{
  "index": "1946-A-6",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "6. A particle of unit mass moves on a straight line under the action of a force which is a function \\( f(v) \\) of the velocity \\( v \\) of the particle, but the form of this function is not known. A motion is observed, and the distance \\( x \\) covered in time \\( t \\) is found to be connected with \\( t \\) by the formula \\( x=a t+b t^{2}+c t^{3} \\), where \\( a, b, c \\) have numerical values determined by observation of the motion. Find the function \\( f(v) \\) for the range of \\( v \\) covered by the experiment.",
  "solution": "Solution. Newton's law of motion for a particle of unit mass takes the form\n\\[\nF=\\text { force }=\\frac{d v}{d t} .\n\\]\n\nSince we are given that\n\\[\nx=a t+b t^{2}+c t^{3}\n\\]\nit follows that\n\\[\n\\begin{array}{c}\nv=\\frac{d x}{d t}=a+2 b t+3 c t^{2} \\\\\n\\frac{d v}{d t}=2 b+6 c t\n\\end{array}\n\\]\n\nWe now express the force in terms of \\( \\boldsymbol{v} \\) :\n\\[\n\\begin{aligned}\nF^{2} & =4 b^{2}+24 b c t+36 c^{2} t^{2} \\\\\n& =4 b^{2}+12 c\\left(2 b t+3 c t^{2}\\right) \\\\\n& =4 b^{2}+12 c(v-a) .\n\\end{aligned}\n\\]\n\nHence\n\\[\nF=f(v)= \\pm \\sqrt{4 b^{2}-12 a c+12 c v}\n\\]\n\nThe sign of the radical is taken to be the sign of \\( 2 b+6 c t \\) which, if the hypotheses of the problem are satisfied, cannot change for the interval of time under consideration, since then \\( v \\) would take the same value twice but \\( d v / d t \\) would not.",
  "vars": [
    "x",
    "t",
    "v",
    "F",
    "f"
  ],
  "params": [
    "a",
    "b",
    "c"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "position",
        "t": "timevar",
        "v": "velocity",
        "F": "forcevar",
        "f": "forcefunc",
        "a": "coeffone",
        "b": "coefftwo",
        "c": "coeffthree"
      },
      "question": "6. A particle of unit mass moves on a straight line under the action of a force which is a function \\( forcefunc(velocity) \\) of the velocity \\( velocity \\) of the particle, but the form of this function is not known. A motion is observed, and the distance \\( position \\) covered in time \\( timevar \\) is found to be connected with \\( timevar \\) by the formula \\( position=coeffone\\,timevar+coefftwo\\,timevar^{2}+coeffthree\\,timevar^{3} \\), where \\( coeffone, coefftwo, coeffthree \\) have numerical values determined by observation of the motion. Find the function \\( forcefunc(velocity) \\) for the range of \\( velocity \\) covered by the experiment.",
      "solution": "Solution. Newton's law of motion for a particle of unit mass takes the form\n\\[\nforcevar=\\text { force }=\\frac{d\\,velocity}{d\\,timevar} .\n\\]\n\nSince we are given that\n\\[\nposition=coeffone\\,timevar+coefftwo\\,timevar^{2}+coeffthree\\,timevar^{3}\n\\]\nit follows that\n\\[\n\\begin{array}{c}\nvelocity=\\frac{d\\,position}{d\\,timevar}=coeffone+2\\,coefftwo\\,timevar+3\\,coeffthree\\,timevar^{2} \\\\\n\\frac{d\\,velocity}{d\\,timevar}=2\\,coefftwo+6\\,coeffthree\\,timevar\n\\end{array}\n\\]\n\nWe now express the force in terms of \\( \\boldsymbol{velocity} \\) :\n\\[\n\\begin{aligned}\nforcevar^{2} & =4\\,coefftwo^{2}+24\\,coefftwo\\,coeffthree\\,timevar+36\\,coeffthree^{2}\\,timevar^{2} \\\\\n& =4\\,coefftwo^{2}+12\\,coeffthree\\left(2\\,coefftwo\\,timevar+3\\,coeffthree\\,timevar^{2}\\right) \\\\\n& =4\\,coefftwo^{2}+12\\,coeffthree(velocity-coeffone) .\n\\end{aligned}\n\\]\n\nHence\n\\[\nforcevar=forcefunc(velocity)= \\pm \\sqrt{4\\,coefftwo^{2}-12\\,coeffone\\,coeffthree+12\\,coeffthree\\,velocity}\n\\]\n\nThe sign of the radical is taken to be the sign of \\( 2\\,coefftwo+6\\,coeffthree\\,timevar \\) which, if the hypotheses of the problem are satisfied, cannot change for the interval of time under consideration, since then \\( velocity \\) would take the same value twice but \\( d\\,velocity / d\\,timevar \\) would not."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "meadowland",
