Scalars and Vectors

Spec mapping: OCR H556 Module 2.3 — Nature of quantities. This lesson defines scalar and vector, classifies the common A-Level quantities, and establishes the tip-to-tail addition and subtraction conventions used throughout Modules 3 (forces and motion), 4 (electricity), 5 (Newtonian world) and 6 (fields). (Refer to the official OCR H556 specification document for exact wording.)

Some physical quantities are fully specified by a single number and a unit — "the mass is $3.0\,\text{kg}$3.0kg", "the temperature is $298\,\text{K}$298K". These are scalars. Others require a direction as well as a magnitude to make physical sense — "a force of $20\,\text{N}$20N east", "a velocity of $5\,\text{m s}^{-1}$5m s−1 downwards". These are vectors. Distinguishing the two, and knowing how to combine vectors correctly, is the foundation of all mechanics, electricity, waves and fields work in the A-Level course.

This lesson covers: the formal definitions; classification of the standard A-Level quantities; vector notation; tip-to-tail (triangle) addition; the parallelogram interpretation; non-perpendicular addition via cosine rule and scale drawing; and subtraction via reversal. The inverse operation — resolving a vector into components — is the subject of the next (and final) lesson.

Key definitions. A scalar is a quantity that is fully specified by a magnitude (with units). A vector is a quantity that requires both magnitude and direction to be specified.

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Scalars

Scalars have magnitude only. Adding two scalars is ordinary arithmetic: $3\,\text{kg} + 2\,\text{kg} = 5\,\text{kg}$3kg+2kg=5kg. It is meaningless to ask "which way does the mass point?".

A-Level scalar quantities

Quantity	Symbol	SI Unit
Mass	$m$m	$\text{kg}$kg
Length, distance	$l, d$l,d	$\text{m}$m
Time	$t$t	$\text{s}$s
Temperature	$T$T	$\text{K}$K
Speed	$v$v	$\text{m s}^{-1}$m s−1
Energy, work	$E, W$E,W	$\text{J}$J
Power	$P$P	$\text{W}$W
Pressure	$p$p	$\text{Pa}$Pa
Electric charge	$Q$Q	$\text{C}$C
Electric potential	$V$V	$\text{V}$V
Density	$\rho$ρ	$\text{kg m}^{-3}$kg m−3
Frequency	$f$f	$\text{Hz}$Hz

Two especially important scalar/vector pairs that catch candidates out:

• Speed (scalar) vs velocity (vector). Speed is the magnitude of velocity. A car driving at $30\,\text{m s}^{-1}$30m s−1 has speed $30\,\text{m s}^{-1}$30m s−1 regardless of direction; its velocity also specifies the direction.
• Distance (scalar) vs displacement (vector). Distance is the total path length travelled; displacement is the straight-line vector from start to finish.

The total energy of a system is a scalar even though energy can take many forms (kinetic, potential, electrical). The scalar nature is what makes the conservation-of-energy bookkeeping so powerful — you can sum across all forms without worrying about direction.

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Vectors

A vector has both magnitude and direction. It is customarily represented as an arrow: the length encodes magnitude, the arrowhead encodes direction.

A-Level vector quantities

Quantity	Symbol	SI Unit
Displacement	$\mathbf{s}, \mathbf{x}$s,x	$\text{m}$m
Velocity	$\mathbf{v}$v	$\text{m s}^{-1}$m s−1
Acceleration	$\mathbf{a}$a	$\text{m s}^{-2}$m s−2
Force	$\mathbf{F}$F	$\text{N}$N
Momentum	$\mathbf{p}$p	$\text{kg m s}^{-1}$kg m s−1
Electric field strength	$\mathbf{E}$E	$\text{N C}^{-1}$N C−1
Magnetic flux density	$\mathbf{B}$B	$\text{T}$T
Gravitational field strength	$\mathbf{g}$g	$\text{N kg}^{-1}$N kg−1
Weight	$\mathbf{W}$W	$\text{N}$N

Note that weight is a vector — it is a force, and forces have direction (downward for weight on Earth). Conflating weight with mass is a classic Mid-band slip; mass is scalar, weight is vector.

Notation for vectors

Form	Example	Use
Bold	$\mathbf{v}$v	Textbooks, typeset documents
Arrow above	$\vec{v}$v	Handwritten work
Underline	$\underline{v}$v​	When bold typesetting is unavailable
Magnitude	$\|\mathbf{v}\|$∥v∥ or just $v$v (italic)	The scalar length of the vector

OCR accepts any clear notation. In handwriting, underline; in typed answers, bold. The key is consistency within a single answer.

