Scalars and vectors — OCR A-Level Physics
Test yourself on Scalars and vectors with OCR A-Level practice questions.
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Scalars and vectors explained
A scalar quantity has magnitude only, while a vector quantity has both magnitude and direction.
Read the full explanation
Mass, time, temperature, speed, distance, energy and power are scalars. Displacement, velocity, acceleration, force, momentum, weight and electric field strength are vectors. Direction matters because vector quantities combine by direction as well as size: walking 3 m east then 4 m north gives a displacement of 5 m on a bearing, not 7 m. When solving problems, first classify each quantity, then decide whether to add magnitudes directly for scalars or resolve and combine directions for vectors. In equations, a scalar such as mass multiplies a vector such as velocity to give the vector momentum, so the direction is retained.
(b) vector addition and subtraction
Vector addition combines two or more vectors to find a resultant, taking direction into account. For perpendicular vectors, resolve into components and use Pythagoras with trigonometry: a 3 N force east plus a 4 N force north gives a resultant of 5 N at an angle where tan θ = 4 ÷ 3, so θ ≈ 53° from the east direction. For non-perpendicular vectors, draw them tip-to-tail and use the cosine rule or scale drawing. Vector subtraction means adding the negative of the vector: a − b = a + (−b), where −b has the same magnitude as b but the opposite direction. Always state the magnitude and direction of the resultant, and check that the answer is sensible relative to the original vectors.
(c) vector triangle to determine the resultant of any two coplanar vectors
A vector triangle determines the resultant of any two coplanar vectors, whether or not they are perpendicular. Draw the first vector to scale in its direction, then draw the second vector starting from the tip of the first, keeping its own direction. The resultant is the single vector drawn from the tail of the first vector to the tip of the second, closing the triangle. Measure its length with a ruler and its angle with a protractor, converting both using the scale. Alternatively, use the cosine rule for the magnitude and the sine rule for an angle. For perpendicular vectors the triangle is right-angled, so Pythagoras and tan θ apply. Always state the scale, the magnitude and the direction of the resultant.
(d) resolving a vector into two perpendicular components; F x = F cos i ; F y = F sin i .
Resolving splits one vector into two perpendicular components that together have the same effect. Draw the vector F as the hypotenuse of a right-angled triangle, with the two components along chosen perpendicular axes. If the angle i is measured from the x-axis to F, then the adjacent component is Fx = F cos i and the perpendicular component is Fy = F sin i. Check with a simple case: at i = 0°, Fx = F and Fy = 0; at i = 90°, Fx = 0 and Fy = F. Always state the angle's reference axis, because swapping sin and cos gives the wrong pair. Components can be recombined using Pythagoras' theorem and trigonometry.
Your focus
- Define scalar and vector quantities and provide accurate examples of each.
- Distinguish between related scalar and vector pairs such as distance and displacement.
- Select the correct method for combining scalars or vectors in a given physical context.
Show all 12 objectives
- Add two or more vectors to find a resultant with magnitude and direction.
- Subtract vectors by adding the reversed vector.
- Use Pythagoras and trigonometry, or a scale drawing, to combine perpendicular and non-perpendicular vectors.
- Construct a vector triangle to find the resultant of two coplanar vectors.
- Measure the magnitude and direction of a resultant from a scale drawing.
- Apply the cosine and sine rules to calculate the resultant of two non-perpendicular vectors.
- Resolve a vector into two perpendicular components using Fx = F cos i and Fy = F sin i.
- Identify the correct trigonometric ratio from a labelled diagram.
- Verify a resolution using limiting cases or recombination.
Scalars and vectors exam tips
Quick Revision Summary (Key Takeaway)
In OCR A-Level Physics, a scalar quantity possesses only magnitude, whereas a vector quantity possesses both magnitude and direction. Mastery of resolving vectors into perpendicular components and calculating resultant vectors using trigonometry or scale drawings is essential for solving mechanics and fields problems.
Topic Overview
Scalars and vectors form the foundational mathematical language of OCR A-Level Physics. This topic introduces the distinction between quantities characterised solely by magnitude versus those requiring both magnitude and spatial direction, providing essential techniques for vector addition, subtraction, and resolution.
A thorough understanding of vector manipulation is vital for every subsequent mechanics module, including projectile motion, equilibrium under coplanar forces, Newton's laws, and momentum. It also underpins electromagnetism, circular motion, and gravitational fields later in the course.
Key Concepts
- →Scalars have magnitude only (e.g. mass, distance, speed, energy, power, temperature, electric charge).
- →Vectors have both magnitude and direction (e.g. displacement, velocity, acceleration, force, momentum).