        "t": "pendulum",
        "v": "latitude",
        "F": "hurricane",
        "f": "zephyrwind",
        "a": "lanterns",
        "b": "crucible",
        "c": "gemstone"
      },
      "question": "6. A particle of unit mass moves on a straight line under the action of a force which is a function \\( zephyrwind(latitude) \\) of the velocity \\( latitude \\) of the particle, but the form of this function is not known. A motion is observed, and the distance \\( meadowland \\) covered in time \\( pendulum \\) is found to be connected with \\( pendulum \\) by the formula \\( meadowland=lanterns pendulum+crucible pendulum^{2}+gemstone pendulum^{3} \\), where \\( lanterns, crucible, gemstone \\) have numerical values determined by observation of the motion. Find the function \\( zephyrwind(latitude) \\) for the range of \\( latitude \\) covered by the experiment.",
      "solution": "Solution. Newton's law of motion for a particle of unit mass takes the form\n\\[ \nhurricane=\\text { force }=\\frac{d latitude}{d pendulum} .\n\\]\n\nSince we are given that\n\\[ \nmeadowland=lanterns pendulum+crucible pendulum^{2}+gemstone pendulum^{3}\n\\]\nit follows that\n\\[ \n\\begin{array}{c}\nlatitude=\\frac{d meadowland}{d pendulum}=lanterns+2 crucible pendulum+3 gemstone pendulum^{2} \\\\\n\\frac{d latitude}{d pendulum}=2 crucible+6 gemstone pendulum\n\\end{array}\n\\]\n\nWe now express the force in terms of \\( \\boldsymbol{latitude} \\) :\n\\[ \n\\begin{aligned}\nhurricane^{2} & =4 crucible^{2}+24 crucible gemstone pendulum+36 gemstone^{2} pendulum^{2} \\\\\n& =4 crucible^{2}+12 gemstone\\left(2 crucible pendulum+3 gemstone pendulum^{2}\\right) \\\\\n& =4 crucible^{2}+12 gemstone(latitude-lanterns) .\n\\end{aligned}\n\\]\n\nHence\n\\[ \nhurricane=zephyrwind(latitude)= \\pm \\sqrt{4 crucible^{2}-12 lanterns gemstone+12 gemstone latitude}\n\\]\n\nThe sign of the radical is taken to be the sign of \\( 2 crucible+6 gemstone pendulum \\) which, if the hypotheses of the problem are satisfied, cannot change for the interval of time under consideration, since then \\( latitude \\) would take the same value twice but \\( d latitude / d pendulum \\) would not."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "closeness",
        "t": "timeless",
        "v": "stillness",
        "F": "weakness",
        "f": "randomness",
        "a": "variable",
        "b": "changing",
        "c": "unstable"
      },
      "question": "6. A particle of unit mass moves on a straight line under the action of a force which is a function \\( randomness(stillness) \\) of the velocity \\( stillness \\) of the particle, but the form of this function is not known. A motion is observed, and the distance \\( closeness \\) covered in time \\( timeless \\) is found to be connected with \\( timeless \\) by the formula \\( closeness=variable\\, timeless+changing\\, timeless^{2}+unstable\\, timeless^{3} \\), where \\( variable, changing, unstable \\) have numerical values determined by observation of the motion. Find the function \\( randomness(stillness) \\) for the range of \\( stillness \\) covered by the experiment.",
