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.
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.
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)
Horizontal axis (x): b: Electron speed v / c
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Electron speed v / c | log₁₀ (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
- 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.
- 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.
- 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.
- The interpolation error in the input is |b_line − b| = 0.149378 c. This is larger than the requested numerical tolerance.
- 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.
- 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.
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.
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
Horizontal axis (x): te: Time t of event E / s
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| te: Time t of event E / s | x″ OF EVENT E IN S″ / m |
|---|---|
| 0.871875 | 8.71875 |
| 1.5609374999999999 | 15.609374999999998 |
| 2.25 | 22.5 |
| 2.9390624999999995 | 29.390624999999993 |
| 3.628125 | 36.28125 |
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
- 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.
- 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.
- 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.
- The interpolation error in the input is |te_line − te| = 0 s. This is within the requested numerical tolerance.
- 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.
- 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.
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.
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
Horizontal axis (x): b: Speed β = v/c
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Speed β = v/c | LORENTZ FACTOR γ |
|---|---|
| 0.2796875 | 1.0415678363642908 |
| 0.42734375 | 1.106085524403928 |
| 0.575 | 1.2222646627042817 |
| 0.72265625 | 1.4467433547506021 |
| 0.8703125 | 2.030457261535957 |
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
- 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.
- 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.
- 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.
- The interpolation error in the input is |b_line − b| = 0.205091. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.2624289 and 1.3403789. Sort the roots even if the response decreases: 0.61036 ≤ b ≤ 0.665881.
- 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.
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.
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
Horizontal axis (x): p: Lab-frame momentum pc / MeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| p: Lab-frame momentum pc / MeV | β = pc/E |
|---|---|
| 0.301875 | 0.4494454708717874 |
| 0.6759375 | 0.7478669678549078 |
| 1.0499999999999998 | 0.8682431421244593 |
| 1.4240624999999998 | 0.9215439910169114 |
| 1.7981249999999998 | 0.9485843377372829 |
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
- 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.
- 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.
- 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.
- The interpolation error in the input is |p_line − p| = 0.427617 MeV. This is larger than the requested numerical tolerance.
- 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.
- 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.