STPM Physics 960 common mistakes
STPM Physics 960 · Understand the traps and how to avoid them.
When combining measured quantities, adding the absolute uncertainties of quantities that are multiplied or divided (for example writing the uncertainty of V = IR as ΔV = ΔI + ΔR in volts) instead of combining fractional uncertainties.
Candidates memorise 'add uncertainties' from addition/subtraction rules and never distinguish the two cases; textbooks state the rule once and students default to the simpler operation. The same slip appears when a quantity is raised to a power and the fractional uncertainty is not multiplied by the exponent.
What to do instead
Before computing, write the algebraic form of the result (e.g. V = IR, T = 2π√(l/g)) and tag each variable as 'added/subtracted' or 'multiplied/divided/powered'. Use absolute uncertainties only for sums and differences; for products, quotients and powers convert each to a percentage uncertainty, multiply by the power, add the percentages, then convert back at the end. Practise with T = 2π√(l/g) so that ΔT/T = ½(Δl/l) + ½(Δg/g) becomes automatic.
Reading a vernier caliper or micrometer screw gauge to the wrong precision, or quoting a final answer with more significant figures than the instrument justifies (for example recording 2.34567 cm from a metre rule).
Candidates confuse the least count of the instrument with the number of decimal places the instrument displays, and they carry the full calculator display into the answer because they believe more digits signal a better answer. Zero error is also forgotten because it is a one-off subtraction that is easy to skip under time pressure.
What to do instead
Fix the reading rule in writing: metre rule to 0.1 cm (estimate one extra digit), vernier to 0.01 cm, micrometer to 0.001 mm, and always state the zero error and apply it before use. Round the final answer to the same number of significant figures as the least precise measurement, and write the uncertainty to one significant figure with the value rounded to match. Do ten timed readings of the same object and check that every one is quoted identically.
Resolving a vector by using the wrong trigonometric ratio or by placing the angle in the wrong triangle, so that a 30° incline produces a component mg cos 30° along the slope.
Candidates memorise 'sin for vertical, cos for horizontal' without drawing the triangle, and they do not check which angle in the diagram is actually given. The error survives because the two ratios differ by only a factor near 1 at small angles.
What to do instead
Always draw the vector, the two perpendicular axes and the angle between the vector and the chosen axis before writing any component. Then state the rule aloud: component along the axis = magnitude × cos(angle to that axis), perpendicular component = magnitude × sin(angle to that axis). Test each resolution with a limiting case (angle 0° should give the full component along the axis, angle 90° should give zero).
Treating a body moving in a circle at constant speed as having zero acceleration, and then applying only the centripetal force while ignoring the tangential weight component on a vertical circle.
Candidates equate 'constant speed' with 'constant velocity' and 'no acceleration', and the circular-motion formula a = v²/r is applied as a number without a direction. On vertical circles the tension is assumed constant because the speed is constant, which is physically impossible once gravity acts.
What to do instead
State explicitly that velocity changes direction every instant, so there is a centripetal acceleration of magnitude v²/r directed towards the centre. For vertical circles, resolve along the radius at the top and bottom separately and write T + mg = mv²/r at the top and T − mg = mv²/r at the bottom, then check that the string tension cannot be negative. Solve one full vertical-circle problem from first principles each revision session.
Using the wrong sign convention in simple harmonic motion, for example writing a = +ω²x, or forgetting that at maximum displacement the velocity is zero while the acceleration is maximum.
The restoring force is directed opposite to the displacement, but candidates copy the magnitude relation a = ω²x and drop the minus sign, then cannot interpret graphs of x, v and a against time. Phase differences between the three quantities are memorised as pictures rather than derived.
What to do instead
Derive the relation from the defining condition: acceleration is proportional to displacement and directed towards equilibrium, hence a = −ω²x. From this, note that when x is maximum, a is maximum in the opposite direction and v = 0; when x = 0, a = 0 and v is maximum. Sketch the three graphs on the same axes and label the phase differences (v leads x by π/2, a leads x by π).
Applying the ideal gas equation pV = nRT with pressure in cmHg or volume in cm³, or using the wrong value of R for the units chosen.
Candidates substitute numbers straight from the question without converting to SI, and they do not check that R = 8.31 J mol⁻¹ K⁻¹ requires p in Pa and V in m³. Temperature in degrees Celsius is also inserted directly into the equation.
What to do instead
Before substituting, write a conversion line for every quantity: cmHg to Pa (multiply by 1333), cm³ to m³ (divide by 10⁶), °C to K (add 273). Then write the equation with units attached and cancel them. If the question is about ratios (p₁V₁/T₁ = p₂V₂/T₂), keep the units consistent on both sides and R cancels, which is a useful check.
Describing the first law of thermodynamics with the wrong sign for work done by and on the gas, so that an isothermal compression is said to have zero heat transfer.
Two conventions exist (ΔU = Q + W and ΔU = Q − W), and candidates mix them within one answer. Isothermal is also confused with adiabatic because both involve 'no temperature change' in the student's memory, despite only one having Q = 0.
What to do instead
Adopt one convention for the whole paper and write it at the top of your working: ΔU = Q − W, where W is work done by the gas. Then classify each process explicitly: isothermal means ΔU = 0 so Q = W; adiabatic means Q = 0 so ΔU = −W; isochoric means W = 0 so ΔU = Q. Apply the classification before calculating, and check the sign against physical intuition (compressing a gas heats it).
Drawing a ray diagram or applying the diffraction grating equation without converting the diffraction angle to the correct order, or confusing the conditions for constructive and destructive interference in thin films and air wedges.
The path-difference conditions (nλ for constructive, (n + ½)λ for destructive) are memorised as a pair and then applied without checking whether a phase reversal occurs on reflection. For gratings, candidates use d sin θ = nλ but forget that d is the slit spacing, not the number of lines per metre.
What to do instead
For any interference problem, first list the sources of phase change: reflection at a denser medium adds π. Then write the total path difference and set it equal to nλ or (n + ½)λ accordingly. For gratings, convert lines per mm to d in metres (d = 1/N) before substituting. Sketch the geometry with the angle marked between the normal and the diffracted ray.
Assuming a capacitor in a DC circuit carries a steady current, or treating the time constant as the time for the capacitor to become fully charged.
Candidates transfer the resistor rules (V = IR, current constant) to capacitors and do not distinguish the transient charging phase from the steady state. The exponential nature of the discharge is memorised as a formula without meaning.
What to do instead
State that in steady state a capacitor blocks direct current, so the current through its branch is zero and it behaves as an open circuit; during charging and discharging the current decays exponentially. Identify the time constant τ = RC as the time to fall to 1/e (about 37%) of the initial value, not to zero. Practise sketching charge, current and voltage against time and marking τ on each curve.
In an operational amplifier circuit, applying the ideal op-amp rules (infinite gain, zero input current, V₊ = V₋) without checking whether negative feedback is present, or ignoring the saturation limits of the supply rails.
Candidates learn 'the two inputs are at the same potential' as a fact and apply it to comparator and Schmitt trigger circuits where there is no negative feedback. They also calculate an output of 15 V from a ±9 V supply because the rail voltage is not considered.
What to do instead
Before using V₊ = V₋, check that the output is connected back to the inverting input through a resistor or wire; if not, the op-amp is being used as a comparator and the output saturates at the supply rails. After calculating the output, always compare it with the supply voltage and state 'saturated at +9 V' if it exceeds the rail. Draw the feedback path explicitly on every circuit diagram.