Scalars and vectors (QCE Physics Unit 2)

What this dot point is asking

QCAA wants you to separate the two families of physical quantity. Scalars are fully described by a magnitude and a unit. Vectors need a magnitude, a unit, and a direction. The same arithmetic rules do not apply to both. This distinction underpins every motion, force and field problem in Units 2, 3 and 4.

Scalars and vectors

Two scalars combine by ordinary arithmetic. Two vectors combine by vector addition, which depends on their directions.

Adding vectors graphically

Place the tail of the second vector at the head of the first. The resultant runs from the tail of the first to the head of the last. The order does not matter (vector addition is commutative).

For perpendicular vectors, magnitude follows Pythagoras and direction follows the inverse tangent. For non-perpendicular vectors, resolve into components or use the cosine rule.

Resolving into components

A vector $\vec{v}$ of magnitude $v$ at angle $\theta$ above the horizontal has components:

Components allow you to add and subtract vectors numerically. Sum the $x$-components, sum the $y$-components, then recombine with Pythagoras and arctan.

Subtracting vectors

$\vec{a} - \vec{b}$ means $\vec{a} + (-\vec{b})$. To subtract $\vec{b}$, reverse its direction and add. This step is critical for change-in-velocity problems (for example $\Delta \vec{v} = \vec{v}_f - \vec{v}_i$ in uniform circular motion or in a collision).

Examples in context

Example 1. A Sunshine Coast tidal current is reported by the buoy as $0.6 \text{ m s}^{-1}$ northeast. The component form is $0.6\cos 45^\circ = 0.424 \text{ m s}^{-1}$ east and $0.424 \text{ m s}^{-1}$ north. A drifter released into a $0.3 \text{ m s}^{-1}$ northward wind-driven current experiences a total velocity vector summed component-wise: $0.424 \text{ m s}^{-1}$ east and $0.724 \text{ m s}^{-1}$ north, magnitude $0.84 \text{ m s}^{-1}$ at bearing $030^\circ$. QCAA EA Unit 2 data-test scenarios use exactly this oceanographic addition.

Example 2. A Cairns crane lifts a $400 \text{ kg}$ load while the boom swings horizontally at $1.5 \text{ m s}^{-1}$. The load's velocity is the vector sum of the boom's horizontal and the hoist's vertical: $\vec{v} = (1.5, 0.5) \text{ m s}^{-1}$ giving magnitude $1.58 \text{ m s}^{-1}$ at $18.4^\circ$ above horizontal. Distinguishing these vectors from the scalar quantity speed ($1.58 \text{ m s}^{-1}$ without direction) is the kind of QCAA-style multi-mark answer markers expect.

Try this

Q1. Distinguish scalar from vector quantities and give two examples of each. [2 marks]

• Cue. Scalar: magnitude only (speed, energy); vector: magnitude and direction (velocity, force).

Q2. A swimmer crosses a river at $1.2 \text{ m s}^{-1}$ perpendicular to the bank; the current is $0.5 \text{ m s}^{-1}$ downstream. Calculate the resultant velocity magnitude and bearing relative to the bank. [3 marks]

• Cue. Magnitude $\sqrt{1.2^2 + 0.5^2} = 1.30 \text{ m s}^{-1}$; angle $\arctan(0.5/1.2) = 22.6^\circ$ downstream of perpendicular.

Q3. A Sunshine Coast drifter experiences a $0.6 \text{ m s}^{-1}$ northeast current and a $0.3 \text{ m s}^{-1}$ north wind-driven drift. (a) Resolve each into east-north components. (b) Sum the components. (c) State the magnitude and bearing of the total velocity. [3+2+2 marks; ISMG: Analysis and interpretation]

• Cue. (a) NE: $(0.424, 0.424)$, wind: $(0, 0.3)$; (b) total $(0.424, 0.724)$; (c) $0.84 \text{ m s}^{-1}$ at $030^\circ$.