}^{-3}$ sinks at constant velocity in viscous oil of density 0.90 g ${\text{cm}}^{-3}$, then … A … no weight force acts on the body. B … the mass of the body equals the mass of the displaced fluid. C … the weight of the body is in equilibrium with the friction force. D … the buoyant force on the body equals the friction force. Solution Calculations and explanations In the fluid, the weight force, the buoyant force and the friction force act on the body. When sinking at constant velocity, these forces add up to zero. Answer A is wrong, since a weight force also acts on the body in the fluid. Answer B is wrong, since then the downward weight force and the buoyant force would cancel, so that the body would not sink because of friction. Answer C is wrong, since then the downward weight force and the upward friction force would cancel, so that the body would decelerate because of the buoyant force and ultimately rise upward. Answer D is correct: When the density of the body is twice as large as the density of the fluid, the buoyant force corresponds to half the weight force. To establish the force equilibrium required for a constant sinking velocity, the friction force must then compensate the other half of the weight force and be equally large as the buoyant force. Correct answer: D Grading - Sinking Body (MC problem) Points 1 Using the three acting forces 1.0 Recognizing the force equilibrium at constant sinking velocity 1.0 Correctly comparing the forces in the answer options 1.0 Stating the correct solution 2.0 5.0 53rd IPhO 2023 - 2nd Round Exam - Sample Solution - 07.12.2022

Topic: Fluid Mechanics, Newtonian Mechanics Metodi: Free-Body Diagram, Physical Modeling, Hydrostatic Equilibrium Competenze: Physical Reasoning, Diagrammatic Reasoning Objects: — Fonte: Testo (PDF) — p.2

Problema 1 Sinking Body (problema MC) (cfr. (idea: gruppo problematico della PhysicsOlympiad - Stefan Petersen) Quando un corpo solido con densità di 1,80 g ${\text{cm}}^{-3}$ si scende a velocità costante in olio viscoso di densità di 0,90 g ${\text{cm}}^{-3}$, allora … A … Non c’è peso che agisce sul corpo. B … la massa del corpo è uguale alla massa del liquido spostato. C … il peso del corpo è in equilibrio con la forza di attrito. D … La forza buoyant sul corpo è uguale alla forza di attrito. Soluzione Calcoli e spiegazioni Nel fluido, la forza di peso, la forza buoyant e la forza di attrito agiscono sul corpo. Quando Sanguando a velocità costante, queste forze si aggiungono a zero. Risposta A è sbagliata, perché una forza di peso agisce anche sul corpo nel fluido. Risposta B è sbagliata, dal momento che la forza di peso in discesa e la forza buoyant cancellerebbe, in modo che il corpo non affondava a causa della frizione. Risposta C è sbagliata, poiché allora la forza di peso verso il basso e la forza di attrito verso l’alto cancelleranno, così che il corpo decelererà a causa della forza buoyant e alla fine salire verso l’alto. Risposta D è corretta: quando la densità del corpo è doppia della densità del fluido, La forza di buoyant corrisponde a metà della forza di peso. Per stabilire l’equilibrio di forza richiesto Per una velocità di sinking costante, la forza di attrito deve quindi compensare l’altra. metà della forza di peso e essere uguale alla forza di buoyant. Risposta corretta: D Grading - Sinking Body (problema MC) Punti 1 Usando le tre forze che agiscono 1.0 Recognizing the force equilibrium at constant sinking velocity 1.0 Corretamente confrontando le forze nelle opzioni di risposta 1.0 Stating the correct solution 2.0 5.0 53° IPhO 2023 - 2° Round Exam - Sample Solution - 07.12.2022

Topic: Fluid Mechanics, Newtonian Mechanics Metodi: Free-Body Diagram, Physical Modeling, Hydrostatic Equilibrium Competenze: Physical Reasoning, Diagrammatic Reasoning Objects: — Fonte: Testo (PDF) — p.2

The problem is that the body is sinking. (five points) (Idea: Problem group of the PhysicsOlympic - Stefan Petersen) When a solid body with density 1.80 g ${\text{cm}}^{-3}$ sinks at constant velocity in viscous oil of density 0.90 g ${\text{cm}}^{-3}$, then … A … No weight force acts on the body. B … The mass of the body is equal to the mass of the displaced fluid. C … The weight of the body is in equilibrium with the friction force. D … The booyant force on the body equals the friction force. The solution Calculations and explanations In the fluid, the weight force, the buoyant force and the friction force act on the body. When Sinking at constant velocity, these forces add up to zero. Answer A is wrong, since a weight force also acts on the body in the fluid. Answer B is wrong, since then the downward weight force and the buoyant force would cancel, so that the body would not sink because of friction. Answer C is wrong, since then the downward weight force and the upward friction force would cancel, so that the body would decelerate because of the buoyant force And ultimately rise upward. Answer D is correct: When the density of the body is twice as large as the density of the fluid, The buoyant force corresponds to half the weight force. To establish the force balance required For a constant sinking velocity, the friction force must then compensate the other half the weight force and be as large as the buoyant force. Correct answer: D Grading - Sinking body (MC problem) Points 1 Using the three acting forces 1.0 Recognizing the force equilibrium at constant sinking velocity 1.0 Correctly comparing the forces in the answer options 1.0 Stating the correct solution 2.0 5.0 The Commission shall take into account the following information:

Topic: Fluid Mechanics, Newtonian Mechanics Metodi: Free-Body Diagram, Physical Modeling, Hydrostatic Equilibrium Competenze: Physical Reasoning, Diagrammatic Reasoning Objects: — Fonte: Testo (PDF) — p.2

Problem 2 Induction in Conducting Loops (MC problem) (5 pts.) (Idea: Problem group of the PhysicsOlympiad - Stefan Petersen) The four conducting loops (a to d) shown in the figure each have edge lengths $l$ or $2l$. They move at constant velocity $v$ into a sharply bounded region containing a homogeneous magnetic field of flux density $B$, oriented into the plane of the drawing. How do the voltages $U_a$ to $U_d$ induced in the loops directly upon entering the region with the magnetic field relate to one another? A $|U_a|=|U_b|=|U_c|=|U_d|$ B $|U_a|<|U_b|<|U_c|<|U_d|$ C $|U_a|=|U_b|<|U_c|=|U_d|$ D $|U_a|<|U_b|=|U_c|<|U_d|$ Region with magnetic field $B$ $v$ d $v$ c $v$ b $v$ a $2l$ $l$ Solution Calculations and explanations The induced voltage is proportional to the change of the magnetic flux through the conducting loop. Since the magnetic field is homogeneous, this is in turn proportional to the change of the area that is in the region with the magnetic field. Directly after entering this region, the induced voltage is therefore proportional to the height of the conducting loop in the plane of the drawing, and the width of the conducting loop plays no role. Therefore the voltages induced in loops a and b are equal and smaller in magnitude than the voltages induced in loops c and d, which are likewise equal to one another. Correct answer: C Note: Alternatively, the problem can also be solved by considering the Lorentz force on a charge in the leading conductor edge. Consider a charge $q$ that moves with the conductor edge at a velocity $v$ into the region with the magnetic field. There it experiences a Lorentz force $F=qvB$ parallel to the conductor edge. Between the two leading corners of the conducting loop, a potential difference of $U=Fl/q=vBl$ or $U=F2l/q=2vBl$ thus arises, which leads to the same answer option. Grading - Induction in Conducting Loops (MC problem) Points 2 Recognizing that the induced voltage is proportional to the flux change 1.0 Recognizing that the flux change is proportional to the area change in the magnetic field region 1.0 Recognizing that therefore only the height of the loop plays a role 1.0 Stating the correct solution 2.0 5.0 53rd IPhO 2023 - 2nd Round Exam - Sample Solution - 07.12.2022

Quattro spire conduttrici in campo magnetico

Topic: Electromagnetic Induction, Magnetism Metodi: Faraday’s Law of Induction, Lorentz Force Analysis, Physical Modeling Competenze: Physical Reasoning, Diagrammatic Reasoning Objects: Coil Fonte: Testo (PDF) — p.3

Problema 2 Induzione nei circuiti di condotta (problema MC) (cfr. (idea: gruppo problematico della PhysicsOlympiad - Stefan Petersen) I quattro circuiti di conduzione (a a d) mostrati nella figura ogni singolo gruppo ha lunghezze di bordo $l$ o $2l$. Si muovono a velocità costante $v$ into a sharply bounded regione contenente un campo magnetico omogeneo di densità di flusso $B$, orientato verso il piano del disegno. Come fare i voltaggi $U_a$ to $U_d$ induciuti nei loops direttamente quando si entra nella regione con il campo magnetico Relatatevi? A $|U_a|=|U_b|=|U_c|=|U_d|$ B $|U_a|<|U_b|<|U_c|<|U_d|$ C $|U_a|=|U_b|<|U_c|=|U_d|$ D $|U_a|<|U_b|=|U_c|<|U_d|$ Regione con campo magnetico $B$ $v$ d $v$ c $v$ b $v$ a $2l$ $l$ Soluzione Calcoli e spiegazioni La tensione indotta è proporzionale al cambiamento del flusso magnetico attraverso il loop di condotta. Poiché il campo magnetico è omogeneo, questo è a sua volta proporzionale al cambiamento dell’area che è nella regione con il campo magnetico. Directly after entering this region, the induced voltage is therefore proportional to the height of the conducting loop in the plane of the drawing, and the width of the drawing. di cui il loop di conduzione non ha alcun ruolo. Pertanto, le tensioni inducite nei cicli a e b sono uguali e minori in magnitudine rispetto alle tensioni inducite nei cicli c e d, che sono uguali l’uno all’altro. Corretta risposta: C Nota: In alternativa, il problema può anche essere risolto considerando la forza di Lorentz su una carica in the leading conductor edge. Consider a charge $q$ that moves with the conductor edge a velocità $v$ nella regione con il campo magnetico. Lì si sperimenta a forza di Lorentz $F=qvB$ parallela all’edge del conduttore. Tra i due principali angoli del • un’interferenza di valore di $U=Fl/q=vBl$ o $U=F2l/q=2vBl$ così si presenta, che porta alla stessa risposta opzione. Grading - Induction in conducting loops (problema MC) Punti 2 Riconoscendo che la tensione indotta è proporzionale al cambiamento del flusso 1.0 Riconoscendo che il cambiamento del flusso è proporzionale al cambiamento di area nella regione del campo magnetico 1.0 Riconoscendo che quindi solo l’altezza del loop gioca un ruolo 1.0 Stating the correct solution 2.0 5.0 53° IPhO 2023 - 2° Round Exam - Sample Solution - 07.12.2022

*Quattro spire conduttrici in campo magnetico *

Topic: Electromagnetic Induction, Magnetism Metodi: Faraday’s Law of Induction, Lorentz Force Analysis, Physical Modeling Competenze: Physical Reasoning, Diagrammatic Reasoning Objects: Coil Fonte: Testo (PDF) — p.3

The following is the list of the types of induction in conducting loops: (five points) (Idea: Problem group of the PhysicsOlympic - Stefan Petersen) The four conducting loops (a to d) shown in the figure Each have edge lengths $l$ or $2l$. They move at constant velocity $v$ into a sharply bounded region containing a homogeneous magnetic field of flux density $B$, oriented into the plane of the drawing. How do the voltages $U_a$ to $U_d$ induced in the loops directly upon entering the region with the magnetic field relate to each other? A $|U_a|=|U_b|=|U_c|=|U_d|$ B $|U_a|<|U_b|<|U_c|<|U_d|$ C $|U_a|=|U_b|<|U_c|=|U_d|$ D $|U_a|<|U_b|=|U_c|<|U_d|$ Region with magnetic field $B$ $v$ d $v$ c $v$ b $v$ a $2l$ $l$ The solution Calculations and explanations The induced voltage is proportional to the change of the magnetic flux through the conducting loop. Since the magnetic field is homogeneous, this is in turn proportional to the change of the area That’s in the region with the magnetic field. Directly after entering this region, the induced voltage is therefore proportional to the height of the conducting loop in the plane of the drawing, and the width of the drawing. of the conducting loop plays no role. Therefore the voltages induced in loops a and b are equal and smaller in magnitude than the voltages induced in loops c and d, which are are equal to each other. Correct answer: C Note: Alternatively, the problem can also be solved by considering the Lorentz force on a charge in the leading conductor edge. Consider a charge $q$ that moves with the conductor edge at a velocity $v$ into the region with the magnetic field. There it experiences a Lorentz force $F=qvB$ parallel to the conductor edge. Between the two leading corners of the conducting loop, a potential difference of $U=Fl/q=vBl$ or $U=F2l/q=2inBl$ thus arises, which leads to the same answer option. Grading - Induction in conducting loops (MC problem) Points 2 Recognizing that the induced voltage is proportional to the flux change 1.0 Recognizing that the flux change is proportional to the area change in the magnetic field region 1.0 Recognizing that therefore only the height of the loop plays a role 1.0 Stating the correct solution 2.0 5.0 The Commission shall take into account the following information:

Quattro spire conduttrici in campo magnetico

Topic: Electromagnetic Induction, Magnetism Metodi: Faraday’s Law of Induction, Lorentz Force Analysis, Physical Modeling Competenze: Physical Reasoning, Diagrammatic Reasoning Objects: Coil Fonte: Testo (PDF) — p.3

Problem 3 Resistor Heating (MC problem) (5 pts.) (Idea: Problem group of the PhysicsOlympiad - Bernd Schade & Stefan Petersen) Two resistors of the same construction are connected in parallel to a voltage source with a voltage of 2.6 V. A total current of 310 mA flows. With an infrared camera, the adjacent image of the circuit is taken. After calibration, the camera can also determine the surface temperatures of the two resistors. They are $33{}^{\circ }\text{C}$ and $67{}^{\circ }\text{C}$. The ambient temperature is $21{}^{\circ }\text{C}$. What values do the two resistors have approximately? Fig. 1. Infrared image of the resistors. A 1.7 $\Omega$ and 6.7 $\Omega$ B 12 $\Omega$ and 30 $\Omega$ C 10 $\Omega$ and 45 $\Omega$ D 20 $\Omega$ and 80 $\Omega$ Solution Calculations and explanations The electrical power $P_{\text{el}}$ dissipated in a resistor with resistance value $R$ follows from the voltage $U$ dropping across the resistor and the current $I$ flowing through the resistor as $P_{\text{el}}=UI=\frac{U^2}{R}.$ (3.1) For the second transformation, Ohm’s law $R=U/I$ was used. The dissipated electrical power is converted into heat, which the resistor releases to the surroundings. For not too high temperatures, the heat release to the surroundings is, to a good approximation, proportional to the temperature difference from the surroundings. For the released thermal power $P_{\text{th}}$ it therefore holds that $P_{\text{th}}=\kappa (\vartheta -{\vartheta }_0)=:\kappa \Delta T,$ (3.2) where $\vartheta$ and ${\vartheta }_0$ denote the temperature of the resistor and of the surroundings, respectively, and $\kappa$ is a proportionality constant that depends on the geometry of the resistor and its coupling to the surroundings. Since the investigated resistors have the same construction, it can be assumed that the proportionality constant is identical for both resistors. By setting (3.1) and (3.2) equal, one obtains $R\Delta T=\frac{U^2}{\kappa }.$ (3.3) Denote by $R_1$ and $R_2$ the sought resistors and by $I_1$ and $I_2$ the currents flowing through the resistors. Since the resistors are connected in parallel, for the total current $I$ in the circuit it holds that $I=I_1+I_2=\frac{U}{R_1}+\frac{U}{R_2}=U\left(\frac{1}{R_1}+\frac{1}{R_2}\right).$ (3.4) 53rd IPhO 2023 - 2nd Round Exam - Sample Solution - 07.12.2022 From (3.3) it follows that $R_1\Delta T_1=R_2\Delta T_2=U^2/\kappa$, where $\Delta T_1$ and $\Delta T_2$ denote the temperature differences from the surroundings of the two resistors. Thus (3.4) can be transformed into $I=U\left(\frac{1}{R_1}+\frac{1}{R_2}\right)=\frac{U}{R_1}\left(1+\frac{\Delta T_2}{\Delta T_1}\right)$ and from this $R_1=\frac{U}{I}\left(1+\frac{\Delta T_2}{\Delta T_1}\right).$ (3.5) Analogously, one obtains an expression for $R_2$ by interchanging the two temperature differences. With the given values, the resistance values come out to $R_1=\frac{U}{I}\left(1+\frac{\Delta T_2}{\Delta T_1}\right)\approx 41\Omega$ and $R_2=\frac{U}{I}\left(1+\frac{\Delta T_1}{\Delta T_2}\right)\approx 11\Omega .$ (3.6) The resistance values thus most closely match those of answer C. Correct answer: C Remark: Answer option A results from assuming a series connection of the two resistors. In this case $R_1=U/I\cdot (1+\Delta T_1/\Delta T_2{)}^{-1}$ holds and analogously for $R_2$. In answer option B, the resistors are chosen such that they are dimensioned roughly in the ratio of the temperatures and the total resistance of the parallel connection matches the given values of $U$ and $I$. One is also led to the same answer option if the ambient temperature is not taken into account and the temperatures in ${}^{\circ }\text{C}$ are substituted into (3.6). In answer option D, the left-hand side of (3.3) is roughly equal for both resistors, but the total resistance of the parallel connection does not match the given values. ${}^a$Alternatively and without approximation, the Stefan-Boltzmann law can also be applied, which then, in place of (3.2), leads to the expression $P_{\text{th}}={\kappa }^{\prime }(T^4-T_0^4)$. Instead of the temperature differences, the differences of the fourth powers of the temperatures (in Kelvin) are then to be used in (3.6), and one obtains $R_1\approx 46\Omega$ and $R_2\approx 10\Omega$. The answer thus remains the same. Grading - Resistor Heating (MC problem) Points 3 Stating an expression for the electrical power (3.1) 0.5 Stating an expression for the thermal power (3.2) (or with Stefan- Boltzmann) 1.0 Using the rules for currents and voltages in parallel connections (3.4) 0.5 Setting the powers equal and stating expressions for the resistors (3.6) 1.0 Stating the correct solution 2.0 5.0 53rd IPhO 2023 - 2nd Round Exam - Sample Solution - 07.12.2022

