AeroLab Reading room

Physics / AS level

Dynamics

Explain force as a rate of change of momentum.

Learning objectives

The big idea

Newton’s second law relates resultant force to rate of change of momentum. For constant mass this becomes F = ma. Draw forces on one body before choosing the direction of acceleration.

During an interaction, the forces between two bodies are equal and opposite and act on different bodies. Momentum conservation concerns an isolated system; kinetic energy is additionally conserved only in an elastic collision.

1. A force diagram belongs to one body

Choose the body and show its interactions with other objects: weight, tension, normal contact and friction where relevant. The resultant is the vector sum of those forces. Do not add an extra arrow labelled resultant alongside its components. Newton’s third-law partner to each interaction acts on another body, so it is absent from this particular single-body diagram.

2. Momentum makes the second law general

Momentum is mass multiplied by velocity, and resultant force is its rate of change. For constant mass this becomes F = ma. For a falling object, weight can stay nearly constant while resistive force increases with speed. When resistance balances weight, acceleration becomes zero and a terminal velocity is possible. The object has not stopped moving and gravity has not vanished.

3. Impulse and impact time

The area under a force–time graph is impulse, equal to momentum change. A ball of mass 0.10 kg reversing from +5 m/s to −3 m/s changes momentum by −0.80 kg m/s. Using speed magnitudes alone would miss the reversal. For the same momentum change, a longer stopping interval reduces average force, explaining one role of cushioning and controlled stopping distances.

4. Select the collision system

Momentum is conserved when external impulse on the chosen system is negligible. Internal collision forces redistribute momentum between bodies without changing its total. An elastic collision also conserves total kinetic energy; an inelastic collision need not. Objects sticking together share a final velocity, but a shared final velocity is not a requirement for every collision. Check energy separately rather than assuming every isolated collision is elastic.

A method you can reuse

  1. Choose one body for a force diagram or a full system for momentum conservation.
  2. Set one positive direction and use signed velocities.
  3. Find resultant force or total momentum as appropriate.
  4. Apply impulse over the actual interaction interval.
  5. Compare kinetic energies only after the momentum calculation.
Do equal and opposite collision forces cancel the acceleration of each object?

No. They act on different objects. Each object can accelerate; their impulses are opposite, so the system’s total momentum remains unchanged when external impulse is negligible.

Equations and relationships

F=ΔpΔtF=\frac{\Delta p}{\Delta t}

Average resultant force over the time interval.

impulse=FΔt=Δp\text{impulse}=F\Delta t=\Delta p

For a constant force, impulse is force times duration.

Worked example

A 0.20 kg ball changes velocity from +10 m/s to −5 m/s. What is its momentum change?

Follow the solution

Δp = 0.20 × (−5 − 10) = −3.0 kg m/s.

Investigate and explain

Compare two simulated collisions with identical starting conditions but different elasticity. Tabulate both momentum and kinetic energy.

Compare elastic and inelastic settings while keeping initial masses and velocities fixed. Idealised friction and collision models omit deformation details and cannot predict injury or protective-equipment performance.

Open the interactive investigation

Check your understanding

1. Which quantity is conserved in an isolated inelastic collision?

  1. Total momentum
  2. Total kinetic energy in every case
  3. Each object’s velocity
Check the answer and reasoning

Total momentum

Internal interactions preserve total momentum while kinetic energy can change form.

2. A 0.20 kg ball changes velocity from +10 m/s to −5 m/s. What is its momentum change?

  1. +1.0 kg m/s
  2. −3.0 kg m/s
  3. −1.0 kg m/s
Check the answer and reasoning

−3.0 kg m/s

Δp = 0.20 × (−5 − 10) = −3.0 kg m/s.

3. Why does a longer stopping time reduce average force for the same momentum change?

  1. The initial momentum becomes zero before stopping begins.
  2. The mass must decrease.
  3. F = Δp/Δt gives a smaller force.
Check the answer and reasoning

F = Δp/Δt gives a smaller force.

The same impulse spread over more time has a lower average force.