Integrate a smooth force without confusing impulse and work
For a fixed-mass particle, net force is the time derivative of momentum. Integrating a signed force over time gives momentum change, not energy change. Here a 2 kg particle initially moves at −1 m/s under the supplied net force F(t) = A sin(πt/T), from t = 0 to T. No measured impact waveform or apparatus is implied.
Learning outcome: Derive both integrations and compare the two equal-impulse packets. Retain signed displacement and kinetic change separately.
Starting recap
Integrate signed force and normalized mass moments. Specify the system before conserving momentum, and independently test energy and observation constraints.
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Try
Change to A = 2 N and T = 2 s. The impulse and final velocity are unchanged, but displacement is −2 + 4/π ≈ −0.726760 m. Doubling duration at half amplitude changes the trajectory.
Open the experimentPredict
Two supplied pulses have equal signed impulse on the same mass and initial velocity. Which endpoint quantity must agree?
Experiment
Derive both integrations and compare the two equal-impulse packets. Retain signed displacement and kinetic change separately.
Integrate a smooth impulse and its trajectory
Worked teaching model · synthetic values · Net force F=Fpeak sin(pi t/T) on0≤t≤T, mass2 kg, initial velocity−1 m/s and position0. Exact smooth packet, not an impact-force measurement. Peak force and duration are independent controls.
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Smooth force pulse over its complete interval
Unlike graph quantities rescale independently; read units and values, not screen angles. 1: supplied sine pulse (solid). Joined curves show the supplied model, not additional observations.
Equivalent numerical coordinates for Smooth force pulse over its complete interval
supplied sine pulse
| time (s) | net force (N) |
|---|---|
| 0 | 0 |
| 0.0125 | 0.1570393 |
| 0.025 | 0.3138364 |
| 0.0375 | 0.4701496 |
| 0.05 | 0.6257379 |
| 0.0625 | 0.7803613 |
| 0.075 | 0.9337815 |
| 0.0875 | 1.085762 |
| 0.1 | 1.236068 |
| 0.1125 | 1.384468 |
| 0.125 | 1.530734 |
| 0.1375 | 1.674639 |
| 0.15 | 1.815962 |
| 0.1625 | 1.954485 |
| 0.175 | 2.089994 |
| 0.1875 | 2.222281 |
| 0.2 | 2.351141 |
| 0.2125 | 2.476376 |
| 0.225 | 2.597792 |
| 0.2375 | 2.715203 |
| 0.25 | 2.828427 |
| 0.2625 | 2.93729 |
| 0.275 | 3.041624 |
| 0.2875 | 3.141268 |
| 0.3 | 3.236068 |
| 0.3125 | 3.325878 |
| 0.325 | 3.410561 |
| 0.3375 | 3.489984 |
| 0.35 | 3.564026 |
| 0.3625 | 3.632573 |
| 0.375 | 3.695518 |
| 0.3875 | 3.752765 |
| 0.4 | 3.804226 |
| 0.4125 | 3.849821 |
| 0.425 | 3.88948 |
| 0.4375 | 3.923141 |
| 0.45 | 3.950753 |
| 0.4625 | 3.972274 |
| 0.475 | 3.987669 |
| 0.4875 | 3.996916 |
| 0.5 | 4 |
| 0.5125 | 3.996916 |
| 0.525 | 3.987669 |
| 0.5375 | 3.972274 |
| 0.55 | 3.950753 |
| 0.5625 | 3.923141 |
| 0.575 | 3.88948 |
| 0.5875 | 3.849821 |
| 0.6 | 3.804226 |
| 0.6125 | 3.752765 |
| 0.625 | 3.695518 |
| 0.6375 | 3.632573 |
| 0.65 | 3.564026 |
| 0.6625 | 3.489984 |
| 0.675 | 3.410561 |
| 0.6875 | 3.325878 |
| 0.7 | 3.236068 |
| 0.7125 | 3.141268 |
| 0.725 | 3.041624 |
| 0.7375 | 2.93729 |
| 0.75 | 2.828427 |
| 0.7625 | 2.715203 |
| 0.775 | 2.597792 |
| 0.7875 | 2.476376 |
| 0.8 | 2.351141 |
| 0.8125 | 2.222281 |
| 0.825 | 2.089994 |
| 0.8375 | 1.954485 |
| 0.85 | 1.815962 |
| 0.8625 | 1.674639 |
| 0.875 | 1.530734 |
| 0.8875 | 1.384468 |
| 0.9 | 1.236068 |
| 0.9125 | 1.085762 |
| 0.925 | 0.9337815 |
| 0.9375 | 0.7803613 |
| 0.95 | 0.6257379 |
| 0.9625 | 0.4701496 |
| 0.975 | 0.3138364 |
| 0.9875 | 0.1570393 |
| 1 | 4.89859e-16 |
At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.
| Quantity | Value | Unit |
|---|---|---|
| Exact net impulse | 2.546479 | N s |
| Average net force | 2.546479 | N |
| Final signed velocity | 0.2732395 | m/s |
| Final signed displacement | -0.3633802 | m |
| Initial momentum | -2 | kg m/s |
| Final momentum | 0.5464791 | kg m/s |
| Kinetic-energy change | -0.9253402 | J |
The force integral fixes the endpoint momentum, while its time placement also matters to displacement.
Integral0toT Fpeak sin(pi t/T)dt=2Fpeak T/pi. Integrating acceleration gives v(t)=−1+[Fpeak T/(2pi)](1−cos(pi t/T)); integrating again yields x(T)=−T+Fpeak T²/(2pi). Equal impulses on the same mass/initial velocity give equal final velocity and kinetic change, but need not give equal displacement or peak force.
Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.
The force integral fixes the endpoint momentum, while its time placement also matters to displacement.
Integral0toT Fpeak sin(pi t/T)dt=2Fpeak T/pi. Integrating acceleration gives v(t)=−1+[Fpeak T/(2pi)](1−cos(pi t/T)); integrating again yields x(T)=−T+Fpeak T²/(2pi). Equal impulses on the same mass/initial velocity give equal final velocity and kinetic change, but need not give equal displacement or peak force.
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Why does a positive final velocity not settle the sign of displacement?
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