Newton's Second Law — AQA GCSE Physics
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 joins two proportional relationships.
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At fixed mass, acceleration is directly proportional to the resultant force, so trebling the force trebles the acceleration. At fixed resultant force, acceleration is inversely proportional to mass, so a trolley loaded to twice its mass accelerates at half the rate. This is why a fully laden lorry pulls away far more slowly than an empty one with the same engine. In symbols, a = F ÷ m, usually written F = m a. The word resultant matters: if a 900 N driving force acts against 300 N of drag on a 200 kg quad bike, the resultant is 600 N and the acceleration is 3 m/s², not 4.5 m/s². Because acceleration is the ratio F ÷ m, a graph of acceleration against resultant force at fixed mass is a straight line through the origin.
resultant force = mass × acceleration
Newton's second law links the resultant force on an object to its mass and acceleration. The equation is resultant force = mass × acceleration, often written F = m × a. Force is measured in newtons (N), mass in kilograms (kg) and acceleration in metres per second squared (m/s²). A larger resultant force produces a larger acceleration for the same mass, and a larger mass needs a larger force for the same acceleration. For example, a 1200 kg car accelerating at 2 m/s² needs a resultant force of 1200 × 2 = 2400 N. If the forces on an object are balanced, the resultant force is zero and acceleration is zero. Rearranged, acceleration = force ÷ mass and mass = force ÷ 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 1200 kg car with a resultant forward force of 3600 N accelerates at 3600 ÷ 1200 = 3 m s⁻². 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 needed to give a 1 kg mass an acceleration of 1 m s⁻², so 1 N = 1 kg m s⁻². Force is a vector: it has size and direction, and the direction of the resultant force determines the direction of the acceleration. When several forces act, F means the resultant force, found by combining them along the line of motion. For example, a 2 kg trolley pushed with 10 N forward while friction acts with 4 N backward has a resultant force of 10 N − 4 N = 6 N, giving a = 6 ÷ 2 = 3 m s⁻². Always record force in newtons, not kilograms, and give direction where a question asks for it.
mass, m, in kilograms, kg
Mass measures how much matter an object contains and how strongly it resists a change in motion. In Newton's Second Law, F = m × a, mass must be in kilograms so that force comes out in newtons when acceleration is in m/s². A newton is defined as the force giving a 1 kg mass an acceleration of 1 m/s², so the kilogram is the coherent SI unit here. Always convert before substituting: 250 g becomes 0.25 kg, and 2.4 tonnes becomes 2400 kg. Mass is not the same as weight; weight is a force in newtons, calculated as W = m × g. On the Moon your mass stays the same while your weight changes. When rearranging, m = F ÷ a, so a larger mass needs a larger force for the same acceleration.
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. An acceleration of 3 m/s² means the velocity increases by 3 m/s every second. In Newton's Second Law, F = m × a, acceleration must be in m/s² so that a mass in kilograms gives a force in newtons. Acceleration is a vector: speeding up, slowing down and changing direction all involve acceleration. Calculate it using a = Δv ÷ t, where Δv is the change in velocity in m/s and t is the time in s. For example, a runner reaching 8 m/s from rest in 4 s has a = 8 ÷ 4 = 2 m/s². Rearranged, a = F ÷ m, so the same force produces greater acceleration on a smaller mass.
(HT only) 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 any change in its velocity, whether that change is a speeding up, a slowing down or a change of direction. A larger inertial mass means a smaller acceleration for the same resultant force, because the force is shared over more mass. For example, a loaded lorry has a much larger inertial mass than an empty one, so the same engine force produces a smaller acceleration and it takes longer to reach a given speed. The idea links directly to Newton's second law, F = m a, where the mass being used is the inertial mass. It is a property of the object itself and does not depend on gravitational field strength, which distinguishes it from weight.
(HT only) Inertial mass is defined as the ratio of force over acceleration.
This statement gives the quantitative definition of inertial mass: inertial mass equals the resultant force divided by the acceleration it produces, written m = F ÷ a. The force must be the resultant force in newtons and the acceleration must be in metres per second squared, so the mass comes out in kilograms. For example, if a resultant force of 12 N gives an acceleration of 3 m/s², the inertial mass is 12 N ÷ 3 m/s² = 4 kg. This definition is simply a rearrangement of Newton's second law, F = m a, and it shows that inertial mass is the constant of proportionality linking force and acceleration for a given object.