- →Resolving vectors involves splitting a single vector into two mutually perpendicular components, typically F_x = F cos(theta) and F_y = F sin(theta).
- →Resultant vectors can be determined either algebraically by summing orthogonal components or graphically using closed vector triangles / polygons tip-to-tail.
Marking Points
- A scalar has magnitude only; examples include mass, time, temperature, speed, distance, energy and power.
- A vector has magnitude and direction; examples include displacement, velocity, acceleration, force, momentum and weight.
- Scalars combine by ordinary arithmetic addition of their magnitudes.
- Vectors must be combined using direction, for example by resolving components or drawing a vector triangle.
- Multiplying or dividing a vector by a scalar changes the magnitude but not the direction, unless the scalar is negative.
- Distance is a scalar while displacement is a vector; speed is a scalar while velocity is a vector.
- Vector addition must account for direction; magnitudes are not simply added unless the vectors are parallel.
- Perpendicular vectors can be combined using Pythagoras' theorem for magnitude and trigonometry for direction.
- Non-perpendicular vectors can be combined by scale drawing or by resolving into perpendicular components.
- Vector subtraction is addition of the reversed vector: a − b = a + (−b).
- The resultant should be stated with both magnitude and direction, for example 5.0 N at 53° north of east.
- A scale drawing must have a stated scale, and the resultant measured with a ruler and protractor.
- A vector triangle is drawn by placing the second vector tip-to-tail with the first, preserving each vector's direction.
- The resultant is the closing side drawn from the start of the first vector to the end of the second.
- A stated scale allows the resultant's magnitude to be found by measuring the closing side.
- The direction of the resultant is measured with a protractor relative to a named reference direction.
- For non-right-angled triangles, the cosine rule gives the resultant magnitude and the sine rule gives an angle.
- For perpendicular vectors, the triangle is right-angled and Pythagoras with trigonometry can be used instead of scale drawing.
- Resolving replaces a single vector by two perpendicular components with the same combined effect.
- For angle i measured from the x-axis, the adjacent component is Fx = F cos i.
- The perpendicular component is Fy = F sin i.
- The components are perpendicular, so they can be treated independently.
- A limiting-case check (i = 0° or i = 90°) confirms which ratio belongs to which axis.
Examiner Tips
- 💡Underline the direction words in a question, such as north, upwards or at 30°, before choosing a method.
- 💡When asked to classify quantities, give the reason: state whether direction is needed to define the quantity.
- 💡Use standard symbols and units consistently, for example velocity $v$ in m s⁻¹ and displacement $s$ in m.
- 💡Sketch the vectors tip-to-tail before calculating so the geometry is clear.
- 💡For perpendicular vectors, calculate the magnitude with Pythagoras first, then find the direction with tan θ = opposite ÷ adjacent.
- 💡State the direction relative to a named reference direction, such as 40° above the horizontal.
- 💡Choose a scale that makes the triangle large enough to measure accurately, for example 1 cm represents 2 N.
- 💡Draw the vectors in pencil with a sharp ruler and protractor, and label each side clearly.
- 💡Check the resultant is longer than the difference and shorter than the sum of the two vector magnitudes.
- 💡Sketch the right-angled triangle and label the angle before choosing sin or cos.
- 💡Check the result at 0° and 90° to confirm the component assignment.
- 💡State the axis from which the angle is measured in your working.
- 💡When adding non-perpendicular vectors, always resolve each vector into horizontal and vertical components first, sum each set independently, then use Pythagoras and inverse tan for the final resultant.
- 💡Draw vector diagrams with a ruler and protractor if a scale drawing is explicitly requested, and clearly label arrowheads showing vector directions.
- 💡State your chosen coordinate system (e.g., taking upwards and to the right as positive) before writing equilibrium or motion equations.
Common Mistakes
- Treating speed as a vector: speed has magnitude only, whereas velocity includes direction.
- Adding vector magnitudes arithmetically regardless of direction: two perpendicular 3 N and 4 N forces give a resultant of 5 N, not 7 N.
- Assuming distance and displacement are interchangeable: displacement is the straight-line distance in a specified direction.
- Believing that a negative sign on a scalar quantity indicates spatial direction: for scalars like temperature or charge, a negative sign indicates a value below zero or a type of charge, not a physical direction.
- Adding magnitudes directly for perpendicular vectors: 3 N and 4 N at right angles give 5 N, not 7 N.
- Forgetting to reverse the second vector in a subtraction: a − b requires the direction of b to be reversed before adding.
- Giving only the magnitude of a resultant: a vector answer needs a direction as well.
