Eduqas A-level Physics (A420QS) · Component 3 Option C: The Physics of Sports
Mini-Lesson · Option C
Option C: The Physics of Sports
This mini-lesson covers Eduqas Component 3, Option C: centre of gravity, stability and toppling; moments in muscle systems; impulse; the coefficient of restitution; moment of inertia, torque, angular momentum and its conservation; rotational kinetic energy; projectiles; and Bernoulli's equation and drag.
⚠️ This is an OPTION. Eduqas Component 3 Section B offers four options — A Alternating Currents, B Medical Physics, C The Physics of Sports, D Energy and the Environment. You study exactly ONE of them. Only work through this mini-lesson if Option C: The Physics of Sports is the one your school teaches. Everyone also needs the Component 3 core (Sections 1–10), which has its own mini-lesson.
The key idea: every rotational quantity is the exact analogue of a linear one. m → I; v → ω; F → τ; p → L; F = ma → τ = Iα; ½mv² → ½Iω². Learn the pairing and half the option comes free.
Press Start when you are ready.
(a)–(b) Stability and moments
Centre of gravity, toppling and muscles
The whole weight of a body acts at its centre of gravity. A body topples when the vertical line through the centre of gravity falls outside the base — beyond that point the weight's moment about the pivoting edge tips it over rather than restoring it.
So a stable stance in sport means a low centre of gravity and a wide base — a rugby player in a scrum, a judoka bracing against a throw. A sprinter in the blocks deliberately does the opposite, pushing the centre of gravity forwards and up so that they fall forwards into the first stride.
Moments in the body. Muscles attach very close to the joint, so their moment arm is tiny. Taking moments about the joint, a small distance must be compensated by a large force.
Worked example — the biceps
Elbow joint at the pivot. Biceps attaches 4.0 cm from the joint. A 50 N weight is held in the hand, 32 cm from the joint.
Moments about the elbow: F × 0.040 = 50 × 0.32 = 16 N m
F = 16 ÷ 0.040 = 400 N — eight times the load. Muscles trade force for speed and range of movement.
Quick check
When does it topple?
?A rugby player is pushed sideways. At what point does he actually topple over?
Calculate
Your turn — impulse in sport
1A footballer strikes a stationary ball of mass 0.44 kg, and it leaves his boot at 25 m s⁻¹. The contact lasts 8.0 ms. Calculate the average force on the ball.
N
Hint: Ft = mv − mu = 0.44 × 25 − 0 = 11 N s. F = 11 ÷ 8.0 × 10⁻³.
(c)–(d) Impulse and restitution
Impulse and the coefficient of restitution
Ft = mv − muImpulse = change of momentum. In sport, this is why a follow-through matters: extending the contact time increases the impulse for a given force, and therefore the final speed.
The same equation, read the other way, is the physics of protection: a boxing glove, a crash mat or a cricketer's "soft hands" all lengthen t for the same Δp, and so reduce the peak force.
e = relative speed after ÷ relative speed beforeThe coefficient of restitution. e = 1 for a perfectly elastic collision; e = 0 for a perfectly inelastic one (the objects move off together).
e = √(h / H)Drop a ball from height H, measure the bounce height h. Very easy to test experimentally — and it is exactly how sports bodies specify a legal ball.
Where the "lost" energy goes: into deforming the ball and the surface, and into internal energy — which is why a squash ball has to be warmed up before it will bounce properly, and why a cold, hard pitch changes a game.
Calculate
Your turn — coefficient of restitution
2A ball is dropped from a height of 1.20 m and bounces back to 0.75 m. Calculate the coefficient of restitution.
(no unit)
Hint: e = √(h/H) = √(0.75 ÷ 1.20) = √0.625.
(e)–(i) Rotation
Moment of inertia, torque and angular momentum
The moment of inertia I is the rotational equivalent of mass: it measures how hard it is to change a body's rotation. It depends not just on the mass but on how far that mass is from the axis.
solid sphere: I = ⅖mr² · thin spherical shell: I = ⅔mr²The shell has the larger I for the same m and r, because all of its mass is out at the maximum radius.
