Add IPhO 2021-2025 theory problems (by_problem)

by_problem/IPhO_2022_3.jsonl

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+{"index": "IPhO_2022_3_A_1", "category": "physics", "source_dataset": "IPhO_2021_2025_extracted", "source_dataset_url": null, "year": 2022, "problem_id": "IPhO_2022_3", "problem_number": 3, "part_id": "A.1", "part_letter": "A", "subquestion_number": 1, "formalization_input_policy": {"previous_parts": "Curated natural-language prerequisites only; do not import or depend on previous Lean outputs.", "images": "Use listed images. Figure choices were curated by semantic review when heuristic PDF alignment was too broad.", "answers": "Answers are official-solution excerpts split to subquestion scope where possible; broad fallbacks are listed in review metadata."}, "context": "Scaling laws (8 points)\nScaling laws describe the functional relationship between two physical quantities that scale with each\nother over a significant interval. This functional relationship can be a power law, but there are other\npossibilities, too. Oftentimes, exact expressions are beyond reach, but scaling laws can still be derived.\nPart A. Spaghetto (2.0 points)", "current_question": "A spaghetto straw of diameter 𝑑is being balanced horizontally from its middle.\nIf 𝑑= 1 mm, the straw breaks under its own weight once its length reaches\n𝑙= 50 cm. What is the maximum length 𝑙′ of the straw of diameter 𝑑′ = 1 cm\nbefore it breaks under its own weight?", "question": "Scaling laws (8 points)\nScaling laws describe the functional relationship between two physical quantities that scale with each\nother over a significant interval. This functional relationship can be a power law, but there are other\npossibilities, too. Oftentimes, exact expressions are beyond reach, but scaling laws can still be derived.\nPart A. Spaghetto (2.0 points)\n\nCurrent subquestion:\nA spaghetto straw of diameter 𝑑is being balanced horizontally from its middle.\nIf 𝑑= 1 mm, the straw breaks under its own weight once its length reaches\n𝑙= 50 cm. What is the maximum length 𝑙′ of the straw of diameter 𝑑′ = 1 cm\nbefore it breaks under its own weight?", "answer": "Task A: Spaghetti (2 pts)\nThis is section 2.2.2 (Statics) of the syllabus.\nConsider only the left half of the spaghetti straw.\nTorque balance at its right endpoint implies that the\ntorque applied to its right endpoint must balance out the\ntorque due to gravity: τ ∝ml ∝d2l2. This torque arises\nfrom the gradient in the horizontal stress. If the typical\nhorizontal stress is σ, then the typical force is F ∝σd2,\nso the torque is τ ∝Fd ∝σd3. Hence, we obtain\nσd3 ∝d2l2 =⇒l ∝\n√\nd,\nso\nl′ =\nr\nd′\nd l =\n√\n10 · 50 cm = 158 cm.\nMarking scheme:\nτ ∝d2l2\n0.4 pts\nF ∝σd2\n0.5 pts\nτ ∝σd3\n0.5 pts\nl ∝\n√\nd\n0.4 pts\nAnswer: 158 cm\n0.2 pts", "answers": ["Task A: Spaghetti (2 pts)\nThis is section 2.2.2 (Statics) of the syllabus.\nConsider only the left half of the spaghetti straw.\nTorque balance at its right endpoint implies that the\ntorque applied to its right endpoint must balance out the\ntorque due to gravity: τ ∝ml ∝d2l2. This torque arises\nfrom the gradient in the horizontal stress. If the typical\nhorizontal stress is σ, then the typical force is F ∝σd2,\nso the torque is τ ∝Fd ∝σd3. Hence, we obtain\nσd3 ∝d2l2 =⇒l ∝\n√\nd,\nso\nl′ =\nr\nd′\nd l =\n√\n10 · 50 cm = 158 cm.\nMarking scheme:\nτ ∝d2l2\n0.4 pts\nF ∝σd2\n0.5 pts\nτ ∝σd3\n0.5 pts\nl ∝\n√\nd\n0.4 pts\nAnswer: 158 cm\n0.2 pts"], "answer_type": ["Official solution excerpt"], "unit": [null], "points": [2.0], "marking": [["Task A: Spaghetti (2 pts)\nThis is section 2.2.2 (Statics) of the syllabus.\nConsider only the left half of the spaghetti straw.\nTorque balance at its right endpoint implies that the\ntorque applied to its right endpoint must balance out the\ntorque due to gravity: τ ∝ml ∝d2l2. This torque arises\nfrom the gradient in the horizontal stress. If the typical\nhorizontal stress is σ, then the typical force is F ∝σd2,\nso the torque is τ ∝Fd ∝σd3. Hence, we obtain\nσd3 ∝d2l2 =⇒l ∝\n√\nd,\nso\nl′ =\nr\nd′\nd l =\n√\n10 · 50 cm = 158 cm.\nMarking scheme:\nτ ∝d2l2\n0.4 pts\nF ∝σd2\n0.5 pts\nτ ∝σd3\n0.5 pts\nl ∝\n√\nd\n0.4 pts\nAnswer: 158 cm\n0.2 pts"]], "modality": "text-only", "field": "Scaling laws", "source": "IPhO_2022", "previous_parts": [], "previous_part_count": 0, "images": [], "image": null, "image_count": 0}
+{"index": "IPhO_2022_3_B_1", "category": "physics", "source_dataset": "IPhO_2021_2025_extracted", "source_dataset_url": null, "year": 2022, "problem_id": "IPhO_2022_3", "problem_number": 3, "part_id": "B.1", "part_letter": "B", "subquestion_number": 1, "formalization_input_policy": {"previous_parts": "Curated natural-language prerequisites only; do not import or depend on previous Lean outputs.", "images": "Use listed images. Figure choices were curated by semantic review when heuristic PDF alignment was too broad.", "answers": "Answers are official-solution excerpts split to subquestion scope where possible; broad fallbacks are listed in review metadata."}, "context": "Scaling laws (8 points)\nScaling laws describe the functional relationship between two physical quantities that scale with each\nother over a significant interval. This functional relationship can be a power law, but there are other\npossibilities, too. Oftentimes, exact expressions are beyond reach, but scaling laws can still be derived.\nPart A. Spaghetto (2.0 points)\nA.1\nA spaghetto straw of diameter 𝑑is being balanced horizontally from its middle.\nIf 𝑑= 1 mm, the straw breaks under its own weight once its length reaches\n𝑙= 50 cm. What is the maximum length 𝑙′ of the straw of diameter 𝑑′ = 1 cm\nbefore it breaks under its own weight?\n2.0pt\nPart B. Sand castle (2.0 points)", "current_question": "The average grain volume of coarse-grained sand is 10 times as large as that\nof fine-grained sand. Wet fine-grained sand and wet coarse-grained sand have\nboth optimal water content (i.e. assuming the maximal strength of the con-\nstructions from it) and are used to build two cylinders of exactly the same shape\nand size. The strength of each cylinder is tested by pressing it between two par-\nallel plates. The cylinder made of coarse-grained sand gets destroyed once the\nforce applied to press the plates reaches 𝐹c = 10 N. How large is the force 𝐹f\nneeded to destroy the cylinder made of fine-grained sand? You may ignore the\neffects of gravity.", "question": "Scaling laws (8 points)\nScaling laws describe the functional relationship between two physical quantities that scale with each\nother over a significant interval. This functional relationship can be a power law, but there are other\npossibilities, too. Oftentimes, exact expressions are beyond reach, but scaling laws can still be derived.\nPart A. Spaghetto (2.0 points)\nA.1\nA spaghetto straw of diameter 𝑑is being balanced horizontally from its middle.\nIf 𝑑= 1 mm, the straw breaks under its own weight once its length reaches\n𝑙= 50 cm. What is the maximum length 𝑙′ of the straw of diameter 𝑑′ = 1 cm\nbefore it breaks under its own weight?\n2.0pt\nPart B. Sand castle (2.0 points)\n\nCurrent subquestion:\nThe average grain volume of coarse-grained sand is 10 times as large as that\nof fine-grained sand. Wet fine-grained sand and wet coarse-grained sand have\nboth optimal water content (i.e. assuming the maximal strength of the con-\nstructions from it) and are used to build two cylinders of exactly the same shape\nand size. The strength of each cylinder is tested by pressing it between two par-\nallel plates. The cylinder made of coarse-grained sand gets destroyed once the\nforce applied to press the plates reaches 𝐹c = 10 N. How large is the force 𝐹f\nneeded to destroy the cylinder made of fine-grained sand? You may ignore the\neffects of gravity.", "answer": "Task B: Sand castle (2 pts)\nThis is Section 2.2.2 (statics) and 2.2.5 (hydrodynamics) of\nthe syllabus\nDue to wetting of the surfaces of the sand grains and\nits large surface tension water acts like a glue for sand.\nThis means that all the grains need to be bound together\nby air-water interface. To achieve this there needs to be\nneither too little nor too much water: if there is too little\nwater, most of the grains are dry with no surface ten-\nsion binding them, and if there is too much water, al-\nmost all the grains are immersed into water, and again,\nthere is no surface tension binding the grains. So, the\noverall strength of the buildings from wet sand depends\non the water content; we assume that for the both types\nof sand, the water content is optimal, and the shape of\nthe grains is statistically similar.