DSE Physics

Course Overview

The Hong Kong Diploma of Secondary Education (DSE) Physics examination assesses candidates' understanding of fundamental physics principles and their ability to apply these principles to solve problems. The syllabus is designed to provide a balanced coverage of both classical and modern physics.

The course is structured into compulsory topics (Papers 1A and 1B) and elective topics (Paper 2). All candidates must study the compulsory topics; candidates then choose ONE elective topic from the four available options.

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Examination Structure

Paper	Component	Duration	Weighting
1A	Compulsory Part -- Multiple-Choice Section	1 h 30 m	30%
1B	Compulsory Part -- Structured Questions	2 h	50%
2	Elective -- Structured Questions	1 h	20%

Total: 100%

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Compulsory Topics

The compulsory syllabus is divided into the following major areas:

I. Mechanics

• Position, displacement, distance, speed, velocity, and acceleration
• Scalars and vectors
• Equations of uniformly accelerated motion (SUVAT)
• Newton's laws of motion
• Weight and mass
• Friction
• Inclined planes
• Projectile motion
• Momentum and impulse
• Work, energy, and power
• Conservation of energy and efficiency

See ./forces-and-motion for detailed notes.

II. Waves

• Longitudinal and transverse waves
• Wave parameters (amplitude, wavelength, frequency, period, wave speed)
• Wave equation $v = f\lambda$
• Reflection, refraction, and diffraction
• Electromagnetic spectrum
• Sound waves
• Standing waves and resonance
• Doppler effect

See ./waves-and-sound for detailed notes.

III. Electricity and Magnetism

• Electric charge, conductors, and insulators
• Electric fields and field lines
• Potential difference, current, and resistance
• Ohm's law ($V = IR$)
• Series and parallel circuits
• Electrical power and energy
• Kirchhoff's laws
• Magnetic fields
• Electromagnetic induction
• Transformers and AC circuits
• Domestic electricity

See ./electrical-circuits and ./magnetism-and-electromagnetism for detailed notes.

IV. Thermal Physics

• Temperature and thermometers
• Specific heat capacity
• Latent heat and specific latent heat of fusion and vaporization
• Gas laws (Boyle's law, Charles's law)
• Ideal gas equation
• Kinetic theory of gases

See ./heat-and-gases for detailed notes.

V. Atomic Physics

• Structure of the atom
• Isotopes
• Radioactivity (alpha, beta, and gamma radiation)
• Half-life and nuclear equations
• Background radiation
• Uses and dangers of radiation
• Nuclear fission and fusion
• Photoelectric effect (basic)
• Energy levels in atoms
• Emission and absorption spectra

See ./nuclear-physics for detailed notes.

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Elective Topics (Choose One)

Elective	Topic
E1	Astronomy and Space Science
E2	Atomic Physics (extended)
E3	Energy and Use of Energy
E4	Medical Physics

Paper 2 consists of structured questions based on the chosen elective. Each elective carries equal weighting (20% of the total mark).

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Assessment Objectives

Candidates are expected to demonstrate the ability to:

1. Recall and understand facts, terminology, principles, and relationships in physics
2. Apply and analyse physics knowledge to familiar and unfamiliar situations
3. Evaluate and synthesise information, form reasoned arguments, and draw conclusions
4. Plan and carry out investigations, interpret experimental data, and evaluate experimental methods

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Key Quantities and Units

Quantity	Symbol	SI Unit
Length	$l$	m
Mass	$m$	kg
Time	$t$	s
Temperature	$T$	K
Current	$I$	A
Amount of substance	$n$	mol
Luminous intensity	$I_v$	cd
Force	$F$	N
Energy	$E$	J
Power	$P$	W
Pressure	$p$	Pa
Electric charge	$Q$	C
Potential difference	$V$	V
Resistance	$R$	$\Omega$

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Worked Examples

Worked Example 1 -- Mechanics (SUVAT)

A car accelerates uniformly from rest to $20 \mathrm{ m/s}$ in $5 \mathrm{ s}$. Find the acceleration and the distance travelled.

Solution

Using $v = u + at$: $a = \frac{v - u}{t} = \frac{20 - 0}{5} = 4.0 \mathrm{ m/s}^2$

Using $s = ut + \frac{1}{2}at^2$: $s = 0 + \frac{1}{2}(4)(25) = 50 \mathrm{ m}$

Worked Example 2 -- Electricity (Circuit)

Two resistors of $6 \Omega$ and $12 \Omega$ are connected in series across a $9 \mathrm{ V}$ battery with negligible internal resistance. Find the current and the power dissipated in each resistor.

