This is how every problem is set out in the paid packs: what is given, the formula, each step with units, and the final answer. Try the question first, then compare your steps.
S1Unit 8: Electric Current and Ohm’s Law
Current through a torch bulb
Question
A torch bulb of resistance 6 Ω is connected to a 3 V battery. Find the current in the bulb and the charge that flows through it in 1 minute.
Given
R = 6 ΩV = 3 Vt = 1 min = 60 s
Formula
I = V/R Q = It
Solution
Current: I = 3 ÷ 6 = 0.5 A
Charge: Q = 0.5 × 60 = 30 C
Answer: I = 0.5 A, Q = 30 C
Watch out: change minutes to seconds before using Q = It.
S2Unit 4: Pressure in Solids and Liquids
Hydraulic lift (Pascal’s principle)
Question
In a hydraulic lift, a force of 50 N is applied to a small piston of area 2 cm². The large piston has an area of 200 cm². What load can the large piston lift?
Given
F₁ = 50 NA₁ = 2 cm²A₂ = 200 cm²
Formula
F₁/A₁ = F₂/A₂ ⇒ F₂ = F₁ × A₂/A₁
Solution
Pressure is transmitted equally through the liquid (Pascal’s principle).
F₂ = 50 × 200 ÷ 2 = 5000 N
Answer: F₂ = 5000 N
Watch out: both areas must be in the same unit; here cm² cancels, so no conversion is needed.
S3Unit 1: Uniform Circular Motion
Maximum safe speed round a bend
Question
A 1000 kg car goes round a flat bend of radius 50 m. The maximum friction between the tyres and the road is 8000 N. What is the fastest speed the car can take the bend without skidding?
Given
m = 1000 kgr = 50 mFmax = 8000 N
Formula
F = mv²/r ⇒ v = √(F r / m)
Solution
Friction provides the centripetal force, so the largest available is 8000 N.
v² = (8000 × 50) ÷ 1000 = 400
v = √400 = 20 m/s
Answer: 20 m/s (72 km/h)
Watch out: above 20 m/s the car slides off along the tangent, not straight outwards.
S4Unit 2: Optical Instruments
Astronomical telescope
Question
An astronomical telescope has an objective of focal length 100 cm and an eyepiece of focal length 5 cm. Find its magnifying power and the length of the tube in normal adjustment.
Given
fo = 100 cmfe = 5 cm
Formula
M = fo / fe L = fo + fe
Solution
Magnifying power: M = 100 ÷ 5 = 20
Tube length: L = 100 + 5 = 105 cm
Answer: M = 20, L = 105 cm
Watch out: in normal adjustment the final image is at infinity, so the lens separation is exactly fo + fe.
S5Unit 1: Simple Harmonic Motion
Maximum speed and acceleration in SHM
Question
A particle moves in simple harmonic motion with amplitude 0.05 m and period 2 s. Find its maximum speed and maximum acceleration.
Given
A = 0.05 mT = 2 s
Formula
ω = 2π/T vmax = Aω amax = ω²A
Solution
Angular frequency: ω = 2π ÷ 2 = π ≈ 3.14 rad/s
Maximum speed: v = 0.05 × 3.14 ≈ 0.157 m/s
Maximum acceleration: a = 3.14² × 0.05 ≈ 0.49 m/s²
Answer: vmax ≈ 0.157 m/s, amax ≈ 0.49 m/s²
Watch out: maximum speed is at the centre of the motion; maximum acceleration is at the ends.
S6Unit 2: Rotational Motion
Flywheel turned by a torque
Question
A flywheel with moment of inertia 4 kg m² starts from rest. A constant torque of 8 N m acts on it for 5 s. Find its angular acceleration, its final angular velocity and its rotational kinetic energy.
Given
I = 4 kg m²τ = 8 N mt = 5 sω₀ = 0
Formula
τ = Iα ω = ω₀ + αt KE = ½Iω²
Solution
Angular acceleration: α = 8 ÷ 4 = 2 rad/s²
Angular velocity: ω = 0 + 2 × 5 = 10 rad/s
Kinetic energy: KE = ½ × 4 × 10² = 200 J
Answer: α = 2 rad/s², ω = 10 rad/s, KE = 200 J
Watch out: check: the torque turns the wheel through θ = ½αt² = 25 rad, and work τθ = 8 × 25 = 200 J, the same answer.
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Every calculation in your level, solved step by step
Each pack will cover all the units of one level in this format, with exam-style questions, marking schemes and common mistakes, as a PDF for your phone or to print. The packs are coming soon.