Choose a momentum system and test conservation across an impact
Identify when a two-body conservation model applies, compare before-and-after totals and explain why individual momentum can change.
Section 4.3 · Momentum
Exam scope and assessment notes
Mechanics (Paper 4, AS & A Level) · Mathematics 9709, 2026–2027, section 4.3. Direct impacts are restricted to one dimension. Choose one positive direction for all velocities and momenta. Conservation applies to the selected system when external effects in that direction are absent or negligible during the impact; it does not make each particle’s momentum constant. Coalescing bodies share a common final velocity and combined mass. Impulse and the coefficient of restitution are not required in this section. Kinetic energy need not be conserved; where used, its expression ½mv² is supplied as an additional model check. Bodies have positive constant masses; stated impact order and interval assumptions must be respected. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Calculate and add directed momentum.
- Distinguish internal interactions from external influences on a selected system.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For a system containing both colliding particles, their mutual contact is what kind of interaction?
What you will learn
- State the system and condition needed for momentum conservation.
- Check a proposed direct-impact outcome using signed totals.
Conservation is a statement about the whole selected system
For a pair of particles undergoing a direct impact, take the same direction as positive before and after. If external effects along the line are absent or negligible over the brief impact, write mₐuₐ + mᵦuᵦ = mₐvₐ + mᵦvᵦ. The contact interaction between the two particles is internal to this pair. It changes the individual momenta, but their changes cancel in the pair’s signed total under the stated model.
A 2 kg particle at +5 m s⁻¹ hits a resting 3 kg particle. Suppose their velocities after impact are −1 and +4 m s⁻¹. Before, total momentum is 2 × 5 + 3 × 0 = 10 kg m s⁻¹. After, it is 2 × (−1) + 3 × 4 = 10. The first particle’s momentum changes from +10 to −2, and the second’s from 0 to +12: opposite changes, with an unchanged system sum.
Check the assumptions before declaring data incompatible
An external force on the pair can change its momentum. A collision with an anchored wall cannot conserve the moving particle’s own momentum simply because the event is brief: the wall acts from outside that one-particle system. In a smooth horizontal two-particle model, gravity and support act vertically while the conserved component is horizontal. State the relevant direction and assumptions rather than making a blanket claim that every collision conserves any chosen system’s momentum.
If the 2 kg and 3 kg pair instead has reported final velocities −1 and +3, the final total is −2 + 9 = 7 kg m s⁻¹. This is incompatible with the stated no-external-effect model, whose initial total is 10. The discrepancy could require corrected data or a different model; it is not proof of a universal law failing. Checking conservation establishes a necessary balance, but alone it does not certify every other physical assumption about a reported impact.
Pause and explain
In an isolated two-particle impact, what quantity is conserved?
Worked example
A 2 kg particle at +5 m s⁻¹ hits a resting 3 kg particle on a smooth horizontal line. External effects along the line are negligible during impact. Test final velocities −1 and +4 m s⁻¹ for conservation.
Show the worked solution
- Choose both particles as the system and keep the same signed horizontal axis.
- Initial momentum is 2 × 5 + 3 × 0 = 10 kg m s⁻¹.
- Final momentum is 2 × (−1) + 3 × 4 = 10 kg m s⁻¹, so the outcome satisfies this balance.
Answer Total momentum is 10 kg m s⁻¹ both before and after; individual changes are −12 and +12 kg m s⁻¹.
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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.
A 1 kg particle at +6 m s⁻¹ hits a resting 2 kg particle. With negligible external effects along the line, test final velocities −2 and +4 m s⁻¹ for momentum conservation.
Give me a hint
Compare the signed sum for the same two particles on both sides.
Compare my reasoning
- Initial momentum is 1 × 6 + 2 × 0 = 6 kg m s⁻¹.
- Final momentum is 1 × (−2) + 2 × 4 = 6 kg m s⁻¹.
- The first momentum change is −8 and the second is +8 kg m s⁻¹, so the pair’s sum is unchanged.
Momentum balance satisfied: 6 kg m s⁻¹ before and after; changes −8 and +8 kg m s⁻¹.
Look for these in your work
- Include the rebound’s negative velocity.
- Check the system sum rather than demanding unchanged individual momentum.
Particles of masses 3 kg and 1 kg have velocities +4 and −2 m s⁻¹ before impact, and +1 and +7 m s⁻¹ afterwards. Test total momentum and calculate each change, assuming external effects along the line are negligible.
Give me a hint
Associate each final velocity with the same mass as its initial velocity.
Compare my reasoning
- Before, momentum is 3 × 4 + 1 × (−2) = 10 kg m s⁻¹.
- After, momentum is 3 × 1 + 1 × 7 = 10 kg m s⁻¹.
- The 3 kg change is 3 − 12 = −9 and the 1 kg change is 7 − (−2) = +9 kg m s⁻¹.
Momentum balance satisfied: total 10 kg m s⁻¹; individual changes −9 and +9 kg m s⁻¹.
Look for these in your work
- Use the negative incoming velocity.
- Retain the association of velocities with their respective masses.
A 2 kg particle at +5 m s⁻¹ and a resting 3 kg particle undergo direct impact. The model neglects external effects along the line, but measured final velocities are −1 and +3 m s⁻¹. Is that outcome compatible with the conservation model? Explain what needs checking.
Give me a hint
A discrepancy can indicate that the data or the model assumptions need investigation.
Compare my reasoning
- Initial signed momentum is 10 kg m s⁻¹.
- The reported final total is 2 × (−1) + 3 × 3 = 7 kg m s⁻¹, a change of −3 kg m s⁻¹.
- The outcome fails this model’s conservation check; verify the measurements and whether outside influences were negligible rather than declaring a universal law invalid.
Incompatible with the stated model: initial 10, final 7, change −3 kg m s⁻¹; check data and external-effect assumptions.
Look for these in your work
- Quantify the signed discrepancy.
- Limit the conclusion to the supplied model and evidence.
Common mistakes
- Conservation applies to the selected system, not automatically to each body.
- State why external effects in the relevant direction can be neglected.
- A matched momentum total is a necessary check, not a complete physical validation.
What can you explain now?
Does a particle hitting an anchored wall conserve its own momentum during a rebound?
Compare with the explanation
No; the wall exerts an external influence on that one-particle system.
A different, larger system requires its own modelling assumptions; the particle’s isolated balance does not apply.
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Make it stick
Choose the system, state the external-effect assumption and compare the same signed momentum total before and after.
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