Physical quantities and measurement techniques

CORE VS EXTENDED GUIDE

• Core students study only the unlabelled sections.
• Extended students must study everything, including Extended Extended points.
• Extended = Core + Supplement.

Measuring Length and Volume

Rulers and measuring cylinders are the standard instruments for determining length and volume in the laboratory.

A ruler measures length in millimetres (mm) or centimetres (cm). Place the object so that one end aligns with the zero mark, then read the other end at eye level to avoid parallax error [the apparent shift in a reading when the eye is not directly in line with the scale]. The resolution of a standard ruler is 1 mm, so the smallest measurable change is 1 mm.

A measuring cylinder measures the volume of a liquid directly. Pour the liquid into the cylinder and read the bottom of the meniscus [the curved surface of a liquid in a tube] at eye level.

The SI unit of volume is m³, but in the laboratory cm³ and ml are commonly used (1 ml = 1 cm³).

For an irregularly shaped solid, submerge the object in water inside the measuring cylinder. The rise in the water level equals the volume of the solid — this is the displacement method.

MisconceptionStudents sometimes read the top of the meniscus instead of the bottom for water. Water curves downward because it is attracted to glass. Always read the bottom of the curved surface. Exam cue: If asked "how to reduce error," state "read the meniscus at eye level."

Measuring Time Intervals

Time intervals can be measured using analogue clocks, digital clocks, and digital timers, depending on the duration and precision required.

An analogue clock displays hours, minutes, and seconds using rotating hands. A digital clock or stopwatch displays time numerically, typically to a resolution of 0.01 s. For very short events, a digital timer triggered by an electronic signal (such as a light gate) removes human reaction time from the measurement.

When using a handheld stopwatch, human reaction time (approximately 0.1–0.3 s) introduces an uncertainty. This reaction time is significant when the interval being measured is short (a few seconds or less), because the error is a large fraction of the total time. For longer intervals, the same reaction time is a smaller fraction, so the percentage uncertainty decreases.

Examiner InsightCIE questions often ask why manual timing is inaccurate for short intervals. The expected answer references human reaction time as the source of error and explains that the error is a significant fraction of a small time interval. Exam cue: Always link reaction time to the size of the interval being measured.

Determining Averages by Measuring Multiples

Measuring multiples of a small distance or a short time interval and then dividing by the number of multiples reduces the effect of uncertainty on each individual measurement.

For a short distance, such as the thickness of a single sheet of paper, measure the thickness of many sheets stacked together (for example, 100 sheets) using a ruler, then divide the total measurement by the number of sheets. This gives a more accurate value for a single sheet because the ruler's resolution error is spread across all 100 measurements.

For a short time interval, such as the period of oscillation of a pendulum, time many complete oscillations (for example, 20) and divide the total time by the number of oscillations. 📌 The period is the time for one complete oscillation — from one extreme position, across to the other extreme, and back to the starting point. Timing 20 oscillations rather than 1 reduces the impact of human reaction time, because the reaction-time error is divided across 20 periods instead of affecting just one.

A fiducial marker [a fixed reference point used to judge when the pendulum passes a specific position] placed at the equilibrium (centre) position helps the observer judge exactly when each oscillation is complete. The pendulum moves fastest at this point, so it is the easiest position at which to start and stop the timing consistently.

MisconceptionStudents sometimes define one oscillation as the swing from one side to the other — this is only half an oscillation. One complete oscillation returns the pendulum to its starting position and direction. Exam cue: If asked to define a complete oscillation, state that the object returns to its original position moving in the same direction.

Examiner InsightCIE frequently asks "How does the student obtain an accurate value for the period?" The expected answer includes: time at least 20 oscillations, divide by the number of oscillations, and repeat to find a mean. Exam cue: Always include the word "divide" and a specific number of oscillations.

Worked Example — Finding the Period of a Pendulum

A student measures the time for 20 complete oscillations of a pendulum and records 18.6 s. Determine the period of one oscillation.