      "solution": "Solution. Newton's law of motion for a particle of unit mass takes the form\n\\[\nweakness=\\text { force }=\\frac{d stillness}{d timeless} .\n\\]\n\nSince we are given that\n\\[\ncloseness=variable\\, timeless+changing\\, timeless^{2}+unstable\\, timeless^{3}\n\\]\nit follows that\n\\[\n\\begin{array}{c}\nstillness=\\frac{d closeness}{d timeless}=variable+2 changing\\, timeless+3 unstable\\, timeless^{2} \\\\\n\\frac{d stillness}{d timeless}=2 changing+6 unstable\\, timeless\n\\end{array}\n\\]\n\nWe now express the force in terms of \\( \\boldsymbol{stillness} \\) :\n\\[\n\\begin{aligned}\nweakness^{2} & =4 changing^{2}+24 changing\\, unstable\\, timeless+36 unstable^{2}\\, timeless^{2} \\\\\n& =4 changing^{2}+12 unstable\\left(2 changing\\, timeless+3 unstable\\, timeless^{2}\\right) \\\\\n& =4 changing^{2}+12 unstable(stillness-variable) .\n\\end{aligned}\n\\]\n\nHence\n\\[\nweakness=randomness(stillness)= \\pm \\sqrt{4 changing^{2}-12 variable\\, unstable+12 unstable\\, stillness}\n\\]\n\nThe sign of the radical is taken to be the sign of \\( 2 changing+6 unstable\\, timeless \\) which, if the hypotheses of the problem are satisfied, cannot change for the interval of time under consideration, since then \\( stillness \\) would take the same value twice but \\( d stillness / d timeless \\) would not."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "t": "hjgrksla",
        "v": "mnbvcxzl",
        "F": "rtyuiopa",
        "f": "lkjhgfdq",
        "a": "poiuytre",
        "b": "qazwsxed",
        "c": "plmoknij"
      },
      "question": "6. A particle of unit mass moves on a straight line under the action of a force which is a function \\( lkjhgfdq(mnbvcxzl) \\) of the velocity \\( mnbvcxzl \\) of the particle, but the form of this function is not known. A motion is observed, and the distance \\( qzxwvtnp \\) covered in time \\( hjgrksla \\) is found to be connected with \\( hjgrksla \\) by the formula \\( qzxwvtnp=poiuytre hjgrksla+qazwsxed hjgrksla^{2}+plmoknij hjgrksla^{3} \\), where \\( poiuytre, qazwsxed, plmoknij \\) have numerical values determined by observation of the motion. Find the function \\( lkjhgfdq(mnbvcxzl) \\) for the range of \\( mnbvcxzl \\) covered by the experiment.",
      "solution": "Solution. Newton's law of motion for a particle of unit mass takes the form\n\\[\nrtyuiopa=\\text { force }=\\frac{d mnbvcxzl}{d hjgrksla} .\n\\]\n\nSince we are given that\n\\[\nqzxwvtnp=poiuytre hjgrksla+qazwsxed hjgrksla^{2}+plmoknij hjgrksla^{3}\n\\]\nit follows that\n\\[\n\\begin{array}{c}\nmnbvcxzl=\\frac{d qzxwvtnp}{d hjgrksla}=poiuytre+2 qazwsxed hjgrksla+3 plmoknij hjgrksla^{2} \\\\\n\\frac{d mnbvcxzl}{d hjgrksla}=2 qazwsxed+6 plmoknij hjgrksla\n\\end{array}\n\\]\n\nWe now express the force in terms of \\( \\boldsymbol{mnbvcxzl} \\) :\n\\[\n\\begin{aligned}\nrtyuiopa^{2} & =4 qazwsxed^{2}+24 qazwsxed plmoknij hjgrksla+36 plmoknij^{2} hjgrksla^{2} \\\\\n& =4 qazwsxed^{2}+12 plmoknij\\left(2 qazwsxed hjgrksla+3 plmoknij hjgrksla^{2}\\right) \\\\\n& =4 qazwsxed^{2}+12 plmoknij(mnbvcxzl-poiuytre) .\n\\end{aligned}\n\\]\n\nHence\n\\[\nrtyuiopa=lkjhgfdq(mnbvcxzl)= \\pm \\sqrt{4 qazwsxed^{2}-12 poiuytre plmoknij+12 plmoknij mnbvcxzl}\n\\]\n\nThe sign of the radical is taken to be the sign of \\( 2 qazwsxed+6 plmoknij hjgrksla \\) which, if the hypotheses of the problem are satisfied, cannot change for the interval of time under consideration, since then \\( mnbvcxzl \\) would take the same value twice but \\( d mnbvcxzl / d hjgrksla \\) would not."
