Dynamics

Overview

Dynamics is the study of how forces cause changes in motion. While Kinematics describes motion using displacement, velocity and acceleration, Dynamics explains why that motion occurs.

This chapter develops the link between force and motion through Newton’s Laws of Motion, and extends to the ideas of momentum, impulse, and collisions.

Dynamics is foundational for later topics such as:

• Work, Energy and Power
• Circular Motion
• Oscillations
• Fields and interactions

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Core Ideas

What Dynamics Studies

Typical questions in Dynamics ask:

• What is the acceleration of a body under given forces?
• What force is needed to produce a certain motion?
• What happens when two bodies interact?
• How do objects move when connected together?
• What happens during collisions or recoil?

To solve such questions, combine:

• force laws
• vector resolution
• Kinematics
• momentum methods

Newton’s Laws of Motion

Newton’s First Law

A body remains:

• at rest, or
• moving with constant velocity in a straight line

unless acted upon by a resultant external force.

Meaning

If:

$$\sum \vec{F}=0$$

then acceleration is zero:

$$\vec{a}=0$$

The object may still be moving at constant velocity.

Inertia

Inertia is the tendency of a body to resist changes in motion.

• Larger mass → greater inertia
• Harder to start moving
• Harder to stop
• Harder to change direction

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Newton’s Second Law

The resultant force acting on a body equals the rate of change of momentum:

$$\sum \vec{F}=\frac{d\vec{p}}{dt}$$

where:

$$\vec{p}=m\vec{v}$$

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Constant Mass Case (H2 Standard Form)

If mass is constant:

$$\sum \vec{F}=\frac{d(m\vec{v})}{dt}=m\frac{d\vec{v}}{dt}=m\vec{a}$$

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Variable Mass Case (General Form)

If mass varies with time:

$$\sum \vec{F}=\frac{d(m\vec{v})}{dt}$$

Applying product rule:

$$\sum \vec{F}=m\frac{d\vec{v}}{dt}+\vec{v}\frac{dm}{dt}$$

or:

$$\sum \vec{F}=m\vec{a}+\vec{v}\frac{dm}{dt}$$

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Important Note (H2 Scope)

• Most H2 Physics problems assume constant mass, so:

$$\sum \vec{F}=m\vec{a}$$

is sufficient.

• The variable-mass form is relevant in systems such as:
  • rockets
  • leaking carts
  • sand falling onto moving belts

but detailed treatment is typically beyond core H2 requirements.

Key meanings

• Force causes acceleration.
• Acceleration is in the same direction as resultant force.
• If resultant force increases, acceleration increases (for fixed mass).

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Newton’s Third Law

If body A exerts a force on body B, then body B exerts a force on A that is:

• equal in magnitude
• opposite in direction
• of the same type

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Figure: Newton’s third-law forces are equal and opposite, but they act on different bodies and therefore do not cancel on a single object.

Properties of Action–Reaction Pairs

They:

• act on different bodies
• are equal in magnitude
• are opposite in direction
• are of the same type of force (e.g. both normal forces, both gravitational forces)
• act along the same line of action
• occur simultaneously

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Important Clarification

Action–reaction forces do not cancel, because they act on different bodies.

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Example

• hand pushes wall
• wall pushes hand

These forces:

• are equal and opposite
• act on different objects
• form an action–reaction pair

For fuller statement and comparison of the three laws, see Newton’s Laws of Motion.

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Mass, Inertia and Weight

Mass

Mass measures inertia.

• scalar quantity
• SI unit: kg
• independent of location

Weight

Weight is the gravitational force acting on a body:

$$\vec{W}=m\vec{g}$$

• vector quantity
• acts vertically downward near the Earth’s surface
• unit: N

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Sign Convention (Scalar Form)

If using a signed scalar approach and choosing upward as positive:

$$\vec{g}=-g$$

where:

$$g=9.81{\text{m s}}^{-2}$$

Hence:

$$W=-mg$$

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Key Idea

• Mass is constant.
• Weight depends on gravitational field strength.

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Resultant Force and Acceleration

To analyse motion:

1. Identify all external forces.
2. Resolve forces into perpendicular directions.
3. Apply:

$$\sum \vec{F}=m\vec{a}$$

Usually:

• horizontal direction
• vertical direction

treated separately.