Infrarosso di resistori su piastra (Abb. 1)

Topic: Circuits, Thermodynamics Metodi: Kirchhoff’s Laws, Energy Conservation Method, Physical Modeling Competenze: Mathematical Modeling, Physical Reasoning Objects: Resistor, Battery Fonte: Testo (PDF) — p.4

Problema 3 Risistore di riscaldamento (problema MC) (cfr. (idea: gruppo problematico della PhysicsOlympiad - Bernd Schade & Stefan Petersen) Due resistori della stessa costruzione sono collegati in parallelo a una fonte di tensione con una tensione di 2,6 V. A corrente totale di 310 mA. Con una fotocamera infrarossa, l’immagine adiacente del circuito è stata presa. Dopo la calibrazione, la macchina fotografica può Quindi, determinare le temperature superficiali dei due resistori. Sono $33{}^{\circ }\text{C}$ e $67{}^{\circ }\text{C}$. Il la temperatura ambientale è $21{}^{\circ }\text{C}$. Quali valori hanno circa i due resistori? Fig. 1. Immagine infrarossa del

• I resistori. A 1.7 $\Omega$ e 6.7 $\Omega$ B 12 $\Omega$ e 30 $\Omega$ C 10 $\Omega$ e 45 $\Omega$ D 20 $\Omega$ e 80 $\Omega$ Soluzione Calcoli e spiegazioni Il potere elettrico $P_{\text{el}}$ dissipatato in una resistore con resistenza $R$ segue dal voltage $U$ che cade attraverso la resistore e dal corrente $I$ che scorre attraverso la resistore as $P_{\text{el}}=UI=\frac{U^2}{R}.$ (3.1) For the second transformation, Ohm’s law $R=U/I$ was used. La potenza elettrica dissipata viene convertita in calore, che il resistore rilascia all’ambiente circostante. Per le temperature non troppo elevate, il rilascio di calore verso l’ambiente circostante è, per un buon Approximation, proporzionale alla differenza di temperatura rispetto all’ambiente circostante. Per la potenza termica rilasciata $P_{\text{th}}$ si ritiene pertanto che $P_{\text{th}}=\kappa (\vartheta -{\vartheta }_0)=:\kappa \Delta T,$ (3.2) dove $\vartheta$ e ${\vartheta }_0$ denotano la temperatura della resistore e dei suoi dintorni, rispettivamente, e $\kappa$ è una costante di proporzionalità che dipende dalla geometria della resistore e dal suo accoppiamento
• Al mondo circostante. Dal momento che le resistori investigate hanno la stessa costruzione, può essere supposto che la costante di proporzionalità sia identica per entrambi i resistori. Per il set (3.1) e (3.2) equal, one obtains $R\Delta T=\frac{U^2}{\kappa }.$ (3.3) Denote by $R_1$ and $R_2$ the sought resistors and by $I_1$ and $I_2$ the currents Flowing attraverso i resistori. Poiché le resistori sono collegate in parallelo, per la totale corrente $I$ in circuito che esso detiene $I=I_1+I_2=\frac{U}{R_1}+\frac{U}{R_2}=U\left(\frac{1}{R_1}+\frac{1}{R_2}\right).$ (3.4) 53° IPhO 2023 - 2° Round Exam - Sample Solution - 07.12.2022 Da (3.3) si segue che $R_1\Delta T_1=R_2\Delta T_2=U^2/\kappa$, dove $\Delta T_1$ e $\Delta T_2$ denotano le differenze di temperatura dal contesto dei due resistori. Così (3.4) può essere trasformato in $I=U\left(\frac{1}{R_1}+\frac{1}{R_2}\right)=\frac{U}{R_1}\left(1+\frac{\Delta T_2}{\Delta T_1}\right)$ e da questo $R_1=\frac{U}{I}\left(1+\frac{\Delta T_2}{\Delta T_1}\right).$ (3.5) Analogamente, one ottiene un’espressione per $R_2$ intercancelando le due differenze di temperatura. Con i valori dati, i valori di resistenza vengono fuori a $R_1=\frac{U}{I}\left(1+\frac{\Delta T_2}{\Delta T_1}\right)\approx 41\Omega$ e $R_2=\frac{U}{I}\left(1+\frac{\Delta T_1}{\Delta T_2}\right)\approx 11\Omega .$ (3.6) I valori di resistenza corrispondono quindi più strettamente a quelli di risposta C. Corretta risposta: C Nota: Risposta opzione A risultati da assumere una connessione serie dei due resistori. In questo caso $R_1=U/I\cdot (1+\Delta T_1/\Delta T_2{)}^{-1}$ è analogamente per $R_2$. In risposta all’opzione B, Le resistori sono scelte in modo che siano dimensionate in proporzione alla temperatura. e la resistenza totale della connessione parallela corrisponde ai valori dati di $U$ e $I$. One è anche portato alla stessa risposta opzione se la temperatura ambiente non è Le temperature in ${}^{\circ }\text{C}$ sono sostituite con (3.6). In risposta all’opzione D, il Il lato sinistro di (3.3) è circa uguale per entrambi i resistori, ma la resistenza totale della connessione parallela non corrisponde ai valori indicati. ${}^a$Alternatively and without approximation, the Stefan-Boltzmann law can also be applied, which then, in place of (3.2), leads to the expression $P_{\text{th}}={\kappa }^{\prime }(T^4-T_0^4)$. Invece delle differenze di temperatura, le differenze di temperatura di cui sono utilizzate le quatri potenze delle temperature (in Kelvin) sono quindi utilizzate in (3.6), e uno ottiene $R_1\approx 46\Omega$ e $R_2\approx 10\Omega$. La risposta rimane così. Classificazione - Risistore di riscaldamento (problema MC) Punti 3 Stating an expression for the electrical power (3.1) 0.5 Stating an expression for the thermal power (3.2) (o con Stefan- Boltzmann) 1.0 Usando le regole per correnti e voltaggi in connessioni parallele (3.4) 0.5 Setting the powers equal and stating expressions for the resistors (3.6) 1.0 Stating the correct solution 2.0 5.0 53° IPhO 2023 - 2° Round Exam - Sample Solution - 07.12.2022

Infrarosso di resistori su piastra (Abb. 1)

Topic: Circuits, Thermodynamics Metodi: Kirchhoff’s Laws, Energy Conservation Method, Physical Modeling Competenze: Mathematical Modeling, Physical Reasoning Objects: Resistor, Battery Fonte: Testo (PDF) — p.4

The following is the list of the types of heating systems used: (five points) (Idea: Problem group of the PhysicsOlympiad - Bernd Schade & Stefan Petersen) Two resistors of the same construction are connected in parallel to a voltage source with a voltage of 2.6 V. A Total current of 310 mA flows. With an infrared camera, the adjacent image of the circuit is taken. After calibration, the camera can So determine the surface temperatures of the two resistors. They are $33{}^{\circ }\text{C}$ and $67{}^{\circ }\text{C}$. The ambient temperature is $21{}^{\circ }\text{C}$. What values do the two resistors have approximately? Fig. 1. Infrared image of the The resistors. A 1.7 $\Omega$ and 6.7 $\Omega$ B 12 $\Omega$ and 30 $\Omega$ C 10 $\Omega$ and 45 $\Omega$ D 20 $\Omega$ and 80 $\Omega$ The solution Calculations and explanations The electrical power $P_{\text{el}}$ dissipated in a resistor with resistance value $R$ follows from the voltage $U$ dropping across the resistor and the current $I$ flowing through the resistor as $P_{\text{el}}=UI=\frac{U^2}{R}.$ (3.1) For the second transformation, Ohm’s law $R=U/I$ was used. The dissipated electrical power is converted into heat, which the resistor releases to the surroundings. For not too high temperatures, the heat release to the surroundings is, to a good approximation, proportional to the temperature difference from the surroundings. For the released thermal power $P_{\text{th}}$ it therefore holds that $P_{\text{th}}=\kappa (\vartheta -{\vartheta }_0)=:\kappa \Delta T,$ (3.2) where $\vartheta$ and ${\vartheta }_0$ denote the temperature of the resistor and of the surroundings, respectively, and $\kappa$ is a proportionality constant that depends on the geometry of the resistor and its coupling to the surroundings. Since the investigated resistors have the same construction, it can be assumed that the proportionality constant is identical for both resistors. By setting (3.1) and (3.2) equal, one obtains $R\Delta T=\frac{U^2}{\kappa }.$ (3.3) Denote by $R_1$ and $R_2$ the sought resistors and by $I_1$ and $I_2$ the currents flowing through the resistors. Since the resistors are connected in parallel, for the Total current $I$ in the circuit it holds that $I=I_1+I_2=\frac{U}{R_1}+\frac{U}{R_2}=U\left(\frac{1}{R_1}+\frac{1}{R_2}\right).$ (3.4) The Commission shall take into account the following information: From (3.3) it follows that $R_1\Delta T_1=R_2\Delta T_2=U^2/\kappa$, where $\Delta T_1$ and $\Delta T_2$ denote the temperature differences from the surroundings of the two resistors. Thus (3.4) can be transformed into $I=U\left(\frac{1}{R_1}+\frac{1}{R_2}\right)=\frac{U}{R_1}\left(1+\frac{\Delta T_2}{\Delta T_1}\right)$ And from this $R_1=\frac{U}{I}\left(1+\frac{\Delta T_2}{\Delta T_1}\right).$ (3.5) Analogously, one obtains an expression for $R_2$ by interchanging the two temperature differences. With the given values, the resistance values come out to $R_1=\frac{U}{I}\left(1+\frac{\Delta T_2}{\Delta T_1}\right)\approx 41\Omega$ and $R_2=\frac{U}{I}\left(1+\frac{\Delta T_1}{\Delta T_2}\right)\approx 11\Omega .$ (3.6) The resistance values thus most closely match those of answer C. Correct answer: C Note: Answer option A results from assuming a series connection of the two resistors. In this case $R_1=U/I\cdot (1+\Delta T_1/\Delta T_2{)}^{-1}$ holds and analogously for $R_2$. In answer option B, The resistors are chosen so that they are dimensioned roughly in the ratio of the temperatures and the total resistance of the parallel connection matches the given values of $U$ and $I$. One is also led to the same answer option if the ambient temperature is not The temperature in ${}^{\circ }\text{C}$ are substituted into (3.6). In answer option D, the The left-hand side of (3.3) is roughly equal for both resistors, but the total resistance of the parallel connection does not match the given values. ${}^a$Alternatively and without approximation, the Stefan-Boltzmann law can also be applied, which then, in place of (3.2), leads to the expression $P_{\text{th}}={\kappa }^{\prime }(T^4-T_0^4)$. Instead of the temperature differences, the differences of the fourth powers of the temperatures (in Kelvin) are then to be used in (3.6), and one obtains $R_1\approx 46\Omega$ and $R_2\approx 10\Omega$. The answer thus remains the same. Grading - Resistor heating (MC problem) Points 3 Stating an expression for the electrical power (3.1) 0.5 Stating an expression for the thermal power (3.2) (or with Stefan- Boltzmann) 1.0 Using the rules for currents and voltages in parallel connections (3.4) 0.5 Setting the powers equal and stating expressions for the resistors (3.6) 1.0 Stating the correct solution 2.0 5.0 The Commission shall take into account the following information:

Infrarosso di resistori su piastra (Abb. 1)

Topic: Circuits, Thermodynamics Metodi: Kirchhoff’s Laws, Energy Conservation Method, Physical Modeling Competenze: Mathematical Modeling, Physical Reasoning Objects: Resistor, Battery Fonte: Testo (PDF) — p.4

Problem 4 Double Spring Pendulum (MC problem) (5 pts.) (Idea: Problem group of the PhysicsOlympiad - Thomas Hellerl) In both spring pendulums shown in the figure, a body of mass $m$ oscillates frictionlessly. The spring constants $D_1$ and $D_2$ of the two Hookean springs are, however, different. Therefore the bodies oscillate after a displacement with different frequencies $f_1$ and $f_2$. $m$ $D_1$ $f_1=1,2$ Hz $m$ $D_2$ $f_2=1,6$ Hz What is the oscillation frequency (natural frequency) of the system shown below, in which the springs are coupled? $m$ $D_1$ $D_2$ $f=?$ A 1.4 Hz B 2.0 Hz C 2.4 Hz D 2.8 Hz Solution Calculations and explanations The natural frequencies of the upper oscillators are given by $f_i=\frac{1}{2\pi }\cdot \sqrt{\frac{D_i}{m}}.$ (4.1) For the spring constants $D_i$ it therefore follows that $D_i=4{\pi }^{2}m{f}_i^2.$ (4.2) The individual spring constants $D_1$ and $D_2$ of the lower system add up to the new spring constant $D$ on account of the following consideration. $m$ $D_1$ $D_2$ $x$ ${\vec{F}}_1$ ${\vec{F}}_2$ At equilibrium, each spring pulls with the force $F_0$. Upon a displacement by $x$ from the equilibrium position, the left spring exerts the force magnitude $F_1=F_0+D_{1}x$ and the right spring the force magnitude $F_2=F_0-D_{2}x$ on $m$. We obtain for the total force acting on the oscillating mass $m$. $F(x)=-F_1+F_2=-(F_0+D_{1}x)+(F_0-D_{2}x)=-(D_1+D_2)x$ (4.3) 53rd IPhO 2023 - 2nd Round Exam - Sample Solution - 07.12.2022 and thus for the system with both springs $D=D_1+D_2$ (4.4) The new system therefore oscillates at the frequency $f=\frac{1}{2\pi }\cdot \sqrt{\frac{D}{m}}=\frac{1}{2\pi }\cdot \sqrt{\frac{D_1+D_2}{m}}=\frac{1}{2\pi }\cdot \sqrt{\frac{4{\pi }^{2}m(f_1^2+f_2^2)}{m}}=\sqrt{f_1^2+f_2^2}.$ (4.5) With the given values, one obtains $f=\sqrt{f_1^2+f_2^2}=\sqrt{1.2^2+1.6^2}\text{Hz}\approx 2,0\text{Hz}.$ (4.6) Correct answer: B Grading - Double Spring Pendulum (MC problem) Points 4 Stating the relation between spring constant, frequency and mass 1.0 Recognizing that the spring constants add up 1.0 Determining an expression for the new frequency 1.0 Stating the correct solution 2.0 5.0 53rd IPhO 2023 - 2nd Round Exam - Sample Solution - 07.12.2022

Pendoli a molla doppio e singolo accoppiato

Topic: Oscillations & Waves Metodi: Simple Harmonic Motion Analysis, Hooke’s Law, Superposition Principle Competenze: Mathematical Modeling, Physical Reasoning Objects: Spring Fonte: Testo (PDF) — p.6

Problema 4 Pendole doppio spruzzatura (problema MC) (cfr. (idea: gruppo problematico della fisica olimpica - Thomas Hellerl) In both spring pendulums shown in the figure, a body of mass $m$ oscilla senza frattura. Le costanti di primavera $D_1$ e $D_2$ delle due sorgenti di Hookean sono tuttavia diverse. Pertanto i corpi oscillate dopo un dislocazione con diverse frequenze $f_1$ e $f_2$. $m$ $D_1$ $f_1=1,2$ Hz $m$ $D_2$ $f_2=1,6$ Hz Qual è la frequenza di oscillazione (natural frequency) del sistema mostrato qui sotto, in cui il sistema oscillazione è le sorgenti sono accoppiate? $m$ $D_1$ $D_2$ $f=?$ A 1.4 Hz B 2.0 Hz C 2.4 Hz D 2.8 Hz Soluzione Calcoli e spiegazioni Le frequenze naturali degli oscillatori superiori sono date da $f_i=\frac{1}{2\pi }\cdot \sqrt{\frac{D_i}{m}}.$ (4.1) Per le costanti di primavera $D_i$ si segue quindi che $D_i=4{\pi }^{2}m{f}_i^2.$ (4.2) Le singole costanti di primavera $D_1$ e $D_2$ del sistema inferiore si aggiungono alla nuova costante di primavera $D$ a causa della seguente considerazione. $m$ $D_1$ $D_2$ $x$ ${\vec{F}}_1$ ${\vec{F}}_2$ All’equilibrio, ogni primavera tira con la forza $F_0$. Dopo un spostamento da $x$ dalla posizione di equilibrio, la primavera sinistra esercita la forza di magnitudine $F_1=F_0+D_{1}x$ e la primavera destra la forza di magnitudine $F_2=F_0-D_{2}x$ on $m$. We obtain for the total force acting on the oscillating mass $m$. $F(x)=-F_1+F_2=-(F_0+D_{1}x)+(F_0-D_{2}x)=-(D_1+D_2)x$ (4.3) 53° IPhO 2023 - 2° Round Exam - Sample Solution - 07.12.2022 e quindi per il sistema con entrambe le sorgenti $D=D_1+D_2$ (4.4) Il nuovo sistema oscilla quindi alla frequenza $f=\frac{1}{2\pi }\cdot \sqrt{\frac{D}{m}}=\frac{1}{2\pi }\cdot \sqrt{\frac{D_1+D_2}{m}}=\frac{1}{2\pi }\cdot \sqrt{\frac{4{\pi }^{2}m(f_1^2+f_2^2)}{m}}=\sqrt{f_1^2+f_2^2}.$ (4.5) Con i valori forniti, si ottiene $f=\sqrt{f_1^2+f_2^2}=\sqrt{1.2^2+1.6^2}\text{Hz}\approx 2,0\text{Hz}.$ (4.6) Risposta corretta: B Classificazione - Pendolo doppio di molla (problema MC) Punti 4 Stating the relation between spring constant, frequency and mass 1.0 Riconoscendo che le costanti di primavera 1.0 Determinare un’espressione per la nuova frequenza 1.0 Stating the correct solution 2.0 5.0 53° IPhO 2023 - 2° Round Exam - Sample Solution - 07.12.2022