Students should be able to estimate the speed, accelerations and forces involved in large accelerations for everyday road transport.
Estimating means choosing sensible everyday figures and carrying them through the equations. A family car has a mass of roughly 1000 to 1500 kg. Motorway speed, 70 mph, is about 31 m/s, and 30 mph is about 13 m/s. A brisk acceleration from rest to 27 m/s in 10 s gives a = 27 ÷ 10 ≈ 2.7 m/s², so the resultant force on a 1200 kg car is about 1200 × 2.7 ≈ 3200 N. Emergency braking is harsher: stopping from 27 m/s in about 3 s gives roughly 9 m/s², needing a force near 11 000 N. Quote estimates to one or two significant figures and show every substitution, because the marks are for the method. Check the size of an estimate against something known: the weight of that car is about 12 000 N, so a braking force near 11 000 N is realistic.
Students should recognise and be able to use the symbol ≈ that indicates an approximate value or approximate answer.
In physics, exact values are not always known or required, particularly when dealing with order-of-magnitude estimations. The symbol ≈ indicates that two values are approximately equal. You will encounter this symbol when a question provides or asks for an estimate rather than a precise calculation. For example, the typical walking speed of a person can be written as v ≈ 1.5 m/s, showing it is a reasonable estimate rather than an exact universal constant. Recognising this symbol helps you understand that a question requires you to apply typical everyday values. Standard calculations rounded to an appropriate number of significant figures should still use the standard equals sign.
Required practical activity 7: 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 practical investigates Newton's Second Law, F = ma. You will use a trolley on a friction-compensated ramp, connected by a string over a pulley to suspended masses. To investigate varying force with constant mass, transfer masses one by one from the trolley to the weight hanger. This increases the accelerating force while keeping the total system mass constant. To investigate varying mass with constant force, keep the hanging mass constant and add masses directly to the trolley. Acceleration is measured using light gates and a data logger, or by timing the motion over a set distance. Plotting graphs of acceleration against force, and acceleration against inverse mass, should yield straight lines through the origin, confirming the relationships.
Your focus
- Calculate the acceleration of a vehicle from its mass and the resultant of the forward and resistive forces.
- Predict how the acceleration changes when a trolley's mass is doubled while the accelerating force is unchanged.
- Explain why a loaded lorry accelerates more slowly than an empty one driven by the same engine force.
Show all 33 objectives
- Recall and apply the equation resultant force = mass × acceleration.
- Rearrange the equation to calculate mass or acceleration.
- Explain how changing mass or resultant force affects the acceleration of an object.
- State and use the equation F = m a to calculate resultant force, mass or acceleration.
- Identify and combine the forces acting on an object to obtain the resultant force before substitution.
- Convert values into consistent SI units and interpret the calculated acceleration in the context of the motion.
- State that force is measured in newtons and describe the newton in terms of mass and acceleration.
- Identify the direction of a resultant force and explain how it determines the direction of acceleration.
- Distinguish between force in newtons and mass in kilograms when solving problems using F = m a.
- State that mass is measured in kilograms and recall the symbol m.
- Convert masses given in grams or tonnes into kilograms before calculation.
- Apply m = F ÷ a to determine an unknown mass from force and acceleration.
- State that acceleration is measured in m/s² and define it as the rate of change of velocity.
- Calculate acceleration from change in velocity and time using a = Δv ÷ t.
- Apply a = F ÷ m to find acceleration from force and mass.
- Describe inertial mass as a measure of how difficult it is to change an object's velocity.
- Explain how a larger inertial mass leads to a smaller acceleration for the same resultant force.
- Distinguish clearly between inertial mass and weight in written answers.
- Define inertial mass as the ratio of force to acceleration.
- Rearrange and apply m = F ÷ a to calculate inertial mass from given values.
- Use correct SI units for force, acceleration and mass in calculations.
- Estimate the acceleration of a car that reaches motorway speed from rest in about ten seconds.
- Estimate the resultant force on a car during that acceleration, choosing a realistic mass and stating your assumptions.
- Judge whether a quoted speed, acceleration or force is realistic for everyday road transport.
- Identify the ≈ symbol and state its mathematical meaning as 'approximately equal to'.