Solution

Total resistance: $R = 6 + 12 = 18 \Omega$

Current: $I = \frac{V}{R} = \frac{9}{18} = 0.5 \mathrm{ A}$

$P_6 = I^2 R_6 = 0.25 \times 6 = 1.5 \mathrm{ W}$

$P_{12} = I^2 R_{12} = 0.25 \times 12 = 3.0 \mathrm{ W}$

Worked Example 3 -- Waves (Wave Equation)

A sound wave of wavelength $0.68 \mathrm{ m}$ travels at $340 \mathrm{ m/s}$. Find the frequency and period.

Solution

$f = \frac{v}{\lambda} = \frac{340}{0.68} = 500 \mathrm{ Hz}$

$T = \frac{1}{f} = \frac{1}{500} = 0.002 \mathrm{ s} = 2.0 \mathrm{ ms}$

Worked Example 4 -- Energy (Conservation)

A $0.5 \mathrm{ kg}$ ball is dropped from a height of $10 \mathrm{ m}$. Find its speed just before it hits the ground, neglecting air resistance.

Solution

$mgh = \frac{1}{2}mv^2 \implies v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 10} = \sqrt{196.2} = 14.0 \mathrm{ m/s}$

Worked Example 5 -- Thermal Physics (Specific Heat)

A $2 \mathrm{ kg}$ block of aluminium (specific heat capacity 900 \mathrm{ J/(kg\cdot}^\circ C)}) is heated from $20^\circ\mathrm{C}$ to $100^\circ\mathrm{C}$. Find the energy required.

Solution

$Q = mc\Delta T = 2 \times 900 \times (100 - 20) = 2 \times 900 \times 80 = 144000 \mathrm{ J} = 144 \mathrm{ kJ}$

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Examination Tips

• Show all working in numerical calculations; method marks are often awarded even if the final answer is incorrect.
• Pay attention to significant figures; generally, answers should be given to the same number of significant figures as the least precise data in the question.
• Draw clear, labelled diagrams where appropriate; free-body diagrams should show forces with correct directions and relative magnitudes.
• For multiple-choice questions (Paper 1A), read each option carefully and eliminate clearly wrong answers before deciding.
• In structured questions (Paper 1B), use the space provided as a guide for the expected length of the answer.
• Memorise key definitions precisely, as examiners look for specific terminology in mark schemes.

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Problem Set

Problem 1. A stone is thrown vertically upward with initial speed $15 \mathrm{ m/s}$. Find the maximum height reached and the total time in the air.

Solution

Maximum height: using $v^2 = u^2 + 2as$ with $v = 0$:

$h = \frac{u^2}{2g} = \frac{225}{19.62} = 11.5 \mathrm{ m}$

Time to reach max height: $t = \frac{u}{g} = \frac{15}{9.81} = 1.53 \mathrm{ s}$

Total time in air: $2t = 3.06 \mathrm{ s}$

If you get this wrong, revise: Mechanics / SUVAT equations

Problem 2. A $5 \mathrm{ kg}$ object is pulled across a rough horizontal surface by a force of $40 \mathrm{ N}$ at $30^\circ$ above the horizontal. If the coefficient of kinetic friction is $0.4$, find the acceleration.

Solution

Normal force: $N = mg - F\sin 30^\circ = 49.05 - 20 = 29.05 \mathrm{ N}$

Friction: $f_k = 0.4 \times 29.05 = 11.62 \mathrm{ N}$

$a = \frac{F\cos 30^\circ - f_k}{m} = \frac{34.64 - 11.62}{5} = \frac{23.02}{5} = 4.60 \mathrm{ m/s}^2$

If you get this wrong, revise: Mechanics / Newton's Laws and Friction

Problem 3. Three resistors of $2 \Omega$, $3 \Omega$, and $6 \Omega$ are connected in parallel. A $6 \mathrm{ V}$ battery with internal resistance $0.5 \Omega$ is connected across the combination. Find the total current from the battery.

Solution

Parallel resistance: $\frac{1}{R_p} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 1 \implies R_p = 1 \Omega$

$I = \frac{\varepsilon}{R_p + r} = \frac{6}{1 + 0.5} = \frac{6}{1.5} = 4.0 \mathrm{ A}$

If you get this wrong, revise: Electricity and Magnetism / Series and Parallel Circuits

Problem 4. A radio wave has a frequency of $3 \times 10^8 \mathrm{ Hz}$. Find its wavelength. Which region of the EM spectrum does it belong to?

