Galicia PAU Physics / Física 2º Bachillerato CIUG common mistakes
Galicia PAU Physics / Física 2º Bachillerato CIUG · Understand the traps and how to avoid them.
When asked to find the gravitational field at a point between two masses, the candidate computes the two field magnitudes, subtracts them, and reports a positive number without stating the direction, or adds them even though the two fields point the same way along the line.
Candidates memorise 'field is attractive, so subtract the magnitudes' and apply it as a rule of thumb instead of drawing the field vectors at the point. They also treat g as a scalar because most earlier problems only ask for its value.
What to do instead
Always draw the line joining the masses, mark the point, and draw one arrow per source pointing toward that source. Then assign a sign convention to the axis (e.g. rightward positive) and add signed components. Write the final answer as a vector or as 'magnitude X N/kg directed toward mass B'. Practise the three zones: outside both masses (fields oppose), between them (fields oppose only if the point is on the segment - they actually point in opposite directions there), and check a numerical case where the null point lies between the masses.
In orbital-mechanics questions the candidate equates gravitational force to centripetal force but substitutes the satellite's height above the surface for r, or uses g = 9.8 m/s² at orbital altitude.
The formula sheet gives F = GMm/r² and the candidate plugs in the radius of the Earth because it is the only radius they remember, forgetting that r is measured from the centre and equals R + h.
What to do instead
Write r = R_T + h explicitly as the first line of every orbital calculation and keep it symbolic until the final substitution. Derive v = sqrt(GM/r) and T = 2π sqrt(r³/GM) once by hand so the r-dependence is visible. Sanity-check: a low-orbit satellite (h ≈ 400 km) should give v ≈ 7.7 km/s and T ≈ 90 min; if your answer is off by a factor of several, you probably used R instead of R + h.
For a system of two or three point charges, the candidate computes the electric potential at a point by adding the magnitudes of the individual potentials, ignoring the sign of each charge, and then reports a positive potential even when the dominant charge is negative.
Candidates conflate potential (a signed scalar, V = kq/r) with field magnitude (always positive), and because they have just done a field problem they carry over the 'take the magnitude' habit.
What to do instead
Write each term with its charge sign included: V = k(q1/r1 + q2/r2 + ...). Do the arithmetic with signs and only take absolute values at the very end if a magnitude is requested. Contrast the two procedures side by side in your notes: field is a vector sum with directions, potential is an algebraic sum with signs. Check the result against intuition: a point closer to a negative charge must have negative potential.
When a charged particle enters a region with both an electric and a magnetic field, or when a wire moves in a magnetic field, the candidate applies the right-hand rule but mixes up which finger is which, or forgets that the magnetic force on a moving charge is qv×B and vanishes when v is parallel to B.
The three-finger rule is taught with different conventions in different textbooks, and candidates rarely practise cases where the angle between v and B is 0° or 90°.
What to do instead
Adopt one convention and write it down: index finger = v (or current I), middle finger = B, thumb = F for a positive charge; reverse the thumb for negative charges. Work through the three angles explicitly: θ = 0° and 180° give F = 0, θ = 90° gives maximum F = qvB. For a straight wire use F = IL×B and check the direction with the same rule, remembering that conventional current is opposite to electron flow.
In a simple-harmonic-motion problem the candidate writes x = A sin(ωt) but then uses the initial condition x(0) = A to solve for the phase, getting an inconsistent result, or confuses angular frequency ω with frequency f.
Candidates pick the sine form by default without checking whether the oscillator starts at the equilibrium position (sine) or at maximum displacement (cosine), and they treat ω and f as interchangeable because both are called 'frequency' in casual speech.
What to do instead
Before choosing the function, read the initial conditions: at t = 0, is x = 0 (use sin) or x = ±A (use cos)? Write ω = 2π/T = 2πf and keep the 2π visible until substitution. Verify by differentiating: v = dx/dt must give the correct initial velocity sign. A quick dimensional check (ω in rad/s, f in Hz = s⁻¹) catches the ω/f swap.
In geometric optics the candidate applies the thin-lens equation 1/f = 1/s' − 1/s with the wrong sign convention for a diverging lens or a virtual image, and reports a real image where the ray diagram shows a virtual one.
There are two common sign conventions (Cartesian and 'real-is-positive'), and the candidate switches between them mid-problem, especially when the lens is diverging and f is negative.
What to do instead
Pick the Cartesian convention and state it at the top of the page: distances measured from the lens in the direction of incident light are positive, so f < 0 for a diverging lens and s' < 0 for a virtual image. Always draw the three principal rays first and use the diagram to predict the sign of s' before calculating. If the algebra gives a sign that contradicts the diagram, recheck the substitution rather than the diagram.
In relativity problems the candidate uses the Galilean velocity addition v = v1 + v2 for speeds close to c, or applies the time-dilation formula with the proper time and the dilated time swapped.
Newtonian kinematics is practised for months and the relativistic formulas appear only in the final topic, so under exam pressure the candidate reverts to the familiar Galilean rule.
What to do instead
Write the Lorentz factor γ = 1/sqrt(1 − v²/c²) first and check whether v/c is large enough to matter (below about 0.1c the difference is under 1%). For velocity addition use u = (u' + v)/(1 + u'v/c²) and verify that the result never exceeds c. For time dilation, identify the proper time as the interval measured in the frame where the two events occur at the same place - the muon's lifetime in its own rest frame, for example - and label it τ₀ before substituting.
In photoelectric-effect and quantum problems the candidate mixes up the work function, the stopping potential, and the maximum kinetic energy, writing eV₀ = hf without subtracting the work function, or using the wavelength in nanometres directly in E = hc/λ.
The three quantities are all energies in electronvolts and the candidate loses track of which is which; unit conversion between nm and m is also skipped under time pressure.
What to do instead
Write the energy balance as a labelled chain: photon energy hf = work function Φ + maximum kinetic energy K_max, and K_max = eV₀. Convert λ to metres before using hc/λ, or use the shortcut E(eV) = 1240/λ(nm). Check that the threshold frequency f₀ = Φ/h is below the incident frequency, otherwise no emission occurs and the answer is zero.
In nuclear-decay problems the candidate balances mass number but not atomic number (or vice versa), or forgets that beta-minus decay increases Z by one while leaving A unchanged.
Alpha decay is drilled, but beta decay is treated as 'an electron comes out' without tracking the change in the nucleus, and the candidate does not write the full balanced equation.
What to do instead
Always write the complete equation with A and Z on both sides and check both sums separately. Memorise the three signatures: alpha (A − 4, Z − 2), beta-minus (A unchanged, Z + 1, antineutrino emitted), beta-plus (A unchanged, Z − 1, neutrino emitted), gamma (no change in A or Z). After balancing, verify that the parent mass exceeds the sum of daughter masses by the Q-value.
In the experimental-skills question the candidate reports a result as 3.14159 m from measurements taken to the nearest millimetre, or adds percentage errors arithmetically when quantities are multiplied or divided.
Candidates treat error propagation as a formula to memorise rather than a consequence of significant figures, and they confuse absolute and relative errors.
What to do instead
Apply the rules explicitly: for sums and differences add absolute errors; for products and quotients add percentage (relative) errors; for a power multiply the percentage error by the exponent. Round the final result to the same number of significant figures as the least precise input, and state the absolute uncertainty with one significant figure (two if the leading digit is 1). Write the answer as (value ± uncertainty) with units and check that the units of the uncertainty match the value.