AQA GCSE Combined Science (8464), Higher tier · Physics › Forces › Forces and motion › Forces, accelerations and Newton's Laws of motion
Practise Newton's Second Law. 19 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Newton's Second Law links resultant force, mass and acceleration: F = m a. You need to use it in calculations, estimate forces in everyday road transport, explain inertial mass (Higher), and describe the required practical on force, mass and acceleration.
Grade by grade
What you need to be able to do, from the first marks up to the top grade.
4
Calculate force using F = m aFor example, 1200 kg × 2.5 m/s2 = 3000 N.
5
Rearrange F = m a for mass or accelerationa = F ÷ m and m = F ÷ a.
5
Describe how acceleration depends on force and massAcceleration is proportional to the resultant force and inversely proportional to the mass.
6
Use the resultant force in F = m aFind the resultant first, e.g. thrust minus drag, then divide by the mass.
6
Describe the acceleration required practicalVary the force (or the mass) on a trolley and measure its acceleration, e.g. with light gates.
7
Estimate forces in everyday road transportFor example, a car of mass ~1000 kg decelerating at ~6 m/s2 needs ~6000 N.
8
Explain inertial massInertial mass is a measure of how difficult it is to change an object's velocity: force ÷ acceleration.
Notes
The law
The acceleration of an object is proportional to the resultant force acting on it, and inversely proportional to its mass.
resultant force = mass × acceleration: F = m a
F in newtons (N), m in kilograms (kg), a in metres per second squared (m/s2).
The acceleration is in the same direction as the resultant force.
Always use the resultant force. Example: thrust 5000 N forwards and drag 1400 N backwards give F = 3600 N.
Inertial mass grade 8+
Inertial mass is a measure of how difficult it is to change the velocity of an object.
Inertial mass is defined as the ratio of force over acceleration: m = F ÷ a.
A larger mass needs a larger force to give it the same acceleration.
Estimating in road transport grade 7+
The symbol ~ means 'approximately'.
Useful rough values: car mass ~1000 kg; a car speeding up at ~3 m/s2; a car braking hard at ~6 m/s2.
Example: to accelerate a car at ~3 m/s2 needs a resultant force of ~1000 × 3 = ~3000 N.
Required practical: force, mass and acceleration
A trolley on a bench is pulled by a string that passes over a pulley to a hanging mass. The weight of the hanging mass is the accelerating force.
Measure the acceleration with light gates and a data logger, or by timing the trolley over marked distances.
To vary the force at constant mass: move masses from the trolley to the hanger, so the total mass being accelerated stays the same.
To vary the mass at constant force: add masses to the trolley and keep the hanging mass the same.
Results: acceleration is proportional to force; acceleration decreases as mass increases.
Cheatsheet
F = m a (resultant force = mass × acceleration)
Units: F in N, m in kg, a in m/s2
a ∝ F (at constant mass)
a ∝ 1/m (at constant force)
Always use the RESULTANT force
Acceleration is in the direction of the resultant force
Inertial mass = force ÷ acceleration grade 8+
~ means 'approximately'; car mass ~1000 kg
How to answer each type of question
Calculate with F = m a
2 marks4
Check the mass is in kg.
Write F = m a, or its rearranged form, and substitute.
Give the unit.
Example. A resultant force of 450 N acts on a motorbike and rider. Their total mass is 300 kg. Calculate their acceleration.
Show the model answer
a = 450 ÷ 300 (1) a = 1.5 m/s2 (1)
Two-step: acceleration from a velocity change, then force
4 marks6
Use a = Δv ÷ t to find the acceleration.
Use F = m a to find the resultant force.
Give each answer with its unit.
Example. A car of mass 1100 kg accelerates from rest to 18 m/s in 6.0 s. Calculate the resultant force on the car.
Show the model answer
a = (18 − 0) ÷ 6.0 (1) a = 3.0 m/s2 (1) F = 1100 × 3.0 (1) F = 3300 N (1)
Describe the required practical
4 to 6 marks6
Describe the apparatus: trolley, string, pulley, hanging masses, light gates.
Say how the force is changed while the mass is kept constant.
Say how the acceleration is measured, and that readings are repeated.
Say what graph is plotted.
Example. Describe how a student could investigate how the acceleration of a trolley depends on the resultant force on it, keeping the mass constant.
Show the model answer
Attach the trolley by a string over a pulley at the end of the bench to a mass hanger (1). Measure the acceleration using light gates and a data logger (or time the trolley over measured distances) (1). The weight of the hanging masses is the force; change it by moving masses from the trolley to the hanger (1) which keeps the total mass constant (1). Use at least five different forces and repeat each reading to calculate a mean (1). Plot a graph of acceleration against force (1).
Estimate a force in road transport
3 marks7
State a sensible value for any quantity not given (e.g. car mass ~1000 kg).
Find the acceleration if needed.
Use F = m a and round sensibly.
Example. Estimate the braking force needed for a car to decelerate from 30 m/s to rest in about 5 s.
