Newton's Second Law — AQA GCSE Combined Science
Test yourself on Newton's Second Law with AQA GCSE practice questions.
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Newton's Second Law explained
Newton’s Second Law links resultant force, mass and acceleration in one relationship.
Read the full explanation
For a fixed mass, doubling the resultant force doubles the acceleration, so acceleration is directly proportional to resultant force. For a fixed resultant force, doubling the mass halves the acceleration, so acceleration is inversely proportional to mass. Combining these gives a = F ÷ m, or F = m × a, where F is resultant force in newtons (N), m is mass in kilograms (kg) and a is acceleration in metres per second squared (m/s²). A worked example: a 2 kg trolley with a 6 N resultant force accelerates at 6 N ÷ 2 kg = 3 m/s². If the resultant force is zero, acceleration is zero and velocity stays constant. Always use the resultant force, not any single applied force, and convert grams to kilograms before calculating.
The acceleration of an object is proportional to the resultant force acting on the object, and inversely proportional to the mass of the object.
This statement describes the two proportionalities behind Newton’s Second Law. Acceleration is proportional to the resultant force: with mass fixed, doubling the resultant force doubles the acceleration, and halving it halves the acceleration. Acceleration is inversely proportional to mass: with resultant force fixed, doubling the mass halves the acceleration, and halving the mass doubles it. These combine as a = F ÷ m, where F is the resultant force in newtons (N), m is mass in kilograms (kg) and a is acceleration in metres per second squared (m/s²). For example, a resultant force of 12 N on a 3 kg object gives a = 12 N ÷ 3 kg = 4 m/s². If the same 12 N acts on 6 kg, a = 2 m/s². The force must be the resultant, found by combining all forces acting along the line of motion.
As an equation:
This statement introduces the mathematical form of Newton's Second Law, showing that the relationship between resultant force, mass and acceleration can be written as a word equation and then as a symbol equation. In AQA GCSE Combined Science: Trilogy, you must recognise that the equation links three quantities: resultant force (F), mass (m) and acceleration (a). The word equation is: resultant force = mass × acceleration. The symbol equation is: F = m × a. For example, a 2 kg trolley accelerating at 3 m/s² has a resultant force of 2 × 3 = 6 N. The equation applies only when the force is the resultant force acting on the object, and the acceleration is in the same direction as that resultant force. You should be able to rearrange the equation to find mass or acceleration, and substitute values with correct units.
resultant force = mass × acceleration
This statement gives the word equation for Newton's Second Law: resultant force = mass × acceleration. It tells you that the acceleration of an object is directly proportional to the resultant force acting on it and inversely proportional to its mass. For example, if a resultant force of 12 N acts on a 3 kg object, the acceleration is 12 ÷ 3 = 4 m/s². If the same 12 N acts on a 6 kg object, the acceleration is only 2 m/s². The equation is usually written as F = m × a, where F is resultant force in newtons (N), m is mass in kilograms (kg) and a is acceleration in metres per second squared (m/s²). You must be able to use this equation to calculate any one of the three quantities when the other two are known, and to describe how changing force or mass affects acceleration.
F = m a
Newton's second law links the resultant force on an object to its mass and acceleration through F = m a. Here F is the resultant force in newtons, m is the mass in kilograms and a is the acceleration in metres per second squared. The equation shows that a larger resultant force produces a larger acceleration for the same mass, while a larger mass needs a larger force for the same acceleration. For example, a 2 kg trolley pushed with a resultant force of 6 N accelerates at 3 m/s² because 6 ÷ 2 = 3. Rearranged, a = F ÷ m and m = F ÷ a. Always use the resultant force, found by adding or subtracting all forces along the line of motion, and keep units consistent before substituting.
force, F, in newtons, N
In F = m a, the symbol F stands for force measured in newtons, N. One newton is the force that gives a 1 kg mass an acceleration of 1 m/s², so 1 N = 1 kg m/s². Force is a vector, so its direction matters: a resultant force along the line of motion changes speed, while a force at an angle can change direction. In calculations, F usually means the resultant force, found by combining all forces acting on the object. For example, if a 3 kg box is pushed with 10 N and friction is 4 N, the resultant force is 10 N − 4 N = 6 N, giving an acceleration of 2 m/s². Always record force in newtons and state its direction where relevant.
