F=ma and Constant Mass Revision Notes | A Level AQA Physics

1. Why F = ma Is the Engine of Mechanics

Every time a car accelerates away from traffic lights, a rocket lifts off, or a footballer kicks a ball, the same relationship governs the motion. That relationship is Newton's second law: $F = ma$. It connects the cause (force) to the effect (acceleration) through the property of the object that resists change (mass).

At A Level, you need more than the formula — you need to understand what it really says, when it applies, and how to use it precisely in multi-step problems. This lesson builds that understanding from the ground up.

2. Newton's Second Law — The Precise Statement

Newton's second law states that the resultant force acting on an object is directly proportional to, and in the same direction as, its acceleration, provided the mass remains constant. Written as an equation: $F = ma$.

The word resultant is doing critical work here. You must find the single net force — accounting for all forces and their directions — before substituting into $F = ma$. Plugging in just one of several forces is one of the most common errors at A Level.

Notice the equation can be rearranged in two other useful forms: $a = \frac{F}{m}$ (find acceleration when force and mass are known) and $m = \frac{F}{a}$ (find mass when force and acceleration are known). All three rearrangements are equally valid — choose whichever has the unknown on the left.

3. Understanding the Relationship — What the Equation Tells You

Think about what $F = ma$ is really saying. For a fixed mass, doubling the resultant force doubles the acceleration — they are directly proportional. For a fixed resultant force, doubling the mass halves the acceleration — mass and acceleration are inversely proportional. These proportionalities are the conceptual heart of the law.

Direction matters throughout. Force and acceleration are both vectors. If the resultant force points to the right, the acceleration points to the right. If the resultant force points backwards (opposing motion), the object decelerates — the acceleration vector points opposite to the velocity vector.

The graph above shows a straight line through the origin — the hallmark of direct proportionality. The gradient of an $a$ against $F$ graph equals $\frac{1}{m}$. Examiners sometimes ask you to interpret such a graph, so recognise that a steeper gradient means a smaller mass.

4. The Constant Mass Condition — When F = ma Applies

The equation $F = ma$ is a special case of Newton's second law that holds when mass is constant. At A Level, the vast majority of problems you encounter — cars, projectiles, blocks on surfaces, athletes — involve constant mass, so this form applies directly.

The more general form of Newton's second law involves the rate of change of momentum: $F = \frac{\Delta p}{\Delta t}$. When mass is constant, this reduces to $F = ma$. You will meet the momentum form in the momentum section of the course, but for this lesson the constant-mass case is the focus.

5. Finding the Resultant Force — The Critical First Step

Before you can use $F = ma$, you must find the resultant force. For forces acting along a single line, this means adding forces in one direction and subtracting forces in the opposite direction. Choose a positive direction at the start and stick to it throughout.

For example, if a 1200 kg car has a driving force of 3000 N forwards and a friction force of 800 N backwards, the resultant force is $3000 - 800 = 2200$ N forwards. That 2200 N is what goes into $F = ma$, not 3000 N.

6. Worked Examples — Building Exam Confidence

Theory is only useful when you can apply it under exam conditions. The following worked examples build from a straightforward single-force calculation to a more demanding multi-force scenario. Work through each step before revealing the next — that active engagement is what builds the skill.

Worked example: Calculating Acceleration from a Single Resultant Force

A resultant force of 360 N acts on a trolley of mass 45 kg. Calculate the acceleration of the trolley. [3 marks]

1. $F = ma$ — [M1] Selecting the correct formula is the first method mark. Always write it before substituting.
2. $a = \frac{F}{m}$ — Rearrange algebraically before substituting numbers — this keeps your working clear and earns the method mark even if you make an arithmetic slip later.
3. $a = \frac{360 \text{ N}}{45 \text{ kg}}$ — [M1] Include units at the substitution stage. The examiner can see your reasoning clearly.
4. $a = 8 \text{ m s}^{-2}$ — [A1] $360 \div 45 = 8$. The unit m s$^{-2}$ is essential for the accuracy mark. A bare '8' will not score the final mark.

Answer: $a = 8 \text{ m s}^{-2}$

Worked example: Multi-Force Problem: Finding Resultant Force First

A 1200 kg car accelerates at 2.5 m s$^{-2}$ along a straight road. The driving force from the engine is 4200 N. Calculate the magnitude of the resistive force acting on the car. [4 marks]

1. $F_{\text{net}} = ma = 1200 \times 2.5$ — [M1] The resultant force equals mass times acceleration. $1200 \times 2.5 = 3000$ N.
2. $F_{\text{net}} = 3000 \text{ N}$ (in the direction of motion) — This is the net force — the combined effect of the driving force and the resistive force.
3. $F_{\text{drive}} - F_{\text{resist}} = F_{\text{net}}$ $4200 - F_{\text{resist}} = 3000$ — [M1] The driving force acts forwards; the resistive force acts backwards. Their difference gives the resultant.
4. $F_{\text{resist}} = 4200 - 3000 = 1200 \text{ N}$ — [A1] The resistive force is 1200 N. Check: $4200 - 1200 = 3000$ N resultant, consistent with $ma = 1200 \times 2.5 = 3000$ N.

Answer: Resistive force $= 1200 \text{ N}$

7. Applying F = ma to Deceleration and Opposing Forces

Deceleration is not a separate concept — it is acceleration in the direction opposite to motion. If you define the direction of motion as positive, a deceleration gives a negative value of $a$, which means the resultant force is also negative (opposing motion). The equation $F = ma$ handles this automatically once you set a consistent sign convention.

Consider a cyclist of total mass 80 kg (including bicycle) travelling at constant velocity. At constant velocity, acceleration is zero, so the resultant force must be zero — the driving force exactly balances the resistive forces. This is a useful check: if $a = 0$, then $F_{\text{net}} = 0$, regardless of how large the individual forces are.

Worked example: Deceleration: Braking Force Calculation

A 900 kg car travelling at 20 m s$^{-1}$ brakes and comes to rest in 5.0 s. Calculate the braking force, assuming it is constant. [4 marks]

1. $a = \frac{\Delta v}{\Delta t} = \frac{0 - 20}{5.0} = \frac{-20}{5.0} = -4.0 \text{ m s}^{-2}$ — [M1] The negative sign shows the acceleration opposes the direction of motion — the car is decelerating. Taking the initial direction of travel as positive.
2. $F = ma = 900 \times (-4.0) = -3600 \text{ N}$ — [M1] The negative resultant force confirms it acts opposite to the direction of travel.
3. Braking force $= 3600 \text{ N}$ opposing the direction of motion — [A1] The question asks for the braking force — quote the magnitude (3600 N) and state its direction. Leaving out the direction loses a mark in many AQA mark schemes.

Answer: Braking force $= 3600 \text{ N}$ (opposing motion)

8. Exam Technique — Securing Every Mark

AQA mark schemes for $F = ma$ questions follow a predictable pattern: one mark for selecting or writing the correct formula, one mark for a correct substitution (including units), and one accuracy mark for the final answer with units. Losing the final mark because you omitted units is avoidable — make it a habit to write units at every stage.

When a question gives you multiple forces, always find the resultant before substituting. Write a clear force equation such as $F_{\text{net}} = F_{\text{drive}} - F_{\text{friction}}$ before doing any arithmetic. This shows the examiner your reasoning and protects your method marks even if you make a numerical slip.

Mastering $F = ma$ for constant mass situations means you can confidently calculate any one of force, mass, or acceleration when the other two are known — and you can handle multi-force scenarios by finding the resultant first. This single equation underpins the entire mechanics section of the AQA A Level course.