\nLet us consider two\nneighbouring grains connected by a water meniscus —\nor “neck”, as we shall be referring to it henceforth. Note\nthat the “neck” may extend perpendicularly to the fig-\nure plane far away; so, more specifically, what the word\n“neck” will refer to is that part of the water-air interface\nfor which the closest two grains are the ones under con-\nsideration.\nThere are two processes binding the sand grains to-\ngether. The first one is the force due to the surface ten-\nsion, F1 = γl, where γ denotes the surface tension coef-\nficient, and l — the perimeter of the “neck”; with l ∼rg,\nwhere rg denotes the length scale of a single grain, we ob-\ntain Fs ∼γr. The second one is the pressure force caused\nby the negative capillary pressure in the neck, Fp = ∆pA,\nwhere A is the cross-sectional area of the “neck”, and\n∆p ∼γ/r. With A ∼r2 we obtain F2 ∼γr. Thus, the both\ncomponents are of the same order of magnitude and us-\ning either of them will lead to the correct scaling law.\nThese forces press the grains against each other, hence\nthe normal force and friction force between the grains is\nalso on the order of Fs and Fp.\nSolution 1:\nBased on what has been said above, the typical force\nneeded to delocate a grain of sand is Fg ∝rg. The force\nneeded to delocate an entire horizontal layer of sand is\nthen ∝FgNl, where Nl ∼A/r2\ng is the number of grains in\na layer. The force of cylinder destruction F thus satisfies\nF ∝FgNl ∝rg/r2\ng = r−1\ng\n∝V −1/3\ng\n,\nso\nFfg = (1/10)−1/3 · Fcg = 21.5 N.\nMarking scheme:\na) Fs ∝rg and Fp ∝rg\n0.5 pts\none of the two missing\n-0.1 pts\nb) Fg ∝rg\n0.5 pts\nc) F ∝FgNl\n0.5 pts\nd) F ∝r−1\ng\n0.3 pts\nAnswer: 21.5 N\n0.2 pts\nNotes:\nIf the student only qualitatively explains the\nmechanism by which the grains of sand are held to-\ngether, a maximum of 0.5 pts are given. Points b)-c) are\ngiven only if derived from a).\nSolution 2:\nWe have seen above that grains to one side of a ficti-\ntious surface exert force per cross-sectional area on the\norder of magnitude as the capillary pressure ∆p ∼γ/rg.\nIn order to get the grains moving, a pressure of the same\norder of magnitude needs to be applied externally. For\nboth cylinders, the surface area where the force is ap-\nplied is the same, hence the force scales as the capillary\npressure, F ∝1/rg ∝V −1/3.\nMarking scheme:\nThe applied pressure must be ∼∆p\n0.6 pts\nThe curvature radius of the interface is ∼rg\n0.6 pts\nCapillary pressure ∆p ∼γ/rg\n0.6 pts\nAnswer: 21.5 N\n0.2 pts\nNote: If the contribution of surface tension is neglected,\n0.1 pts are subtracted.\nSolution 3:\nThe compression force serves to break the surface ten-\nsion bonds between sand grains.\nConsider the energy E required to push a single layer\nof sand into the layer beneath it. E ∝Frg, where F is the\nIPhO 2022\nTheoretical problems: solutions. Language: English\nforce required and rg is the typical height of a layer (i.e.,\nthe typical length scale of a grain).\nOn the other hand, E = γ∆A, where γ is the surface\ntension of water and ∆A is the total amount by which\nthe surface of the water in the layer stretches before all\nthe “water bonds” between the sand grains are broken.\nHere, ∆A is proportional to the area A of a layer and is\nthus a constant between the two cylinders. Hence, E ∝\nFrg is a constant between the two cylinders, i.e., F ∝r−1\ng .\nMarking scheme:\nE ∝Frg\n0.5 pts\nE ∝γ∆A\n0.5 pts\n∆A ∝A\n0.5 pts\nF ∝r−1\ng\n0.3 pts\nAnswer: 21.5 N\n0.2 pts\nNote: If the student only qualitatively explains the mech-\nanism by which the grains of sand are held together, a\nmaximum of 0.5 pts are given.\nSolution 4:\nFirst of all, the force F should be proportional to the\ncylinder’s base area A. The force required to destroy a\ncylinder with base area A = nA0 is equal to the force\nrequired to destroy n cylinders each with base area A0.\nAs a result, F ∝n ∝A.\nIn addition, F depends on the grain’s length scale rg\nand the water’s surface tension γ. Dimensional analysis\nthus gives\nF ∝Aγ\nrg\n∝r−1\ng\nfor fixed A and γ.\nMarking scheme:\nF ∝A\n0.6 pts\nF = F(A, rg, γ)\n0.6 pts\nF ∝Aγ\nrg\n0.6 pts\nAnswer: 21.5 N\n0.2 pts\nNote: If the student only qualitatively explains the mech-\nanism by which the grains of sand are held together, a\nmaximum of 0.5 pts are given.", "answers": ["Task B: Sand castle (2 pts)\nThis is Section 2.2.2 (statics) and 2.2.5 (hydrodynamics) of\nthe syllabus\nDue to wetting of the surfaces of the sand grains and\nits large surface tension water acts like a glue for sand.\nThis means that all the grains need to be bound together\nby air-water interface. To achieve this there needs to be\nneither too little nor too much water: if there is too little\nwater, most of the grains are dry with no surface ten-\nsion binding them, and if there is too much water, al-\nmost all the grains are immersed into water, and again,\nthere is no surface tension binding the grains. So, the\noverall strength of the buildings from wet sand depends\non the water content; we assume that for the both types\nof sand, the water content is optimal, and the shape of\nthe grains is statistically similar.\nLet us consider two\nneighbouring grains connected by a water meniscus —\nor “neck”, as we shall be referring to it henceforth. Note\nthat the “neck” may extend perpendicularly to the fig-\nure plane far away; so, more specifically, what the word\n“neck” will refer to is that part of the water-air interface\nfor which the closest two grains are the ones under con-\nsideration.\nThere are two processes binding the sand grains to-\ngether. The first one is the force due to the surface ten-\nsion, F1 = γl, where γ denotes the surface tension coef-\nficient, and l — the perimeter of the “neck”; with l ∼rg,\nwhere rg denotes the length scale of a single grain, we ob-\ntain Fs ∼γr. The second one is the pressure force caused\nby the negative capillary pressure in the neck, Fp = ∆pA,\nwhere A is the cross-sectional area of the “neck”, and\n∆p ∼γ/r. With A ∼r2 we obtain F2 ∼γr. Thus, the both\ncomponents are of the same order of magnitude and us-\ning either of them will lead to the correct scaling law.\nThese forces press the grains against each other, hence\nthe normal force and friction force between the grains is\nalso on the order of Fs and Fp.\nSolution 1:\nBased on what has been said above, the typical force\nneeded to delocate a grain of sand is Fg ∝rg. The force\nneeded to delocate an entire horizontal layer of sand is\nthen ∝FgNl, where Nl ∼A/r2\ng is the number of grains in\na layer. The force of cylinder destruction F thus satisfies\nF ∝FgNl ∝rg/r2\ng = r−1\ng\n∝V −1/3\ng\n,\nso\nFfg = (1/10)−1/3 · Fcg = 21.5 N.\nMarking scheme:\na) Fs ∝rg and Fp ∝rg\n0.5 pts\none of the two missing\n-0.1 pts\nb) Fg ∝rg\n0.5 pts\nc) F ∝FgNl\n0.5 pts\nd) F ∝r−1\ng\n0.3 pts\nAnswer: 21.5 N\n0.2 pts\nNotes:\nIf the student only qualitatively explains the\nmechanism by which the grains of sand are held to-\ngether, a maximum of 0.5 pts are given. Points b)-c) are\ngiven only if derived from a).\nSolution 2:\nWe have seen above that grains to one side of a ficti-\ntious surface exert force per cross-sectional area on the\norder of magnitude as the capillary pressure ∆p ∼γ/rg.\nIn order to get the grains moving, a pressure of the same\norder of magnitude needs to be applied externally. For\nboth cylinders, the surface area where the force is ap-\nplied is the same, hence the force scales as the capillary\npressure, F ∝1/rg ∝V −1/3.\nMarking scheme:\nThe applied pressure must be ∼∆p\n0.6 pts\nThe curvature radius of the interface is ∼rg\n0.6 pts\nCapillary pressure ∆p ∼γ/rg\n0.6 pts\nAnswer: 21.5 N\n0.2 pts\nNote: If the contribution of surface tension is neglected,\n0.1 pts are subtracted.\nSolution 3:\nThe compression force serves to break the surface ten-\nsion bonds between sand grains.\nConsider the energy E required to push a single layer\nof sand into the layer beneath it. E ∝Frg, where F is the\nIPhO 2022\nTheoretical problems: solutions. Language: English\nforce required and rg is the typical height of a layer (i.e.,\nthe typical length scale of a grain).\nOn the other hand, E = γ∆A, where γ is the surface\ntension of water and ∆A is the total amount by which\nthe surface of the water in the layer stretches before all\nthe “water bonds” between the sand grains are broken.\nHere, ∆A is proportional to the area A of a layer and is\nthus a constant between the two cylinders. Hence, E ∝\nFrg is a constant between the two cylinders, i.e., F ∝r−1\ng .