Pendoli a molla doppio e singolo accoppiato

Topic: Oscillations & Waves Metodi: Simple Harmonic Motion Analysis, Hooke’s Law, Superposition Principle Competenze: Mathematical Modeling, Physical Reasoning Objects: Spring Fonte: Testo (PDF) — p.6

The problem is that the two-dimensional pendulum is a double spring pendulum. (five points) (Idea: Problem group of the PhysicsOlympic - Thomas Hellerl) In both spring pendulums shown in the figure, a body of mass $m$ It oscillates frictionlessly. The spring constants $D_1$ and $D_2$ of the two Hookean springs are, however, different. Therefore the bodies oscillate after a displacement with different frequencies $f_1$ and $f_2$. $m$ $D_1$ $f_1=1,2$ Hz $m$ $D_2$ $f_2=1,6$ Hz What is the oscillation frequency (natural frequency) of the system shown below, in which the

• springs are coupled? $m$ $D_1$ $D_2$ $f=?$ A 1.4 Hz B 2.0 Hz C 2.4 Hz D 2.8 Hz The solution Calculations and explanations The natural frequencies of the upper oscillators are given by $f_i=\frac{1}{2\pi }\cdot \sqrt{\frac{D_i}{m}}.$ (4.1) For the spring constants $D_i$ it therefore follows that $D_i=4{\pi }^{2}m{f}_i^2.$ (4.2) The individual spring constants $D_1$ and $D_2$ of the lower system add up to the new spring constant $D$ on account of the following consideration. $m$ $D_1$ $D_2$ $x$ ${\vec{F}}_1$ ${\vec{F}}_2$ At equilibrium, each spring pulls with the force $F_0$. Upon a displacement by $x$ from the equilibrium position, the left spring exerts the force magnitude $F_1=F_0+D_{1}x$ and the right spring the force magnitude $F_2=F_0-D_{2}x$ on $m$. We obtain for the total force acting on the oscillating mass $m$. $F(x)=-F_1+F_2=-(F_0+D_{1}x)+(F_0-D_{2}x)=-(D_1+D_2)x$ (4.3) The Commission shall take into account the following information: and thus for the system with both springs $D=D_1+D_2$ (4.4) The new system therefore oscillates at the frequency $f=\frac{1}{2\pi }\cdot \sqrt{\frac{D}{m}}=\frac{1}{2\pi }\cdot \sqrt{\frac{D_1+D_2}{m}}=\frac{1}{2\pi }\cdot \sqrt{\frac{4{\pi }^{2}m(f_1^2+f_2^2)}{m}}=\sqrt{f_1^2+f_2^2}.$ (4.5) With the given values, one obtains $f=\sqrt{f_1^2+f_2^2}=\sqrt{1.2^2+1.6^2}\text{Hz}\approx 2,0\text{Hz}.$ (4.6) Correct answer: B Grading - Double spring pendulum (MC problem) Points 4 Stating the relation between spring constant, frequency and mass 1.0 Recognizing that the spring constants add up 1.0 Determining an expression for the new frequency 1.0 Stating the correct solution 2.0 5.0 The Commission shall take into account the following information:

Pendoli a molla doppio e singolo accoppiato

Topic: Oscillations & Waves Metodi: Simple Harmonic Motion Analysis, Hooke’s Law, Superposition Principle Competenze: Mathematical Modeling, Physical Reasoning Objects: Spring Fonte: Testo (PDF) — p.6

Problem 5 Tidal Heating (MC problem) (5 pts.) (Problem group of the PhysicsOlympiad - Tim Pokart) Although a thick ice layer reflects most of the sunlight incident on Saturn’s moon Enceladus, the space probe Cassini was able to photograph water fountains several hundred kilometers high on its surface. The moon obtains the energy required for this from tidal forces, which heat it through their conversion into frictional work. Consider a celestial body with radius $r$ that orbits a planet of mass $M_P$ on a path with semi-major axis $a$ and eccentricity $e$. The eccentricity is, for closed orbits, a value with $0\le e<1$ that indicates how strongly the orbit deviates from a circular orbit. The heating power that the body experiences can be expressed by $P\approx \frac{21}{100}{r}^5{e}^2\frac{G^{\alpha }{M}_P^{\beta }}{a^{\gamma }}.$ What values do the exponents $\alpha$, $\beta$ and $\gamma$ have? A $\alpha =-3/2$, $\beta =5/2$ and $\gamma =-15/2$. B $\alpha =3/2$, $\beta =5/2$ and $\gamma =-15/2$. C $\alpha =3/2$, $\beta =-5/2$ and $\gamma =15/2$. D $\alpha =-3/2$, $\beta =5/2$ and $\gamma =15/2$. Solution Calculations and explanations The result can be derived from a dimensional analysis. Denote by $M$, $L$ and $T$ the dimensions mass, length and time. Then the quantities in the formula have the following dimensions: $[P]=M{L}^2{T}^{-3}$ $[G]=L^3{T}^{-2}{M}^{-1}$ $[M_P]=M$ $[r]=L$ $[a]=L.$ (5.1) Accordingly, for the dimensions in the given formula it holds that $M{L}^2{T}^{-3}=L^5{L}^{3\alpha }{T}^{-2\alpha }{M}^{-\alpha }{M}^{\beta }{L}^{\gamma }=M^{\beta -\alpha }{L}^{5+3\alpha +\gamma }{T}^{-2\alpha }.$ For the exponents this yields the system of equations $1=\beta -\alpha$ $2=5+3\alpha +\gamma$ $-3=-2\alpha .$ (5.2) This is solved by $\alpha =3/2\beta =5/2\gamma =-15/2.$ (5.3) Correct answer: B Remark: For an alternative solution, one can use the expected physical behavior of the formula for the power. The heating power should increase with increasing planet mass at otherwise 53rd IPhO 2023 - 2nd Round Exam - Sample Solution - 07.12.2022 constant parameters. Therefore $\beta >0$ must hold. Conversely, the heating power should decrease with increasing semi-major axis, which requires $\gamma <0$. The only answer option that fulfills these two conditions is B. Grading - Tidal Heating (MC problem) Points 5 Stating the relevant units/dimensions 1.0 Using a dimensional analysis 1.0 Setting up and solving the system of equations (5.2) 1.0 Stating the correct solution 2.0 5.0 53rd IPhO 2023 - 2nd Round Exam - Sample Solution - 07.12.2022

Topic: Astrophysics, Gravitation Metodi: Dimensional Analysis, Newton’s Law of Gravitation, Kepler’s Laws Competenze: Estimation & Approximation, Physical Reasoning Objects: Planet Fonte: Testo (PDF) — p.8

Problema 5 Risoluzione del problema del riscaldamento del mare (cfr. (Problema group of the PhysicsOlympiad - Tim Pokart) Sebbene un’épace strata di ghiaccio rifletta la maggior parte dell’incidente di luce solare sulla luna di Saturno Enceladus, la sonda spaziale Cassini è stata in grado di fotografare fonti d’acqua diverse centinaia di chilometri di altezza sulla sua superficie. La luna ottiene l’energia necessaria per questo da Le forze di marea, che lo calano attraverso la loro conversione in lavoro fratturoso. Consider a celestial body with radius $r$ that orbits a planet of mass $M_P$ on a path con semi-major axis $a$ e eccentricità $e$. The eccentricity is, for closed Orbit, un valore con $0\le e<1$ che indica quanto forte l’orbita deviasse da un’orbita circolare. Il potere di riscaldamento che il corpo sperimenta può essere espresso da $P\approx \frac{21}{100}{r}^5{e}^2\frac{G^{\alpha }{M}_P^{\beta }}{a^{\gamma }}.$ Quali valori hanno gli esponenti $\alpha$, $\beta$ e $\gamma$? A $\alpha =-3/2$, $\beta =5/2$ e $\gamma =-15/2$. B $\alpha =3/2$, $\beta =5/2$ e $\gamma =-15/2$. C $\alpha =3/2$, $\beta =-5/2$ e $\gamma =15/2$. D $\alpha =-3/2$, $\beta =5/2$ e $\gamma =15/2$. Soluzione Calcoli e spiegazioni Il risultato può essere derivato da un’analisi dimensionale. Denote by $M$, $L$ and $T$ the dimensions mass, length and time. Quindi le quantità nella formula hanno le seguenti dimensioni: $[P]=M{L}^2{T}^{-3}$ $[G]=L^3{T}^{-2}{M}^{-1}$ $[M_P]=M$ $[r]=L$ $[a]=L.$ (5.1) Pertanto, per le dimensioni nella formula data si ritiene che $M{L}^2{T}^{-3}=L^5{L}^{3\alpha }{T}^{-2\alpha }{M}^{-\alpha }{M}^{\beta }{L}^{\gamma }=M^{\beta -\alpha }{L}^{5+3\alpha +\gamma }{T}^{-2\alpha }.$ Per gli esponenti questo rende il sistema di equazioni $1=\beta -\alpha$ $2=5+3\alpha +\gamma$ $-3=-2\alpha .$ (5.2) This is solved by $\alpha =3/2\beta =5/2\gamma =-15/2.$ (5.3) Risposta corretta: B Nota: Per una soluzione alternativa, si può usare il comportamento fisico atteso della formula Per il potere. Il potere di riscaldamento dovrebbe aumentare con l’aumento del massa del pianeta 53° IPhO 2023 - 2° Round Exam - Sample Solution - 07.12.2022

• parametri costanti. Pertanto $\beta >0$ must hold. Invece, il potere di riscaldamento dovrebbe essere decrease with increasing semi-major axis, which requires $\gamma <0$. L’unica risposta che “che soddisfa queste due condizioni è B”. Classificazione - riscaldamento del mare (problema MC) Punti 5 Stating the relevant units/dimensions 1.0 Usando un’analisi dimensionale 1.0 Setting up and solving the system of equations (5.2) 1.0 Stating the correct solution 2.0 5.0 53° IPhO 2023 - 2° Round Exam - Sample Solution - 07.12.2022

Topic: Astrophysics, Gravitation Metodi: Dimensional Analysis, Newton’s Law of Gravitation, Kepler’s Laws Competenze: Estimation & Approximation, Physical Reasoning Objects: Planet Fonte: Testo (PDF) — p.8

Problem 5 Tidal heating (MC problem) (five points) (Problem group of the PhysicsOlympic - Tim Pokart) Although a thick ice layer reflects most of the sunlight incident on Saturn’s moon Enceladus, the space probe Cassini was able to photograph water fountains several hundred kilometers high on its surface. The moon gets the energy required for this from tidal forces, which heat it through their conversion into frictional work. Consider a celestial body with radius $r$ that orbits a planet of mass $M_P$ on a path with semi-major axis $a$ and eccentricity $e$. The eccentricity is, for closed orbits, a value with $0\le e<1$ that indicates how strongly the orbit deviates from a circular orbit. The heating power that the body experiences can be expressed by $P\approx \frac{21}{100}{r}^5{e}^2\frac{G^{\alpha }{M}_P^{\beta }}{a^{\gamma }}.$ What values do the exponents $\alpha$, $\beta$ and $\gamma$ have? A $\alpha =-3/2$, $\beta =5/2$ and $\gamma =-15/2$. B $\alpha =3/2$, $\beta =5/2$ and $\gamma =-15/2$. C $\alpha =3/2$, $\beta =-5/2$ and $\gamma =15/2$. D $\alpha =-3/2$, $\beta =5/2$ and $\gamma =15/2$. The solution Calculations and explanations The result can be derived from a dimensional analysis. Denote by $M$, $L$ and $T$ the dimensions mass, length and time. Then the quantities in the formula have the following dimensions: $[P]=M{L}^2{T}^{-3}$ $[G]=L^3{T}^{-2}{M}^{-1}$ $[M_P]=M$ $[r]=L$ $[a]=L.$ (5.1) Accordingly, for the dimensions in the given formula it holds that $M{L}^2{T}^{-3}=L^5{L}^{3\alpha }{T}^{-2\alpha }{M}^{-\alpha }{M}^{\beta }{L}^{\gamma }=M^{\beta -\alpha }{L}^{5+3\alpha +\gamma }{T}^{-2\alpha }.$ For the exponents this yields the system of equations $1=\beta -\alpha$ $2=5+3\alpha +\gamma$ $-3=-2\alpha .$ (5.2) This is solved by $\alpha =3/2\beta =5/2\gamma =-15/2.$ (5.3) Correct answer: B Note: For an alternative solution, one can use the expected physical behavior of the formula For the power. The heating power should increase with increasing planet mass at otherwise The Commission shall take into account the following information: The parameters are constant. Therefore $\beta >0$ must hold. Conversely, the heating power should be decrease with increasing semi-major axis, which requires $\gamma <0$. The only answer option that If these two conditions are met, B is. Grading - Tidal heating (MC problem) Points 5 Stating the relevant units/dimensions 1.0 Using a dimensional analysis 1.0 Setting up and solving the system of equations (5.2) 1.0 Stating the correct solution 2.0 5.0 The Commission shall take into account the following information:

Topic: Astrophysics, Gravitation Metodi: Dimensional Analysis, Newton’s Law of Gravitation, Kepler’s Laws Competenze: Estimation & Approximation, Physical Reasoning Objects: Planet Fonte: Testo (PDF) — p.8

Problem 6 Coaxial Cable (MC problem) (5 pts.) (Problem group of the PhysicsOlympiad - Arne Wolf) A coaxial cable consists, as shown in the left cross section alongside, of a long narrow cylinder with specific resistance ${\rho }_1$ encased by a hollow cylinder with specific resistance ${\rho }_2>{\rho }_1$. Through the cable flows a current of magnitude $I$. A second coaxial cable, shown on the right, looks the same from the outside as the first, but on the inside consists of only one material. The specific resistance of this material is $\rho$ and the current in the second cable is likewise $I$. ${\rho }_1$ ${\rho }_2$ A B C $\rho$ A B C Fig. 2. Cross section of the first (left) and second (right) coaxial cable. At how many of the marked points A, B and C do the magnetic fields produced by the respective cable differ? A 0 B 1 C 2 D 3 Solution Calculations and explanations Since the total current is the same in both cables and the specific resistance of the core in the first cable is smaller than that of its sheath, more current flows in the core of the first cable than in the second. According to Ampère’s law, the magnetic field generated by a straight wire at distance $r$ from the wire axis is proportional to the current that flows at a distance less than or equal to $r$ from the wire axis. Since this current is increased for points A and B in the first cable and is equal at point C for both cables, the magnetic field differs at points A and B. Thus C is the correct answer. Correct answer: C Grading - Coaxial Cable (MC problem) Points 6 Recognizing where more current flows 1.5 Using Ampère’s law 1.5 Stating the correct solution 2.0 5.0 53rd IPhO 2023 - 2nd Round Exam - Sample Solution - 07.12.2022

Sezione trasversale cavi coassiali (Abb. 2)

Topic: Magnetism, Electromagnetism Metodi: Ampère’s Law, Physical Modeling, Symmetry Argument Competenze: Physical Reasoning, Diagrammatic Reasoning Objects: Wire, Cylinder Fonte: Testo (PDF) — p.10

Problema 6 Cable coaxial (problema MC) (cfr. (Problema group of the PhysicsOlympiad - Arne Wolf) Un cavo coaxial è costituito, come mostrato nel di un lungo cilindro con resistenza specifica ${\rho }_1$ incassato da un cilindro vuoto con resistenza specifica ${\rho }_2>{\rho }_1$. Attraverso il cavo flussi a corrente di magnitudo $I$. Un secondo cavo coaxial, mostrato sulla destra, Sembra lo stesso dall’esterno come il primo, ma sul inside è composto da un solo materiale. La resistenza specifica di questo materiale è $\rho$ e il corrente nel secondo cavo è di tipo simile $I$. ${\rho }_1$ ${\rho }_2$ A B C $\rho$ A B C Fig. 2. Sezione trasversale del primo (sinistra) e del secondo (destra) cavo coaxial. A how many of the marked points A, B e C fanno i campi magnetici prodotti dal rispettivo

