University Physics I · topic UPI-08.02
Impulse and force–time graphs
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.
Further practice
Work through a fresh case before tackling the extension.
Impulse and force–time graphs: predict a finite change
Impulse is the time integral of a force, J = ∫F dt, a vector measured in N s; the net impulse on a body equals its change in momentum, and for each component it is the signed area under the force–time graph. Use the supplied topic model Y_out = ct*acos((r-1)/(r+1))/pi, where Y_out is BALL AT REST AT t (ms). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; unmarked trigonometric functions use radians. r is Rebound speed ÷ impact speed r, dimensionless. Keep v (Impact speed v (0.40 kg ball)) = 6 m s⁻¹; ct (Contact time τ) = 8 ms fixed. For this exercise use only 0.10625 ≤ r ≤ 0.89375 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. Calculate Y_out at r = 0.2795 and 0.38975. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δr across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
BALL AT REST AT t against r
Predictions from the supplied Impulse and force–time graphs 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): BALL AT REST AT t / ms
Horizontal axis (x): r: Rebound speed ÷ impact speed r
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r: Rebound speed ÷ impact speed r | BALL AT REST AT t / ms |
|---|---|
| 0.10625 | 6.3952086242465125 |
| 0.303125 | 5.436820564230729 |
| 0.5 | 4.865387583755142 |
| 0.6968749999999999 | 4.457350696107122 |
| 0.8937499999999999 | 4.1429468267039296 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of r separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At r = 0.2795, substitute into Y_out = ct*acos((r-1)/(r+1))/pi: Y₀ = 5.52316 ms.
- At r = 0.38975, the same fixed-parameter model gives Y₁ = 5.15764 ms.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.365521 ms; the input change is Δr = 0.11025.
- Average sensitivity is ΔY_out/Δr = -3.31538 ms per unit input. Its negative sign means that this response decreases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 5.52316 ms, Y₁ = 5.15764 ms; ΔY_out = -0.365521 ms; average sensitivity = -3.31538 ms per unit input.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
Impulse and force–time graphs: infer an input with uncertainty
Impulse is the time integral of a force, J = ∫F dt, a vector measured in N s; the net impulse on a body equals its change in momentum, and for each component it is the signed area under the force–time graph. Use the supplied topic model Y_out = ct*acos((r-1)/(r+1))/pi, where Y_out is BALL AT REST AT t (ms). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; unmarked trigonometric functions use radians. r is Rebound speed ÷ impact speed r, dimensionless. Keep v (Impact speed v (0.40 kg ball)) = 6 m s⁻¹; ct (Contact time τ) = 8 ms fixed. For this exercise use only 0.10625 ≤ r ≤ 0.89375 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 = 4.6711976 ms, with a bounded reading uncertainty ±0.079238 ms. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown r by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00007875, 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.
BALL AT REST AT t against r
Predictions from the supplied Impulse and force–time graphs 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): BALL AT REST AT t / ms
Horizontal axis (x): r: Rebound speed ÷ impact speed r
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r: Rebound speed ÷ impact speed r | BALL AT REST AT t / ms |
|---|---|
| 0.10625 | 6.3952086242465125 |
| 0.303125 | 5.436820564230729 |
| 0.5 | 4.865387583755142 |
| 0.6968749999999999 | 4.457350696107122 |
| 0.8937499999999999 | 4.1429468267039296 |
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(r) = 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(0.10625) = 6.39521 ms and Y_out(0.89375) = 4.14295 ms. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives r_line = 0.10625 + (4.6712 − 6.39521)(0.89375 − 0.10625)/(4.14295 − 6.39521) = 0.709048. 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 r = 0.586625; substituting back gives Y_out = 4.6711976 ms.
- The interpolation error in the input is |r_line − r| = 0.122423. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 4.5919596 ms and 4.7504356 ms. Sort the roots even if the response decreases: 0.549849 ≤ r ≤ 0.625522.
- The lower and upper input deviations are 0.0367761 and 0.038897. 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 r ≈ 0.586625; compatible interval [0.549849, 0.625522]. Endpoint interpolation gives 0.709048.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.