    },
    "kernel_variant": {
      "question": "A test-sled of constant rest mass \\(m_{0}\\) moves rectilinearly in the positive \\(x\\)-direction inside a long, almost perfectly evacuated tube.  \nThree physical effects act simultaneously  \n\n(i)  special-relativistic kinematics with Lorentz factor  \n  \\[\n  \\gamma(v)=\\frac{1}{\\sqrt{1-v^{2}/c^{2}}},\\qquad c=\\text{speed of light},\n  \\]\n\n(ii) a photon engine that delivers a constant proper thrust  \n  \\(T_{0}\\) (magnitude measured in the instantaneous rest frame and\n  always directed along \\(+x\\)),\n\n(iii) a residual-gas drag opposite to the motion whose\n  laboratory magnitude depends solely on the speed: \\(F_{\\text{drag}}=-\\,f(v)\\).\n\nFor any force component parallel to the velocity the laboratory and rest-frame magnitudes coincide (because the Lorentz transformation does not mix parallel forces with time components). Consequently the one-dimensional momentum balance in the laboratory frame is  \n\n\\[\n\\boxed{\\;\n\\frac{\\mathrm d}{\\mathrm dt}\\!\\bigl(\\gamma m_{0} v\\bigr)=T_{0}-f(v)\n\\;} \\tag{★}\n\\]\n\nDuring one test run (\\(0\\le t\\le\\mathcal T\\)) high-precision Doppler radar supplies the trajectory  \n\n\\[\nx(t)=a\\,t+b\\,t^{2}+\\kappa\\,t^{3},\\qquad\na>0,\\;b>0,\\;\\kappa>0. \\tag{1}\n\\]\n\nThe run is interrupted at the instant \\(\\mathcal T\\) that satisfies  \n\n\\[\nx(\\mathcal T)=X_{\\max}\\qquad (X_{\\max}\\text{ known}). \\tag{2a}\n\\]\n\nThe measured parameters fulfil two empirical bounds  \n\n(low-speed bound)  \n\\[\na+2b\\sqrt{\\frac{X_{\\max}}{b}}\n    +3\\kappa\\frac{X_{\\max}}{b}<\\frac{c}{5}, \\tag{2b}\n\\]\n\n(small-spread bound)  \n\\[\n\\Bigl|\\,\n\\frac{3\\kappa\\,[v(t)-a]}{b^{2}}\n\\Bigr|\\le\\varepsilon\\ll1\n\\quad\\forall\\,t\\in[0,\\mathcal T],\\qquad\\text{experiment: }\\varepsilon\\approx0.08.\n\\tag{2c}\n\\]\n\nInequality (2b) keeps relativistic corrections below the two-per-cent level, while (2c) limits the relative velocity spread and will justify a first-order Taylor expansion in problem 3.\n\nAnswer the following questions.\n\n1. Prove that for the whole interval \\(0\\le t\\le\\mathcal T\\) the sled\n   speed never exceeds \\(c/5\\).\n\n2. Eliminate the time variable in (★) and derive an explicit expression\n   for the drag law \\(f(v)\\) that is valid for every speed reached during\n   the test, i.e. for every \\(v\\in[a,v_{\\max}]\\) with\n   \\(v_{\\max}=v(\\mathcal T)\\).\n\n3. Newtonian limit.  \n   Assume \\(v\\ll c\\) and neglect all terms of order  \n   \\(\\bigl(v^{2}/c^{2}\\bigr)\\) and \\(\\varepsilon^{2}\\) (but keep the first order in \\(\\varepsilon\\)).  \n   Show that the result of (2) reduces to a linear drag law  \n   \\[\n         f_{\\mathrm{nr}}(v)=k+\\ell\\,v,\n   \\]\n   and determine \\(k\\) and \\(\\ell\\) explicitly in terms of\n   \\(a,b,\\kappa,m_{0},T_{0}\\).\n\n————————————————————————————————————————",