• Cable differ? A 0 B 1 C 2 D 3 Soluzione Calcoli e spiegazioni Dal momento che la corrente totale è la stessa in entrambi i cavi e la resistenza specifica del nucleo in primo cavo è più piccolo di quello della sua tenda, più corrente scorre nel nucleo del primo cavo che
• In secondo. Secondo la legge di Ampere, il campo magnetico generato da un filo dritto a distance $r$ from the wire axis is proportional to the current that flows at a distance less than or equal to $r$ from the
• Il filo. Poiché questo corrente è aumentato per i punti A e B nel primo cavo ed è uguale al punto C per entrambi I cavi, il campo magnetico differisce ai punti A e B. Quindi C è il corretto
• Risposta. Corretta risposta: C Cable a coassi (problema MC) Punti 6 Riconoscere dove più corrente scorre 1.5 Usando la legge di Ampere 1.5 Stating the correct solution 2.0 5.0 53° IPhO 2023 - 2° Round Exam - Sample Solution - 07.12.2022

Sezione trasversale cavi coassiali (Fig. 2)*

Topic: Magnetism, Electromagnetism Metodi: Ampère’s Law, Physical Modeling, Symmetry Argument Competenze: Physical Reasoning, Diagrammatic Reasoning Objects: Wire, Cylinder Fonte: Testo (PDF) — p.10

The problem is that the coaxing cable is not a coaxing cable. (five points) (Problem group of the PhysicsOlympiad - Arne Wolf) A coaxial cable consists of a Left cross section alongside, of a long narrow cylinder with specific resistance ${\rho }_1$ encased by a hollow cylinder with specific resistance ${\rho }_2>{\rho }_1$. Through the cable The current of magnitude $I$. A second coaxial cable, shown on the right, looks the same from the outside as the first, but on the Inside consists of only one material. The specific resistance of this material is $\rho$ and The current in the second cable is likewise $I$. ${\rho }_1$ ${\rho }_2$ A B C $\rho$ A B C Fig. 2. Cross section of the first (left) and second (right) coaxial cable. At how many of the marked points A, B and C do the magnetic fields produced by the respective cable differ? A 0 B 1 C 2 D 3 The solution Calculations and explanations Since the total current is the same in both cables and the specific resistance of the core In the first cable is smaller than that of its sheath, more current flows in the core of the first cable than In the second. According to Ampère’s law, the magnetic field generated by a straight wire at distance $r$ from the wire axis is proportional to the current that flows at a distance less than or equal to $r$ from the The wire axis. Since this current is increased for points A and B in the first cable and is equal at point C for both The magnetic field differs at points A and B. So C is the correct Answer. Correct answer: C Grading - Coaxial cable (MC problem) Points 6 Recognizing where more current flows 1.5 Using Ampère’s law 1.5 Stating the correct solution 2.0 5.0 The Commission shall take into account the following:

Sezione trasversale cavi coassiali (Abb. 2)

Topic: Magnetism, Electromagnetism Metodi: Ampère’s Law, Physical Modeling, Symmetry Argument Competenze: Physical Reasoning, Diagrammatic Reasoning Objects: Wire, Cylinder Fonte: Testo (PDF) — p.10

Problem 7 Glass Block (5 pts.) (Idea: Problem group of the PhysicsOlympiad - Thomas Hellerl & Titus Bornträger) A laser beam running in the plane of the drawing strikes a glass block (refractive index $n=1,5$) with side lengths $a$ and $4a$ from the left at the angle of incidence $\alpha =30^{\circ }$. As indicated in the not-to-scale sketch in Figure 3, it finally strikes exactly the lower right corner inside the glass block. $4a$ $a$ $n=1,5$ $\alpha$ $x$ Fig. 3. Not-to-scale sketch of the laser beam in the glass block in side view. What is the distance $x$ of the entry point from the upper boundary surface of the block? A $a\cdot \left(\sqrt{2}-1\right)$ B $a\cdot \left(2-\sqrt{3}\right)$ C $a\cdot \left(1-\frac{\sqrt{2}}{2}\right)$ D $a\cdot \left(1-\frac{\sqrt{3}}{3}\right)$ Solution Calculations and explanations $a$ $n=1,5$ $\alpha$ $x$ $y$ $4a-y$ ${\alpha }^{\prime }$ Fig. 4. Complete sketch of the beam path in the glass block in side view By the law of refraction: $\sin \alpha =n\cdot \sin {\alpha }^{\prime }.$ (7.1) The beam running inside the glass is totally internally reflected at the upper side. In the left and the right triangle, one finds on account of the similarity $\frac{a}{4a-y}=\frac{x}{y}=\tan {\alpha }^{\prime }.$ (7.2) The value of $\tan {\alpha }^{\prime }$ can be determined directly by means of $\tan {\alpha }^{\prime }=\frac{\sin {\alpha }^{\prime }}{\cos {\alpha }^{\prime }}=\frac{\frac{1}{n}\cdot \sin \alpha }{\sqrt{1-\frac{{\sin }^2\alpha }{n^2}}}=\frac{\frac{1}{1,5}\cdot \sin 30^{\circ }}{\sqrt{1-\frac{{\sin }^{2}30^{\circ }}{1,5^2}}}=\frac{1}{2\sqrt{2}}.$ (7.3) Here $\sin \alpha =\sin 30^{\circ }=0,5$ was used. It follows immediately from (7.2): $y=2\sqrt{2}x.$ (7.4) 53rd IPhO 2023 - 2nd Round Exam - Sample Solution - 07.12.2022 Substituting into equation (7.2) and solving for $x$ gives: $\frac{a}{4a-2\sqrt{2}x}=\frac{1}{2\sqrt{2}}\text{bzw.}2\sqrt{2}a=4a-2\sqrt{2}x$ (7.5) and thus $x=2\sqrt{2}a-a=a\cdot \left(\sqrt{2}-1\right).$ (7.6) Correct answer: A Grading - Glass Block Points 7 Using the similarity of the triangles (7.2) 1.0 Using the correct value for $\tan {\alpha }^{\prime }$ with (7.3) 1.0 Deriving the result for $x$ from (7.2) 1.0 Stating the correct solution 2.0 5.0 53rd IPhO 2023 - 2nd Round Exam - Sample Solution - 07.12.2022 Long-answer problems Work on the following three problems likewise in the boxes provided for this. Unlike for the multiple-choice problems, no answer options are given. Describe your solution method in such a way that it is easy to follow but not unnecessarily long. If, for example, you use the law of conservation of energy, write this down briefly.

Schizzo raggio laser nel blocco di vetro (Abb. 3)

Percorso completo raggio laser vetro (Abb. 4)

Topic: Geometric Optics Metodi: Snell’s Law, Ray Tracing, Physical Modeling Competenze: Mathematical Modeling, Diagrammatic Reasoning Objects: — Fonte: Testo (PDF) — p.11

Problema 7 Blocco di vetro (cfr. (idea: gruppo di problemi della PhysicsOlympiad - Thomas Hellerl & Titus Bornträger) Un fascio laser che corre nel piano del disegno colpisce un blocco di vetro (indice refraettivo $n=1,5$) con lunghezze laterali $a$ e $4a$ da sinistra all’angolo di incidenza $\alpha =30^{\circ }$. Come indicato nello schema non a scala nella figura 3, infine colpisce esattamente il Cortile inferiore a destra all’interno del blocco di vetro. $4a$ $a$ $n=1,5$ $\alpha$ $x$ Fig. 3. Sketch non a scala del raggio laser nel blocco di vetro in vista laterale. Qual è la distanza $x$ del punto di ingresso dalla superficie di confine superiore del blocco? A $a\cdot \left(\sqrt{2}-1\right)$ B $a\cdot \left(2-\sqrt{3}\right)$ C $a\cdot \left(1-\frac{\sqrt{2}}{2}\right)$ D $a\cdot \left(1-\frac{\sqrt{3}}{3}\right)$ Soluzione Calcoli e spiegazioni $a$ $n=1,5$ $\alpha$ $x$ $y$ $4a-y$ ${\alpha }^{\prime }$ Fig. 4. Complete sketch of the beam path in the glass block in side view Per la legge della refrazione: $\sin \alpha =n\cdot \sin {\alpha }^{\prime }.$ (7.1) Il fascio che corre all’interno del vetro è totalmente riflesso internamente sul lato superiore. In quello sinistro e il triangolo destro, uno trova a causa della somiglianza $\frac{a}{4a-y}=\frac{x}{y}=\tan {\alpha }^{\prime }.$ (7.2) Il valore di $\tan {\alpha }^{\prime }$ può essere determinato direttamente $\tan {\alpha }^{\prime }=\frac{\sin {\alpha }^{\prime }}{\cos {\alpha }^{\prime }}=\frac{\frac{1}{n}\cdot \sin \alpha }{\sqrt{1-\frac{{\sin }^2\alpha }{n^2}}}=\frac{\frac{1}{1,5}\cdot \sin 30^{\circ }}{\sqrt{1-\frac{{\sin }^{2}30^{\circ }}{1,5^2}}}=\frac{1}{2\sqrt{2}}.$ (7.3) Here $\sin \alpha =\sin 30^{\circ }=0,5$ was used. Il seguente è stato risposto immediatamente a (7.2): $y=2\sqrt{2}x.$ (7.4) 53° IPhO 2023 - 2° Round Exam - Sample Solution - 07.12.2022 Substituing into equation (7.2) and solving for $x$ dà: $\frac{a}{4a-2\sqrt{2}x}=\frac{1}{2\sqrt{2}}\text{bzw.}2\sqrt{2}a=4a-2\sqrt{2}x$ (7.5) e così $x=2\sqrt{2}a-a=a\cdot \left(\sqrt{2}-1\right).$ (7.6) Risposta corretta: A Grading - Blocco di vetro Punti 7 Usando la similitudine dei triangoli (7.2) 1.0 Usando il valore corretto per $\tan {\alpha }^{\prime }$ with (7.3) 1.0 Deriving the result for $x$ from (7.2) 1.0 Stating the correct solution 2.0 5.0 53° IPhO 2023 - 2° Round Exam - Sample Solution - 07.12.2022 Problemi di risposta lunga La Commissione ha inoltre presentato una serie di proposte di risoluzione. Un’altra cosa è la problemi di scelta multipla, non sono state indicate le opzioni di risposta. Descrivere il metodo di soluzione in un modo simile che è facile da seguire ma non troppo lungo. Se, per esempio, usi la legge della conservazione dell’energia, scrivi questo brevemente.

Schizzo raggio laser nel blocco di vetro (Abb. 3)

Percorso completo raggio laser vetro (Abb. 4)

Topic: Geometric Optics Metodi: Snell’s Law, Ray Tracing, Physical Modeling Competenze: Mathematical Modeling, Diagrammatic Reasoning Objects: — Fonte: Testo (PDF) — p.11

Problem 7 Glass block (five points) (Idea: Problem group of the PhysicsOlympic - Thomas Hellerl & Titus Bornträger) A laser beam running in the plane of the drawing strikes a glass block (refractive index $n=1,5$) with side lengths $a$ and $4a$ from the left at the angle of incidence $\alpha =30^{\circ }$. As indicated in the not-to-scale sketch in Figure 3, it finally strikes exactly the Lower right corner inside the glass block. $4a$ $a$ $n=1,5$ $\alpha$ $x$ Fig. 3. Not-to-scale sketch of the laser beam in the glass block in side view. What is the distance $x$ of the entry point from the upper boundary surface of the block? A $a\cdot \left(\sqrt{2}-1\right)$ B $a\cdot \left(2-\sqrt{3}\right)$ C $a\cdot \left(1-\frac{\sqrt{2}}{2}\right)$ D $a\cdot \left(1-\frac{\sqrt{3}}{3}\right)$ The solution Calculations and explanations $a$ $n=1,5$ $\alpha$ $x$ $y$ $4a-y$ ${\alpha }^{\prime }$ Fig. 4. Complete sketch of the beam path in the glass block in side view By the law of refraction: $\sin \alpha =n\cdot \sin {\alpha }^{\prime }.$ (7.1) The beam running inside the glass is totally internally reflected at the upper side. In the left and The right triangle, one finds on account of the similarity $\frac{a}{4a-y}=\frac{x}{y}=\tan {\alpha }^{\prime }.$ (7.2) The value of $\tan {\alpha }^{\prime }$ can be determined directly by means of $\tan {\alpha }^{\prime }=\frac{\sin {\alpha }^{\prime }}{\cos {\alpha }^{\prime }}=\frac{\frac{1}{n}\cdot \sin \alpha }{\sqrt{1-\frac{{\sin }^2\alpha }{n^2}}}=\frac{\frac{1}{1,5}\cdot \sin 30^{\circ }}{\sqrt{1-\frac{{\sin }^{2}30^{\circ }}{1,5^2}}}=\frac{1}{2\sqrt{2}}.$ (7.3) Here $\sin \alpha =\sin 30^{\circ }=0,5$ was used. It follows immediately from (7.2): $y=2\sqrt{2}x.$ (7.4) The Commission shall take into account the following information: Substituting into equation (7.2) and solving for $x$ gives: $\frac{a}{4a-2\sqrt{2}x}=\frac{1}{2\sqrt{2}}\text{bzw.}2\sqrt{2}a=4a-2\sqrt{2}x$ (7.5) and thus $x=2\sqrt{2}a-a=a\cdot \left(\sqrt{2}-1\right).$ (7.6) Correct answer: A Grading - Glass block Points 7 Using the similarity of the triangles (7.2) 1.0 Using the correct value for $\tan {\alpha }^{\prime }$ with (7.3) 1.0 Deriving the result for $x$ from (7.2) 1.0 Stating the correct solution 2.0 5.0 The Commission shall take into account the following information: Long-response problems Work on the following three problems also in the boxes provided for this. Unlike for the Multiple-choice problems, no answer options are given. Describe your solution method in such a way That it’s easy to follow but not unnecessarily long. If, for example, you use the law of conservation of energy, write this down briefly.

Schizzo raggio laser nel blocco di vetro (Abb. 3)

Percorso completo raggio laser vetro (Abb. 4)

Topic: Geometric Optics Metodi: Snell’s Law, Ray Tracing, Physical Modeling Competenze: Mathematical Modeling, Diagrammatic Reasoning Objects: — Fonte: Testo (PDF) — p.11

