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University Physics IV · course UPIV

University Physics IV

Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.

Extra challenging questions

A selection from this pathway. Open a topic below for its complete practice and challenge set.

Extra challenging · 01

The classical synthesis and its limits: infer an input with uncertainty

The classical synthesis is the closed predictive account formed by Newtonian dynamics, Maxwell's electrodynamics and statistical mechanics; its domain of validity is the region in which the ratios v/c, hν/kT and λ/d are all small. Use the supplied topic model Y_out = log(b), where Y_out is log₁₀ (v/c) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; log(x) means log₁₀x. b is Electron speed v, measured in c. Keep s (Cavity size, log₁₀(d / nm)) = 3; t (Temperature T) = 300 K fixed. For this exercise use only 0.106081 c ≤ b ≤ 0.884919 c on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = -0.23569522, with a bounded reading uncertainty ±0.02739. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown b by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000778837 c, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.

Explore the question data

log₁₀ (v/c) against b

Predictions from the supplied The classical synthesis and its limits model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.

Vertical axis (y): log₁₀ (v/c)

log₁₀ (v/c) against bPredictions from the supplied The classical synthesis and its limits model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law. Horizontal axis: b: Electron speed v / c. Vertical axis: log₁₀ (v/c). Exact values are available in the data table below.-1.011-0.7625-0.5137-0.265-0.016250.074930.28520.49550.70580.9161xy

Horizontal axis (x): b: Electron speed v / c

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1b: Electron speed v / c: 0.1061log₁₀ (v/c): -0.9744
Read the exact data as a table
log₁₀ (v/c) against b — supplied values
b: Electron speed v / clog₁₀ (v/c)
0.10608125-0.9743613714345515
0.300790625-0.521735703898937
0.4955-0.30495634117870585
0.690209375-0.16101914605421552
0.88491875-0.053096602801090194

These are the supplied values, not a worked answer. No fitted trend is drawn.

Show a hint

The model's endpoint predictions bracket the reading. Solve Y_out(b) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.

Reveal the worked solution
  1. Endpoint predictions are Y_out(0.106081) = -0.974361 and Y_out(0.884919) = -0.0530966. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives b_line = 0.106081 + (-0.235695 − -0.974361)(0.884919 − 0.106081)/(-0.0530966 − -0.974361) = 0.73055 c. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
  3. Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives b = 0.581172 c; substituting back gives Y_out = -0.23569522.
  4. The interpolation error in the input is |b_line − b| = 0.149378 c. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = -0.26308522 and -0.20830522. Sort the roots even if the response decreases: 0.545651 c ≤ b ≤ 0.619006 c.
  6. The lower and upper input deviations are 0.0355214 c and 0.0378338 c. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.

Answer b ≈ 0.581172 c; compatible interval [0.545651, 0.619006] c. Endpoint interpolation gives 0.73055 c.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

Extra challenging · 02

Inertial frames and the Galilean group: infer an input with uncertainty

An inertial frame is one in which a body free of net force keeps a constant velocity, and the Galilean group is the ten-parameter set of coordinate changes — boosts, rotations, spatial displacements and clock shifts — that carries one such frame to another. Use the supplied topic model Y_out = (30-v-w)*te, where Y_out is x″ OF EVENT E IN S″ (m). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. te is Time t of event E, measured in s. Keep v (Boost v of S′ relative to S) = 12 m s⁻¹; w (Further boost w of S″ vs S′) = 8 m s⁻¹ fixed. For this exercise use only 0.871875 s ≤ te ≤ 3.62812 s on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 25.531875 m, with a bounded reading uncertainty ±1.6124 m. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown te by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000275625 s, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.

Explore the question data

x″ OF EVENT E IN S″ against te

Predictions from the supplied Inertial frames and the Galilean group model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.

Vertical axis (y): x″ OF EVENT E IN S″ / m

x″ OF EVENT E IN S″ against tePredictions from the supplied Inertial frames and the Galilean group model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law. Horizontal axis: te: Time t of event E / s. Vertical axis: x″ OF EVENT E IN S″ / m. Exact values are available in the data table below.7.61615.0622.529.9437.380.76161.5062.252.9943.738xy

Horizontal axis (x): te: Time t of event E / s

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1te: Time t of event E / s: 0.8719x″ OF EVENT E IN S″ / m: 8.719
Read the exact data as a table
x″ OF EVENT E IN S″ against te — supplied values
te: Time t of event E / sx″ OF EVENT E IN S″ / m
0.8718758.71875
1.560937499999999915.609374999999998
2.2522.5
2.939062499999999529.390624999999993
3.62812536.28125

These are the supplied values, not a worked answer. No fitted trend is drawn.

Show a hint

The model's endpoint predictions bracket the reading. Solve Y_out(te) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.

Reveal the worked solution
  1. Endpoint predictions are Y_out(0.871875) = 8.71875 m and Y_out(3.62812) = 36.2813 m. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives te_line = 0.871875 + (25.5319 − 8.71875)(3.62812 − 0.871875)/(36.2813 − 8.71875) = 2.55319 s. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
  3. Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives te = 2.55319 s; substituting back gives Y_out = 25.531875 m.
  4. The interpolation error in the input is |te_line − te| = 0 s. This is within the requested numerical tolerance.
  5. Invert both limits Y_out = 23.919475 m and 27.144275 m. Sort the roots even if the response decreases: 2.39195 s ≤ te ≤ 2.71443 s.
  6. The lower and upper input deviations are 0.16124 s and 0.16124 s. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.