Solution

$\lambda = \frac{c}{f} = \frac{3.0 \times 10^8}{3.0 \times 10^8} = 1.0 \mathrm{ m}$

A wavelength of $1.0 \mathrm{ m}$ places this in the radio wave region of the EM spectrum.

If you get this wrong, revise: Waves / Electromagnetic Spectrum

Problem 5. A $3 \mathrm{ kg}$ trolley moving at $4 \mathrm{ m/s}$ collides with a $1 \mathrm{ kg}$ stationary trolley and they stick together. Find the velocity after collision and the kinetic energy lost.

Solution

Conservation of momentum: $3 \times 4 + 1 \times 0 = (3 + 1)v \implies v = 3.0 \mathrm{ m/s}$

Initial KE: $\frac{1}{2}(3)(16) = 24.0 \mathrm{ J}$

Final KE: $\frac{1}{2}(4)(9) = 18.0 \mathrm{ J}$

Energy lost: $24.0 - 18.0 = 6.0 \mathrm{ J}$

If you get this wrong, revise: Mechanics / Momentum and Collisions

Problem 6. A $200 \mathrm{ g}$ block of copper at $100^\circ\mathrm{C}$ is placed in $300 \mathrm{ g}$ of water at $20^\circ\mathrm{C}$. Find the final temperature, assuming no heat loss to the surroundings. (Specific heat capacity of copper = 390 \mathrm{ J/(kg\cdot}^\circ C)}, water = 4200 \mathrm{ J/(kg\cdot}^\circ C)})

Solution

Heat lost by copper = Heat gained by water:

$m_c c_c (100 - T_f) = m_w c_w (T_f - 20)$

$0.2 \times 390 \times (100 - T_f) = 0.3 \times 4200 \times (T_f - 20)$

$78(100 - T_f) = 1260(T_f - 20)$

$7800 - 78T_f = 1260T_f - 25200$

$33000 = 1338T_f$

$T_f = 24.7^\circ\mathrm{C}$

If you get this wrong, revise: Thermal Physics / Specific Heat Capacity

Problem 7. A coil of 100 turns and area $0.005 \mathrm{ m}^2$ is placed in a magnetic field of $0.4 \mathrm{ T}$. The coil is removed from the field in $0.02 \mathrm{ s}$. Find the average induced EMF.

Solution

$\Delta\Phi = BA = 0.4 \times 0.005 = 0.002 \mathrm{ Wb}$

$\varepsilon = N\frac{\Delta\Phi}{\Delta t} = 100 \times \frac{0.002}{0.02} = 100 \times 0.1 = 10.0 \mathrm{ V}$

If you get this wrong, revise: Electricity and Magnetism / Electromagnetic Induction

Problem 8. A sound source produces an intensity level of $70 \mathrm{ dB}$ at $3 \mathrm{ m}$. What is the intensity level at $12 \mathrm{ m}$ from the source?

Solution

Intensity follows inverse square law: $I_2 = I_1 \times (3/12)^2 = I_1/16$

$\beta_2 = 10\log_{10}(I_2/I_0) = 10\log_{10}(I_1/(16I_0)) = 10\log_{10}(I_1/I_0) - 10\log_{10}(16)$

$\beta_2 = 70 - 10 \times 1.204 = 70 - 12.0 = 58.0 \mathrm{ dB}$

If you get this wrong, revise: Waves / Sound Intensity

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Data Analysis Techniques for DSE Physics

Significant Figures

All final answers should be given to an appropriate number of significant figures (s.f.):

• Raw data is recorded to the precision of the measuring instrument.
• Calculated results should have the same number of s.f. as the least precise input value.
• Intermediate calculations should keep at least one extra s.f. to avoid rounding errors.

Examples:

Measurement	Significant Figures
$12.5 \mathrm{ cm}$	3
$0.0040 \mathrm{ kg}$	2 (leading zeros are not significant)
$1500 \mathrm{ m}$	Ambiguous (2, 3, or 4). Use scientific notation: $1.5 \times 10^3$ (2 s.f.)

Uncertainty and Error Analysis

Absolute uncertainty: The range within which the true value is expected to lie. e.g., $L = (25.0 \pm 0.5) \mathrm{ cm}$.

Percentage uncertainty: $\frac{\Delta x}{x} \times 100\%$.

Combining uncertainties:

For addition/subtraction: add absolute uncertainties.

