AP Inter Physics (BIEAP) common mistakes
AP Inter Physics (BIEAP) · Understand the traps and how to avoid them.
When computing percentage error in a derived quantity such as density from mass and volume, the candidate adds the percentage errors of mass and volume directly even though the formula involves division (ρ = m/V).
The candidate memorises 'errors always add' without applying the rule that for division the fractional errors add, and for powers they are multiplied by the exponent. They also forget that a volume computed from a length cubed carries three times the percentage error of the length.
What to do instead
Before substituting numbers, write the working formula, then apply the propagation rule symbolically: for z = A^p B^q / C^r, Δz/z = |p|ΔA/A + |q|ΔB/B + |r|ΔC/C. Practise by deriving the percentage error of density from a cube's side and mass, showing the factor 3 on the side explicitly before any arithmetic.
The candidate reports a final answer to five or six decimal places from a calculator, or writes a result with more significant figures than the least precise given data allows.
Calculator output is mistaken for physical precision, and the candidate has not internalised that a measurement quoted as 2.5 cm carries two significant figures. They also misapply the rule that leading zeros are not significant while trailing zeros after a decimal point are.
What to do instead
Adopt a fixed reporting routine: identify the data value with the fewest significant figures, round the final result to that count, and state the absolute uncertainty to one significant figure with the value rounded to the same decimal place. Drill by taking ten mixed measurements and writing each result in the form (value ± uncertainty) unit.
In projectile and relative-velocity problems, the candidate mixes horizontal and vertical equations, using a single velocity in both directions, or applies the range formula R = u²sin2θ/g to a case where launch and landing heights differ.
The candidate treats projectile motion as one-dimensional motion with a single acceleration, forgetting that horizontal acceleration is zero while vertical acceleration is g. The range formula is memorised without its condition of equal launch and landing heights.
What to do instead
Always resolve the initial velocity into u cosθ and u sinθ, then write the two equation sets separately: x = u cosθ·t with constant horizontal velocity, and y = u sinθ·t − ½gt² with vertical acceleration. Use the range formula only after checking the heights match; otherwise solve for time from the vertical equation first and substitute into the horizontal one.
The candidate draws a free-body diagram that omits a force, typically the normal reaction on an inclined plane, the tension in a string connecting two blocks, or the pseudo-force in a non-inertial frame, and then applies F = ma to the wrong body.
Forces are copied from a similar textbook diagram rather than derived from the contact and field interactions of the actual situation. Connected-body problems are attempted as a single system without deciding whether to analyse the system or an individual block.
What to do instead
Use a written checklist for every diagram: weight, each contact normal, each tension, friction with its direction opposing relative motion, and any applied force. For two-block problems, solve once for the system acceleration and once for the connecting tension by isolating one block, and label which body each equation refers to.
The candidate equates the work done by a variable force with force times displacement using the initial or final force value, and cannot compute work from the area under a force-displacement graph.
The constant-force formula W = Fs is over-applied because the candidate has not connected work to the integral of force over displacement. Graph-based questions are skipped because area counting feels unfamiliar.
What to do instead
For any force that varies, write W = ∫F dx and evaluate it, or find the area under the F–x curve by splitting it into rectangles and triangles. Practise with a spring force F = kx to show that the work is ½kx², the triangular area, not kx².
In SHM, the candidate uses the same value of acceleration, velocity or energy at every point of the motion, for example claiming maximum acceleration at the mean position or maximum velocity at the extreme.
The phase relationship between displacement, velocity and acceleration is memorised as a list of formulas without the picture of a body moving between extremes. Energy conversion between potential and kinetic is treated as if both peak together.
What to do instead
Anchor every SHM answer to the reference circle or the sine curve: displacement is zero at the mean and maximum at the extreme, velocity leads displacement by a quarter cycle, and acceleration is always opposite to displacement. Sketch displacement, velocity and acceleration curves on one set of axes for each problem before answering.
The candidate applies the mirror or lens formula with a sign convention chosen ad hoc, producing a virtual image on the wrong side or a wrong magnification sign, and misplaces the object distance as positive for a concave mirror.
The Cartesian sign convention is learned as a set of rules in isolation and abandoned under time pressure. The candidate also confuses the focal length sign of a concave mirror with that of a convex lens.
What to do instead
Fix one convention and state it at the top of the answer: distances measured from the pole in the direction of incident light are positive. Write the focal length sign explicitly for each element (concave mirror negative, convex lens positive) before substituting, and check that a virtual image emerges with a positive image distance on the same side as the object.
The candidate adds electric potentials or fields from two charges as scalars without direction, or computes the field at a point using the total charge rather than the field contributions of each charge.
Potential is a scalar and field is a vector, but the candidate treats both the same way. The inverse-square dependence on distance from each charge is dropped in favour of a single distance to the midpoint.
What to do instead
For fields, draw the field vector from each charge at the point, then add them as vectors with their angles; for potentials, add the scalar quantities kq/r with their signs. Solve one problem both ways for two equal charges to see why the field can vanish at the midpoint while the potential does not.
In current electricity problems, the candidate applies the terminal voltage as equal to the emf while current flows, or ignores internal resistance when finding the current in a circuit with a battery and external resistor.
The ideal-battery model from earlier circuit work is carried into problems where internal resistance matters, and the candidate does not distinguish emf from potential difference across the terminals.
What to do instead
Write the circuit equation in the form ε = I(R + r) and the terminal voltage as V = ε − Ir every time, then check the limiting cases: open circuit gives V = ε, and short circuit gives I = ε/r. Practise a two-cell circuit with one cell charging to see the sign of the internal drop.
The candidate answers questions on photoelectric effect by claiming the maximum kinetic energy of photoelectrons depends on the intensity of light, or states that there is no emission below the threshold frequency regardless of intensity.
Wave intuition suggests brighter light gives faster electrons, and the particle picture of photons with energy hν is not applied quantitatively. The candidate also confuses stopping potential with the kinetic energy in joules.
What to do instead
Write Einstein's equation hν = φ + K_max and the stopping-potential relation eV₀ = K_max, then reason from them: intensity changes the number of photons and hence the saturation current, while frequency changes the photon energy and hence K_max. Work through a graph of K_max against frequency to identify the slope as h and the intercept as −φ.