α = (ω₂ − ω₁) / t · τ = Iα · L = IωAngular acceleration · torque (the rotational F = ma) · angular momentum (the rotational p = mv).
rotational KE = ½Iω²A rolling ball has BOTH ½mv² (translation) and ½Iω² (rotation) — which is why a rolling ball reaches the bottom of a slope more slowly than a sliding one.
Conservation of angular momentum: with no external torque, Iω is constant. A spinning skater pulls her arms in, cutting I — so ω must rise. A diver tucks to somersault fast, then opens out to slow the rotation for a clean entry.
The energy question examiners love: the skater's rotational KE (½Iω²) increases when she pulls her arms in. Where does the energy come from? From the work she does pulling her arms in against the outward (centripetal-requiring) force. Angular momentum is conserved; kinetic energy is not.
Calculate
Your turn — moment of inertia
3A cricket ball of mass 0.16 kg and radius 0.036 m is treated as a uniform solid sphere. Calculate its moment of inertia, in units of 10⁻⁵ kg m².
× 10⁻⁵ kg m²
Hint: I = ⅖mr² = 0.4 × 0.16 × 0.036² = 0.4 × 0.16 × 1.296 × 10⁻³ = 8.29 × 10⁻⁵ kg m². Enter just the number in front.
Calculate
Your turn — the spinning skater
4A skater spins with moment of inertia 4.0 kg m² at an angular velocity of 2.0 rad s⁻¹. She pulls her arms in, reducing her moment of inertia to 1.6 kg m². Calculate her new angular velocity.
rad s⁻¹
Hint: Angular momentum is conserved: I₁ω₁ = I₂ω₂, so 4.0 × 2.0 = 1.6 × ω₂.
Calculate
Your turn — rotational kinetic energy
5Using the skater above after she has pulled her arms in (I = 1.6 kg m², ω = 5.0 rad s⁻¹), calculate her rotational kinetic energy.
J
Hint: KE = ½Iω² = 0.5 × 1.6 × 5.0². (Compare with 8 J before — she did 12 J of work pulling her arms in.)
Sort it
Linear, rotational or aerodynamic?
Tap an equation, then tap the family it belongs to.
➡️ Linear motion
🔄 Rotational motion
💨 Aerodynamics
Quick check
Why does she speed up?
?A spinning skater pulls her arms in and spins faster. Which statement is correct?
(m)–(o) Projectiles, Bernoulli and drag
Flight: the range, the swerve and the drag
Projectiles. Horizontal and vertical motion are independent. Ignoring air resistance, the maximum range for a given launch speed is at 45° — but real balls are launched at less than 45°, because drag punishes a long, slow, high flight.
p + ½ρv² = constant (Bernoulli)Eduqas writes it as p = p₀ − ½ρv²: where the fluid moves faster, the pressure is lower.
That single line explains a great deal of sport. A spinning ball drags a layer of air round with it, so the air moves faster on one side than the other. Lower pressure on the fast side means a net sideways force — the swerve of a free kick, the lift on a topspin tennis shot, the curve of a spinning cricket ball. A racing car's inverted wing uses the same effect to push down.
FD = ½ρv²ACDDrag depends on the square of the speed, on the frontal area A, and on the shape (through the drag coefficient CD). Doubling your speed quadruples the drag force — and it takes eight times the power to overcome it (since P = Fv).
Why cyclists crouch: at racing speeds nearly all a cyclist's power goes into overcoming aerodynamic drag. Reducing the frontal area A and the drag coefficient CD — a low tuck, a skinsuit, a teardrop helmet, or simply sitting in another rider's slipstream — is by far the cheapest way to go faster.
Calculate
Your turn — drag on a cyclist
6A cyclist rides at 12 m s⁻¹. Her frontal area is 0.40 m², her drag coefficient is 0.90, and the density of air is 1.2 kg m⁻³. Calculate the drag force.