Érettségi Physics common mistakes
Érettségi Physics · Understand the traps and how to avoid them.
When a problem asks for the time, distance, or speed of a uniformly accelerating body, the candidate substitutes numbers directly into s = v·t or v = s/t (the formulas for uniform motion) without checking whether acceleration is zero, and so obtains a plausible but wrong result.
In earlier studies the formula 'speed = distance ÷ time' was drilled in as the default, and the candidate does not first list the known quantities and decide which kinematic model applies. Hungarian exam problems often give the acceleration implicitly, through phrases such as 'starting from rest' or 'braking uniformly', and this implicit information gets overlooked.
What to do instead
Before writing any equation, write down the known quantities (v₀, v, a, s, t) and mark which one is zero or unknown. If a ≠ 0, use the set v = v₀ + a·t, s = v₀t + a·t²/2, and v² = v₀² + 2as. After solving, do a dimensional check and a check of the sign of the acceleration direction, and confirm that the result is physically plausible (e.g. braking distance is positive and finite).
In momentum and energy problems involving collisions or explosions, the candidate applies conservation of kinetic energy to an inelastic collision, or applies conservation of momentum to a process where an external force (friction, a fixed pivot, gravity during a long process) acts.
The two conservation laws are learned as a pair, and the candidate does not examine the conditions under which each holds. The phrase 'collision' triggers 'energy is conserved' automatically, and the impulse of external forces is ignored.
What to do instead
For each problem, state explicitly: (1) the system and the time interval; (2) whether the net external force is zero during that interval (only then is momentum conserved); (3) whether the collision is elastic (kinetic energy also conserved) or inelastic (only momentum conserved, kinetic energy is lost). Write the conservation equation for the chosen quantity and check the result against the extreme cases (perfectly inelastic collision, equal masses).
In statics and torque problems the candidate adds the magnitudes of the forces and ignores the line of action, or chooses a torque axis that passes through no unknown force, so the lever arms and the signs of the torques are mishandled and the equilibrium equations give a wrong result.
The candidate treats force as a scalar and does not draw a free-body diagram. The choice of axis is made arbitrarily, and the sign convention for clockwise and counterclockwise torques is not fixed in advance.
What to do instead
Always draw a free-body diagram with the forces at their points of application. Write ΣF = 0 for two perpendicular directions and ΣM = 0 about a chosen point, and choose that point to pass through as many unknown forces as possible. Fix the sign convention in writing (e.g. counterclockwise positive), compute each lever arm as the perpendicular distance from the axis to the line of action, and verify with a second axis.
In gas-process problems the candidate uses pV = nRT with the wrong temperature (Celsius instead of kelvin), or applies the adiabatic relation pV^κ = constant to a process that is actually isothermal or isobaric, and mixes up the molar heat capacities c_v and c_p.
The candidate memorises the formulas without linking them to the process name. The kelvin conversion is skipped because the numbers 'look small'. The distinction between c_p and c_v is not tied to whether the process is at constant pressure or constant volume.
What to do instead
At the start of every thermodynamics problem, convert the temperature to kelvin and write down which quantity is constant (p, V, T, or Q = 0). Use pV = nRT for the states and the first law ΔU = Q − W (or Q = ΔU + W) for the process; for an ideal gas ΔU = n·c_v·ΔT always. Use c_p only for constant-pressure heating, and use the adiabatic relation only when Q = 0. Check the sign of W against the direction of the volume change.
In phase-change and calorimetry problems the candidate adds the heat of fusion or vaporisation to a term m·c·ΔT using the wrong specific heat (e.g. using the specific heat of water while the substance is still ice), or forgets that during the phase change the temperature does not change.
The candidate treats Q = m·c·ΔT as a universal formula and does not draw the heating curve. The latent heat and the sensible heat are confused because both have the form 'mass × constant'.
