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SCIENCE · INVESTIGATE AND EXPLAIN

Account for mass crossing a momentum boundary

The identity F = ma assumes a particle of fixed mass. A subsystem that loses mass needs an additional momentum-flux account. Consider an abstract one-dimensional retained body with no external force, initially at rest. An infinitesimal positive mass dμ leaves with velocity v − u relative to the inertial frame, where u is its fixed backward speed relative to the body.

Learning outcome: Derive the differential boundary balance before integrating it. Compare two initial masses at equal fractions, and then compare constant thrust with increasing acceleration.

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 the retained fraction to 0.8 at any supported initial mass. The exact change is 100 ln(1.25) ≈ 22.3144 m/s, whereas the fixed-initial-mass shortcut gives 20 m/s.

Open the experiment

Predict

At final retained fraction 0.6 and u = 100 m/s, the ideal velocity change is:

Experiment

Derive the differential boundary balance before integrating it. Compare two initial masses at equal fractions, and then compare constant thrust with increasing acceleration.

Keep expelled mass in the momentum boundary

Worked teaching model · synthetic values · Pure mathematical one-dimensional mass-expulsion model in an inertial force-free environment. Relative expelled speed100 m/s, initial rest and constant mass-loss rate over10 s. No fuel, device, pressure, design, trajectory targeting or physical propulsion instructions. Total retained-plus-expelled mass is closed; retained subsystem is not.

Reset keeps your prediction, captures and practice history.

Retained speed from the changing-mass balance

retained speed (m/s)elapsed time (s)01051.082560

Unlike graph quantities rescale independently; read units and values, not screen angles. 1: 100 ln(m0/m(t)) (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Retained speed from the changing-mass balance

100 ln(m0/m(t))

elapsed time (s)retained speed (m/s)
00
0.1250.5012542
0.251.005034
0.3751.511364
0.52.020271
0.6252.531781
0.753.045921
0.8753.562718
14.082199
1.1254.604394
1.255.129329
1.3755.657035
1.56.18754
1.6256.720875
1.757.257069
1.8757.796154
28.338161
2.1258.883121
2.259.431068
2.3759.982034
2.510.53605
2.62511.09316
2.7511.65338
2.87512.21676
312.78334
3.12513.35314
3.2513.92621
3.37514.50258
3.515.08229
3.62515.66538
3.7516.25189
3.87516.84187
417.43534
4.12518.03236
4.2518.63296
4.37519.23719
4.519.84509
4.62520.45672
4.7521.0721
4.87521.6913
522.31436
5.12522.94132
5.2523.57223
5.37524.20716
5.524.84614
5.62525.48922
5.7526.13648
5.87526.78794
627.44368
6.12528.10375
6.2528.76821
6.37529.43711
6.530.11051
6.62530.78848
6.7531.47107
6.87532.15836
732.85041
7.12533.54727
7.2534.24903
7.37534.95575
7.535.66749
7.62536.38434
7.7537.10637
7.87537.83364
838.56625
8.12539.30426
8.2540.04776
8.37540.79682
8.541.55154
8.62542.312
8.7543.07829
8.87543.8505
944.62871
9.12545.41303
9.2546.20355
9.37547.00036
9.547.80358
9.62548.6133
9.7549.42963
9.87550.25268
1051.08256

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.

Calculated quantities under the declared measurement model
QuantityValueUnit
Final retained mass9kg
Expelled total mass6kg
Constant mass-loss magnitude0.6kg/s
Momentum-flux thrust magnitude60N
Final retained velocity change51.08256m/s
Final retained acceleration6.666667m/s²
Fixed-initial-mass shortcut velocity change40m/s

The expelled mass carries momentum across the retained subsystem boundary; its contribution cannot be omitted.

For positive expelled amount dmu=−dm, momentum balance gives m dv=ve dmu=−ve dm to first order. Integrate dv=−ve dm/m to obtain ve ln(m0/mf). Constant mass-loss rate gives constant momentum-flux thrust but increasing acceleration as retained mass falls. This is not d(mv)/dt=0 for the retained subsystem and not a fuel or hardware model.

Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.

The expelled mass carries momentum across the retained subsystem boundary; its contribution cannot be omitted.

For positive expelled amount dmu=−dm, momentum balance gives m dv=ve dmu=−ve dm to first order. Integrate dv=−ve dm/m to obtain ve ln(m0/mf). Constant mass-loss rate gives constant momentum-flux thrust but increasing acceleration as retained mass falls. This is not d(mv)/dt=0 for the retained subsystem and not a fuel or hardware model.

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Account for mass crossing a momentum boundary

The identity F = ma assumes a particle of fixed mass. A subsystem that loses mass needs an additional momentum-flux account. Consider an abstract one-dimensional retained body with no external force, initially at rest. An infinitesimal positive mass dμ leaves with velocity v − u relative to the inertial frame, where u is its fixed backward speed relative to the body.

Before the separation the relevant mass moves at v. After separation the retained part changes velocity by dv while the expelled part carries its own momentum. Retaining first-order terms gives m dv = u dμ = −u dm. The whole retained-plus-expelled system is closed, but the retained part alone is not.

Integrating between positive masses gives Δv = u ln(m₀/mf). With u = 100 m/s and final fraction mf/m₀ = 0.6, the result is about 51.0826 m/s, independent of the common mass scale. Replacing the mass throughout by m₀ would instead predict 40 m/s and omit the changing inertia.

The supplied mathematical packet loses mass at a constant rate over 10 s. Momentum-flux thrust is then constant, while acceleration increases as retained mass decreases. At final fraction 1 there is no expelled mass and both thrust and velocity change are zero. These are academic boundary equations, not a fuel, device, construction, pressure or trajectory-targeting model.

Changed-case check

Predict before reading the comparison. Change the retained fraction to 0.8 at any supported initial mass. The exact change is 100 ln(1.25) ≈ 22.3144 m/s, whereas the fixed-initial-mass shortcut gives 20 m/s.

Retained investigation

Derive the differential boundary balance before integrating it. Compare two initial masses at equal fractions, and then compare constant thrust with increasing acceleration. Which matter must be included for the zero-external-force momentum conservation statement?

Source and boundary

Primary conceptual reference checked 30 October 2026: OpenStax University Physics Volume 1. All packets, derivations and prose are original. These bounded mathematical models are not measured experiments, physical procedures, practical certification or institutional credit.

Apply

At final retained fraction 0.6 and u = 100 m/s, the ideal velocity change is:

Public practice checks a specific claim. It does not establish independent mastery.

Which matter must be included for the zero-external-force momentum conservation statement?

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