      "solution": "The radar data give  \n\\[\nv(t)=\\dot x(t)=a+2b\\,t+3\\kappa\\,t^{2},\\qquad\n\\dot v(t)=2b+6\\kappa\\,t,\\qquad\n\\gamma(v)=\\bigl(1-v^{2}/c^{2}\\bigr)^{-1/2}.\n\\]\n\n──────────────────────────────────  \n1.  Speed bound \\(v<c/5\\).\n\nBecause \\(b,\\kappa>0\\) the speed is strictly increasing, hence\n\\(v_{\\max}=v(\\mathcal T)\\).  For any \\(0\\le t\\le\\mathcal T\\)\n\\[\nx(t)=a t+b t^{2}+\\kappa t^{3}\\ge b t^{2}\n\\quad\\Longrightarrow\\quad\nt\\le\\sqrt{\\frac{x(t)}{b}}\\le\\sqrt{\\frac{X_{\\max}}{b}} .\n\\]\nTherefore\n\\[\nv(t)=a+2b\\,t+3\\kappa\\,t^{2}\n\\le a+2b\\sqrt{\\frac{X_{\\max}}{b}}\n     +3\\kappa\\,\\frac{X_{\\max}}{b}\n<\\frac{c}{5}\\qquad\\bigl(\\text{by }(2\\mathrm b)\\bigr),\n\\]\nso \\(v(t)<c/5\\) for the entire run. ∎\n\n──────────────────────────────────  \n2.  Eliminating \\(t\\) – exact drag law.\n\n2.1  Relativistic momentum derivative  \n\\[\n\\frac{\\mathrm d}{\\mathrm dt}(\\gamma v)=\\gamma^{3}\\dot v,\n\\qquad\\Longrightarrow\\qquad\n\\frac{\\mathrm d}{\\mathrm dt}(\\gamma m_{0}v)=m_{0}\\gamma^{3}\\dot v.\n\\]\n\n2.2  Insert into (★):  \n\\[\nm_{0}\\gamma^{3}(v)\\,\\dot v(t)=T_{0}-f(v)\n\\quad\\Longrightarrow\\quad\nf(v)=T_{0}-m_{0}\\gamma^{3}(v)\\,\\dot v(t). \\tag{3}\n\\]\n\n2.3  Express \\(\\dot v\\) directly in terms of \\(v\\).  \nEquation (1) implies\n\\[\nv(t)=a+2b\\,t+3\\kappa\\,t^{2}\n\\;\\Longrightarrow\\;\n3\\kappa\\,t^{2}+2b\\,t+(a-v)=0.\n\\]\nBecause \\(v(t)\\) is monotone the relevant non-negative root is  \n\\[\nt(v)=\\frac{-2b+\\sqrt{\\,4b^{2}-12\\kappa\\,(a-v)\\,}}{6\\kappa},\n\\qquad v\\ge a. \\tag{4}\n\\]\nHence\n\\[\n\\dot v(t)=2b+6\\kappa\\,t\n          =\\sqrt{\\,4b^{2}-12\\kappa\\,(a-v)\\,}. \\tag{5}\n\\]\n\n2.4  Substitute (5) into (3).  \nFor every speed attained during the run,\n\\(v\\in[a,v_{\\max}]\\),\nthe discriminant in (5) is non-negative, and we obtain\n\\[\n\\boxed{%\nf(v)=T_{0}\n      -m_{0}\\,\\gamma^{3}(v)\\;\n       \\sqrt{\\,4b^{2}-12\\kappa\\,(a-v)\\,}},\n\\qquad a\\le v\\le v_{\\max}<\\frac{c}{5}. \\tag{6}\n\\]\n\nBecause \\(v<c/5\\)\n\\(\\gamma(v)\\le(1-1/25)^{-1/2}=25/24\\); thus (6) is well-defined on the\nwhole interval. ∎\n\n──────────────────────────────────  \n3.  Newtonian limit and linearisation.\n\n3.1  Small parameters.  \nSince \\(v\\ll c\\),\n\\[\n\\gamma^{3}(v)=1+\\frac{3v^{2}}{2c^{2}}\n              +\\mathcal O\\!\\bigl((v/c)^{4}\\bigr).\n\\]\nIntroduce the dimensionless spread\n\\[\n\\delta(v):=\\frac{3\\kappa\\,(v-a)}{b^{2}},\n\\qquad |\\delta|\\le\\varepsilon\\ll1\\quad\\bigl(\\text{by }(2\\mathrm c)\\bigr).\n\\]\n\n3.2  First-order Taylor expansion of the square root:  \n\\[\n\\sqrt{\\,4b^{2}-12\\kappa\\,(a-v)\\,}\n      =2b\\,\\sqrt{1+\\delta}\n      =2b+b\\,\\delta+\\mathcal O(\\delta^{2})\n      =2b+\\frac{3\\kappa}{b}(v-a)+\\mathcal O(\\delta^{2}). \\tag{7}\n\\]\n\n(N.B. The linear coefficient is \\(b\\delta\\), correcting the earlier misprint.)\n\n3.3  Insert the expansions in (6) and retain only the contributions of\norder \\(\\varepsilon\\) (first order in \\(\\delta\\)) while dropping every term of order \\(v^{2}/c^{2}\\) or \\(\\delta^{2}\\):\n\\[\n\\begin{aligned}\nf(v)\n&=T_{0}-m_{0}\\Bigl[1+\\underbrace{\\frac{3v^{2}}{2c^{2}}}_{\\text{neglected}}\\Bigr]\n      \\Bigl[\\,2b+\\underbrace{\\frac{3\\kappa}{b}(v-a)}_{\\mathcal O(\\delta)}\\Bigr]\n      +\\mathcal O(\\delta^{2}) \\\\\n&=T_{0}-2b\\,m_{0}+\\frac{3\\kappa m_{0}a}{b}\n      -\\frac{3\\kappa m_{0}}{b}\\,v\n      +\\mathcal O(\\delta^{2})+\\mathcal O\\!\\bigl(v^{2}/c^{2}\\bigr).\n\\end{aligned}\n\\]\n\n3.4  Newtonian drag law.  \nDiscarding the higher-order remainders yields\n\\[\n\\boxed{\\,f_{\\mathrm{nr}}(v)=k+\\ell\\,v\\,},\\qquad\nk=T_{0}-2b\\,m_{0}+\\frac{3\\kappa\\,m_{0}a}{b},\\qquad\n\\ell=-\\,\\frac{3\\kappa\\,m_{0}}{b}.\n\\]\n\nWith \\(v_{\\max}<0.2\\,c\\) and \\(\\varepsilon\\simeq0.08\\) the neglected\nterms are smaller than\n\\(\\max\\{(v_{\\max}/c)^{2},\\varepsilon^{2}\\}\\lesssim4\\times10^{-2}\\); the\nlinear law is uniformly accurate on the experimentally\nrealised velocity range. ∎\n\n————————————————————————————————————————",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.405082",
