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MECHANICS 1 • DYNAMICS • COMPLETE ATLAS

Dynamics of a Particle

A complete Edexcel IAL Mechanics 1 guide to Newton's laws, resultant forces, weight, friction, connected particles, tension, inclined planes and dynamic modelling.

F = ma Newton II
W = mg Weight
ΣF = ma Resultant
100 Exam Questions

LESSON 01

Dynamics vs Kinematics

Dynamics is the part of mechanics concerned with the relationship between forces and motion. While kinematics describes how an object moves, dynamics investigates why its motion changes.

Kinematics

Deals with displacement, velocity, acceleration and time without considering the forces responsible for the motion.

Dynamics

Deals with forces and uses Newton's laws to determine acceleration and hence the resulting motion.

Particle model

In M1, an object is often modelled as a particle. Its size and shape are ignored, so all relevant forces can be treated as acting on a single point.

LESSON 02

Newton's First Law

Newton's First Law describes what happens when the resultant force on a particle is zero.

Newton's First Law

A particle remains at rest or continues to move with constant velocity unless acted upon by a non-zero resultant force.

Since acceleration is the rate of change of velocity, constant velocity means zero acceleration.

ΣF = 0 → a = 0

Equilibrium

A particle is in equilibrium when the vector sum of all forces acting on it is zero.

Equilibrium does not necessarily mean that the particle is stationary. A particle travelling with constant velocity can also be in equilibrium.

Exam trap:

Do not automatically write "the particle is stationary" when the resultant force is zero. The correct conclusion is that its acceleration is zero, so its velocity is constant.

LESSON 03

Newton's Second Law

Newton's Second Law is the central equation used throughout M1 dynamics.

ΣF = ma

The symbol ΣF means the vector sum, or resultant, of all forces acting on the particle.

F

Resultant force, measured in newtons (N).

m

Mass, measured in kilograms (kg).

a

Acceleration, measured in m/s².

Direction

The acceleration is in the direction of the resultant force.

Vector principle

If several forces act along one straight line, choose a positive direction and assign signs consistently. Forces acting in the opposite direction are negative.

LESSON 04

Force Diagrams and Free-Body Diagrams

Before applying Newton's Second Law, identify every force acting on the particle.

Common forces

Free-body diagram rule

Draw only forces acting ON the chosen particle. Do not draw forces that the particle exerts on other bodies.

LESSON 05

Weight and Gravity

The weight of a particle is the gravitational force acting on it.

W = mg

where m is mass and g is the acceleration due to gravity.

Unless another value is specified, Edexcel mechanics questions commonly use:

g = 9.8 m/s²

Mass vs Weight

Mass Weight
Amount of matter in an object. Force caused by gravity.
Measured in kg. Measured in N.
Does not depend on gravitational field strength. Depends on g.

Worked Example

A particle has mass 7 kg. Find its weight. Take g = 9.8 m/s².

W = mg

W = 7 × 9.8

W = 68.6 N

LESSON 06

Newton's Third Law

Newton's Third Law

When two bodies interact, the force exerted by the first body on the second is equal in magnitude and opposite in direction to the force exerted by the second body on the first.

The three important conditions

Common mistake:

The normal reaction and weight of an object are NOT automatically a Newton's Third Law pair because they act on the same body.

LESSON 07

Resultant Force

The resultant force is the single force that has the same effect on the motion of the particle as all the individual forces combined.

Collinear forces

Forces acting along one straight line can be combined algebraically after selecting a positive direction.

Rightward forces − Leftward forces = ma

Worked Example

A 5 kg particle is pulled horizontally by a 32 N force to the right and experiences a 12 N resistance to the left.

Resultant force = 32 − 12 = 20 N

20 = 5a

a = 4 m/s²

LESSON 08

Horizontal Motion

For horizontal motion on a surface, the vertical forces may balance while the horizontal forces produce acceleration.

For example, on a horizontal surface:

R = mg

provided there are no vertical accelerations and no other vertical forces.

Horizontally:

ΣF_horizontal = ma
Important distinction

A balanced vertical pair of forces does not imply that the particle has zero horizontal acceleration.

LESSON 09

Vertical Motion

When a particle moves vertically, weight acts downward. Other vertical forces may include tension or an applied force.

Choose a positive direction before writing Newton's Second Law.

ΣF_vertical = ma

Moving upward

If upward is positive, weight is negative:

T − mg = ma

Moving downward

If downward is positive:

mg − T = ma
Remember:

The direction of motion does not determine the direction of acceleration. Acceleration depends on the resultant force.

LESSON 10

Acceleration, Deceleration and Direction

A particle accelerates whenever its velocity changes.

It is important to distinguish between velocity and acceleration.

Same direction

If velocity and acceleration have the same direction, the particle's speed increases.

Opposite directions

If velocity and acceleration are in opposite directions, the particle slows down.

Therefore, a particle can be moving to the right while accelerating to the left.

LESSON 11

Normal Reaction

The normal reaction is the force exerted by a surface on a particle in contact with it.

It acts perpendicular to the surface.

R ⟂ surface

For a particle resting on a horizontal surface with no vertical acceleration and no additional vertical forces:

R = mg

However, this equation should not be used automatically on inclined planes or when other vertical forces are present.

LESSON 12

Inclined Planes

When a particle lies on a plane inclined at angle θ to the horizontal, resolve forces parallel and perpendicular to the plane.

