NSSCAS Physics (8225) common mistakes
NSSCAS Physics (8225) · Understand the traps and how to avoid them.
When resolving a vector, the candidate labels the horizontal component as sin θ and the vertical as cos θ without checking which angle the diagram actually defines, then carries the swapped pair through the whole calculation.
The candidate memorizes 'vertical is cos' from the standard inclined-plane picture and applies it to every triangle, including ones where the angle is measured from the vertical or where the slope angle is at the top of the incline. The habit survives because the final numeric answer looks plausible.
What to do instead
Before substituting, redraw the right triangle and mark which side is adjacent to the given angle. Write the component relation explicitly as 'component along direction = magnitude × cos(angle between vector and that direction)', then substitute. Sanity-check with a limiting case: at θ = 0 the component along the surface must be zero and the perpendicular component maximal.
The candidate uses v = u + at, s = ut + ½at², and v² = u² + 2as on a motion where acceleration is not constant, such as a ball falling through a fluid reaching terminal velocity or a spring decelerating a trolley.
These equations are drilled so heavily in kinematics practice that they become the default response to any motion problem. The candidate does not pause to ask whether a is uniform, and the question often supplies u, v, s, and t, which makes a constant-a formula look like the intended route.
What to do instead
Make 'is a constant?' the first written step for every motion question. If acceleration varies, switch tools: gradient of a velocity–time graph for instantaneous acceleration, area under it for displacement, or energy methods (work done against drag = loss of kinetic plus potential energy) for terminal-velocity problems. Write the chosen principle before any numbers.
The candidate treats a two-body contact-force or connected-pulley problem by applying F = ma to the whole system and then assuming the tension or contact force equals the driving force, rather than solving each body separately.
The whole-system shortcut gives the correct acceleration quickly, and the candidate then stops, forgetting that internal forces cancel in the system equation and must be recovered from a single-body equation. The error is invisible because the acceleration is right.
What to do instead
After finding the system acceleration, always write one additional equation for a single body, e.g. T − m₁g = m₁a for the hanging mass, and solve for the internal force. Check the answer against limits: if the two masses are equal the tension should equal the weight, and if one mass is zero the tension should vanish.
The candidate resolves forces on an object on an incline and includes a force 'ma' or 'centripetal force' as an extra arrow on the free-body diagram, then double-counts it in the resultant equation.
Textbooks often draw 'the centripetal force' as a labelled arrow, so the candidate believes it is a separate physical force like weight or tension. The same confusion appears when a lift accelerates and the candidate adds a fictitious 'ma' arrow alongside the normal reaction.
What to do instead
Adopt the rule that a free-body diagram contains only forces exerted by identifiable objects (gravity from Earth, contact from a surface, tension from a string). The resultant of those forces equals ma; ma is never an arrow. When marking a diagram, name the object exerting each force before accepting it.
In a collision or explosion problem, the candidate calculates total kinetic energy before and after, finds it is not conserved, and concludes momentum is therefore not conserved either, or vice versa.
The two conservation laws are learned together and the candidate conflates them. Elastic and inelastic collisions are also taught as a single topic, so the candidate applies 'energy is always conserved' without distinguishing mechanical kinetic energy from total energy.
What to do instead
Separate the two statements explicitly in every solution: momentum is conserved in all collisions with no external resultant force; kinetic energy is conserved only in elastic collisions. For inelastic cases, compute the kinetic energy lost and account for it as heat, sound, or deformation. Never use energy conservation to find a velocity unless the collision is stated elastic.
The candidate calculates power as force × velocity using the average speed, or uses P = W/t with a time interval during which the force or speed changed, and reports a value that is neither the average nor the peak power.
The formula P = Fv is remembered without the condition that F and v must be instantaneous and parallel. Candidates also mix average and instantaneous quantities when a graph is provided, reading a gradient as if it were an average.
What to do instead
State whether the question asks for average or maximum power before choosing a formula. Use P = Fv only for instantaneous values with F parallel to v; for a changing force, find the gradient of a work–energy or energy–time graph. For average power, use total work divided by total time and confirm the work was actually done over that whole interval.
The candidate reads a stress–strain graph and treats the gradient of the whole curve as Young's modulus, or takes the area under the entire curve as the elastic strain energy, even beyond the limit of proportionality.
The linear region and the curved region look continuous, and candidates do not mark the limit of proportionality. The area under the curve is also taught as 'energy per unit volume' without the caveat that beyond yielding some energy is dissipated rather than stored elastically.
What to do instead
On every stress–strain graph, mark the limit of proportionality and the elastic limit first. Take Young's modulus only from the straight section's gradient. Use area under the curve up to the point of interest for energy density, and state that beyond the elastic limit the recovered energy is less than the area because of plastic deformation.