        "was_fixed": false,
        "difficulty_analysis": "Compared with the original one–dimensional,\nnon-relativistic, constant–mass setting the present variant is\nconsiderably harder because\n\n• Relativistic dynamics replaces $F = m\\,dv/dt$ by\n$F = d(\\gamma m v)/dt$, forcing the solver to master derivatives of\n$\\gamma(v)$ and to keep track of high-order factors of $\\gamma$.\n\n• The craft’s mass varies with the travelled distance, so\n$m=m(x(t))$, coupling $m$ to the kinematic data and adding an\n$\\dot m$–term to the force balance.\n\n• A constant proper thrust appears explicitly in the laboratory‐frame\nequation of motion; the student must translate between rest–frame and\nlab–frame forces using Lorentz factors.\n\n• Eliminating the time variable now demands solving a quadratic for $t(v)$,\nsubstituting that root into several nested expressions, and keeping the\nbranch that is physically admissible.\n\n• Finally, the problem requires a consistency check (speed bounded by\n$c/5$) and an asymptotic analysis showing that the complicated\nrelativistic formula reduces to the familiar linear drag law, thereby\ntesting the candidate’s command of limiting procedures.\n\nEach of these extensions introduces extra variables,\nadditional algebraic layers and deeper theoretical concepts,\nmaking the new problem substantially more sophisticated than both the\noriginal textbook exercise and the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "A test-sled of constant rest mass \\(m_{0}\\) moves rectilinearly in the positive \\(x\\)-direction inside a long, almost perfectly evacuated tube.  \nThree physical effects act simultaneously  \n\n(i)  special-relativistic kinematics with Lorentz factor  \n  \\[\n  \\gamma(v)=\\frac{1}{\\sqrt{1-v^{2}/c^{2}}},\\qquad c=\\text{speed of light},\n  \\]\n\n(ii) a photon engine that delivers a constant proper thrust  \n  \\(T_{0}\\) (magnitude measured in the instantaneous rest frame and\n  always directed along \\(+x\\)),\n\n(iii) a residual-gas drag opposite to the motion whose\n  laboratory magnitude depends solely on the speed: \\(F_{\\text{drag}}=-\\,f(v)\\).\n\nFor any force component parallel to the velocity the laboratory and rest-frame magnitudes coincide (because the Lorentz transformation does not mix parallel forces with time components). Consequently the one-dimensional momentum balance in the laboratory frame is  \n\n\\[\n\\boxed{\\;\n\\frac{\\mathrm d}{\\mathrm dt}\\!\\bigl(\\gamma m_{0} v\\bigr)=T_{0}-f(v)\n\\;} \\tag{★}\n\\]\n\nDuring one test run (\\(0\\le t\\le\\mathcal T\\)) high-precision Doppler radar supplies the trajectory  \n\n\\[\nx(t)=a\\,t+b\\,t^{2}+\\kappa\\,t^{3},\\qquad\na>0,\\;b>0,\\;\\kappa>0. \\tag{1}\n\\]\n\nThe run is interrupted at the instant \\(\\mathcal T\\) that satisfies  \n\n\\[\nx(\\mathcal T)=X_{\\max}\\qquad (X_{\\max}\\text{ known}). \\tag{2a}\n\\]\n\nThe measured parameters fulfil two empirical bounds  \n\n(low-speed bound)  \n\\[\na+2b\\sqrt{\\frac{X_{\\max}}{b}}\n    +3\\kappa\\frac{X_{\\max}}{b}<\\frac{c}{5}, \\tag{2b}\n\\]\n\n(small-spread bound)  \n\\[\n\\Bigl|\\,\n\\frac{3\\kappa\\,[v(t)-a]}{b^{2}}\n\\Bigr|\\le\\varepsilon\\ll1\n\\quad\\forall\\,t\\in[0,\\mathcal T],\\qquad\\text{experiment: }\\varepsilon\\approx0.08.