This mini-lesson covers all eight areas of Eduqas Component 1: basic physics (units, vectors, density, moments), kinematics, dynamics, energy concepts, circular motion, vibrations (SHM), kinetic theory and thermal physics.
Where it sits: Component 1 is a 2 h 15 min written paper worth 31.25% of the A level, and it also contains synoptic questions that draw on the rest of the specification.
Press Start when you are ready.
1. Basic physics
Units, homogeneity, vectors and moments
Eduqas names six base SI units: kg, m, s, A, mol, K. Every other unit is built from these — 1 N = 1 kg m s⁻², 1 J = 1 kg m² s⁻², 1 W = 1 kg m² s⁻³.
Checking homogeneity: an equation can only be correct if both sides have the same base units. Test v² = u² + 2as: left is (m s⁻¹)² = m² s⁻²; right is m² s⁻² + (m s⁻²)(m) = m² s⁻². Homogeneous ✓. (Careful: homogeneity proves an equation is not obviously wrong — it cannot prove a dimensionless factor like ½ is right.)
ρ = m / V · Fx = F cos θ, Fy = F sin θVectors have magnitude and direction (displacement, velocity, force); scalars have magnitude only (mass, speed, energy, density).
Principle of moments: for a body in equilibrium, the total clockwise moment about any point equals the total anticlockwise moment. Equilibrium also needs zero resultant force. Take moments about a point where an unknown force acts and it drops out of the equation.
Centre of gravity & stability: the whole weight acts at the centre of gravity — the centre for a uniform cylinder, sphere or cuboid. A body topples when the vertical line through its centre of gravity falls outside its base. Specified practicals: measuring the density of solids; finding unknown masses using the principle of moments.
Quick check
Which is a base unit?
?Which of the following is a base SI unit?
2. Kinematics
Motion graphs, suvat and projectiles
v = u + at · x = ut + ½at² · v² = u² + 2axDerive them from the velocity–time graph: gradient = acceleration, area = displacement.
Falling with air resistance: as speed rises, drag rises. When drag equals weight, the resultant force is zero, so the acceleration is zero and the object falls at a constant terminal velocity — it does not decelerate (unless it was already going faster).
Projectiles. The vertical and horizontal motions are independent; the only thing they share is the time. Horizontally, a = 0 so x = uxt. Vertically, a = −g and the suvat equations apply.
Worked example — a projectile launched at an angle
Launch at speed u and angle θ: ux = u cos θ, uy = u sin θ.
At the top of the flight the vertical velocity is zero (the horizontal velocity is not).
Maximum height: 0 = uy² − 2g·h, so h = (u sin θ)² / 2g.
Specified practical: measurement of g by free fall — release a ball through light gates or use a timer circuit, plot s against t², and take g = 2 × gradient.
Calculate
Your turn — projectile
1A ball is launched at 20 m s⁻¹ at 30° above the horizontal. Take g = 9.81 m s⁻² and ignore air resistance. Calculate the maximum height reached.
m
Hint: u(vertical) = 20 sin 30° = 10 m s⁻¹. At the top, v = 0: h = u² / (2g) = 10² ÷ (2 × 9.81).
Quick check
Terminal velocity
?A skydiver has reached terminal velocity. Which statement is correct?
3. Dynamics
Newton's laws, momentum and collisions
Newton's third law: if A exerts a force on B, B exerts an equal and opposite force on A — the same type of force, acting on a different body. Draw a free-body diagram for each body separately.
F = ma · p = mv · F = Δp / ΔtForce is the rate of change of momentum. For constant mass this reduces to F = ma.
Conservation of momentum: with no external resultant force, the total momentum of a system is unchanged by a collision or explosion. This applies to every collision.
Elastic — kinetic energy is also conserved (relative speed of approach = relative speed of separation).
Inelastic — kinetic energy is lost (to internal energy, sound, deformation). Momentum is still conserved.
Crumple zones: Δp is fixed by the crash. Extending the time of the collision reduces the average force (F = Δp/Δt) — the whole point of crumple zones, airbags and seat belts.
Calculate
Your turn — an inelastic collision
2A trolley of mass 2.0 kg moving at 3.0 m s⁻¹ collides with a stationary trolley of mass 4.0 kg, and they move off together. Calculate the kinetic energy lost in the collision.
J
Hint: Momentum: 2.0 × 3.0 = 6.0 kg m s⁻¹ = (2.0 + 4.0)v, so v = 1.0 m s⁻¹. KE before = ½ × 2.0 × 3.0² = 9.0 J. KE after = ½ × 6.0 × 1.0² = 3.0 J.
Sort it
Elastic or inelastic?
Tap a statement, then tap the type of collision it applies to.
⚪ Elastic only
💥 Inelastic only
🔁 True of both
4. Energy concepts
Work, energy, power and efficiency
W = Fx cos θOnly the component of the force along the displacement does work. A force perpendicular to the motion (like the centripetal force) does no work at all.
