M3 • SECTION 01
Further Kinematics
M3 extends the constant-acceleration model by allowing acceleration, velocity and displacement to be related through functions of time, position and other variables.
Lesson 1 — The Fundamental Definitions
For one-dimensional motion, displacement x is measured from a chosen origin. Velocity is the rate of change of displacement, while acceleration is the rate of change of velocity.
These equations are the starting point for almost every variable acceleration problem.
Lesson 2 — Integrating Acceleration
If acceleration is known as a function of time, integrate to obtain velocity. A constant of integration must then be determined from the initial conditions.
Then use
and integrate again when displacement is required.
Reliable method
- Write a = dv/dt.
- Integrate with respect to t.
- Use a known value of v to find the constant.
- Write v = dx/dt if x is required.
- Integrate again.
- Use the displacement condition to determine the second constant.
WORKED EXAMPLE 01
A particle moves in a straight line with acceleration a = 6t + 4. At t = 0 its velocity is 3 m/s. Find its velocity at t = 5 s.
a = dv/dt
dv/dt = 6t + 4
Integrating:
v = 3t² + 4t + C
At t = 0, v = 3, so C = 3.
Therefore:
v = 3t² + 4t + 3
At t = 5:
v = 75 + 20 + 3
v = 98 m/s
Lesson 3 — Finding Displacement from a(t)
If the particle's initial displacement is known, integration may be performed twice.
WORKED EXAMPLE 02
A particle has acceleration a = 12t − 6. Initially it has velocity 5 m/s and displacement 2 m. Find x in terms of t.
a = dv/dt
dv/dt = 12t − 6
v = 6t² − 6t + C
At t = 0, v = 5, so C = 5.
Thus:
v = 6t² − 6t + 5
Since v = dx/dt:
x = 2t³ − 3t² + 5t + D
At t = 0, x = 2, so D = 2.
x = 2t³ − 3t² + 5t + 2
Lesson 4 — Acceleration as a Function of Displacement
A particularly important M3 technique occurs when acceleration is given in terms of displacement rather than time.
Since
the chain rule gives
Therefore:
This allows time to be eliminated from the problem.
WORKED EXAMPLE 03
A particle has acceleration a = 4x. Its velocity is 3 m/s when x = 2. Find v² in terms of x.
a = v dv/dx
v dv/dx = 4x
v dv = 4x dx
Integrating:
½v² = 2x² + C
Therefore:
v² = 4x² + C'
When x = 2 and v = 3:
9 = 16 + C'
C' = −7
v² = 4x² − 7
Lesson 5 — Useful Kinematic Identities
Choose the form that removes the variable you do not need. If the question gives a(t), work with t. If it gives a(x), the chain-rule form is usually more efficient.
M3 • SECTION 02
Elastic Strings & Springs
This section develops the mathematical model of elastic strings and springs, including Hooke's law, modulus, energy, equilibrium and oscillatory motion.
Lesson 1 — Extension and Natural Length
Let l be the natural length and L be the current length. The extension x is
Extension is not the same as the total length. This distinction is essential in M3 questions.
Lesson 2 — Hooke's Law for a Spring
For an ideal spring obeying Hooke's law, the magnitude of the force is proportional to its extension or compression.
where k is the spring stiffness.
For a string, the standard elastic model is expressed using the modulus λ.
where:
- T is the tension.
- λ is the modulus.
- x is the extension.
- l is the natural length.
WORKED EXAMPLE 04
An elastic string has natural length 2.5 m and modulus 80 N. It is stretched to a length of 3.0 m. Find the tension.
Extension:
x = 3.0 − 2.5 = 0.5 m
Using:
T = λx/l
T = 80(0.5)/2.5
T = 16 N
Lesson 3 — Elastic Potential Energy
The work done in stretching an elastic string is stored as elastic potential energy.
For a spring obeying F = kx:
The energy may also be interpreted as the area under a force-extension graph.
WORKED EXAMPLE 05
An elastic string has modulus 100 N and natural length 2 m. It is extended by 0.4 m. Find the elastic energy stored.
E = ½λx²/l
E = ½ × 100 × 0.4² / 2
E = 4 J
Elastic potential energy = 4 J
Lesson 4 — Elastic Strings in Equilibrium
When an elastic string supports a particle, its tension can be balanced against weight or other forces.
Vertical equilibrium method
- Draw the particle and identify the direction of tension.
- Write the extension from the natural length.
- Use the elastic law to calculate T.
- Apply Newton's second law or equilibrium.
- Only then solve for the required variable.
for a particle hanging at rest from a single vertical elastic string.
Lesson 5 — Energy in Elastic Systems
When a particle moves while attached to an elastic string or spring, energy can transfer between gravitational potential energy, kinetic energy and elastic potential energy.
If no non-conservative work is done, the total remains constant.
M3 • SECTION 03
Further Dynamics & SHM
M3 dynamics introduces variable forces, gravitation and simple harmonic motion, requiring a strong connection between Newton's second law and differential equations.
Lesson 1 — Newton's Second Law
The fundamental equation remains
but the acceleration may now depend on displacement, velocity or time. The correct positive direction must be chosen before resolving forces.
