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.

  • A-level H2 Physics topic extensions
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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).
Optional extension

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.

Why this matters

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

Core

Problem

A rocket ejects fuel at 2.0 kg s⁻¹. The exhaust speed relative to the rocket is 500 m s⁻¹. Estimate the thrust magnitude.
Study the worked solution
  1. 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 vₑ
  2. Calculate

    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 N
  3. Check 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)

About 4 min

Problem

A 1000 kg rocket in deep space has thrust 1.20 × 10⁴ N. Ignore gravity and drag. Find its acceleration at that instant.

Try this before viewing the solution

Unit: m s^-2

Hints

Hint 1: use the instantaneous force balance
With gravity and drag ignored, ∑ F = T at this instant.
View solution step by step
  1. Set the resultant

    Method

    ∑ F = T = 1.20 × 10⁴ N.

    Reason

    No other forces are retained in the stated deep-space model.

    Working

    ∑ F = T
  2. Use 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)

Find and correct the mistake

Learner claim

A rocket produces constant 9000 N thrust with drag and weight ignored. A learner says its acceleration stays constant. Compare acceleration at masses 1500 kg and 1000 kg.

Try this before viewing the solution

Unit: m s^-2
Unit: m s^-2

View solution step by step
  1. Write the dependence

    Method

    a = T/m at each instant.

    Reason

    Thrust is constant but mass is not.

    Working

    a ∝ 1/m for fixed T
  2. Calculate 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⁻²
  3. Calculate 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)

3 marks

Examination question

A belt moves at 3.0 m s⁻¹. Sand falls vertically onto it at 0.50 kg s⁻¹ and reaches belt speed. Estimate the extra horizontal driving force, ignoring friction losses. [3 marks]

Try this before viewing the solution

Unit: N

View solution step by step
  1. Set horizontal velocity change

    1 mark

    Method

    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⁻¹
  2. Use momentum flux

    1 mark

    Method

    F = m dot Δ vₓ.

    Reason

    Force equals horizontal momentum change per unit time.

    Working

    F = (Δ pₓ)/(Δ t) = m dot Δ vₓ
  3. Calculate

    1 mark

    Method

    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

Challenge 5

Water jet on a plate (momentum flux force)

Minimal support

Independent transfer

Water at 0.20 kg s⁻¹ strikes a stationary plate at 8.0 m s⁻¹ and is brought to rest horizontally. Estimate the horizontal force exerted on the plate.

Try this before viewing the solution

Unit: N

Hints

Hint 1: track the water first
For the water, vₓ changes from 8.0 to 0 m s⁻¹; find the force on water, then apply Newton’s third law.
View solution step by step
  1. 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 N
  2. Use 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.

Continue with the next resource in this course.

Course and syllabus information
Course
A-level H2 Physics topic extensions
Syllabus scope
Beyond the syllabus
Edition
A-level H2 Physics topic extensions