Momentum conservation and elastic interactions
Key idea: Momentum is conserved for a closed system even when the bodies deform or stick. Elasticity requires an additional kinetic-energy or relative-speed check.
Continue where you stopped
The core idea
H1 Physics 8867 · Lesson 2 of 3
Check your understandingBy the end of this lesson, you should be able to
- State the principle of conservation of momentum with its closed-system condition.
- Solve one-dimensional elastic and inelastic collisions using signed velocities.
- Use relative speed of approach and separation to identify a perfectly elastic collision.
Learn the idea
Big question: What exactly must be isolated before momentum can be conserved in a collision?
Define the system and the condition
Total momentum is conserved when the net external impulse on the chosen system is zero or negligible during the interaction. Internal collision forces change each body's momentum but cancel in the system total.
Choose one positive direction and give every velocity a sign. A rebounding body's sign changes; kinetic energy still uses speed squared and cannot be negative.
Check your understanding: Why can each cart's momentum change while total momentum stays constant?
The carts exert equal and opposite internal impulses, so their individual changes cancel in the total.
Require an extra test for elasticity
Momentum conservation alone does not make a collision elastic. A perfectly elastic collision also conserves total kinetic energy.
In one dimension you may instead compare relative speeds: speed of separation equals speed of approach. Use signed velocities carefully, then report the relative speeds as positive magnitudes.
Check your understanding: Two carts stick together. Can the collision be perfectly elastic?
Not unless there was no relative motion initially; sticking normally transfers kinetic energy and is perfectly inelastic.
Key ideas
- Define the system before claiming conservation.
- Momentum is always conserved for the closed system; kinetic energy need not be.
- Relative speed is a positive closing or separating speed, formed from signed velocities carefully.
Relationships to know
Σp before = Σp afterelastic: relative speed of separation = relative speed of approach
Follow the reasoning
Worked example
Solve and test a one-dimensional collision
Question: A 2.0 kg trolley moving at +5.0 m s⁻¹ collides with a 3.0 kg trolley at rest. Afterwards the 2.0 kg trolley moves at −1.0 m s⁻¹. Find the other velocity and decide whether the collision is perfectly elastic.
Step 1: Conserve signed momentum
Why: The short collision has negligible external impulse.
Working: 2.0(5.0) + 3.0(0) = 2.0(−1.0) + 3.0v, giving v = 4.0 m s⁻¹.
Step 2: Test kinetic energy
Why: Elasticity needs a condition beyond momentum conservation.
Working: Initial Ek = ½(2)(5²) = 25 J. Final Ek = ½(2)(1²) + ½(3)(4²) = 25 J.
Step 3: Cross-check relative speeds
Why: This offers an independent one-dimensional elastic test.
Working: Approach speed = 5 − 0 = 5 m s⁻¹; separation speed = 4 − (−1) = 5 m s⁻¹.
Answer: The 3.0 kg trolley moves at +4.0 m s⁻¹ and the collision is perfectly elastic.
Check: Momentum and kinetic energy totals match, and the bodies separate rather than continue overlapping.
Now try it with support
Practise with support
A 1.0 kg trolley at +6.0 m s⁻¹ sticks to a 2.0 kg trolley at rest. Find their common velocity.
Hints
- The combined mass after collision is 3.0 kg.
- Conserve total momentum, not each trolley’s momentum.
View the guided answer
1.0(6.0) + 2.0(0) = 3.0v, so v = +2.0 m s⁻¹.
Your turn
Practise independently
A 2.0 kg trolley at 5.0 m s⁻¹ hits a stationary 3.0 kg trolley and rebounds at 1.0 m s⁻¹. Find the other velocity and test whether the collision is elastic.
Check your answer
Momentum conservation gives 2.0(5.0) = 2.0(−1.0) + 3.0v, hence v = 4.0 m s⁻¹. Initial kinetic energy is 25 J; final kinetic energy is ½(2)(1²) + ½(3)(4²) = 25 J. Relative approach speed and separation speed are both 5.0 m s⁻¹, so the collision is perfectly elastic.
Common mistakes and exam guidance
Watch out for
- Conserving the momentum of each body separately.
- Declaring a collision elastic simply because total momentum is conserved.
In an exam
- Write the signed momentum equation before substituting numbers.
- Show the kinetic-energy or relative-speed test when the question asks whether a collision is elastic.
Put the ideas together
Exam-style practice [7 marks]
A 0.40 kg cart at +6.0 m s⁻¹ collides with a 0.60 kg cart at −2.0 m s⁻¹. They stick. Find their common velocity, calculate the external impulse on the two-cart system if the measured common velocity is +1.1 m s⁻¹ instead, and state what that measurement implies.
Plan before you answer
- First assume a closed system.
- For the measurement, compare final and initial total momentum.
- Interpret non-zero system impulse.
View the marking points and model answer
Marking points
- Initial momentum = +1.20 kg m s⁻¹.
- Total mass = 1.00 kg.
- Closed-system common velocity = +1.20 m s⁻¹.
- Measured final momentum = +1.10 kg m s⁻¹.
- External impulse = −0.10 N s.
- Gives its negative direction.
- States an external influence/measurement limitation means the system was not perfectly isolated.
Model answer
Initial total momentum = 0.40(6.0) + 0.60(−2.0) = +1.20 kg m s⁻¹. If isolated, v = 1.20/1.00 = +1.20 m s⁻¹. The measured final momentum is +1.10 kg m s⁻¹, so external impulse = 1.10 − 1.20 = −0.10 N s. A small negative external impulse, such as track friction, acted or the measurements have uncertainty.
Finish from memory
Three-question recap
State the condition for momentum conservation.
Check
The net external impulse on the chosen system is zero or negligible.
What extra quantity is conserved in a perfectly elastic collision?
Check
Total kinetic energy.
What sign should a rebounding velocity have?
Check
The opposite sign from its initial direction under one fixed convention.
Continue with the next resource in this course.
Course and syllabus information
- Course
- GCE A-Level H1 Physics
- Edition
- GCE A-Level H1 Physics 2027