Lesson 2 — Variable Forces
If a force varies with displacement, Newton's second law may be combined with
to obtain a relationship between velocity and position.
Variable-force strategy
- Choose a positive direction.
- Write the resultant force.
- Use F = ma.
- Replace a with v dv/dx if the force depends on x.
- Integrate.
- Apply the given initial condition.
Lesson 3 — Newton's Law of Gravitation
Two point masses attract one another with a force proportional to the product of their masses and inversely proportional to the square of their separation.
Here G is the universal gravitational constant and r is the distance between the centres of the two masses.
Lesson 4 — Gravitational Field Strength
The gravitational field strength at distance r from a point mass M is
Therefore the gravitational force on a mass m is
WORKED EXAMPLE 06
A satellite of mass m is at distance r from the centre of a planet of mass M. Show that its orbital speed for a circular orbit is
For circular motion:
GMm/r² = mv²/r
Cancel m:
GM/r² = v²/r
Multiply by r:
v² = GM/r
Therefore:
v = √(GM/r)
Lesson 5 — Definition of Simple Harmonic Motion
If x is measured from the equilibrium position, then
or
The negative sign is essential: it indicates that acceleration is directed towards the equilibrium position.
Lesson 6 — SHM Displacement
A general displacement equation may be written as
or equivalently using a sine form. A is the amplitude and φ is the phase constant.
The amplitude is the maximum displacement from equilibrium.
Lesson 7 — SHM Velocity
A very useful velocity-displacement relationship is
Therefore the maximum speed occurs at x = 0:
At an extreme position x = ±A, the velocity is zero.
Lesson 8 — SHM Acceleration
Hence maximum acceleration occurs at maximum displacement:
Lesson 9 — Period and Frequency
The period is the time for one complete oscillation.
WORKED EXAMPLE 07
A particle performs SHM with amplitude 0.25 m and angular frequency 8 rad/s. Find its maximum speed and maximum acceleration.
Maximum speed:
vmax = ωA
vmax = 8 × 0.25
vmax = 2 m/s
Maximum acceleration:
amax = ω²A
amax = 8² × 0.25
amax = 16 m/s²
Lesson 10 — SHM Energy
For SHM, total mechanical energy remains constant. The kinetic energy is greatest at equilibrium and zero at the extreme positions.
At displacement x:
M3 • SECTION 04
Motion in a Circle
Circular-motion problems require careful resolution towards the centre of the circle. The radial equation is one of the most important tools in this part of M3.
Lesson 1 — Angular Speed
For one complete revolution:
so
Lesson 2 — Radial Acceleration
A particle moving in a circle has an acceleration directed towards the centre.
Therefore the resultant inward force must satisfy
WORKED EXAMPLE 08
A particle of mass 2 kg moves in a circle of radius 1.5 m at speed 6 m/s. Find the resultant radial force.
F = mv²/r
F = 2 × 36 / 1.5
F = 48 N
Lesson 3 — Horizontal Circular Motion
For a horizontal circle, the vertical and horizontal force balances must be considered separately where appropriate.
For a conical pendulum, the tension has:
- a vertical component balancing the weight;
- a horizontal component providing the centripetal force.
Conical pendulum method
- Draw the tension along the string.
- Resolve vertically: vertical component balances mg.
- Resolve horizontally towards the centre.
- Set the horizontal resultant equal to mv²/r.
- Use the geometry of the cone.
Lesson 4 — Vertical Circular Motion
For a vertical circle, gravity changes its component relative to the radial direction as the particle moves.
At every point:
Energy conservation is often used to relate speeds at different points of the circle.
WORKED EXAMPLE 09
A particle moves in a vertical circle of radius r. At the lowest point its speed is v. Taking the highest point as 2r above the lowest point, find its speed V at the top if no energy is lost.
Loss of kinetic energy = gain in gravitational potential energy.
½mv² − ½mV² = mg(2r)
Cancel m and multiply by 2:
v² − V² = 4gr
Therefore:
V² = v² − 4gr
M3 • SECTION 05
Statics of Rigid Bodies
Rigid-body statics combines forces, moments, centres of mass and equilibrium. Diagram quality becomes especially important here.
Lesson 1 — Rigid Bodies
A rigid body is an idealised body whose shape does not change under the forces being considered.
Unlike a particle, a rigid body can experience rotational effects, so moments must be considered.
Lesson 2 — Moment of a Force
The moment of a force about a point measures its turning effect.
The unit is N m.
A clockwise and anticlockwise sign convention should be chosen and used consistently.
WORKED EXAMPLE 10
A 30 N force acts perpendicular to a rod at a distance of 0.8 m from a pivot. Find its moment.
Moment = Fd
= 30 × 0.8
= 24 N m
Lesson 3 — Equilibrium of a Rigid Body
For a rigid body to remain in equilibrium:
These conditions ensure there is neither translational nor rotational acceleration.
Lesson 4 — Choosing a Point for Moments
Take moments about a point that eliminates as many unknown forces as possible.
Exam technique
- Draw the complete force diagram.
- Resolve horizontal and vertical forces.
