# Forces, movement, shape and momentum

> Edexcel International GCSE Physics · 4PH1 2017 Spec
> Source: https://www.owlsprep.com/study/edexcel-igcse-physics-s1-forces-movement-shape-and-momentum/

This guide covers all Edexcel IGCSE Physics (4PH1) content for forces, movement, shape and momentum, including core and higher Paper 2-only content, worked examples and exam-specific tips.

**Prerequisites:** [Basic kinematics (speed, velocity, acceleration)](https://www.owlsprep.com/study/edexcel-igcse-physics-s1-motion-and-kinematics/)

## Learning objectives

- Describe effects of forces: changes to speed, shape, direction
- Identify common force types and distinguish scalar vs vector quantities
- Calculate resultant collinear forces, use F=ma and W=mg relationships
- Explain stopping distance factors and forces on falling objects (including terminal velocity for higher tier)
- Apply Hooke's law and interpret force-extension graphs for springs, wires and rubber bands
- Solve higher tier problems on momentum conservation, force-momentum change, Newton's third law and principle of moments

## Core Forces Fundamentals: Types, Vectors & Resultant Forces

**Scalar vs Vector Quantities** — Scalar quantities have magnitude only, while vector quantities have both magnitude and direction. Force is a vector quantity.

Common force types you will encounter include gravitational (weight), electrostatic, magnetic, friction, air resistance, tension and normal contact force. Friction always acts to oppose motion.

To calculate the resultant of forces acting along a straight line, assign a positive sign to one direction and negative to the opposite direction, then calculate the algebraic sum of the forces.

**Worked example:** A cyclist applies a forward driving force of 320 N, while air resistance and friction create a total opposing force of 170 N. Calculate the resultant force on the cyclist and state its direction.

1. Step 1: Assign forward direction as positive, so driving force = +320 N, opposing force = -170 N.
2. $$\text{Resultant force} = 320 - 170 = 150 \text{ N}$$
3. Step 2: Resultant force is 150 N in the forward direction.

> **Exam tip:** Always state direction when asked for a vector quantity like force, you will lose 1 mark per question if you only provide magnitude.

*Calculator:* allowed

## Core Force Calculations: F=ma, Weight & Stopping Distance

**Weight** — Weight is the gravitational force acting on an object, calculated as $W = m \times g$, where $g = 10$ N/kg unless stated otherwise in a question.

Recall and use $F = m \times a$ (force = mass × acceleration) to calculate unknown force, mass or acceleration for moving objects.

Stopping distance = thinking distance + braking distance. Key factors affecting stopping distance include vehicle speed, vehicle mass, road surface conditions (wet/icy) and driver reaction time (affected by tiredness, alcohol or distractions).

For falling objects: core content requires you to identify the two forces acting: weight downward, and air resistance upward. For higher tier, explain that as speed increases, air resistance increases until it balances weight, resultant force becomes zero, and the object falls at constant terminal velocity.

**Worked example:** Calculate the resultant force required to accelerate a 1400 kg car at 1.5 m/s².

1. Step 1: Use the formula $F = m \times a$.
2. $$F = 1400 \times 1.5 = 2100 \text{ N}$$
3. Step 2: Resultant force = 2100 N.

> **Exam tip:** Never confuse mass (measured in kg) and weight (measured in N): this is one of the most common mark-losing errors in this topic.

*Calculator:* allowed

## Force, Shape & Hooke's Law Practical

**Hooke's Law** — The extension of an elastic object is directly proportional to the force applied, up to the limit of proportionality (the end of the linear region of the force-extension graph). Extension = final length - original length.

*Example:* A spring stretched from 4 cm to 7 cm has an extension of 3 cm.

Elastic behaviour means an object returns to its original shape once the deforming force is removed. In the required practical, you investigate extension vs applied force for helical springs, metal wires and rubber bands: only springs and wires follow Hooke's law in their linear region, rubber bands have non-linear force-extension graphs.

**Worked example:** A spring has an original length of 6 cm. When a 3 N weight is hung from it, its length becomes 10.5 cm. Calculate the extension, and state if it follows Hooke's law if its limit of proportionality is 4 N.

1. Step 1: Calculate extension = final length - original length.
2. $$\text{Extension} = 10.5 - 6 = 4.5 \text{ cm}$$
3. Step 2: The applied force is 3 N, which is less than the 4 N limit of proportionality, so the spring follows Hooke's law for this load.

> **Exam tip:** Always subtract the original length of the object to get extension: never use total stretched length in Hooke's law calculations.

*Calculator:* allowed

## Paper 2 (Higher Only): Momentum & Newton's Third Law

**Momentum** — Momentum is a vector quantity, calculated as $p = m \times v$, with units kg m/s. In a closed system with no external forces, total momentum is conserved: total momentum before a collision/explosion = total momentum after.

*Example:* A 2 kg ball moving at 4 m/s has momentum of 8 kg m/s.

Use $F = (mv - mu)/t$ (force = change in momentum / time) to explain safety features: crumple zones, seatbelts and airbags increase the time of impact, reducing the force exerted on passengers. Newton's third law states that for every action force, there is an equal and opposite reaction force, acting on two different bodies.