\nMarking scheme:\nE ∝Frg\n0.5 pts\nE ∝γ∆A\n0.5 pts\n∆A ∝A\n0.5 pts\nF ∝r−1\ng\n0.3 pts\nAnswer: 21.5 N\n0.2 pts\nNote: If the student only qualitatively explains the mech-\nanism by which the grains of sand are held together, a\nmaximum of 0.5 pts are given.\nSolution 4:\nFirst of all, the force F should be proportional to the\ncylinder’s base area A. The force required to destroy a\ncylinder with base area A = nA0 is equal to the force\nrequired to destroy n cylinders each with base area A0.\nAs a result, F ∝n ∝A.\nIn addition, F depends on the grain’s length scale rg\nand the water’s surface tension γ. Dimensional analysis\nthus gives\nF ∝Aγ\nrg\n∝r−1\ng\nfor fixed A and γ.\nMarking scheme:\nF ∝A\n0.6 pts\nF = F(A, rg, γ)\n0.6 pts\nF ∝Aγ\nrg\n0.6 pts\nAnswer: 21.5 N\n0.2 pts\nNote: If the student only qualitatively explains the mech-\nanism by which the grains of sand are held together, a\nmaximum of 0.5 pts are given."], "answer_type": ["Official solution excerpt"], "unit": [null], "points": [2.0], "marking": [["Task B: Sand castle (2 pts)\nThis is Section 2.2.2 (statics) and 2.2.5 (hydrodynamics) of\nthe syllabus\nDue to wetting of the surfaces of the sand grains and\nits large surface tension water acts like a glue for sand.\nThis means that all the grains need to be bound together\nby air-water interface. To achieve this there needs to be\nneither too little nor too much water: if there is too little\nwater, most of the grains are dry with no surface ten-\nsion binding them, and if there is too much water, al-\nmost all the grains are immersed into water, and again,\nthere is no surface tension binding the grains. So, the\noverall strength of the buildings from wet sand depends\non the water content; we assume that for the both types\nof sand, the water content is optimal, and the shape of\nthe grains is statistically similar.\nLet us consider two\nneighbouring grains connected by a water meniscus —\nor “neck”, as we shall be referring to it henceforth. Note\nthat the “neck” may extend perpendicularly to the fig-\nure plane far away; so, more specifically, what the word\n“neck” will refer to is that part of the water-air interface\nfor which the closest two grains are the ones under con-\nsideration.\nThere are two processes binding the sand grains to-\ngether. The first one is the force due to the surface ten-\nsion, F1 = γl, where γ denotes the surface tension coef-\nficient, and l — the perimeter of the “neck”; with l ∼rg,\nwhere rg denotes the length scale of a single grain, we ob-\ntain Fs ∼γr. The second one is the pressure force caused\nby the negative capillary pressure in the neck, Fp = ∆pA,\nwhere A is the cross-sectional area of the “neck”, and\n∆p ∼γ/r. With A ∼r2 we obtain F2 ��γr. Thus, the both\ncomponents are of the same order of magnitude and us-\ning either of them will lead to the correct scaling law.\nThese forces press the grains against each other, hence\nthe normal force and friction force between the grains is\nalso on the order of Fs and Fp.\nSolution 1:\nBased on what has been said above, the typical force\nneeded to delocate a grain of sand is Fg ∝rg. The force\nneeded to delocate an entire horizontal layer of sand is\nthen ∝FgNl, where Nl ∼A/r2\ng is the number of grains in\na layer. The force of cylinder destruction F thus satisfies\nF ∝FgNl ∝rg/r2\ng = r−1\ng\n∝V −1/3\ng\n,\nso\nFfg = (1/10)−1/3 · Fcg = 21.5 N.\nMarking scheme:\na) Fs ∝rg and Fp ∝rg\n0.5 pts\none of the two missing\n-0.1 pts\nb) Fg ∝rg\n0.5 pts\nc) F ∝FgNl\n0.5 pts\nd) F ∝r−1\ng\n0.3 pts\nAnswer: 21.5 N\n0.2 pts\nNotes:\nIf the student only qualitatively explains the\nmechanism by which the grains of sand are held to-\ngether, a maximum of 0.5 pts are given. Points b)-c) are\ngiven only if derived from a).\nSolution 2:\nWe have seen above that grains to one side of a ficti-\ntious surface exert force per cross-sectional area on the\norder of magnitude as the capillary pressure ∆p ∼γ/rg.\nIn order to get the grains moving, a pressure of the same\norder of magnitude needs to be applied externally. For\nboth cylinders, the surface area where the force is ap-\nplied is the same, hence the force scales as the capillary\npressure, F ∝1/rg ∝V −1/3.\nMarking scheme:\nThe applied pressure must be ∼∆p\n0.6 pts\nThe curvature radius of the interface is ∼rg\n0.6 pts\nCapillary pressure ∆p ∼γ/rg\n0.6 pts\nAnswer: 21.5 N\n0.2 pts\nNote: If the contribution of surface tension is neglected,\n0.1 pts are subtracted.\nSolution 3:\nThe compression force serves to break the surface ten-\nsion bonds between sand grains.\nConsider the energy E required to push a single layer\nof sand into the layer beneath it. E ∝Frg, where F is the\nIPhO 2022\nTheoretical problems: solutions. Language: English\nforce required and rg is the typical height of a layer (i.e.,\nthe typical length scale of a grain).\nOn the other hand, E = γ∆A, where γ is the surface\ntension of water and ∆A is the total amount by which\nthe surface of the water in the layer stretches before all\nthe “water bonds” between the sand grains are broken.\nHere, ∆A is proportional to the area A of a layer and is\nthus a constant between the two cylinders. Hence, E ∝\nFrg is a constant between the two cylinders, i.e., F ∝r−1\ng .\nMarking scheme:\nE ∝Frg\n0.5 pts\nE ∝γ∆A\n0.5 pts\n∆A ∝A\n0.5 pts\nF ∝r−1\ng\n0.3 pts\nAnswer: 21.5 N\n0.2 pts\nNote: If the student only qualitatively explains the mech-\nanism by which the grains of sand are held together, a\nmaximum of 0.5 pts are given.\nSolution 4:\nFirst of all, the force F should be proportional to the\ncylinder’s base area A. The force required to destroy a\ncylinder with base area A = nA0 is equal to the force\nrequired to destroy n cylinders each with base area A0.\nAs a result, F ∝n ∝A.\nIn addition, F depends on the grain’s length scale rg\nand the water’s surface tension γ. Dimensional analysis\nthus gives\nF ∝Aγ\nrg\n∝r−1\ng\nfor fixed A and γ.\nMarking scheme:\nF ∝A\n0.6 pts\nF = F(A, rg, γ)\n0.6 pts\nF ∝Aγ\nrg\n0.6 pts\nAnswer: 21.5 N\n0.2 pts\nNote: If the student only qualitatively explains the mech-\nanism by which the grains of sand are held together, a\nmaximum of 0.5 pts are given."]], "modality": "text-only", "field": "Scaling laws", "source": "IPhO_2022", "previous_parts": [], "previous_part_count": 0, "images": [], "image": null, "image_count": 0}
+{"index": "IPhO_2022_3_C_1", "category": "physics", "source_dataset": "IPhO_2021_2025_extracted", "source_dataset_url": null, "year": 2022, "problem_id": "IPhO_2022_3", "problem_number": 3, "part_id": "C.1", "part_letter": "C", "subquestion_number": 1, "formalization_input_policy": {"previous_parts": "Curated natural-language prerequisites only; do not import or depend on previous Lean outputs.", "images": "Use listed images. Figure choices were curated by semantic review when heuristic PDF alignment was too broad.", "answers": "Answers are official-solution excerpts split to subquestion scope where possible; broad fallbacks are listed in review metadata."}, "context": "Scaling laws (8 points)\nScaling laws describe the functional relationship between two physical quantities that scale with each\nother over a significant interval. This functional relationship can be a power law, but there are other\npossibilities, too. Oftentimes, exact expressions are beyond reach, but scaling laws can still be derived.\nPart A. Spaghetto (2.0 points)\nA.1\nA spaghetto straw of diameter 𝑑is being balanced horizontally from its middle.\nIf 𝑑= 1 mm, the straw breaks under its own weight once its length reaches\n𝑙= 50 cm. What is the maximum length 𝑙′ of the straw of diameter 𝑑′ = 1 cm\nbefore it breaks under its own weight?\n2.0pt\nPart B. Sand castle (2.0 points)\nB.1\nThe average grain volume of coarse-grained sand is 10 times as large as that\nof fine-grained sand. Wet fine-grained sand and wet coarse-grained sand have\nboth optimal water content (i.e. assuming the maximal strength of the con-\nstructions from it) and are used to build two cylinders of exactly the same shape\nand size. The strength of each cylinder is tested by pressing it between two par-\nallel plates. The cylinder made of coarse-grained sand gets destroyed once the\nforce applied to press the plates reaches 𝐹c = 10 N. How large is the force 𝐹f\nneeded to destroy the cylinder made of fine-grained sand? You may ignore the\neffects of gravity.\n2.0pt\nPart C. Interstellar travel (2.0 points)", "current_question": "The spaceship of an interstellar expedition travels at a constant magnitude of\nthe proper acceleration 𝑔= 10 m/s2, i.e., this is the acceleration of the space-\nship in the inertial frame of reference where it is instantaneously at rest. The\npassengers must be able to return to Earth within their remaining expected life-\ntime of 50 years. The maximum distance from Earth reached by the spaceship\nis 𝑑. If the acceleration is increased to 𝑔′ = 15 m/s2 , the spaceship can reach a\nfarther distance 𝑑′. What is the ratio 𝑑′/𝑑?