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IGCSE Physics notes · Mechanics

Momentum

Momentum is mass multiplied by velocity, including direction. Total momentum is conserved when the resultant external impulse is zero or negligible; kinetic energy need not be conserved in a collision.

By Dr Desouky Physics AcademyUpdated 25 September 202613 min reading + practice

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In this lesson
  1. 01The key idea
  2. 02Choose the sign convention before substituting
  3. 03Define the system and conserve its total momentum
  4. 04A change of direction makes a large momentum change
  5. 05Check what is conserved and what is transferred
  6. 06Key definitions
  7. 07Equations & units
  8. 08Worked examples
  9. 09Practice & solutions
  10. 10Common mistakes
  11. 11Questions, explained
  12. 12Learning checklist
  13. 13Sources & citation

The key idea

Use signed velocities to calculate momentum, apply conservation to a clearly defined system and connect impulse with average force. Cambridge assesses this topic in Supplement content; Pearson International GCSE includes its own physics-only requirements.

Understand it · 01

Choose the sign convention before substituting↗

Momentum describes motion using both mass and velocity. Choose one direction as positive before putting numbers into p = mv. An object moving the opposite way has negative momentum in that convention. This negative sign does not mean a smaller mass or a negative speed; it records direction. Adding momentum magnitudes without checking direction can reverse the predicted outcome of a collision.

For a system containing two trolleys, add their momenta to find the total. Opposite momenta can cancel even while both objects are moving. A system with zero total momentum can therefore still contain kinetic energy, because kinetic energy depends on speed squared and has no directional sign. Write a before-and-after sketch with masses and signed velocities to keep the distinction clear throughout a calculation.

Pause & explain

Two equal masses move at equal speeds in opposite directions. Is their total kinetic energy zero because their total momentum is zero?

Check your reasoning

No. Their signed momenta cancel, but each has positive kinetic energy. The kinetic energies add.

Understand it · 02

Define the system and conserve its total momentum↗

During a short collision, forces between the colliding objects are internal to the chosen system. They transfer momentum from one object to another. If the net impulse from external forces such as friction is negligible during the collision, the system's total momentum before and after is approximately equal. The individual objects' momenta usually change, so do not apply conservation separately to each object.

If the objects stick together, they share a single final velocity and the combined mass multiplies that velocity. If they separate, retain a separate final momentum term for each. State the direction associated with the answer. A negative final velocity is useful information: the combined object moves opposite the positive direction you chose, rather than the calculation having automatically failed.

Before collision a 3 kilogram trolley moves right at 4 metres per second towards a stationary 1 kilogram trolley. After they stick together, the combined 4 kilogram mass moves right at 3 metres per second. Total momentum remains 12 kilogram metres per second.
Figure 2. Before: 3 × 4 + 1 × 0 = 12 kg m/s. After: (3 + 1) × 3 = 12 kg m/s. External impulse is assumed negligible during the collision.Open full-size SVG diagram ↗

Understand it · 03

A change of direction makes a large momentum change↗

Impulse is the change in momentum. For a constant force, impulse equals force multiplied by time; for a changing force, use the average force over the interaction. Always calculate final momentum minus initial momentum using signed velocities. A ball that rebounds can have a greater momentum change than a ball that merely stops, because its final momentum points the other way.

For the same change in momentum, spreading the interaction over a longer time reduces the average force. This is the key explanation for padded catching surfaces, crumple zones and airbags. The actual force during an impact changes with time, so a calculation of Δp/Δt gives the average, not necessarily the peak force. Safety also depends on factors such as contact area and restraint design; the momentum argument explains one physical mechanism.

Pause & explain

Why must signs be used when a ball approaches a wall and then rebounds?

Check your reasoning

The initial and final velocities have opposite directions. Subtracting their signed momenta gives the full reversal; subtracting their speeds would underestimate the momentum change.

Understand it · 04

Check what is conserved and what is transferred↗

An elastic collision conserves both total momentum and total kinetic energy in an isolated system. In an inelastic collision, total momentum can still be conserved while some kinetic energy becomes internal energy, sound or deformation. Objects sticking together are a familiar inelastic case. Conservation of total energy remains valid; a reduction in kinetic energy is not destruction of energy.

A practical collision investigation can measure masses with a balance and velocities using light gates or a motion sensor. Compare total momentum before and after, allowing for measurement uncertainty and friction. A small discrepancy does not by itself disprove conservation: the chosen system may receive an external impulse, or the measurements may not be precise enough. Explain the assumptions when using the ideal conservation equation in a real experiment.

