Scalars and vectors

Distinguishing Scalars and Vectors

A physical quantity is classified as either a scalar or a vector, depending on whether direction is needed to describe it fully. A scalar has only magnitude, while a vector has both magnitude and direction. This classification controls how the quantity is added, subtracted, and used in equations.

A scalar is fully described by a single number with a unit. Two scalars of the same type are combined by ordinary arithmetic. A vector requires both a numerical size and a direction, so two vectors of the same type can give very different resultants depending on their relative directions.

The syllabus expects you to recognise which quantities are vectors. A common pairing trap is distance/displacement and speed/velocity — the second of each pair is the vector.

MisconceptionA negative scalar does not mean “opposite direction.” Negative temperature or negative charge are signs of a property, not a vector direction. Only vectors carry directional meaning through their sign or arrow. Exam cue: Always ask “does this quantity need a direction to make sense?” before classifying it.

Adding and Subtracting Coplanar Vectors

Two or more vectors lying in the same plane combine to give a single resultant vector that produces the same physical effect as the originals together. Coplanar means all the vectors lie flat in one two-dimensional plane, so they can be drawn on a sheet of paper.

Key Equations

Resultant of two perpendicular vectors:

$$R=\sqrt{{A}^{2}+{B}^{2}}$$

Direction of resultant relative to vector A:

$$\theta ={tan}^{-1}\left(\frac{B}{A}\right)$$

Variables: $R$ — magnitude of resultant vector $A$, $B$ — magnitudes of the two perpendicular component vectors $\theta $ — angle between the resultant and vector A SI unit: depends on the quantity (e.g. N for force, m s⁻¹ for velocity) Proportionality: when $A$ and $B$ are equal, $R$ is $\sqrt{2}$ times either one; doubling both $A$ and $B$ doubles $R$.

Two graphical methods are used to add vectors. The tip-to-tail (triangle) method places the tail of the second vector at the tip of the first; the resultant is drawn from the tail of the first to the tip of the second. The parallelogram method places both vectors tail-to-tail; the resultant is the diagonal of the parallelogram drawn from the common tail.

For perpendicular vectors, the resultant is calculated using Pythagoras’ theorem and trigonometry. For non-perpendicular vectors, either resolve each vector into components first, or use a scaled drawing.

To subtract vector B from vector A, reverse the direction of B and then add it to A. So A − B = A + (−B), where −B has the same magnitude as B but points in the opposite direction. This is essential for finding change in velocity or change in momentum.

Examiner InsightWhen stating the resultant of vectors, always give both magnitude AND direction. A single number without direction earns at most half marks. State direction relative to a clear reference, such as “30° above the horizontal” or “north of east.” Exam cue: A vector answer without a direction is incomplete.

Worked Example: Change in Velocity

Scenario

A ball moving east at 6.0 m s⁻¹ strikes a wall and rebounds west at 4.0 m s⁻¹. Calculate the change in velocity.

Equation used — change in velocity as a vector subtraction

$$\Delta v={v}_{f}-{v}_{i}$$

Taking east as positive:

$${v}_{f}=-4.0 m {s}^{-1}$$

$${v}_{i}=+6.0 m {s}^{-1}$$

Working

$$\Delta v=(-4.0)-(+6.0)$$

$$\Delta v=-10 m {s}^{-1}$$

Answer

$$\Delta v=10 m {s}^{-1}$$

Interpretation

The negative sign shows the change in velocity is directed west, because the ball reversed direction and lost speed.

Resolving a Vector into Perpendicular Components

A single vector can be replaced by two perpendicular components that, when added together, reproduce the original vector exactly. This technique converts an awkward angled vector into two independent vectors along chosen axes, which is the standard approach for solving force, motion, and projectile problems.

Key Equations

Horizontal and vertical components of a vector V at angle θ to the horizontal:

$${V}_{x}=Vcos\theta $$

$${V}_{y}=Vsin\theta $$

Reconstruction of original vector from components:

$$V=\sqrt{{V}_{x}^{2}+{V}_{y}^{2}}$$

$$\theta ={tan}^{-1}\left(\frac{{V}_{y}}{{V}_{x}}\right)$$

Variables: $V$ — magnitude of the original vector ${V}_{x}$ — component along the chosen horizontal axis ${V}_{y}$ — component along the chosen vertical axis $\theta $ — angle between V and the horizontal axis SI unit: same as the original vector Proportionality: ${V}_{x}$ and ${V}_{y}$ are each directly proportional to $V$; ${V}_{x}$ is proportional to $cos\theta $ and ${V}_{y}$ is proportional to $sin\theta $.

The axes do not have to be horizontal and vertical — they only have to be perpendicular to each other. For an object on an inclined plane, the two perpendicular axes are usually chosen parallel and perpendicular to the slope, because that simplifies the analysis of motion along the slope.

When θ is measured from the horizontal:

• The component along the reference axis uses cos θ.
• The component perpendicular to the reference axis uses sin θ.

When θ is measured from the vertical, these roles swap. Always check which angle is given before applying cos or sin.

MisconceptionResolving a vector does not split it into two new physical vectors that act in addition to the original. The components replace the original; using all three together would double-count the effect. Exam cue: Use either the original vector OR its components, never both at once.

Worked Example: Resolving Weight on an Incline

Scenario

A 4.0 kg block rests on a frictionless slope inclined at 30° to the horizontal. Calculate the component of the block’s weight acting parallel to the slope (down the slope) and the component acting perpendicular to the slope.

Equation used — weight

$$W=mg$$

Given

$$m=4.0 kg$$

$$g=9.81 m {s}^{-2}$$

$$\theta =30^{\circ}$$

Working — Weight magnitude

$$W=4.0\times 9.81$$

$$W=39.24 N$$

Working — Component parallel to the slope (down the slope)

$${W}_{∥}=Wsin\theta $$

$${W}_{∥}=39.24\times sin30^{\circ}$$

$${W}_{∥}=19.62 N$$

Working — Component perpendicular to the slope (into the slope)

$${W}_{⟂}=Wcos\theta $$

$${W}_{⟂}=39.24\times cos30^{\circ}$$

$${W}_{⟂}=33.98 N$$

Answer

$${W}_{∥}=20 N$$

$${W}_{⟂}=34 N$$

Interpretation

The 20 N component along the slope causes the block to accelerate down the slope. The 34 N component into the slope is balanced by the normal contact force from the surface.