Finding the period

Equation used

$$T=\frac{t}{n}$$

where T = period (s), t = total time (s), n = number of oscillations.

Given

$$t=18.6\text{ s}$$

$$n=20$$

Substitution

$$T=\frac{18.6}{20}$$

$$T=0.93\text{ s}$$

Answer

$$T=0.93\text{ s}$$

Extended Scalars and Vectors

A scalar quantity has magnitude (size) only. A vector quantity has both magnitude and direction.

This distinction matters because two vectors of the same magnitude can produce different effects depending on their directions, while scalars are fully described by a single number and a unit.

The following common quantities must be classified correctly:

Distance vs Displacement: Distance is a scalar — it measures the total length of the path travelled. Displacement is a vector — it measures the straight-line distance from start to finish in a stated direction.

Speed vs Velocity: Speed is distance travelled per unit time (scalar). Velocity is speed in a stated direction (vector).

MisconceptionStudents often treat speed and velocity as interchangeable. A car travelling in a circle at constant speed has a changing velocity because its direction changes continuously. Exam cue: If a question asks about velocity, always include a direction in your answer.

Extended Resultant of Two Vectors at Right Angles

The resultant of two vectors is the single vector that has the same effect as the two original vectors acting together. When two vectors act at right angles, the resultant is found using Pythagoras' theorem for magnitude and trigonometry for direction.

Reading vector diagrams: An arrow represents a vector — the length of the arrow represents the magnitude (drawn to scale), and the direction of the arrow represents the direction of the vector. When two vectors act at right angles, they form two sides of a right-angled triangle, and the resultant is the hypotenuse.

Key Equations

Resultant magnitude (two vectors at right angles):

$$R=\sqrt{{F}_{1}^{2}+{F}_{2}^{2}}$$

Variables:

• R = resultant magnitude (N for forces, m s⁻¹ for velocities)
• F₁ = first vector component
• F₂ = second vector component (perpendicular to F₁)

SI unit: same as the component vectors

Direction of resultant:

$$tan\theta =\frac{{F}_{2}}{{F}_{1}}$$

Variables:

• θ = angle between the resultant and F₁

ProportionalityThe resultant magnitude increases as either component increases. Because of the squared terms, the resultant is always less than the simple arithmetic sum of the two components (unless they act in the same direction).

Graphical method: Draw the two vectors to scale at right angles, tip-to-tail. The resultant is the straight line from the start of the first vector to the tip of the second vector. Measure this line with a ruler and convert using the scale. Measure the angle with a protractor.

Worked Example — Resultant of Two Forces at Right Angles

A box is pushed with a horizontal force of 3.0 N to the east and a vertical force of 4.0 N to the north. Determine the magnitude and direction of the resultant force.

Taking east as the reference direction for the angle.

Finding the resultant magnitude

Equation used

$$R=\sqrt{{F}_{1}^{2}+{F}_{2}^{2}}$$

Given

$${F}_{1}=3.0\text{ N (east)}$$

$${F}_{2}=4.0\text{ N (north)}$$

Substitution

$$R=\sqrt{{3.0}^{2}+{4.0}^{2}}$$

$$R=\sqrt{9.0+16.0}$$

$$R=\sqrt{25.0}$$

$$R=5.0\text{ N}$$

Finding the direction

Equation used

$$tan\theta =\frac{{F}_{2}}{{F}_{1}}$$

Substitution

$$tan\theta =\frac{4.0}{3.0}$$

$$\tan\theta = 1.333\ldots$$

$$\theta = \tan^{-1}(1.333\ldots)$$

$$\theta =53.1^{\circ}$$

Answer

$$R=5.0\text{ N at }53^{\circ}\text{ north of east (2 s.f.)}$$

Examiner InsightCIE questions require both the magnitude and the direction of the resultant. Students who give only the magnitude lose a mark. Exam cue: Always state the angle relative to one of the original vectors, with a compass or reference direction.