OCR A-level Physics A (H556) · Module 3: Forces and Motion
Mini-Lesson
Forces & Motion
Module 3 is the mechanics backbone of H556: motion, forces in action, work, energy and power, materials, and Newton's laws with momentum.
v = u + at · s = ut + ½at² · v² = u² + 2as · F = ma · W = Fx cos θ · p = mvfive chapters, one set of tools
Newton's second law, properly stated: the resultant force is equal to the rate of change of momentum. F = ma is only the special case for constant mass. OCR asks for the momentum form, so learn it that way round.
Work through each screen, answer the questions as you go (several are full A-level calculations) and collect ⭐ stars. Press Start when you're ready.
Module 3 · motion
Kinematics and motion graphs
The four SUVAT equations apply only when acceleration is uniform. Choose the one missing the variable you neither know nor want.
v = u + at (no s) · s = ut + ½at² (no v) · v² = u² + 2as (no t) · s = ½(u + v)t (no a)
Displacement–time graph: gradient = velocity.
Velocity–time graph: gradient = acceleration; area under = displacement.
Acceleration–time graph: area under = change in velocity.
Projectiles split into two independent problems, joined only by time: horizontal velocity is constant (no horizontal force), while vertical motion has a = 9.81 m s⁻² downwards.
Worked example — braking
A car decelerates uniformly from 30 m s⁻¹ to rest in 60 m.
v² = u² + 2as → 0 = 30² + 2a(60) → a = −900 ÷ 120 = −7.5 m s⁻² (a deceleration of 7.5 m s⁻²)
Calculate
Your turn — deceleration
1A car travelling at 30 m s⁻¹ brakes uniformly to rest in 60 m. Calculate the magnitude of its deceleration. Give your answer in m s⁻².
m s⁻²
Hint: v² = u² + 2as: 0 = 30² + 2a(60), so a = −900 ÷ 120. Quote the magnitude.
Calculate
Your turn — projectile time of flight
2A ball is thrown horizontally from a height of 20 m at 15 m s⁻¹. Taking g = 9.81 m s⁻², calculate the time it takes to hit the ground. Give your answer in s to 3 significant figures.
s
Hint: Vertical only: 20 = ½ × 9.81 × t², so t² = 40 ÷ 9.81 = 4.077. Take the square root. (The horizontal speed is irrelevant here.)
Module 3 · Newton
Newton's laws, moments and equilibrium
N1: a body remains at rest or at constant velocity unless acted on by a resultant force.
N2: resultant force = rate of change of momentum, F = Δp/Δt → F = ma when mass is constant.
N3: if A exerts a force on B, then B exerts an equal and opposite force on A — same type, different bodies.
moment = F × perpendicular distance from the pivota COUPLE is two equal, antiparallel forces: torque = F × separation of the lines of action
A body in equilibrium satisfies both: resultant force = 0 and resultant moment = 0 about any point.
Terminal velocity is a straight application of N1: as speed rises, drag rises, until drag = weight, the resultant force is zero, and the acceleration becomes zero. The object keeps moving — fast — but stops speeding up.
Calculate
Your turn — principle of moments
3A uniform plank is pivoted at its centre. A 30 N weight is placed 0.80 m to the left of the pivot. What force, applied 0.40 m to the right of the pivot, will balance it? Give your answer in N.
N
Hint: Clockwise moment = anticlockwise moment: F × 0.40 = 30 × 0.80.
Sort it
Which of Newton's laws?
Tap a statement, then tap the law it illustrates.
1️⃣ First law
2️⃣ Second law
3️⃣ Third law
Quick check
Third-law pairs
?A book rests on a table. Its weight acts downwards and the table pushes up with a normal contact force. Are these a Newton's third-law pair?
Module 3 · energy
Work, energy, power and efficiency
W = Fx cos θ · Ek = ½mv² · Ep = mgh · P = W/t = Fvefficiency = useful output ÷ total input
Work is done only by the component of force along the displacement — a force at 90° does no work.
The principle of conservation of energy: energy cannot be created or destroyed, only transferred.
P = Fv is enormously useful for vehicles and lifts: a motor pulling with force F at steady speed v delivers power Fv.
Worked example — a lift motor
A motor raises a 50 kg load at a steady 0.40 m s⁻¹.
