University Physics I · unit UPI-01
Mathematical & Physical Foundations
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.
SI Quantities, Units & Prefixes: infer an input with uncertainty
An SI measurement is a number multiplied by a unit, where every coherent derived unit is a product of powers of seven base units with no extra numerical factor, and a prefix multiplies a unit by an exact power of ten. Use the supplied topic model Y_out = log(n)+k*p, where Y_out is LOG₁₀ OF VALUE IN SI (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; log(x) means log₁₀x. n is Number n in front of the unit, dimensionless. Keep k (Prefix 10ᵏ (milli −3, centi −2, kilo 3)) = -2; p (Power p on the unit (2 area, 3 volume)) = 2 fixed. For this exercise use only 1.90312 ≤ n ≤ 8.59688 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 = -3.2228406, with a bounded reading uncertainty ±0.023577. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown n by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000669375, 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 OF VALUE IN SI against n
Predictions from the supplied SI Quantities, Units & Prefixes 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 OF VALUE IN SI
Horizontal axis (x): n: Number n in front of the unit
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| n: Number n in front of the unit | LOG₁₀ OF VALUE IN SI |
|---|---|
| 1.903125 | -3.7205326856870307 |
| 3.5765625 | -3.446534181309344 |
| 5.25 | -3.279840696594043 |
| 6.9234375 | -3.1596782239522754 |
| 8.596875 | -3.0656593879302223 |
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(n) = 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(1.90312) = -3.72053 and Y_out(8.59688) = -3.06566. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives n_line = 1.90312 + (-3.22284 − -3.72053)(8.59688 − 1.90312)/(-3.06566 − -3.72053) = 6.99026. 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 n = 5.98631; substituting back gives Y_out = -3.2228406.
- The interpolation error in the input is |n_line − n| = 1.00394. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -3.2464176 and -3.1992636. Sort the roots even if the response decreases: 5.66999 ≤ n ≤ 6.32028.
- The lower and upper input deviations are 0.316321 and 0.333968. 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 n ≈ 5.98631; compatible interval [5.66999, 6.32028]. Endpoint interpolation gives 6.99026.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Dimensional Analysis & Scaling: infer an input with uncertainty
Dimensional analysis tracks the powers of mass, length and time inside a quantity, so equations can be tested and simple relationships inferred before a single number is substituted. Use the supplied topic model Y_out = a, where Y_out is M RESIDUAL a (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. a is Exponent a on mass m, dimensionless. Keep b (Exponent b on length L) = 1; c (Exponent c on g) = 0 fixed. For this exercise use only -0.7875 ≤ a ≤ 0.7875 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 = 0.17325, with a bounded reading uncertainty ±0.092137. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown a by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0001575, 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.
M RESIDUAL a against a
Predictions from the supplied Dimensional Analysis & Scaling 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): M RESIDUAL a
Horizontal axis (x): a: Exponent a on mass m
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: Exponent a on mass m | M RESIDUAL a |
|---|---|
| -0.7875 | -0.7875 |
| -0.39375000000000004 | -0.39375000000000004 |
| -1.1102230246251565e-16 | -1.1102230246251565e-16 |
| 0.39374999999999993 | 0.39374999999999993 |
| 0.7874999999999998 | 0.7874999999999998 |
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(a) = 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(-0.7875) = -0.7875 and Y_out(0.7875) = 0.7875. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives a_line = -0.7875 + (0.17325 − -0.7875)(0.7875 − -0.7875)/(0.7875 − -0.7875) = 0.17325. 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 a = 0.17325; substituting back gives Y_out = 0.17325.
- The interpolation error in the input is |a_line − a| = 5.55112e-17. This is within the requested numerical tolerance.
- Invert both limits Y_out = 0.081113 and 0.265387. Sort the roots even if the response decreases: 0.081113 ≤ a ≤ 0.265387.