\nHint 1. You may wish to use the relativistic velocity addition formula, however,\nthere are also other approaches.\nHint 2.\nYou may need to deal with hyperbolic functions defined as follows:\ncosh 𝑥= 1\n2(e𝑥+ e−𝑥), sinh 𝑥= 1\n2(e𝑥−e−𝑥), tanh 𝑥= e𝑥−e−𝑥\ne𝑥+e−𝑥.\nHint 3. Depending on your approach, you may need one or more of these in-\ntegrals: ∫\nd𝑥\n1−𝑥2 = atanh 𝑥+ 𝐶, ∫\nd𝑥\n√\n1+𝑥2 = asinh 𝑥+ 𝐶, ∫sinh 𝑥d𝑥= cosh 𝑥+ 𝐶,\nwhere asinh 𝑥and atanh 𝑥are the inverse functions of the respective hyperbolic\nfunctions.", "question": "Scaling laws (8 points)\nScaling laws describe the functional relationship between two physical quantities that scale with each\nother over a significant interval. This functional relationship can be a power law, but there are other\npossibilities, too. Oftentimes, exact expressions are beyond reach, but scaling laws can still be derived.\nPart A. Spaghetto (2.0 points)\nA.1\nA spaghetto straw of diameter 𝑑is being balanced horizontally from its middle.\nIf 𝑑= 1 mm, the straw breaks under its own weight once its length reaches\n𝑙= 50 cm. What is the maximum length 𝑙′ of the straw of diameter 𝑑′ = 1 cm\nbefore it breaks under its own weight?\n2.0pt\nPart B. Sand castle (2.0 points)\nB.1\nThe average grain volume of coarse-grained sand is 10 times as large as that\nof fine-grained sand. Wet fine-grained sand and wet coarse-grained sand have\nboth optimal water content (i.e. assuming the maximal strength of the con-\nstructions from it) and are used to build two cylinders of exactly the same shape\nand size. The strength of each cylinder is tested by pressing it between two par-\nallel plates. The cylinder made of coarse-grained sand gets destroyed once the\nforce applied to press the plates reaches 𝐹c = 10 N. How large is the force 𝐹f\nneeded to destroy the cylinder made of fine-grained sand? You may ignore the\neffects of gravity.\n2.0pt\nPart C. Interstellar travel (2.0 points)\n\nCurrent subquestion:\nThe spaceship of an interstellar expedition travels at a constant magnitude of\nthe proper acceleration 𝑔= 10 m/s2, i.e., this is the acceleration of the space-\nship in the inertial frame of reference where it is instantaneously at rest. The\npassengers must be able to return to Earth within their remaining expected life-\ntime of 50 years. The maximum distance from Earth reached by the spaceship\nis 𝑑. If the acceleration is increased to 𝑔′ = 15 m/s2 , the spaceship can reach a\nfarther distance 𝑑′. What is the ratio 𝑑′/𝑑?\nHint 1. You may wish to use the relativistic velocity addition formula, however,\nthere are also other approaches.\nHint 2.\nYou may need to deal with hyperbolic functions defined as follows:\ncosh 𝑥= 1\n2(e𝑥+ e−𝑥), sinh 𝑥= 1\n2(e𝑥−e−𝑥), tanh 𝑥= e𝑥−e−𝑥\ne𝑥+e−𝑥.\nHint 3. Depending on your approach, you may need one or more of these in-\ntegrals: ∫\nd𝑥\n1−𝑥2 = atanh 𝑥+ 𝐶, ∫\nd𝑥\n√\n1+𝑥2 = asinh 𝑥+ 𝐶, ∫sinh 𝑥d𝑥= cosh 𝑥+ 𝐶,\nwhere asinh 𝑥and atanh 𝑥are the inverse functions of the respective hyperbolic\nfunctions.", "answer": "Task C: Interstellar travel (2 pts)\nThis is Section 2.5 (Relativity) of the syllabus\nLet T\n= 50 yrs be the astronauts’ total travel time.\nFor maximal travel distance, the spaceship accelerates\nat constant proper acceleration a = g for proper time\nT/4, during which a distance of d is traveled. The space-\nship then decelerates at a = −g for proper time T/4 to\ncome to a rest, during which another distance d is trav-\neled. The spaceship then returns to Earth using the same\nprocedure.\nNotes:\nFormula relating acceleration to proper ac-\nceleration is not considered as a basic SR formula and\ntherefore if the formula is written without motivation,\n0.2 pts are subtracted.\nSolution 0: (incorrect)\nIf we ignore relativity, then d ∝1\n2gt2 ∝g, which gives\nan answer of 1.5.\nMarking scheme:\nd ∝gt2\n0.2 pts\nAnswer: 1.5\n0.1 pts\nSolution 1:\nOne way to approach the problem is to notice that con-\nstant acceleration in spaceship’s frame means a constant\nforce in the Earth’s frame. This follows directly from the\nLorentz transform for the electromagnetic field, more\nspecifically from the fact that when going to a frame\nmoving parallel to the x-axis, the x-directional electric\nfield Ex remains unchanged. Hence, on the one hand,\nthe force Fx = eEx exerted on an accelerating particle of\nrest mass m0 and carrying a charge e remains constant\nin the lab frame. On the other hand, the acceleration of\nthat particle in an inertial frame moving with velocity v,\nwhere v denotes the particle’s velocity at a certain mo-\nment of time t, is always equal to eEx/m0, regardless of\nthe value of t, i.e. constant in time.\nThose who are not familiar with the Lorenz transform\nfor electromagnetic field can derive the above described\nproperty from the Lorenz transform for momentum and\ncoordinates. We use again (i) the lab frame, and (ii) an in-\nertial frame moving with velocity v, where v denotes the\nspaceship’s velocity at a certain moment of time which\nwill be used as the origin, t = t′ = 0; let primes denote\nquantities in the second frame. Assuming a very short\ntime period t, we can neglect terms quadratic in time so\nthat in the frame (ii), the momentum, coordinate and the\nrelativistic mass can be expressed as p′ = F ′t′, x′ = 0,\nm′ = m0, respectively; applying the Lorenz transform\nyields t = γt′ and p = γ(F ′t′ + m0v) = tF ′ + γm0v. On the\nother hand, in the frame (i), p = γm0v + Ft; comparing\nthis with the previous result yields F = F ′.\nIt appears that in either case, the spaceship’s speed will\nreach almost c much faster than the travel time. Hence,\nusing for convenience the system of units where c = 1,\nthe travel distance x equals with a very good precision\nthe travel time t, x = t.\nWhat is left to do is to relate t to the proper time τ,\ndτ = dt\nγ = dt\nm0\np\nm2\n0 + m2\n0g2t2 ;\nupon integration we obtain\nτ = asinh(gt)/g ⇒x ≈t = sinh(gτ)/g ≈exp(gτ)/2g.\nSo we conclude that the ratio of the travel distances is\nd2\nd1\n=\ng\n1.5g exp(1.5gτ −gτ) = 2\n3 exp(gT/8) ≈480.\nNote that an exact relationship between x and t could\nhave been obtained by expressing the energy of the\nspaceship as m = m0 + m0gx, and the momentum as\np = m0gt. Then the Lorenz invariant (m0 + m0gx)2 −\n(m0gt)2 = m2\n0 yields x(x + 2/g) = t2 = sinh2(gτ)/g2, hence\nx = [cosh(gτ) −1]/g.\nIPhO 2022\nTheoretical problems: solutions. Language: English\nFx is Lorentz invariant\n0.4 pts\nx ≈t\n0.4 pts\ndτ = dt\nγ\n0.2 pts\nγ−1 = m0/m\n0.2 pts\nm =\np\nm2\n0 + p2\n0.2 pts\np = m0gt\n0.2 pts\nt = sinh(gτ)/g\n0.2 pts\nAnswer: 480\n0.2 pts\nRemark: if integration boundaries for distance or proper\ntime are wrong by a factor of 0.5, 2, 4, etc., -0.1 pts.\nSolution 2:\nLet w be the rapidity of the spaceship, defined as w ≡\ntanh−1(β), where β is the spaceship’s velocity. Then β =\ntanh w, the Lorentz factor γ = cosh w, and its momentum\np = m0 sinh w.\nAs shown by Solution 1, a spaceship experiencing a\nconstant proper acceleration g experiences a constant\nthree-force\nF = m0g = dp\ndt = m0 cosh wdw\ndt\n=⇒dw\ndt =\ng\ncosh w.\nMeanwhile, time dilation relates t to the spaceship’s\nproper time τ as\ndt\ndτ = γ = cosh w =⇒dw\ndτ = dw\ndt\ndt\ndτ = g.\nIntegrating yields w = gτ. Recalling that dt = γdτ, we get\nthe following as the total distance traveled over a quar-\nter of the spaceship’s trip:\nd =\nZ T /4\n0\nβγ dτ =\nZ T /4\n0\ntanh w cosh w dτ\n=\nZ T /4\n0\nsinh gτ dτ = 1\ng (cosh(gT/4) −1).\nThe answer is thus\ng1\ng2\ncosh(g2T/4c) −1\ncosh(g1T/4c) −1 = 10\n15\ncosh(19.72) −1\ncosh(13.15) −1\n≈2\n3e19.72−13.15 = 480.\nMarking scheme:\nd\ndt(m0 sinh w) = m0g\n0.5 pts\ndw\ndt =\ng\ncosh w\n0.1 pts\ndt\ndτ = cosh w\n0.4 pts\ndw\ndτ = g\n0.1 pts\nw = gτ\n0.1 pts\nd\n2 =\nR T /4\n0\nβγ dτ\n0.3 pts\nd\n2 =\nR T /4\n0\ntanh w cosh w dτ\n0.2 pts\nd\n2 = 1\ng(cosh(gT/4) −1)\n0.1 pts\nAnswer: 480\n0.2 pts\nRemark: if integration boundaries for distance or proper\ntime are wrong by a factor of 0.5, 2, 4, etc., -0.1 pts.\nSolution 3: The problem can be also solved by using the\ntrick introduced in 1905 by Henri Poincaré [Poincaré,\nM.H. Sur la dynamique de l’électron. Rend. Circ. Matem.