- Interpret exam questions containing the ≈ symbol to recognise when an approximate or typical value is being used.
- Differentiate between the use of the ≈ symbol for estimations and the standard equals sign for rounded calculations.
- Describe a method to investigate how the acceleration of an object depends on the applied force for a constant mass.
- Describe a method to investigate how the acceleration of an object depends on its mass for a constant applied force.
- Analyse experimental data to determine the relationship between force, mass, and acceleration.
Newton's Second Law exam tips
Marking Points
- one mark for identifying the resultant force before applying the law
- one mark for a proportional statement such as doubling the resultant force doubles the acceleration
- one mark for stating that doubling the mass at constant force halves the acceleration
- one mark for a correct calculation using resultant force = mass × acceleration with consistent units
- State the equation resultant force = mass × acceleration and recall the correct units: newtons, kilograms and metres per second squared.
- Substitute known values of mass and acceleration into the equation correctly.
- Calculate the resultant force, or rearrange the equation to find mass or acceleration when required.
- Interpret a zero resultant force as producing zero acceleration, and relate larger forces or masses to changes in acceleration.
- Give the final answer with the correct unit and a sensible size for the situation described.
- State the equation as F = m a and identify F as the resultant force in newtons, m as mass in kilograms and a as acceleration in metres per second squared.
- Substitute values into F = m a with consistent SI units, converting grams to kilograms and kilometres per hour to metres per second where needed.
- Rearrange the equation correctly to find acceleration using a = F ÷ m or mass using m = F ÷ a.
- Calculate the resultant force by combining all forces acting along the line of motion before using F = m a.
- Interpret a numerical answer in context, for example stating that doubling the resultant force doubles the acceleration when mass is unchanged.
- State that force is measured in newtons, symbol N, and that 1 N is the force giving a 1 kg mass an acceleration of 1 m s⁻².
- Recognise force as a vector quantity with both magnitude and direction, and state the direction of the resultant force when required.
- Distinguish force in newtons from mass in kilograms, explaining that weight is a force while mass is not.
- Combine multiple forces to find the resultant force in newtons before using it in F = m a.
- Use the relationship 1 N = 1 kg m s⁻² to check that units in a calculation are consistent.
- State that mass is measured in kilograms (kg) and is the quantity of matter in an object.
- Convert grams to kilograms by dividing by 1000, and tonnes to kilograms by multiplying by 1000, before using F = m × a.
- Explain that the kilogram is the SI base unit of mass and is required for force to be calculated in newtons.
- Use m = F ÷ a to find mass when force in newtons and acceleration in m/s² are known.
- Distinguish mass (kg, scalar, unchanged by location) from weight (N, a force, dependent on gravitational field strength).
- State that acceleration is measured in metres per second squared (m/s²) and is the rate of change of velocity.
- Calculate acceleration using a = Δv ÷ t, ensuring velocity change is in m/s and time is in s.
- Use a = F ÷ m to find acceleration from force in newtons and mass in kilograms.
- Interpret an acceleration such as 5 m/s² as a velocity increase of 5 m/s each second.
- Recognise acceleration as a vector quantity, so deceleration and direction change are also accelerations.
- State that inertial mass measures how strongly an object resists a change in its velocity, including speeding up, slowing down or changing direction.
- Link a larger inertial mass to a smaller acceleration for the same resultant force, using F = m a to support the reasoning.
- Distinguish inertial mass from weight, noting that inertial mass is a property of the object and is measured in kilograms.
- Apply the idea to a comparison, such as a loaded versus an unloaded vehicle, and explain the difference in acceleration.
- Use the equation in the form m = F ÷ a when a numerical value of inertial mass is required.
- State the definition as inertial mass = resultant force ÷ acceleration, or m = F ÷ a.
- Identify that force must be the resultant force measured in newtons and acceleration in metres per second squared.
- Rearrange F = m a correctly to make mass the subject when carrying out a calculation.
- Substitute values accurately and give the mass in kilograms with an appropriate unit.
- Explain that the ratio is constant for a given object, which is why it defines its inertial mass.