Key definitions

Momentum
The product of mass and velocity: p = mv. A vector quantity, measured in kg m/s.
Impulse
The change in momentum produced by a force: impulse = FΔt = Δp, where F is the constant force or its average over the interaction.
Principle of conservation of momentum
When bodies interact, their total momentum before equals their total momentum after if the resultant external impulse on the system is zero. A negligible external impulse gives an approximate conservation model.

Equations, units & conditions

Momentum

p = m × v

Units: kg m/s

Impulse

F × t = Δp = mv − mu

Units: N s

Conservation

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Open the IGCSE Physics formula sheet →

Worked examples, step by step

Example 1

A 2.0 kg trolley moving at 3.0 m/s collides with a stationary 1.0 kg trolley and they stick together. Calculate their common velocity after the collision.

Follow the method

  1. Total momentum before = (2.0 × 3.0) + (1.0 × 0) = 6.0 kg m/s.
  2. After: momentum = (2.0 + 1.0) × v = 3.0v.
  3. Conservation: 3.0v = 6.0, so v = 2.0.

v = 2.0 m/s in the original direction

Practice questions with worked solutions

Try each question on paper before opening the solution. These are original academy questions; the suggested marks are a self-checking guide, not an official exam-board mark scheme. Difficulty labels describe this practice set, not predicted grades.

  1. Build · 4 suggested marks

    1. A collision with opposing velocities

    A 1.5 kg trolley moving right at 4.0 m/s collides with a 0.50 kg trolley moving left at 2.0 m/s. They stick together. Neglect external impulse and find their final velocity.

    Show solution and self-check: question 1
    1. Choose right as positive: initial total momentum = 1.5 × 4.0 + 0.50 × (−2.0) = 5.0 kg m/s.
    2. Combined mass = 1.5 + 0.50 = 2.0 kg.
    3. Final velocity = 5.0/2.0 = 2.5 m/s, positive.

    2.5 m/s to the right.

    Self-check · one mark per point

    • Uses a negative velocity for the left-moving trolley.
    • Finds total initial momentum 5.0 kg m/s.
    • Uses combined mass 2.0 kg.
    • Finds 2.5 m/s to the right.
  2. Stretch · 4 suggested marks

    2. Average force on a rebounding ball

    A 0.20 kg ball moving right at 10 m/s hits a wall and rebounds left at 6.0 m/s. Contact lasts 0.040 s. Find the average resultant force on the ball during contact.

    Show solution and self-check: question 2
    1. Choose right as positive: u = +10 m/s and v = −6.0 m/s.
    2. Change in momentum = 0.20 × (−6.0 − 10) = −3.2 kg m/s.
    3. Average force = −3.2/0.040 = −80 N.

    Average resultant force = 80 N to the left.

    Self-check · one mark per point

    • Assigns opposite signs to initial and final velocities.
    • Calculates momentum change −3.2 kg m/s.
    • Divides by 0.040 s.
    • States 80 N to the left.

Common mistakes & how to avoid them

  • Ignoring direction. Momentum is a vector: choose a positive direction, and give anything moving the other way a negative velocity before you add.
  • Forgetting that objects moving together after a collision share one velocity — the combined mass (m₁ + m₂) multiplies the single final velocity.
  • Explaining airbags only with 'they absorb the impact'. Increasing the time for the same change in momentum reduces the average force (F = Δp ÷ Δt); it does not necessarily make the force constant throughout the impact.
  • Quoting momentum in newtons. The unit is kg m/s (or the equivalent N s).

Understand the why

Questions, explained

Start with the short answer, then follow the reasoning. Each explanation has a permanent link you can share with a classmate or return to when revising.

  1. How do crumple zones reduce injury?
  2. Is momentum always conserved?
  3. What is the connection between impulse and momentum?
  4. How do I calculate the final velocity when two objects stick together?
  5. How can momentum be conserved when kinetic energy is not?
  6. Why is a ball's momentum change larger when it rebounds than when it stops?
Browse all physics questions →

How do crumple zones reduce injury?

Crumple zones increase the time over which a vehicle and its occupants are brought to rest. For the same change in momentum, a longer stopping time reduces the average force because F = Δp/Δt. Their controlled deformation also transfers kinetic energy into other forms.

Compare collisions with the same initial and final momentum. The required impulse is then unchanged; the design alters how that impulse is delivered over time. A smaller average force does not mean the momentum change is smaller, nor does it require the actual force to remain constant during the collision.