mass, m, in kilograms, kg
Mass measures the quantity of matter in an object and is measured in kilograms (kg). In Newton's second law, F = m × a, mass is the constant that links force in newtons to acceleration in m/s². A larger mass accelerates less for the same resultant force. For example, a 2 kg trolley pushed with a resultant force of 6 N accelerates at 3 m/s² because 6 N ÷ 2 kg = 3 m/s². If the mass were 3 kg, the same 6 N would give only 2 m/s². Always convert grams to kilograms by dividing by 1000 before substituting into F = m × a, and remember that mass is not the same as weight; weight is a force measured in newtons and depends on gravitational field strength.
acceleration, a, in metres per second squared, m/s²
Acceleration is the rate of change of velocity and is measured in metres per second squared (m/s²). In Newton's second law, F = m × a, acceleration is directly proportional to the resultant force when mass is constant, and inversely proportional to mass when force is constant. For example, a resultant force of 10 N acting on a 2 kg mass gives a = F ÷ m = 10 N ÷ 2 kg = 5 m/s². Acceleration can be positive (speeding up) or negative (slowing down, often called deceleration). Always use the resultant force, not any single force, and ensure mass is in kilograms before dividing. A common method is to write the equation, rearrange to a = F ÷ m, substitute values with units, then calculate and state m/s².
inertial mass is a measure of how difficult it is to change the velocity of an object
Inertial mass describes an object's resistance to a change in velocity. A larger inertial mass means that for a given resultant force, the acceleration is smaller, so the velocity changes more slowly. Alternatively, to produce the same acceleration, a larger resultant force is needed. You can investigate this by applying the same resultant force to trolleys of different masses and measuring acceleration; the more massive trolley accelerates less. Inertial mass is measured in kilograms and is defined through the equation F = m a, where m is the inertial mass. It is not the same as gravitational mass, although in everyday situations the two have the same value. Explaining this idea means linking mass, resultant force and acceleration quantitatively and qualitatively.
inertial mass is defined as the ratio of force over acceleration.
Inertial mass measures how strongly an object resists a change in its velocity. It is defined as the ratio of force over acceleration, so m = F ÷ a, where F is the resultant force in newtons and a is the acceleration in metres per second squared. A larger inertial mass gives a smaller acceleration for the same resultant force. For example, a 1500 kg car pushed by a resultant force of 4500 N accelerates at 4500 N ÷ 1500 kg = 3 m/s². Rearranging gives F = m × a and a = F ÷ m. The ratio definition shows that mass is not the same as weight: weight is a force measured in newtons, while inertial mass is measured in kilograms.
Students should be able to estimate the speed, accelerations and forces involved in large accelerations for everyday road transport.
Everyday road transport involves large accelerations, so you should estimate speeds, accelerations and forces rather than measure them precisely. A car may accelerate from rest to about 27 m/s (roughly 60 mph) in about 10 s, giving an average acceleration of a = Δv ÷ t = 27 m/s ÷ 10 s = 2.7 m/s². For a 1200 kg car, the resultant force is F = m × a = 1200 kg × 2.7 m/s² = 3240 N. Braking gives a larger deceleration, perhaps 8 m/s², so the braking force is about 9600 N. Use sensible values, show the equation, substitute, and round the answer to one or two significant figures.
Students should recognise and be able to use the symbol that indicates an approximate value or approximate answer, ̴
In physics, measurements and calculated answers are rarely exact, so scientists show when a value is approximate. The symbol ̴ (a single wavy line, read as 'is approximately equal to' or 'approximately') can be placed between two quantities that are close in value, or used as a prefix before a number. For example, if a trolley of mass 0.50 kg is pushed with a force of 1.0 N, the acceleration is a = F ÷ m = 1.0 N ÷ 0.50 kg = 2.0 m/s² exactly. If the force is 1.1 N, then a = 1.1 N ÷ 0.50 kg = 2.2 m/s², but if the mass is only known to be about 0.5 kg, you would write a ̴ 2 m/s² or simply state the acceleration is ̴ 2 m/s². Use ̴ when rounding a calculated answer, reading a value from a graph, or when a measurement has uncertainty. Do not use it for exact definitions.
