IGCSE Physics: Motion, Forces and Energy

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IGCSE Physics: Motion, Forces and Energy

TL;DR

You'll learn to calculate speed, acceleration, and the forces behind them using Newton's laws, plus how energy transforms and stays conserved in any system. You'll master distance-time and velocity-time graphs, F = ma, and the work-energy connection. By the end, you can solve any motion or energy problem IGCSE throws at you, numbers included.

1. The Mental Model

Motion is just how position changes with time; forces are what change motion. Energy is the currency that gets spent whenever something moves, heats up, or changes shape — it never disappears, only moves between accounts. Every problem in this topic is really asking one question: where did the energy go, and what force caused the change in motion? If something speeds up, slows down, or changes direction, a resultant force is acting on it — no exceptions.

2. The Core Material

2.1 Describing Motion: Speed, Velocity, Acceleration

Speed is distance covered per unit time; velocity is speed with a direction. Acceleration is the rate at which velocity changes.

$$\text{speed} = \frac{\text{distance}}{\text{time}}, \qquad a = \frac{v - u}{t}$$

where $u$ is initial velocity, $v$ is final velocity, $t$ is time taken. Units: m/s for speed/velocity, m/s² for acceleration.

Reading graphs is where marks live:

• Distance-time graph: the gradient = speed. A flat line means stationary. A curve means changing speed (acceleration).
• Velocity-time graph: the gradient = acceleration. The area under the graph = distance travelled. A horizontal line means constant velocity (zero acceleration). A line crossing zero means the object has reversed direction.

For constant acceleration, four equations connect your five motion variables ($u$, $v$, $a$, $t$, $s$ for displacement):

$$v = u + at \qquad s = ut + \tfrac{1}{2}at^2 \qquad v^2 = u^2 + 2as \qquad s = \tfrac{(u+v)}{2}t$$

You don't need to derive these for IGCSE — just know which one to use. Rule of thumb: identify the three values you're given, then pick the equation containing those plus the one you want.

2.2 Newton's Laws and Force

Newton's First Law: an object stays at rest or moves at constant velocity unless a resultant (unbalanced) force acts on it. This is why a car moving at constant speed on a flat road has balanced forces — driving force equals resistive forces (friction + air resistance) exactly.

Newton's Second Law gives you the calculation examiners love:

$$F = ma$$

where $F$ is the resultant force in newtons (N), $m$ is mass in kg, $a$ is acceleration in m/s². If multiple forces act, find the resultant first (forces in the same direction add, opposite directions subtract) before applying $F=ma$.

Newton's Third Law: every force has an equal and opposite reaction force, acting on a different object. A rocket pushes gas down; the gas pushes the rocket up. These two forces are the same type, same size, opposite direction, acting on different bodies — never on the same object, so they never cancel each other out.

Terminal velocity ties these together. A skydiver falling:
1. Starts with weight >> air resistance → accelerates downward.
2. As speed increases, air resistance increases.
3. Eventually air resistance = weight → resultant force = 0 → constant velocity (terminal velocity).

flowchart TD
    A["Skydiver jumps: weight greater than air resistance"] --> B["Resultant force downward, accelerates"]
    B --> C["Speed increases, air resistance increases"]
    C --> D{"Air resistance equals weight?"}
    D -- No --> B
    D -- Yes --> E["Resultant force = 0, terminal velocity reached"]
    E --> F["Parachute opens: air resistance now greater than weight"]
    F --> G["Decelerates until new, slower terminal velocity"]

2.3 Work, Energy and Power

Work done is the transfer of energy by a force acting over a distance:

$$W = Fd$$

where $W$ is work in joules (J), $F$ is force in newtons, $d$ is distance moved in the direction of the force, in metres. 1 joule = 1 newton-metre.

Two key energy stores you must calculate:

$$E_k = \tfrac{1}{2}mv^2 \qquad \qquad E_p = mgh$$

$E_k$ is kinetic energy (moving objects), $E_p$ is gravitational potential energy, $g \approx 9.8\,\text{m/s}^2$ (sometimes given as 10 for easier numbers), $h$ is height above a reference point.