- one mark for a sensible estimate of speed, such as 30 m/s for motorway driving
- one mark for a sensible estimate of mass, such as 1000 kg to 1500 kg for a car
- one mark for using change in velocity ÷ time to estimate the acceleration
- one mark for multiplying the estimated mass by the estimated acceleration to give a force in newtons
- one mark for a resultant force of the right order, in the range about 2000 N to 4000 N, for a car of 1000 kg to 1500 kg reaching about 27 m/s in 10 s
- Identify the ≈ symbol as meaning 'approximately equal to' in the context of estimations.
- Interpret given data containing the ≈ symbol as an indication to use an estimated or typical value.
- Apply the ≈ symbol correctly when stating an estimated physical quantity, such as writing v ≈ 1.5 m/s for walking speed.
- Distinguish between the approximate equals sign in an expression like x ≈ 5 and the proportionality symbol in an expression like a ∝ F.
- Describe a valid method to measure the acceleration of the trolley, such as using two light gates to find initial and final velocities over a known distance.
- Explain how to vary the accelerating force while keeping the total mass of the system constant by transferring masses between the trolley and the weight hanger.
- Explain how to vary the mass of the system while keeping the accelerating force constant by adding masses to the trolley but leaving the hanging mass unchanged.
- Identify the independent, dependent, and control variables for both parts of the investigation.
- Describe how to use the collected data to plot appropriate graphs to verify that acceleration is directly proportional to force and inversely proportional to mass.
Examiner Tips
- 💡When a question changes two quantities at once, deal with the force first, then the mass.
- 💡Show the subtraction that gives the resultant force as a separate line of working.
- 💡If the acceleration is negative, say the vehicle is decelerating rather than dropping the sign.
- 💡Write the equation, then substitute values with units before calculating to reduce errors.
- 💡Check that the unit of the answer matches the quantity asked for, such as N for force or m/s² for acceleration.
- 💡If forces are given in different directions, resolve them to find the resultant force before using the equation.
- 💡Write the equation, then the substituted values with units, then the answer with its unit; this makes method clear even if the final arithmetic slips.
- 💡Underline the word resultant in a question and show the addition or subtraction of forces before using F = m a.
- 💡Check the size and unit of your answer: a car accelerating at 3 m s⁻² is reasonable, while 3000 m s⁻² suggests a unit or rearrangement error.
- 💡Write the unit N after every force value in your working, and convert any value given in kilonewtons by multiplying by 1000.
- 💡When a question asks for the resultant force, show the addition or subtraction of the individual forces before stating the final value and direction.
- 💡Use the definition of the newton to justify unit conversions, for example explaining that kg m s⁻² is equivalent to N.
- 💡Write the unit kg next to every mass value in your working to keep the quantity clear.
- 💡Convert all masses to kilograms in a single step at the start of a calculation before substituting.
- 💡Check the size of your answer: a car of mass 1200 kg accelerating at 2 m/s² needs 2400 N, so a force of 2.4 N signals a conversion error.
- 💡Show the change in velocity calculation separately before dividing by time.
- 💡Include the direction when an acceleration is negative or opposes motion, describing it as deceleration.
- 💡Check that the unit of your answer matches m/s², since a unit of m/s indicates velocity was calculated instead.
- 💡Read the question carefully to see whether it asks about difficulty of changing velocity or about the ratio of force to acceleration, and answer the version asked.
- 💡When comparing two objects, quote the equation F = m a and explain the effect on acceleration rather than only stating that one is heavier.
- 💡Use the term resultant force, not just force, because an unbalanced resultant force is what causes the acceleration.
- 💡Write down m = F ÷ a before substituting numbers so the rearrangement is clear and easy to follow.
- 💡Check that the force used is the resultant force by considering whether forces are balanced or unbalanced.
- 💡Include the unit kg with the final answer and round sensibly to match the data given.
- 💡Write down each estimated value with its unit before you calculate anything.
- 💡Use the rough conversion that 1 mph is a little under 0.5 m/s to turn road speeds into m/s.
- 💡Say clearly that the answer is an estimate, and use the approximation symbol when you quote it.
- 💡Look out for the ≈ symbol in question stems, as it often hints that you should use typical estimated values rather than performing a precise calculation.
- 💡When asked to estimate a value from your own knowledge, you can write the ≈ symbol before your numerical answer to demonstrate your understanding of approximation.
- 💡Clearly state the equipment used to measure each variable, for example using a balance to measure mass and light gates to measure velocity and time.