\nPalermo 21, 129–175 (1906)] of depicting things in x −it-\ndiagram. The benefit of using this diagram is that the\nrelativistic invariant x2 −t2 transforms into Euclidean\nsquared distance x2 + θ2 with θ = it. This means that\nin that diagram, we can use the knowledge of Euclidean\ngeometry. In particular, the Lorentz transform is now\nthe rotation of the Euclidean x −it-space by an angle\nα = arctan v\nic. Now, consider the trajectory of the space\nship; its infinitesimal arc length is icdτ, where dτ is the\ndifferential of the proper time, and the infintesimal rota-\ntion angle of its tangent is dα = arctan(dv/ic) = dv/ic =\ngdτ/ic. Therefore, the curvature radius R = icdτ/dα =\n−c2/g is constant, i.e. the trajectory is a circle of radius\nR. Now we can easily relate the travel distance x to the\narc length icτ:\nx = R(1 −cos α) = R\n\u0012\n1 −cos icτ\nR\n\u0013\n= c2\ng\n\u0010\ncosh gτ\nc −1\n\u0011\n.\nMarking scheme:\nR = const in x-ict-diag.\n0.5 pts\nR = −g2/c\n0.5 pts\nmissing ‘−’\n-0.2 pts\npartial credit for R = icτ\ndα\n0.2 pts\nx = R(1 −cos α)\n0.5 pts\nc2\ng\n\u0000cosh gτ\nc −1\n\u0001\n0.3 pts\nAnswer: 480\n0.2 pts\nSolution 4: The problem can be solved by using the ve-\nlocity addition formula. Let v = βc be the speed of the\nspaceship in the lab frame, t be the lab time, and τ —\nthe proper time. Also, we consider a frame which moves\nwith constant speed v in which the spaceship accelerates\nfrom rest:\nβ + dβ = β + gdτ\nc\n1 + βgdτ\nc\n= β + gdτ\nc (1 −β2).\nThus,\ndβ\n1 −β2 = gdτ\nc\n⇒β = tanh(gτ\nc ).\nFrom relativistic time dilation formula we obtain\ndt =\ndτ\np\n1 −β2 = cosh(gτ\nc )dτ\nso that the travel distance\nd\n2 =\nZ\nvdt = c\nZ T /4\n0\nsinh(gτ\nc )dτ = c2\ng\n\u0014\ncosh\n\u0012gT\n4c\n\u0013\n−1\n\u0015\nwhich leads to the same answer as before.\na) β + dβ =\nβ+ gdτ\nc\n1+ βgdτ\nc\n0.3 pts\nb)\ndβ\n1−β2 = gdτ\nc\n0.2 pts\nc) β = tanh( gτ\nc )\n0.2 pts\nd) dt =\ndτ\n√\n1−β2\n0.3 pts\ne) dt = cosh( gτ\nc )dτ\n0.2 pts\nf) d\n2 =\nR\nvdt\n0.2 pts\ng) d\n2 = c\nR T /4\n0\nsinh( gτ\nc )dτ\n0.2 pts\nh) d\n2 = c2\ng\nh\ncosh\n\u0010\ngT\n4c\n\u0011\n−1\ni\n0.2 pts\ni) Answer: 480\n0.2 pts\nIPhO 2022\nTheoretical problems: solutions. Language: English\nRemark: if integration in f) is done over proper time,\nno points are given for f). If integration boundaries for\ndistance or proper time are wrong by a factor of 0.5, 2, 4,\netc., -0.1 pts.", "answers": ["Task C: Interstellar travel (2 pts)\nThis is Section 2.5 (Relativity) of the syllabus\nLet T\n= 50 yrs be the astronauts’ total travel time.\nFor maximal travel distance, the spaceship accelerates\nat constant proper acceleration a = g for proper time\nT/4, during which a distance of d is traveled. The space-\nship then decelerates at a = −g for proper time T/4 to\ncome to a rest, during which another distance d is trav-\neled. The spaceship then returns to Earth using the same\nprocedure.\nNotes:\nFormula relating acceleration to proper ac-\nceleration is not considered as a basic SR formula and\ntherefore if the formula is written without motivation,\n0.2 pts are subtracted.\nSolution 0: (incorrect)\nIf we ignore relativity, then d ∝1\n2gt2 ∝g, which gives\nan answer of 1.5.\nMarking scheme:\nd ∝gt2\n0.2 pts\nAnswer: 1.5\n0.1 pts\nSolution 1:\nOne way to approach the problem is to notice that con-\nstant acceleration in spaceship’s frame means a constant\nforce in the Earth’s frame. This follows directly from the\nLorentz transform for the electromagnetic field, more\nspecifically from the fact that when going to a frame\nmoving parallel to the x-axis, the x-directional electric\nfield Ex remains unchanged. Hence, on the one hand,\nthe force Fx = eEx exerted on an accelerating particle of\nrest mass m0 and carrying a charge e remains constant\nin the lab frame. On the other hand, the acceleration of\nthat particle in an inertial frame moving with velocity v,\nwhere v denotes the particle’s velocity at a certain mo-\nment of time t, is always equal to eEx/m0, regardless of\nthe value of t, i.e. constant in time.\nThose who are not familiar with the Lorenz transform\nfor electromagnetic field can derive the above described\nproperty from the Lorenz transform for momentum and\ncoordinates. We use again (i) the lab frame, and (ii) an in-\nertial frame moving with velocity v, where v denotes the\nspaceship’s velocity at a certain moment of time which\nwill be used as the origin, t = t′ = 0; let primes denote\nquantities in the second frame. Assuming a very short\ntime period t, we can neglect terms quadratic in time so\nthat in the frame (ii), the momentum, coordinate and the\nrelativistic mass can be expressed as p′ = F ′t′, x′ = 0,\nm′ = m0, respectively; applying the Lorenz transform\nyields t = γt′ and p = γ(F ′t′ + m0v) = tF ′ + γm0v. On the\nother hand, in the frame (i), p = γm0v + Ft; comparing\nthis with the previous result yields F = F ′.\nIt appears that in either case, the spaceship’s speed will\nreach almost c much faster than the travel time. Hence,\nusing for convenience the system of units where c = 1,\nthe travel distance x equals with a very good precision\nthe travel time t, x = t.\nWhat is left to do is to relate t to the proper time τ,\ndτ = dt\nγ = dt\nm0\np\nm2\n0 + m2\n0g2t2 ;\nupon integration we obtain\nτ = asinh(gt)/g ⇒x ≈t = sinh(gτ)/g ≈exp(gτ)/2g.\nSo we conclude that the ratio of the travel distances is\nd2\nd1\n=\ng\n1.5g exp(1.5gτ −gτ) = 2\n3 exp(gT/8) ≈480.\nNote that an exact relationship between x and t could\nhave been obtained by expressing the energy of the\nspaceship as m = m0 + m0gx, and the momentum as\np = m0gt. Then the Lorenz invariant (m0 + m0gx)2 −\n(m0gt)2 = m2\n0 yields x(x + 2/g) = t2 = sinh2(gτ)/g2, hence\nx = [cosh(gτ) −1]/g.\nIPhO 2022\nTheoretical problems: solutions. Language: English\nFx is Lorentz invariant\n0.4 pts\nx ≈t\n0.4 pts\ndτ = dt\nγ\n0.2 pts\nγ−1 = m0/m\n0.2 pts\nm =\np\nm2\n0 + p2\n0.2 pts\np = m0gt\n0.2 pts\nt = sinh(gτ)/g\n0.2 pts\nAnswer: 480\n0.2 pts\nRemark: if integration boundaries for distance or proper\ntime are wrong by a factor of 0.5, 2, 4, etc., -0.1 pts.\nSolution 2:\nLet w be the rapidity of the spaceship, defined as w ≡\ntanh−1(β), where β is the spaceship’s velocity. Then β =\ntanh w, the Lorentz factor γ = cosh w, and its momentum\np = m0 sinh w.\nAs shown by Solution 1, a spaceship experiencing a\nconstant proper acceleration g experiences a constant\nthree-force\nF = m0g = dp\ndt = m0 cosh wdw\ndt\n=⇒dw\ndt =\ng\ncosh w.\nMeanwhile, time dilation relates t to the spaceship’s\nproper time τ as\ndt\ndτ = γ = cosh w =⇒dw\ndτ = dw\ndt\ndt\ndτ = g.\nIntegrating yields w = gτ. Recalling that dt = γdτ, we get\nthe following as the total distance traveled over a quar-\nter of the spaceship’s trip:\nd =\nZ T /4\n0\nβγ dτ =\nZ T /4\n0\ntanh w cosh w dτ\n=\nZ T /4\n0\nsinh gτ dτ = 1\ng (cosh(gT/4) −1).\nThe answer is thus\ng1\ng2\ncosh(g2T/4c) −1\ncosh(g1T/4c) −1 = 10\n15\ncosh(19.72) −1\ncosh(13.15) −1\n≈2\n3e19.72−13.15 = 480.\nMarking scheme:\nd\ndt(m0 sinh w) = m0g\n0.5 pts\ndw\ndt =\ng\ncosh w\n0.1 pts\ndt\ndτ = cosh w\n0.4 pts\ndw\ndτ = g\n0.1 pts\nw = gτ\n0.1 pts\nd\n2 =\nR T /4\n0\nβγ dτ\n0.3 pts\nd\n2 =\nR T /4\n0\ntanh w cosh w dτ\n0.2 pts\nd\n2 = 1\ng(cosh(gT/4) −1)\n0.1 pts\nAnswer: 480\n0.2 pts\nRemark: if integration boundaries for distance or proper\ntime are wrong by a factor of 0.5, 2, 4, etc., -0.1 pts.\nSolution 3: The problem can be also solved by using the\ntrick introduced in 1905 by Henri Poincaré [Poincaré,\nM.H. Sur la dynamique de l’électron. Rend. Circ. Matem.\nPalermo 21, 129–175 (1906)] of depicting things in x −it-\ndiagram. The benefit of using this diagram is that the\nrelativistic invariant x2 −t2 transforms into Euclidean\nsquared distance x2 + θ2 with θ = it. This means that\nin that diagram, we can use the knowledge of Euclidean\ngeometry. In particular, the Lorentz transform is now\nthe rotation of the Euclidean x −it-space by an angle\nα = arctan v\nic. Now, consider the trajectory of the space\nship; its infinitesimal arc length is icdτ, where dτ is the\ndifferential of the proper time, and the infintesimal rota-\ntion angle of its tangent is dα = arctan(dv/ic) = dv/ic =\ngdτ/ic. Therefore, the curvature radius R = icdτ/dα =\n−c2/g is constant, i.e. the trajectory is a circle of radius\nR. Now we can easily relate the travel distance x to the\narc length icτ:\nx = R(1 −cos α) = R\n\u0012\n1 −cos icτ\nR\n\u0013\n= c2\ng\n\u0010\ncosh gτ\nc −1\n\u0011\n.\nMarking scheme:\nR = const in x-ict-diag.\n0.5 pts\nR = −g2/c\n0.5 pts\nmissing ‘−’\n-0.2 pts\npartial credit for R = icτ\ndα\n0.2 pts\nx = R(1 −cos α)\n0.5 pts\nc2\ng\n\u0000cosh gτ\nc −1\n\u0001\n0.3 pts\nAnswer: 480\n0.2 pts\nSolution 4: The problem can be solved by using the ve-\nlocity addition formula. Let v = βc be the speed of the\nspaceship in the lab frame, t be the lab time, and τ —\nthe proper time. Also, we consider a frame which moves\nwith constant speed v in which the spaceship accelerates\nfrom rest:\nβ + dβ = β + gdτ\nc\n1 + βgdτ\nc\n= β + gdτ\nc (1 −β2).