Solve one-dimensional collisions when objects stick
Use the combined mass and one shared final velocity after a stated sticking collision.
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 initial momenta.
- State negligible net external impulse for the chosen pair.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is initial total for 2 kg at +3 m/s and 1 kg at −1 m/s?
What you will learn
- Apply total momentum conservation to a sticking collision.
- Interpret a signed shared final velocity.
Sticking supplies one common final velocity
For two objects that stick together after a one-dimensional collision, the final moving mass is m1 + m2 and both share velocity v. With negligible net external impulse, m1u1 + m2u2 = (m1 + m2)v. This shared-velocity condition comes from the stated sticking outcome; it is not valid for every collision.
Taking right positive, a 2 kg cart at +3 m/s meets a 1 kg cart at −1 m/s and they stick. Initial total = 2 × 3 + 1 × (−1) = +5 kg m/s. Their combined mass is 3 kg, so v = 5/3 m/s right, approximately 1.67 m/s. Check: 3 × (5/3) restores +5 kg m/s.
Use the mass total and the sign
A 1.5 kg cart at +4 m/s sticking to a stationary 0.50 kg cart has v = (1.5 × 4)/(1.5 + 0.50) = +3 m/s. Dividing the initial total by only the moving cart's mass ignores the other object's mass in the final shared motion. A stationary object contributes zero initial momentum but still contributes final mass.
For two 1 kg carts initially at +2 and −2 m/s, the total is zero and the stuck pair ends at rest in the stated isolated model. That final-rest result follows from zero total and the supplied common-velocity condition. Momentum conservation alone does not imply that every other zero-total collision ends with both objects stationary, nor that kinetic energy is conserved in a sticking collision.
Pause and explain
What mass multiplies the shared final velocity after two objects stick?
Worked example
A 2 kg cart at +3 m/s and a 1 kg cart at −1 m/s stick with negligible external impulse. Find shared final velocity.
Show the worked solution
- Initial total = 2 × 3 + 1 × (−1) = +5 kg m/s.
- Combined mass = 3 kg; 3v = 5.
- v = +5/3 m/s, approximately 1.67 m/s right.
Answer +5/3 m/s, approximately 1.67 m/s right.
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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 shared final velocity when 2 kg at +3 m/s and 1 kg at −1 m/s stick without external impulse.
Give me a hint
Add signed initial momenta, then divide by total final mass.
Compare my reasoning
- Initial total = +5 kg m/s.
- Final mass = 3 kg.
- v = +5/3 m/s, approximately 1.67 m/s right.
+5/3 m/s, approximately 1.67 m/s right.
Look for these in your work
- Use the negative opposing velocity.
- Divide by combined mass.
A 1.5 kg cart at +4 m/s sticks to a stationary 0.50 kg cart with negligible external impulse. Find v.
Give me a hint
The resting partner has zero initial momentum but non-zero final mass.
Compare my reasoning
- Initial total = 1.5 × 4 = +6 kg m/s.
- Final mass = 1.5 + 0.50 = 2.0 kg.
- v = 6/2 = +3 m/s.
+3 m/s.
Look for these in your work
- Include the stationary partner's mass.
- Check the conserved total.
Two 1 kg carts at +2 and −2 m/s stick. Explain their final velocity and why this does not mean all zero-total collisions end at rest.
Give me a hint
Use both the zero total and the specifically supplied shared-velocity condition.
Compare my reasoning
- Initial total is +2 − 2 = 0 kg m/s.
- Sticking gives final momentum 2v, so v = 0.
- A non-sticking pair can instead have equal opposite non-zero final momenta.
The stuck pair has v = 0; a non-sticking zero-total pair may still move oppositely.
Look for these in your work
- Use the stated sticking outcome.
- Distinguish a shared velocity from two independent final velocities.
Common mistakes
- Use the combined mass after sticking.
- Include an opposing initial velocity sign.
- Do not impose shared final velocity unless sticking is stated.
What can you explain now?
Does a stationary collision partner contribute to the final combined mass?
Compare with the explanation
Yes.
It has zero initial momentum but remains part of the stuck moving pair.
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Make it stick
Derive the shared-velocity equation and explain why stationary partner mass cannot be omitted.
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.