Newton's First and Second Laws: F = m a and the Required Practical — GCSE Physics — Physics with Kate
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4.5.6.2.1 · 4.5.6.2.2
AQA GCSE Physics · Topic 4.5 Forces · Combined and Triple · Required practical

Newton's First and Second Laws: F = m a and the Required Practical

In this lesson you'll learn to: decide whether Newton's first or second law is acting, and whether the speed, the direction or neither is changing; use F = m a in all three rearrangements; run the force and acceleration required practical; and read the mass out of the gradient of an acceleration–force graph.
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Start here — have a play

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

m/s²

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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't
The 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 yourself
3

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 axis
Horizontal axis (x) — along the bottom
Vertical axis (y) — up the side

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Resultant force zero → the velocity stays the same. Not zero → the velocity changes: speed, direction, or both.

The resultant force decides whether the velocity changes. F = m a decides by how much.

The notes below cover both laws, the equation, and exactly what to write about the practical.

Read the notes ↓
✎

Revision notes

Newton's First Law

⭐ Learn this

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 changes in two ways

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.

ExampleResultant forceLawWhat changes
car at a steady 20 m/s on a straight road0 Nfirstneither
book resting on a table0 Nfirstneither (stays still)
van speeding upforwardssecondspeed
cyclist brakingbackwardssecondspeed (slows down)
car going round a bend at a steady 12 m/stowards the centreseconddirection

The tendency of an object to stay at rest or keep moving at a steady velocity is called its inertia.

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 constantWhat happensIn wordsGraph 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.

✍

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

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.

💡 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.

Mass kept constant a ∝ F — directly proportional Δx Δy through the origin resultant force / N acceleration / m/s² gradient = 1 ÷ mass Force kept constant a ∝ 1/m — inversely proportional m 2m double the mass → half the acceleration mass / kg acceleration / m/s² a curve, never a straight line
The required practical produces the graph on the left. The gradient is 1 ÷ mass, so a steeper line means a lighter system.

📘 Now do it in your workbook

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 13–14

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Prefer to watch? Here's the whole thing

Video walkthrough — coming soon

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.

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.

Now put it into practice

Pages 13–14 of the Forces workbook are free: Newton's first and second laws with F = m a, then the required practical.

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