A-Level Maths Mechanics Formula Guide
Cambridge A-Level Mathematics (9709) | Paper 4 Mechanics
Mathematical modelling of motion and forces
Focus on setting up equations correctly - the maths follows naturally
Get the Complete Mechanics Guide
Enter your email to unlock all formulas with worked examples
No spam. Unsubscribe anytime.
Enter your email above to view the full guide
Mechanics Problem-Solving Strategy
- Draw a diagram - show all forces with arrows
- Choose axes - usually positive in direction of motion
- Resolve forces - into components if needed
- Apply Newton's laws - F = ma in each direction
- Use kinematics - to link displacement, velocity, acceleration, time
- Check units and signs - especially for direction
Kinematics (Constant Acceleration)
SUVAT equations - use when acceleration is constant. Choose based on which variable you DON'T need.
Final velocity from initial velocity, acceleration, time
Use when: You don't need displacement (s)
Variables: v = final velocity (m/s), u = initial velocity (m/s), a = acceleration (m/s²), t = time (s)
This is just the definition of acceleration rearranged: a = (v-u)/t
Worked Example:
A car accelerates from 10 m/s at 2 m/s² for 5 seconds. Find the final velocity.
u = 10 m/s, a = 2 m/s², t = 5 s
v = u + at = 10 + (2)(5)
v = 20 m/s
Practice:
A ball is thrown upward at 15 m/s. How long until it stops rising? (g = 10 m/s²)
Show Answer
v = 0, u = 15, a = -10
0 = 15 + (-10)t → t = 1.5 s
Displacement from initial velocity, acceleration, time
Use when: You don't need final velocity (v)
Key insight: This is a quadratic in t - can have two solutions
For objects dropped from rest: s = ½at² (since u = 0)
Worked Example:
A stone is dropped from a cliff. How far does it fall in 3 seconds? (g = 10 m/s²)
u = 0 (dropped), a = 10 m/s², t = 3 s
s = ut + ½at² = 0 + ½(10)(3)²
s = 45 m
Practice:
A car travelling at 20 m/s brakes at 4 m/s². How far does it travel in 3 s?
Show Answer
s = (20)(3) + ½(-4)(3)² = 60 - 18 = 42 m
Links velocities and displacement without time
Use when: You don't need time (t)
Rearrangements: s = (v² - u²)/2a, a = (v² - u²)/2s
Very useful for "stopping distance" problems where v = 0
Worked Example:
A car brakes from 30 m/s to rest over 45 m. Find the deceleration.
u = 30 m/s, v = 0 (stops), s = 45 m
v² = u² + 2as → 0 = 900 + 2a(45)
a = -900/90
a = -10 m/s² (deceleration of 10 m/s²)
Practice:
A ball is thrown upward at 20 m/s. Find the maximum height. (g = 10 m/s²)
Show Answer
At max height v = 0: 0 = 400 + 2(-10)s → s = 20 m
Displacement using average velocity
Use when: You don't need acceleration (a)
Key insight: Average velocity = (u + v)/2 for constant acceleration
Worked Example:
A train slows from 25 m/s to 15 m/s in 20 seconds. Find the distance travelled.
u = 25 m/s, v = 15 m/s, t = 20 s
s = ½(u + v)t = ½(25 + 15)(20)
s = 400 m
Newton's Laws of Motion
The foundation of mechanics - always draw a force diagram first!
Newton's Second Law
Use when: Relating force, mass, and acceleration
Variables: F = resultant force (N), m = mass (kg), a = acceleration (m/s²)
F is the RESULTANT (net) force - add all forces as vectors first
Worked Example:
A 1200 kg car accelerates at 2.5 m/s². The resistance force is 400 N. Find the driving force.
Resultant F = ma = 1200 × 2.5 = 3000 N
Resultant = Driving force - Resistance
3000 = D - 400
Driving force D = 3400 N
Practice:
A 50 kg box is pushed with 200 N. Friction is 80 N. Find the acceleration.
Show Answer
F = 200 - 80 = 120 N
a = F/m = 120/50 = 2.4 m/s²
Weight
Use when: Finding the gravitational force on an object
Variables: W = weight (N), m = mass (kg), g = 9.8 or 10 m/s²
Weight always acts DOWNWARD (towards center of Earth)
Worked Example:
Find the weight of a 75 kg person. (g = 10 m/s²)
W = mg = 75 × 10
W = 750 N
Friction
Use when: Object is on a rough surface
Variables: F = friction force (N), μ = coefficient of friction, R = normal reaction (N)
F = μR at limiting equilibrium (about to slip). F < μR when stationary and not about to slip.