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Concrete Worked Distinctions

Distance vs displacement — the walking example

A student walks $3\,\text{km}$3km east, then $4\,\text{km}$4km north.

• Distance = $3 + 4 = 7\,\text{km}$3+4=7km (the total path length — scalar).
• Displacement = the straight line from start to finish.

Magnitude:

$|\mathbf{s}| = \sqrt{3^{2} + 4^{2}} = \sqrt{25} = 5\,\text{km}.$∣s∣=32+42​=25​=5km.

Direction: $\arctan(4/3) = 53.1^{\circ}$arctan(4/3)=53.1∘ north of east. So $\mathbf{s} = 5\,\text{km}$s=5km at $53^{\circ}\text{N of E}$53∘N of E.

Speed vs velocity — the racetrack example

A car drives once around a $60\,\text{km}$60km circular racetrack in $1\,\text{h}$1h, ending at its starting point.

• Average speed = total distance / time = $60/1 = 60\,\text{km h}^{-1}$60/1=60km h−1.
• Average velocity = displacement / time = $0/1 = 0\,\text{km h}^{-1}$0/1=0km h−1 (the displacement is zero — start and end coincide).

A car that returns home has zero average velocity but non-zero average speed. This is not a trick — it is the precise statement of why velocity is a vector and speed is its magnitude.

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Adding Vectors: the Tip-to-Tail (Triangle) Method

To add vectors $\mathbf{A}$A and $\mathbf{B}$B, draw $\mathbf{A}$A from a starting point, then draw $\mathbf{B}$B starting from the tip of $\mathbf{A}$A (tip-to-tail). The resultant $\mathbf{R} = \mathbf{A} + \mathbf{B}$R=A+B is the arrow drawn from the start of $\mathbf{A}$A to the tip of $\mathbf{B}$B — the closing side of the triangle.

Properties of vector addition

• Commutative: $\mathbf{A} + \mathbf{B} = \mathbf{B} + \mathbf{A}$A+B=B+A. Drawing $\mathbf{B}$B first then $\mathbf{A}$A gives the same closing arrow.
• Associative: $(\mathbf{A} + \mathbf{B}) + \mathbf{C} = \mathbf{A} + (\mathbf{B} + \mathbf{C})$(A+B)+C=A+(B+C). Grouping is free.
• Parallelogram equivalent: drawing both vectors from a common origin, the resultant is the diagonal of the parallelogram they form. Same answer, different visualisation.

Worked Example 1 — perpendicular vectors

A boat sails at $3\,\text{m s}^{-1}$3m s−1 due east across a river flowing south at $4\,\text{m s}^{-1}$4m s−1. Find the resultant velocity (magnitude and direction relative to east).

Solution. The two vectors are perpendicular, so Pythagoras applies directly:

$|\mathbf{v}| = \sqrt{3^{2} + 4^{2}} = \sqrt{25} = 5\,\text{m s}^{-1}.$∣v∣=32+42​=25​=5m s−1.

Direction (measured from east towards south):

$\tan\theta = \frac{4}{3} \Longrightarrow \theta = \arctan\frac{4}{3} = 53.1^{\circ}.$tanθ=34​⟹θ=arctan34​=53.1∘.

The resultant is $5\,\text{m s}^{-1}$5m s−1 at $53^{\circ}\text{S of E}$53∘S of E.

Worked Example 2 — two perpendicular forces

A box on a frictionless surface is pulled by two ropes: $30\,\text{N}$30N north and $40\,\text{N}$40N east. Calculate the magnitude and direction of the resultant force.

Solution.

$|\mathbf{F}| = \sqrt{30^{2} + 40^{2}} = \sqrt{2500} = 50\,\text{N}.$∣F∣=302+402​=2500​=50N.

$\theta = \arctan\frac{30}{40} = \arctan(0.75) = 36.9^{\circ}\,\text{N of E}.$θ=arctan4030​=arctan(0.75)=36.9∘N of E.

Resultant: $50\,\text{N}$50N at $37^{\circ}\text{N of E}$37∘N of E.

