Canary PAU Physics common mistakes
Canary PAU Physics · Understand the traps and how to avoid them.
When computing gravitational potential energy or potential of a system of point masses, summing each pair once but forgetting the sign convention, or adding the potential energy of a mass with itself.
Students memorize U = -GMm/r and then apply a single-pair formula to a many-body configuration, losing track of which pairs exist and that gravitational potential energy is always negative (zero at infinity).
What to do instead
Before substituting numbers, draw the configuration and list every unordered pair of masses exactly once. Write U_total = -G·Σ(m_i·m_j/r_ij) and check the sign is negative. For potential (not energy), sum V = -G·Σ(m_i/r_i) over the sources only, never including the test point.
Equating gravitational field strength g at a height h to GM/(R+h)² but then using g = 9.8 m/s² in the same problem, or treating orbital speed as independent of radius.
Textbook surface value of g is over-learned and applied at altitudes where it is invalid; students also confuse angular velocity (which decreases with radius) with linear speed (which also decreases for circular orbits, v = √(GM/r)).
What to do instead
State explicitly whether the point is at Earth's surface or at altitude. If altitude is given, recompute g = GM/r² with r = R + h. For circular orbits, derive v = √(GM/r) and T = 2π√(r³/GM) from the centripetal-force balance each time rather than recalling them, and check that larger r gives smaller v and larger T.
Using Coulomb's law with r measured between surfaces of charged spheres, or adding electric field magnitudes from two charges without vector addition.
Point-charge formula is drilled, but extended objects and two-dimensional geometry are treated as scalar sums; students plug in the gap between spheres or add E values algebraically.
What to do instead
For spheres, use centre-to-centre distance and note that a uniformly charged shell behaves as a point charge at its centre outside the shell. For multiple charges, draw the field vectors at the point, resolve into components, and add vectorially; verify direction points away from positive and toward negative charges.
Treating electric potential as a vector and adding potentials with direction, or assuming zero field implies zero potential (or vice versa).
Potential energy and field are taught together, so students transfer vector rules to the scalar potential; the relationship E = -dV/dr is often memorized as a magnitude only.
What to do instead
Add potentials algebraically with signs (V = kΣq_i/r_i). Test the distinction with a concrete case: at the midpoint between equal opposite charges, V = 0 but E ≠ 0; at the midpoint between equal like charges, E = 0 but V ≠ 0. Use E = -dV/dr to check that field points from high to low potential.
Applying F = BIL or F = qv×B without checking that the current or velocity is perpendicular to B, and getting the direction wrong by ignoring the right-hand rule for negative charges.
Formulas are recalled in magnitude form; the sin θ factor and the sign flip for electrons are dropped under time pressure, and the three-finger rule is applied inconsistently.
What to do instead
Always write F = BIL sin θ and F = qvB sin θ, then substitute θ explicitly (zero force when parallel). For direction, use the right-hand rule on conventional current or positive charge, then reverse the result for electrons. Sanity-check with a known case: a charge moving along B experiences no magnetic force.
Using Faraday's law as ε = -N·ΔΦ/Δt with Φ = BA only, forgetting that flux can change because the area changes, the field changes, or the orientation changes, and mishandling the Lenz sign.
Most drill problems vary B only, so students default to that; when a rod slides on rails or a coil rotates, they still differentiate B and get zero or the wrong magnitude.
What to do instead
Write Φ = B·A·cos θ symbolically first, then identify which factor varies and differentiate only that one. For a sliding rod, ε = Bℓv; for rotation, ε = NBAω sin(ωt). Determine current direction by Lenz's law (oppose the change in flux), and confirm the induced current produces a force opposing the motion.
Mixing up the wave equation quantities: using frequency instead of period, or substituting wavelength in centimetres while speed is in m/s, and confusing a standing wave's node spacing with wavelength.
v = fλ is simple enough to apply mechanically; unit conversion and the node/antinode geometry of standing waves are the actual stumbling blocks.
What to do instead
Convert every length to metres before substitution and write units through the calculation. For standing waves, derive the geometry: adjacent nodes are λ/2 apart, node-to-antinode is λ/4, and a string fixed at both ends has L = nλ/2. Check the answer's order of magnitude against a familiar sound speed (~340 m/s) or light speed.
Applying the thin-lens or mirror equation with a sign convention chosen ad hoc, so real/virtual images and magnification signs come out inconsistent.
Different textbooks use different sign conventions, and students blend them; they also forget that a virtual image has a negative image distance in the Cartesian convention.
What to do instead
Commit to one convention and state it at the top of the solution (Cartesian: distances measured from the lens, positive in the direction of incident light). Use 1/f = 1/v - 1/u consistently, then interpret the sign of v: positive means real image on the far side, negative means virtual on the same side. Verify magnification m = v/u matches the expected orientation.
Plugging everyday speeds into relativistic formulas and expecting measurable effects, or forgetting the Lorentz factor in time dilation and length contraction.
Students apply γ = 1/√(1 - v²/c²) without checking v/c, and they confuse proper time (measured in the rest frame of the event) with dilated time.
What to do instead
First compute v/c and γ numerically; if γ ≈ 1, note that classical results are adequate. Identify the proper time as the interval measured by a clock at rest relative to the two events (it is the shortest), then apply Δt = γΔτ. For lengths, the proper length is measured in the object's rest frame and is the longest; contracted length is L = L₀/γ.
Computing nuclear mass defect using atomic masses without accounting for electron masses, or confusing activity units and half-life arithmetic (e.g., using t½ instead of the decay constant in N = N₀e^(-λt)).
Mass-energy problems mix nuclear and atomic data, and decay problems invite formula-shopping; the relationship λ = ln2/t½ is often inverted.
What to do instead
For nuclear reactions, use nuclear masses or, if using atomic masses, verify electron counts cancel on both sides. For decay, always convert half-life to λ = ln2/t½ first, then use N = N₀e^(-λt) or the equivalent N = N₀(1/2)^(t/t½). Check that after one half-life the remaining fraction is exactly 0.5.