\nThus,\ndβ\n1 −β2 = gdτ\nc\n⇒β = tanh(gτ\nc ).\nFrom relativistic time dilation formula we obtain\ndt =\ndτ\np\n1 −β2 = cosh(gτ\nc )dτ\nso that the travel distance\nd\n2 =\nZ\nvdt = c\nZ T /4\n0\nsinh(gτ\nc )dτ = c2\ng\n\u0014\ncosh\n\u0012gT\n4c\n\u0013\n−1\n\u0015\nwhich leads to the same answer as before.\na) β + dβ =\nβ+ gdτ\nc\n1+ βgdτ\nc\n0.3 pts\nb)\ndβ\n1−β2 = gdτ\nc\n0.2 pts\nc) β = tanh( gτ\nc )\n0.2 pts\nd) dt =\ndτ\n√\n1−β2\n0.3 pts\ne) dt = cosh( gτ\nc )dτ\n0.2 pts\nf) d\n2 =\nR\nvdt\n0.2 pts\ng) d\n2 = c\nR T /4\n0\nsinh( gτ\nc )dτ\n0.2 pts\nh) d\n2 = c2\ng\nh\ncosh\n\u0010\ngT\n4c\n\u0011\n−1\ni\n0.2 pts\ni) Answer: 480\n0.2 pts\nIPhO 2022\nTheoretical problems: solutions. Language: English\nRemark: if integration in f) is done over proper time,\nno points are given for f). If integration boundaries for\ndistance or proper time are wrong by a factor of 0.5, 2, 4,\netc., -0.1 pts."], "answer_type": ["Official solution excerpt"], "unit": [null], "points": [2.0], "marking": [["Task C: Interstellar travel (2 pts)\nThis is Section 2.5 (Relativity) of the syllabus\nLet T\n= 50 yrs be the astronauts’ total travel time.\nFor maximal travel distance, the spaceship accelerates\nat constant proper acceleration a = g for proper time\nT/4, during which a distance of d is traveled. The space-\nship then decelerates at a = −g for proper time T/4 to\ncome to a rest, during which another distance d is trav-\neled. The spaceship then returns to Earth using the same\nprocedure.\nNotes:\nFormula relating acceleration to proper ac-\nceleration is not considered as a basic SR formula and\ntherefore if the formula is written without motivation,\n0.2 pts are subtracted.\nSolution 0: (incorrect)\nIf we ignore relativity, then d ∝1\n2gt2 ∝g, which gives\nan answer of 1.5.\nMarking scheme:\nd ∝gt2\n0.2 pts\nAnswer: 1.5\n0.1 pts\nSolution 1:\nOne way to approach the problem is to notice that con-\nstant acceleration in spaceship’s frame means a constant\nforce in the Earth’s frame. This follows directly from the\nLorentz transform for the electromagnetic field, more\nspecifically from the fact that when going to a frame\nmoving parallel to the x-axis, the x-directional electric\nfield Ex remains unchanged. Hence, on the one hand,\nthe force Fx = eEx exerted on an accelerating particle of\nrest mass m0 and carrying a charge e remains constant\nin the lab frame. On the other hand, the acceleration of\nthat particle in an inertial frame moving with velocity v,\nwhere v denotes the particle’s velocity at a certain mo-\nment of time t, is always equal to eEx/m0, regardless of\nthe value of t, i.e. constant in time.\nThose who are not familiar with the Lorenz transform\nfor electromagnetic field can derive the above described\nproperty from the Lorenz transform for momentum and\ncoordinates. We use again (i) the lab frame, and (ii) an in-\nertial frame moving with velocity v, where v denotes the\nspaceship’s velocity at a certain moment of time which\nwill be used as the origin, t = t′ = 0; let primes denote\nquantities in the second frame. Assuming a very short\ntime period t, we can neglect terms quadratic in time so\nthat in the frame (ii), the momentum, coordinate and the\nrelativistic mass can be expressed as p′ = F ′t′, x′ = 0,\nm′ = m0, respectively; applying the Lorenz transform\nyields t = γt′ and p = γ(F ′t′ + m0v) = tF ′ + γm0v. On the\nother hand, in the frame (i), p = γm0v + Ft; comparing\nthis with the previous result yields F = F ′.\nIt appears that in either case, the spaceship’s speed will\nreach almost c much faster than the travel time. Hence,\nusing for convenience the system of units where c = 1,\nthe travel distance x equals with a very good precision\nthe travel time t, x = t.\nWhat is left to do is to relate t to the proper time τ,\ndτ = dt\nγ = dt\nm0\np\nm2\n0 + m2\n0g2t2 ;\nupon integration we obtain\nτ = asinh(gt)/g ⇒x ≈t = sinh(gτ)/g ≈exp(gτ)/2g.\nSo we conclude that the ratio of the travel distances is\nd2\nd1\n=\ng\n1.5g exp(1.5gτ −gτ) = 2\n3 exp(gT/8) ≈480.\nNote that an exact relationship between x and t could\nhave been obtained by expressing the energy of the\nspaceship as m = m0 + m0gx, and the momentum as\np = m0gt. Then the Lorenz invariant (m0 + m0gx)2 −\n(m0gt)2 = m2\n0 yields x(x + 2/g) = t2 = sinh2(gτ)/g2, hence\nx = [cosh(gτ) −1]/g.\nIPhO 2022\nTheoretical problems: solutions. Language: English\nFx is Lorentz invariant\n0.4 pts\nx ≈t\n0.4 pts\ndτ = dt\nγ\n0.2 pts\nγ−1 = m0/m\n0.2 pts\nm =\np\nm2\n0 + p2\n0.2 pts\np = m0gt\n0.2 pts\nt = sinh(gτ)/g\n0.2 pts\nAnswer: 480\n0.2 pts\nRemark: if integration boundaries for distance or proper\ntime are wrong by a factor of 0.5, 2, 4, etc., -0.1 pts.\nSolution 2:\nLet w be the rapidity of the spaceship, defined as w ≡\ntanh−1(β), where β is the spaceship’s velocity. Then β =\ntanh w, the Lorentz factor γ = cosh w, and its momentum\np = m0 sinh w.\nAs shown by Solution 1, a spaceship experiencing a\nconstant proper acceleration g experiences a constant\nthree-force\nF = m0g = dp\ndt = m0 cosh wdw\ndt\n=⇒dw\ndt =\ng\ncosh w.\nMeanwhile, time dilation relates t to the spaceship’s\nproper time τ as\ndt\ndτ = γ = cosh w =⇒dw\ndτ = dw\ndt\ndt\ndτ = g.\nIntegrating yields w = gτ. Recalling that dt = γdτ, we get\nthe following as the total distance traveled over a quar-\nter of the spaceship’s trip:\nd =\nZ T /4\n0\nβγ dτ =\nZ T /4\n0\ntanh w cosh w dτ\n=\nZ T /4\n0\nsinh gτ dτ = 1\ng (cosh(gT/4) −1).\nThe answer is thus\ng1\ng2\ncosh(g2T/4c) −1\ncosh(g1T/4c) −1 = 10\n15\ncosh(19.72) −1\ncosh(13.15) −1\n≈2\n3e19.72−13.15 = 480.\nMarking scheme:\nd\ndt(m0 sinh w) = m0g\n0.5 pts\ndw\ndt =\ng\ncosh w\n0.1 pts\ndt\ndτ = cosh w\n0.4 pts\ndw\ndτ = g\n0.1 pts\nw = gτ\n0.1 pts\nd\n2 =\nR T /4\n0\nβγ dτ\n0.3 pts\nd\n2 =\nR T /4\n0\ntanh w cosh w dτ\n0.2 pts\nd\n2 = 1\ng(cosh(gT/4) −1)\n0.1 pts\nAnswer: 480\n0.2 pts\nRemark: if integration boundaries for distance or proper\ntime are wrong by a factor of 0.5, 2, 4, etc., -0.1 pts.\nSolution 3: The problem can be also solved by using the\ntrick introduced in 1905 by Henri Poincaré [Poincaré,\nM.H. Sur la dynamique de l’électron. Rend. Circ. Matem.\nPalermo 21, 129���175 (1906)] of depicting things in x −it-\ndiagram. The benefit of using this diagram is that the\nrelativistic invariant x2 −t2 transforms into Euclidean\nsquared distance x2 + θ2 with θ = it. This means that\nin that diagram, we can use the knowledge of Euclidean\ngeometry. In particular, the Lorentz transform is now\nthe rotation of the Euclidean x −it-space by an angle\nα = arctan v\nic. Now, consider the trajectory of the space\nship; its infinitesimal arc length is icdτ, where dτ is the\ndifferential of the proper time, and the infintesimal rota-\ntion angle of its tangent is dα = arctan(dv/ic) = dv/ic =\ngdτ/ic. Therefore, the curvature radius R = icdτ/dα =\n−c2/g is constant, i.e. the trajectory is a circle of radius\nR. Now we can easily relate the travel distance x to the\narc length icτ:\nx = R(1 −cos α) = R\n\u0012\n1 −cos icτ\nR\n\u0013\n= c2\ng\n\u0010\ncosh gτ\nc −1\n\u0011\n.\nMarking scheme:\nR = const in x-ict-diag.\n0.5 pts\nR = −g2/c\n0.5 pts\nmissing ‘−’\n-0.2 pts\npartial credit for R = icτ\ndα\n0.2 pts\nx = R(1 −cos α)\n0.5 pts\nc2\ng\n\u0000cosh gτ\nc −1\n\u0001\n0.3 pts\nAnswer: 480\n0.2 pts\nSolution 4: The problem can be solved by using the ve-\nlocity addition formula. Let v = βc be the speed of the\nspaceship in the lab frame, t be the lab time, and τ —\nthe proper time. Also, we consider a frame which moves\nwith constant speed v in which the spaceship accelerates\nfrom rest:\nβ + dβ = β + gdτ\nc\n1 + βgdτ\nc\n= β + gdτ\nc (1 −β2).\nThus,\ndβ\n1 −β2 = gdτ\nc\n⇒β = tanh(gτ\nc ).\nFrom relativistic time dilation formula we obtain\ndt =\ndτ\np\n1 −β2 = cosh(gτ\nc )dτ\nso that the travel distance\nd\n2 =\nZ\nvdt = c\nZ T /4\n0\nsinh(gτ\nc )dτ = c2\ng\n\u0014\ncosh\n\u0012gT\n4c\n\u0013\n−1\n\u0015\nwhich leads to the same answer as before.\na) β + dβ =\nβ+ gdτ\nc\n1+ βgdτ\nc\n0.3 pts\nb)\ndβ\n1−β2 = gdτ\nc\n0.2 pts\nc) β = tanh( gτ\nc )\n0.2 pts\nd) dt =\ndτ\n√\n1−β2\n0.3 pts\ne) dt = cosh( gτ\nc )dτ\n0.2 pts\nf) d\n2 =\nR\nvdt\n0.2 pts\ng) d\n2 = c\nR T /4\n0\nsinh( gτ\nc )dτ\n0.2 pts\nh) d\n2 = c2\ng\nh\ncosh\n\u0010\ngT\n4c\n\u0011\n−1\ni\n0.2 pts\ni) Answer: 480\n0.2 pts\nIPhO 2022\nTheoretical problems: solutions. Language: English\nRemark: if integration in f) is done over proper time,\nno points are given for f). If integration boundaries for\ndistance or proper time are wrong by a factor of 0.5, 2, 4,\netc., -0.1 pts."]], "modality": "text-only", "field": "Scaling laws", "source": "IPhO_2022", "previous_parts": [], "previous_part_count": 0, "images": [], "image": null, "image_count": 0}