Weight components

Component down the plane = mg sin θ
Component perpendicular to plane = mg cos θ

For a particle on a smooth plane with no other perpendicular force:

R = mg cos θ

and, if gravity is the only force parallel to the plane:

mg sin θ = ma
therefore:
a = g sin θ
Exam memory aid:

For an inclined plane measured from the horizontal, the component down the plane is mg sin θ and the component into the plane is mg cos θ.

LESSON 13

Friction

Friction acts between surfaces in contact and opposes relative motion or the tendency for relative motion.

Limiting friction

When a particle is on the point of sliding, friction has its maximum possible value, called limiting friction.

F = μR

where μ is the coefficient of friction and R is the normal reaction.

Coefficient of friction

The coefficient μ is dimensionless and describes the relationship between limiting friction and normal reaction.

If the surface is rough and the particle is moving or is on the point of moving, friction acts opposite to the direction of motion or impending motion.

LESSON 14

Smooth vs Rough Surfaces

Smooth surface Rough surface
No friction is considered. Friction acts between surfaces.
Only relevant forces such as R, W and T are considered. Include friction where appropriate.
Useful for simplified models. Used when resistance is important.
Key phrase:

If a question says the surface is smooth, do not introduce a friction force.

LESSON 15

Tension in a String

A taut light string transmits tension between connected particles.

For an ideal string and ideal pulley, the tension is the same throughout the string.

T = same throughout a light inextensible string

Important assumptions

LESSON 16

Connected Particles

When two or more particles are connected by a light inextensible string, their accelerations are related.

Key principle

If connected particles move under a taut inextensible string, their accelerations have equal magnitudes while the string remains taut.

Typical procedure

  1. Draw a separate force diagram for each particle.
  2. Choose a positive direction for each particle.
  3. Apply Newton's Second Law separately.
  4. Use the common acceleration condition.
  5. Solve the simultaneous equations.
  6. Check whether the result makes physical sense.

LESSON 17

Particle on a Rough Horizontal Surface

Suppose a particle of mass m is pulled horizontally by force P and the surface has coefficient of friction μ.

If there is no vertical acceleration:

R = mg
At limiting friction:
F = μmg
Therefore, horizontally:
P − μmg = ma
so:
a = (P − μmg) / m

This equation is only valid when the assumed friction state is appropriate.

LESSON 18

Particle on a Rough Inclined Plane

For a particle on an inclined plane, resolve forces parallel and perpendicular to the plane.

If the plane is inclined at θ:

Weight component down plane = mg sin θ
Normal component = mg cos θ
If friction acts up the plane:
mg sin θ − F = ma
If the particle is on the point of sliding:
F = μR
and, in the simplest case:
F = μmg cos θ

LESSON 19

Limiting Equilibrium

A particle is in limiting equilibrium when it is in equilibrium but is on the point of moving.

At limiting equilibrium, friction has reached its limiting value:

F = μR

This is different from ordinary equilibrium, where friction may be less than μR.

Very important:

Do not write F = μR for every rough-surface problem. It applies when the particle is moving or is on the point of slipping, according to the model being used.

LESSON 20

Modelling Assumptions

Mechanics questions often describe an idealised mathematical model. Understanding the assumptions is essential.

Phrase Meaning
Particle Size and shape are ignored.
Light string String mass is negligible.
Inextensible string String length remains constant.
Smooth pulley Negligible friction at pulley; tension is unchanged.
Smooth surface No friction.
Rough surface Friction may act.
Uniform gravitational field g is treated as constant.

WORKED EXAMPLES

High-Value Exam Problems

Example 1 — Horizontal Dynamics

A 3 kg particle is acted upon by forces of 25 N to the right and 7 N to the left. Find its acceleration.

Resultant = 25 − 7 = 18 N

18 = 3a

a = 6 m/s²

Example 2 — Vertical Motion

A 4 kg particle is pulled vertically upward by a tension of 60 N. Find its acceleration.

Take upward as positive.

T − mg = ma

60 − 4(9.8) = 4a

60 − 39.2 = 4a

20.8 = 4a

a = 5.2 m/s² upward

Example 3 — Smooth Inclined Plane

A particle rests on a smooth plane inclined at 30° to the horizontal. Find its acceleration down the plane.

Parallel component of weight:

mg sin 30°

ma = mg sin 30°

a = g sin 30°

a = 4.9 m/s²

Example 4 — Rough Horizontal Surface

A 5 kg particle is pulled horizontally by 35 N. The coefficient of friction is 0.2. Find its acceleration, assuming it is moving.

R = mg = 49 N

F = μR = 0.2 × 49 = 9.8 N

35 − 9.8 = 5a

25.2 = 5a

a = 5.04 m/s²

Example 5 — Connected Particles

Particles of masses 3 kg and 5 kg are connected by a light inextensible string passing over a smooth pulley. The 5 kg particle moves downward. Find the acceleration.

For the 5 kg particle:

49 − T = 5a

For the 3 kg particle:

T − 29.4 = 3a

Adding:

19.6 = 8a

a = 2.45 m/s²

FINAL EXAM

100-Question Dynamics Exam

A comprehensive M1 Dynamics assessment covering definitions, Newton's laws, forces, weight, friction, inclined planes, connected particles, tension and modelling.

100 Questions
100 Marks
MCQ Format
M1 Level
Question 1 of 100
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