The candidate adds or subtracts sound intensities or amplitudes directly, for example adding two 60 dB sources to get 120 dB, or assuming doubling the amplitude doubles the intensity.
Decibels are logarithmic, but candidates treat them as ordinary linear quantities because the arithmetic looks like addition. The intensity–amplitude relationship is also frequently misremembered as proportional rather than squared.
What to do instead
Convert decibels to intensity ratios before combining sources: I/I₀ = 10^(L/10). Add intensities, then convert back with L = 10 log₁₀(I/I₀). Remember intensity is proportional to amplitude squared, so doubling amplitude quadruples intensity and adds about 6 dB. Check the answer is physically sensible: two equal sources give a 3 dB rise, not a doubling of the level.
When drawing a ray diagram or applying Snell's law at a boundary, the candidate measures the angle to the boundary rather than to the normal, or forgets that the frequency is unchanged on refraction and recalculates it in the second medium.
Diagrams often show the boundary prominently and the normal as a faint dashed line, so the eye picks the wrong angle. The frequency rule is stated once in lessons and then overshadowed by the wavelength and speed changes.
What to do instead
Draw and label the normal as the first action in any refraction problem, then mark the angles from it. Write the invariant list explicitly: frequency constant across a boundary; speed and wavelength change in proportion to 1/n. Use n₁ sin θ₁ = n₂ sin θ₂ and check that the ray bends towards the normal when entering the higher-index medium.
The candidate analyses a circuit by treating a parallel branch as if it carried the total current, or assumes the current divides equally between parallel resistors of different resistance.
Series and parallel rules are memorized as slogans ('current is the same in series') without the underlying conservation of charge and Kirchhoff's laws, so the candidate cannot reconstruct the rule in an unfamiliar circuit with mixed combinations.
What to do instead
Reduce mixed networks in stages, redrawing after each step and labelling the current in every branch. Apply Kirchhoff's junction rule explicitly at each node and the loop rule around each loop. For parallel branches, use the inverse-resistance rule and verify that the larger current flows through the smaller resistance and that branch currents sum to the total.
The candidate calculates the e.m.f. induced in a rotating coil or a rod moving in a field using the peak flux linkage rather than the rate of change of flux linkage, and omits the number of turns N.
Faraday's law is quoted as 'e.m.f. equals flux', losing the 'rate of change' and the N. In sinusoidal cases the candidate also forgets that the average and peak e.m.f. differ by a factor involving π or √2.
What to do instead
Write Faraday's law in full as ε = −N dΦ/dt before substituting, and identify what is changing: B, area, or angle. For a coil rotating at angular frequency ω, use ε = NBAω sin(ωt) and distinguish peak from r.m.s. For a rod, use ε = Bℓv with v perpendicular to both B and the rod. State which quantity the question wants.
The candidate applies the ideal gas equation or a thermodynamic relation without converting temperature to kelvin, or uses Celsius differences and absolute temperatures interchangeably in pV = nRT.
Laboratory data are recorded in degrees Celsius, and the candidate substitutes them directly because the numbers look reasonable for pressures and volumes in SI units. The error is systematic and produces answers that are wrong by a factor of several hundred.
What to do instead
Make kelvin conversion the first line of any thermal calculation: T/K = θ/°C + 273.15. Note that temperature differences are numerically the same in kelvin and Celsius, but absolute temperatures in pV = nRT and in efficiency expressions must be in kelvin. Check the order of magnitude: room temperature is about 300 K, not 20.
The candidate reports a calculated quantity to more significant figures than the data justify, or quotes an uncertainty as a percentage of a value without stating the unit and confidence level.
Calculators return many digits and candidates equate precision with correctness. Uncertainties are often tacked on as an afterthought rather than propagated from the raw readings, so the quoted figure is inconsistent with the measurements.
What to do instead
Decide the significant figures from the least precise input before writing the final answer, and give the absolute uncertainty to one significant figure with the value rounded to match. Propagate uncertainties through the actual operations: add absolute uncertainties for sums and differences, add percentage uncertainties for products and quotients. State the unit and, where relevant, the confidence level.
When asked to improve an experiment, the candidate writes 'repeat the experiment and take an average' as the sole suggestion, without identifying a specific source of error or a change to the apparatus or procedure.
This phrase is learned as a universal answer and is accepted in lower-level marking, so the candidate never learns to diagnose the dominant uncertainty. It also fails to distinguish random from systematic error.
What to do instead
For each experiment, name the dominant uncertainty and classify it. For random error, propose repeats, a wider range, or a more sensitive instrument. For systematic error, propose a zero check, a calibration against a known value, or a method change such as using a fiducial mark or subtracting a background reading. Link the improvement to the specific measurement it affects.