+{"index": "IPhO_2022_3_D_1", "category": "physics", "source_dataset": "IPhO_2021_2025_extracted", "source_dataset_url": null, "year": 2022, "problem_id": "IPhO_2022_3", "problem_number": 3, "part_id": "D.1", "part_letter": "D", "subquestion_number": 1, "formalization_input_policy": {"previous_parts": "Curated natural-language prerequisites only; do not import or depend on previous Lean outputs.", "images": "Use listed images. Figure choices were curated by semantic review when heuristic PDF alignment was too broad.", "answers": "Answers are official-solution excerpts split to subquestion scope where possible; broad fallbacks are listed in review metadata."}, "context": "Scaling laws (8 points)\nScaling laws describe the functional relationship between two physical quantities that scale with each\nother over a significant interval. This functional relationship can be a power law, but there are other\npossibilities, too. Oftentimes, exact expressions are beyond reach, but scaling laws can still be derived.\nPart A. Spaghetto (2.0 points)\nA.1\nA spaghetto straw of diameter 𝑑is being balanced horizontally from its middle.\nIf 𝑑= 1 mm, the straw breaks under its own weight once its length reaches\n𝑙= 50 cm. What is the maximum length 𝑙′ of the straw of diameter 𝑑′ = 1 cm\nbefore it breaks under its own weight?\n2.0pt\nPart B. Sand castle (2.0 points)\nB.1\nThe average grain volume of coarse-grained sand is 10 times as large as that\nof fine-grained sand. Wet fine-grained sand and wet coarse-grained sand have\nboth optimal water content (i.e. assuming the maximal strength of the con-\nstructions from it) and are used to build two cylinders of exactly the same shape\nand size. The strength of each cylinder is tested by pressing it between two par-\nallel plates. The cylinder made of coarse-grained sand gets destroyed once the\nforce applied to press the plates reaches 𝐹c = 10 N. How large is the force 𝐹f\nneeded to destroy the cylinder made of fine-grained sand? You may ignore the\neffects of gravity.\n2.0pt\nPart C. Interstellar travel (2.0 points)\nC.1\nThe spaceship of an interstellar expedition travels at a constant magnitude of\nthe proper acceleration 𝑔= 10 m/s2, i.e., this is the acceleration of the space-\nship in the inertial frame of reference where it is instantaneously at rest. The\npassengers must be able to return to Earth within their remaining expected life-\ntime of 50 years. The maximum distance from Earth reached by the spaceship\nis 𝑑. If the acceleration is increased to 𝑔′ = 15 m/s2 , the spaceship can reach a\nfarther distance 𝑑′. What is the ratio 𝑑′/𝑑?\nHint 1. You may wish to use the relativistic velocity addition formula, however,\nthere are also other approaches.\nHint 2.\nYou may need to deal with hyperbolic functions defined as follows:\ncosh 𝑥= 1\n2(e𝑥+ e−𝑥), sinh 𝑥= 1\n2(e𝑥−e−𝑥), tanh 𝑥= e𝑥−e−𝑥\ne𝑥+e−𝑥.\nHint 3. Depending on your approach, you may need one or more of these in-\ntegrals: ∫\nd𝑥\n1−𝑥2 = atanh 𝑥+ 𝐶, ∫\nd𝑥\n√\n1+𝑥2 = asinh 𝑥+ 𝐶, ∫sinh 𝑥d𝑥= cosh 𝑥+ 𝐶,\nwhere asinh 𝑥and atanh 𝑥are the inverse functions of the respective hyperbolic\nfunctions.\n2.0pt\nPart D. That sinking feeling (2.0 points)", "current_question": "A solid wooden ball of radius 𝑟0 is floating in the water. Ignoring frictional ef-\nfects, the frequency of small oscillations would be 𝜔0, but because of viscous\nfriction, after being displaced vertically, the frequency of decaying oscillations\nis actually 0.99 𝜔0. What is the minimum radius 𝑟min of a wooden ball floating in\nwater that undergos small oscillations when displaced? Hint: the viscous drag\nforce acting on a given body is proportional to its speed relative to the bulk\nof the fluid, and to the viscosity 𝜂of the fluid it is moving in. The unit of the\nviscosity is kg/(m ⋅s).", "question": "Scaling laws (8 points)\nScaling laws describe the functional relationship between two physical quantities that scale with each\nother over a significant interval. This functional relationship can be a power law, but there are other\npossibilities, too. Oftentimes, exact expressions are beyond reach, but scaling laws can still be derived.\nPart A. Spaghetto (2.0 points)\nA.1\nA spaghetto straw of diameter 𝑑is being balanced horizontally from its middle.\nIf 𝑑= 1 mm, the straw breaks under its own weight once its length reaches\n𝑙= 50 cm. What is the maximum length 𝑙′ of the straw of diameter 𝑑′ = 1 cm\nbefore it breaks under its own weight?\n2.0pt\nPart B. Sand castle (2.0 points)\nB.1\nThe average grain volume of coarse-grained sand is 10 times as large as that\nof fine-grained sand. Wet fine-grained sand and wet coarse-grained sand have\nboth optimal water content (i.e. assuming the maximal strength of the con-\nstructions from it) and are used to build two cylinders of exactly the same shape\nand size. The strength of each cylinder is tested by pressing it between two par-\nallel plates. The cylinder made of coarse-grained sand gets destroyed once the\nforce applied to press the plates reaches 𝐹c = 10 N. How large is the force 𝐹f\nneeded to destroy the cylinder made of fine-grained sand? You may ignore the\neffects of gravity.\n2.0pt\nPart C. Interstellar travel (2.0 points)\nC.1\nThe spaceship of an interstellar expedition travels at a constant magnitude of\nthe proper acceleration 𝑔= 10 m/s2, i.e., this is the acceleration of the space-\nship in the inertial frame of reference where it is instantaneously at rest. The\npassengers must be able to return to Earth within their remaining expected life-\ntime of 50 years. The maximum distance from Earth reached by the spaceship\nis 𝑑. If the acceleration is increased to 𝑔′ = 15 m/s2 , the spaceship can reach a\nfarther distance 𝑑′. What is the ratio 𝑑′/𝑑?\nHint 1. You may wish to use the relativistic velocity addition formula, however,\nthere are also other approaches.\nHint 2.\nYou may need to deal with hyperbolic functions defined as follows:\ncosh 𝑥= 1\n2(e𝑥+ e−𝑥), sinh 𝑥= 1\n2(e𝑥−e−𝑥), tanh 𝑥= e𝑥−e−𝑥\ne𝑥+e−𝑥.\nHint 3. Depending on your approach, you may need one or more of these in-\ntegrals: ∫\nd𝑥\n1−𝑥2 = atanh 𝑥+ 𝐶, ∫\nd𝑥\n√\n1+𝑥2 = asinh 𝑥+ 𝐶, ∫sinh 𝑥d𝑥= cosh 𝑥+ 𝐶,\nwhere asinh 𝑥and atanh 𝑥are the inverse functions of the respective hyperbolic\nfunctions.\n2.0pt\nPart D. That sinking feeling (2.0 points)\n\nCurrent subquestion:\nA solid wooden ball of radius 𝑟0 is floating in the water. Ignoring frictional ef-\nfects, the frequency of small oscillations would be 𝜔0, but because of viscous\nfriction, after being displaced vertically, the frequency of decaying oscillations\nis actually 0.99 𝜔0. What is the minimum radius 𝑟min of a wooden ball floating in\nwater that undergos small oscillations when displaced? Hint: the viscous drag\nforce acting on a given body is proportional to its speed relative to the bulk\nof the fluid, and to the viscosity 𝜂of the fluid it is moving in. The unit of the\nviscosity is kg/(m ⋅s).", "answer": "Task D: That sinking feeling (2 pts)\n(This is Sections 2.2.5 (Hydrodynamics) and 2.4.1 (Single\noscillator) of the syllabus\nSolution 1: The oscillation of the half-sunk sphere is\ndriven by gravity. The non-damped angular frequency\ndepends on the gravitational acceleration and a charac-\nteristic length, which is, for a sphere, its radius r, so\nω0 ∝\np\ng/r\nis the only dimensionally correct possible function.\nThe drag force Fd depends on the sphere’s speed v\n[m/s], its size r [m], and viscosity of the liquid η [Pa·s].\nDimensional analysis thus gives Fd ∝ηrv. The damping\nfactor is thus\nβ = Fd\n2mv ∝ηr\nm .\nSince the mass scales with r3, we have\nβ ∝1\nr2 .\nThen the relation\nβ2\nω2\n0\n= 1 −ω2\nω2\n0\nscales as\nβ2\nω2\n0\n∝1\nr3\nOscillations only occur if β/ω0 < 1, so solve\nr\nr0\n=\n3p\n1 −(0.99)2 = 0.271\nNotes:\n1. To obtain ω0 ∝1/√r without dimensional analysis,\nnote that a small displacement y changes the sub-\nmerged volume of the ball by ∆V\n∝r2y, so the\nchange in buoyant force F ∝r2y, which gives ω0 =\np\nk/m ∝\np\nr2/r3 =\np\n1/r.\n2. To obtain Fd ∝ηrv without dimensional analysis,\nnote that the typical length scale l in the variations\nin the velocity field of the water is proportional to r.\nThus, the viscous shear σ ∝ηv/l ∝ηv/r. The total\ndrag force is thus Fd ∼Aσ ∝ηrv, where A is ball’s\narea of contact with the water.\n3. Alternatively, to obtain Fd ∝ηrv, make use of the\nStokes drag relation Fd = 6πηKrv, where K is a di-\nmensionless constant that takes into account that\nthe ball is not in infinite homogeneous fluid.