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Don't read anything yet. Watch each object: is it Newton's first law or the second? Is the speed changing, the direction, or neither? Where there's a resultant force, work out the acceleration. Then switch to the second tab and run the required practical.
1 · Which law is acting?
2 · What is changing?
3 · Calculate with F = m a
Physics With Kate · physicswithkate.com — AQA GCSE Physics 4.5.6.2
The mass in F = m a is the mass of everything that is accelerating — the trolley and the string and every mass, whether it is sitting on the trolley or hanging on the end. That is why you move masses across instead of adding new ones: the total stays at 1.00 kg for every single reading.
The force is only the weight of the masses hanging vertically downwards (F = mhanging × g). A mass sitting on the trolley is held up by the trolley, so it pulls on nothing.
| total mass / kg |
hanging mass / kg |
force / N |
initial speed (u) / m/s |
final speed (v) / m/s |
time between start and light gate / s |
acceleration / m/s² |
|---|
Physics With Kate · physicswithkate.com
Resultant force zero → the velocity stays the same. Not zero → the velocity changes: speed, direction, or both.
The notes below cover both laws, the equation, and exactly what to write about the practical.
If the resultant force on an object is zero, the forces are balanced. A stationary object stays stationary, and a moving object carries on at the same speed in the same direction (constant velocity).
So a car on a straight, level road at a steady speed has driving force = resistive forces. A skydiver at terminal velocity has air resistance = weight. Balanced doesn't mean stopped: it means the velocity doesn't change.
Velocity is speed in a given direction, so it changes if the speed changes, if the direction changes, or both. Any change in velocity needs a resultant force: that's Newton's second law.
| Example | Resultant force | Law | What changes |
|---|---|---|---|
| car at a steady 20 m/s on a straight road | 0 N | first | neither |
| book resting on a table | 0 N | first | neither (stays still) |
| van speeding up | forwards | second | speed |
| cyclist braking | backwards | second | speed (slows down) |
| car going round a bend at a steady 12 m/s | towards the centre | second | direction |
The tendency of an object to stay at rest or keep moving at a steady velocity is called its inertia.
A resultant force makes an object accelerate. How much it accelerates depends on how big the force is and how much mass is being pushed.
resultant force = mass × acceleration · F = m a
F in newtons (N) · m in kilograms (kg) · a in metres per second squared (m/s²)
Rearranged: a = F ÷ m and m = F ÷ a. You need all three.
It is the resultant force that goes in the equation — the single force left over once you have combined everything acting on the object. If the resultant force is zero, the acceleration is zero, and the object stays still or carries on at constant velocity.
Read a = F ÷ m twice, holding one thing constant each time:
| Keep this constant | What happens | In words | Graph shape |
|---|---|---|---|
| Mass constant | Double F → double a | Acceleration is directly proportional to resultant force | Straight line through the origin |
| Force constant | Double m → half a | Acceleration is inversely proportional to mass | Curve, falling away |
The required practical tests the first row: keep the mass fixed, change the force, and see whether the acceleration really is proportional to it.
A trolley on a bench is pulled by a string that runs over a pulley, with masses hanging on the end. Here is the trap:
If you added new masses to the hanger, you would increase the force and the total mass at the same time — two variables changing at once, so the experiment would prove nothing.
By moving a mass from the trolley to the hanger, the force increases while the total mass being accelerated stays exactly the same. That is what makes it a fair test.
The force — the weight of the hanging masses. This is what you change.
Goes on the x-axis (along the bottom).The acceleration, worked out from the light gate readings. This is what you measure.
Goes on the y-axis (up the side).The total mass of trolley + string + all the masses, and the release point.
Kept the same every run — that is why it goes on neither axis.The trolley is released from rest at the same point every time, so u = 0. The light gate gives the speed v as the card cuts the beam, and the timer gives t, the time from release to the gate.
a = (v − u) ÷ t, and because it starts from rest this becomes a = v ÷ t.
You do not need the distance to the gate for this — only the speed and the time.
Plot acceleration (y) against force (x) and draw a straight line of best fit through the origin. Because a = (1/m) × F, that line has:
gradient = 1 ÷ mass, so mass = 1 ÷ gradient.
If the gradient is 1.00 (m/s²)/N, the mass of the whole system is 1 ÷ 1.00 = 1.00 kg — which should match the trolley plus all the masses you started with. That is your check that the experiment worked.
