The four-step method
Every numerical, at every level, yields to the same sequence. Write it out fully even when the problem looks easy - the habit is what saves you when the problem is not.
- ✓Step 1 - List what is given, with units, and what is asked. Convert everything to SI now.
- ✓Step 2 - Name the concept, not the formula. 'This is conservation of energy' or 'this is a series circuit' narrows the formula choice to one or two.
- ✓Step 3 - Write the formula symbolically, then rearrange for the unknown before substituting numbers. Rearranging with numbers in place is where algebra errors creep in.
- ✓Step 4 - Substitute with units, compute, and sanity-check the magnitude and sign. A car accelerating at 400 m/s² is a wrong answer regardless of the arithmetic.
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Kinematics and Newton's laws
A car starting from rest reaches 20 m/s in 5 s. Find its acceleration and the distance covered.
Given u = 0, v = 20 m/s, t = 5 s. From v = u + at, a = 20/5 = 4 m/s². Then s = ut + ½at² = 0 + ½ × 4 × 25 = 50 m. You could also use s = (u+v)t/2 = 10 × 5 = 50 m as a check.
A force of 10 N acts on a 5 kg body at rest on a frictionless surface. Find the acceleration and the velocity after 3 s.
From F = ma, a = 10/5 = 2 m/s². Then v = u + at = 0 + 2 × 3 = 6 m/s. Note that Newton's second law gives acceleration, never velocity directly - a step students often skip.
A ball is projected at 20 m/s at 30° to the horizontal. Find its maximum height and horizontal range (g = 10 m/s²).
H = u²sin²θ/2g = 400 × 0.25 / 20 = 5 m. R = u²sin2θ/g = 400 × sin60° / 10 = 400 × 0.866 / 10 ≈ 34.6 m. Note that sin2θ uses the doubled angle - substituting sinθ here is the standard error.
A 2 kg body moving at 5 m/s is brought to rest in 4 s. Find the retarding force.
a = (v − u)/t = (0 − 5)/4 = −1.25 m/s², so F = ma = 2 × (−1.25) = −2.5 N. The negative sign means the force opposes the motion, and stating that is worth a mark.
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Work, energy, power and circular motion
A force of 10 N moves a body 5 m along a direction at 60° to the force. Find the work done.
W = Fs cosθ = 10 × 5 × cos60° = 10 × 5 × 0.5 = 25 J. Only the component of force along the displacement does work, which is why a force perpendicular to motion - like the centripetal force in uniform circular motion - does none.
Find the kinetic energy of a 2 kg body moving at 10 m/s, and the work needed to double its speed.
KE = ½mv² = ½ × 2 × 100 = 100 J. Doubling the speed quadruples the kinetic energy to 400 J, so the extra work required is 300 J. Kinetic energy scales with the square of speed - the source of many wrong intuitions.
A body of mass 1 kg falls freely from a height of 20 m. Find its speed on reaching the ground (g = 10 m/s²).
By conservation of energy, mgh = ½mv², so v = √(2gh) = √400 = 20 m/s. The mass cancels, which is why all bodies fall alike in the absence of air resistance.
A stone moves in a circle of radius 5 m at 10 m/s. Find its centripetal acceleration.
a_c = v²/r = 100/5 = 20 m/s², directed towards the centre. The speed may be constant while the velocity is not, since the direction keeps changing - which is precisely why there is an acceleration at all.
A pump raises 100 kg of water through 10 m in 20 s. Find its power output (g = 10 m/s²).
Work = mgh = 100 × 10 × 10 = 10,000 J, so power = W/t = 10,000/20 = 500 W. Power is always work divided by time, and mixing up the two is a common one-mark loss.
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Electricity
Three resistors of 2 Ω, 3 Ω and 6 Ω are connected in parallel across a 6 V battery. Find the equivalent resistance and total current.
1/R = 1/2 + 1/3 + 1/6 = (3 + 2 + 1)/6 = 1, so R = 1 Ω. Then I = V/R = 6/1 = 6 A. Sanity check: the parallel equivalent must be smaller than the smallest resistor, and 1 Ω is.