- Using the wrong trigonometric ratio when finding the angle: check which sides are opposite and adjacent to the required angle.
- Drawing the second vector from the tail of the first instead of from its tip: the vectors must be placed tip-to-tail.
- Measuring the resultant from the wrong end of the triangle: the resultant runs from the start of the first vector to the end of the second.
- Omitting the scale or the direction: a scale drawing answer needs both a magnitude and a direction.
- Using the cosine rule with the wrong included angle: the angle between the two vectors must be identified correctly.
- Swapping the ratios so that Fx = F sin i and Fy = F cos i: the component adjacent to the stated angle uses cos i.
- Using the wrong angle, such as the angle to the y-axis, without adjusting the ratios: identify the reference axis first.
- Treating components as extra forces that add to the original magnitude: the components replace the original vector, not add to it.
- Thinking work done is a vector: Although work is calculated from force and displacement (both vectors), work done is energy transferred and is strictly a scalar quantity.
- Assuming the horizontal component always uses cos(theta): The choice between cos and sin depends on where the reference angle theta is situated relative to the component line.
- Confusing negative signs in 1D kinematics: A negative sign on a vector indicates opposite direction relative to a chosen positive direction, not a magnitude below zero.
Revision Plan
- 1Day 1-2: Memorise which standard physics quantities are scalars and vectors; practice distinguishing scalar-vector pairs (e.g. distance/displacement, speed/velocity).
- 2Day 3-5: Practice resolving single forces along horizontal/vertical axes and parallel/perpendicular to inclined planes.
- 3Day 6-8: Solve two-dimensional resultant vector problems using vector triangles and component addition.
- 4Day 9-10: Complete OCR past exam questions involving coplanar equilibrium and projectile initial velocity vectors.
Exam Question Types
- 📋Multiple choice classification questions: Quickly categorising a list of physical quantities as scalar or vector combinations.
- 📋Inclined plane equilibrium: Structured 3-4 mark calculations resolving gravitational forces into parallel and normal contact components.
- 📋Navigation and crosswind/current problems: Calculating resultant velocities and bearings using vector addition.
- 📋Closed triangle of forces: Proving a system is in equilibrium by showing that three coplanar forces form a closed polygon.
Command Word Expectations (OCR)
Provide a complete algebraic or numerical progression from first principles or standard formulas to the given value, including intermediate steps and unrounded working.
Obtain an answer by calculating, measuring, or using a graphical technique, showing all necessary working clearly.
Give a concise, factual answer without needing lengthy explanation or mathematical derivation.
How Students Lose Marks (Examiner Pitfalls)
Step-by-Step Worked Solutions
Question: An aircraft flies with an engine airspeed of 120 m s^-1 due North. A steady crosswind of 35 m s^-1 blows from West to East. Calculate the resultant velocity (magnitude and bearing) of the aircraft.
- 1.Step 1: Sketch a vector triangle showing the 120 m s^-1 vector pointing North and the 35 m s^-1 vector pointing East placed tip-to-tail.
- 2.Step 2: Use Pythagoras' theorem to find the magnitude of the resultant velocity: v = sqrt((120)^2 + (35)^2) = sqrt(14400 + 1225) = sqrt(15625) = 125 m s^-1.
- 3.Step 3: Use trigonometry to find the angle theta East of North: tan(theta) = opposite / adjacent = 35 / 120, so theta = arctan(35 / 120) = 16.3 degrees.
- 4.Step 4: Express the direction as a three-figure bearing: Bearing = 016 degrees (or 16.3 degrees East of North).
Question: A crate of mass 45 kg rests on a frictionless ramp inclined at 28 degrees to the horizontal. It is held in equilibrium by a cable running parallel to the ramp. Calculate the tension in the cable and the normal contact force exerted by the ramp on the crate. (Take g = 9.81 m s^-2)
- 1.Step 1: Identify all forces acting on the crate: weight W acting vertically downwards (W = mg = 45 * 9.81 = 441.45 N), normal contact force R perpendicular to the ramp, and tension T acting upwards parallel to the ramp.
- 2.Step 2: Resolve weight parallel to the ramp: W_parallel = W * sin(28) = 441.45 * sin(28) = 207.27 N.
- 3.Step 3: Since the crate is in equilibrium along the plane, T = W_parallel = 207 N (to 3 s.f.).
- 4.Step 4: Resolve weight perpendicular to the ramp: W_perp = W * cos(28) = 441.45 * cos(28) = 389.74 N.
- 5.Step 5: Since the crate is in equilibrium perpendicular to the plane, R = W_perp = 390 N (to 3 s.f.).