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AQA GCSE Physics · Topic 4.5 Forces · Required Practical
Force, Mass and Acceleration
In this lesson you'll learn to: use F = m a in all three rearrangements, explain why
acceleration is proportional to the resultant force and inversely proportional to the
mass, run the required practical and know exactly which quantity is changed, measured
and controlled, and read the mass out of the gradient of an acceleration–force graph.
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Prefer to watch? Scroll to the bottom for the video on this topic
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Start here — run the practical
Don't read anything yet. Move a mass from the trolley onto the hanger and press release. Watch the two panels: the total mass never changes, but the force goes up every time. Read the light gate, work out the acceleration yourself, then plot the graph and let it draw the line of best fit.
4.5.6.2.2 · AQA GCSE PHYSICS · REQUIRED PRACTICAL
Force, Mass and Acceleration
Aim: investigate how the force on an object of constant mass
affects its acceleration.
1
Move a mass from the trolley to the hanger
— watch what changes, and what doesn'tThe system — trolley + string + masses
This never changes — the masses stay in the system, they just move.
The force — hanging masses only
Only the masses hanging on the string pull the trolley along.
The bit everyone gets wrong
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.
2
Release the trolley and read the light gate
— then work out the acceleration yourself3
Your results
— a row appears each time you get the acceleration right| 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 |
|---|
4
Plot the graph
— you choose what goes on each axisHorizontal axis (x) — along the bottom
Vertical axis (y) — up the side
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▲ The masses never leave the system — that is the whole trick of this practical.
Got the graph? Now learn why it works.
That straight line through the origin is Newton's Second Law, and its gradient hands you the mass. The notes below explain the equation, why you move masses across instead of adding them, and exactly what to write when the examiner asks about this practical.
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Revision notes
Newton's Second Law
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.
⭐ Key equation — Newton's Second Law
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.
The two things the equation is telling you
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.
The required practical — the one thing everyone gets wrong
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:
- The force pulling the system along is the weight of the hanging masses only (F = mhanging × g). A mass sitting on the trolley is held up by the trolley, so it pulls on nothing.
- The mass in F = m a is everything that is accelerating — the trolley, the string, and every mass, whether it is on the trolley or on the hanger.
⭐ Key technique — why you MOVE masses instead of ADDING them
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 three variables
Learn these words→
Independent
The force — the weight of the hanging masses. This is what you change.
Goes on the x-axis (along the bottom).↑
Dependent
The acceleration, worked out from the light gate readings. This is what you measure.
Goes on the y-axis (up the side).=
Control
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.Getting the acceleration out of the light gate
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.
⭐ Key equation — acceleration
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.
The graph, and where the mass is hiding
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:
⭐ Key result — the gradient
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, and how to compensate for it
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.
Inertial mass
Rearranged the other way, m = F ÷ a defines the inertial mass of an object.
⭐ Key definition — inertial mass
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.
💡 Exam tip — four things students get wrong
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.
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How to write it in the exam
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.
The PWK sentence frame
Equation: F = m a
Rearrange: a = F ÷ m (do this before you put numbers in)
Substitute: a = 24 ÷ 6
Answer: a = 4 m/s² — with the unit
Rearrange: a = F ÷ m (do this before you put numbers in)
Substitute: a = 24 ÷ 6
Answer: a = 4 m/s² — with the unit
name the step the algebra, before the numbers never leave off the unit
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.
💡 Exam tip — the four-part check
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.
📘 Now do it in your workbook
The full method with the gaps to fill, a results table to complete, a grid to plot the graph on, and the model conclusion and evaluation — with the answers at the back.
AQA GCSE Physics Workbook · FORCES · page 14Free sample = the page for this lesson. The full workbook covers the whole of Topic 5 Forces with exam-style questions and worked answers.
✅ Can you do it? Tick as you go
0% complete🎉 Nice work! You've ticked off every objective for this spec point.
Don't forget to hit “Mark complete” at the bottom of the lesson.
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Prefer to watch? Here's the whole thing
Everything above, explained out loud — useful for a last-minute recap, or if you'd rather hear it than read it.
▲ In GHL you can also use the lesson's built-in video field instead of this embed.
Now put it into practice
Required practical questions are worth a lot of marks and they are very predictable — but only if you have written the method out yourself at least once. Page 14 of the workbook is free, and the full book covers every spec point in Topic 5 Forces with worked answers.
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