Ek = ½mv² · ΔEp = mgΔh · Eelastic = ½kx²Work–energy relationship: Fx = ½mv² − ½mu²
P = ΔE / Δt = Fv · efficiency = (useful energy out ÷ total energy in) × 100%Dissipative forces (friction, drag) transfer energy out of the system as internal energy, which is why efficiency is always below 100%.
Quick check
How efficient?
?A motor is supplied with 600 J of electrical energy and does 240 J of useful work lifting a load. What is its efficiency?
5. Circular motion
Radians, angular velocity and centripetal force
One radian is the angle subtended at the centre of a circle by an arc equal in length to the radius; a full circle is 2π rad.
ω = 2π / T = 2πf · v = ωrω is the angular velocity, in rad s⁻¹.
a = v²/r = ω²r · F = mv²/r = mω²rThe centripetal force is the resultant force on the body, directed towards the centre. It is provided by a real force — tension, friction, gravity or a normal contact force.
Constant speed, changing velocity: the speed does not change, but the direction of the velocity does — so the body is accelerating, and a resultant force is needed. That force does no work (it is perpendicular to the motion), which is why the kinetic energy stays constant.
Calculate
Your turn — circular motion
3A fairground ride carries riders in a horizontal circle of radius 8.0 m, completing one revolution every 4.0 s. Calculate the centripetal acceleration.
m s⁻²
Hint: ω = 2π/T = 2π/4.0 = 1.57 rad s⁻¹. a = ω²r = 1.57² × 8.0.
6. Vibrations
Simple harmonic motion, damping and resonance
SHM definition: the acceleration is proportional to the displacement from a fixed point and always directed towards that point.
a = −ω²xSolution: x = A cos(ωt + ε), and the velocity is v = −Aω sin(ωt + ε), so vmax = ωA.
T = 2π√(m/k) · T = 2π√(l/g)Spring–mass system (stiffness k) and simple pendulum (length l). Neither period depends on the amplitude.
Energy in SHM: kinetic energy is maximum at the centre (where x = 0 and v is greatest); potential energy is maximum at the extremes (where v = 0). The total stays constant if there is no damping.
Damping — dissipative forces remove energy, so the amplitude decays. Critical damping returns the system to equilibrium in the shortest time without oscillating (vehicle suspension).
Resonance — a driver at the natural frequency transfers energy most efficiently, so the amplitude becomes large. Increased damping lowers and broadens the resonance peak.
Useful (circuit tuning, microwave cooking) or dangerous (bridges, buildings) depending on the context.
Specified practicals: measurement of g with a pendulum (plot T² against l — gradient = 4π²/g); investigation of the damping of a spring.
Calculate
Your turn — the simple pendulum
4Calculate the period of a simple pendulum of length 0.80 m. Take g = 9.81 m s⁻².
s
Hint: T = 2π√(l/g) = 2π√(0.80/9.81) = 2π × 0.286.
Quick check
Damping the resonance curve
?How does increasing the damping change the resonance curve of a forced oscillator?
7. Kinetic theory · 8. Thermal physics
Ideal gases, internal energy and the first law
pV = nRT · pV = NkTR = 8.31 J K⁻¹ mol⁻¹; k = R/NA = 1.38 × 10⁻²³ J K⁻¹; NA = 6.02 × 10²³ mol⁻¹. T is always in kelvin.
Kinetic theory treats a gas as point molecules in random motion making elastic collisions with the walls. Analysing the momentum change per collision gives
pV = ⅓ N m ⟨c²⟩Combining this with pV = NkT gives the mean molecular kinetic energy: ½m⟨c²⟩ = (3/2)kT — temperature is mean molecular kinetic energy. The internal energy of an ideal monatomic gas is therefore wholly kinetic: U = (3/2)nRT.
First law of thermodynamics — conservation of energy for a gas:
ΔU = Q − WΔU = increase in internal energy; Q = heat supplied to the gas; W = work done by the gas. At constant pressure W = pΔV; in general W is the area under the p–V graph.
For a solid or liquid the volume change is tiny, so W ≈ 0 and Q = ΔU. That is why Q = mcΔθ works — it is the first law with the work term dropped. Specified practicals: estimating absolute zero using the gas laws; measuring the specific heat capacity of a solid.
Calculate
Your turn — mean molecular kinetic energy
5Calculate the mean kinetic energy of a molecule of an ideal gas at 300 K. (k = 1.38 × 10⁻²³ J K⁻¹.) Give your answer in units of 10⁻²¹ J.
× 10⁻²¹ J
Hint: Mean KE = (3/2)kT = 1.5 × 1.38 × 10⁻²³ × 300 = 6.21 × 10⁻²¹ J. Enter just the number in front.
Quick check
Applying the first law
?500 J of heat is supplied to a gas, which then does 200 J of work as it expands. What is the change in the gas's internal energy?
Match it
Match each equation to its topic
Tap an item on the left, then its partner on the right.
Equation
Topic
Recap
Component 1 — the big ideas
Basic physics: six base units; check homogeneity; ρ = m/V; moments and stability