Topic 2 builds the machinery you will use for the rest of the course: vectors, kinematics (SUVAT), projectiles, Newton's laws, and work, energy and power.
v = u + at · s = ut + ½at² · v² = u² + 2as · s = ½(u + v)tthe four SUVAT equations — each one is missing a different variable
The one rule that never breaks: horizontal and vertical motion are independent. In a projectile the horizontal velocity is constant while the vertical velocity changes at g. Treat them as two separate SUVAT problems joined only by time.
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.
Topic 2 · vectors
Scalars, vectors and resolving
A scalar has magnitude only (mass, speed, energy, power). A vector has magnitude and direction (displacement, velocity, acceleration, force, momentum).
horizontal = F cos θ · vertical = F sin θθ measured from the horizontal — always draw the triangle first
Any vector splits into two perpendicular components that are completely independent of each other.
Adding vectors: for two perpendicular vectors, resultant = √(x² + y²) and direction = tan⁻¹(y/x). For non-perpendicular ones, resolve both into x and y first, add the components, then recombine.
Calculate
Your turn — resolving
1A rope pulls a sledge with a force of 25 N at 30° above the horizontal. Calculate the horizontal component of the force. Give your answer in N to 3 significant figures.
N
Hint: Horizontal component = F cos θ = 25 × cos 30° = 25 × 0.866.
Sort it
Scalar, vector — or just a unit?
Students lose easy marks by confusing a quantity with its unit. Sort them out.
📏 Scalar quantity
➡️ Vector quantity
🔤 A unit, not a quantity
Topic 2 · moments
Moments, couples and equilibrium
moment = F × perpendicular distance from the pivotunit: N m — and it is NOT a joule, even though the units look identical
An object is in equilibrium when two conditions hold at once:
The resultant force is zero (it will not accelerate).
The resultant moment about any point is zero (it will not rotate).
The principle of moments: for a body in equilibrium, total clockwise moment = total anticlockwise moment about the same point.
Perpendicular is the word that matters. If a 20 N force acts at 60° to a 0.50 m spanner, the moment is 20 × 0.50 × sin 60° = 8.7 N m, not 10 N m.
Topic 2 · kinematics
The SUVAT equations
The four equations only apply when acceleration is uniform. Choose the one that is 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)
Take one direction as positive and stick to it. If up is positive, then a = −9.81 m s⁻².
On a velocity–time graph: gradient = acceleration, area under = displacement.
On a displacement–time graph: gradient = velocity.
Worked example — vertical throw
A ball is thrown straight up at u = 12 m s⁻¹. At the top, v = 0.
v² = u² + 2as → 0 = 12² + 2(−9.81)s → s = 144 ÷ 19.62 = 7.34 m
Calculate
Your turn — maximum height
2A ball is thrown vertically upwards at 12 m s⁻¹. Taking g = 9.81 m s⁻², calculate the maximum height reached. Give your answer in m to 3 significant figures.
m
Hint: At the top v = 0. Use v² = u² + 2as: 0 = 12² − 2(9.81)s, so s = 144 ÷ 19.62.
Topic 2 · projectiles
Projectile motion
A projectile is any object moving freely under gravity alone (air resistance neglected). Split the motion in two:
Horizontal: a = 0, so velocity is constant → s = vx t
Vertical: a = 9.81 m s⁻² downwards → full SUVAT applies
Horizontal launch: the vertical drop is exactly the same as if the ball had simply been dropped.
At the highest point of a projectile the vertical velocity is zero but the acceleration is still 9.81 m s⁻² downwards, and the horizontal velocity is unchanged. The object is never momentarily weightless.
Calculate
Your turn — horizontal projectile
3A stone is thrown horizontally at 20 m s⁻¹ from a cliff 45 m high. Taking g = 9.81 m s⁻², calculate how far from the base of the cliff it lands. Give your answer in m to 3 significant figures.
m
Hint: Vertical: 45 = ½ × 9.81 × t², so t = √(90 ÷ 9.81) = 3.03 s. Horizontal: s = 20 × t.
