HKDSE Physics common mistakes
HKDSE Physics · Understand the traps and how to avoid them.
When a trolley moves down a runway or a ball is thrown vertically, the candidate substitutes the given speed or the time of flight directly into s = ut + ½at² without first deciding which direction is positive, so a downward acceleration and an upward initial velocity are given the same sign and the displacement comes out wrong (often a negative distance reported as a positive answer).
HKDSE mechanics questions almost always ask for a vector quantity but the paper gives magnitudes only. Candidates memorise the four equations of uniformly accelerated motion as symbol-matching recipes and never write down a sign convention, so the sign of g or a is decided by habit (always positive) rather than by the chosen axis.
What to do instead
Before touching an equation, draw the motion on one line, mark the positive direction with an arrow, and write u, v, a, s, t with explicit signs beside the arrow. Practise a fixed drill: vertical throw up with up positive gives a = −9.81 m s⁻²; the same throw with down positive gives u negative. Check the sign of the answer against the sketch before writing it down.
On a block on an inclined plane or a lift problem, the candidate writes N = mg regardless of the situation, then uses f = μN with that value, so the friction force and hence the acceleration are wrong whenever the surface is tilted or accelerating vertically.
The normal reaction is taught early as 'the weight' and the free-body diagram is often skipped under time pressure. Candidates also confuse 'apparent weight' in a lift with true weight and cannot separate the forces when the lift accelerates.
What to do instead
Always draw a free-body diagram with every force labelled, then resolve perpendicular to the surface (or vertically for a lift) and apply ΣF = ma in that direction. On a slope of angle θ with no other vertical force, N = mg cos θ; in a lift accelerating upward at a, N = m(g + a). Use the value of N from that resolution, not mg, in f = μN.
In a collision or explosion question, the candidate applies conservation of kinetic energy to an inelastic collision, or applies conservation of momentum to a situation where an external force (friction, a wall, gravity during the impact) acts, and then reports a speed that violates one of the two laws.
Momentum and energy conservation are learned together and candidates treat them as interchangeable. They do not check the words 'perfectly elastic', 'sticks together', or 'the wall is fixed', and they forget that momentum is a vector while kinetic energy is a scalar.
What to do instead
Adopt a two-question test before calculating: (1) Is there a net external impulse during the interaction? If no, momentum is conserved - and use signed velocities. (2) Does the question say the collision is elastic or give 'they move off together'? Only then use ½mv², and only for an elastic collision can you equate total KE before and after. For a perfectly inelastic collision, calculate the KE lost and state it.
When asked why a gas pressure changes, the candidate writes 'the molecules move faster so pressure increases' without linking it to either the rate of collisions with the wall or the momentum change per collision, and when asked about a fixed mass of gas heated at constant volume, they use p₁V₁ = p₂V₂ and get no change.
Kinetic theory explanations are memorised as a phrase rather than as two separate mechanisms, and the gas laws are memorised as one combined equation without the conditions attached to each.
What to do instead
For any kinetic-theory explanation, write both links explicitly: pressure ∝ (number of collisions per unit area per unit time) × (average momentum change per collision). Then say which one changes. For the laws, write the condition next to each: Boyle (T constant), Charles (p constant), pressure law (V constant), and only use the combined form when the amount of gas is fixed.
In a specific heat capacity or latent heat calculation involving a mixture (hot metal dropped into water, or ice added to water), the candidate equates mcΔT for the two bodies but forgets the latent heat term for melting ice, or uses the final temperature as ΔT for the cold body and the initial temperature for the hot one.
Candidates treat 'heat lost = heat gained' as a single formula and do not list the stages a substance passes through. They also mix up 'temperature change' with 'temperature' when substituting into ΔT.
What to do instead
Before calculating, write the energy path of each body as a sequence of stages (e.g. ice at −10 °C: warm to 0 °C, melt at 0 °C, warm to T). Give each stage its own term (mcΔT or mL) and sum. For each term, write ΔT = (higher temperature − lower temperature) explicitly so the sign is never ambiguous.
On a ray diagram or a lens/mirror question, the candidate draws rays that do not obey the rules (a ray through the centre of a convex lens bent, or a ray parallel to the axis not passing through the focal point after refraction) and then reads a magnification off the diagram that contradicts the calculated value.
Ray-diagram rules are learned as pictures rather than as statements, and candidates draw from memory of a textbook diagram instead of applying the three standard rays. They also forget that a virtual image has a negative image distance in the lens formula.
What to do instead
Learn and always use exactly three rays: parallel-to-axis → through F; through optical centre → undeviated; through F → parallel to axis. Then verify with 1/f = 1/u + 1/v and m = v/u, remembering v is negative for a virtual image. If the diagram and the calculation disagree, trust the calculation and redraw.
In a stationary-wave or sound question, the candidate labels a node as an antinode (or vice versa), or states that the distance between adjacent nodes is one wavelength instead of half a wavelength, and then calculates the frequency from the wrong wavelength.
The pattern is memorised visually without the underlying superposition idea, and the relationship between the length of the string/air column and the wavelength depends on the boundary conditions (fixed–fixed, open–open, closed–open), which candidates do not distinguish.
What to do instead
Anchor on the definition: a node is a point of zero amplitude (destructive superposition), an antinode is maximum amplitude (constructive). The node-to-node distance is always λ/2. Then derive the length condition for the specific system: string fixed at both ends, L = nλ/2; pipe open at both ends, L = nλ/2; pipe closed at one end, L = (2n−1)λ/4. Write the condition down before substituting.
When a question asks for the current or potential difference in a circuit with two resistors in parallel and one in series with them, the candidate adds all three resistances, or applies the potential-divider formula across a parallel combination as if it were a single series resistor.
Candidates recognise the circuit shape but not the topology, and they apply a memorised formula without checking whether the components share the same current (series) or the same potential difference (parallel).
What to do instead
Redraw the circuit, colour-code each node, and mark which components are between the same pair of nodes - those are in parallel. Reduce the parallel block to one equivalent resistance first, then treat the result as a series element. Only use the potential divider V₁/V₂ = R₁/R₂ for two resistors in series carrying the same current.
On an electromagnetic induction question (a magnet dropped through a coil, or a coil rotated in a field), the candidate says 'a current is induced because the coil is in a magnetic field' instead of identifying the changing flux, and gets the direction of the induced current wrong by ignoring Lenz's law.
The statement 'a conductor in a magnetic field has an induced current' is a common over-generalisation; candidates do not connect induction to a change in flux linkage, and they treat Lenz's law as a slogan rather than a direction rule.
What to do instead
State the chain explicitly every time: flux linkage Φ = NBA cos θ changes → an e.m.f. is induced (Faraday) → the induced current opposes the change in flux (Lenz). To find the direction, ask what the induced current's own field must do (oppose the approach of a N pole, or oppose the fall in flux) and use the right-hand grip rule to get the current direction.
In a radioactive decay question, the candidate balances the mass number but not the atomic number (or vice versa) in a decay equation, and in a half-life problem they divide the initial activity by the number of half-lives instead of by 2 raised to that power.
The two conservation rules (nucleon number and proton number) are learned as one, and half-life is confused with a linear decrease. Candidates also forget that beta-minus decay increases the proton number by one while leaving the nucleon number unchanged.
What to do instead
Write every decay equation with both totals shown above and below the arrow and check them separately: nucleon numbers balance, proton numbers balance. For half-life, always write the number of half-lives n = t/t½ first, then activity or mass remaining = initial × (1/2)ⁿ. Practise one alpha and one beta equation side by side until the proton-number shift is automatic.