- The lower and upper input deviations are 0.092137 and 0.092137. 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 a ≈ 0.17325; compatible interval [0.081113, 0.265387]. Endpoint interpolation gives 0.17325.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Significant Figures & Uncertainty: infer an input with uncertainty
The uncertainty of a measurement is a number in the same unit as the value, saying how far the true value could reasonably sit from it — by convention one standard deviation of the spread, unless a coverage factor is stated. Use the supplied topic model Y_out = (s/b)^2, where Y_out is BREAK-EVEN N (readings). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. b is Systematic uncertainty b, measured in ms. Keep s (Scatter of one reading s) = 80 ms; N (Readings averaged N) = 25 fixed. For this exercise use only 5.14375 ms ≤ b ≤ 35.8562 ms 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 = 11.224589 readings, with a bounded reading uncertainty ±0.937 readings. 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 0.00307125 ms, 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.
BREAK-EVEN N against b
Predictions from the supplied Significant Figures & Uncertainty 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): BREAK-EVEN N / readings
Horizontal axis (x): b: Systematic uncertainty b / ms
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Systematic uncertainty b / ms | BREAK-EVEN N / readings |
|---|---|
| 5.14375 | 241.8913113125232 |
| 12.821874999999999 | 38.92932707852021 |
| 20.499999999999996 | 15.229030339083884 |
| 28.178124999999998 | 8.060385095180033 |
| 35.856249999999996 | 4.977946669310391 |
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(5.14375) = 241.891 readings and Y_out(35.8562) = 4.97795 readings. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives b_line = 5.14375 + (11.2246 − 241.891)(35.8562 − 5.14375)/(4.97795 − 241.891) = 35.0465 ms. 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 = 23.8784 ms; substituting back gives Y_out = 11.224589 readings.
- The interpolation error in the input is |b_line − b| = 11.1681 ms. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 10.287589 readings and 12.161589 readings. Sort the roots even if the response decreases: 22.9401 ms ≤ b ≤ 24.9421 ms.
- The lower and upper input deviations are 0.9383 ms and 1.06373 ms. 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 ≈ 23.8784 ms; compatible interval [22.9401, 24.9421] ms. Endpoint interpolation gives 35.0465 ms.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Coordinate Systems & Components: infer an input with uncertainty
A Cartesian coordinate system is a chosen origin plus perpendicular axes, each with a positive direction and a unit; a vector's Cartesian components are the signed lengths of its projections onto those axes. Use the supplied topic model Y_out = (500*sind(theta-phi))/100, where Y_out is COMPONENT A_y′ (before display rounding) (m). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. phi is Axis rotation φ, measured in °. Keep theta (Direction θ of A) = 125 °; s (Origin shift s) = 0 m fixed. For this exercise use only 37.9125 ° ≤ phi ≤ 80.4375 ° 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 = 4.3793139 m, with a bounded reading uncertainty ±0.092134 m. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown phi by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0042525 °, 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.
COMPONENT A_y′ (before display rounding) against phi
Predictions from the supplied Coordinate Systems & Components 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): COMPONENT A_y′ (before display rounding) / m
Horizontal axis (x): phi: Axis rotation φ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| phi: Axis rotation φ / ° | COMPONENT A_y′ (before display rounding) / m |
|---|---|
| 37.9125 | 4.993541478746162 |
| 48.54375 | 4.860956911846637 |
| 59.175 | 4.561494460023537 |
| 69.80625 | 4.105434745277949 |
| 80.4375 | 3.508434413405297 |
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(phi) = 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(37.9125) = 4.99354 m and Y_out(80.4375) = 3.50843 m. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives phi_line = 37.9125 + (4.37931 − 4.99354)(80.4375 − 37.9125)/(3.50843 − 4.99354) = 55.5005 °. 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 phi = 63.8527 °; substituting back gives Y_out = 4.3793139 m.
- The interpolation error in the input is |phi_line − phi| = 8.35227 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 4.2871799 m and 4.4714479 m. Sort the roots even if the response decreases: 61.5827 ° ≤ phi ≤ 65.9701 °.
- The lower and upper input deviations are 2.27007 ° and 2.11735 °. 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 phi ≈ 63.8527 °; compatible interval [61.5827, 65.9701] °. Endpoint interpolation gives 55.5005 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.