EASA Module 2 Physics common mistakes
EASA Module 2 Physics · Understand the traps and how to avoid them.
Takes a density given in g/cm³ straight into ρ = m/V or into a buoyancy/pressure calculation, converting cm³ to m³ by multiplying by 100 (or not converting at all), and gives an answer that is a hundred or a thousand times out.
The g→kg factor is applied correctly but the cm³→m³ factor is treated as a single length conversion instead of being cubed; the number '1' for water also feels like it needs no conversion because students remember 'water is 1'.
What to do instead
Convert every quantity to kg and metres on the line above the calculation, before substituting. Anchor the conversion with 1 g/cm³ = 1000 kg/m³ and sanity-check any density against water (1000 kg/m³); a solid metal must come out in the thousands of kg/m³.
Substitutes a deceleration into v = u + at or v² = u² + 2as as a positive number while also treating the initial velocity as positive, then reports a final speed larger than the initial one or takes the square root of a negative number and abandons the question.
Deceleration is read as 'a number to add' rather than as a direction; the student never fixes which way is positive, so the sign of a is decided by feel at the moment of substitution.
What to do instead
On the working, draw an arrow and write '+ →' or 'up is positive' before any equation. Then write the sign of every vector explicitly (u = +12 m/s, a = −3 m/s²). For v² = u² + 2as, check the discriminant is positive and that the answer is physically sensible before moving on.
On a slope of angle θ, writes the component of weight down the incline as mg cos θ and the normal reaction as mg sin θ (or picks sin/cos by habit for a tow-rope at an angle), and then carries the error through the friction calculation.
A rule such as 'sin is for the slope' is memorised from one worked example without a diagram, so it is applied even when the angle is defined differently (to the vertical, between two forces, or at the top of the wedge).
What to do instead
Never choose sin or cos without drawing the right-angled triangle. Mark the angle between the force vector and the axis you are resolving onto; the component along that axis is F cos(angle between them). Check the limiting cases: at θ = 0 the down-slope component must be zero, at θ = 90° it must be the full weight.
Reports the pressure at a depth h in a tank as p = ρgh and calls it the absolute pressure, or in a horizontal pipe states that the pressure is highest where the fluid moves fastest.
The formula is recalled without its atmospheric term, and 'fast flowing water has high pressure' comes from everyday intuition about hoses and taps rather than from the energy equation.
What to do instead
Always write p_abs = p_atm + ρgh and state which pressure the question asks for. For flow, write p + ½ρv² + ρgh = constant and reason from it: along a horizontal streamline, higher speed means lower static pressure. Cross-check with the continuity equation A₁v₁ = A₂v₂ to get the speed direction right first.
Puts a Celsius temperature into the general gas law, into Charles's law, or into a thermal expansion calculation, e.g. using '20' and '40' as the volume ratio when the gas goes from 20 °C to 40 °C.
The thermometer in the question stem is in °C, the numbers are small and convenient, and the kelvin conversion feels like an optional extra step rather than a condition for the law to be valid.
What to do instead
Convert on the first line of working, every time: T(K) = t(°C) + 273. Write the converted values in the equation, not the originals. Also check a ratio answer against common sense - doubling the kelvin temperature roughly doubles the volume, doubling the Celsius number does not.
Uses Q = mcΔT for a block of ice being warmed from −10 °C to steam at 120 °C, skipping the melting and boiling stages, or adds a latent heat term where no change of state occurs.
The question is read as 'one heating problem, one formula', and the flat portions of the temperature–time graph are not linked to the physical change happening there.
What to do instead
Before calculating, sketch the temperature–time path and split it into segments. Use Q = mcΔT only where the temperature changes and Q = mL only where it is constant. Sum the segments and label each one with its state so a missed stage is visible.
Substitutes a wavelength in centimetres (or a period in milliseconds) into v = fλ, or inverts period and frequency the wrong way round, producing a sound speed of 34 000 m/s or 0.34 m/s that is written down without comment.
The formula is treated as pure arithmetic and the units in the stem are whatever the examiner chose; there is no habit of checking the answer against a known value.
What to do instead
Convert to SI (m, s, Hz) before substitution and keep the unit attached to each number. Then check the result: sound in air is about 340 m/s, light about 3 × 10⁸ m/s. If the answer is not of that order, the conversion or the f = 1/T inversion is wrong.
States that the frequency of a sound changes when it passes from air into water (or that the wavelength stays the same), and gets Doppler shift questions wrong by shifting the frequency in the wrong direction for an approaching source.
The equation v = fλ is treated as if any of the three quantities may adjust, with no rule about which one is fixed by the source and which by the medium.
What to do instead
Learn the sentence: the source sets f, the medium sets v, and λ = v/f follows. For Doppler, reason physically - an approaching source crowds the wavefronts, so the observed frequency rises. Test the answer by asking whether the listener would hear a higher or lower pitch.
Puts distances and focal lengths into 1/f = 1/v + 1/u (or the mirror form) without agreeing a sign convention, so a real object is entered as negative on one line and positive on the next, and the sign of the magnification is then used to claim a real image is virtual.
Several conventions exist in textbooks, and the student switches between a 'real is positive' rule and a Cartesian rule part-way through a multi-step question.
What to do instead
Write the convention at the top of the working ('real object distances positive, virtual image distances negative') and apply it to every term. At the end, interpret the sign in words: negative image distance = virtual image, and negative magnification = inverted image, consistent with the same convention.
Measures the angle of incidence from the glass or water surface rather than from the normal when applying n₁ sin θ₁ = n₂ sin θ₂, and then uses that same wrong angle in the critical-angle formula sin C = 1/n.
Ray diagrams in the question often draw the surface horizontally as the visually obvious reference line, and the normal is a faint dashed line that is easy to overlook.
What to do instead
Draw the normal at the point of incidence first and mark both angles against it before writing Snell's law. Check the limits: at grazing incidence (θ₁ ≈ 90°) the refracted ray must bend towards the normal in the denser medium, and sin C must come out less than 1 - if it does not, an angle was measured from the surface.