**Worked example:** A 3 kg trolley moving right at 4 m/s collides with a stationary 1 kg trolley, and they stick together after impact. Calculate their combined velocity after the collision.

1. Step 1: Calculate total momentum before collision (right = positive direction).
2. $$\text{Total before} = (3 \times 4) + (1 \times 0) = 12 \text{ kg m/s}$$
3. Step 2: Total momentum after collision = 12 kg m/s, combined mass = 4 kg.
4. $$v = p/m = 12 / 4 = 3 \text{ m/s}$$
5. Step 3: Combined velocity is 3 m/s to the right.

> **Exam tip:** Always assign positive and negative signs to opposite directions in momentum problems to avoid sign errors.

*Calculator:* allowed

## Paper 2 (Higher Only): Moments & Centre of Gravity

**Moment of a Force** — The turning effect of a force, calculated as moment = force × perpendicular distance from the pivot, units Nm. The weight of an object acts through its centre of gravity.

*Example:* A 10 N force applied 0.5 m from a pivot creates a moment of 5 Nm.

For a body in equilibrium, the sum of clockwise moments about a pivot equals the sum of anticlockwise moments about the same pivot (principle of moments). For a light beam supported at both ends, upward forces change as a heavy object moves along the beam: the support closer to the object exerts a larger upward force.

**Worked example:** A uniform seesaw has a pivot at its centre. A 500 N child sits 1.2 m to the left of the pivot. Where must a 300 N child sit to balance the seesaw?

1. Step 1: Calculate anticlockwise moment from the left child.
2. $$\text{Anticlockwise turn} = 500 \times 1.2 = 600 \text{ N m}$$
3. Step 2: Clockwise moment from the right child must equal 600 Nm for equilibrium.
4. $$d = moment / F = 600 / 300 = 2\m$$
5. Step 3: The 300 N child must sit 2 m to the right of the pivot.

> **Exam tip:** Always use the perpendicular distance from the pivot to the line of action of the force, not the total length of the lever.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Using total stretched length instead of extension for Hooke's law calculations
  - Why it fails: Hooke's law applies to the change in length, not the total length of the object
  - Correct: Always calculate extension as final length minus original length before using Hooke's law
- **Wrong:** Stating weight is measured in kilograms
  - Why it fails: Mass is measured in kg, weight is a force measured in Newtons
  - Correct: Use $W = mg$ to convert mass to weight, using $g = 10$ N/kg unless stated otherwise
- **Wrong:** Forgetting to assign direction signs to opposite velocities in momentum problems
  - Why it fails: Momentum is a vector, so opposite directions require opposite signs to get an accurate total momentum
  - Correct: Choose a positive direction (e.g. right = +) and assign negative values to velocities in the opposite direction
- **Wrong:** Using $g = 9.81$ m/s² in calculations
  - Why it fails: The Edexcel 4PH1 specification explicitly requires you use $g = 10$ m/s² unless a question states otherwise
  - Correct: Default to $g = 10$ N/kg for all calculations unless given a different value in the question
- **Wrong:** Adding perpendicular forces to calculate resultant force
  - Why it fails: The specification only requires you to calculate resultant forces along a single straight line, no vector resolution is required
  - Correct: For collinear forces, assign signs to opposite directions and calculate the algebraic sum
- **Wrong:** Omitting direction when answering for vector quantities (force, momentum, velocity)
  - Why it fails: Vectors require both magnitude and direction for full marks
  - Correct: Always state direction (e.g. forward, 2 m/s to the left) when giving answers for vector quantities

## Cheatsheet

| Formula | Variables | Units | Notes |
| --- | --- | --- | --- |
| $F = m \times a$ | F = force, m = mass, a = acceleration | F: N, m: kg, a: m/s² | Core, recall required |
| $W = m \times g$ | W = weight, g = gravitational field strength | W: N, g: N/kg | Core, use $g=10$ N/kg default |
| Stopping distance = thinking + braking distance |  |  | Core, factors: speed, mass, road condition, reaction time |
| $p = m \times v$ | p = momentum, v = velocity | p: kg m/s | Higher Paper 2 only, vector quantity |
| $F = (mv - mu)/t$ | $\Delta p$ = change in momentum, t = time | t: s | Higher Paper 2 only, explains safety features |
| Moment = F × perpendicular distance from pivot |  | Moment: Nm | Higher Paper 2 only, equilibrium: clockwise = anticlockwise moments |
| Hooke's Law: $F \propto$ extension |  |  | Core, extension = final length - original length, applies up to limit of proportionality |

## What's next

Now that you have mastered forces, movement, shape and momentum, you are ready to progress to the next topics in the Edexcel IGCSE Physics Forces and Motion unit. This topic forms the foundation for work, energy and power, as well as further mechanics concepts if you go on to study A-level Physics. Ensure you practice regular formula recall, as no formula sheet is provided in the exam, and focus on Paper 2-only content if you are sitting the higher tier assessment. Work through past paper questions for this topic to familiarise yourself with exam phrasing and mark scheme requirements, especially for Hooke's law practical and calculation questions on momentum and moments.

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