Required practical activity 19: investigate the effect of varying the force on the acceleration of an object of constant mass, and the effect of varying the mass of an object on the acceleration produced by a constant force.
This required practical tests Newton's Second Law, a = F ÷ m, in two ways. In the first, keep the mass constant and change the force, then measure the acceleration; a graph of acceleration against force should be a straight line through the origin, showing that acceleration is directly proportional to force. In the second, keep the force constant and change the mass, then measure the acceleration; a graph of acceleration against mass curves downwards, and a graph of acceleration against 1 ÷ mass is a straight line through the origin, showing that acceleration is inversely proportional to mass. A typical method uses a trolley on a runway, a string over a pulley, and hanging masses to provide the force, with light gates or a motion sensor to time the trolley. Control variables include the same runway, the same trolley and the same starting position.
Your focus
- Describe how acceleration depends on resultant force and mass.
- Apply F = m × a to calculate force, mass or acceleration.
- Use correct units and convert mass to kilograms before substitution.
Show all 39 objectives
- Explain the proportional and inverse proportional relationships in Newton’s Second Law.
- Use a = F ÷ m to solve numerical problems involving resultant force, mass and acceleration.
- Interpret changes in resultant force or mass and predict the effect on acceleration.
- State the word and symbol equations linking resultant force, mass and acceleration.
- Use the equation F = m × a to calculate resultant force, mass or acceleration.
- Apply correct SI units and convert values where necessary before substitution.
- Use the word equation resultant force = mass × acceleration to perform calculations.
- Describe the relationship between resultant force, mass and acceleration.
- Rearrange the equation to find mass or acceleration and apply correct units.
- State the equation F = m a and identify what each symbol represents, with its unit.
- Rearrange F = m a to find force, mass or acceleration and substitute values correctly.
- Apply the equation to a described situation and interpret the result in terms of the motion.
- Define force and state that it is measured in newtons, N.
- Explain the meaning of the newton in terms of mass and acceleration.
- Determine the resultant force on an object from the forces acting along the line of motion.
- State that mass is measured in kilograms and identify the symbol m.
- Convert masses between grams and kilograms accurately before calculation.
- Apply m = F ÷ a to calculate mass from resultant force and acceleration.
- State that acceleration is measured in m/s² and define it as the rate of change of velocity.
- Calculate acceleration using a = F ÷ m with force in newtons and mass in kilograms.
- Explain how acceleration changes when resultant force or mass is altered.
- Define inertial mass as a measure of resistance to a change in velocity.
- Explain how inertial mass affects acceleration for a given resultant force.
- Use F = m a to calculate or compare inertial mass, force and acceleration.
- State the definition of inertial mass as the ratio of force over acceleration.
- Apply m = F ÷ a to calculate mass, force or acceleration from given data.
- Explain how inertial mass affects the acceleration produced by a resultant force.
- Estimate realistic speeds, accelerations and forces for everyday road transport.
- Calculate acceleration from a speed change and time, and force from mass and acceleration.
- Justify estimated values using sensible assumptions and appropriate units.
- State that ̴ means 'is approximately equal to' and identify it in a written expression.
- Select correctly between = and ̴ when writing the result of a calculation or a measurement.
- Write a rounded or estimated value from a Newton's Second Law calculation using ̴ in the correct position.
- Plan and carry out a practical to measure how acceleration depends on force at constant mass and on mass at constant force.
- Plot and interpret graphs of acceleration against force and acceleration against 1 ÷ mass to identify direct and inverse proportionality.
- Evaluate the method by identifying sources of error and suggesting realistic improvements.
Newton's Second Law exam tips
Marking Points
- State that acceleration is directly proportional to the resultant force when mass is constant.
- State that acceleration is inversely proportional to mass when the resultant force is constant.
- Recall and apply the relationship F = m × a, or its rearrangements a = F ÷ m and m = F ÷ a.