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Adding Non-Perpendicular Vectors

When two vectors are not perpendicular, Pythagoras does not apply directly. There are three options:

1. Scale drawing with ruler and protractor — direct but limited in precision.
2. Resolve into components (next lesson) — adds in the $x$x- and $y$y-directions separately, then recombine. The universal A-Level tool.
3. Cosine rule on the general triangle — elegant but heavy algebraically.

The cosine rule method

For two vectors $\mathbf{A}$A and $\mathbf{B}$B with angle $\theta$θ between them (when drawn from a common origin),

$|\mathbf{R}|^{2} = |\mathbf{A}|^{2} + |\mathbf{B}|^{2} + 2|\mathbf{A}||\mathbf{B}|\cos\theta.$∣R∣2=∣A∣2+∣B∣2+2∣A∣∣B∣cosθ.

(The $+$+ sign comes from the geometry — the angle in the resulting triangle is $180^{\circ} - \theta$180∘−θ, and $\cos(180^{\circ} - \theta) = -\cos\theta$cos(180∘−θ)=−cosθ.) The direction is found via the sine rule.

Worked Example 3 — two forces at $60^{\circ}$60∘

Two forces, $10\,\text{N}$10N and $15\,\text{N}$15N, act with $60^{\circ}$60∘ between them. Find the magnitude of the resultant.

Solution.

$|\mathbf{R}|^{2} = 10^{2} + 15^{2} + 2 \times 10 \times 15 \times \cos 60^{\circ} = 100 + 225 + 150 = 475.$∣R∣2=102+152+2×10×15×cos60∘=100+225+150=475.

$|\mathbf{R}| = \sqrt{475} \approx 21.8\,\text{N} \approx 22\,\text{N}.$∣R∣=475​≈21.8N≈22N.

The component method (next lesson) gives the same answer with simpler arithmetic — and is more general.

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Scale-Drawing Method

For full marks on a scale-drawing question:

1. Choose a sensible scale and state it explicitly (e.g. $1\,\text{cm} = 2\,\text{N}$1cm=2N).
2. Draw the first vector with a ruler; set its direction with a protractor.
3. From the tip of the first vector, draw the second.
4. Draw the resultant from the start of the first to the tip of the second.
5. Measure the length and convert with the scale.
6. Measure the angle with the protractor.

State the answer to appropriate precision — typically 2 s.f. for scale-drawing, since the ruler resolution is $\sim 1\,\text{mm}$∼1mm on a $10\,\text{cm}$10cm vector ($1\%$1%).

Worked Example 4 — scale drawing

Three forces act on a point: $5\,\text{N}$5N east, $3\,\text{N}$3N at $60^{\circ}\text{N of E}$60∘N of E, $4\,\text{N}$4N north. Find the resultant by scale drawing.

Solution. Choose $1\,\text{cm} = 1\,\text{N}$1cm=1N. Draw:

1. $5\,\text{cm}$5cm east.
2. From its tip, $3\,\text{cm}$3cm at $60^{\circ}$60∘ above horizontal.
3. From its tip, $4\,\text{cm}$4cm due north.

Measuring the closing arrow: $\approx 10\,\text{cm}$≈10cm at $\approx 45^{\circ}\text{N of E}$≈45∘N of E. So $\mathbf{R} \approx 10\,\text{N}$R≈10N at $45^{\circ}\text{N of E}$45∘N of E.

A component calculation (next lesson) gives the more precise $9.3\,\text{N}$9.3N at $45.4^{\circ}\text{N of E}$45.4∘N of E — agreement to within the scale-drawing resolution.

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Subtracting Vectors

To subtract vector $\mathbf{B}$B from $\mathbf{A}$A, reverse $\mathbf{B}$B and add:

$\mathbf{A} - \mathbf{B} = \mathbf{A} + (-\mathbf{B}).$A−B=A+(−B).

A "minus" in front of a vector means "reverse the arrow" — equivalent to flipping the sign of every component.

Worked Example 5 — change in velocity (1D)

A ball moves east at $8\,\text{m s}^{-1}$8m s−1, bounces off a wall, and moves west at $6\,\text{m s}^{-1}$6m s−1. Find $\Delta\mathbf{v}$Δv.

Solution. Take east as positive. $v_1 = +8\,\text{m s}^{-1}$v1​=+8m s−1, $v_2 = -6\,\text{m s}^{-1}$v2​=−6m s−1.

$\Delta v = v_2 - v_1 = -6 - 8 = -14\,\text{m s}^{-1}.$Δv=v2​−v1​=−6−8=−14m s−1.