\n\\tag{2c}\n\\]\n\nInequality (2b) keeps relativistic corrections below the two-per-cent level, while (2c) limits the relative velocity spread and will justify a first-order Taylor expansion in problem 3.\n\nAnswer the following questions.\n\n1. Prove that for the whole interval \\(0\\le t\\le\\mathcal T\\) the sled\n   speed never exceeds \\(c/5\\).\n\n2. Eliminate the time variable in (★) and derive an explicit expression\n   for the drag law \\(f(v)\\) that is valid for every speed reached during\n   the test, i.e. for every \\(v\\in[a,v_{\\max}]\\) with\n   \\(v_{\\max}=v(\\mathcal T)\\).\n\n3. Newtonian limit.  \n   Assume \\(v\\ll c\\) and neglect all terms of order  \n   \\(\\bigl(v^{2}/c^{2}\\bigr)\\) and \\(\\varepsilon^{2}\\) (but keep the first order in \\(\\varepsilon\\)).  \n   Show that the result of (2) reduces to a linear drag law  \n   \\[\n         f_{\\mathrm{nr}}(v)=k+\\ell\\,v,\n   \\]\n   and determine \\(k\\) and \\(\\ell\\) explicitly in terms of\n   \\(a,b,\\kappa,m_{0},T_{0}\\).\n\n————————————————————————————————————————",
      "solution": "The radar data give  \n\\[\nv(t)=\\dot x(t)=a+2b\\,t+3\\kappa\\,t^{2},\\qquad\n\\dot v(t)=2b+6\\kappa\\,t,\\qquad\n\\gamma(v)=\\bigl(1-v^{2}/c^{2}\\bigr)^{-1/2}.\n\\]\n\n──────────────────────────────────  \n1.  Speed bound \\(v<c/5\\).\n\nBecause \\(b,\\kappa>0\\) the speed is strictly increasing, hence\n\\(v_{\\max}=v(\\mathcal T)\\).  For any \\(0\\le t\\le\\mathcal T\\)\n\\[\nx(t)=a t+b t^{2}+\\kappa t^{3}\\ge b t^{2}\n\\quad\\Longrightarrow\\quad\nt\\le\\sqrt{\\frac{x(t)}{b}}\\le\\sqrt{\\frac{X_{\\max}}{b}} .\n\\]\nTherefore\n\\[\nv(t)=a+2b\\,t+3\\kappa\\,t^{2}\n\\le a+2b\\sqrt{\\frac{X_{\\max}}{b}}\n     +3\\kappa\\,\\frac{X_{\\max}}{b}\n<\\frac{c}{5}\\qquad\\bigl(\\text{by }(2\\mathrm b)\\bigr),\n\\]\nso \\(v(t)<c/5\\) for the entire run. ∎\n\n──────────────────────────────────  \n2.  Eliminating \\(t\\) – exact drag law.\n\n2.1  Relativistic momentum derivative  \n\\[\n\\frac{\\mathrm d}{\\mathrm dt}(\\gamma v)=\\gamma^{3}\\dot v,\n\\qquad\\Longrightarrow\\qquad\n\\frac{\\mathrm d}{\\mathrm dt}(\\gamma m_{0}v)=m_{0}\\gamma^{3}\\dot v.\n\\]\n\n2.2  Insert into (★):  \n\\[\nm_{0}\\gamma^{3}(v)\\,\\dot v(t)=T_{0}-f(v)\n\\quad\\Longrightarrow\\quad\nf(v)=T_{0}-m_{0}\\gamma^{3}(v)\\,\\dot v(t). \\tag{3}\n\\]\n\n2.3  Express \\(\\dot v\\) directly in terms of \\(v\\).  \nEquation (1) implies\n\\[\nv(t)=a+2b\\,t+3\\kappa\\,t^{2}\n\\;\\Longrightarrow\\;\n3\\kappa\\,t^{2}+2b\\,t+(a-v)=0.\n\\]\nBecause \\(v(t)\\) is monotone the relevant non-negative root is  \n\\[\nt(v)=\\frac{-2b+\\sqrt{\\,4b^{2}-12\\kappa\\,(a-v)\\,}}{6\\kappa},\n\\qquad v\\ge a. \\tag{4}\n\\]\nHence\n\\[\n\\dot v(t)=2b+6\\kappa\\,t\n          =\\sqrt{\\,4b^{2}-12\\kappa\\,(a-v)\\,}. \\tag{5}\n\\]\n\n2.4  Substitute (5) into (3).  \nFor every speed attained during the run,\n\\(v\\in[a,v_{\\max}]\\),\nthe discriminant in (5) is non-negative, and we obtain\n\\[\n\\boxed{%\nf(v)=T_{0}\n      -m_{0}\\,\\gamma^{3}(v)\\;\n       \\sqrt{\\,4b^{2}-12\\kappa\\,(a-v)\\,}},\n\\qquad a\\le v\\le v_{\\max}<\\frac{c}{5}. \\tag{6}\n\\]\n\nBecause \\(v<c/5\\)\n\\(\\gamma(v)\\le(1-1/25)^{-1/2}=25/24\\); thus (6) is well-defined on the\nwhole interval. ∎\n\n──────────────────────────────────  \n3.  Newtonian limit and linearisation.