- Use the correct units: resultant force in newtons (N), mass in kilograms (kg) and acceleration in metres per second squared (m/s²).
- Calculate a resultant force by vector addition or subtraction of forces acting along the same line before using F = m × a.
- Interpret proportionality statements, for example recognising that doubling F doubles a, and doubling m halves a.
- Explain that acceleration is directly proportional to the resultant force when mass is constant.
- Explain that acceleration is inversely proportional to mass when the resultant force is constant.
- Combine the two proportionalities into the relationship a = F ÷ m, or equivalently F = m × a.
- Identify the resultant force as the vector sum of all forces acting along the line of motion.
- Use consistent SI units: newtons for force, kilograms for mass and m/s² for acceleration.
- Predict the effect on acceleration of changing resultant force or mass by a given factor.
- States the word equation: resultant force = mass × acceleration.
- States the symbol equation: F = m × a, with F, m and a defined.
- Uses the equation to calculate resultant force when mass and acceleration are given.
- Rearranges the equation to calculate mass as m = F ÷ a or acceleration as a = F ÷ m.
- Applies correct units: force in newtons (N), mass in kilograms (kg), acceleration in metres per second squared (m/s²).
- Recognises that the force in the equation is the resultant force, not any single applied force.
- States the word equation: resultant force = mass × acceleration.
- Identifies F as resultant force in newtons (N), m as mass in kilograms (kg) and a as acceleration in metres per second squared (m/s²).
- Calculates acceleration using a = F ÷ m when resultant force and mass are known.
- Calculates mass using m = F ÷ a when resultant force and acceleration are known.
- Explains that for a constant mass, acceleration is directly proportional to resultant force.
- Explains that for a constant resultant force, acceleration is inversely proportional to mass.
- State that F is the resultant force in newtons, m is the mass in kilograms and a is the acceleration in metres per second squared.
- Use the equation in the form F = m a, or rearrange it to a = F ÷ m or m = F ÷ a, as the question requires.
- Calculate the resultant force by combining all forces acting along the line of motion before substituting into the equation.
- Substitute values with consistent units, for example convert grams to kilograms and show the arithmetic clearly.
- Interpret a numerical answer in context, such as stating the direction of the acceleration or explaining what a larger mass does to the acceleration for a fixed force.
- Identify F as force and state that it is measured in newtons, N.
- Explain that the newton is the SI unit of force and that 1 N = 1 kg m/s².
- Recognise that force is a vector quantity, so both magnitude and direction are needed to describe it fully.
- Distinguish the resultant force from individual applied forces by combining forces along the line of motion.
- Use force values correctly in F = m a, keeping the unit N with the answer.
- State that mass is measured in kilograms (kg) and is a scalar quantity representing the amount of matter.
- Use the correct unit kg in calculations and final answers, converting grams to kilograms where necessary.
- Rearrange F = m × a correctly to m = F ÷ a when mass is the unknown.
- Substitute numerical values with consistent units before calculating, for example 500 g = 0.5 kg.
- Interpret a larger mass as producing a smaller acceleration for the same resultant force.
- Distinguish mass (kg) from weight (N) and explain that weight = mass × gravitational field strength.
- State that acceleration is measured in metres per second squared (m/s²) and is the rate of change of velocity.
- Rearrange F = m × a correctly to a = F ÷ m when acceleration is the unknown.
- Use the resultant force, found by combining all forces acting along the line of motion, before calculating acceleration.
- Substitute mass in kilograms and force in newtons to obtain acceleration in m/s².
- Interpret acceleration as directly proportional to resultant force for constant mass.
- Interpret acceleration as inversely proportional to mass for constant resultant force.
- Recognise negative acceleration as deceleration when the resultant force opposes motion.
- Define inertial mass as a measure of how difficult it is to change an object's velocity, or its resistance to acceleration.
- State that inertial mass is measured in kilograms and is the m in the equation F = m a.
- Explain that for a given resultant force, a larger inertial mass produces a smaller acceleration, so velocity changes more slowly.