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Free-Body Diagrams Overview

A free-body diagram (FBD) isolates one object and shows all external forces acting on it.

Common forces:

• Weight $(W)$
• Normal contact force $(N)$
• Tension $(T)$
• Friction $(f)$
• Resistive force
• Applied force
• Upthrust (if relevant)

See full treatment: Free-Body Diagrams and Force Analysis

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Force Resolution Overview

When a force acts at an angle, resolve it into perpendicular components. Each component is treated as a signed scalar along a chosen 1D axis.

Example:

For a force of magnitude $F$ making an angle $\theta$ above the positive horizontal direction:

Horizontal component:

$$F_x=F\cos \theta$$

Vertical component:

$$F_y=F\sin \theta$$

If the force points opposite to the positive direction of an axis, the corresponding component is negative.

Use Vectors carefully.

Standard Dynamics Applications

1. Horizontal Motion

Choose a positive direction (usually the direction of motion or acceleration).

Apply Newton’s Second Law along the horizontal direction:

$$\sum F_x=ma$$

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Frictionless Case

If no resistive forces are present and taking the direction of the applied force as positive:

$$F=ma$$

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With Friction

If friction is present (acting opposite to motion), and taking the direction of motion as positive:

$$F-f=ma$$

where:

• $F$ is the driving force
• $f$ is the frictional force

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2. Vertical Motion / Lifts

Figure: $\vec{R}$ is the upward pushing force exerted by the scale on the person, while $\vec{W}$ is the downward weight. The scale reading is the scalar $R=|\vec{R}|$, so it represents apparent weight and changes when the lift accelerates.

For vertical motion, first choose a positive direction.

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Lift Accelerating Upward (upward positive)

$$R-W=ma$$

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Lift Accelerating Downward (downward positive)

$$W-R=ma$$

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Where:

• $R$ is the normal reaction (scale reading)
• $W=mg$ is the weight

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Free Fall Case

If the lift accelerates downward with $a=g$:

$$R=0$$

The person is in free fall and experiences apparent weightlessness.

3. Connected Bodies

Bodies connected by a light, inextensible string share the same magnitude of acceleration along the direction of the string, provided the string remains taut.

Figure: For two bodies connected by a light inextensible string over a smooth pulley, the tension has the same magnitude throughout the string and both bodies share the same acceleration magnitude.

This figure illustrates a system of two masses connected by a light, inextensible string passing over a smooth pulley.

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(A) Physical Setup

• Two masses, $m_1$ and $m_2$, are connected by a single string over a pulley.
• The string is light (mass negligible) and inextensible (length constant).
• The pulley is smooth, so tension is the same throughout the string.

If $m_2>m_1$:

• $m_2$ accelerates downward
• $m_1$ accelerates upward

Both masses have the same magnitude of acceleration, denoted by $a$.

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(B) Free-Body Diagram for $m_1$

For mass $m_1$:

• Upward force: tension $T$
• Downward force: weight $m_{1}g$
• Acceleration: upward with magnitude $a$

Applying Newton’s Second Law (taking upward as positive):

$$T-m_{1}g=m_{1}a$$

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(C) Free-Body Diagram for $m_2$

For mass $m_2$:

• Upward force: tension $T$
• Downward force: weight $m_{2}g$
• Acceleration: downward with magnitude $a$

Applying Newton’s Second Law (taking downward as positive):

$$m_{2}g-T=m_{2}a$$

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Key Ideas

• The tension $T$ is the same in both diagrams (ideal string and pulley).
• The accelerations of $m_1$ and $m_2$ have the same magnitude but opposite directions.
• Each mass must be analysed using its own free-body diagram.
• Equations are solved simultaneously to find $a$ and $T$.

These results rely on the ideal assumptions that the string is light and inextensible, and that the pulley is smooth and light.

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Method

• draw a separate free-body diagram for each body
• apply:

$$\sum \vec{F}=m\vec{a}$$

to each body

• solve the resulting simultaneous equations

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Important Note

$N=mg\cos \theta$ only holds when there are no additional forces with components perpendicular to the plane (e.g. no pushing force or tension acting into or away from the plane).

