Choose a system and state momentum conservation conditions
Distinguish internal interactions from the external impulse that can change a system total.
Section 1.6 · Momentum
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
- Add signed momenta of several objects.
- Distinguish force on one object from force on another.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For a two-cart system, are their mutual collision forces internal?
What you will learn
- State when total momentum is conserved.
- Explain system boundaries and negligible net external impulse.
Conservation belongs to the selected system
For a chosen system with negligible net external impulse over the interaction, its total momentum before equals its total momentum after. Choose both colliding carts as the system for a collision calculation. Their mutual contact forces are internal to that pair and transfer momentum between its parts; they do not themselves change the pair's total momentum.
Selecting only one cart gives a different boundary: the other cart's force is now external to that single-cart system, so its momentum can change. The conservation statement does not say that each object's momentum must remain unchanged. State the direction, included objects and time interval before writing the total equation.
Check external contributions rather than saying no forces act
A horizontal cart collision may approximately conserve the pair's horizontal momentum when net external horizontal impulse, such as friction over the short interval, is negligible. Weight and support can still act vertically. The useful condition is negligible net external impulse in the selected direction, rather than the absence of every individual force.
If the pair initially has +5 kg m/s and receives external impulse −2 N s, its final total is +3 kg m/s. For instance, a constant external −4 N resultant for 0.50 s supplies that change. Large external forces that cancel as a resultant can also give zero net impulse; identify the net contribution instead of treating every contact as internal or every force as forbidden.
Pause and explain
What condition justifies conserving a system's horizontal momentum over a collision?
Worked example
Two colliding carts start with total momentum +5 kg m/s. Their net external horizontal impulse is negligible. What is their final total, and must each cart keep its own momentum?
Show the worked solution
- Use the two-cart system and one signed horizontal direction.
- Negligible net external impulse gives total momentum after = +5 kg m/s.
- Internal contact can change each cart's momentum while preserving that total.
Answer Total remains +5 kg m/s; each cart may change its own momentum.
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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.
A two-cart system has total +5 kg m/s and negligible external horizontal impulse. State its final total and explain internal contact.
Give me a hint
Use conservation for the sum over the chosen two objects.
Compare my reasoning
- The external horizontal impulse is negligible.
- Total momentum remains +5 kg m/s.
- The contact can still change individual cart momenta in opposite compensating ways.
Total +5 kg m/s; internal contact may change each cart's momentum.
Look for these in your work
- Name the selected system.
- Separate total conservation from individual changes.
A system initially has +5 kg m/s and receives constant external resultant −4 N for 0.50 s. Find final total.
Give me a hint
Use the external impulse to update the total rather than assuming isolation.
Compare my reasoning
- External impulse = −4 × 0.50 = −2 N s.
- Final total = +5 − 2.
- Final total = +3 kg m/s.
+3 kg m/s.
Look for these in your work
- Include external impulse.
- Preserve the signed update.
A student says a collision conserves the momentum of cart A alone because the contact force is internal to the two-cart pair. Explain the boundary error.
Give me a hint
Internal or external depends on which objects the chosen system contains.
Compare my reasoning
- For the pair, the mutual contact is internal.
- For cart A alone, cart B lies outside and its contact force is external.
- A's individual momentum can change; only the suitably isolated pair total is conserved.
Cart B's contact is external to A alone; conservation of the pair total does not conserve each cart separately.
Look for these in your work
- Compare the two boundaries.
- State which momentum can change.
Common mistakes
- State the system boundary.
- Conserve the sum, not every object's momentum.
- Check net external impulse in the chosen direction.
What can you explain now?
What happens to a +5 kg m/s system total after external impulse −2 N s?
Compare with the explanation
It becomes +3 kg m/s.
Add the external impulse to the initial total.
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Make it stick
Draw a boundary around one cart and then both carts; classify the same contact and state the conservation condition.
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.