Occupant restraints and vehicle deformation must work together for the stopping-time argument to apply to the occupant. In an exam explanation, identify the relevant object and connect its momentum change to the time and average force. Avoid the vague phrase absorbs impact without explaining a physical quantity, and do not treat a simplified average-force calculation as a complete real-world safety assessment.

On force-time axes, a rectangle of height 800 newtons and width 0.050 second has the same area as a rectangle of height 200 newtons and width 0.20 second. Each area is an impulse of 40 newton seconds.
These idealised constant-force rectangles have equal area, representing equal impulse. Real collision forces vary with time; the calculated values are averages.Open full-size SVG diagram ↗

Same impulse over different times

Two simplified impacts each change momentum by 40 kg m/s. One lasts 0.050 s and the other 0.20 s. Calculate the average force magnitude in each.

  1. Short impact: average force = 40/0.050 = 800 N.
  2. Longer impact: average force = 40/0.20 = 200 N.
  3. The time is four times larger, so the average force is one quarter as large.

Average forces are 800 N and 200 N; both impacts deliver an impulse of 40 N s.

  • Read the supporting explanation ↑
  • Continue with forces →
  • Continue with energy, work and power →

Is momentum always conserved?

A chosen system's total momentum is conserved when the resultant external impulse is zero, and approximately conserved when it is negligible over the interval. Internal forces can transfer momentum between objects without changing the system total. One object's momentum alone generally changes during a collision.

The word system matters. A ball striking a wall changes momentum, so momentum is not conserved for the ball considered alone. If the wall and Earth are included, the interaction transfers momentum to that larger system. The Earth's speed change is tiny because its mass is enormous, rather than because the exchanged momentum has ceased to exist.

For two trolleys on a nearly level low-friction track, collision forces can dominate over the short contact time, making external impulse negligible. Over a much longer interval, resistance may no longer be negligible. State the time interval and identify which forces are external instead of applying conservation automatically to every collection of moving objects.

  • Read the supporting explanation ↑
  • Continue with forces →
  • Continue with motion →

What is the connection between impulse and momentum?

Impulse equals the change in momentum: FΔt = mv − mu for constant mass, using the average force if it varies. Include velocity directions when subtracting. Impulse and momentum share equivalent units because one newton second equals one kilogram metre per second.

The subtraction is final momentum minus initial momentum, not simply the difference between speed magnitudes. A change from rightward motion to leftward motion can therefore have a large magnitude even when the two speeds are similar. The sign of the impulse gives the direction of the momentum change in your chosen convention.

A force–time graph gives the same information through its signed area. A rectangular area applies to a constant force; a triangular pulse uses half the base times height. If positive and negative forces occur during an interval, their signed impulses can partly cancel. The average force is the net impulse divided by the full interval, not necessarily the largest force shown.

Read a triangular force pulse

A force rises linearly from zero to 120 N and then falls linearly to zero over a total of 0.10 s. The force stays in the same direction. Find the impulse and average force.

  1. Impulse = area of triangle = ½ × 0.10 × 120 = 6.0 N s.
  2. Average force = 6.0/0.10 = 60 N.

Impulse = 6.0 N s in the force direction; average force = 60 N.

  • Read the supporting explanation ↑
  • Continue with forces →
  • Continue with motion →

How do I calculate the final velocity when two objects stick together?

Add their signed momenta before the collision and divide by their combined mass, assuming external impulse is negligible. Sticking together means both objects share one final velocity, so the final momentum is (m₁ + m₂)v. A negative result indicates motion opposite your chosen positive direction.

Choose a direction before substituting. A stationary object contributes zero initial momentum but still contributes mass after joining the other object. Forgetting that added mass predicts a final speed that is too large. If both objects are initially moving, keep both initial momentum terms even when their directions differ.

Do not average the velocities unless the masses happen to be equal. Momentum weights each velocity by mass, so a heavier object's motion contributes more strongly to the final result. Check your answer by multiplying the combined mass by the calculated final velocity and confirming that it reproduces the signed total momentum you started with.

Unequal masses moving towards each other

A 2.0 kg trolley moves right at 5.0 m/s and a 3.0 kg trolley moves left at 1.0 m/s. They stick together; neglect external impulse. Find the final velocity.