Inclined Plane (Force Conditions)

Figure: On an inclined plane, the weight $mg$ is resolved into components parallel and perpendicular to the slope, while the normal reaction acts perpendicular to the surface and friction acts along the surface opposing motion or impending motion.

Resolve the weight $mg$:

• Parallel to slope (down the plane):

$$mg\sin \theta$$

• Perpendicular to slope (into the plane):

$$mg\cos \theta$$

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Normal Reaction

If no other forces act perpendicular to the surface:

$$|N|=mg\cos \theta$$

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Friction (Static or Constant Velocity)

If the object is:

• at rest, or
• moving with constant velocity (i.e. $a=0$),

then along the slope:

$$|f|=mg\sin \theta$$

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Important Notes

• The direction of friction is opposite to motion or impending motion.
• The relation $|f|=mg\sin \theta$ only holds when:
  • acceleration is zero, and
  • no other forces act along the slope.

General Case (taking downslope as positive)

$$mg\sin \theta -f=ma$$

where $f$ is the magnitude of the frictional force acting upslope. More generally, you may treat friction as a signed scalar if you define the positive direction first and keep the sign convention consistent.

For more worked applications involving lifts, connected bodies, and inclined planes, see Newtonian Dynamics Applications.

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Momentum Overview

Momentum:

$$\vec{p}=m\vec{v}$$

• vector quantity
• same direction as velocity

Large momentum may arise from:

• large mass
• high speed
• both

See: Momentum and Impulse

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Impulse Overview

Impulse is the change in momentum:

$$\vec{J}=\Delta \vec{p}$$

For constant force:

$$\vec{J}=\vec{F}\Delta t$$

More generally:

$$\vec{J}={\int }_0^{\Delta t}\vec{F}dt$$

For constant force, this reduces to:

$$\vec{J}=\vec{F}\Delta t$$

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1D Scalar Form

In 1D, with a chosen positive direction and signed scalar convention:

$$J={\int }_0^{\Delta t}Fdt$$

Impulse is equal to the area under the force–time graph.

Figure: The impulse delivered over a time interval equals the signed area under the force-time graph and is equal to the change in momentum.

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Key Idea

For a given change in momentum:

• increasing the collision time reduces the average force

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Applications

• airbags
• crumple zones
• bending knees when landing

See: Momentum and Impulse

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Conservation of Momentum Overview

For an isolated system with no resultant external force:

$$\text{Total momentum before}=\text{Total momentum after}$$

or equivalently,

$$\sum {\vec{p}}_{\text{before}}=\sum {\vec{p}}_{\text{after}}$$

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Two-Body System in general

$$m_1{\vec{u}}_1+m_2{\vec{u}}_2=m_1{\vec{v}}_1+m_2{\vec{v}}_2$$

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1D Scalar Form

For motion in one dimension, using a signed scalar convention:

$$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$$

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Applications

• recoil
• explosions
• collisions

See: Momentum Conservation and Collisions

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Collisions Overview

Elastic Collision

In an elastic collision, both momentum and kinetic energy are conserved.

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Conservation Laws

• momentum:

$$m_1{\vec{u}}_1+m_2{\vec{u}}_2=m_1{\vec{v}}_1+m_2{\vec{v}}_2$$

• kinetic energy:

$$\frac{1}{2}{m}_1|{\vec{u}}_1{|}^2+\frac{1}{2}{m}_2|{\vec{u}}_2{|}^2=\frac{1}{2}{m}_1|{\vec{v}}_1{|}^2+\frac{1}{2}{m}_2|{\vec{v}}_2{|}^2$$

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1D Collision Result

Figure: In a one-dimensional elastic collision, the relative speed of approach equals the relative speed of separation when velocities are treated as signed scalars along the same line.

For motion in one dimension, combining conservation of momentum and kinetic energy gives the signed-scalar relation:

$$u_1-u_2=v_2-v_1$$

Interpretation

• relative speed of approach = relative speed of separation

Inelastic Collision

In an inelastic collision, momentum is conserved but kinetic energy is not.