The change in velocity is $14\,\text{m s}^{-1}$14m s−1 westwards. This is more than either the initial or final speed — counterintuitive, but it is precisely why elastic bounces transmit such large forces. The magnitude of $\Delta\mathbf{v}$Δv can exceed the magnitudes of $\mathbf{v}_1$v1​ and $\mathbf{v}_2$v2​ when the directions reverse.

Worked Example 6 — change in momentum (2D)

A $0.20\,\text{kg}$0.20kg ball travels east at $5.0\,\text{m s}^{-1}$5.0m s−1 and, after being struck, travels at $6.0\,\text{m s}^{-1}$6.0m s−1 at $30^{\circ}\text{N of W}$30∘N of W. Find $|\Delta\mathbf{p}|$∣Δp∣.

Solution.

$\mathbf{p}_1 = m\mathbf{v}_1 = 1.0\,\text{kg m s}^{-1}\,\text{(east)}.$p1​=mv1​=1.0kg m s−1(east).

$\mathbf{p}_2 = m\mathbf{v}_2 = 1.2\,\text{kg m s}^{-1}\,\text{at}\,30^{\circ}\,\text{N of W}.$p2​=mv2​=1.2kg m s−1at30∘N of W.

Take east as $+x$+x, north as $+y$+y. Components of $\mathbf{p}_2$p2​:

$p_{2x} = -1.2\cos 30^{\circ} = -1.039, \qquad p_{2y} = +1.2\sin 30^{\circ} = +0.600.$p2x​=−1.2cos30∘=−1.039,p2y​=+1.2sin30∘=+0.600.

Components of $-\mathbf{p}_1$−p1​ (reverse the original east-pointing vector):

$(-p_1)_x = -1.0, \qquad (-p_1)_y = 0.$(−p1​)x​=−1.0,(−p1​)y​=0.

Sum to get $\Delta\mathbf{p} = \mathbf{p}_2 - \mathbf{p}_1 = \mathbf{p}_2 + (-\mathbf{p}_1)$Δp=p2​−p1​=p2​+(−p1​):

$\Delta p_x = -1.039 - 1.0 = -2.039\,\text{kg m s}^{-1}, \qquad \Delta p_y = +0.600\,\text{kg m s}^{-1}.$Δpx​=−1.039−1.0=−2.039kg m s−1,Δpy​=+0.600kg m s−1.

Magnitude:

$|\Delta\mathbf{p}| = \sqrt{(-2.039)^{2} + 0.600^{2}} = \sqrt{4.158 + 0.360} = \sqrt{4.518} \approx 2.13\,\text{kg m s}^{-1}.$∣Δp∣=(−2.039)2+0.6002​=4.158+0.360​=4.518​≈2.13kg m s−1.

Direction: $\arctan(0.600/2.039) = 16.4^{\circ}\text{N of W}$arctan(0.600/2.039)=16.4∘N of W. The bat delivered an impulse of $\approx 2.1\,\text{kg m s}^{-1}$≈2.1kg m s−1 at $16^{\circ}\text{N of W}$16∘N of W.

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Forces in Equilibrium

If three or more vectors sum to zero, they form a closed polygon when drawn tip-to-tail. This is the geometric statement of equilibrium: all forces on a body cancel.

For three forces, the polygon is a closed triangle. You can solve such problems by scale drawing, by resolving into components, or (rarely) by sine/cosine-rule arithmetic.

Worked Example 7 — three forces in equilibrium

Three forces act on a body at rest: $10\,\text{N}$10N west, $15\,\text{N}$15N at $30^{\circ}\text{E of N}$30∘E of N, and a third force $\mathbf{F}$F. Find $\mathbf{F}$F.

Solution. For equilibrium, the sum of components is zero. Take east as $+x$+x, north as $+y$+y.

$F_{1x} = -10, \quad F_{1y} = 0.$F1x​=−10,F1y​=0.

$F_{2x} = +15\sin 30^{\circ} = +7.5, \quad F_{2y} = +15\cos 30^{\circ} = +12.99.$F2x​=+15sin30∘=+7.5,F2y​=+15cos30∘=+12.99.

For equilibrium:

$F_x = -(F_{1x} + F_{2x}) = -(-10 + 7.5) = +2.5\,\text{N}\,\text{(east)}.$Fx​=−(F1x​+F2x​)=−(−10+7.5)=+2.5N(east).

