Myanmar Matriculation Physics common mistakes
Myanmar Matriculation Physics · Understand the traps and how to avoid them.
When a body moves on an inclined plane, the candidate resolves the weight into components using the angle of the incline incorrectly, writing the component along the plane as mg·cosθ and the normal component as mg·sinθ, then computes friction or acceleration from those swapped components.
The candidate memorises the pair 'sin and cos' without drawing the right-angled triangle in which the angle θ appears between the weight vector and the normal to the surface, so the geometric relation is never verified. Under exam time pressure the swap is not noticed because both expressions look plausible.
What to do instead
Before substituting numbers, always draw the free-body diagram and mark the angle θ at the base of the incline. Then state explicitly: the component of weight perpendicular to the plane is mg·cosθ (adjacent to θ) and the component along the plane is mg·sinθ (opposite to θ). Check two limiting cases: at θ = 0 the along-plane component must vanish (sin 0 = 0) and at θ = 90° the normal component must vanish (cos 90° = 0). If your formulas fail these limits, they are swapped.
The candidate uses v = u + at, s = ut + ½at², and v² = u² + 2as for motion with non-uniform acceleration, for example applying them to a body falling through a fluid where air resistance grows with speed, or to a motion whose velocity–time graph is curved.
The three kinematic equations are drilled as universal tools, and the condition 'constant acceleration' printed in the textbook is treated as a formality rather than a precondition. Candidates also confuse 'deceleration' with 'negative acceleration that changes', not realising constant deceleration is still uniform and the equations do apply.
What to do instead
Before choosing an equation, ask one question: is the acceleration constant over the interval? If the velocity–time graph is a straight line, yes; if curved, no. For non-uniform cases use the graph area (distance) and gradient (acceleration), or energy methods (work done by gravity minus work done against resistance equals change in kinetic energy). Write 'a = constant' at the top of any solution that uses the three equations, so the assumption is visible and examinable.
In momentum and collision problems the candidate adds speeds of bodies moving in opposite directions, or forgets that momentum is a vector, writing m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ with all quantities taken as positive magnitudes regardless of direction.
Candidates treat momentum as a scalar because the formula 'p = mv' looks scalar, and they are used to adding magnitudes in energy calculations. Sign conventions are rarely written down, so a body rebounding after collision is given a positive velocity instead of a negative one.
What to do instead
At the start of every momentum problem, draw an arrow for each velocity, choose one direction as positive, and write each velocity with an explicit sign before substituting. For a ball rebounding off a wall, compute change of momentum as m(v_final − v_initial) with v_final negative, and state that the impulse is directed away from the wall. Cross-check with kinetic energy: if kinetic energy is unchanged in a perfectly elastic collision, the relative speed of approach must equal the relative speed of separation.
When calculating the efficiency or coefficient of performance of a heat engine or refrigerator, the candidate uses temperatures in degrees Celsius directly in η = 1 − T₂/T₁ instead of converting to kelvin, or mixes Celsius and kelvin in the same ratio.
Thermodynamic efficiency formulas are often quoted with T, and candidates do not connect the absolute-temperature requirement to the second law. Because a ratio of Celsius temperatures sometimes gives a numerically similar answer, the error escapes detection.
What to do instead
Adopt a habit: write 'T(K) = θ(°C) + 273' at the top of every thermodynamics calculation and convert every temperature before substitution. Sanity-check the result: an efficiency must lie between 0 and 1, and a Carnot efficiency computed from Celsius values can exceed 1 or become negative, which immediately reveals the error. Also remember that the efficiency of a real engine must be less than the Carnot value between the same two reservoirs.
The candidate applies the ideal gas equation pV = nRT using pressure in cmHg or atm and volume in cm³ without converting to SI, or uses R = 8.31 J mol⁻¹ K⁻¹ with non-SI pressure and volume, producing answers wrong by powers of ten.
Gas problems in textbooks are often set in atmospheres and litres, and candidates copy the numbers without tracking units. The gas constant has several numerical values depending on units, and candidates do not realise R must match the unit system used.
What to do instead
Standardise on SI for gas calculations: pressure in Pa, volume in m³, temperature in K, n in mol, R = 8.31 J mol⁻¹ K⁻¹. Convert 1 atm = 1.01 × 10⁵ Pa and 1 litre = 10⁻³ m³ explicitly on the answer line. If a problem is naturally in atm and litres, either convert first or use R = 0.0821 L atm mol⁻¹ K⁻¹ and say so; never mix the two conventions in one calculation.