A wire of resistance 4 Ω is stretched to twice its original length. Find its new resistance.
16 Ω. Volume is conserved, so doubling the length halves the area, and R = ρl/A therefore increases by a factor of 2 × 2 = 4. Stretching questions always turn on this simultaneous change in both l and A.
An electric heater draws 5 A from a 220 V supply. Find its power and the energy consumed in 2 hours.
P = VI = 220 × 5 = 1100 W = 1.1 kW. Energy = 1.1 × 2 = 2.2 kWh, or 2.2 units. Convert watts to kilowatts before multiplying by hours, or the unit count comes out a thousand times too large.
Find the heat produced when a current of 2 A flows through a 5 Ω resistor for 30 s.
H = I²Rt = 4 × 5 × 30 = 600 J. Note that heat depends on the square of the current, which is why a small increase in current causes a disproportionate rise in heating - the basis of both fuses and transmission losses.
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Optics: get the signs right first
Optics numericals are the most mechanical in physics and the most error-prone, for one reason: the sign convention. Fix it before you write the formula. Distances are measured from the pole or optical centre, positive along the direction of incident light and negative against it - so a real object always has a negative u.
An object is placed 20 cm from a concave mirror of focal length 15 cm. Find the image distance and magnification.
With u = −20 cm and f = −15 cm: 1/v = 1/f − 1/u = −1/15 + 1/20 = (−4 + 3)/60 = −1/60, so v = −60 cm - a real image 60 cm in front. m = −v/u = −(−60)/(−20) = −3: inverted and three times larger.
A concave lens of focal length 20 cm has an object 30 cm away. Find the image position.
With u = −30 cm and f = −20 cm: 1/v = 1/f + 1/u = −1/20 − 1/30 = (−3 − 2)/60 = −1/12, so v = −12 cm. A concave lens always forms a virtual, diminished image on the same side as the object, whatever the object distance - a useful check on your answer.
Two lenses of power +3 D and −5 D are placed in contact. Find the power and focal length of the combination.
Powers in contact simply add: P = 3 + (−5) = −2 D. Then f = 1/P = −0.5 m = −50 cm, so the combination behaves as a diverging lens.
Light passes from air into glass of refractive index 1.5 at an angle of incidence of 30°. Find the angle of refraction.
From Snell's law, sin r = sin i / n = 0.5/1.5 = 0.333, so r ≈ 19.5°. The ray bends towards the normal on entering a denser medium, which the smaller angle confirms.
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Modern physics
The work function of a metal is 2 eV. Light of energy 5 eV falls on it. Find the maximum kinetic energy of the emitted photoelectrons.
By Einstein's photoelectric equation, KE_max = hν − φ = 5 − 2 = 3 eV. Photons below the work function eject no electrons regardless of how intense the light is - the observation that established the particle nature of light.
Find the energy of an electron in the second orbit of a hydrogen atom.
E_n = −13.6/n² eV, so E₂ = −13.6/4 = −3.4 eV. The negative sign indicates a bound state, and its magnitude is the energy needed to remove the electron from that orbit.
A radioactive sample has a half-life of 10 days. What fraction remains after 40 days?
1/16. Forty days is four half-lives, and each halves the remaining quantity: 1/2, 1/4, 1/8, 1/16. Count half-lives rather than reaching for the exponential formula when the time is a whole multiple.
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Where the marks actually go
- ✓Unit conversion done late or not at all - km/h left unconverted, centimetres in a formula expecting metres, grams where kilograms belong.
- ✓Sign convention errors in optics, which flip the nature of the image and lose every subsequent mark.
- ✓Substituting before rearranging, which turns a clean algebra step into an arithmetic mess.
- ✓Not writing the formula. Marking schemes usually award a separate mark for it, so a correct formula with a wrong final answer still scores.
- ✓Skipping the sanity check. Physically absurd magnitudes are the cheapest errors to catch and the most embarrassing to leave.
- ✓Practising untimed, so speed never develops. Solve in sets against a clock once the method is secure.
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