IGCSE Physics 1.1 — Measurement Numericals (Solutions)

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IGCSE Physics 0625 · Unit 1.1 · Solutions (companion to the Question Bank)

1.1 Physical Quantities and Measurement Techniques — Solutions

Full worked solutions in the same order and labelling (B/I/D/P/M) as the Question Bank.

Basic (Recall)

B1 Core

Name the piece of apparatus most commonly used to measure the volume of a liquid.

✓A measuring cylinder
B2 Core

State why the period of a pendulum is usually found by timing many oscillations rather than a single one.

✓Because human reaction time introduces a roughly fixed error each time the stopwatch is started and stopped; spreading this fixed error across many oscillations, then dividing, makes its effect on the calculated period much smaller.
B3 Extended

Define a scalar quantity.

✓A scalar quantity has magnitude (size) only — no direction.
B4 Extended

Define a vector quantity.

✓A vector quantity has both magnitude and direction.
B5 Extended

State two examples of a vector quantity, other than force.

✓Any two of: velocity, acceleration, momentum, electric field strength, gravitational field strength, weight

Intermediate (Direct Application)

I1 Core

A student measures a pencil with a ruler. The left end aligns with the 2.3 cm mark and the right end aligns with the 15.8 cm mark. Calculate the length of the pencil.

1
Subtract the start reading from the end reading
15.8 − 2.3
✓Length = 13.5 cm
I2 Core

A stack of 60 identical sheets of cardboard is 45 mm thick. Calculate the thickness of one sheet.

1
Divide total thickness by the number of sheets
45 ÷ 60
✓Thickness of one sheet = 0.75 mm
I3 Extended

Classify each of the following as a scalar or a vector: energy, velocity, temperature, momentum, mass, gravitational field strength.

✓Energy — scalar; velocity — vector; temperature — scalar; momentum — vector; mass — scalar; gravitational field strength — vector
I4 Core

A student records the time for 25 oscillations of a pendulum as 40.0 s. Calculate the period of one oscillation.

1
Divide total time by the number of oscillations
40.0 ÷ 25
✓Period = 1.6 s
I5 Core

Explain why a length measurement should be read to the nearest half of the smallest division on the scale, using a millimetre-graduated ruler as your example.

✓Between two adjacent mm marks, the position of the object’s edge can be judged to about half a division by eye — e.g. reading 12.5 mm rather than rounding to 12 or 13 mm. This half-division estimate captures genuine extra precision the instrument can support, without overstating the accuracy beyond what the scale can actually justify.

Difficult (Multi-Step / Unfamiliar Context)

D1 Extended

Two forces act on a point object: 9 N acting due east and 12 N acting due north. Calculate the magnitude and direction of the resultant force.

1
Find the magnitude using Pythagoras’ theorem
R = √(9² + 12²) = √(81+144) = √225
2
Find the direction using tan θ = opposite/adjacent
tan θ = 12/9 → θ = tan⁻¹(1.33) = 53.1°
✓Resultant = 15 N, at 53.1° north of east
D2 Core

A student wants to determine the diameter of a ball bearing more precisely than a millimetre ruler alone would allow. Describe how the “measuring multiples” technique could be adapted for this purpose.

✓Line up a known number of identical ball bearings (e.g. 10) touching each other in a straight row against the ruler. Measure the total length of the row, then divide by the number of ball bearings to find the average diameter of one — this turns a small, hard-to-measure single diameter into a much larger, easily measured total length.
D3 Extended

An aircraft flies due north at 150 km/h through air that is itself moving due east at 30 km/h (a crosswind). Calculate the aircraft’s resultant speed and direction relative to the ground.

1
Find the magnitude using Pythagoras’ theorem
R = √(150² + 30²) = √(22500+900) = √23400
2
Find the direction using tan θ = opposite/adjacent
tan θ = 30/150 = 0.2 → θ = tan⁻¹(0.2) = 11.3°
✓Resultant ≈ 153.0 km/h, at 11.3° east of north
D4 Extended

Explain why two forces of equal magnitude, acting at right angles to each other, produce a resultant force that is less than the arithmetic sum of the two magnitudes.