Problem 8 Laser Rangefinder (15 pts.) (Idea: Problem group of the PhysicsOlympiad - Bernd Schade & Jörg Steiper) For measuring rooms, laser rangefinders are often used. Laser rangefinders available in the hardware store can typically determine distances in the range from a few centimeters up to about 50 m with an accuracy of a few millimeters. For the distance measurement, the device emits a laser beam and receives the beam reflected from an object. 8.a) Calculate the travel time of laser light at a measurement distance of 50.0 cm. Determine how accurately this travel-time measurement would have to be carried out in order to achieve a measurement accuracy of $\pm 2$ mm. (4.0 pts.) Such a high time resolution is not achieved by ordinary laser rangefinders. Instead, the distance is determined via the phase shift of the emitted and received signal. However, this is not relevant for the following problems. The accuracy of the measurement is, however, also influenced by what is located in the light path. 8.b) You wish to measure the length of a thin-walled aquarium filled with water. Explain qualitatively why a measurement through the aquarium yields different values than a measurement with a ruler. (2.0 pts.) This effect can be used to determine the refractive index of a transparent material with a laser rangefinder. In the experiment sketched alongside, the distance to the bottom of a glass cylinder partially filled with a liquid is measured with a fixedly mounted laser rangefinder. The distance values displayed by the laser rangefinder for various liquid volumes are shown in the table below. The inner diameter of the glass cylinder is 8.00 cm. 8.c) Using the measurement series, determine the refractive index $n$ of the liquid. To do so, create a suitable graph. (9.0 pts.) Measured values of the distance $l$ measured by the laser rangefinder as a function of the liquid volume $V$ present in the cylinder 53rd IPhO 2023 - 2nd Round Exam - Sample Solution - 07.12.2022 $V$ / L $l$ / m $V$ / L $l$ / m 0.00 0.602 0.95 0.668 0.07 0.607 1.22 0.687 0.15 0.611 1.42 0.697 0.27 0.619 1.56 0.705 0.37 0.622 1.64 0.709 0.51 0.636 1.76 0.718 0.67 0.647 1.85 0.727 0.77 0.657 1.93 0.733 Solution 8.a) Calculations and explanations Since the light traverses the distance to be measured twice, the distance traveled by the light is $d=100,0$ cm. The speed of light in air, with $c\approx 3,00\cdot 10^8{\text{m s}}^{-1}$, corresponds approximately to the value in vacuum. Therefore the light needs, to traverse the distance $d$, the time $t=\frac{d}{c}=\frac{1,00\text{m}}{3,00\cdot 10^8{\text{m s}}^{-1}}\approx 3,33\cdot 10^{-9}\text{s}.$ (8.1) The light thus needs only a few nanoseconds to traverse the path. If the measurement accuracy of the distance measurement is to be 2 mm, the laser rangefinder must be able to temporally resolve a distance of $\Delta d=4$ mm. It must therefore be able to resolve a time difference $\Delta t$ with $\Delta t=\frac{\Delta d}{c}=\frac{0,004\text{m}}{3,00\cdot 10^8{\text{m s}}^{-1}}\approx 1\cdot 10^{-11}\text{s},$ (8.2) that is, in the range of ten picoseconds. 8.b) Calculations and explanations For the propagation of the laser beam, the medium being traversed must also be considered. Since the refractive index of water ($n_{\text{Wasser}}\approx 1,3$) is different from that of air ($n_{\text{Luft}}\approx 1,0$), the speed of light in the two media also differs. As a result, the light needs different travel times for the same path in the media and the laser rangefinder measures different distances. 8.c) Calculations and explanations The measured distance is composed of a path of length $l_{\text{Luft}}$ in air and a path traveled in the liquid of length $l_{\text{Fl}}$, which can be calculated from the liquid volume $V$ and the inner radius $r=4,00$ cm of the cylinder as $l_{\text{Fl}}=\frac{V}{\pi r^2}.$ (8.3) In the liquid, the light propagates at the velocity $c/n$, where $n$ denotes the refractive index of the liquid. As a result, the light needs, for the passage through 53rd IPhO 2023 - 2nd Round Exam - Sample Solution - 07.12.2022 the liquid, $n$ times as long as in air. Correspondingly, the length measured by the laser rangefinder for the partial path in the liquid increases by exactly this factor. The total path $l$ measured by the laser rangefinder is thus given by $l=l_{\text{Luft}}+n{l}_{\text{Fl}}=l_{\text{ges}}+(n-1)l_{\text{Fl}}=l_{\text{ges}}+\frac{n-1}{\pi r^2}V.$ (8.4) Here $l_{\text{ges}}=l_{\text{Luft}}+l_{\text{Fl}}$ denotes the constant distance between the laser rangefinder and the bottom of the glass cylinder. Equation (8.4) describes an (affine) linear relationship between the liquid volume $V$ and the measured distance $l$. If one therefore plots $l$ as a function of $V$ in a graph, a straight line with slope $b=\frac{n-1}{\pi r^2}$ must result as the best-fit curve. From the slope, the refractive index of the liquid can then be determined by means of $n=1+\pi r^{2}b$ (8.5) . In Figure 5, the data given in the problem are represented accordingly. 0.5 1.0 1.5 2.0 0.60 0.65 0.70 0.75 1.70 L 0.115 m $V$ / L $l$ / m Fig. 5. Graph of the distance $l$ measured by the laser rangefinder as a function of the liquid volume $V$ in the cylinder with best-fit line. From the best-fit line in the graph one obtains $b=\frac{0,115\text{m}}{1,70\text{L}}=67,6{\text{m}}^{-2}\text{mit Unsicherheit}\Delta b=2,6{\text{m}}^{-2}.$ (8.6) From this, for the refractive index of the liquid one obtains $n=1+\pi r^{2}b=1,34\pm 0,02.$ (8.7) The liquid could therefore be water. 53rd IPhO 2023 - 2nd Round Exam - Sample Solution - 07.12.2022 Grading - Laser Rangefinder Points 8.a) Using time equals distance divided by velocity 0.5 Accounting for the doubled distance for the light path 0.5 Calculating the time (8.1) 1.0 Recognizing the relationship between accuracy and time difference 0.5 Accounting for the doubled distance for the light path 0.5 Calculating the time difference (8.2) 1.0 8.b) Naming the different speeds of light 1.0 Stating that different travel times lead to different measurement results 1.0 8.c) Decomposing the distance into a part in air and one in the liquid 1.0 Expressing the path in the liquid via volume (8.3) 1.0 Recognizing that the optical path length in the liquid is longer by a factor $n$ 1.0 Setting up a linear relationship (8.4) 1.0 Creating a suitable graph from the measured values 2.0 Evaluating the slope 1.0 Result for the refractive index with $1,32\le n\le 1,36$ 2.0 15.0 53rd IPhO 2023 - 2nd Round Exam - Sample Solution - 07.12.2022

Misuratore laser su cilindro con liquido

Grafico distanza misurata vs volume (Abb. 5)

Topic: Geometric Optics Metodi: Physical Modeling, Graph Linearization, Experimental Data Analysis Competenze: Experimental Data Analysis, Graph Linearization, Mathematical Modeling Objects: Cylinder Fonte: Testo (PDF) — p.13

Problema 8 Laser Rangefinder 15 punti) (idea: gruppo problematico della PhysicsOlympiad - Bernd Schade & Jörg Steiper) Per le camere di misurazione, i laser sono spesso utilizzati. Laser rangefinders available in the hardware store può determinare distanze nella gamma da pochi centimetri fino a circa 50 m con un’accuratezza di pochi millimetri. Per la misurazione della distanza, il dispositivo emette un fascio laser e riceve il fascio riflesso da un oggetto. 8.a) Calcolare il tempo di viaggio della luce laser a una distanza di misurazione di 50,0 cm. Determine come Accurately this travel-time measurement would have to be carried out in order to raggiungere una misurazione accurata di $\pm 2$ mm. (4,0 p.) Una risoluzione di tempo così elevata non è raggiunta con i normali laser. Invece, la distanza è determinata tramite il phase shift del segnale emesso e ricevuto. Tuttavia, questo non è rilevante per i seguenti problemi. L’accuratezza della misurazione è, tuttavia, influenzata anche da ciò che è situato nel

• Il sentiero della luce. 8.b) Si desidera misurare la lunghezza di un acquario a sottili pareti pieno di acqua. Esprimi qualitativamente perché un’azione attraverso l’acquario produce valori diversi che una misurazione con un ruler. (punto 2.0) Questo effetto può essere utilizzato per determinare l’indice di refraczione di un materiale trasparente con un rangefinder laser. Nell’esperimento disegnato al fianco, la distanza al fondo di un cilindro di vetro parzialmente filled with a liquid è misurato con un rangefinder laser fissamente montato. I valori di distanza visualizzati dal laser rangefinder per vari volumi di liquidi sono mostrati nella tabella seguente. Il diametro interno del cilindro di vetro è di 8,00 cm. 8.c) Usando la serie di misurazione, determinare l’indice di refraczione $n$ del liquido. Per farlo, creare un grafico appropriato. (9,0 p.s.) Valori misurati della distanza $l$ measured by the laser rangefinder as a function of the liquid volume $V$ present in the cylinder 53° IPhO 2023 - 2° Round Exam - Sample Solution - 07.12.2022 $V$ / L $l$ / m $V$ / L $l$ / m 0.00 0.602 0.95 0.668 0.07 0.607 1.22 0.687 0.15 0.611 1.42 0.697 0.27 0.619 1.56 0.705 0.37 0.622 1.64 0.709 0.51 0.636 1.76 0.718 0.67 0.647 1.85 0.727 0.77 0.657 1.93 0.733 Soluzione 8.a) Calcoli e spiegazioni Dal momento che la luce attraversa la distanza da misurare due volte, la distanza percorsa dalla luce is $d=100,0$ cm. La velocità della luce in aria, con $c\approx 3,00\cdot 10^8{\text{m s}}^{-1}$, corrisponde approssimativamente al valore in vuoto. Pertanto la luce ha bisogno di attraversare la distanza $d$, il tempo $t=\frac{d}{c}=\frac{1,00\text{m}}{3,00\cdot 10^8{\text{m s}}^{-1}}\approx 3,33\cdot 10^{-9}\text{s}.$ (8.1) La luce ha quindi bisogno di pochi nanosegondi per attraversare il percorso. Se l’accuratezza di misurazione della distanza è di 2 mm, il laser rangefinder deve essere in grado di risolvere temporalmente una distanza di $\Delta d=4$ mm. Il programma deve quindi essere in grado di risolve a time difference $\Delta t$ with $\Delta t=\frac{\Delta d}{c}=\frac{0,004\text{m}}{3,00\cdot 10^8{\text{m s}}^{-1}}\approx 1\cdot 10^{-11}\text{s},$ (8.2) Cioè, nell’intervallo di dieci picosessoni. 8.b) Calcoli e spiegazioni Per la propagazione del fascio laser, il mezzo attraversato deve essere considerato. Poiché l’indice di refrazione dell’acqua ($n_{\text{Wasser}}\approx 1,3$) è diverso da quello dell’aria ($n_{\text{Luft}}\approx 1,0$), la velocità della luce nei due media è quindi diversa. Di conseguenza, la luce ha bisogno di diversi tempi di viaggio per lo stesso percorso nei media e il rangefinder laser misura distanze diverse. 8.c) Calcoli e spiegazioni La distanza misurata è composta da un percorso di lunghezza $l_{\text{Luft}}$ in aria e un percorso percorso in liquido di lunghezza $l_{\text{Fl}}$, che può essere calcolato dal volume liquido $V$ e dal raggio interno $r=4,00$ cm del cilindro come $l_{\text{Fl}}=\frac{V}{\pi r^2}.$ (8.3) In liquido, la luce si propaga alla velocità $c/n$, dove $n$ indica l’indice refraettivo del liquido. Come risultato, la luce ha bisogno, per il passaggio attraverso 53° IPhO 2023 - 2° Round Exam - Sample Solution - 07.12.2022 il liquido, $n$ volte più a lungo che in aria. Corrispondentemente, la lunghezza misurata dal laser rangefinder per il percorso parziale nel liquido aumenta esattamente questo fattore. Il percorso totale $l$ misurato dal laser rangefinder è così dato da $l=l_{\text{Luft}}+n{l}_{\text{Fl}}=l_{\text{ges}}+(n-1)l_{\text{Fl}}=l_{\text{ges}}+\frac{n-1}{\pi r^2}V.$ (8.4) Here $l_{\text{ges}}=l_{\text{Luft}}+l_{\text{Fl}}$ denotes the constant distance between the laser rangefinder e il fondo del cilindro di vetro. L’equazione (8.4) descrive una relazione lineare (affine) tra il volume liquido $V$ e la distanza misurata $l$. Se quindi si fa $l$ come funzione di $V$ in un grafico, una linea retta con slope $b=\frac{n-1}{\pi r^2}$ deve risultare come la curva migliore. Dal slope, l’indice refrazionale del liquido può quindi essere determinato $n=1+\pi r^{2}b$ (8.5) . In Figura 5, i dati forniti nel problema sono rappresentati in base a ciò. 0.5 1.0 1.5 2.0 0.60 0.65 0.70 0.75 1.70 L 0.115 m $V$ / L $l$ / m Fig. 5. Grafico della distanza $l$ misurata dal laser rangefinder come funzione di volume liquido $V$ nel cilindro con linea migliore. Dal linee migliore del grafico si ottiene $b=\frac{0,115\text{m}}{1,70\text{L}}=67,6{\text{m}}^{-2}\text{mit Unsicherheit}\Delta b=2,6{\text{m}}^{-2}.$ (8.6) From this, for the refractive index of the liquid one obtains $n=1+\pi r^{2}b=1,34\pm 0,02.$ (8.7) Il liquido potrebbe quindi essere acqua. 53° IPhO 2023 - 2° Round Exam - Sample Solution - 07.12.2022 Classificazione - Laser Rangefinder Punti 8.a) Usando il tempo uguale alla distanza dividuta dalla velocità 0.5 Accounting for the doubled distance for the light path 0.5 Calcolatore del tempo (8.1) 1.0 Recognizing the relationship between accuracy and time difference (Riconoscere la relazione tra accurazione e differenza di tempo) 0.5 Accounting for the doubled distance for the light path 0.5 Calcolatore della differenza di tempo (8.2) 1.0 8.b) Naming the different speeds of light 1.0 Stating that different travel times lead to different measurement results 1.0 8.c) Decomposizione della distanza in una parte in aria e una in liquido 1.0 Expressing the path in the liquid via volume (8.3) 1.0 Recognizing that the optical path length in the liquid is longer by a factor $n$ 1.0 Setting up a linear relationship (8.4) 1.0 Creating a suitable graph from the measured values 2.0 Evaluando la penetrazione 1.0 Result for the refractive index with $1,32\le n\le 1,36$ 2.0 15.0 53° IPhO 2023 - 2° Round Exam - Sample Solution - 07.12.2022

*Laser su cilindro con liquido *

Il numero di unità di misurazione è di circa un milione di unità di misurazione. 5)*

Topic: Geometric Optics Metodi: Physical Modeling, Graph Linearization, Experimental Data Analysis Competenze: Experimental Data Analysis, Graph Linearization, Mathematical Modeling Objects: Cylinder Fonte: Testo (PDF) — p.13

The problem is that the laser rangefinder (Figure 15) (Idea: problem group of the PhysicsOlympiad - Bernd Schade & Jörg Steiper) For measuring rooms, laser rangefinders are often used. Laser rangefinders available in the hardware store can typically determine distances in the range from a few centimeters up to about 50 m with an accuracy of a few millimeters. For the distance measurement, the device emits a laser beam and receives the beam reflected from an object. 8. (a) Calculate the travel time of laser light at a measurement distance of 50.0 cm. Determine how accurately this travel time measurement would have to be carried out in order to achieve a measurement accuracy of $\pm 2$ mm. (4.0 p.m.) Such a high time resolution is not achieved by ordinary laser rangefinders. Instead, the distance is determined via the phase shift of the emitted and received signal. However, this is not relevant for the following problems. The accuracy of the measurement is, however, also influenced by what is located in the Light path. 8.b) You wish to measure the length of a thin-walled aquarium filled with water. Explain qualitatively why a measurement through the aquarium yields different values than a measurement with a ruler. (b) the number of persons who have been This effect can be used to determine the refractive index of a transparent material with a laser rangefinder. In the experiment sketched alongside, the distance to the bottom of a glass cylinder partially filled with a liquid is measured with a fixedly mounted laser rangefinder. The distance values displayed by the laser rangefinder for various liquid volumes are shown in the table below. The inner diameter of the glass cylinder is 8.00 cm. 8.c) Using the measurement series, determine the refractive index $n$ of the liquid. To do that, create a suitable graph. (9.0 pts.) Measured values of the distance $l$ measured by the laser rangefinder as a function of the liquid volume $V$ present in the cylinder The Commission shall take into account the following: $V$ / L $l$ / m $V$ / L $l$ / m 0.00 0.602 0.95 0.668 0.07 0.607 1.22 0.687 0.15 0.611 1.42 0.697 0.27 0.619 1.56 0.705 0.37 0.622 1.64 0.709 0.51 0.636 1.76 0.718 0.67 0.647 1.85 0.727 0.77 0.657 1.93 0.733 The solution 8.a) Calculations and explanations Since the light traverses the distance to be measured twice, the distance traveled by the light is $d=100,0$ cm. The speed of light in air, with $c\approx 3,00\cdot 10^8{\text{m s}}^{-1}$, corresponds approximately to the value in vacuum. Therefore the light needs to cross the distance $d$, the time $t=\frac{d}{c}=\frac{1,00\text{m}}{3,00\cdot 10^8{\text{m s}}^{-1}}\approx 3,33\cdot 10^{-9}\text{s}.$ (8.1) The light thus needs only a few nanoseconds to cross the path. If the measurement accuracy of the distance measurement is to be 2 mm, the laser rangefinder must be able to temporarily resolve a distance of $\Delta d=4$ mm. It must therefore be able to resolve a time difference $\Delta t$ with $\Delta t=\frac{\Delta d}{c}=\frac{0,004\text{m}}{3,00\cdot 10^8{\text{m s}}^{-1}}\approx 1\cdot 10^{-11}\text{s},$ (8.2) That is, in the range of ten picoseconds. 8.b) Calculations and explanations For the propagation of the laser beam, the medium being traversed must also be considered. Since the refractive index of water ($n_{\text{Wasser}}\approx 1,3$) is different from that of air ($n_{\text{Luft}}\approx 1,0$), the speed of light in the two media is also different. As a result, the light needs different travel times for the same path in the media and the laser rangefinder measures different distances. 8.c) Calculations and explanations The measured distance is composed of a path of length $l_{\text{Luft}}$ in air and a path traveled in the liquid of length $l_{\text{Fl}}$, which can be calculated from the liquid volume $V$ and the inner radius $r=4,00$ cm of the cylinder as $l_{\text{Fl}}=\frac{V}{\pi r^2}.$ (8.3) In the liquid, the light propagates at the velocity $c/n$, where $n$ denotes the refractive index of the liquid. As a result, the light needs, for the passage through The Commission shall take into account the following information: The liquid, $n$ times as long as in air. Correspondingly, the length measured by the laser rangefinder for the partial path in the liquid increases by exactly this The Commission will take the necessary measures. The total path $l$ measured by the laser rangefinder is thus given by $l=l_{\text{Luft}}+n{l}_{\text{Fl}}=l_{\text{ges}}+(n-1)l_{\text{Fl}}=l_{\text{ges}}+\frac{n-1}{\pi r^2}V.$ (8.4) Here $l_{\text{ges}}=l_{\text{Luft}}+l_{\text{Fl}}$ denotes the constant distance between the laser rangefinder And the bottom of the glass cylinder. Equation (8.4) describes a (fine) linear relationship between the liquid volume $V$ and the measured distance $l$. If one plots $l$ as a function of $V$ in a graph, a straight line with slope $b=\frac{n-1}{\pi r^2}$ must result as the best-fit curve. From the slope, the refractive index of the liquid can then be determined by means of $n=1+\pi r^{2}b$ (8.5) . In Figure 5, the data given in the problem are represented accordingly. 0.5 1.0 1.5 2.0 0.60 0.65 0.70 0.75 1.70 L 0.115 m $V$ / L $l$ / m Fig. 5. Graph of the distance $l$ measured by the laser rangefinder as a function of the liquid volume $V$ in the cylinder with best-fit line. From the best-fit line in the graph one gets $b=\frac{0,115\text{m}}{1,70\text{L}}=67,6{\text{m}}^{-2}\text{mit Unsicherheit}\Delta b=2,6{\text{m}}^{-2}.$ (8.6) From this, for the refractive index of the liquid one obtains $n=1+\pi r^{2}b=1,34\pm 0,02.$ (8.7) The liquid could therefore be water. The Commission shall take into account the following information: Grading - Laser rangefinder Points 8.a) Using time equals distance divided by velocity 0.5 Accounting for the doubled distance for the light path 0.5 Calculating the time (8.1) 1.0 Recognizing the relationship between accuracy and time difference 0.5 Accounting for the doubled distance for the light path 0.5 Calculating the time difference (8.2) 1.0 8.b) Naming the different speeds of light 1.0 Stating that different travel times lead to different measurement results 1.0 8.c) Decomposing the distance into a part in air and one in the liquid 1.0 Expressing the path in the liquid via volume (8.3) 1.0 Recognizing that the optical path length in the liquid is longer by a factor $n$ 1.0 Setting up a linear relationship (8.4) 1.0 Creating a suitable graph from the measured values 2.0 Evaluating the slope 1.0 Result for the refractive index with $1,32\le n\le 1,36$ 2.0 15.0 The Commission shall take into account the following information:

The manufacturer shall provide the manufacturer with the following information:

The following table shows the number of units in the unit of measurement. 5)*

Topic: Geometric Optics Metodi: Physical Modeling, Graph Linearization, Experimental Data Analysis Competenze: Experimental Data Analysis, Graph Linearization, Mathematical Modeling Objects: Cylinder Fonte: Testo (PDF) — p.13

Problem 9 Rocket Launches and Satellites (20 pts.) The number of rocket launches has increased sharply in recent years - in 2021 there were more than 140 launches that aimed to reach an Earth orbit. During the launch phase, the rockets and their payloads are exposed to enormous loads. The aerodynamic load due to friction in the atmosphere plays an essential role. As a simple model, consider a rocket with a cone-shaped tip that has a diameter $d$ and an opening angle of $\alpha$ at the cone tip. The rocket flies at a velocity $v$ through the atmosphere, which at the current altitude has a density of ${\rho }_{\text{atm}}$. You may assume that the motion of the air molecules in the atmosphere is negligible compared to the rocket velocity. Through collisions of the rocket tip with the air molecules, treated as elastic for simplicity, the rocket experiences a friction force. 9.a) Derive an expression for the friction force acting on the rocket as a function of the parameters $d$, $\alpha$, $v$ and ${\rho }_{\text{atm}}$. Determine the magnitude of the friction force for the values $d=3,7$ m, $\alpha =90^{\circ }$, $v=2,0{\text{km s}}^{-1}$ and ${\rho }_{\text{atm}}=1,0\cdot 10^{-3}{\text{kg m}}^{-3}$. (4.0 pts.) R O C K E T S C I E N C E $v$ $d$ $\alpha$ The friction force acting on a rocket changes during the rocket flight. The following figures show the velocity of a rocket after launch as a function of the flight altitude (left) as well as the atmospheric pressure as a function of the altitude above the ground (right). For simplicity, it is assumed that the temperature of the atmosphere is constant. 10 20 30 40 50 0,5 1,0 1,5 2,0 $h$ / km $v$ / km ${\text{s}}^{-1}$ 10 20 30 40 50 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1,0 $h$ / km $p_{\text{atm}}$ / $10^5$ Pa Fig. 6. Velocity $v$ of the rocket (left) as well as air pressure $p_{\text{atm}}$ of the atmosphere (right) as a function of the altitude $h$ above the ground. 9.b) Using the data from the graphs, estimate the altitude above the ground at which the friction force on the rocket is maximal. (6.0 pts.) This point, which is critical during a launch, is called Max Q and denotes the location and time of greatest aerodynamic load on the rocket. To bring satellites into an Earth orbit, the rocket must accelerate further. Denote by $m_E=5,97\cdot 10^{24}$ kg the mass of the Earth and by $R_E=6,37\cdot 10^6$ m the Earth radius. 53rd IPhO 2023 - 2nd Round Exam - Sample Solution - 07.12.2022 9.c) Determine the velocity to which the rocket must accelerate before switching off the engines in order to be able to orbit the Earth in a low Earth orbit outside the atmosphere without crashing onto the Earth. Also state the orbital period on the orbit. (3.0 pts.) 9.d) Determine likewise the velocity to which the rocket must accelerate at minimum before switching off the engines in order to escape the influence of the Earth completely. State the ratio of this velocity to the one determined in the previous part of the problem. (3.0 pts.) Now assume that a satellite orbits the Sun on a path whose radius corresponds to the mean Earth orbital radius around the Sun of about $1,5\cdot 10^{11}$ m. Let the satellite be far away from the Earth and all other celestial bodies. The mass of the Sun is about $m_S=1,99\cdot 10^{30}$ kg and the radius of the Sun can be assumed to be very small compared to the Earth orbital radius. Suddenly the satellite stops completely relative to the Sun. 9.e) Estimate how long it takes until the satellite crashes into the Sun. Depending on the solution approach, Kepler’s laws can be helpful for this. (4.0 pts.) Solution 9.a) Calculations and explanations In the rocket’s frame, the air molecules strike the tip of the rocket head-on at velocity $v$. In the collision with the rocket tip, assumed to be elastic, they are deflected by an angle $\alpha$ relative to their original direction of motion, as sketched alongside, but retain their speed in magnitude. The momentum $\Delta p_{\text{Moleku¨l}}$ transferred in this process by an air molecule of mass $m_{\text{Moleku¨l}}$ to the rocket in the direction of the molecule’s original direction of motion is $\alpha$ $v$ $\alpha$ $d$ $\Delta p_{\text{Moleku¨l}}=m_{\text{Moleku¨l}}v(1-\cos \alpha ).$ (9.1) During a small time interval $\Delta t$, $\Delta n=Av\Delta t\frac{{\rho }_{\text{atm}}}{m_{\text{Moleku¨l}}}=\frac{\pi d^{2}v{\rho }_{\text{atm}}}{4m_{\text{Moleku¨l}}}\Delta t$ (9.2) air molecules strike the rocket. Here $A=\pi d^2/4$ denotes the cross-sectional area of the rocket. Thus the total momentum transferred per unit time by the air molecules to the rocket is $F=\frac{\Delta p}{\Delta t}=\frac{\Delta n\Delta p_{\text{Moleku¨l}}}{\Delta t}=\frac{\pi }{4}{d}^2{v}^2{\rho }_{\text{atm}}(1-\cos \alpha ).$ (9.3) The momentum change per unit time corresponds, according to Newton’s second law, exactly to the sought friction force $F$ on the rocket. With the given values $d=3,7$ m, $\alpha =90^{\circ }$, $v=2,0{\text{km s}}^{-1}$ and ${\rho }_{\text{atm}}=1,0\cdot 10^{-3}{\text{kg m}}^{-3}$, the value of the friction force comes out to $F\approx 4,3\cdot 10^4\text{N}.$ (9.4) 53rd IPhO 2023 - 2nd Round Exam - Sample Solution - 07.12.2022 9.b) Calculations and explanations According to (9.3), the friction force is proportional to the square of the rocket velocity and to the density of the air. If the temperature of the atmosphere is constant, the density of the air is, according to the equation of state of ideal gases, proportional to the pressure of the air. Thus the friction force on the rocket is proportional to the product $v^2\cdot p_{\text{atm}}$. All other dependencies in (9.3) are fixed by the geometry of the rocket and do not change during the flight . To determine when the friction force on the rocket becomes maximal, it is therefore sufficient to find the maximum of $v^2\cdot p_{\text{atm}}$ as a function of the altitude $h$. With the help of the data from the graph, this can be done e.g. graphically, as shown alongside. From the graph, the altitude at which the maximum friction force acts on the rocket comes out to about $h_{\text{Max Q}}\approx (15\pm 1)\text{km}.$ (9.5) Note: For the estimate in the exam, it is sufficient to evaluate the data pointwise and from this determine the altitude approximately. 10 20 30 40 50 10 20 30 40 50 $h$ / km $v^2{p}_{\text{atm}}$ / $10^8$ kg m ${\text{s}}^{-4}$ Fig. 7. Product $v^2\cdot p_{\text{atm}}$ at rocket launch as a function of the altitude $h$ above the ground. Alternatively, the altitude can also be estimated by the following consideration: The velocity $v$ is, with the exception of the first roughly 7 km, to a good approximation proportional to the altitude $h$. Thus the friction force is approximately proportional to $h^2{e}^{-h/h_{\text{scale}}}$ with a scale height $h_{\text{scale}}$ of 8.4 km determinable from the pressure graph. By setting the derivative of this function to zero, the altitude for the maximum load due to friction force can be estimated to be twice the scale height, that is, about 16.8 km. 9.c) Calculations and explanations In order to orbit the Earth at velocity $v$ outside the atmosphere and thus almost frictionlessly, the centripetal force acting on the rocket must be supplied by the gravitational force. If $m$ denotes the mass of the rocket and $R$ the radius of the circular orbit, it must therefore hold that: $F_{\text{Zentripetal}}=\frac{m{v}^2}{R}=\frac{Gm{m}_E}{R^2}=F_{\text{Gravitation}}.$ (9.6) Here $G=6,674\cdot 10^{-11}{\text{m}}^3{\text{kg}}^{-1}{\text{s}}^{-2}$ is the gravitational constant and $m_E=5,97\cdot 10^{24}$ kg is the mass of the Earth. The Earth’s atmosphere is very thin compared to the diameter of the Earth (cf. also the graph for the atmospheric pressure in the problem statement). Therefore, for a low Earth orbit, the orbital radius can be assumed to be about $R_E=6,37\cdot 10^6$ m. By 53rd IPhO 2023 - 2nd Round Exam - Sample Solution - 07.12.2022 rearranging (9.6), the velocity $v_1$ required for the circular orbit can thus be determined to be $v_1=\sqrt{\frac{G{m}_E}{R_E}}=7,91{\text{km s}}^{-1}.$ (9.7) This velocity is also called the first cosmic velocity. For somewhat larger assumed radii, the value becomes somewhat smaller. The corresponding orbital period is $T_1=\frac{2\pi R_E}{v_1}=2\pi R_E\sqrt{\frac{R_E}{G{m}_E}}=5,06\cdot 10^3\text{s}\approx 84,3\text{min}.$ (9.8) 9.d) Calculations and explanations In order to escape the influence of the Earth and thus its gravitational field, the velocity of the rocket at a great distance from the Earth must be at least zero. Otherwise the rocket would aga

Illustrazione razzo con angolo cono e velocità

Velocità e pressione atmosferica vs quota (Abb. 6)

Schema cono razzo con molecole incidenti

Prodotto v²·p_atm vs quota (Abb. 7)

Topic: Newtonian Mechanics, Gravitation, Kinetic Theory Metodi: Conservation of Momentum, Kepler’s Laws, Newton’s Law of Gravitation, Ideal Gas Law Competenze: Mathematical Modeling, Estimation & Approximation, Diagrammatic Reasoning Objects: Satellite, Planet, Star Fonte: Testo (PDF) — p.17

Problema 9 Lanci di razzi e satelliti (cfr. Il numero di lanci di razzi è aumentato notevolmente negli ultimi anni. Nel 2021 ci sono stati più di 140 lanci che hanno lo scopo di raggiungere l’orbita terrestre. Durante la fase di lancio, i razzi e i loro carichi sono esposti a enormi

• Loads. Il carico aerodinamico dovuto a La friczione nell’atmosfera gioca un ruolo essenziale. Come modello semplice, considerate un razzo con una punta conica che ha un diametro $d$ e un angolo di apertura di $\alpha$ alla punta del cono. Il razzo vola a velocità $v$ attraverso il atmosfera, che all’altitudine corrente ha una densità di ${\rho }_{\text{atm}}$. Tu può supporre che il movimento delle molecole d’aria nell’atmosfera è negligible rispetto alla velocità del razzo. Attraverso collisioni del filo di razzo con le molecole d’aria, trattate come elastic per semplicità, Il razzo sperimenta una forza di attrito. 9.a) Derivo di un’espressione per la forza di frizione che agisce sul razzo come a function of the parameters $d$, $\alpha$, $v$ and ${\rho }_{\text{atm}}$. Determinazione la magnitudine della forza di frizione per i valori $d=3,7$ m, $\alpha =90^{\circ }$, $v=2,0{\text{km s}}^{-1}$ e ${\rho }_{\text{atm}}=1,0\cdot 10^{-3}{\text{kg m}}^{-3}$. (4,0 p.) R O C K E T S C I E N C E $v$ $d$ $\alpha$ La forza di attrito che agisce su un razzo cambia durante il volo del razzo. Il seguente Figure mostrano la velocità di un razzo dopo il lancio come funzione del L’altitudine di volo (a sinistra) e la pressione atmosferica a funzione dell’altitudine sopra il suolo (Ritto) Per semplicità, si presume che la temperatura dell’atmosfera sia costante. 10 20 30 40 50 0,5 1,0 1,5 2,0 $h$ / km $v$ / km ${\text{s}}^{-1}$ 10 20 30 40 50 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1,0 $h$ / km $p_{\text{atm}}$ / $10^5$ Pa Fig. 6. Velocity $v$ of the rocket (left) as well as air pressure $p_{\text{atm}}$ of the atmosphere (right) as a funzione dell’altitudine $h$ sopra il suolo. 9.b) Usando i dati dei grafici, stimare l’altitudine sopra il terreno a cui il La forza di attrito sul razzo è massima. (6,0 p.p.) Questo punto, che è critico durante un lancio, è chiamato Max Q e denota la posizione e il tempo di Il più grande carico aerodinamico del razzo. Per portare i satelliti in orbita terrestre, il razzo deve accelerare ulteriormente. Denote per $m_E=5,97\cdot 10^{24}$ kg la massa della Terra e per $R_E=6,37\cdot 10^6$ m il raggio della Terra. 53° IPhO 2023 - 2° Round Exam - Sample Solution - 07.12.2022 9.c) Determina la velocità alla quale il razzo deve accelerare prima di spegnere i motori per poter orbitare la Terra in una bassa orbita terrestre al di fuori dell’atmosfera senza schiantarsi sulla Terra. Quindi, state il periodo orbitale in orbita. (Punto di riferimento) 9.d) Determina anche la velocità alla quale il razzo deve accelerare al minimo prima di spegnere i motori per sfuggire completamente all’influenza della Terra. Detti il rapporto di questa velocità con quella determinata nella parte precedente del problema. (Punto di riferimento) Ora supponiamo che un satellite orbita il Sole su un percorso il cui raggio corrisponde al medio Radius orbitale della Terra intorno al Sole di circa $1,5\cdot 10^{11}$ m. Lasciate che il satellite sia lontano dal Terra e tutti gli altri corpi celesti. La massa del Sole è di circa $m_S=1,99\cdot 10^{30}$ kg e Il raggio del Sole può essere considerato molto piccolo rispetto al raggio orbitale terrestre. All’improvviso il satellite si ferma completamente rispetto al Sole. 9.e) Estimare quanto tempo ci vorrà prima che il satellite crolla nel Sole. In base all’approccio di soluzione, Le leggi di Kepler possono essere utili per questo. (4,0 p.) Soluzione 9.a) Calcoli e spiegazioni Nel quadro del razzo, le molecole d’aria colpiscono la punta del razzo a velocità $v$. In collisione con la punta del razzo, presunto essere elastico, they are deflected by an angle $\alpha$ relative to their original direzione di movimento, come disegnato insieme, ma mantenere la loro velocità in grandezza. Il momento $\Delta p_{\text{Moleku¨l}}$ trasferito in questo processo da un’aria molecola di massa $m_{\text{Moleku¨l}}$ al razzo nella direzione della direzione originale del movimento della molecola is $\alpha$ $v$ $\alpha$ $d$ $\Delta p_{\text{Moleku¨l}}=m_{\text{Moleku¨l}}v(1-\cos \alpha ).$ (9.1) Durante un piccolo intervallo di tempo $\Delta t$, $\Delta n=Av\Delta t\frac{{\rho }_{\text{atm}}}{m_{\text{Moleku¨l}}}=\frac{\pi d^{2}v{\rho }_{\text{atm}}}{4m_{\text{Moleku¨l}}}\Delta t$ (9.2) Molcole di aria colpiscono il razzo. Qui $A=\pi d^2/4$ indica l’area cross-sectional del razzo. Così il momento totale trasferito per unità di tempo dalle molecole d’aria al razzo is $F=\frac{\Delta p}{\Delta t}=\frac{\Delta n\Delta p_{\text{Moleku¨l}}}{\Delta t}=\frac{\pi }{4}{d}^2{v}^2{\rho }_{\text{atm}}(1-\cos \alpha ).$ (9.3) Il cambiamento di momentum per unità di tempo corrisponde, secondo la seconda legge di Newton, esattamente alla sought friction force $F$ on the rocket. Con i dati $d=3,7$ m, $\alpha =90^{\circ }$, $v=2,0{\text{km s}}^{-1}$ e ${\rho }_{\text{atm}}=1,0\cdot 10^{-3}{\text{kg m}}^{-3}$, il valore della forza di frattura viene fuori a $F\approx 4,3\cdot 10^4\text{N}.$ (9.4) 53° IPhO 2023 - 2° Round Exam - Sample Solution - 07.12.2022 9.b) Calcoli e spiegazioni Secondo (9.3), la forza di attrito è proporzionale al quadrato della velocità del razzo e alla densità dell’aria. Se la temperatura dell’atmosfera è costante, la densità dell’atmosfera è L’aria è, secondo l’equazione dello stato di gas ideali, proporzionale alla pressione dell’aria. Così la La forza di frattura sul razzo è proporzionale al prodotto $v^2\cdot p_{\text{atm}}$. Tutti gli altri dipendenti in (9.3) sono fissati dalla geometria del razzo e non cambiano durante il volo . Per determinare quando la forza di attrito sul razzo diventa massima, è sufficiente per trovare il massimo di $v^2\cdot p_{\text{atm}}$ come funzione dell’altitudine $h$. Con l’aiuto dei dati del grafico, questo può essere fatto, per esempio. graficamente, Come mostrato al fianco. Dal grafico, dall’altitudine at which the maximum friction force acts on il razzo viene fuori a circa $h_{\text{Max Q}}\approx (15\pm 1)\text{km}.$ (9.5) Nota: Per l’estimo in Exam, it is sufficient to evaluate the data punto e da questo determinare il Altitudine circa. 10 20 30 40 50 10 20 30 40 50 $h$ / km $v^2{p}_{\text{atm}}$ / $10^8$ kg m ${\text{s}}^{-4}$ Fig. 7. Prodotto $v^2\cdot p_{\text{atm}}$ at rocket launch as a funzione dell’altitudine $h$ sopra il suolo. In alternativa, l’altitudine può anche essere stimata con la seguente considerazione: la velocità $v$ è, con l’eccezione del primo circa 7 km, a una buona approssimativa proporzionale alla Altezza $h$. Così la forza di frattura è approssimativamente proporzionale a $h^2{e}^{-h/h_{\text{scale}}}$ con un scale height $h_{\text{scale}}$ of 8.4 km determinable from the pressure graph. Setting the derivative of this function to zero, l’altitudine per il massimo carico dovuto alla forza di attrito può essere stimata per essere due volte più alto della scala, cioè circa 16 chilometri. 9.c) Calcoli e spiegazioni Per orbitare la Terra a velocità $v$ al di fuori dell’atmosfera e quindi quasi senza frizione, la forza centripetal che agisce sul razzo deve essere fornita dalla forza gravitazionale. Se $m$ indica la massa del razzo e $R$ il raggio del orbit circolare, deve pertanto contenere: $F_{\text{Zentripetal}}=\frac{m{v}^2}{R}=\frac{Gm{m}_E}{R^2}=F_{\text{Gravitation}}.$ (9.6) Qui $G=6,674\cdot 10^{-11}{\text{m}}^3{\text{kg}}^{-1}{\text{s}}^{-2}$ è la costante gravitazionale e $m_E=5,97\cdot 10^{24}$ kg è la massa della Terra. L’atmosfera terrestre è molto sottile rispetto al diametro della Terra (cfr. Quindi la grafica per la pressione atmosferica nella dichiarazione del problema). Pertanto, per un low Earth orbit, il raggio orbitale può essere presunto essere di circa $R_E=6,37\cdot 10^6$ m. By 53° IPhO 2023 - 2° Round Exam - Sample Solution - 07.12.2022 rearranging (9.6), the velocity $v_1$ required for the circular orbit can thus be determinato ad essere $v_1=\sqrt{\frac{G{m}_E}{R_E}}=7,91{\text{km s}}^{-1}.$ (9.7) Questa velocità è anche chiamata la prima velocità cosmica. In un certo senso Se si assume un raggio più grande, il valore diventa un po’ più piccolo. Il periodo orbitale corrispondente is $T_1=\frac{2\pi R_E}{v_1}=2\pi R_E\sqrt{\frac{R_E}{G{m}_E}}=5,06\cdot 10^3\text{s}\approx 84,3\text{min}.$ (9.8) 9.d) Calcoli e spiegazioni Per sfuggire all’influenza della Terra e quindi al suo campo gravitazionale, la velocità del razzo a grande distanza dalla Terra deve essere almeno zero. Altrimenti Il razzo sarebbe andato

