Relate momentum change per unit time to resultant force
Interpret the rate of change and connect it to F = ma only for the stated constant-mass model.
Exam scope and assessment notes
Extended only (Supplement) · Physics 0972, 2026–2028, section 1.6. Define and use signed momentum p = mv, impulse FΔt = Δ(mv), conservation in simple one-dimensional systems, and resultant force as momentum change per unit time. F is the relevant resultant, with a constant force or stated average over the interval. Conservation requires negligible net external impulse on the chosen system; collision outcomes need enough supplied information. Written practice is self-checked. Supervised practical experience, cumulative assessment and human subject-teacher review remain pending. Cambridge states that IGCSE 0972 has the same subject content as IGCSE 0625, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Calculate signed differences between momentum states.
- Read a straight graph gradient with units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the change from +5 to +2 kg m/s?
What you will learn
- Define resultant force as momentum change per unit time.
- Connect a constant momentum-time gradient to a constant resultant.
A rate describes change rather than a final amount
Resultant force is the rate of change of momentum. Over a stated interval, average resultant Faverage = Δp/Δt. A straight momentum-time trend from +5 to +2 kg m/s over 3 s has gradient (2 − 5)/3 = −1 N. Its constant gradient represents a constant resultant −1 N in this model, even though the momentum remains positive throughout.
A horizontal momentum-time line has zero gradient and zero resultant in the chosen direction. The object may still move with non-zero constant momentum. A curved trend has a changing rate: a whole-interval quotient gives an average rather than every instantaneous force value. Do not replace the change by the final momentum alone.
Connect the constant-mass equations carefully
For a fixed mass, Δp = mΔv, so Faverage = mΔv/Δt = maaverage. The original 0.50 kg cart changes velocity from +10 to +4 m/s in 3 s: acceleration is −2 m/s² and force is 0.50 × (−2) = −1 N. This agrees with momentum changing from +5 to +2 kg m/s.
The simple connection assumes the same mass and reference direction over the interval. If material enters or leaves a selected object, use the correctly defined system and initial and final momenta instead of blindly applying one fixed m to Δv. These one-dimensional exercises specify their masses and intervals; they do not infer a changing-mass propulsion model from an incomplete description.
Pause and explain
A momentum-time graph is horizontal at a non-zero positive value. What does its gradient imply?
Worked example
Momentum decreases linearly from +5 to +2 kg m/s in 3 s. Find the model resultant force.
Show the worked solution
- Change = +2 − 5 = −3 kg m/s.
- Rate = −3/3 = −1 kg m/s².
- Resultant is −1 N, constant for the stated straight trend.
Answer −1 N, constant in the stated model.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find resultant for a straight momentum trend from +5 to +2 kg m/s over 3 s.
Give me a hint
Find the signed vertical change divided by the time interval.
Compare my reasoning
- Δp = 2 − 5 = −3 kg m/s.
- Δt = 3 s.
- F = −3/3 = −1 N.
−1 N.
Look for these in your work
- Use final minus initial.
- Report the rate in force units.
A fixed 0.50 kg cart changes velocity from +10 to +4 m/s in 3 s. Find acceleration, momentum change and average resultant.
Give me a hint
Check force once through momentum rate and once through fixed-mass acceleration.
Compare my reasoning
- aaverage = (4 − 10)/3 = −2 m/s².
- Δp = 0.50(4 − 10) = −3 kg m/s.
- Faverage = −3/3 = −1 N = 0.50 × (−2).
−2 m/s²; Δp = −3 kg m/s; average resultant −1 N.
Look for these in your work
- Keep the mass fixed and units consistent.
- Check agreement of the two force methods.
A curved momentum-time graph has different local slopes. A student uses the full-interval quotient as the force at every instant. Explain the limitation.
Give me a hint
Distinguish an interval average from a changing local rate.
Compare my reasoning
- The full-interval quotient gives average momentum change per time.
- Different local slopes show changing momentum-change rates.
- The average need not equal every instantaneous resultant.
The quotient gives an average; changing local slopes imply changing instantaneous resultant force.
Look for these in your work
- Name the average quantity.
- Use the supplied curve evidence without inventing one constant force.
Common mistakes
- Use momentum change rather than final momentum.
- A whole-interval rate can be an average.
- The direct F = ma connection assumes fixed mass.
What can you explain now?
Can a body have positive momentum and a negative resultant force?
Compare with the explanation
Yes.
A negative momentum-change rate can reduce a still-positive momentum.
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Make it stick
Explain a non-zero horizontal momentum graph, then connect a straight changing trend to F = ma for fixed mass.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.