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Conservation of Momentum

$$m_1{\vec{u}}_1+m_2{\vec{u}}_2=m_1{\vec{v}}_1+m_2{\vec{v}}_2$$

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Energy Change

• kinetic energy decreases
• the loss is converted into:
  • heat
  • sound
  • deformation

Perfectly Inelastic Collision

In a perfectly inelastic collision, the bodies stick together after impact and move with a common velocity.

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Common Final Velocity

$${\vec{v}}_1={\vec{v}}_2=\vec{v}$$

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Conservation of Momentum

$$m_1{\vec{u}}_1+m_2{\vec{u}}_2=(m_1+m_2)\vec{v}$$

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Short Worked Examples

Example 1: Horizontal Force

A 4.0 kg block experiences a resultant force of 10 N.

$$a=\frac{F}{m}=\frac{10}{4.0}=2.5{\text{m s}}^{-2}$$

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Example 2: Momentum

A 2.0 kg trolley moves at 3.0 m s${}^{-1}$.

$$p=mv=(2.0)(3.0)=6.0{\text{kg m s}}^{-1}$$

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Example 3: Impulse

A force of 20 N acts for 0.50 s.

$$J=F\Delta t=(20)(0.50)=10\text{N s}$$

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Exam Relevance

Dynamics questions test whether you can:

• identify forces correctly
• choose a sensible system
• apply Newton’s laws consistently
• use momentum methods when force-based methods are awkward

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Formula Summary

Newtonian Motion

$$\sum \vec{F}=m\vec{a}$$ $$\sum \vec{F}=\frac{d\vec{p}}{dt}$$

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Weight

$$\vec{W}=m\vec{g}$$

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Momentum

$$\vec{p}=m\vec{v}$$

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Impulse

$$\vec{J}=\Delta \vec{p}$$ $$\vec{J}=\vec{F}\Delta t\text{(constant force)}$$ $$\vec{J}=\int \vec{F}dt$$

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Conservation of Momentum

$$m_1{\vec{u}}_1+m_2{\vec{u}}_2=m_1{\vec{v}}_1+m_2{\vec{v}}_2$$

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Elastic Collision (1D)

$$u_1-u_2=v_2-v_1$$

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Common Exam Pitfalls

1. Forgetting vector directions

Momentum, force, acceleration are vectors.

2. Mixing action-reaction pair with balanced forces

Balanced forces act on same body.
Third-law pair act on different bodies.

3. Assuming normal force always equals weight

Only true in special cases.

4. Using momentum conservation when external force exists

Must analyse system carefully.

5. Wrong sign convention

Choose positive direction clearly.

6. Confusing mass and weight

Mass in kg, weight in N.

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Revision Strategy

Master this order:

1. Newton’s laws
2. Free-body diagrams
3. Connected-body systems
4. Inclined planes
5. Momentum
6. Impulse
7. Collisions

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Learning Path

• Newton’s Laws of Motion
• Free-Body Diagrams and Force Analysis
• Newtonian Dynamics Applications
• Momentum and Impulse
• Momentum Conservation and Collisions
• Dynamics Methods and Non-Constant Forces (optional enrichment)
• Forces
• Kinematics
• Vectors
• Work, Energy and Power
• Circular Motion

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Final Takeaway

Dynamics turns motion description into physical explanation. If you can:

• draw accurate free-body diagrams
• apply $\sum \vec{F}=m\vec{a}$ consistently
• choose between force, impulse, and momentum methods sensibly
• handle collisions with the correct conservation laws

you will be strong in one of the most important H2 Physics chapters.

Links

• Prerequisite: kinematics
• Prerequisite: forces
• Prerequisite: vectors
• Related: newtons laws of motion
• Related: force diagrams and resolution
• Related: newtonian dynamics applications
• Related: momentum and impulse
• Related: momentum conservation and collisions
• Related: dynamics methods and non constant forces
• Related: work energy and power
• Related: circular motion
• Related: centre of gravity and stability
• Misconception: newtons laws distinction
• Misconception: force identification errors
• Misconception: vector direction consistency
• Misconception: momentum conservation system
• Misconception: inelastic ke conservation
• Misconception: elastic relative speed application
• Misconception: apparent weightlessness gravity
• Misconception: force resolution angles
• Misconception: force time graph area
• Misconception: scalars vs vectors

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