Illustrazione razzo con angolo cono e velocità

Velocità e pressione atmosferica vs quota (Abb. 6)

Schema cono razzo con molecole incidenti

Prodotto v2·p_atm vs quota (Fig. 7)

Topic: Newtonian Mechanics, Gravitation, Kinetic Theory Metodi: Conservation of Momentum, Kepler’s Laws, Newton’s Law of Gravitation, Ideal Gas Law Competenze: Mathematical Modeling, Estimation & Approximation, Diagrammatic Reasoning Objects: Satellite, Planet, Star Fonte: Testo (PDF) — p.17

Problem 9 Rocket launches and satellites The Commission shall adopt implementing acts in accordance with Article 21 of this Regulation. The number of rocket launches has increased sharply in recent years - In 2021, there were more than 140 launches aimed at reaching Earth orbit. During the launch phase, the rockets and their payloads are exposed to enormous Loads. The aerodynamic load due to The effects of friction in the atmosphere are essential. As a simple model, consider a rocket with a cone-shaped tip that has a diameter $d$ and an opening angle of $\alpha$ at the cone tip. The rocket flies at a velocity $v$ through the atmosphere, which at the current altitude has a density of ${\rho }_{\text{atm}}$. You may assume that the motion of the air molecules in the atmosphere is negligible compared to the rocket velocity. Through collisions of the rocket tip with the air molecules, treated as elastic for simplicity, The rocket experiences a friction force. 9. (a) Derive an expression for the friction force acting on the rocket as a function of the parameters $d$, $\alpha$, $v$ and ${\rho }_{\text{atm}}$. Determine the magnitude of the friction force for the values $d=3,7$ m, $\alpha =90^{\circ }$, $v=2,0{\text{km s}}^{-1}$ and ${\rho }_{\text{atm}}=1,0\cdot 10^{-3}{\text{kg m}}^{-3}$. (4.0 p.m.) R O C K E T S C I E N C E $v$ $d$ $\alpha$ The friction force acting on a rocket changes during the rocket flight. The following Figures show the velocity of a rocket after launch as a function of the flight altitude (left) as well as the atmospheric pressure as a function of the altitude above the ground (right) For simplicity, it is assumed that the temperature of the atmosphere is constant. 10 20 30 40 50 0,5 1,0 1,5 2,0 $h$ / km $v$ / km ${\text{s}}^{-1}$ 10 20 30 40 50 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1,0 $h$ / km $p_{\text{atm}}$ / $10^5$ Pa Fig. 6. Velocity $v$ of the rocket (left) as well as air pressure $p_{\text{atm}}$ of the atmosphere (right) as a function of the altitude $h$ above the ground. 9.b) Using the data from the graphs, estimate the altitude above the ground at which the friction force on the rocket is maximum. (6.0 pts) This point, which is critical during a launch, is called Max Q and denotes the location and time of The greatest aerodynamic load on the rocket. To bring satellites into Earth orbit, the rocket must accelerate further. Other by $m_E=5,97\cdot 10^{24}$ kg the mass of the Earth and by $R_E=6,37\cdot 10^6$ m the Earth radius. The Commission shall take into account the following information: 9.c) Determine the velocity at which the rocket must accelerate before switching off the engines in order to be able to orbit the Earth in a low Earth orbit outside the atmosphere without crashing onto the Earth. So state the orbital period on the orbit. (including the following) 9. (d) Determine also the velocity at which the rocket must accelerate at least before switching off the engines in order to escape the Earth’s influence completely. State the ratio of this velocity to the one determined in the previous part of the problem. (including the following) Now suppose that a satellite orbits the Sun on a path whose radius corresponds to the mean Earth orbital radius around the Sun of about $1,5\cdot 10^{11}$ m. Let the satellite be far away from the Earth and all other celestial bodies. The mass of the Sun is about $m_S=1,99\cdot 10^{30}$ kg and The radius of the Sun can be assumed to be very small compared to the Earth orbital radius. Suddenly the satellite stops completely relative to the Sun. 9.e) Estimate how long it will take until the satellite crashes into the Sun. Depending on the solution approach, Kepler’s laws can be helpful for this. (4.0 p.m.) The solution 9.a) Calculations and explanations In the rocket’s frame, the air molecules strike the tip of the rocket head-on at velocity $v$. In the collision with the rocket tip, assumed to be elastic, they are deflected by an angle $\alpha$ relative to their original direction of motion, as sketched alongside, but retain their speed In magnitude. The momentum $\Delta p_{\text{Moleku¨l}}$ transferred in this process by an air molecule of mass $m_{\text{Moleku¨l}}$ to the rocket in the direction of the molecule’s original direction of motion is $\alpha$ $v$ $\alpha$ $d$ $\Delta p_{\text{Moleku¨l}}=m_{\text{Moleku¨l}}v(1-\cos \alpha ).$ (9.1) During a small time interval $\Delta t$, $\Delta n=Av\Delta t\frac{{\rho }_{\text{atm}}}{m_{\text{Moleku¨l}}}=\frac{\pi d^{2}v{\rho }_{\text{atm}}}{4m_{\text{Moleku¨l}}}\Delta t$ (9.2) Air molecules strike the rocket. Here $A=\pi d^2/4$ denotes the cross-sectional area of the rocket. Thus the total momentum transferred per unit time by the air molecules to the rocket is $F=\frac{\Delta p}{\Delta t}=\frac{\Delta n\Delta p_{\text{Moleku¨l}}}{\Delta t}=\frac{\pi }{4}{d}^2{v}^2{\rho }_{\text{atm}}(1-\cos \alpha ).$ (9.3) The momentum change per unit time corresponds, according to Newton’s second law, exactly to the sought friction force $F$ on the rocket. With the given values $d=3,7$ m, $\alpha =90^{\circ }$, $v=2,0{\text{km s}}^{-1}$ and ${\rho }_{\text{atm}}=1,0\cdot 10^{-3}{\text{kg m}}^{-3}$, the value of the friction force comes out to $F\approx 4,3\cdot 10^4\text{N}.$ (9.4) The Commission shall take into account the following information: 9.b) Calculations and explanations According to (9.3), the friction force is proportional to the square of the rocket velocity and to the density of the air. If the temperature of the atmosphere is constant, the density of the Air is, according to the equation of state of ideal gases, proportional to the pressure of the air. Thus the friction force on the rocket is proportional to the product $v^2\cdot p_{\text{atm}}$. All other dependencies in (9.3) are fixed by the geometry of the rocket and do not change during the flight . To determine when the friction force on the rocket becomes maximum, it is therefore sufficient to find the maximum of $v^2\cdot p_{\text{atm}}$ as a function of the altitude $h$. With the help of the data from the graph, this can be done e.g. graphically, As shown next to. From the graph, the altitude at which the maximum friction force acts on The rocket comes out to about $h_{\text{Max Q}}\approx (15\pm 1)\text{km}.$ (9.5) Note: For the estimate in the The data is evaluated by the pointwise and from this determine the altitude approximately. 10 20 30 40 50 10 20 30 40 50 $h$ / km $v^2{p}_{\text{atm}}$ / $10^8$ kg m ${\text{s}}^{-4}$ Fig. 7. Product $v^2\cdot p_{\text{atm}}$ at rocket launch as a function of the altitude $h$ above the ground. Alternatively, the altitude can also be estimated by the following consideration: The velocity $v$ is, with the exception of the first roughly 7 km, to a good approximation proportional to the altitude $h$. Thus the friction force is approximately proportional to $h^2{e}^{-h/h_{\text{scale}}}$ with a scale height $h_{\text{scale}}$ of 8.4 km determinable from the pressure graph. By setting the derivative of this function to zero, the altitude for the maximum load due to friction force can be estimated to be twice the height of the scale, that is, about 16.8 km. 9.c) Calculations and explanations In order to orbit the Earth at velocity $v$ outside the atmosphere and thus almost frictionlessly, the centripetal force acting on the rocket must be supplied by the gravitational force. If $m$ denotes the mass of the rocket and $R$ the radius of the rocket circular orbit, it must therefore hold that: $F_{\text{Zentripetal}}=\frac{m{v}^2}{R}=\frac{Gm{m}_E}{R^2}=F_{\text{Gravitation}}.$ (9.6) Here $G=6,674\cdot 10^{-11}{\text{m}}^3{\text{kg}}^{-1}{\text{s}}^{-2}$ is the gravitational constant and $m_E=5,97\cdot 10^{24}$ kg is the mass of the Earth. The Earth’s atmosphere is very thin compared to the diameter of the Earth (cf. So, what do you mean? The graph for the atmospheric pressure in the problem statement). Therefore, for a low Earth orbit, the orbital radius can be assumed to be about $R_E=6,37\cdot 10^6$ m. By The Commission shall take into account the following information: The speed $v_1$ required for the circular orbit can thus be determined to be $v_1=\sqrt{\frac{G{m}_E}{R_E}}=7,91{\text{km s}}^{-1}.$ (9.7) This velocity is also called the first cosmic velocity. For some reason If the radii are larger, the value becomes somewhat smaller. The corresponding orbital period is $T_1=\frac{2\pi R_E}{v_1}=2\pi R_E\sqrt{\frac{R_E}{G{m}_E}}=5,06\cdot 10^3\text{s}\approx 84,3\text{min}.$ (9.8) 9.d) Calculations and explanations In order to escape the influence of the Earth and thus its gravitational field, the velocity of the rocket at a great distance from the Earth must be at least zero. Otherwise The rocket would go

Illustrazione razzo con angolo cono e velocità

Velocità e pressione atmosferica vs quota (Abb. 6)

Schema cono razzo con molecole incidenti

Prodotto v²·p_atm vs quota (Abb. 7)

Topic: Newtonian Mechanics, Gravitation, Kinetic Theory Metodi: Conservation of Momentum, Kepler’s Laws, Newton’s Law of Gravitation, Ideal Gas Law Competenze: Mathematical Modeling, Estimation & Approximation, Diagrammatic Reasoning Objects: Satellite, Planet, Star Fonte: Testo (PDF) — p.17