$z = x + y \implies \Delta z = \Delta x + \Delta y$

For multiplication/division and powers: add percentage uncertainties in quadrature.

$z = x^a y^b \implies \frac{\Delta z}{z} = \sqrt{\left(a\frac{\Delta x}{x}\right)^2 + \left(b\frac{\Delta y}{y}\right)^2}$

Example: $A = \frac{1}{2}mv^2$ with $m = (0.200 \pm 0.002) \mathrm{ kg}$, $v = (3.0 \pm 0.1) \mathrm{ m/s}$:

$A = \frac{1}{2}(0.200)(3.0)^2 = 0.90 \mathrm{ J}$

$\frac{\Delta A}{A} = \sqrt{\left(\frac{0.002}{0.200}\right)^2 + \left(2 \times \frac{0.1}{3.0}\right)^2} = \sqrt{0.0001 + 0.0044} = \sqrt{0.0045} = 6.7\%$

$\Delta A = 0.067 \times 0.90 = 0.06 \mathrm{ J}$

$A = (0.90 \pm 0.06) \mathrm{ J}$

Graphical Analysis

Line of best fit: Draw a straight line that passes through as many data points as possible, with approximately equal numbers of points above and below the line.

Determining uncertainty from graphs:

1. Draw the line of best fit and calculate its gradient.
2. Draw the steepest and shallowest reasonable lines (worst-fit lines) through the error bars.
3. The uncertainty in the gradient is half the difference between the worst-fit gradients.

Linearising data: Many physical relationships are not linear. Plot transformed variables to obtain a straight line:

Relationship	Plot (y vs x)	Gradient	Intercept
$y = ax + b$	$y$ vs $x$	$a$	$b$
$y = ax^2$	$y$ vs $x^2$	$a$	$0$
$y = a/x$	$y$ vs $1/x$	$a$	$0$
$y = a\sqrt{x}$	$y$ vs $\sqrt{x}$	$a$	$0$
$v^2 = u^2 + 2as$	$v^2$ vs $s$	$2a$	$u^2$
$T^2 = \frac{4\pi^2}{g}L$	$T^2$ vs $L$	$\frac{4\pi^2}{g}$	$0$

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Practical Skills

Measuring Instruments

Instrument	Precision	Notes
Metre rule	$\pm 1 \mathrm{ mm}$	Parallax error possible
Vernier calipers	$\pm 0.05 \mathrm{ mm}$	For internal/external dimensions
Micrometer screw gauge	$\pm 0.01 \mathrm{ mm}$	Zero error must be checked
Stopwatch	$\pm 0.01 \mathrm{ s}$	Reaction time dominates; use electronic
Thermometer	$\pm 0.5^\circ\mathrm{C}$	Mercury or alcohol
Ammeter	$\pm$ half smallest division	Connected in series
Voltmeter	$\pm$ half smallest division	Connected in parallel
Protractor	$\pm 0.5^\circ$	Align carefully with normal

Systematic vs Random Errors

Type	Cause	Effect	Reduction
Systematic	Faulty instrument, zero error	All readings shifted same way	Calibrate, use different method
Random	Fluctuations, human judgement	Readings spread above and below	Repeat and average

Zero Error Correction

A zero error occurs when an instrument does not read zero when it should. Always check and correct:

Positive zero error: Instrument reads too high. Subtract the zero error from all readings.

Negative zero error: Instrument reads too low. Add the magnitude of the zero error to all readings.

Example: A micrometer reads $0.03 \mathrm{ mm}$ when fully closed (positive zero error). A measurement of $5.47 \mathrm{ mm}$ gives a corrected value of $5.47 - 0.03 = 5.44 \mathrm{ mm}$.

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Exam Technique for DSE Physics

Structured Questions

DSE Physics Paper 1 contains structured questions. Each question typically has 3-8 parts, progressing from basic recall to calculation and explanation.

Approach:

1. Read the entire question before starting. Later parts may give hints for earlier parts.
2. Show all working clearly. Method marks are awarded even if the final answer is wrong.
3. Use the correct number of significant figures in final answers (usually 3 s.f.).
4. Include units in every calculated answer.
5. For explanations, use precise physics terminology.

