Two perpendicular forces of 5.0\ N and 12\ N act on a point. Find the magnitude of the resultant. [2 marks]

A velocity of 20\ m s^-1 acts at 30^ above the horizontal. Find its vertical component. [2 marks]

OCR OCR 2019-style practice: A ship travels 6.0\ km due east and then 8.0\ km due north. Calculate the magnitude and direction of its resultant displacement from the start.

The two displacements are perpendicular, so use Pythagoras for the magnitude: R = sqrt(6.0^2 + 8.0^2) = sqrt(36 + 64) = sqrt(100) = 10\ km. Find the direction using trigonometry. The angle east of north is θ = tan^-1((6.0)/(8.0)) = tan^-1(0.75) = 37^. So the resultant is 10\ km at a bearing of 037^ (37 degrees east of north). Markers reward the magnitude 10\ km, a correct trig ratio, and a clearly stated direction.

EnglandPhysicsSyllabus dot point

How do we add, subtract and resolve vector quantities to analyse physical situations?

Scalars and vectors: distinguishing the two, combining vectors by scale drawing and by calculation, and resolving a vector into two perpendicular components.

A focused answer to the OCR H556 foundations content on scalars and vectors, covering the distinction between them, combining vectors by scale drawing and by calculation using Pythagoras and trigonometry, and resolving a vector into perpendicular components.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

OCR wants you to distinguish scalar from vector quantities, combine two or more vectors both by accurate scale drawing and by calculation, and resolve a single vector into two perpendicular components using trigonometry.

The answer

Scalars versus vectors

The distinction matters physically. Two $5\ \text{N}$ forces can give a resultant of anything from $0\ \text{N}$ (opposite) to $10\ \text{N}$ (same direction), depending on their relative directions, whereas two $5\ \text{kg}$ masses always total $10\ \text{kg}$ .

Combining vectors

For vectors that are not perpendicular, resolve each into components along the same two axes, add the components separately, then recombine. This component method is more accurate than scale drawing and is the standard approach in calculation questions.

Resolving a vector

Choosing convenient axes is half the skill. On a slope, resolving weight along and perpendicular to the incline (rather than horizontal and vertical) lines one component up with the motion and isolates the normal contact force, which simplifies the algebra dramatically.

Walking across a moving walkway

A person walks at $1.5\ \text{m s}^{-1}$ directly across an airport walkway that itself moves at $2.0\ \text{m s}^{-1}$ along its length. Find the magnitude and direction of the resultant velocity.

step 1: Identify perpendicular components

The walking velocity ( $1.5\ \text{m s}^{-1}$ ) is perpendicular to the walkway velocity ( $2.0\ \text{m s}^{-1}$ ).

step 2: Magnitude by Pythagoras

$v = \sqrt{1.5^2 + 2.0^2} = \sqrt{2.25 + 4.0} = \sqrt{6.25} = 2.5\ \text{m s}^{-1}$ .

step 3: Direction

The angle from the walkway direction is $\theta = \tan^{-1}\left(\frac{1.5}{2.0}\right) = \tan^{-1}(0.75) = 37^{\circ}$ , so the resultant is $2.5\ \text{m s}^{-1}$ at $37^{\circ}$ to the walkway.

Examples in context

A pilot resolves the aircraft velocity and the crosswind velocity to find the true ground track. A climber checks equilibrium on a slope by resolving the rope tension and weight along and perpendicular to the surface. Projectile motion treats horizontal and vertical velocity components independently. Force tables in the lab let you confirm experimentally that three forces in equilibrium have components that cancel in both directions.

Try this

Q1. State one difference between a scalar and a vector, with an example of each. [2 marks]

Cue. A vector has direction as well as magnitude (force); a scalar has magnitude only (mass).

Q2. Two perpendicular forces of $5.0\ \text{N}$ and $12\ \text{N}$ act on a point. Find the magnitude of the resultant. [2 marks]

Cue. $R = \sqrt{5.0^2 + 12^2} = \sqrt{169} = 13\ \text{N}$ .

Q3. A velocity of $20\ \text{m s}^{-1}$ acts at $30^{\circ}$ above the horizontal. Find its vertical component. [2 marks]

Cue. $v_y = 20\sin 30 = 10\ \text{m s}^{-1}$ .

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20194 marksA ship travels

6.0\ \text{km}

due east and then

8.0\ \text{km}

due north. Calculate the magnitude and direction of its resultant displacement from the start.

Show worked answer →

The two displacements are perpendicular, so use Pythagoras for the magnitude: $R = \sqrt{6.0^2 + 8.0^2} = \sqrt{36 + 64} = \sqrt{100} = 10\ \text{km}$ .

Find the direction using trigonometry. The angle east of north is $\theta = \tan^{-1}\left(\frac{6.0}{8.0}\right) = \tan^{-1}(0.75) = 37^{\circ}$ .

So the resultant is $10\ \text{km}$ at a bearing of $037^{\circ}$ (37 degrees east of north). Markers reward the magnitude $10\ \text{km}$ , a correct trig ratio, and a clearly stated direction.

OCR 20213 marksA force of

24\ \text{N}

acts at an angle of

40^{\circ}

above the horizontal. Calculate its horizontal and vertical components.

Show worked answer →

Resolve along each axis. The horizontal component is $F_x = 24\cos 40 = 24 \times 0.766 = 18.4\ \text{N}$ .

The vertical component is $F_y = 24\sin 40 = 24 \times 0.643 = 15.4\ \text{N}$ .

Markers reward $\cos$ for the component adjacent to the angle, $\sin$ for the opposite component, and the two values (about $18\ \text{N}$ and $15\ \text{N}$ ). A quick check: $\sqrt{18.4^2 + 15.4^2} \approx 24\ \text{N}$ , recovering the original force.

Related dot points

Sources & how we know this

OCR AS and A Level Physics A (H156, H556) specification — OCR (2015)