Problem 10 Goethe Barometer (15 pts.) (Idea: Problem group of the PhysicsOlympiad - Thomas Hellerl) With a Goethe barometer, air pressure changes can be measured. It consists of a vessel closed at the top, which is filled with air in the upper part and with water in the lower part. This lower part is connected to the atmosphere via a vertical riser tube open at the top. Through changes in the external air pressure, the water level in the “spout” of the barometer falls or rises. At constant ambient temperature, the air pressure change $\Delta p$ is the sole cause of a water level change $\Delta h$ in the riser tube. Consider a simple Goethe barometer, as shown in the sketch. The cross-sectional areas of the riser tube and of the vessel are $A=0,30{\text{cm}}^2$ and $A_i=20{\text{cm}}^2$, respectively. At an air pressure of $p=1000$ hPa, the difference between the water levels in the riser tube and the vessel is $h=3,0$ cm, and the air volume enclosed in the vessel at pressure $p_i$ is $V_i=100{\text{cm}}^3$. The vapor pressure of the water is not to be taken into account. When the air pressure increases by $\Delta p$, the water level in the riser tube falls. Denote by $p^{\prime }$, $V_i^{\prime }$, $p_i^{\prime }$ and $h^{\prime }$ the quantities established at the changed air pressure according to the figure. (a) Photo of a Goethe barometer, CC BY-SA 3.0) $V_i$, $p_i$ $V_i^{\prime }$, $p_i^{\prime }$ $A_i$ $A$ $\Delta h<0$ $\Delta h_i>0$ $p$, $p^{\prime }$, $\Delta p=p^{\prime }-p>0$ $h$ $h^{\prime }$ (b) Simple Goethe barometer with constant cross sections. The red arrows indicate the displacement upon a pressure increase. Fig. 8. Weather glass or Goethe barometer 10.a) Derive a relationship between the air pressure change $\Delta p$ and the corresponding water level change $\Delta h$ in the riser tube. Calculate the value of the air pressure change when the water level in the riser tube falls by 1.0 cm. Use for the density of water $\rho =1,00\cdot 10^3{\text{kg m}}^{-3}$ and for the gravitational acceleration $g=9,81{\text{m s}}^{-2}$. (11.0 pts.) 10.b) Determine approximately how large the water level change in the riser tube would be for the same pressure change in a Goethe barometer scaled down by a factor 1:2. (4.0 pts.) Note: For $x\ll 1$ you may use the approximation $(1-x{)}^n\approx 1-nx$. 53rd IPhO 2023 - 2nd Round Exam - Sample Solution - 07.12.2022 Solution 10.a) Calculations and explanations For the enclosed gas volume, the law of Boyle and Mariotte holds, since the temperature of the system is assumed to be constant. $p_i\cdot V_i=p_i^{\prime }\cdot V_i^{\prime }.$ (10.1) The pressure of the inner air volume is always composed of the air pressure and the additional hydrostatic pressure of the water column above the inner water level. $(p+\rho gh)V_i=(p+\Delta p+\rho g(h+\Delta h-\Delta h_i))V_i^{\prime }.$ (10.2) Here, for an air pressure increase with $\Delta p>0$, the height changes are $\Delta h<0$ and $\Delta h_i>0$. The inner volume is reduced by exactly the volume of the water level decrease in the riser tube. $(p+\rho gh)V_i=(p+\Delta p+\rho g(h+\Delta h-\Delta h_i))(V_i+A\Delta h).$ (10.3) With the volume conservation of the incompressible water it follows that $-\Delta h_i=\frac{A}{A_i}\Delta h.$ (10.4) Substituted into (10.3), one obtains $(p+\rho gh)V_i=\left(p+\Delta p+\rho g\left(h+\Delta h+\frac{A}{A_i}\Delta h\right)\right)(V_i+A\Delta h).$ (10.5) Solving for $\Delta p$ finally yields $\Delta p=\frac{(p+\rho gh)V_i}{V_i+A\Delta h}-\left(p+\rho gh+\rho g\Delta h\right)\left(1+\frac{A}{A_i}\right)=(p+\rho gh)\left(\frac{1}{1+\frac{A}{V_i}\Delta h}-1\right)-\rho g\Delta h\left(1+\frac{A}{A_i}\right).$ (10.6) With the approximation $\frac{A}{A_i}=0,015\ll 1$ and $\frac{1}{1+x}\approx 1-x$ for $x\ll 1$, one obtains $\Delta p\approx -(p+\rho gh)\frac{A}{V_i}\Delta h-\rho g\Delta h\approx -\left(\frac{p\cdot A}{V_i}+\rho g\right)\Delta h.$ (10.7) In the second step, the approximation $p\gg \rho gh$ was additionally used. With (10.6) or (10.7), the value for the pressure difference can be determined to be $\Delta p\approx 4,0\text{hPa}.$ (10.8) 53rd IPhO 2023 - 2nd Round Exam - Sample Solution - 07.12.2022 10.b) Calculations and explanations In the approximate expression (10.7), the quotient $\frac{A}{V_i}$ is the only remaining vessel parameter that influences the relationship between pressure and height change. For a version of the Goethe barometer scaled down by a factor 2, $A$ is only $\frac{1}{4}$ of the original value and $V_i$ only $\frac{1}{8}$. Thus the quotient $\frac{A}{V_i}$ is then twice as large. Thus, for the height change $\Delta h_{\text{skal}}$ in the scaled Goethe barometer, with the help of (10.7), one obtains $\Delta h_{\text{skal}}\approx -\frac{\Delta p}{2\frac{pA}{V_i}+\rho g}\approx -0,57\text{cm}.$ (10.9) Grading - Goethe Barometer Points 10.a) Using the gas equation (law of Boyle and Mariotte) (10.1) 1.0 Accounting for the hydrostatic pressure 1.0 Formulating the change of state (10.3) with correct signs 3.0 Accounting for the volume conservation (10.5) 1.0 Substituting and solving for $\Delta p$ (10.6) 3.0 Calculating the result (10.8) (exactly or approximately) 2.0 10.b) Recognizing that $\frac{A}{V_i}$ is the only relevant vessel parameter 1.0 Correctly scaling the geometric quantities 2.0 Calculating the rise height in the scaled thermometer (10.9) 1.0 15.0

Foto e schema Goethebarometro (Abb. 8)

Topic: Thermodynamics, Fluid Mechanics Metodi: Ideal Gas Law, Hydrostatic Equilibrium, Approximation & Series Expansion Competenze: Mathematical Modeling, Physical Reasoning Objects: Manometer, Gas, Tube Fonte: Testo (PDF) — p.23

Problema 10 Barometro di Goethe 15 punti) (idea: gruppo di problemi della PhysicsOlympiad - Thomas Hellerl) Con un barometro di Goethe, si possono misurare i cambiamenti di pressione dell’aria. Consiste in una nave chiuso in cima, che è pieno di aria nella parte superiore e di acqua nella parte inferiore. Questo la parte inferiore è collegata all’atmosfera attraverso un tubo riser verticale aperto in cima. La pressione dell’aria esterna, il livello dell’acqua nello sput del Barometro di cassa o di rissi. At constant ambient temperature, the air pressure change $\Delta p$ is the sole cause di un cambiamento del livello di acqua $\Delta h$ nel tubo riser. Considerate un semplice barometro Goethe, come mostrato nello schizzo. Le aree cross-sectionali del riser tube and of the vessel are $A=0,30{\text{cm}}^2$ and $A_i=20{\text{cm}}^2$, respectively. Al di sotto della pressione dell’aria $p=1000$ hPa, la differenza tra i livelli di acqua nel tubo riser e il recipiente è $h=3,0$ cm, e il volume di aria chiuso nel recipiente a pressione $p_i$ è $V_i=100{\text{cm}}^3$. Il La pressione di vapore dell’acqua non deve essere presa in considerazione. Quando la pressione dell’aria aumenta di $\Delta p$, il livello dell’acqua nel tubo riser scende. Denote by $p^{\prime }$, $V_i^{\prime }$, $p_i^{\prime }$ e $h^{\prime }$ le quantità stabilite alla pressione dell’aria cambiata secondo la figura. (a) Photo of a Goethe barometer, CC BY-SA 3.0) $V_i$, $p_i$ $V_i^{\prime }$, $p_i^{\prime }$ $A_i$ $A$ $\Delta h<0$ $\Delta h_i>0$ $p$, $p^{\prime }$, $\Delta p=p^{\prime }-p>0$ $h$ $h^{\prime }$ (b) Barometro Goethe semplice con sezioni incrociate costanti. Le frecce rosse indicano il spostamento a causa di un aumento di pressione. Fig. 8. Meteo e barometri 10.a) Derivare una relazione tra il cambiamento di pressione dell’aria $\Delta p$ e il corrispondente cambiamento di livello dell’acqua $\Delta h$ nel tubo riser. Calcolare il valore del cambiamento di pressione dell’aria quando Il livello di acqua nel tubo riser scende di 1.0 cm. Use for the density of water $\rho =1,00\cdot 10^3{\text{kg m}}^{-3}$ e per l’accelerazione gravitazionale $g=9,81{\text{m s}}^{-2}$. (cfr. 10.b) Determina approssimativamente quanto grande sarebbe il cambiamento del livello dell’acqua nel tubo riser per lo stesso Cambiamento di pressione in un barometro di Goethe ridotto da un fattore 1:2. (4,0 p.) Nota: per $x\ll 1$ si può usare l’approssimazione $(1-x{)}^n\approx 1-nx$. 53° IPhO 2023 - 2° Round Exam - Sample Solution - 07.12.2022 Soluzione 10.a) Calcoli e spiegazioni Per il volume di gas chiuso, la legge di Boyle e Mariotte si mantiene, dal momento che la temperatura del sistema è presunta essere costante. $p_i\cdot V_i=p_i^{\prime }\cdot V_i^{\prime }.$ (10.1) La pressione del volume di aria interna è sempre composta dalla pressione dell’aria e dalla pressione idrostatica aggiuntiva della colonna d’acqua sopra il livello dell’acqua interna. $(p+\rho gh)V_i=(p+\Delta p+\rho g(h+\Delta h-\Delta h_i))V_i^{\prime }.$ (10.2) Qui, per un aumento di pressione con $\Delta p>0$, le variazioni di altezza sono $\Delta h<0$ e $\Delta h_i>0$. Il volume interno è ridotto esattamente dal volume del livello dell’acqua diminuito nel

• Il tubo riser. $(p+\rho gh)V_i=(p+\Delta p+\rho g(h+\Delta h-\Delta h_i))(V_i+A\Delta h).$ (10.3) Con il volume di conservazione dell’acqua incompressibile si segue che $-\Delta h_i=\frac{A}{A_i}\Delta h.$ (10.4) Substitute into (10.3), one obtains $(p+\rho gh)V_i=\left(p+\Delta p+\rho g\left(h+\Delta h+\frac{A}{A_i}\Delta h\right)\right)(V_i+A\Delta h).$ (10.5) Solving for $\Delta p$ finally yields $\Delta p=\frac{(p+\rho gh)V_i}{V_i+A\Delta h}-\left(p+\rho gh+\rho g\Delta h\right)\left(1+\frac{A}{A_i}\right)=(p+\rho gh)\left(\frac{1}{1+\frac{A}{V_i}\Delta h}-1\right)-\rho g\Delta h\left(1+\frac{A}{A_i}\right).$ (10.6) Con l’approssimazione $\frac{A}{A_i}=0,015\ll 1$ e $\frac{1}{1+x}\approx 1-x$ per $x\ll 1$, uno ottiene $\Delta p\approx -(p+\rho gh)\frac{A}{V_i}\Delta h-\rho g\Delta h\approx -\left(\frac{p\cdot A}{V_i}+\rho g\right)\Delta h.$ (10.7) In questo passo, si utilizza l’approssimazione $p\gg \rho gh$. Con (10.6) O (10.7), il valore per la differenza di pressione può essere determinato a $\Delta p\approx 4,0\text{hPa}.$ (10.8) 53° IPhO 2023 - 2° Round Exam - Sample Solution - 07.12.2022 10.b) Calcoli e spiegazioni In espressione approssimativa (10.7), il quotiente $\frac{A}{V_i}$ è l’unico vessel rimanente parametri che influenzano la relazione tra pressione e altezza di cambiamento. Per un version of the Goethe barometer scaled down by a factor 2, $A$ is only $\frac{1}{4}$ of the il valore originale e $V_i$ solo $\frac{1}{8}$. Quindi il quotiente $\frac{A}{V_i}$ è quindi due volte più grande. Così, per il cambiamento di altezza $\Delta h_{\text{skal}}$ nel barometro di Goethe a scala, con l’aiuto di di (10.7), one obtains $\Delta h_{\text{skal}}\approx -\frac{\Delta p}{2\frac{pA}{V_i}+\rho g}\approx -0,57\text{cm}.$ (10.9) Grading - Barometro di Goethe Punti 10.a) Utilizzo dell’equazione del gas (legge di Boyle e Mariotte) (10.1) 1.0 Contabilità della pressione idrostatica 1.0 Formulare il cambiamento di stato (10.3) with correct signs 3.0 Contabilità del volume di conservazione (10.5) 1.0 Substituzione e soluzione per $\Delta p$ (10.6) 3.0 Calcolatore del risultato (10.8) (exactly or approximately) 2.0 10.b) Recognizing that $\frac{A}{V_i}$ is the only relevant vessel parameter 1.0 Corretto scaling delle quantità geometriche 2.0 Calcolare l’altezza di aumento nel termometro scalato (10.9) 1.0 15.0

Foto e schema Goethebarometro (Abb. 8)

Topic: Thermodynamics, Fluid Mechanics Metodi: Ideal Gas Law, Hydrostatic Equilibrium, Approximation & Series Expansion Competenze: Mathematical Modeling, Physical Reasoning Objects: Manometer, Gas, Tube Fonte: Testo (PDF) — p.23

Problem 10 Goethe barometer (Figure 15) (Idea: Problem group of the PhysicsOlympic - Thomas Hellerl) With a Goethe barometer, air pressure changes can be measured. It consists of a vessel closed at the top, which is filled with air in the upper part and with water in the lower part. This Lower part is connected to the atmosphere via a vertical riser tube open at the top. Through changes in the external air pressure, the water level in the “spout” of the Barometer falls or rises. At constant ambient temperature, the air pressure change $\Delta p$ is the sole cause of a water level change $\Delta h$ in the riser tube. Consider a simple Goethe barometer, as shown in the sketch. The cross-sectional areas of the riser tube and of the vessel are $A=0,30{\text{cm}}^2$ and $A_i=20{\text{cm}}^2$, respectively. At an air pressure of $p=1000$ hPa, the difference between the water levels in the riser tube and the vessel is $h=3,0$ cm, and the air volume enclosed in the vessel at pressure $p_i$ is $V_i=100{\text{cm}}^3$. The The vapor pressure of the water is not to be taken into account. When the air pressure increases by $\Delta p$, the water level in the riser tube falls. Denote by $p^{\prime }$, $V_i^{\prime }$, $p_i^{\prime }$ and $h^{\prime }$ the quantities established at the changed air pressure according to the figure. (a) Photo of a Goethe barometer, CC BY-SA 3.0) $V_i$, $p_i$ $V_i^{\prime }$, $p_i^{\prime }$ $A_i$ $A$ $\Delta h<0$ $\Delta h_i>0$ $p$, $p^{\prime }$, $\Delta p=p^{\prime }-p>0$ $h$ $h^{\prime }$ (b) Simple Goethe barometer with constant cross sections. The red arrows indicate the displacement upon a pressure increase. Fig. 8. Weather glass or Goethe barometer 10. (a) Derive a relationship between the air pressure change $\Delta p$ and the corresponding water level change $\Delta h$ in the riser tube. Calculate the value of the air pressure change when The water level in the riser tube drops by 1.0 cm. Use for the density of water $\rho =1,00\cdot 10^3{\text{kg m}}^{-3}$ and for the gravitational acceleration $g=9,81{\text{m s}}^{-2}$. (including the European Parliament and the Council) 10.b) Determine approximately how large the water level change in the riser tube would be for the same Pressure change in a Goethe barometer scaled down by a factor of 1:2. (4.0 p.m.) Note: For $x\ll 1$ you may use the approximation $(1-x{)}^n\approx 1-nx$. The Commission shall take into account the following: The solution 10.a) Calculations and explanations For the enclosed gas volume, the law of Boyle and Mariotte holds, since the temperature of the system is assumed to be constant. $p_i\cdot V_i=p_i^{\prime }\cdot V_i^{\prime }.$ (10.1) The pressure of the inner air volume is always composed of the air pressure and the additional hydrostatic pressure of the water column above the inner water level. $(p+\rho gh)V_i=(p+\Delta p+\rho g(h+\Delta h-\Delta h_i))V_i^{\prime }.$ (10.2) Here, for an air pressure increase with $\Delta p>0$, the height changes are $\Delta h<0$ and $\Delta h_i>0$. The internal volume is reduced by exactly the volume of the water level decrease in the The reservoir is closed. $(p+\rho gh)V_i=(p+\Delta p+\rho g(h+\Delta h-\Delta h_i))(V_i+A\Delta h).$ (10.3) With the volume conservation of the incompressible water it follows that $-\Delta h_i=\frac{A}{A_i}\Delta h.$ (10.4) Substituted into (10.3), one obtains $(p+\rho gh)V_i=\left(p+\Delta p+\rho g\left(h+\Delta h+\frac{A}{A_i}\Delta h\right)\right)(V_i+A\Delta h).$ (10.5) Solving for $\Delta p$ finally yields $\Delta p=\frac{(p+\rho gh)V_i}{V_i+A\Delta h}-\left(p+\rho gh+\rho g\Delta h\right)\left(1+\frac{A}{A_i}\right)=(p+\rho gh)\left(\frac{1}{1+\frac{A}{V_i}\Delta h}-1\right)-\rho g\Delta h\left(1+\frac{A}{A_i}\right).$ (10.6) With the approximation $\frac{A}{A_i}=0,015\ll 1$ and $\frac{1}{1+x}\approx 1-x$ for $x\ll 1$, one obtains $\Delta p\approx -(p+\rho gh)\frac{A}{V_i}\Delta h-\rho g\Delta h\approx -\left(\frac{p\cdot A}{V_i}+\rho g\right)\Delta h.$ (10.7) In the second step, the approximation $p\gg \rho gh$ was additionally used. With (10.6) or (10.7), the value for the pressure difference can be determined to be $\Delta p\approx 4,0\text{hPa}.$ (10.8) The Commission shall take into account the following: 10.b) Calculations and explanations In the approximate expression (10.7), the quotient $\frac{A}{V_i}$ is the only remaining vessel Parameters that influence the relationship between pressure and height change. For a version of the Goethe barometer scaled down by a factor 2, $A$ is only $\frac{1}{4}$ of the The value of the original value and $V_i$ only $\frac{1}{8}$. Thus the quotient $\frac{A}{V_i}$ is then twice as large. Thus, for the height change $\Delta h_{\text{skal}}$ in the scaled Goethe barometer, with the help of (10.7), one obtains $\Delta h_{\text{skal}}\approx -\frac{\Delta p}{2\frac{pA}{V_i}+\rho g}\approx -0,57\text{cm}.$ (10.9) Grading - Goethe barometer Points 10.a) Using the gas equation (law of Boyle and Mariotte) (10.1) 1.0 Accounting for hydrostatic pressure 1.0 Formulating the change of state (10.3) with correct signs 3.0 Accounting for the volume conservation (10.5) 1.0 Substitution and solving for $\Delta p$ (10.6) 3.0 Calculating the result (10.8) (exactly or approximately) 2.0 10.b) Recognizing that $\frac{A}{V_i}$ is the only relevant vessel parameter 1.0 Correctly scaling the geometric quantities 2.0 Calculating the rise height in the scaled thermometer (10.9) 1.0 15.0

Foto e schema Goethebarometro (Abb. 8)

Topic: Thermodynamics, Fluid Mechanics Metodi: Ideal Gas Law, Hydrostatic Equilibrium, Approximation & Series Expansion Competenze: Mathematical Modeling, Physical Reasoning Objects: Manometer, Gas, Tube Fonte: Testo (PDF) — p.23