- Explain that to give two objects the same acceleration, the object with greater inertial mass needs the greater resultant force.
- Use a practical example, such as comparing a loaded and an unloaded trolley pushed with the same force, to show the effect of inertial mass.
- Rearrange F = m a to m = F ÷ a and describe how this relationship defines inertial mass from force and acceleration measurements.
- State that inertial mass is the ratio of force to acceleration, written as m = F ÷ a.
- Identify F as the resultant force in newtons and a as the acceleration in metres per second squared.
- Explain that a larger inertial mass produces a smaller acceleration for the same resultant force.
- Substitute values into m = F ÷ a and give the unit of mass as kilograms.
- Rearrange the relationship to find force using F = m × a or acceleration using a = F ÷ m.
- Distinguish inertial mass in kilograms from weight, which is a force in newtons.
- Estimate a realistic speed for road transport, such as about 13 m/s for a town car or about 30 m/s for a motorway car.
- Calculate acceleration from a speed change and time using a = Δv ÷ t, including deceleration as a negative acceleration.
- Use F = m × a with an estimated vehicle mass to estimate the resultant force during acceleration or braking.
- Choose sensible approximate values for speed, time and mass, and state the assumptions made.
- Round estimated answers to one or two significant figures and give correct units such as m/s, m/s² and N.
- Recognise that braking produces a larger deceleration and therefore a larger force than gentle acceleration.
- Recognises the symbol ̴ as meaning 'is approximately equal to' or 'approximately' and reads it correctly.
- Uses ̴ correctly, either between two numerical values or expressions, or as a prefix before a number, for example writing a ̴ 2 m/s² or ̴ 2 m/s² after rounding a calculated acceleration.
- Distinguishes between situations that need an equals sign, such as exact arithmetic like 1.0 N ÷ 0.50 kg = 2.0 m/s², and situations that need ̴, such as a rounded or measured value.
- Applies the symbol when reporting results from Newton's Second Law calculations where force, mass or acceleration have been measured to limited precision or read from a graph.
- Interprets a value written with ̴ as an approximation, so a later calculation using that value should also be treated as approximate.
- Uses the symbol consistently in written answers, tables and graph labels without confusing it with the equals sign or the 'proportional to' symbol.
- Identifies the independent variable, dependent variable and control variables for each part of the investigation, for example force changed and acceleration measured while mass is fixed in part one.
- Describes a workable method to apply a known force to a trolley of constant mass, such as hanging masses on a string over a pulley, and to measure the acceleration, such as using light gates or a motion sensor.
- Explains how to reduce the effect of friction, for example by tilting the runway slightly until the trolley moves at constant velocity before the force is applied.
- Processes results by plotting a graph of acceleration against force and recognising a straight line through the origin as evidence that acceleration is directly proportional to force.
- Processes results by plotting acceleration against mass and, where appropriate, acceleration against 1 ÷ mass, recognising the straight line through the origin as evidence that acceleration is inversely proportional to mass.
- Evaluates the investigation by identifying sources of uncertainty, such as timing errors or friction, and suggesting improvements such as repeating readings and calculating a mean.
Examiner Tips
- 💡Write the equation, substitute values with units, then calculate and give the unit with your answer.
- 💡When a question says constant speed, check whether the resultant force is zero before calculating acceleration.
- 💡For proportional reasoning questions, compare ratios rather than recalculating from scratch, and state clearly which quantity is constant.
- 💡Underline the words resultant force and mass in the question to decide which proportionality is being tested.
- 💡Show the rearrangement you use, especially when finding mass or force, so your method is clear.
- 💡Check whether your final acceleration is sensible: a small force on a large mass should give a small acceleration.
- 💡Write the equation you are using before substituting numbers, so the examiner can see your method.
- 💡Check that mass is in kilograms and acceleration is in metres per second squared before calculating.
- 💡When rearranging, use inverse operations carefully: divide both sides by mass to find acceleration, or by acceleration to find mass.
- 💡Underline the values given in the question and identify which quantity you need to find.
- 💡Show the rearranged equation before substituting numbers, especially when finding mass or acceleration.