$F_y = -(F_{1y} + F_{2y}) = -(0 + 12.99) = -12.99\,\text{N}\,\text{(south)}.$Fy​=−(F1y​+F2y​)=−(0+12.99)=−12.99N(south).

Magnitude:

$|\mathbf{F}| = \sqrt{2.5^{2} + 12.99^{2}} = \sqrt{175} \approx 13.2\,\text{N}.$∣F∣=2.52+12.992​=175​≈13.2N.

Direction: $\arctan(12.99/2.5) = 79.1^{\circ}\text{S of E}$arctan(12.99/2.5)=79.1∘S of E. So $\mathbf{F} \approx 13\,\text{N}$F≈13N at $79^{\circ}\text{S of E}$79∘S of E.

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Classification Table — Scalar vs Vector at a Glance

The most common Mid-band slip on Module 2.3 questions is mis-classifying a borderline quantity. The table below collects every A-Level quantity you may be asked to classify, with a one-line "why" for those students routinely confuse.

Quantity	Type	Why (where confusion arises)
Mass	Scalar	Quantity of matter; no preferred direction. Often confused with weight (vector).
Time	Scalar	A reading on a clock; no spatial direction.
Distance	Scalar	Total path length; no direction. Confused with displacement.
Displacement	Vector	Straight line from start to finish; direction matters.
Speed	Scalar	Magnitude of velocity; "$30\,\text{m s}^{-1}$30m s−1" alone.
Velocity	Vector	Speed with direction; "$30\,\text{m s}^{-1}$30m s−1 east".
Acceleration	Vector	Rate of change of velocity; even in uniform circular motion it is non-zero, pointing centripetally.
Force	Vector	Push or pull along a specific line of action.
Weight	Vector	A specific force — vertically downward; not the same as mass.
Momentum	Vector	$\mathbf{p} = m\mathbf{v}$p=mv; direction inherited from velocity.
Impulse	Vector	$\mathbf{J} = \Delta\mathbf{p}$J=Δp; direction inherited from momentum change.
Energy, work	Scalar	Scalar by definition; "work done against gravity" sounds directional but the directional information lives in the displacement and force separately.
Power	Scalar	Rate of energy transfer; no direction.
Temperature	Scalar	A thermodynamic state; no direction.
Electric charge	Scalar	A signed scalar (positive or negative), but not a vector. Sign is not direction.
Electric potential	Scalar	Energy per unit charge; scalar. Electric field $\mathbf{E}$E is the (vector) gradient of potential.
Electric field	Vector	Force per unit positive test charge; direction matters.
Magnetic flux density	Vector	$\mathbf{B}$B-field has direction (by the right-hand rule).
Gravitational field strength	Vector	$\mathbf{g}$g points toward the source mass.
Density	Scalar	Mass per unit volume; no direction.
Pressure	Scalar	Force per unit area; isotropic in a static fluid — the same magnitude in every direction at a point. The associated force on a surface element is a vector (perpendicular to the surface).
Frequency	Scalar	Cycles per second; no direction.

The borderline pairs that trip candidates most often:

• Speed vs velocity — speed is $|\mathbf{v}|$∣v∣; velocity is $\mathbf{v}$v. A car driving north at $30\,\text{m s}^{-1}$30m s−1 and one driving south at $30\,\text{m s}^{-1}$30m s−1 have the same speed but opposite velocities.
• Distance vs displacement — distance is the total length of the path; displacement is the straight-line vector from start to finish. The two coincide only when the motion is in a straight line.
• Mass vs weight — mass is a scalar property of the body (kilograms); weight is the gravitational force on it (newtons, vector, downward).
• Charge vs current — both are scalars at A-Level. The "direction" of current is a sign convention (conventional current flows in the direction positive charges would flow), but $I$I in a circuit equation is treated as a signed scalar, not a vector.
• Electric potential vs electric field — $V$V is a scalar (joules per coulomb); $\mathbf{E}$E is the vector field whose magnitude is $-\mathrm{d}V/\mathrm{d}x$−dV/dx. Mid-band candidates routinely call $V$V a "vector quantity" because of its association with $\mathbf{E}$E.
• Pressure vs force — pressure $p$p is a scalar (isotropic in a static fluid); the force $p\,\Delta\mathbf{A}$pΔA on a small surface element is a vector perpendicular to the surface.