The Principle of Conservation of Energy: energy cannot be created or destroyed, only transferred between stores or dissipated (usually as heat due to friction/air resistance). In a falling or swinging object with no resistive forces, loss in $E_p$ = gain in $E_k$:

$$mgh = \tfrac{1}{2}mv^2$$

Power is the rate of energy transfer (or rate of doing work):

$$P = \frac{E}{t} = \frac{W}{t}$$

measured in watts (W), where 1 W = 1 J/s.

3. Worked Example

Question: A 0.5 kg ball is dropped from rest from a height of 4.0 m. Ignore air resistance. Take $g = 10\,\text{m/s}^2$.

(a) Calculate the gravitational potential energy lost as it falls.
(b) Calculate its speed just before hitting the ground.
(c) If instead a resistive force acted and the ball hit the ground at 8.0 m/s, calculate the energy dissipated as heat.

Solution:

(a)
$$E_p = mgh = 0.5 \times 10 \times 4.0 = 20\,\text{J}$$

(b) With no air resistance, all $E_p$ converts to $E_k$:
$$E_k = 20\,\text{J} = \tfrac{1}{2}mv^2 = \tfrac{1}{2}(0.5)v^2$$
$$v^2 = \frac{20}{0.25} = 80 \quad \Rightarrow \quad v = \sqrt{80} \approx 8.9\,\text{m/s}$$

(c) With air resistance, actual $E_k$ at ground:
$$E_k = \tfrac{1}{2}(0.5)(8.0)^2 = \tfrac{1}{2}(0.5)(64) = 16\,\text{J}$$

Energy dissipated as heat = energy lost that didn't become kinetic energy:
$$20\,\text{J} - 16\,\text{J} = 4.0\,\text{J}$$

This last step is exactly how IGCSE tests conservation of energy with resistive forces — you always compare the "ideal" energy transfer to the "actual" one, and the gap is the dissipated energy.

4. Examiner's Breakdown

4.1 What Examiners Actually Reward

1. "State the correct SI unit alongside every numerical answer" — losing the unit loses the mark even if the number is right.
2. "Show the equation used before substituting numbers" — mark schemes award a method mark for writing $F=ma$ before plugging in values.
3. "Identify the resultant force by combining forces in the same and opposite directions before applying F=ma."
4. "State that energy is conserved, not destroyed, and is dissipated as heat/sound due to friction/air resistance."
5. "Read the gradient of a velocity-time graph as acceleration and the area under it as distance travelled."
6. "Give the direction as well as the magnitude when the quantity is a vector (velocity, force, acceleration, displacement)."

4.2 Trapdoor Mistakes

1. MISTAKE: Using "weight" and "mass" interchangeably, writing $F = mg$ as $9.8\,\text{kg}$. → FIX: Mass is in kg (amount of matter); weight is in N (force due to gravity, $W = mg$). Always separate them in your answer.
2. MISTAKE: Calculating distance from a velocity-time graph by just reading the final velocity value. → FIX: Distance is the area under the graph — split it into triangles/rectangles and sum the areas.
3. MISTAKE: Forgetting that deceleration is just negative acceleration, and writing it as a separate formula. → FIX: Use the same $a = (v-u)/t$; a negative result simply means the object is slowing down.
4. MISTAKE: Saying "the energy is lost" when describing friction. → FIX: Energy is never lost — say "energy is transferred/dissipated to thermal energy store," because conservation of energy must be stated correctly for full marks.

4.3 Score-Boosting Comparisons

5. Now Try It

Exam-style question: A 1200 kg car accelerates uniformly from rest to 20 m/s in 8 seconds along a flat road, then travels at constant velocity for 10 seconds, then brakes to rest in 4 seconds.

(a) Sketch the velocity-time graph for this journey. (b) Calculate the acceleration during the first 8 seconds. (c) Calculate the resultant force acting on the car during this acceleration phase. (d) Calculate the total distance travelled over the whole 22 seconds.

A top-band answer must: show $a = (v-u)/t$ with substitution before the answer; correctly apply $F=ma$ with units in newtons; calculate distance as the sum of three separate areas (triangle, rectangle, triangle) under the velocity-time graph, not a single formula; and state units at every step.

Success looks like: you get $a = 2.5\,\text{m/s}^2$, $F = 3000\,\text{N}$, and a total distance of 320 m, with every step shown and units included.

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