  1. Taking right as positive, total initial momentum = 2.0 × 5.0 + 3.0 × (−1.0) = 7.0 kg m/s.
  2. Combined mass = 5.0 kg, so v = 7.0/5.0 = 1.4 m/s.
  3. Check: 5.0 × 1.4 = 7.0 kg m/s.

Final velocity = 1.4 m/s to the right.

  • Read the supporting explanation ↑
  • Continue with motion →
  • Continue with energy, work and power →

How can momentum be conserved when kinetic energy is not?

Momentum and kinetic energy are different quantities with different conservation conditions. An inelastic collision can conserve a system's total momentum while transferring some kinetic energy into deformation, internal energy and sound. Total energy is still conserved; only the kinetic-energy part has decreased.

Momentum is a vector and depends on signed velocity, whereas kinetic energy depends on speed squared. You cannot infer that one is unchanged just because the other is unchanged. A collision described as objects sticking together requires a momentum calculation first; assuming their total kinetic energy also stays constant would usually give an incompatible answer.

To check energy, calculate each object's kinetic energy before and after using the velocities obtained from the momentum model. Add those positive energies, then compare the totals. The decrease identifies energy transferred into other forms within the wider system. An elastic collision is the special case where both total momentum and total kinetic energy are conserved under the isolated-system assumptions.

Compare both totals

A 1.0 kg trolley at 4.0 m/s hits a stationary 1.0 kg trolley and they stick. Neglect external impulse. Compare initial and final momentum and kinetic energy.

  1. Initial momentum = 1.0 × 4.0 = 4.0 kg m/s, so final velocity = 4.0/2.0 = 2.0 m/s.
  2. Final momentum = 2.0 × 2.0 = 4.0 kg m/s.
  3. Initial KE = ½ × 1.0 × 4.0² = 8.0 J; final KE = ½ × 2.0 × 2.0² = 4.0 J.

Momentum stays at 4.0 kg m/s. Kinetic energy falls from 8.0 J to 4.0 J, with 4.0 J transferred to other forms.

  • Read the supporting explanation ↑
  • Continue with energy, work and power →
  • Continue with forces →

Why is a ball's momentum change larger when it rebounds than when it stops?

Stopping removes the ball's initial momentum, while rebounding also gives it momentum in the opposite direction. With signed velocities, final minus initial includes both contributions. For the same contact time, the larger momentum-change magnitude gives a larger average force magnitude.

Suppose right is positive. A ball initially moving right has positive momentum; if it rebounds left, its final momentum is negative. Subtracting the positive initial value from the negative final value gives a change whose magnitude is the sum of the two momentum magnitudes, not their difference.

This is a directional argument, not a claim that rebound always produces a greater force in every real collision. The contact time may change too, so compare both Δp and Δt before comparing average force. Keep separate the force on the ball and the opposite force it exerts on the wall; their directions belong to different objects.

Assign signs before subtracting

A 0.10 kg ball travels right at 12 m/s and rebounds left at 8.0 m/s. Find its momentum change and compare with simply stopping.

  1. Rebound change = 0.10 × (−8.0 − 12) = −2.0 kg m/s.
  2. Stopping change = 0.10 × (0 − 12) = −1.2 kg m/s.

Rebound changes momentum by 2.0 kg m/s leftwards; stopping changes it by 1.2 kg m/s leftwards.

  • Read the supporting explanation ↑
  • Continue with forces →
  • Continue with motion →

Your learning checklist

Use these goals to check your understanding. Requirements vary by specification, tier and exam year; the official references below define your full course.

  • Define momentum as mass × velocity and calculate it
  • Define impulse as force × time and relate it to the change in momentum
  • Apply the principle of conservation of momentum to collisions in one line
  • Explain safety features (crumple zones, airbags) using impulse

Course & model notes

Momentum and impulse are Supplement content in Cambridge 0625. Pearson 4PH1 includes momentum calculations in physics-only content, identified by P in the specification.

Check which exam board you study

Sources & how to cite these notes

Original explanations, illustrations and practice by Dr Desouky Physics Academy. Official specifications guide the course scope; questions and suggested marks on this page are our own revision material.

  • Cambridge IGCSE Physics 0625 — 2026–2028 syllabus
  • Pearson Edexcel International GCSE Physics 4PH1 — specification

Reference this page

Dr Desouky Physics Academy. IGCSE Physics: Momentum. Updated 2026-09-25. https://www.drdesouky.com/revision/igcse-physics/momentum

Use the arrow beside an explanation heading to link straight to that section. Spotted a problem? Contact the academy with the topic and the step you want us to check.

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