Conserve momentum in one-dimensional interactions
Choose a system and direction, conserve total momentum, and distinguish elastic collisions from interactions where kinetic energy changes.
Section 3 · Dynamics
AS Level · Cambridge International Physics 9702, 2025–2027 syllabus, Topic 3. These original explanations and practice support theory study; they do not establish supervised practical competence or replace timed assessment and teacher review. Drag is treated qualitatively, and coefficient of restitution is outside this syllabus section.
Before you begin
- Use vectors and signed values consistently, and distinguish mass from weight.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
When can total momentum be conserved for the interacting objects?
What you will learn
- State conservation of momentum for an isolated system
- Solve one-dimensional elastic and inelastic interaction problems
- Compare kinetic energy before and after an interaction
- Use the elastic-collision relative-speed condition
Momentum before equals momentum after for an isolated system
During an interaction, the objects exert forces on one another. For a system with no resultant external force over the interaction, the vector sum of momentum is conserved: total momentum before equals total momentum after.
For motion along one line, choose a positive direction and include signs for every velocity. Write the total momentum before and after before solving for an unknown. Objects that join after impact share one final velocity.
Momentum conservation does not guarantee kinetic-energy conservation
In an inelastic interaction, momentum is conserved for an isolated system but some kinetic energy may be transferred to deformation, sound or internal energy. If objects stick together, their kinetic energy after impact is usually smaller.
For an elastic collision, total kinetic energy is also conserved. In one dimension, the relative speed of approach equals the relative speed of separation. Do not apply that speed condition to an inelastic collision.
Pause and explain
What extra condition identifies an elastic collision?
Worked example
A 2.0 kg cart moving at 3.0 m s⁻¹ collides with a stationary 1.0 kg cart and they stick together. Find their common velocity after impact.
Show the worked solution
- Take the original direction as positive.
- Momentum before = 2.0(3.0) + 1.0(0) = 6.0 kg m s⁻¹.
- After impact the combined mass is 3.0 kg, so v = 6.0/3.0.
Answer 2.0 m s⁻¹ in the original direction. The interaction is inelastic; kinetic energy is not conserved.
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.
A 0.40 kg trolley moves at 5.0 m s⁻¹ and collides with a stationary 0.60 kg trolley. They lock together. Find their final velocity.
Give me a hint
Conserve signed momentum for the two-trolley system.
Compare my reasoning
- Initial momentum = 0.40(5.0) + 0.60(0) = 2.0 kg m s⁻¹.
- Combined mass after impact = 1.00 kg.
- Final velocity v = 2.0/1.00.
2.0 m s⁻¹ in the original direction.
Look for these in your work
- System identified
- Momentum before and after equated
- Shared final velocity found
A 1.0 kg cart moving at +4.0 m s⁻¹ collides with a 3.0 kg cart moving at −1.0 m s⁻¹. They stick. Find their final velocity.
Give me a hint
Use the signs to represent opposite initial directions.
Compare my reasoning
- Initial total momentum = 1.0(4.0) + 3.0(−1.0) = +1.0 kg m s⁻¹.
- Combined mass is 4.0 kg.
- v = +1.0/4.0.
+0.25 m s⁻¹, in the first cart's original direction.
Look for these in your work
- Both signed momenta included
- Net momentum calculated
- Direction inferred from sign
Two 0.50 kg pucks collide elastically in one dimension. The first approaches at +3.0 m s⁻¹ and the second is at rest. After collision, the first moves at +1.0 m s⁻¹. Find the second puck's velocity and check the elastic speed condition.
Give me a hint
First use momentum conservation; then compare relative speeds.
Compare my reasoning
- Momentum before = 0.50(3.0) = 1.5 kg m s⁻¹.
- After: 0.50(1.0) + 0.50v₂ = 1.5, so v₂ = 2.0 m s⁻¹.
- Approach speed = 3.0 m s⁻¹; separation speed = 2.0 − 1.0 = 1.0 m s⁻¹. This contradicts the elastic condition, so the stated post-collision value is inconsistent with an elastic collision.
Momentum would give v₂ = +2.0 m s⁻¹, but the relative speed of separation is 1.0 m s⁻¹, not the 3.0 m s⁻¹ approach speed. Therefore the supplied post-collision velocity cannot describe an elastic collision. For equal masses, the elastic outcome is +0.0 m s⁻¹ for the first puck and +3.0 m s⁻¹ for the second.
Look for these in your work
- Momentum equation solved
- Relative speeds compared
- Inconsistent given data identified
Common mistakes
- Conserving momentum for one object instead of the chosen isolated system
- Dropping negative velocities
- Assuming kinetic energy is conserved in every collision
What can you explain now?
When can total momentum be conserved for the interacting objects?
Compare with the explanation
When the system has no resultant external force during the interaction
Internal forces exchange momentum within the system; an external resultant can change total momentum.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Which conservation law applies to an isolated inelastic collision, and what extra condition applies to an elastic one?
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.