Calculate signed linear momentum and combine moving particles
Use p = mv with consistent mass and velocity units, and distinguish a system’s signed total from the momentum of each moving body.
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
- Use signed velocities on a one-dimensional axis.
- Convert grams to kilograms and multiply quantities with units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A 250 g particle moves left at 8 m s⁻¹, with right positive. Which momentum is correct?
What you will learn
- Calculate linear momentum with its direction and SI units.
- Add momenta on a shared signed axis without confusing zero total with stationary bodies.
Momentum combines mass with directed velocity
Linear momentum is p = mv, where m is the positive mass and v is velocity. With mass in kg and velocity in m s⁻¹, momentum is measured in kg m s⁻¹. It is directed: on a line, take right as positive and keep a leftward velocity negative. A 3 kg particle moving right at 4 m s⁻¹ has p = +12 kg m s⁻¹. A 2 kg particle moving left at 3 m s⁻¹ has p = −6 kg m s⁻¹, not +6.
Speed is a non-negative magnitude; velocity supplies the sign in p = mv. Mass remains positive even for leftward motion. A resting particle has zero momentum, but a light fast particle and a heavy slow particle can have the same momentum. Momentum is neither force, measured in newtons, nor kinetic energy, measured in joules. Writing units beside the calculation helps identify whether the question asks for the right physical quantity.
A total uses signed addition over the chosen system
For the 3 kg and 2 kg particles above, system momentum is +12 − 6 = +6 kg m s⁻¹. The positive result means a net rightward momentum, even though one body moves left. Adding magnitudes to obtain 18 would describe neither the signed total nor a conservation equation. State which bodies belong to the system before adding their momenta.
Two bodies can move while their total momentum is zero: 2 kg at +3 m s⁻¹ and 3 kg at −2 m s⁻¹ give +6 − 6 = 0. This does not mean each velocity is zero, and it does not tell you that the bodies will stop without an impact condition. A change in a particle’s momentum uses final minus initial signed momentum; a reversal can produce a larger change than the change of speed alone suggests.
Pause and explain
Two particles have momenta +6 and −6 kg m s⁻¹. What follows from the total being zero?
Worked example
Take right as positive. A 3 kg particle moves right at 4 m s⁻¹ and a 2 kg particle moves left at 3 m s⁻¹. Find each momentum and the total.
Show the worked solution
- For the first particle, pₐ = 3 × (+4) = +12 kg m s⁻¹.
- For the second, pᵦ = 2 × (−3) = −6 kg m s⁻¹.
- Add signed momenta: pₐ + pᵦ = +6 kg m s⁻¹, directed right.
Answer Momenta +12 and −6 kg m s⁻¹; total +6 kg m s⁻¹, rightwards.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
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.
With right positive, find the momentum of a 250 g particle moving left at 8 m s⁻¹. State both signed value and direction.
Give me a hint
Convert the mass first and use a negative leftward velocity.
Compare my reasoning
- The mass is 0.25 kg and velocity is −8 m s⁻¹.
- Momentum is p = 0.25 × (−8) = −2 kg m s⁻¹.
- Its magnitude is 2 kg m s⁻¹ and its direction is leftwards.
p = −2 kg m s⁻¹; magnitude 2 kg m s⁻¹, leftwards.
Look for these in your work
- Use the converted kilogram mass.
- Keep the sign consistent with the chosen rightward positive direction.
A 3 kg particle has velocity −2 m s⁻¹ and a 2 kg particle has velocity +5 m s⁻¹ on the same axis. Find their momenta and total.
Give me a hint
Multiply each mass by its own directed velocity before adding.
Compare my reasoning
- The first momentum is 3 × (−2) = −6 kg m s⁻¹.
- The second is 2 × 5 = +10 kg m s⁻¹.
- Their total is −6 + 10 = +4 kg m s⁻¹, in the positive direction.
Momenta −6 and +10 kg m s⁻¹; total +4 kg m s⁻¹.
Look for these in your work
- Keep both individual signed values.
- Interpret the direction of the net rather than adding speed magnitudes.
A 500 g particle reverses from 6 m s⁻¹ leftwards to 4 m s⁻¹ rightwards. Take right as positive. Find its initial momentum, final momentum and signed momentum change.
Give me a hint
Use final minus initial momentum, retaining the initial negative sign.
Compare my reasoning
- With m = 0.5 kg, initial momentum is 0.5 × (−6) = −3 kg m s⁻¹.
- Final momentum is 0.5 × 4 = +2 kg m s⁻¹.
- The change is +2 − (−3) = +5 kg m s⁻¹; it is not obtained by subtracting unsigned speeds.
Initial −3; final +2; change +5 kg m s⁻¹.
Look for these in your work
- Convert 500 g correctly and describe the reversal with signed velocities.
- Subtract the initial momentum rather than its magnitude.
Common mistakes
- Use velocity rather than unsigned speed when directions differ.
- Convert the mass to kilograms before reporting SI momentum.
- Zero system momentum does not imply that every particle is at rest.
What can you explain now?
Can two moving particles have zero total momentum?
Compare with the explanation
Yes, if their signed momenta are equal in magnitude and opposite in direction.
The cancellation concerns the selected system’s sum, not each particle’s velocity.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Choose one direction, convert mass units and calculate each mv before adding signed momenta for a system.
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.