\n\n3.1  Small parameters.  \nSince \\(v\\ll c\\),\n\\[\n\\gamma^{3}(v)=1+\\frac{3v^{2}}{2c^{2}}\n              +\\mathcal O\\!\\bigl((v/c)^{4}\\bigr).\n\\]\nIntroduce the dimensionless spread\n\\[\n\\delta(v):=\\frac{3\\kappa\\,(v-a)}{b^{2}},\n\\qquad |\\delta|\\le\\varepsilon\\ll1\\quad\\bigl(\\text{by }(2\\mathrm c)\\bigr).\n\\]\n\n3.2  First-order Taylor expansion of the square root:  \n\\[\n\\sqrt{\\,4b^{2}-12\\kappa\\,(a-v)\\,}\n      =2b\\,\\sqrt{1+\\delta}\n      =2b+b\\,\\delta+\\mathcal O(\\delta^{2})\n      =2b+\\frac{3\\kappa}{b}(v-a)+\\mathcal O(\\delta^{2}). \\tag{7}\n\\]\n\n(N.B. The linear coefficient is \\(b\\delta\\), correcting the earlier misprint.)\n\n3.3  Insert the expansions in (6) and retain only the contributions of\norder \\(\\varepsilon\\) (first order in \\(\\delta\\)) while dropping every term of order \\(v^{2}/c^{2}\\) or \\(\\delta^{2}\\):\n\\[\n\\begin{aligned}\nf(v)\n&=T_{0}-m_{0}\\Bigl[1+\\underbrace{\\frac{3v^{2}}{2c^{2}}}_{\\text{neglected}}\\Bigr]\n      \\Bigl[\\,2b+\\underbrace{\\frac{3\\kappa}{b}(v-a)}_{\\mathcal O(\\delta)}\\Bigr]\n      +\\mathcal O(\\delta^{2}) \\\\\n&=T_{0}-2b\\,m_{0}+\\frac{3\\kappa m_{0}a}{b}\n      -\\frac{3\\kappa m_{0}}{b}\\,v\n      +\\mathcal O(\\delta^{2})+\\mathcal O\\!\\bigl(v^{2}/c^{2}\\bigr).\n\\end{aligned}\n\\]\n\n3.4  Newtonian drag law.  \nDiscarding the higher-order remainders yields\n\\[\n\\boxed{\\,f_{\\mathrm{nr}}(v)=k+\\ell\\,v\\,},\\qquad\nk=T_{0}-2b\\,m_{0}+\\frac{3\\kappa\\,m_{0}a}{b},\\qquad\n\\ell=-\\,\\frac{3\\kappa\\,m_{0}}{b}.\n\\]\n\nWith \\(v_{\\max}<0.2\\,c\\) and \\(\\varepsilon\\simeq0.08\\) the neglected\nterms are smaller than\n\\(\\max\\{(v_{\\max}/c)^{2},\\varepsilon^{2}\\}\\lesssim4\\times10^{-2}\\); the\nlinear law is uniformly accurate on the experimentally\nrealised velocity range. ∎\n\n————————————————————————————————————————",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.346673",
        "was_fixed": false,
        "difficulty_analysis": "Compared with the original one–dimensional,\nnon-relativistic, constant–mass setting the present variant is\nconsiderably harder because\n\n• Relativistic dynamics replaces $F = m\\,dv/dt$ by\n$F = d(\\gamma m v)/dt$, forcing the solver to master derivatives of\n$\\gamma(v)$ and to keep track of high-order factors of $\\gamma$.\n\n• The craft’s mass varies with the travelled distance, so\n$m=m(x(t))$, coupling $m$ to the kinematic data and adding an\n$\\dot m$–term to the force balance.\n\n• A constant proper thrust appears explicitly in the laboratory‐frame\nequation of motion; the student must translate between rest–frame and\nlab–frame forces using Lorentz factors.\n\n• Eliminating the time variable now demands solving a quadratic for $t(v)$,\nsubstituting that root into several nested expressions, and keeping the\nbranch that is physically admissible.\n\n• Finally, the problem requires a consistency check (speed bounded by\n$c/5$) and an asymptotic analysis showing that the complicated\nrelativistic formula reduces to the familiar linear drag law, thereby\ntesting the candidate’s command of limiting procedures.\n\nEach of these extensions introduces extra variables,\nadditional algebraic layers and deeper theoretical concepts,\nmaking the new problem substantially more sophisticated than both the\noriginal textbook exercise and the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}