Quick check
At the top of the arc
?A ball is launched at an angle and follows a parabolic path. At the highest point of its flight, which statement is correct?
Topic 2 · Newton
Newton's three laws
First law: an object stays at rest or at constant velocity unless acted on by a resultant force. Constant velocity ⇒ resultant force = 0.
Second law: resultant force = rate of change of momentum. For constant mass this becomes F = ma.
Third law: if A exerts a force on B, B exerts an equal and opposite force on A — same type, different bodies, acting simultaneously.
Fresultant = maF in newtons, m in kilograms, a in m s⁻²
Third-law trap: the weight of a book on a table and the normal contact force from the table are not a third-law pair — they act on the same body and are different types of force. The partner of the book's weight is the gravitational pull the book exerts on the Earth.
Calculate
Your turn — Newton's second law
4A car of mass 1500 kg accelerates uniformly from rest to 27 m s⁻¹ in 9.0 s. Calculate the resultant force on the car. Give your answer in N.
N
Hint: a = (27 − 0) ÷ 9.0 = 3.0 m s⁻². Then F = ma = 1500 × 3.0.
Match it
Match each SUVAT equation to the variable it leaves out
Tap an item on the left, then its partner on the right.
Equation
Missing variable
Topic 2 · energy
Work, energy, power and efficiency
W = Fs cos θ · ΔEk = ½mv² · ΔEgrav = mgΔh · P = W/t = Fvwork and energy in joules; power in watts (J s⁻¹)
Work is only done when the force has a component along the displacement. A force at 90° to the motion does zero work — which is why the centripetal force in circular motion does no work.
Conservation of energy: energy cannot be created or destroyed. In a frictionless drop, all the lost gravitational potential energy appears as kinetic energy: mgh = ½mv².
Efficiency = useful output ÷ total input (× 100 for a percentage). It is always less than 1 in any real machine.
Worked example — power of a climber
A 70 kg student climbs 4.0 m of stairs in 5.0 s.
Work done against gravity = mgh = 70 × 9.81 × 4.0 = 2746.8 J
P = W ÷ t = 2746.8 ÷ 5.0 = 549 W
Calculate
Your turn — power
5A student of mass 70 kg climbs stairs of vertical height 4.0 m in 5.0 s. Taking g = 9.81 m s⁻², calculate their useful output power. Give your answer in W to 3 significant figures.
W
Hint: Work done = mgh = 70 × 9.81 × 4.0 = 2746.8 J. Then P = W ÷ t = 2746.8 ÷ 5.0.
Quick check
Zero work
?A satellite moves in a circular orbit at constant speed. How much work does the gravitational force do on it during one complete orbit?
Quick check
Constant velocity
?A skydiver falls at a constant velocity of 55 m s⁻¹. What is true of the forces acting?
Quick check
Efficiency
?A motor draws 800 W of electrical power and lifts a load, doing useful work at 600 W. What is its efficiency, and where does the rest go?
Recap
The big ideas to know
Vectors: components F cos θ and F sin θ; resultant √(x² + y²); scalars have no direction
Moments: moment = F × perpendicular distance; equilibrium needs zero resultant force AND zero resultant moment
SUVAT: v = u + at · s = ut + ½at² · v² = u² + 2as · s = ½(u+v)t — uniform acceleration only
Projectiles: horizontal velocity constant, vertical acceleration 9.81 m s⁻²; time is the shared variable
Newton: N1 zero resultant → constant velocity · N2 F = ma · N3 equal, opposite, same type, different bodies
Energy: W = Fs cos θ · Ek = ½mv² · ΔE = mgΔh · P = W/t = Fv · efficiency = useful ÷ total
That is the full sweep of Edexcel Topic 2 — the foundation for Further Mechanics and Fields. Press Finish to see your score.
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