IGCSE Physics: Motion, Forces and Energy

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IGCSE Physics: Motion, Forces and Energy

TL;DR

Motion is how position changes over time; forces make motion happen; energy is the capacity to do work, and it transforms but never disappears. These three ideas connect everything in mechanics—learn how to measure them, predict what happens next, and solve real problems with equations that actually work.

1. The Mental Model

Think of a ball you throw. It starts at rest (no motion). Your hand pushes it (force). It flies through the air, slowing down (motion changing). Gravity pulls it back (another force). When it hits the ground, its motion energy converts into sound and heat (energy transforms). Three things happened: position changed, forces acted, and energy moved around. That's this entire topic.

2. The Core Material

Kinematics: Describing Motion

Motion isn't magic—it's just position changing. To describe it precisely, you need three things:

Displacement is how far an object moved in a specific direction. Distance is how far it actually travelled (total path length). They're different. If you walk 5 m north, then 3 m south, your distance is 8 m, but your displacement is 2 m north.

Velocity is displacement per unit time (a direction matters). Speed is distance per unit time (no direction). So:

$$v = \frac{\Delta s}{\Delta t}$$

where $\Delta s$ is displacement and $\Delta t$ is time. If you're asked "how fast", they probably want speed. If they ask "velocity", include direction.

Acceleration is how velocity changes:

$$a = \frac{\Delta v}{\Delta t}$$

Acceleration can be speeding up, slowing down (called deceleration or negative acceleration), or changing direction. All three count.

For motion in a straight line at constant acceleration, use these four equations (called the kinematic equations):

$$v = u + at$$
$$s = ut + \frac{1}{2}at^2$$
$$v^2 = u^2 + 2as$$
$$s = \frac{u + v}{2} \cdot t$$

Here $u$ is initial velocity, $v$ is final velocity, $a$ is acceleration, $s$ is displacement, $t$ is time. Use whichever one lets you find the unknown with the knowns you have.

Graphs matter. On a displacement–time graph, the gradient (slope) is velocity. On a velocity–time graph, the gradient is acceleration, and the area under the line is displacement. Flat lines mean zero gradient (no change). Steeper lines mean faster change.

Forces: Why Motion Changes

A force is a push or pull. Newton's first law says: an object stays at rest or moving in a straight line unless a force acts on it. That's why you slide forward when a car brakes—nothing pushed you back.

Newton's second law is the big one:

$$F = ma$$

Force (in newtons, N) equals mass (in kg) times acceleration (in m/s²). This is why a heavier object needs a bigger push to accelerate the same way. It also means if you apply the same force to two objects, the lighter one accelerates faster.

Newton's third law: forces come in pairs. If A pushes B with 10 N to the right, B pushes A with 10 N to the left. Equal and opposite, always.

Weight is the force of gravity on an object: $W = mg$, where $g = 9.8$ m/s² (or 10 m/s² for rough calculations). Weight is a force (measured in newtons), not mass.

When forces on an object cancel (net force = 0), it's in equilibrium—no acceleration. When they don't cancel, the net force causes acceleration in the direction of the net force.

Friction opposes motion. It acts parallel to a surface, against the direction of movement. The maximum static friction (before something starts moving) is different from kinetic friction (while moving). Both depend on the materials and the normal force (the force pushing the surfaces together).

Energy: What Makes Change Possible

Energy is the capacity to do work. Work happens when a force moves an object:

$$W = F \cdot s \cdot \cos\theta$$

where $F$ is force, $s$ is displacement, and $\theta$ is the angle between them. If the force and displacement are in the same direction, $\cos\theta = 1$, so $W = Fs$ (measured in joules, J).

There are many forms of energy:

• Kinetic energy (motion): $E_k = \frac{1}{2}mv^2$
• Gravitational potential energy (height): $E_p = mgh$ (where $h$ is height above a reference point)
• Elastic potential energy (stretched or compressed springs): $E_e = \frac{1}{2}kx^2$ (where $k$ is stiffness and $x$ is extension)
• Thermal energy (heat), chemical energy (in fuel), electrical energy, nuclear energy, light energy, sound energy

The law of conservation of energy says energy can't be created or destroyed, only converted. A falling ball loses gravitational PE and gains KE. A speeding car doing work against friction converts KE into thermal energy (heat in the brakes and tyres). A battery converts chemical energy into electrical energy, then into heat or light or motion.

When you pull a spring and let it go, elastic PE becomes KE. When you lift a book, you do work against gravity, storing gravitational PE. Whenever energy transforms, the total amount stays the same (ignoring losses to friction or heat for an ideal system).

Power is energy transferred per unit time: $P = \frac{E}{t}$ (measured in watts, W). A 60 W light bulb uses 60 joules every second.

graph TB
  A["Object at rest<br/>v = 0"] -->|"Force applied<br/>F = ma"| B["Object accelerates<br/>v increases"]
  B -->|"Velocity increases"| C["Kinetic energy<br/>E_k = ½mv²"]
  D["Height h<br/>Gravitational PE<br/>E_p = mgh"] -->|"Object falls<br/>Force = weight"| E["PE decreases"]
  E -->|"Force moves object"| F["Work done<br/>W = Fs"]
  C -->|"Friction acts"| G["Thermal energy<br/>lost to heat"]
  F -->|"Energy conservation"| C
  H["Energy stored<br/>in spring<br/>E_e = ½kx²"] -->|"Spring released"| B

3. Worked Example

A car of mass 1000 kg accelerates from rest at 2 m/s² for 5 seconds on a level road. Then it brakes with a deceleration of 3 m/s² until it stops. Find:
(a) The distance travelled during acceleration.
(b) The maximum velocity reached.
(c) The distance travelled during braking.
(d) The total work done by friction during braking.