- Choose a convenient moment point.
- Write clockwise moments = anticlockwise moments.
- Use the force equations to determine remaining unknowns.
Lesson 5 — Centre of Mass
The centre of mass is the point at which the mass of a body may be considered to be concentrated when considering translational motion and gravitational effects.
For a uniform rod, the centre of mass lies at its midpoint.
For composite systems, moments about a convenient origin can be used to determine the centre of mass.
WORKED EXAMPLE 11
Two particles of masses 2 kg and 6 kg are positioned at x = 1 m and x = 5 m respectively. Find the x-coordinate of their centre of mass.
x̄ = Σ(mx)/Σm
x̄ = (2×1 + 6×5)/(2+6)
x̄ = 32/8
x̄ = 4 m
Lesson 6 — Ladders and Contact Forces
For a ladder or rod resting against surfaces, identify whether contacts are smooth or rough.
- A smooth contact provides a normal reaction only.
- A rough contact may provide both normal reaction and friction.
- The weight acts through the centre of mass.
- Moments should be taken about a point that removes unknown reactions.
M3 • FORMULA ATLAS
M3 Formula System
A compact reference for the major mathematical relationships used throughout M3.
Velocity
Rate of change of displacement.
Acceleration
Rate of change of velocity.
Chain Rule
Useful when acceleration depends on displacement.
Elastic String
Hooke-type elastic string model.
Elastic Energy
Energy stored in a stretched string.
Spring
Linear spring law.
Spring Energy
Elastic potential energy of a spring.
Gravitation
Newton's inverse-square law.
SHM
Acceleration towards equilibrium.
SHM Velocity
Speed at displacement x.
SHM Maximum Speed
Occurs at equilibrium.
SHM Maximum Acceleration
Occurs at an extreme position.
SHM Period
Time for one oscillation.
Angular Speed
Relation between linear and angular speed.
Radial Acceleration
Directed towards the centre.
Radial Force
Resultant inward force.
Moment
d is perpendicular distance to the line of action.
Centre of Mass
One-dimensional centre of mass.
- Variable acceleration: a = dv/dt or a = v dv/dx.
- Elastic energy: force-extension area becomes stored energy.
- SHM: a = −ω²x links displacement, velocity and acceleration.
- Circular motion: the inward resultant force produces mv²/r.
- Rigid statics: zero resultant force and zero resultant moment.
M3 • EXAM METHODS
How to Attack M3 Problems
Variable Kinematics
- Identify whether the acceleration depends on t, x or another variable.
- Use a = dv/dt for time-based acceleration.
- Use a = v dv/dx when displacement is involved.
- Integrate carefully.
- Apply initial conditions immediately after integration.
Elastic Systems
- Find natural length and extension separately.
- Use T = λx/l or F = kx.
- Check whether the string is taut.
- Use Newton's laws for force questions.
- Use energy when several positions and speeds are involved.
SHM
- Measure x from the equilibrium position.
- Establish that a is proportional to −x.
- Identify ω² from the acceleration equation.
- Use T = 2π/ω.
- Use v² = ω²(A² − x²) for speed at a given displacement.
Circular Motion
- Draw the circle.
- Mark the centre.
- Choose the inward direction as positive for the radial equation.
- Resolve actual forces towards the centre.
- Set the resultant equal to mv²/r.
- Use energy separately when heights change.
Rigid Bodies
- Draw every force.
- Resolve horizontally and vertically.
- Choose a convenient pivot.
- Use perpendicular distances for moments.
- Apply ΣFx = 0, ΣFy = 0 and ΣM = 0.
M3 • COMBINED PRACTICE
Exam-Style Practice
These questions deliberately combine methods from different parts of M3.
Show answer
Integrate: v = 4t² − 4t + C. At t = 0, C = 6. At t = 3: v = 30 m/s.
Show answer
v dv/dx = 6x. Hence v² = 6x² + C. Using x = 1, v² = 4, so C = −2. v² = 6x² − 2.
Show answer
T = 90(0.6)/3 = 18 N. E = ½(90)(0.6²)/3 = 5.4 J.
Show answer
vmax = ωA = 5(0.4) = 2 m/s.
Show answer
a = v²/r = 81/3 = 27 m/s².
Show answer
M = Fd = 12(2.5) = 30 N m.
Show answer
GMm/r² = mv²/r. Therefore: v = √(GM/r).
Show answer
A = 0.3 m, ω = 4 rad/s, T = 2π/4 = π/2 s.
Show answer
x̄ = (3×2 + 5×8)/8 = 46/8. x̄ = 5.75 m.
Show answer
Newton's second law gives the radial force condition ΣF = mv²/r, while energy conservation can relate speeds at different heights.
- Can you derive a = v dv/dx?
- Can you distinguish natural length from current length?
- Can you recognise SHM from an acceleration equation?
- Can you construct a radial equation?
- Can you take moments using perpendicular distances?
- Can you use several M3 ideas in the same problem?
M3 • 100 QUESTION EXAM
M3 Mastery Examination
A 100-question interactive examination covering the major M3 methods and relationships. Questions progress from fundamentals to integrated applications.