Calculate momentum with signed velocity
Use mass times velocity, preserve direction and distinguish momentum from force.
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
- Distinguish speed from signed velocity.
- Convert grams to kilograms.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
With right positive, what velocity represents 3 m/s left?
What you will learn
- Define momentum and use p = mv.
- Convert mass units and state momentum direction.
Momentum includes direction
Momentum is mass multiplied by velocity: p = mv. Its unit is kg m/s. For one-dimensional motion choose a positive direction first, keep mass positive, and put the direction sign in velocity. A 2 kg cart at +3 m/s has p = +6 kg m/s; the same cart at −3 m/s has p = −6 kg m/s. Equal speeds in opposite directions are not identical momenta.
A 500 g cart is 0.500 kg. At +4 m/s its momentum is +2.0 kg m/s. Momentum is a property of the moving body at that instant; force describes an interaction that can change it. A non-zero momentum does not by itself imply a non-zero resultant force or acceleration.
Compare the product under stated controls
At the same velocity, twice the mass gives twice the momentum. At the same mass, twice the velocity gives twice the signed momentum. A stationary object's momentum is zero in the stated reference frame even when its mass is large. Name the reference direction or frame whenever a sign or comparison could otherwise be ambiguous.
For two carts, calculate each signed momentum before adding them. A 2 kg cart at +3 m/s and a 1 kg cart at −1 m/s have total momentum +6 − 1 = +5 kg m/s. Adding their momentum magnitudes would incorrectly give 7 kg m/s. A zero total can also arise from two moving objects with equal opposite momenta; it does not prove that each is stationary.
Pause and explain
Does zero total momentum prove both objects in a system are stationary?
Worked example
Find the momentum of a 2 kg cart at 3 m/s right, then at the same speed left, taking right positive.
Show the worked solution
- Use p = mv with mass 2 kg.
- Rightward velocity +3 gives p = +6 kg m/s.
- Leftward velocity −3 gives p = −6 kg m/s.
Answer +6 kg m/s right; −6 kg m/s left.
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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.
Calculate momentum of a 2 kg cart at 3 m/s right and then 3 m/s left, with right positive.
Give me a hint
Use the same positive mass and change the velocity sign.
Compare my reasoning
- Rightward p = 2 × (+3) = +6 kg m/s.
- Leftward p = 2 × (−3) = −6 kg m/s.
- The magnitudes match but the directions differ.
+6 kg m/s and −6 kg m/s.
Look for these in your work
- Keep mass positive.
- State the direction sign.
Find momentum of a 500 g cart at 4 m/s right, then total momentum of 2 kg at 3 m/s right and 1 kg at 1 m/s left.
Give me a hint
Convert the first mass and add signed contributions in the second system.
Compare my reasoning
- 0.500 × (+4) = +2.0 kg m/s.
- The pair contributes +6 and −1 kg m/s.
- Total = +5 kg m/s.
+2.0 kg m/s; total +5 kg m/s, for right positive.
Look for these in your work
- Convert grams.
- Subtract the opposing contribution in the system total.
Two equal 1 kg carts move at 2 m/s in opposite directions. Explain their total momentum and why it does not prove either cart is at rest.
Give me a hint
Calculate each signed contribution before describing the sum.
Compare my reasoning
- Their momenta are +2 and −2 kg m/s.
- The total is zero.
- Each cart still has a non-zero velocity; cancellation concerns the system sum.
Total zero; both carts are still moving with equal opposite momenta.
Look for these in your work
- Distinguish a system total from its parts.
- Retain both non-zero velocities.
Common mistakes
- Use signed velocity rather than speed alone.
- Convert mass to kg.
- Momentum is not a force or proof of acceleration.
What can you explain now?
What momentum does a 500 g cart have at 4 m/s right?
Compare with the explanation
+2.0 kg m/s for right positive.
Use mass 0.500 kg in p = mv.
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Make it stick
Define momentum, give its unit and explain a zero-total example involving two moving carts.
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.