In simple harmonic motion the candidate confuses angular frequency ω with frequency f, substituting f directly into a = −ω²x or v = ω√(A² − x²), or uses T = 2π√(m/k) with the mass in grams.
The symbols ω and f look interchangeable in quick reading, and the relation ω = 2πf is treated as a definition to memorise rather than a conversion to apply. The period formulas for springs and pendulums are memorised without the SI requirement on m.
What to do instead
Write the conversion ω = 2π/T = 2πf explicitly before substituting into any SHM equation, and keep masses in kilograms. Check the dimensions: ω must have units of rad s⁻¹ (equivalently s⁻¹), so a quantity in Hz must be multiplied by 2π before entering a = −ω²x. For a spring–mass system, verify the period is independent of amplitude; if your answer changes when amplitude changes, the formula used is wrong.
When solving refraction problems the candidate applies Snell's law as n₁ sin θ₁ = n₂ sin θ₂ but measures the angles from the surface rather than from the normal, or uses the refractive index of the wrong medium as n₁.
Diagrams in textbooks show the angle to the normal, but in a quickly drawn sketch the candidate marks the angle between the ray and the glass surface because that is the angle visible in the drawing. The convention that all angles in optics are measured from the normal is not stated often enough.
What to do instead
In every ray diagram, draw the normal as a dashed line at the point of incidence first, then mark the angles between the ray and the normal. Label n₁ as the medium the light comes from and n₂ as the medium it enters, and write 'n₁ sin θ₁ = n₂ sin θ₂' with the subscripts matching the direction of travel. Check the direction of bending: entering a denser medium (larger n), the ray must bend towards the normal, so θ₂ < θ₁.
In two-source interference problems the candidate uses the path difference alone to decide between constructive and destructive interference, without checking whether the sources are in phase, or forgets to count the half-wavelength phase change on reflection in thin-film and Newton's-rings questions.
The rule 'path difference = nλ for maxima' is learned as unconditional, and the in-phase condition of the sources is not treated as an assumption. The reflection phase change is a subtle rule that is easily dropped because it does not appear in the basic formula.
What to do instead
State the source condition explicitly at the start: if the sources are coherent and in phase, maxima occur at path difference nλ and minima at (n + ½)λ; if they are in antiphase, swap the two. For thin films and Newton's rings, note whether the reflection at the second surface is from a medium of higher refractive index (giving an extra λ/2) and add it before applying the condition. Verify with the central fringe: in Newton's rings in reflected light the centre is dark, which confirms the half-wavelength shift has been included.
When finding the equivalent resistance of a network, the candidate adds series and parallel resistors by pattern-matching the drawing rather than by tracing the current path, for example treating two resistors as in series because they are drawn side by side, or applying 1/R = 1/R₁ + 1/R₂ to a branch that actually carries the same current throughout.
Circuit diagrams in exams are drawn in many layouts, and candidates recognise shapes rather than nodes. The definitions (same current for series, same potential difference for parallel) are known verbally but not applied to the diagram.
What to do instead
Redraw the circuit marking each node with a letter. Two components are in series if the same current must pass through both with no junction between them; they are in parallel if both are connected between the same pair of nodes. Then reduce the network step by step from the innermost parallel or series group, writing the reduced value at each stage. Check the final equivalent resistance is smaller than the smallest parallel resistor and larger than the largest series resistor.
In questions on the photoelectric effect or atomic energy levels, the candidate equates the photon energy hf directly to the maximum kinetic energy of the emitted electron, omitting the work function, or uses the photon wavelength in nanometres in E = hc/λ without converting to metres.
The photoelectric equation is often memorised as 'energy = hf' and the work function is treated as an optional extra. Nanometre wavelengths are common in exam data, and the conversion to metres is skipped because the numbers look manageable.
What to do instead
Always write the full equation hf = φ + K_max (or hc/λ = φ + eV_s) and identify each term before substituting. Convert λ to metres (1 nm = 10⁻⁹ m) and check the photon energy is greater than the work function, otherwise no emission occurs and K_max should be reported as zero, not negative. For energy levels, use hf = E₂ − E₁ with the energy difference in joules (or convert electron-volts with 1 eV = 1.6 × 10⁻¹⁹ J) and state which level is the upper one.