Unit 2: Linear motion and waves

QLDPhysicsSyllabus dot point

Topic 1: Linear motion and force

Distinguish between scalar and vector quantities, including identifying examples and applying operations of addition and subtraction in one and two dimensions

A focused answer to the QCE Physics Unit 2 dot point on scalar and vector quantities. Defines the distinction with examples, walks through vector addition (head-to-tail and component methods), subtraction as adding the opposite, and the standard QCAA component resolution students use throughout Unit 2 motion and Unit 3 fields.

Generated by Claude OpusReviewed by Better Tuition Academy6 min answer

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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

Type Definition Examples
Scalar Magnitude only mass, time, distance, speed, energy, temperature, work, power
Vector Magnitude and direction displacement, velocity, acceleration, force, momentum, impulse

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 v\vec{v} of magnitude vv at angle θ\theta above the horizontal has components:

vx=vcosθ,vy=vsinθv_x = v \cos\theta, \quad v_y = v \sin\theta

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

Subtracting vectors

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

Worked example

A car is travelling east at 2020 m s1^{-1} then turns and travels north at 2020 m s1^{-1}. Find the change in velocity.

vi=20ı^\vec{v}_i = 20\hat{\imath} (east), vf=20ȷ^\vec{v}_f = 20\hat{\jmath} (north).

Δv=vfvi=20ı^+20ȷ^\Delta \vec{v} = \vec{v}_f - \vec{v}_i = -20\hat{\imath} + 20\hat{\jmath}.

Magnitude: Δv=202+202=28.3|\Delta \vec{v}| = \sqrt{20^2 + 20^2} = 28.3 m s1^{-1}.

Direction: 45°45° north of west.

The speed did not change but the velocity did. This is exactly the source of centripetal acceleration in circular motion, which you will meet in Unit 3.

Common traps

Calling speed and velocity the same thing. Speed is the magnitude of velocity. They have the same SI unit but velocity is a vector. A constant-speed object turning a corner has changing velocity.

Adding magnitudes when directions differ. A 33 m walk east followed by 44 m north is a 77 m total distance but a 55 m displacement. Magnitudes do not add unless the vectors are collinear.

Forgetting to specify a reference for direction. "Magnitude 55 N at 30°30°" is ambiguous. State the reference axis (30°30° above the horizontal, 30°30° east of north).

Mixing degrees and radians on a calculator. Trig functions return different values depending on calculator mode. Set degrees for QCAA Physics unless a problem explicitly uses radians.

In one sentence

A scalar has only magnitude (mass, time, distance, speed, energy); a vector has magnitude and direction (displacement, velocity, acceleration, force, momentum) and is added or subtracted graphically (head to tail) or by resolving into perpendicular components.

Past exam questions, worked

Real questions from past QCAA papers on this dot point, with our answer explainer.

Year 11 SAC4 marksA hiker walks $4.0$ km east, then $3.0$ km north. Calculate (a) the total distance travelled and (b) the magnitude and direction of the resultant displacement.
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(a) Total distance is a scalar sum along the path: 4.0+3.0=7.04.0 + 3.0 = 7.0 km.

(b) Resultant displacement is the vector from start to finish. With the legs at right angles, use Pythagoras and trig.

Magnitude: s=4.02+3.02=25=5.0|s| = \sqrt{4.0^2 + 3.0^2} = \sqrt{25} = 5.0 km.

Direction: θ=tan1(3.0/4.0)=36.9°\theta = \tan^{-1}(3.0 / 4.0) = 36.9° measured north of east.

Markers reward the explicit distinction between scalar distance and vector displacement, the right-angle Pythagoras step, and a direction given as an angle from a stated reference axis.

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