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 0625, 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.
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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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.