\nMarking scheme:\na) ω0 ∝\np\ng/r\n0.4 pts\nstated without justification\n-0.2 pts\neffective mass ∝r3\n0.2 pts\njust the mass of the ball considered\n-0.1 pts\neffective returning force ∝r2\n0.1 pts\nω0 ∝\np\ng/r\n0.1 pts\nb) Fd ∝ηrv\n0.6 pts\nno justification\n-0.3 pts\nStokes without constant K\n-0.1 pts\nc) β ∝1/r2\n0.3 pts\nd) β2\nω2\n0 = 1 −ω2\nω2\n0\n0.4 pts\ne) β2\nω2\n0 ∝\n1\nr3\n0.2 pts\nf) Answer: 0.271\n0.1 pts\nSolution 2: The oscillation of the half-sunk sphere is\ndriven by the change in buoyancy force, which is pro-\nportional to the change in displaced water volume. Thus,\nthe restoring force Fr ∝r2x, where x is the displacement\nof the sphere.\nAs discussed in Solution 1, the drag force Fd ∝rv = r ˙x.\nThe effective mass of the oscillation m ∝r3. This leads\nto the equation of motion\nk1r2¨x + k2 ˙x + k3rx = 0,\nwhere k1, k2 and k3 are constant. In the case with no\nviscous drag, k2 = 0, the motion is at frequency\nω0 =\nr\nk3\nk1r.\nWith viscous drag, we can get the frequency ω by sub-\nstituting trial solution x = eαt and using ω = Imα. This\nleads to\nω =\ns\nω2\n0 −\nk2\n2\n4k2\n1r4 ,\nω2 = ω2\n0\n\u0012\n1 −\nk2\n2\n4k2\n1r4ω2\n0\n\u0013\n= ω2\n0\n\u0012\n1 −\nk2\n2\n4k1k3r3\n\u0013\n.\nAt r = rmin, ω = 0, so\nk2\n2\n4k1k3\n= r3\nmin,\nand\nω2 = ω2\n0\n\u0012\n1 −r3\nmin\nr3\n0\n\u0013\n,\ngiving rmin\nr0\n= 0.271.\nMarking scheme:\na) effective mass ∝r3\n0.2 pts\njust the mass of the ball considered\n-0.1 pts\nb) effective returning force ∝r2\n0.1 pts\nc) ω0 ∝\np\n1/r\n0.1 pts\nd) Fd ∝rv\n0.6 pts\nno justification\n-0.3 pts\nStokes without constant K\n-0.1 pts\ne) k1r2¨x + k2 ˙x + k3rx = 0\n0.3 pts\nf) ω in terms of r and ω0\n0.6 pts\nif not expressed in terms of ω0\n-0.3 pts\ng) Answer: 0.271\n0.1 pts", "answers": ["Task D: That sinking feeling (2 pts)\n(This is Sections 2.2.5 (Hydrodynamics) and 2.4.1 (Single\noscillator) of the syllabus\nSolution 1: The oscillation of the half-sunk sphere is\ndriven by gravity. The non-damped angular frequency\ndepends on the gravitational acceleration and a charac-\nteristic length, which is, for a sphere, its radius r, so\nω0 ∝\np\ng/r\nis the only dimensionally correct possible function.\nThe drag force Fd depends on the sphere’s speed v\n[m/s], its size r [m], and viscosity of the liquid η [Pa·s].\nDimensional analysis thus gives Fd ∝ηrv. The damping\nfactor is thus\nβ = Fd\n2mv ∝ηr\nm .\nSince the mass scales with r3, we have\nβ ∝1\nr2 .\nThen the relation\nβ2\nω2\n0\n= 1 −ω2\nω2\n0\nscales as\nβ2\nω2\n0\n∝1\nr3\nOscillations only occur if β/ω0 < 1, so solve\nr\nr0\n=\n3p\n1 −(0.99)2 = 0.271\nNotes:\n1. To obtain ω0 ∝1/√r without dimensional analysis,\nnote that a small displacement y changes the sub-\nmerged volume of the ball by ∆V\n∝r2y, so the\nchange in buoyant force F ∝r2y, which gives ω0 =\np\nk/m ∝\np\nr2/r3 =\np\n1/r.\n2. To obtain Fd ∝ηrv without dimensional analysis,\nnote that the typical length scale l in the variations\nin the velocity field of the water is proportional to r.\nThus, the viscous shear σ ∝ηv/l ∝ηv/r. The total\ndrag force is thus Fd ∼Aσ ∝ηrv, where A is ball’s\narea of contact with the water.\n3. Alternatively, to obtain Fd ∝ηrv, make use of the\nStokes drag relation Fd = 6πηKrv, where K is a di-\nmensionless constant that takes into account that\nthe ball is not in infinite homogeneous fluid.\nMarking scheme:\na) ω0 ∝\np\ng/r\n0.4 pts\nstated without justification\n-0.2 pts\neffective mass ∝r3\n0.2 pts\njust the mass of the ball considered\n-0.1 pts\neffective returning force ∝r2\n0.1 pts\nω0 ∝\np\ng/r\n0.1 pts\nb) Fd ∝ηrv\n0.6 pts\nno justification\n-0.3 pts\nStokes without constant K\n-0.1 pts\nc) β ∝1/r2\n0.3 pts\nd) β2\nω2\n0 = 1 −ω2\nω2\n0\n0.4 pts\ne) β2\nω2\n0 ∝\n1\nr3\n0.2 pts\nf) Answer: 0.271\n0.1 pts\nSolution 2: The oscillation of the half-sunk sphere is\ndriven by the change in buoyancy force, which is pro-\nportional to the change in displaced water volume. Thus,\nthe restoring force Fr ∝r2x, where x is the displacement\nof the sphere.\nAs discussed in Solution 1, the drag force Fd ∝rv = r ˙x.\nThe effective mass of the oscillation m ∝r3. This leads\nto the equation of motion\nk1r2¨x + k2 ˙x + k3rx = 0,\nwhere k1, k2 and k3 are constant. In the case with no\nviscous drag, k2 = 0, the motion is at frequency\nω0 =\nr\nk3\nk1r.\nWith viscous drag, we can get the frequency ω by sub-\nstituting trial solution x = eαt and using ω = Imα. This\nleads to\nω =\ns\nω2\n0 −\nk2\n2\n4k2\n1r4 ,\nω2 = ω2\n0\n\u0012\n1 −\nk2\n2\n4k2\n1r4ω2\n0\n\u0013\n= ω2\n0\n\u0012\n1 −\nk2\n2\n4k1k3r3\n\u0013\n.\nAt r = rmin, ω = 0, so\nk2\n2\n4k1k3\n= r3\nmin,\nand\nω2 = ω2\n0\n\u0012\n1 −r3\nmin\nr3\n0\n\u0013\n,\ngiving rmin\nr0\n= 0.271.\nMarking scheme:\na) effective mass ∝r3\n0.2 pts\njust the mass of the ball considered\n-0.1 pts\nb) effective returning force ∝r2\n0.1 pts\nc) ω0 ∝\np\n1/r\n0.1 pts\nd) Fd ∝rv\n0.6 pts\nno justification\n-0.3 pts\nStokes without constant K\n-0.1 pts\ne) k1r2¨x + k2 ˙x + k3rx = 0\n0.3 pts\nf) ω in terms of r and ω0\n0.6 pts\nif not expressed in terms of ω0\n-0.3 pts\ng) Answer: 0.271\n0.1 pts"], "answer_type": ["Official solution excerpt"], "unit": [null], "points": [2.0], "marking": [["Task D: That sinking feeling (2 pts)\n(This is Sections 2.2.5 (Hydrodynamics) and 2.4.1 (Single\noscillator) of the syllabus\nSolution 1: The oscillation of the half-sunk sphere is\ndriven by gravity. The non-damped angular frequency\ndepends on the gravitational acceleration and a charac-\nteristic length, which is, for a sphere, its radius r, so\nω0 ∝\np\ng/r\nis the only dimensionally correct possible function.\nThe drag force Fd depends on the sphere’s speed v\n[m/s], its size r [m], and viscosity of the liquid η [Pa·s].\nDimensional analysis thus gives Fd ∝ηrv. The damping\nfactor is thus\nβ = Fd\n2mv ∝ηr\nm .\nSince the mass scales with r3, we have\nβ ∝1\nr2 .\nThen the relation\nβ2\nω2\n0\n= 1 −ω2\nω2\n0\nscales as\nβ2\nω2\n0\n∝1\nr3\nOscillations only occur if β/ω0 < 1, so solve\nr\nr0\n=\n3p\n1 −(0.99)2 = 0.271\nNotes:\n1. To obtain ω0 ∝1/√r without dimensional analysis,\nnote that a small displacement y changes the sub-\nmerged volume of the ball by ∆V\n∝r2y, so the\nchange in buoyant force F ∝r2y, which gives ω0 =\np\nk/m ∝\np\nr2/r3 =\np\n1/r.\n2. To obtain Fd ∝ηrv without dimensional analysis,\nnote that the typical length scale l in the variations\nin the velocity field of the water is proportional to r.\nThus, the viscous shear σ ∝ηv/l ∝ηv/r. The total\ndrag force is thus Fd ∼Aσ ∝ηrv, where A is ball’s\narea of contact with the water.\n3. Alternatively, to obtain Fd ∝ηrv, make use of the\nStokes drag relation Fd = 6πηKrv, where K is a di-\nmensionless constant that takes into account that\nthe ball is not in infinite homogeneous fluid.\nMarking scheme:\na) ω0 ∝\np\ng/r\n0.4 pts\nstated without justification\n-0.2 pts\neffective mass ∝r3\n0.2 pts\njust the mass of the ball considered\n-0.1 pts\neffective returning force ∝r2\n0.1 pts\nω0 ∝\np\ng/r\n0.1 pts\nb) Fd ∝ηrv\n0.6 pts\nno justification\n-0.3 pts\nStokes without constant K\n-0.1 pts\nc) β ∝1/r2\n0.3 pts\nd) β2\nω2\n0 = 1 −ω2\nω2\n0\n0.4 pts\ne) β2\nω2\n0 ∝\n1\nr3\n0.2 pts\nf) Answer: 0.271\n0.1 pts\nSolution 2: The oscillation of the half-sunk sphere is\ndriven by the change in buoyancy force, which is pro-\nportional to the change in displaced water volume. Thus,\nthe restoring force Fr ∝r2x, where x is the displacement\nof the sphere.\nAs discussed in Solution 1, the drag force Fd ∝rv = r ˙x.\nThe effective mass of the oscillation m ∝r3. This leads\nto the equation of motion\nk1r2¨x + k2 ˙x + k3rx = 0,\nwhere k1, k2 and k3 are constant. In the case with no\nviscous drag, k2 = 0, the motion is at frequency\nω0 =\nr\nk3\nk1r.\nWith viscous drag, we can get the frequency ω by sub-\nstituting trial solution x = eαt and using ω = Imα. This\nleads to\nω =\ns\nω2\n0 −\nk2\n2\n4k2\n1r4 ,\nω2 = ω2\n0\n\u0012\n1 −\nk2\n2\n4k2\n1r4ω2\n0\n\u0013\n= ω2\n0\n\u0012\n1 −\nk2\n2\n4k1k3r3\n\u0013\n.\nAt r = rmin, ω = 0, so\nk2\n2\n4k1k3\n= r3\nmin,\nand\nω2 = ω2\n0\n\u0012\n1 −r3\nmin\nr3\n0\n\u0013\n,\ngiving rmin\nr0\n= 0.271.\nMarking scheme:\na) effective mass ∝r3\n0.2 pts\njust the mass of the ball considered\n-0.1 pts\nb) effective returning force ∝r2\n0.1 pts\nc) ω0 ∝\np\n1/r\n0.1 pts\nd) Fd ∝rv\n0.6 pts\nno justification\n-0.3 pts\nStokes without constant K\n-0.1 pts\ne) k1r2¨x + k2 ˙x + k3rx = 0\n0.3 pts\nf) ω in terms of r and ω0\n0.6 pts\nif not expressed in terms of ω0\n-0.3 pts\ng) Answer: 0.271\n0.1 pts"]], "modality": "text-only", "field": "Scaling laws", "source": "IPhO_2022", "previous_parts": [], "previous_part_count": 0, "images": [], "image": null, "image_count": 0}