Variable-Mass Systems (Optional Extension)
Key idea: Optional extension on momentum flux, thrust and variable-mass system boundaries; this material is not required in H2 Physics 9478.
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The core idea
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Learning objectives
- Explore friction coefficients, variable-mass systems and the fundamental interactions beyond the H2 syllabus.
1. Definitions (Must Know)
- Variable-mass system: a system whose mass changes with time because mass enters or leaves (e.g. a rocket ejecting fuel, sand falling onto a trolley).
- Momentum, vector p: vector p = m vector v (vector).
- Mass flow rate, m dot: rate at which mass changes, in kg s⁻¹.
- Exhaust speed, vₑ: speed of expelled mass relative to the rocket (direction matters).
- Thrust, T: the force produced by expelling mass; in simple models, magnitude T ≈ m dot vₑ (given in the question or derived from a model).
Variable-mass systems are not a stated H2 Physics 9478 outcome. Complete the core momentum and one-dimensional collision lessons first.
Most core mechanics questions assume constant mass, so you can use: ∑ F = ma
For variable mass, start from Newton’s 2nd law in momentum form:
∑ vector Fₑₓₜ = (d vector p)/dt
where vector p = m vector v.
See also: Momentum & Impulse.
If mass changes, you must be careful: d(mv)/dt is not just mdv/dt. Real rocket problems depend on your system choice and the relative velocity of the expelled/collected mass.
2. Key Ideas (What Earns Marks)
- Use momentum ideas and state your system clearly (“rocket only”, or “rocket + fuel”, etc.).
- Most A Level questions will either:
- give the thrust T, or
- give m dot and vₑ so you can find T.
- Once you have a thrust force T (already accounting for the mass ejection), you can usually write the rocket’s equation of motion at that instant as: ∑ F = ma (for the rocket’s instantaneous mass m)
- Track signs carefully (choose a positive direction and apply it consistently).
3. Detailed Explanations
A. Why F = ma needs care when mass changes
Momentum is: p = mv
If both m and v can change: dp/dt = d(mv)/dt = mdv/dt + vdm/dt
So you cannot blindly replace dp/dt with mdv/dt unless mass is constant.
B. Thrust as “momentum per second”
When a rocket ejects mass backwards, it gives that mass momentum. By Newton’s 3rd law, the rocket gets an equal and opposite effect: a forward thrust.
In a simple model (constant exhaust speed relative to the rocket), the thrust magnitude is: T ≈ m dot vₑ
Many questions treat T as given, so you can focus on forces on the rocket (thrust, weight, drag).
4. Common Mistakes
- Writing ∑ F = d(mv)/dt and then replacing it with ∑ F = mdv/dt without stating “mass constant”.
- Mixing up speed relative to the ground with speed relative to the rocket (vₑ).
- Using m dot without a sign convention (is mass decreasing or increasing?).
- Double-counting thrust (e.g. using both T and a separate “momentum flux” term when T already represents it).
5. Exam Tips
- Start with: “take forward as positive” (or similar).
- Write down what mass is changing and why (ejected/collected).
- If thrust is given, treat it as an external force on the rocket and use ∑ F = ma with the rocket’s instantaneous mass.
- Always check units: m dot vₑ has units kg s⁻¹ · m s⁻¹ = N.
