Scalars and coplanar vector operations

Key idea: A complete H2 Physics lesson on SI quantities, estimation, errors, uncertainty and coplanar vectors.

  • GCE A-Level H2 Physics 2027

Learn the idea

Big question: How do components make vector addition and subtraction reliable?

A scalar has magnitude only; a vector also has direction. Choose perpendicular positive axes, resolve every vector with signed sine or cosine components, combine corresponding components and only then recover magnitude and direction. Subtracting a vector means adding its reverse.

Decide whether direction belongs in the answer

A scalar has magnitude only: mass, time, temperature, distance, speed and energy are examples. A vector has both magnitude and direction: displacement, velocity, acceleration, force and momentum are examples. Changing direction changes a vector even if its magnitude stays constant.

Check your understanding: A car keeps a constant speed while turning. Is its velocity constant?

No. Velocity is a vector, so its changing direction means the velocity changes.

Subtraction means adding a reversed vector

To find A − B, reverse B and add the reversed vector to A. This definition works in a scale drawing and in components. It also explains relative velocity: velocity of P relative to Q is vP − vQ.

A component is a signed projection, not part of the magnitude left over after subtraction. Once axes are chosen, every vector in the problem must use the same positive directions.

Check your understanding: If A is 5 N east and B is 2 N east, what is A − B?

3 N east. Reversing B gives 2 N west, which is then added to 5 N east.

Use geometry and components as cross-checks

A head-to-tail or parallelogram drawing reveals the approximate direction of the answer. Components then give precision. If the calculated resultant points outside the region suggested by the sketch, a sign or angle has probably been mishandled.

Check your understanding: Can a resultant of two non-zero vectors be zero?

Yes, but only when the two vectors have equal magnitudes and opposite directions.

Resolving a vector into perpendicular componentsA 10 newton force points 30 degrees above the positive x-axis. Its horizontal component is 8.66 newtons and its vertical component is 5.00 newtons. Dashed projection lines form a right triangle, and arrows show that the components add head-to-tail to recover the original force.Perpendicular components+x+y30°Fₓ = 8.66 NFᵧ = 5.00 NF = 10.0 Nresultant of Fₓ and Fᵧ
For a force at angle θ from +x, the signed components are Fx = F cos θ and Fy = F sin θ. Here they add head-to-tail to recover the 10 N force.

Key ideas to keep

  • A component sign comes from direction, not from the calculator.
  • The resultant magnitude is not normally the sum of magnitudes.
  • State the final direction with an angle and reference direction.

Worked example

Combine two forces at an obtuse angle

Question: An 8.0 N force acts east. A 6.0 N force acts at 120° anticlockwise from east. Find the resultant magnitude and direction.

  1. Step 1: Choose axes and resolve

    Why: The obtuse angle makes the second force’s horizontal component negative.

    Working: F₁ = (8.0, 0) N; F₂ = (6 cos120°, 6 sin120°) = (−3.0, 5.20) N.

  2. Step 2: Add like components

    Why: Only components along the same axis can be added directly.

    Working: R = (8.0 − 3.0, 0 + 5.20) = (5.0, 5.20) N.

  3. Step 3: Recover magnitude and direction

    Why: The perpendicular component triangle contains the complete resultant.

    Working: |R| = √(5.0² + 5.20²) = 7.21 N; θ = tan⁻¹(5.20/5.0) = 46.1°.

Answer: The resultant is 7.2 N at 46° north of east.

Check: Its magnitude lies between the 2 N difference and 14 N sum, and its direction lies between the two original directions.

Question

Add 8 N east to 6 N north, then subtract the 6 N north vector from the 8 N east vector.

Check the worked solution

With east as +x and north as +y, the sum is (8, 6) N: magnitude 10 N at 36.9° north of east. The difference is (8, −6) N: magnitude 10 N at 36.9° south of east.

Practise with support

Try this

Resolve a 20 m displacement at 30° west of north into east and north components.

Hint: Declare east +x and north +y before assigning signs.

Check your answer

East component = −20 sin 30° = −10 m. North component = 20 cos 30° = 17.3 m.

Practise independently

Your turn

Two coplanar forces are A = (−3, 4) N and B = (5, −2) N. Find A + B and A − B, giving component and magnitude forms.

Check your answer

A + B = (2, 2) N with magnitude 2.83 N. A − B = (−8, 6) N with magnitude 10.0 N.

Common mistakes

Common mistake

Vector magnitudes may be added or subtracted without directions.

What is wrong with this reasoning?

Show better thinking

Choose axes, resolve each vector with signs, combine corresponding components, then recover magnitude and direction.

Exam guidance

Draw a quick arrow sketch and declare the axes before writing component equations.

Exam-style practice [4 marks]

Three forces act on a point: 12 N north, 5 N east and 9 N west. Determine the resultant force, including direction.

Plan before you answer

  • Combine the east–west components with signs.
  • Keep the north component separate.
  • Use Pythagoras and state the angle reference.
Mark your answer and compare the model

Marking points

Tick each point only if your answer states it clearly.

Model answer

Taking east and north as positive, Rₓ = 5 − 9 = −4 N and Rᵧ = 12 N. Hence R = √(4² + 12²) = 12.6 N. The angle west of north is tan⁻¹(4/12) = 18.4°, so the resultant is 12.6 N, 18.4° west of north.

Check what stayed with you

Recall question 1

How is A − B constructed graphically?

Check the answer

Reverse B, then add the reversed vector to A head to tail.

Recall question 2

What does a negative x-component mean?

Check the answer

The component points opposite the chosen positive x-direction.

Recall question 3

Why must a vector answer include an angle reference?

Check the answer

An angle alone does not identify the direction from which it was measured.

Try this next

Use the longer mixed questions to connect the ideas, calculations and diagrams from this topic.

Open Measurement structured practice

Syllabus and review details

This lesson covers the listed H2 Physics 9478 outcomes.

  • GCE A-Level H2 PhysicsTopic 1(h) / Topic 1(i) / Topic 1(j) · 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