Answer te ≈ 2.55319 s; compatible interval [2.39195, 2.71443] s. Endpoint interpolation gives 2.55319 s.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

Extra challenging · 03

Proper time and time dilation: infer an input with uncertainty

Proper time is the interval between two events measured by a single clock that is present at both; time dilation is the statement that every other inertial frame assigns a larger interval, Δt = γΔτ. Use the supplied topic model Y_out = 1/sqrt(max(1-b*b,0.0001)), where Y_out is LORENTZ FACTOR γ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value. b is Speed β = v/c, dimensionless. Keep t0 (Proper time Δτ) = 2.2 μs fixed. For this exercise use only 0.279687 ≤ b ≤ 0.870313 on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 1.3014039, with a bounded reading uncertainty ±0.038975. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown b by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000590625, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.

Explore the question data

LORENTZ FACTOR γ against b

Predictions from the supplied Proper time and time dilation model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.

Vertical axis (y): LORENTZ FACTOR γ

LORENTZ FACTOR γ against bPredictions from the supplied Proper time and time dilation model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law. Horizontal axis: b: Speed β = v/c. Vertical axis: LORENTZ FACTOR γ. Exact values are available in the data table below.1.0021.2691.5361.8032.070.25610.41550.5750.73450.8939xy

Horizontal axis (x): b: Speed β = v/c

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1b: Speed β = v/c: 0.2797LORENTZ FACTOR γ: 1.042
Read the exact data as a table
LORENTZ FACTOR γ against b — supplied values
b: Speed β = v/cLORENTZ FACTOR γ
0.27968751.0415678363642908
0.427343751.106085524403928
0.5751.2222646627042817
0.722656251.4467433547506021
0.87031252.030457261535957

These are the supplied values, not a worked answer. No fitted trend is drawn.

Show a hint

The model's endpoint predictions bracket the reading. Solve Y_out(b) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.

Reveal the worked solution
  1. Endpoint predictions are Y_out(0.279687) = 1.04157 and Y_out(0.870313) = 2.03046. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives b_line = 0.279687 + (1.3014 − 1.04157)(0.870313 − 0.279687)/(2.03046 − 1.04157) = 0.434877. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
  3. Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives b = 0.639969; substituting back gives Y_out = 1.3014039.
  4. The interpolation error in the input is |b_line − b| = 0.205091. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 1.2624289 and 1.3403789. Sort the roots even if the response decreases: 0.61036 ≤ b ≤ 0.665881.
  6. The lower and upper input deviations are 0.0296083 and 0.0259126. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.

Answer b ≈ 0.639969; compatible interval [0.61036, 0.665881]. Endpoint interpolation gives 0.434877.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

Extra challenging · 04

Photon energy, momentum and the massless limit: infer an input with uncertainty

A photon is the massless quantum of the electromagnetic field: because m = 0, its energy and momentum are locked together as E = pc, so fixing E = hf fixes p = h/λ, and neither can change without the other. Use the supplied topic model Y_out = p/max(sqrt(p^2+m^2),0.01), where Y_out is β = pc/E (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value. p is Lab-frame momentum pc, measured in MeV. Keep m (Rest energy mc²) = 0.6 MeV; b (Observer boost β = v/c) = 0.3 fixed. For this exercise use only 0.301875 MeV ≤ p ≤ 1.79812 MeV on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 0.89657043, with a bounded reading uncertainty ±0.0078021. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown p by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000149625 MeV, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.

Explore the question data

β = pc/E against p

Predictions from the supplied Photon energy, momentum and the massless limit model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.

Vertical axis (y): β = pc/E

β = pc/E against pPredictions from the supplied Photon energy, momentum and the massless limit model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law. Horizontal axis: p: Lab-frame momentum pc / MeV. Vertical axis: β = pc/E. Exact values are available in the data table below.0.42950.56420.6990.83380.96850.2420.6461.051.4541.858xy

Horizontal axis (x): p: Lab-frame momentum pc / MeV

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1p: Lab-frame momentum pc / MeV: 0.3019β = pc/E: 0.4494
Read the exact data as a table
β = pc/E against p — supplied values
p: Lab-frame momentum pc / MeVβ = pc/E
0.3018750.4494454708717874
0.67593750.7478669678549078
1.04999999999999980.8682431421244593
1.42406249999999980.9215439910169114
1.79812499999999980.9485843377372829

These are the supplied values, not a worked answer. No fitted trend is drawn.

Show a hint

The model's endpoint predictions bracket the reading. Solve Y_out(p) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.

Reveal the worked solution
  1. Endpoint predictions are Y_out(0.301875) = 0.449445 and Y_out(1.79812) = 0.948584. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives p_line = 0.301875 + (0.89657 − 0.449445)(1.79812 − 0.301875)/(0.948584 − 0.449445) = 1.6422 MeV. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
  3. Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives p = 1.21459 MeV; substituting back gives Y_out = 0.89657043.
  4. The interpolation error in the input is |p_line − p| = 0.427617 MeV. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 0.88876833 and 0.90437253. Sort the roots even if the response decreases: 1.16342 MeV ≤ p ≤ 1.27154 MeV.
  6. The lower and upper input deviations are 0.0511683 MeV and 0.0569563 MeV. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.

Answer p ≈ 1.21459 MeV; compatible interval [1.16342, 1.27154] MeV. Endpoint interpolation gives 1.6422 MeV.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

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