What to do instead
Draw the heating curve (temperature against heat added) and mark the segments. For each segment use only one formula: m·c·ΔT on the sloped parts and m·L on the horizontal parts. Check whether the available heat is enough to complete the phase change before assuming the final state; if not, the final temperature is the phase-change temperature and part of the substance remains in the other phase.
In direct-current circuits the candidate applies the formula for parallel resistors to a series connection (or vice versa), or uses the terminal voltage of the source in place of the EMF when computing the current, and so ignores the internal resistance.
The two combination rules are memorised as 'reciprocals for parallel' without a picture of the circuit. The internal resistance r is treated as a detail, although Hungarian problems often make it decisive.
What to do instead
Redraw the circuit and label the nodes. For series: R = R₁ + R₂, the current is the same everywhere, and the voltages add. For parallel: 1/R = 1/R₁ + 1/R₂, the voltage is the same, and the currents add. Use the complete circuit I = ε/(R + r), and the terminal voltage U = ε − I·r. Check the power balance: the power delivered by the source equals the sum of the powers dissipated.
In induction problems the candidate computes the induced EMF from the flux itself (ε ∝ Φ) rather than from the rate of change of flux, or ignores Lenz's law and gives the direction of the induced current the wrong way round.
Faraday's law is remembered only as 'EMF is proportional to flux', and the derivative is dropped. Lenz's law is treated as a separate rule to memorise rather than as a consequence of energy conservation.
What to do instead
Write Faraday's law in the form ε = −N·ΔΦ/Δt, and identify explicitly what is changing (B, A, or the angle). For the direction, apply Lenz's law: the induced current opposes the change in flux, so determine whether the flux is increasing or decreasing and choose the direction that opposes that change. Verify with the energy argument: the induced current must do negative work against the motion that produces it.
In geometric optics the candidate applies the thin-lens or mirror formula with the wrong sign convention (e.g. a virtual image treated as real), and in refraction problems measures the angle from the surface instead of from the normal.
The sign convention is not written down, and the candidate relies on intuition about 'where the image looks like it is'. The angle of incidence is confused with the angle between the ray and the surface because the diagram is drawn casually.
What to do instead
Write the sign convention at the top of the solution (e.g. real object distance positive, virtual image distance negative, focal length positive for a converging lens). Draw the ray diagram with the normal at the point of incidence and measure all angles from the normal. After computing, check the result against the ray diagram and against the limiting cases (object at the focal point, object at infinity).
In atomic and nuclear problems the candidate balances a nuclear equation by mass number only, or ignores the conservation of charge (atomic number), and in decay chains confuses alpha and beta decay and their effect on Z and A.
The candidate treats the equation as an arithmetic puzzle and does not use the two conservation laws (mass number and charge) as a check. The properties of the alpha and beta particles are memorised separately from the decay equation.
What to do instead
For every nuclear equation, write the conservation of mass number A and of charge Z as two separate equations and solve them for the unknown particle. Remember: alpha decay decreases A by 4 and Z by 2; beta-minus decay leaves A unchanged and increases Z by 1; gamma emission changes neither. Check the result by adding up both sides of the equation.
In gravitational and celestial-mechanics problems the candidate uses g = 9.81 m/s² at a height comparable to the Earth's radius, or applies the formula for circular orbital speed to an elliptical orbit, and mixes up the gravitational field strength with the gravitational potential.
The constant g is used as universal, and the inverse-square law is not connected to the idea that g is the field strength at the surface. The distinction between field strength (a vector) and potential (a scalar) is not clear.
What to do instead
Decide first whether the height is negligible compared with the Earth's radius; if not, use F = G·m₁m₂/r² and g(r) = G·M/r² with r measured from the centre of the Earth. For orbits, use the circular-orbit relation v = √(GM/r) only for circular orbits; for elliptical orbits use conservation of energy and of angular momentum (Kepler's second law). Keep the sign of the potential (negative, zero at infinity) separate from the field strength.