Solution:

(a) During acceleration: $u = 0$, $a = 2$ m/s², $t = 5$ s. Use $s = ut + \frac{1}{2}at^2$:
$$s = 0 + \frac{1}{2}(2)(5^2) = 25 \text{ m}$$

(b) Maximum velocity (at end of acceleration): $v = u + at = 0 + 2(5) = 10$ m/s.

(c) During braking: $u = 10$ m/s, $v = 0$, $a = -3$ m/s². Use $v^2 = u^2 + 2as$:
$$0 = 100 + 2(-3)s$$
$$s = \frac{100}{6} = 16.67 \text{ m}$$

(d) During braking, kinetic energy is lost to friction. The work done by friction equals the change in KE:
$$W = \Delta E_k = 0 - \frac{1}{2}(1000)(10^2) = -50,000 \text{ J}$$

The negative sign shows friction removes energy. Friction does 50 kJ of work.

4. Examiner's Breakdown

4.1 What Examiners Actually Reward

1. Always include a direction when stating velocity or displacement. Examiners mark "10 m/s to the right" correct; "10 m/s" alone is incomplete for velocity (though it works for speed).

2. Show your choice of equation in kinematics problems. Write "Using $v^2 = u^2 + 2as$ because we know $u$, $v$, and $a$, and we need $s$" before you substitute. This proves you can choose the right tool.

3. Distinguish weight from mass explicitly. Write "mass = 50 kg" and "weight = 50 × 9.8 = 490 N" as separate statements, even if they ask for one of them.

4. In energy conservation questions, state the initial and final forms of energy clearly. Write "Initially, the spring stores elastic PE. Finally, the ball has KE and the surroundings have gained thermal energy due to friction." This shows you understand the transformation.

5. When friction or air resistance is mentioned, acknowledge the energy loss. Write "Some energy is lost to friction, so the final KE is less than the initial PE" rather than ignoring it and using perfect conservation.

6. Quote the relevant law or equation by name when it applies. For example: "By Newton's second law, $F = ma$, so..." or "By the law of conservation of energy, the total energy remains constant."

4.2 Trapdoor Mistakes

MISTAKE: Writing $F = ma$ without identifying what mass and acceleration you're using (which object? which direction?).
FIX: Always state "The net force on the car is $F_{\text{net}} = 1000 \times 2 = 2000$ N" and label what object and direction the equation applies to.

MISTAKE: Using $v = \frac{\Delta s}{\Delta t}$ for average velocity and then treating it as instantaneous velocity in the kinematic equations.
FIX: In kinematic equations, $v$ and $u$ are instantaneous velocities (at a specific moment). For average velocity over a time interval, use $v_{\text{avg}} = \frac{s}{t}$.

MISTAKE: Forgetting to include the $\frac{1}{2}$ in $E_k = \frac{1}{2}mv^2$ or $E_e = \frac{1}{2}kx^2$. Writing $E_k = mv^2$ costs marks.
FIX: The $\frac{1}{2}$ is non-negotiable. Check your formula sheet and write it every time.

MISTAKE: Assuming friction only acts when an object is sliding. Applying friction calculations to a stationary object that's about to move.
FIX: Static friction (before motion) can be any value up to $\mu_s N$; kinetic friction (during motion) is $\mu_k N$. Know which applies to your problem.

4.3 Score-Boosting Comparisons

5. Now Try It

Exam-style question (20 minutes):

A skier of mass 70 kg starts from rest at the top of a slope and accelerates down a frictionless section for 4 seconds. The acceleration is 1.5 m/s².

(a) Calculate the displacement down the slope during these 4 seconds. (2 marks)

(b) Calculate the skier's velocity at the end of this section. (2 marks)

(c) The skier then enters a section with friction. The friction force is 200 N. Using Newton's second law, calculate the new acceleration. (3 marks)

(d) The skier comes to rest after travelling a further 50 m in the friction section. Using $v^2 = u^2 + 2as$, calculate the velocity at which the skier entered the friction section. Does this match your answer to part (b)? Explain any difference. (4 marks)

(e) Calculate the thermal energy generated by friction in part (d). (3 marks)

What a top-band answer must contain:

• Part (a): State $s = ut + \frac{1}{2}at^2$; substitute $u = 0$, $a = 1.5$, $t = 4$; arrive at 12 m.
• Part (b): State $v = u + at$; substitute and arrive at 6 m/s.
• Part (c): Calculate net force from $F_{\text{net}} = mg\sin\theta - \text{friction}$ or show that the friction force alone (200 N) acts against motion; use $F = ma$ to find $a = F/m = 200/70 = 2.86$ m/s² deceleration.
• Part (d): Rearrange $v^2 = u^2 + 2as$ with $v = 0$, $a = -2.86$ m/s², $s = 50$ m; solve for $u \approx 15.1$ m/s. Explain the mismatch: the skier accelerated further on the slope before entering friction, or friction wasn't present immediately.
• Part (e): Use $W = \Delta E_k = 0 - \frac{1}{2}(70)(15.1)^2 \approx 8000$ J, or state that friction force × distance = 200 × 50 = 10,000 J and note the discrepancy comes from gravitational PE conversion.

Success looks like: all five parts attempted with correct equations stated, correct substitution of numbers, and final answers in the right units with at least three significant figures.

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