Momentum conservation and one-dimensional interactions
Key idea: H2 Physics lessons on force–time impulse, one-dimensional momentum conservation, elastic interactions and kinetic-energy change.
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The core idea
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Big question: When is total momentum conserved during an interaction?
For a system with negligible external impulse, total vector momentum before an interaction equals total momentum after it. Internal third-law impulses cancel within the system. In one dimension, choose a positive direction and keep velocity signs; sticking together is only one possible final condition.
Define the system before conserving momentum
Total linear momentum is conserved when the resultant external force on the chosen system is zero. During a short collision, this is often used as the equivalent condition that external impulse is zero or negligible.
The forces between colliding bodies are internal to the combined system. Their equal and opposite impulses change each body's momentum but cancel in the total.
Check your understanding: Why can each trolley's momentum change while their total remains constant?
They receive equal and opposite internal impulses, so the two momentum changes cancel in the system total.
Use one sign convention throughout
Choose one positive direction and attach a sign to every velocity. Then write total momentum before equals total momentum after. A rebound has a velocity opposite in sign to the original direction.
The momentum equation applies to inelastic and perfectly elastic interactions. It does not decide the final arrangement by itself, so use information such as 'stick together', a stated final velocity or another physical condition to supply the remaining relationship.
Check your understanding: Two trolleys stick after collision. What extra relationship can you use?
They share one common final velocity because they move together.
Key ideas to keep
- Momentum is conserved for the whole isolated system, not necessarily for each object.
- A rebound needs a negative velocity if the original direction is positive.
- Kinetic energy need not be conserved even when momentum is.
See the reasoning
Worked example
Find a shared velocity and account for kinetic energy
Question: A 2.0 kg body at +5.0 m s⁻¹ and a 3.0 kg body at −1.0 m s⁻¹ stick together. Find their final velocity and the kinetic-energy decrease.
Step 1: Declare the system condition
Why: Momentum conservation needs negligible external impulse during the collision.
Working: Treat both bodies as one system; take right as positive.
Step 2: Use the sticking condition
Why: Both masses share one final velocity.
Working: 2.0(5.0) + 3.0(−1.0) = (2.0 + 3.0)v, so v = +1.4 m s⁻¹.
Step 3: Compare kinetic energies
Why: Momentum conservation does not imply unchanged kinetic energy.
Working: Kᵢ = 26.5 J and Kf = ½(5.0)(1.4²) = 4.9 J, so the decrease is 21.6 J.
Answer: The joined bodies move at 1.4 m s⁻¹ to the right, and kinetic energy decreases by 21.6 J.
Check: The positive result matches the system's initial momentum; the missing kinetic energy is transferred to deformation, internal energy and sound.
Use a hint if needed
Practise with support
Try this
A stationary object explodes into 0.80 kg and 1.20 kg fragments. The 0.80 kg fragment moves at +6.0 m s⁻¹. Find the other velocity.
Hint: The closed system starts with zero total momentum.
Check your answer
0 = 0.80(6.0) + 1.20v, so v = −4.0 m s⁻¹. The negative sign means the second fragment moves in the opposite direction.
Now work without the hint
Practise independently
Your turn
A 0.30 kg body at +10 m s⁻¹ collides with a 0.20 kg body at −5.0 m s⁻¹. Afterwards the first moves at −2.0 m s⁻¹. Find the second body’s velocity, assuming negligible external impulse.
Check your answer
Initial momentum = 0.30(10) + 0.20(−5.0) = 2.0 kg m s⁻¹. Thus 2.0 = 0.30(−2.0) + 0.20v, giving v = +13 m s⁻¹.
Avoid these traps
Common mistakes
Common mistake
Momentum is conserved separately for each body in a collision.
What is wrong with this reasoning?
Show better thinking
Each body receives an impulse and changes momentum. Total momentum is conserved only for the chosen closed system when external impulse is negligible.
Common mistake
Speeds can be inserted as positive values because momentum conservation is a scalar equation in one dimension.
What is wrong with this reasoning?
Show better thinking
Momentum is a vector. Declare a positive direction and use signed velocities throughout the one-dimensional equation.
Write for the examiner
Exam guidance
Write the symbolic before-and-after momentum equation before inserting signed velocities.
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. If the measured common velocity is instead +1.1 m s⁻¹, calculate the external impulse on the two-cart system and state what it implies.
Plan before you answer
- First assume zero external impulse.
- For the measurement, compare final and initial total momentum.
- Interpret the sign and cause of any difference.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
Initial momentum is 0.40(6.0) + 0.60(−2.0) = +1.20 kg m s⁻¹. With negligible external impulse, v = 1.20/(0.40 + 0.60) = +1.20 m s⁻¹. The measured final momentum would be +1.10 kg m s⁻¹, so the system impulse is 1.10 − 1.20 = −0.10 N s. This indicates a small leftward external impulse, such as friction, or a measurement limitation.
Come back in three days
Check what stayed with you
Recall question 1
State the condition for momentum conservation.
Check the answer
The resultant external force, and hence external impulse, on the chosen system is zero or negligible.
Recall question 2
Why do internal collision forces not change total momentum?
Check the answer
Their equal and opposite impulses cancel within the system.
Recall question 3
What sign should a rebound velocity have?
Check the answer
The sign opposite to its original direction under the chosen convention.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. The coefficient of restitution is not required. Use momentum conservation, the perfectly elastic relative-speed condition and a direct comparison of total kinetic energy.
- GCE A-Level H2 PhysicsTopic 6(b) / Topic 6(c) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027