Worked Example:
A 20 kg box on a floor (μ = 0.4) is pushed horizontally. Find the maximum friction force.
R = W = mg = 20 × 10 = 200 N (on horizontal surface)
F_max = μR = 0.4 × 200
F_max = 80 N
Practice:
A 5 kg block is on a slope at 30°. It's about to slip down. Find μ.
Show Answer
R = mg cos30° = 50 × 0.866 = 43.3 N
F = mg sin30° = 50 × 0.5 = 25 N
μ = F/R = 25/43.3 = 0.577 (or tan30°)
Connected Particles & Pulleys
For systems with strings, pulleys, or particles moving together.
Key Principles for Connected Particles
- Inextensible string: Both particles have the SAME acceleration magnitude
- Light string: Tension is the SAME throughout
- Smooth pulley: Tension is the SAME on both sides
Apply F = ma to EACH particle separately, then solve simultaneous equations
Worked Example:
Two particles (3 kg and 5 kg) are connected by a string over a smooth pulley. Find the acceleration and tension.
For 5 kg (moving down): 5g - T = 5a → 50 - T = 5a ... (1)
For 3 kg (moving up): T - 3g = 3a → T - 30 = 3a ... (2)
Adding: 50 - 30 = 8a → a = 20/8 = 2.5 m/s²
From (2): T = 30 + 3(2.5) = 37.5 N
a = 2.5 m/s², T = 37.5 N
Practice:
A 2 kg mass on a smooth table is connected to a 3 kg mass hanging over the edge. Find the acceleration.
Show Answer
For 3 kg: 30 - T = 3a
For 2 kg: T = 2a
Adding: 30 = 5a → a = 6 m/s²
Momentum & Impulse
Momentum is conserved in collisions when no external forces act.
Momentum
Use when: Analysing collisions or explosions
Variables: p = momentum (kg m/s or Ns), m = mass (kg), v = velocity (m/s)
Momentum is a VECTOR - direction matters! Use +/- for opposite directions.
Worked Example:
Find the momentum of a 1500 kg car travelling at 20 m/s.
p = mv = 1500 × 20
p = 30,000 kg m/s
Conservation of Momentum
Use when: Two objects collide or separate (explosion)
Key insight: Total momentum before = Total momentum after
For objects sticking together: m₁u₁ + m₂u₂ = (m₁ + m₂)v
Worked Example:
A 2 kg ball moving at 5 m/s collides with a stationary 3 kg ball. They stick together. Find their velocity.
Before: p = 2(5) + 3(0) = 10 kg m/s
After: p = (2 + 3)v = 5v
Conservation: 10 = 5v
v = 2 m/s
Practice:
A 4 kg trolley at 3 m/s collides with a 2 kg trolley at -2 m/s (opposite direction). They stick. Find v.
Show Answer
4(3) + 2(-2) = 6v
12 - 4 = 6v → v = 4/3 m/s
Impulse
Use when: Force acts for a short time (impact)
Variables: I = impulse (Ns), F = force (N), t = time (s)
Impulse = Change in momentum. Units: Ns or kg m/s
Worked Example:
A 0.4 kg ball hits a wall at 10 m/s and rebounds at 8 m/s. Find the impulse.
Taking toward wall as positive:
u = 10 m/s, v = -8 m/s (rebounds)
I = mv - mu = 0.4(-8) - 0.4(10) = -3.2 - 4
I = -7.2 Ns (away from wall)
Energy, Work and Power
Energy is conserved - track where it goes!
Kinetic Energy
Use when: Object is moving
Variables: KE = kinetic energy (J), m = mass (kg), v = speed (m/s)
KE is always positive (v² is always positive)
Worked Example:
Find the kinetic energy of a 1200 kg car travelling at 15 m/s.
KE = ½mv² = ½ × 1200 × 15²
KE = ½ × 1200 × 225
KE = 135,000 J = 135 kJ
Gravitational Potential Energy
Use when: Object is at height h above reference level
Variables: PE = potential energy (J), m = mass (kg), g = 10 m/s², h = height (m)
Choose a convenient reference level where PE = 0
Worked Example:
A 2 kg ball is dropped from 5 m. Find its speed just before hitting the ground (using energy).
