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Free samples · Senior 1 – Senior 6

Worked examples, one for each level

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
  1. Current: I = 3 ÷ 6 = 0.5 A
  2. 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
  1. Pressure is transmitted equally through the liquid (Pascal’s principle).
  2. 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
  1. Friction provides the centripetal force, so the largest available is 8000 N.
  2. v² = (8000 × 50) ÷ 1000 = 400
  3. 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
  1. Magnifying power: M = 100 ÷ 5 = 20
  2. 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
  1. Angular frequency: ω = 2π ÷ 2 = π ≈ 3.14 rad/s
  2. Maximum speed: v = 0.05 × 3.14 ≈ 0.157 m/s
  3. 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
  1. Angular acceleration: α = 8 ÷ 4 = 2 rad/s²
  2. Angular velocity: ω = 0 + 2 × 5 = 10 rad/s
  3. 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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