6. Worked Examples
Modelled example 1
Thrust from mass flow rate and exhaust speed
Problem
Study the worked solution
Identify the model quantities
Method
m dot = 2.0 kg s⁻¹ and vₑ = 500 m s⁻¹ relative to the rocket.Reason
The simple thrust model uses the rate of expelled mass and its relative exhaust speed.Working
T ≈ m dot vₑCalculate
Method
T = 1.0 × 10³ N.Reason
Thrust magnitude equals the exhaust momentum carried away per second in this model.Working
T ≈ (2.0)(500) = 1000 NCheck units
Method
kg m s⁻² = N.Reason
Mass flow rate times speed has force units.Working
kg s⁻¹ · m s⁻¹ = N
Guided practice 2
Acceleration of a rocket in deep space (thrust given)
Problem
Try this before viewing the solution
Hints
Hint 1: use the instantaneous force balance
View solution step by step
Set the resultant
Method
∑ F = T = 1.20 × 10⁴ N.Reason
No other forces are retained in the stated deep-space model.Working
∑ F = TUse instantaneous mass
Method
a = 12.0 m s⁻².Reason
Once thrust accounts for mass ejection, use a = T/m for the rocket’s instantaneous mass.Working
a = (1.20 × 10⁴)/1000 = 12.0 m s⁻²
Common misconception 3
Acceleration increases as mass decreases (same thrust)
Learner claim
Try this before viewing the solution
View solution step by step
Write the dependence
Method
a = T/m at each instant.Reason
Thrust is constant but mass is not.Working
a ∝ 1/m for fixed TCalculate the first value
Method
a₁ = 6.0 m s⁻².Reason
Use the 1500 kg instantaneous mass.Working
a₁ = 9000/1500 = 6.0 m s⁻²Calculate and compare
Method
a₂ = 9.0 m s⁻², so acceleration increases.Reason
The same force acts on a smaller mass.Working
a₂ = 9000/1000 = 9.0 m s⁻²
Examiner practice 4
Sand falling onto a moving belt (mass increasing)
Examination question
Try this before viewing the solution
View solution step by step
Set horizontal velocity change
1 markMethod
The sand changes horizontally from 0 to 3.0 m s⁻¹.Reason
Its incoming motion is vertical, then the belt brings it to belt speed horizontally.Working
Δ vₓ = 3.0 m s⁻¹Use momentum flux
1 markMethod
F = m dot Δ vₓ.Reason
Force equals horizontal momentum change per unit time.Working
F = (Δ pₓ)/(Δ t) = m dot Δ vₓCalculate
1 markMethod
The extra driving force is 1.5 N horizontally.Reason
The belt must supply the sand’s momentum gain while maintaining constant speed.Working
F = (0.50)(3.0) = 1.5 N
Self-mark with the mark scheme
Compare your response with each mark point. Select a point only when your response contains that evidence.
Self-mark the horizontal velocity change, momentum-flux relation and force.
Challenge 5
Water jet on a plate (momentum flux force)
Independent transfer
Try this before viewing the solution
Hints
Hint 1: track the water first
View solution step by step
Find water momentum change
Method
The water loses horizontal momentum at 1.6 kg m s⁻².Reason
Each second, 0.20 kg loses 8.0 m s⁻¹ horizontal velocity.Working
|dpₓ/dt| = (0.20)(8.0) = 1.6 NUse the interaction pair
Method
The water exerts a 1.6 N force on the plate in the incoming-jet direction.Reason
The plate’s force on the water and water’s force on the plate are equal and opposite.Working
|F_(water on plate)| = 1.6 N
7. Mind Stretchers
Mind stretcher 1: Why does “just use F = ma” fail?Extension
In one sentence, explain why you must be careful using F = ma when a system’s mass changes.
Show Answer
Because Newton’s 2nd law is ∑ F = dp/dt and if p = mv with changing m, then d(mv)/dt is not simply mdv/dt.
Mind stretcher 2: Choosing the system boundary (rocket)Extension
In a rocket problem, why is it important to state whether your “system” is the rocket alone or the rocket plus the ejected fuel?
Show Answer
Because momentum conservation/force equations depend on what crosses the system boundary. If you choose “rocket only”, the ejected mass leaving carries momentum away and appears as thrust (momentum flux). If you choose “rocket + ejected fuel”, you must track internal forces and the momentum of both parts. Stating the system prevents sign errors and double-counting.
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Course and syllabus information
- Course
- A-level H2 Physics topic extensions
- Syllabus scope
- Beyond the syllabus
- Edition
- A-level H2 Physics topic extensions