PE lost = KE gained
mgh = ½mv²
v² = 2gh = 2 × 10 × 5 = 100
v = 10 m/s
Practice:
A ball is thrown upward at 20 m/s. Find the maximum height using energy.
Show Answer
KE = PE: ½mv² = mgh
h = v²/2g = 400/20 = 20 m
Work Done
Use when: Force moves through a distance
Variables: W = work (J), F = force (N), d = distance (m), θ = angle between F and d
If F is parallel to motion: W = Fd. If perpendicular: W = 0.
Worked Example:
A force of 50 N pulls a box 10 m at 30° to the horizontal. Find the work done.
W = Fd cos θ = 50 × 10 × cos30°
W = 500 × 0.866
W = 433 J
Power (force × velocity)
Use when: Object moving at constant velocity under a force
Also: P = W/t = Energy transferred / time
At maximum speed: Driving force = Resistance (acceleration = 0)
Worked Example:
A car has maximum power 60 kW. At top speed, resistance is 2000 N. Find the top speed.
At top speed: Driving force = Resistance = 2000 N
P = Fv → v = P/F = 60000/2000
v = 30 m/s
Practice:
A 1000 kg car accelerates at 2 m/s² while travelling at 20 m/s. Resistance is 500 N. Find the power.
Show Answer
F = ma + R = 1000(2) + 500 = 2500 N
P = Fv = 2500 × 20 = 50,000 W = 50 kW
Projectile Motion
Treat horizontal and vertical motion SEPARATELY - they're independent!
Key Principles for Projectiles
Horizontal:
- No acceleration (a = 0)
- Constant velocity: vₓ = u cos θ
- Distance: x = (u cos θ)t
Vertical:
- Acceleration = g (downward)
- Initial: vᵧ = u sin θ
- Use SUVAT equations
At maximum height: vertical velocity = 0
Worked Example:
A ball is projected at 20 m/s at 30° to horizontal. Find: (a) time of flight, (b) range, (c) max height.
uₓ = 20 cos30° = 17.3 m/s, uᵧ = 20 sin30° = 10 m/s
(a) At landing, y = 0: 0 = 10t - ½(10)t² → t(10 - 5t) = 0
t = 2 s (time of flight)
(b) Range: x = uₓ × t = 17.3 × 2
Range = 34.6 m
(c) At max height, vᵧ = 0: 0 = 10² - 2(10)h
Max height h = 5 m
Practice:
A ball is thrown horizontally at 15 m/s from a 20 m cliff. How far from the base does it land?
Show Answer
Vertical: 20 = ½(10)t² → t = 2 s
Horizontal: x = 15 × 2 = 30 m
Variable Acceleration (Calculus)
When acceleration changes with time or position - use calculus!
Differentiation relationships
Use when: Given s(t) or v(t), need to find velocity or acceleration
Key insight: Differentiate displacement → velocity → acceleration
Worked Example:
A particle moves with s = 2t³ - 3t² + t. Find the velocity and acceleration at t = 2.
v = ds/dt = 6t² - 6t + 1
At t = 2: v = 6(4) - 6(2) + 1 = 24 - 12 + 1
v = 13 m/s
a = dv/dt = 12t - 6
At t = 2: a = 12(2) - 6
a = 18 m/s²
Integration relationships
Use when: Given a(t) or v(t), need to find velocity or displacement
Key insight: Integrate acceleration → velocity → displacement. Don't forget +C!
Worked Example:
A particle has a = 6t - 4 m/s². At t = 0, v = 5 m/s and s = 2 m. Find s when t = 3.
v = ∫(6t - 4)dt = 3t² - 4t + C₁
At t = 0, v = 5: 5 = 0 + C₁ → C₁ = 5
So v = 3t² - 4t + 5
s = ∫(3t² - 4t + 5)dt = t³ - 2t² + 5t + C₂
At t = 0, s = 2: 2 = 0 + C₂ → C₂ = 2
So s = t³ - 2t² + 5t + 2
At t = 3: s = 27 - 18 + 15 + 2
s = 26 m
Practice:
A particle has v = t² - 4t + 3. Find when the particle is at rest.
Show Answer
At rest: v = 0
t² - 4t + 3 = 0 → (t-1)(t-3) = 0
t = 1 s and t = 3 s
More A-Level Resources
LumiExams.com | Cambridge A-Level Mathematics (9709) | Paper 4 Mechanics