- 💡Give the unit with your final answer; for acceleration use m/s², for force use N and for mass use kg.
- 💡Write the equation, then the rearranged form, then substitute values, so the examiner can follow your method.
- 💡Check the unit of every value before calculating and convert grams to kilograms or centimetres to metres where needed.
- 💡Give the unit with your final answer and add a short sentence explaining what the result means in the context of the question.
- 💡Write the unit N after every force value in your working and final answer.
- 💡When several forces act, draw a simple free-body diagram and label the direction of each force before calculating the resultant.
- 💡If a question asks for the resultant force, show the addition or subtraction of the individual forces, not just the final value.
- 💡Write the equation, substitute values with units, then calculate and give the unit kg in your final answer.
- 💡Check whether a mass is given in grams and convert to kilograms before using it in any calculation.
- 💡If asked to compare accelerations, refer to the inverse relationship: doubling mass halves acceleration for the same force.
- 💡Underline the resultant force in the question and calculate it before using a = F ÷ m.
- 💡Show the rearranged equation, substitution and answer with unit m/s² to gain method and accuracy credit.
- 💡For graph questions, recall that acceleration is the gradient of a velocity–time graph, so read values carefully.
- 💡Use the phrase 'resistance to change in velocity' or 'difficulty in changing velocity' to show you understand inertial mass rather than just quoting the word mass.
- 💡When comparing objects, refer explicitly to a constant resultant force or a constant acceleration so your proportional reasoning is clear.
- 💡Support explanations with the equation F = m a, identifying each symbol and its unit, to connect the definition to quantitative work.
- 💡Write the equation m = F ÷ a before substituting numbers so the examiner can see the relationship you are using.
- 💡Check the unit of every quantity: force in newtons, acceleration in metres per second squared and mass in kilograms.
- 💡When a question gives a velocity change and a time, calculate acceleration first using a = Δv ÷ t, then use m = F ÷ a.
- 💡State your estimated values clearly, for example 'assume the car reaches 30 m/s in 12 s', so your calculation can be followed.
- 💡Show the equation, the substitution and the answer with its unit; an estimate with clear working earns more credit than a bare number.
- 💡Sanity-check the answer: a family car's driving force is typically a few thousand newtons, not a few hundred or a few million.
- 💡When a question asks for an answer to a given number of significant figures, show the unrounded value first and then the rounded value with ̴, so the examiner can see both the calculation and the approximation.
- 💡If you read a value from a graph, quote it with ̴ and state the precision you used, for example 'from the graph, a ̴ 2.2 m/s²'.
- 💡Check every symbol in your final line: use = only when the two sides are exactly equal, and use ̴ when rounding, estimating or reporting a measurement with uncertainty.
- 💡State clearly which quantity you change, which you measure and which you keep the same; examiners look for this control of variables in practical questions.
- 💡When describing the graph, name both axes and the shape you expect, and link the shape to the relationship between the variables rather than just saying 'it goes up'.
- 💡Include one realistic improvement, such as repeating each reading and taking a mean, and explain how it reduces the effect of random error.
Common Mistakes
- Using an individual applied force instead of the resultant force; correct this by finding the vector sum of all forces along the line of motion first.
- Substituting mass in grams into F = m × a; correct this by converting grams to kilograms by dividing by 1000.
- Writing the relationship as a = m ÷ F or F = m ÷ a; correct this by checking that a larger mass gives a smaller acceleration for the same force.
- Treating acceleration as proportional to mass; correct this by recalling that greater mass gives smaller acceleration for the same resultant force.
- Forgetting to combine opposing forces before applying the relationship; correct this by subtracting a smaller opposing force from a larger driving force.
- Mixing units, such as using grams with newtons; correct this by converting all masses to kilograms before calculating.
- Writing the equation as force = mass ÷ acceleration. Correction: force = mass × acceleration, so F = m × a.
- Using grams instead of kilograms for mass without converting. Correction: convert grams to kilograms by dividing by 1000 before substituting.
- Treating any single force as the resultant force. Correction: add forces in the same direction and subtract forces in opposite directions to find the resultant force first.