The line goes through the origin because no force means no acceleration. If your line misses the origin, something systematic is wrong — usually friction.
Friction between the wheels and the bench opposes the motion, so the real resultant force is smaller than the weight you hung on the string. The measured acceleration comes out too low and the line does not pass through the origin.
The fix is to compensate for friction: tilt the runway by raising the far end slightly, until the trolley rolls at a constant speed when given a gentle push with nothing hanging on the string. At that tilt, a component of the trolley's weight exactly cancels friction, so the only resultant force left is the one you add on the string.
Rearranged the other way, m = F ÷ a defines the inertial mass of an object.
Inertial mass is a measure of how difficult it is to change the velocity of an object. It is defined as the ratio of force over acceleration.
1. "Add more masses to the hanger." That changes the total mass too — move them across instead.
2. "The mass is just the trolley." It is the trolley and the string and every mass, hanging or not.
3. "The gradient is the mass." The gradient is 1 ÷ mass, so you have to do one more step.
4. Joining the dots instead of drawing a single straight line of best fit through the origin.
Calculation questions want the same four lines every time. Write them out even when you can do it in your head — the marks are for the working, not the number.
name the step the algebra, before the numbers never leave off the unit
A car of mass 1500 kg is driven along a straight, level road. The driving force is 5.4 kN and the total resistive force is 1.65 kN. Calculate the resultant force and the acceleration. (5 marks)
Opposite directions, so subtract: 5.4 − 1.65 = 3.75 kN = 3750 N. a = F ÷ m → a = 3750 ÷ 1500 → a = 2.5 m/s²
A car goes round a bend at a steady 12 m/s. Is there a resultant force on it? Explain. (2 marks)
Yes. Its direction is changing, so its velocity is changing, and a change in velocity needs a resultant force (Newton’s second law).
A 1200 kg car accelerates at 2.5 m/s². Find the resultant force.
F = m a → F = 1200 × 2.5 → F = 3000 N
A resultant force of 24 N acts on a 6 kg mass. Find the acceleration.
a = F ÷ m → a = 24 ÷ 6 → a = 4 m/s²
Light gate data: released from rest, v = 1.25 m/s, t = 1.28 s. Find the acceleration.
It started from rest so u = 0. a = (v − u) ÷ t → a = (1.25 − 0) ÷ 1.28 → a = 0.98 m/s²
The graph of acceleration against force has a gradient of 1.00 (m/s²)/N. Find the mass. [2 marks]
Since a = (1/m) × F, the gradient = 1 ÷ m, so m = 1 ÷ gradient → m = 1 ÷ 1.00 = 1.00 kg.
Explain why the masses are moved from the trolley to the hanger rather than added to the hanger. [3 marks]
The force is the weight of the hanging masses, so moving one across increases the force. The masses are still part of the system, so the total mass being accelerated stays the same. This keeps the mass a control variable, so any change in acceleration must be caused by the force alone.
The line of best fit does not pass through the origin. Suggest why. [2 marks]
Friction acts on the trolley, so the real resultant force is less than the weight hung on the string and the acceleration is too low. Tilt the runway to compensate for friction — raise one end until the trolley moves at constant speed with nothing hanging on the string.
Every calculation answer should show: 1) the equation, 2) the rearrangement, 3) the substitution, 4) the answer with its unit. Four lines, four chances to pick up a mark even if the arithmetic slips.
Page 13: both laws in words, which law is acting in four pictures, and an F = m a exam question. Page 14: the required practical, with the method to fill in, a results table and a graph to plot.
AQA GCSE Physics Workbook · FORCES · pages 13–14Both laws, F = m a and the practical explained out loud — useful for a last-minute recap, or if you'd rather hear it than read it.
Both laws, F = m a and the practical explained out loud — useful for a last-minute recap, or if you'd rather hear it than read it. The video is on its way.
Video coming soon
The walkthrough for this lesson is being filmed and will appear right here. Until then, the notes and the free workbook pages above cover every mark.
Pages 13–14 of the Forces workbook are free: Newton's first and second laws with F = m a, then the required practical.
Spec-aligned revision resources, group courses, and 1:1 tutoring for GCSE and A Level Physics — built by an experienced teacher and examiner.