Force needed = weight = mg = 50 × 9.81 = 490.5 N (steady speed → zero resultant force)
P = Fv = 490.5 × 0.40 = 196 W
Calculate
Your turn — power
4A motor raises a 50 kg load at a constant 0.40 m s⁻¹. Taking g = 9.81 m s⁻², calculate the useful output power. Give your answer in W to 3 significant figures.
W
Hint: At constant speed the motor's force equals the weight: F = mg = 490.5 N. Then P = Fv = 490.5 × 0.40.
Module 3 · materials
Hooke's law, the Young modulus and material behaviour
F = kx · Eelastic = ½Fx = ½kx²elastic potential energy = the area under a force–extension graph
σ = F/A · ε = x/L · E = σ/εstress in Pa · strain is dimensionless · Young modulus in Pa (usually GPa)
The Young modulus is the gradient of the linear region of a stress–strain graph — a property of the material, independent of the sample's shape.
Elastic deformation is fully recoverable; plastic deformation leaves a permanent extension.
Brittle (glass): fractures with almost no plastic deformation. Ductile (copper): a long plastic region, so it can be drawn into a wire.
The area under a stress–strain graph is the energy stored per unit volume (J m⁻³).
Springs in combination: in series, each spring feels the full load, so the extensions add and the combination is less stiff (1/k = 1/k₁ + 1/k₂). In parallel, the load is shared, so the combination is stiffer (k = k₁ + k₂).
Calculate
Your turn — the Young modulus
5A wire under test has a stress of 2.0 × 10⁸ Pa and a strain of 1.0 × 10⁻³. Calculate its Young modulus. Give your answer in GPa.
GPa
Hint: E = σ/ε = 2.0 × 10⁸ ÷ 1.0 × 10⁻³ = 2.0 × 10¹¹ Pa. Divide by 10⁹ to convert to GPa.
Module 3 · momentum
Momentum, impulse and collisions
p = mv · F = Δp/Δt · impulse = FΔt = Δpmomentum in kg m s⁻¹ (= N s) — and it is a VECTOR
Conservation of linear momentum: in a closed system, total momentum before = total momentum after — in every collision and explosion.
Perfectly elastic: kinetic energy is also conserved. Inelastic: kinetic energy is not — some becomes internal energy, sound and deformation.
Impulse = area under a force–time graph. Extending the contact time reduces the peak force for the same change in momentum — airbags, crumple zones, bending your knees.
Worked example — a tennis serve
A 0.058 kg ball is accelerated from rest to 40 m s⁻¹ during a 5.0 ms contact.
Δp = mΔv = 0.058 × 40 = 2.32 kg m s⁻¹
F = Δp ÷ Δt = 2.32 ÷ 5.0 × 10⁻³ = 464 N — over 800 times the ball's weight
Calculate
Your turn — average force
6A tennis ball of mass 0.058 kg is served from rest to 40 m s⁻¹. The racket is in contact for 5.0 ms. Calculate the average force on the ball. Give your answer in N to 3 significant figures.
N
Hint: Δp = 0.058 × 40 = 2.32 kg m s⁻¹. F = Δp ÷ Δt = 2.32 ÷ 0.0050.
Quick check
Terminal velocity
?A skydiver has reached terminal velocity. Which statement is correct?
Quick check
Collisions
?Two trolleys collide and stick together. Which quantity is definitely conserved?
Quick check
Equilibrium
?A rigid body is in equilibrium. Which pair of conditions must hold?
Match it
Match the quantity to its equation
Tap an item on the left, then its partner on the right.
Quantity
Equation
Recap
The big ideas to know
SUVAT: uniform acceleration only; v–t graph gradient = a, area = s
Projectiles: horizontal velocity constant; vertical a = 9.81 m s⁻²; time links the two
Newton: N1 zero resultant → constant velocity · N2 F = Δp/Δt (= ma) · N3 equal, opposite, different bodies
Equilibrium: zero resultant force AND zero resultant moment; a couple has torque = F × separation
Energy: W = Fx cos θ · Ek = ½mv² · Ep = mgh · P = W/t = Fv · efficiency = useful ÷ total
Materials: F = kx · E_elastic = ½Fx · σ = F/A · ε = x/L · E = σ/ε (gradient of stress–strain)
Momentum: p = mv, always conserved; impulse = FΔt = Δp = area under a force–time graph
That is OCR Module 3 — the largest mechanics module on the specification. Press Finish to see your score.
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