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.
Read the full explanation
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.
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.
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
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.
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