Show the model answer
mass of car ~1000 kg (1) deceleration = 30 ÷ 5 = 6 m/s2 (1) F ≈ 1000 × 6 ≈ 6000 N (1)
Explain using inertial mass
2 marks8
Say that a greater mass is more difficult to accelerate (greater inertial mass).
Use a = F ÷ m to compare the accelerations.
Example. The same resultant force is applied to an empty shopping trolley and to a full one. Use the idea of inertial mass to explain why the full trolley has a smaller acceleration.
Show the model answer
The full trolley has a greater (inertial) mass, so it is more difficult to change its velocity (1). a = F ÷ m, so with the same force, the greater mass has a smaller acceleration (1).
Shortcuts and memory tricks
Formula triangle: F on top, m and a underneath.
Direction check: the acceleration always points the same way as the resultant force.
In the practical, moving masses from the trolley to the hanger changes the force without changing the total mass, so it is a fair test.
Rough values for estimates: car ~1000 kg, person ~70 kg.
Where marks are lost
Using one of the forces instead of the resultant force.
Using a weight in N as if it were a mass in kg. If a weight is given, use m = W ÷ g.
Using a mass in grams: convert to kg first.
In the practical, adding new masses to the hanger, which changes the total mass as well as the force.
Saying acceleration is proportional to mass. It is inversely proportional.
Exam technique
F = m a must be recalled.
In estimates, state the values you assumed (e.g. 'mass of car ≈ 1000 kg') and use ~ or 'about' in your answer.
In practical questions, name the independent, dependent and control variables clearly.
Required practical:Acceleration (method, variables and exam tips)
Quick recall
Cover the answers and test yourself. The app has these as flashcards that come back just before you'd forget them.
Write down the equation that links acceleration (a), mass (m) and resultant force (F).
F = m a
The device exerts a resultant force of 60 N on an astronaut sitting on a sliding seat. The acceleration of the astronaut and the seat is 0.80 m/s2. Calculate the total mass of the astronaut and the seat.
75 kg
A student of mass 60 kg stands on a set of bathroom scales in a lift. The scales measure the normal contact force on the student’s feet. gravitational field strength = 9.8 N/kg Name the two forces acting on the student.
Weight and the normal contact force.
Sample questions
Written for this site in the style of AQA exam questions. They are not taken from real past papers.
Question 1Easy4 marks
(a) A car of mass 1200 kg accelerates at 2.5 m/s2. Calculate the resultant force on the car. Use the equation: resultant force = mass × acceleration[2]
(b) The same resultant force acts on a van with a greater mass than the car. How does the acceleration of the van compare with the acceleration of the car? Tick (✓) one box.[1]
It is smaller
It is the same
It is greater
(c) Complete the sentence. The acceleration of an object is ............ to the resultant force acting on the object. Tick (✓) one box.[1]
directly proportional
inversely proportional
equal
unrelated
Show the answer and mark scheme
(a)Answer: 3000 N
F = 1200 × 2.5
3000 (N)
(b)Answer: It is smaller
(c)Answer: directly proportional
Question 2Medium7 marks
(a) Write down the equation that links acceleration (a), mass (m) and resultant force (F).[1]
(b) A sprinter of mass 64 kg accelerates from rest to 8.0 m/s in 2.0 s. Calculate the average resultant force on the sprinter.[4]
(c) A second sprinter of the same mass experiences an average resultant force of 320 N. Calculate the average acceleration of the second sprinter.[2]
Show the answer and mark scheme
(a)Answer: F = m a
F = m a / resultant force = mass × acceleration
(b)Answer: 256 N
a = 8.0 ÷ 2.0
a = 4.0 (m/s2)
F = 64 × 4.0
256 (N)
(c)Answer: 5.0 m/s2
a = 320 ÷ 64
5.0 (m/s2)
Question 3Hard6 marks
Astronauts in orbit cannot measure their mass using bathroom scales. Instead, a device pushes the astronaut with a known force and measures the astronaut’s acceleration.
(a) Explain what is meant by inertial mass.[2]
(b) The device exerts a resultant force of 60 N on an astronaut sitting on a sliding seat. The acceleration of the astronaut and the seat is 0.80 m/s2. Calculate the total mass of the astronaut and the seat.[2]
(c) The mass of the sliding seat is 5.0 kg. What is the mass of the astronaut?[1]
(d) Explain why this method gives the same value for the astronaut’s mass as it would on the Earth.[1]
Show the answer and mark scheme
(a)Answer: It measures how difficult it is to change an object’s velocity; inertial mass = force ÷ acceleration.
a measure of how difficult it is to change the velocity of an object
(defined as) force ÷ acceleration
(b)Answer: 75 kg
m = 60 ÷ 0.80
75 (kg)
(c)Answer: 70 kg
70 (kg)
(d)Answer: Inertial mass does not depend on gravitational field strength.
inertial mass (force ÷ acceleration) does not depend on the gravitational field strength / on weight