Common Error Patterns in DSE Exams

Error Type	Example	Correction
Unit conversion	$1 \mathrm{ cm} = 0.01 \mathrm{ m}$	Always convert to SI before calculating
Wrong formula	Using $v = f\lambda$ for standing waves	Check which formula applies to the situation
Sign errors	Forgetting negative sign in $v = u + at$	Define positive direction clearly
Confusing quantities	Speed vs velocity, mass vs weight	Check definitions before answering
Missing steps	Jumping from $F = ma$ to numerical answer	Show substitution clearly
Incomplete explanations	"Because of gravity"	State the principle and link to the scenario

Multiple Choice Strategy

DSE Physics Paper 2 consists of multiple choice questions. Key strategies:

1. Eliminate obviously wrong answers first.
2. Check dimensions: If the answer should be a force (N) and an option has units of energy (J), eliminate it.
3. Estimate: Quick mental arithmetic can often eliminate wrong options.
4. Check extreme cases: What happens if $m \to 0$, $R \to \infty$, etc.?
5. Do not leave blanks: There is no penalty for wrong answers in DSE.

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Problem Set (Mixed Topics)

Problem 9. A steel wire of length $2.0 \mathrm{ m}$ and cross-sectional area $2.0 \times 10^{-6} \mathrm{ m}^2$ is stretched by $1.0 \mathrm{ mm}$. The Young's modulus of steel is $2.0 \times 10^{11} \mathrm{ Pa}$. Calculate the tension in the wire and the elastic potential energy stored.

Solution

$\mathrm{Stress} = E \times \mathrm{strain} = 2.0 \times 10^{11} \times \frac{1.0 \times 10^{-3}}{2.0} = 1.0 \times 10^{8} \mathrm{ Pa}$

$F = \mathrm{Stress} \times A = 1.0 \times 10^{8} \times 2.0 \times 10^{-6} = 200 \mathrm{ N}$

$E_p = \frac{1}{2}Fx = \frac{1}{2} \times 200 \times 1.0 \times 10^{-3} = 0.10 \mathrm{ J}$

If you get this wrong, revise: Mechanics / Elasticity

Problem 10. A radioactive source has a half-life of $5.0 \mathrm{ years}$. The initial activity is $800 \mathrm{ Bq}$. Calculate the activity after $20 \mathrm{ years}$ and the time for the activity to fall to $50 \mathrm{ Bq}$.

Solution

After $20 \mathrm{ years}$: number of half-lives $= 20/5.0 = 4$

$A = A_0 \times \left(\frac{1}{2}\right)^4 = 800 \times \frac{1}{16} = 50 \mathrm{ Bq}$

For $A = 50 \mathrm{ Bq}$: $50 = 800 \times (1/2)^{t/5}$

$(1/2)^{t/5} = 50/800 = 1/16 = (1/2)^4$

$t/5 = 4 \implies t = 20 \mathrm{ years}$

(This is consistent: after 4 half-lives, the activity is $1/16$ of the original.)

If you get this wrong, revise: Nuclear Physics / Radioactive Decay

Problem 11. In a Young's double-slit experiment, the fringe spacing is $0.80 \mathrm{ mm}$ when light of wavelength $600 \mathrm{ nm}$ is used. The screen is $1.5 \mathrm{ m}$ from the slits. Find the slit separation.

Solution

$\Delta y = \frac{\lambda D}{d} \implies d = \frac{\lambda D}{\Delta y} = \frac{600 \times 10^{-9} \times 1.5}{0.80 \times 10^{-3}} = \frac{9.0 \times 10^{-7}}{8.0 \times 10^{-4}} = 1.125 \times 10^{-3} \mathrm{ m} = 1.13 \mathrm{ mm}$

If you get this wrong, revise: Waves / Interference

Problem 12. A convex mirror has a focal length of $15 \mathrm{ cm}$. An object of height $4.0 \mathrm{ cm}$ is placed $25 \mathrm{ cm}$ from the mirror. Find the image position, magnification, and nature.

Solution

For a convex mirror, $f = -15 \mathrm{ cm}$ (real-is-positive convention).

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \implies \frac{1}{v} = \frac{1}{-15} - \frac{1}{25} = -\frac{1}{15} - \frac{1}{25} = \frac{-5 - 3}{75} = -\frac{8}{75}$

$v = -9.38 \mathrm{ cm}$

The image is virtual ($v < 0$), $9.38 \mathrm{ cm}$ behind the mirror.

$m = -\frac{v}{u} = -\frac{-9.38}{25} = 0.375$

Image height: $h_i = 0.375 \times 4.0 = 1.50 \mathrm{ cm}$

The image is virtual, upright, and diminished.

If you get this wrong, revise: Optics / Curved Mirrors