✓Simple addition only applies when two vectors act along exactly the same line (the same direction). When two forces act at an angle to each other, only part of each force contributes toward the direction of the resultant — combining them via Pythagoras’ theorem (rather than simple addition) accounts for this partial contribution, and this always gives a magnitude smaller than the sum of the two individual magnitudes (except in the special case where one force is zero).
D5 Core

A student times 20 oscillations of a pendulum on three separate attempts, obtaining 36.4 s, 36.0 s, and 36.2 s. Calculate the best estimate for the period of the pendulum, and explain your method.

1
Average the three timings to reduce the effect of random timing error
(36.4 + 36.0 + 36.2) ÷ 3 = 36.2 s
2
Divide by the number of oscillations to find the period
36.2 ÷ 20
✓Period ≈ 1.81 s

Practical / Experimental Design

P1 Core

Describe an experiment to determine the average diameter of a marble, using a ruler and a number of identical marbles placed in a row. Identify one source of error in this method and suggest an improvement.

✓Method: Place a known number of marbles (e.g. 10) touching each other in a straight row. Measure the total length of the row with a ruler, viewed at eye level to avoid parallax error. Divide the total length by the number of marbles to find the average diameter of one.

Source of error: small gaps may exist between marbles if they aren’t pushed firmly together, making the measured total length (and hence the calculated diameter) too large.
Improvement: gently push the marbles together against a fixed block or wall before measuring, or use a straight-edged channel/groove to keep them aligned in a perfectly straight line with no gaps.

P2 Core

Describe an experiment to determine the period of a simple pendulum. Identify the independent variable, the dependent variable, and one variable that should be controlled, and explain why timing multiple oscillations improves accuracy.

✓Method: Set up a pendulum of a chosen length, and displace it through a small angle. Release it and use a stopwatch to time a fixed number of complete oscillations (e.g. 20), then divide the total time by 20 to calculate the period.

Independent variable: the length of the pendulum (if investigating how period depends on length).
Dependent variable: the period, T.
Controlled variable: the angle of release (amplitude) should be kept small and the same each time, since a large amplitude can affect the period.

Why time multiple oscillations: timing just one oscillation makes the fixed error from human reaction time a large fraction of the (short) measured time. Timing many oscillations means this same fixed error is divided across a much longer total time, making its effect on the calculated period much smaller and the result more accurate.

Mastery Test — Answer Key

  1. 17.2 cm (18.7 − 1.5)
  2. 1.55 g (62 ÷ 40)
  3. Distance — scalar; weight — vector; speed — scalar; acceleration — vector
  4. 1.8 s (27.0 ÷ 15)
  5. 13 N — R = √(5² + 12²) = √(25+144) = √169 = 13 N
  6. Reaction time introduces a roughly fixed error each time the stopwatch is started/stopped. Timing many oscillations spreads this same fixed error across a much longer total time, so once divided by the number of oscillations, its effect on the calculated period is much smaller.
  7. 1.7 m/s — R = √(1.5² + 0.8²) = √(2.25+0.64) = √2.89 = 1.7 m/s
  8. A digital timer removes human reaction-time error from starting and stopping the measurement (it can also give a more precise, easily read digital display).
  9. Speed is a scalar — it only states magnitude (e.g. “20 m/s”). Velocity is a vector — it states magnitude AND direction (e.g. “20 m/s due north”). Two objects can have the same speed but different velocities if moving in different directions.
  10. A ruler cannot be used to calculate the volume of an irregularly shaped object using a standard geometric formula, since it has no consistent, measurable dimensions. The displacement method instead finds the volume from the rise in liquid level when the object is submerged. Volume = 58 − 42 = 16 cm³.

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