- Confusing the rearrangement and writing a = m ÷ F. Correction: a = F ÷ m, because acceleration increases with force and decreases with mass.
- Forgetting to find the resultant force when several forces act. Correction: calculate the vector sum of forces first, then substitute into F = m × a.
- Mixing units, such as using mass in grams with acceleration in m/s². Correction: convert mass to kilograms before calculating.
- Using a single applied force instead of the resultant force: add or subtract all forces along the line of motion first, then substitute.
- Mixing units, such as using grams with newtons: convert mass to kilograms before calculating.
- Rearranging incorrectly, for example writing a = m ÷ F: check by substituting simple numbers, since a larger mass must give a smaller acceleration for the same force.
- Writing the unit as N m or N/m: force is measured in newtons, N, and the unit is written after the number.
- Treating force as a scalar and ignoring direction: state the direction of the resultant force or acceleration when the question requires it.
- Confusing mass with force: mass is in kilograms and measures the amount of matter, while force is in newtons and is a push or pull.
- Using grams directly in F = m × a without converting to kilograms; correct by dividing grams by 1000 first.
- Confusing mass with weight and giving newtons as the unit of mass; correct by stating mass is in kg and weight is a force in N.
- Rearranging F = m × a incorrectly, for example writing m = a ÷ F; correct by dividing force by acceleration, m = F ÷ a.
- Using a single applied force instead of the resultant force; correct by subtracting opposing forces such as friction before dividing by mass.
- Forgetting to convert grams to kilograms, which gives an acceleration 1000 times too small; correct by converting mass to kg first.
- Writing the unit as m/s instead of m/s²; correct by remembering acceleration is the change in velocity per second, so the unit is metres per second squared.
- Confusing inertial mass with weight; correct this by stating that weight is a force in newtons caused by gravity, while inertial mass is a property in kilograms resisting velocity change.
- Thinking that a more massive object cannot accelerate; correct this by explaining that it can accelerate, but a greater resultant force is needed for the same acceleration.
- Believing that inertial mass changes with location; correct this by stating that inertial mass is constant for an object, whereas weight changes with gravitational field strength.
- Using weight in newtons in place of mass in kilograms: correct this by dividing weight by gravitational field strength, for example 600 N ÷ 10 N/kg = 60 kg, before using m = F ÷ a.
- Dividing acceleration by force instead of force by acceleration: correct this by writing m = F ÷ a and checking that a larger force gives a larger mass for the same acceleration.
- Forgetting to use the resultant force when several forces act: correct this by adding or subtracting forces along the line of motion first, then dividing by acceleration.
- Mixing units, such as using speed in km/h with time in seconds: correct this by converting to m/s first, for example 72 km/h ÷ 3.6 = 20 m/s.
- Using the final speed instead of the change in speed: correct this by calculating Δv = final velocity − initial velocity before dividing by time.
- Forgetting that deceleration is a negative acceleration: correct this by keeping the sign or by stating the magnitude and direction of the braking force.
- Writing an equals sign for a rounded answer, for example 1.1 N ÷ 0.50 kg = 2 m/s². Correction: use ̴ because 2 m/s² is a rounded value, so write 1.1 N ÷ 0.50 kg ̴ 2 m/s².
- Confusing ̴ with the proportionality symbol ∝ or with a minus sign. Correction: ̴ is a single wavy line meaning 'approximately equal to', while ∝ means 'is proportional to' and links quantities whose ratio is constant.
- Using ̴ instead of a decimal point. Correction: ̴ must be used to indicate approximation, either between values or as a prefix before a complete value, for example ̴ 2 m/s², not 2̴ 0 m/s².
- Changing both force and mass in the same set of readings. Correction: change only one variable at a time, keeping the other quantity constant so the effect on acceleration can be attributed to the chosen variable.
- Ignoring friction and treating the hanging weight as the only force on the trolley. Correction: compensate for friction by tilting the runway, or discuss friction as a source of error that makes the measured acceleration smaller than expected.
- Plotting acceleration against mass and expecting a straight line through the origin. Correction: acceleration is inversely proportional to mass, so the straight-line graph is acceleration against 1 ÷ mass.