- 💡When describing the method, explicitly state which variables are kept constant in each part of the experiment to ensure a fair test.
- 💡Remember that a graph of acceleration against mass produces a curve, but a graph of acceleration against the inverse of mass produces a straight line through the origin.
Common Mistakes
- writing that acceleration is proportional to mass rather than inversely proportional to it
- using the driving force rather than the resultant force in the calculation
- substituting the weight of an object on a horizontal surface as the resultant force
- reading 'inversely proportional' as simply 'decreases', then predicting any smaller value
- Using grams instead of kilograms for mass; correct this by converting grams to kilograms by dividing by 1000 before substituting.
- Confusing resultant force with a single applied force when several forces act; correct this by finding the vector sum of all forces first.
- Rearranging the equation incorrectly, for example dividing force by acceleration to find force; correct this by using acceleration = force ÷ mass or mass = force ÷ acceleration as appropriate.
- Using one individual force instead of the resultant force; correct this by adding forces in the same direction and subtracting forces in the opposite direction first.
- Substituting mass in grams or force in kilonewtons without converting; correct this by converting to kilograms and newtons before calculating.
- Rearranging incorrectly, such as dividing mass by force when finding acceleration; correct this by using a = F ÷ m and checking the rearrangement by substituting simple numbers.
- Treating newtons and kilograms as interchangeable; correct this by remembering that mass is measured in kilograms and force in newtons, and that weight equals mass times gravitational field strength.
- Giving a force without direction when the question asks for the resultant force; correct this by stating the direction, such as 6 N forwards.
- Adding all quoted forces regardless of direction; correct this by subtracting forces that act in the opposite direction along the line of motion.
- Substituting grams directly into F = m × a, giving a force 1000 times too large; correct by converting to kilograms first.
- Treating mass and weight as interchangeable and quoting mass in newtons; correct by giving mass in kg and weight in N.
- Forgetting to convert tonnes, so 3 tonnes is used as 3 kg instead of 3000 kg; correct by multiplying tonnes by 1000.
- Writing the unit as m/s, which is a velocity unit; correct by writing m/s² for acceleration.
- Dividing final velocity by time without subtracting the initial velocity; correct by using the change in velocity, Δv.
- Mixing units, such as using time in minutes with velocity in m/s; correct by converting time to seconds first.
- Confusing inertial mass with weight: weight is a force measured in newtons and depends on gravitational field strength, whereas inertial mass is measured in kilograms and resists velocity change.
- Thinking that a larger mass always means a larger acceleration: the correct relationship is that for a fixed resultant force, a larger mass gives a smaller acceleration.
- Treating a change of direction as unrelated to inertial mass: any change in velocity, including direction, involves acceleration and is resisted by inertial mass.
- Using a single applied force instead of the resultant force: the correct approach is to find the vector sum of all forces before dividing by acceleration.
- Dividing acceleration by force instead of force by acceleration: the correct form is m = F ÷ a.
- Forgetting to convert units, such as using grams or centimetres per second squared: the correct units are newtons and metres per second squared, giving kilograms.
- estimating the mass of a car as 100 kg, or quoting it in grams
- leaving the speed in mph inside a calculation that needs metres per second
- copying the full calculator display into an estimate that is only good to one significant figure
- writing a bare number with no working, so the method marks cannot be awarded
- Believing that the ≈ symbol must be used every time a calculated answer is rounded to significant figures; correction: the standard equals sign is perfectly acceptable and expected when providing a rounded calculated answer.
- Confusing the approximate equals sign with the proportionality symbol; correction: recognise that an expression like y ≈ 10 means approximately equal to, whereas y ∝ x indicates how one variable changes with respect to another.
- Writing the ≈ symbol without a numerical value or variable on the right-hand side; correction: ensure the symbol always sits between two quantities, such as writing v ≈ 3 m/s rather than leaving the relation empty.
- Changing the accelerating force by adding new masses to the hanger from an external supply; correction: transfer masses from the trolley to the hanger so the total mass of the system remains constant.
- Failing to compensate for friction; correction: slightly tilt the ramp until the trolley moves at a constant velocity when given a small push, ensuring the hanging weight is the only resultant force.
- Confusing the mass of the trolley with the total mass of the system; correction: remember that the total mass being accelerated includes the trolley, the string, and the hanging masses.