Newton's Laws of Motion Revision Notes — 9709 Paper 4

topic	marks/paper
Kinematics of Motion in a Straight Line	20.9
Forces and Equilibrium	13.8
Newton's Laws of Motion	13.2
Energy, Work and Power	12.9
Momentum	4.6

sub-topic	parts	marks	sections
Applying Newton's second law to linear motion under constant forces	89	354	01–04, 12–15
Motion on inclined planes with constant acceleration	61	285	05–08, 10, 11, 14, 15
Motion of connected particles (e.g. pulley systems)	35	167	09–12, 14
Relationship between mass and weight ( $W = mg$ )	6	23	01, 04

Newton's three laws, F = ma, and W = mg

Syllabus requirement · §4.4

“

apply Newton's laws of motion to the linear motion of a particle of constant mass moving under the action of constant forces, which may include friction, tension in an inextensible string and thrust in a connecting rod; use the relationship between mass and weight (W = mg; in this component, questions are mainly numerical, and use of the approximate numerical value 10 (m s⁻²) for g is expected).

”

In §4.1 you drew force diagrams for particles that were not accelerating, and the forces always added up to zero. Real questions are mostly about particles that speed up, slow down, or are pulled along by something. Newton's three laws are the rules that connect the forces on a particle to the way it moves.

First law. A particle stays at rest, or keeps moving in a straight line at constant speed, unless a resultant force acts on it.

So "at rest" and "moving at constant velocity" are the same situation as far as forces are concerned: in both, the resultant force is zero, and the equilibrium methods of §4.1 apply. Whenever a question says a car moves at a constant speed or a steady speed along a straight road, read it as "the forces balance".

Second law. When the resultant force is not zero, the particle accelerates in the direction of the resultant, and

F = ma

where $F$ is the resultant force in newtons (N), $m$ is the mass in kilograms (kg) and $a$ is the acceleration in $\text{m s}^{-2}$ . One newton is defined as the force that gives a mass of 1 kg an acceleration of $1\text{ m s}^{-2}$ , which is why the units fit: $1\text{ N} = 1\text{ kg m s}^{-2}$ .

The word resultant is the whole of this topic. $F$ is not "the force in the question"; it is every force along the line of motion added together with signs. Forces in the direction of the acceleration count as positive, forces against it count as negative.

Third law. If body $A$ pushes or pulls on body $B$ , then $B$ pushes or pulls on $A$ with a force of the same size in the opposite direction. The two forces act on different bodies, so they never appear on the same force diagram. You will use this for a crate on a lift floor (the floor pushes the crate up, the crate pushes the floor down), for a string (it pulls both particles it joins, each towards the other) and for a tow-bar (it pulls the trailer forward and the car backward).

F = ma

Newton's second law

F is the resultant force along the direction of the acceleration; m is constant throughout this topic

Mass and weight. Mass (kg) measures how much matter a body contains, and it does not change. Weight (N) is the force of gravity on the body. A body falling freely has only its weight acting on it, and it accelerates at $g$ , so the second law with $a = g$ gives the weight:

W = mg, \qquad g = 10\text{ m s}^{-2} \text{ on Paper 4}

Two slips come up again and again. First, a question may give a weight ("a particle of weight $30\text{ N}$ "): its mass is $30 \div 10 = 3\text{ kg}$ , and it is the $3$ , not the $30$ , that goes on the right-hand side of $F = ma$ . Second, the right-hand side of $F = ma$ always uses the mass, never the weight: writing $30a$ for a $3\text{ kg}$ particle multiplies the answer by 10.

Given in the question	Mass $m$ (for $ma$ )	Weight $mg$ (for the force diagram)
a block of mass $4\text{ kg}$	$4\text{ kg}$	$40\text{ N}$
a particle of mass $0.35\text{ kg}$	$0.35\text{ kg}$	$3.5\text{ N}$
a particle of weight $12\text{ N}$	$1.2\text{ kg}$	$12\text{ N}$
a lorry of mass $15\,000\text{ kg}$	$15\,000\text{ kg}$	$150\,000\text{ N}$

Weight goes on the force diagram; mass goes in ma. Convert once, at the start.

Applying Newton's second law, every time

1
Draw a force diagram for the particle, showing every force: weight, normal reaction $R$ , friction $F$ , tensions, driving forces, resistances.
A force missing from the diagram is missing from the equation, and the mark scheme's first method mark always checks the number of terms.
2
Mark the direction of the acceleration (usually the direction of motion) and call it positive.
Then every force pointing that way is added and every force pointing the other way is subtracted.
3
Perpendicular to the motion, there is no acceleration, so the forces balance. Use this to find the normal reaction $R$ if friction is involved.
A particle sliding along a floor does not accelerate into the floor or off it. This is the §4.1 equilibrium step, applied in the one direction where it still holds.
4
Along the motion, write resultant $= ma$ : (forces in the positive direction) $-$ (forces against it) $= ma$ .
This is the line that earns the 'N2L' method mark: correct number of terms, correct mass.
5
Solve for the unknown, then use suvat (§4.2) if the question asks about times, distances or speeds.
Forces give the acceleration; suvat turns the acceleration into motion.

Pulled forwards by 30 N, held back by 10 N of friction: the resultant is 20 N, so 20 = 5a and a = 4 m s⁻². Vertically, R = 50 N balances the weight.

Finding the acceleration from the forces

A block of mass $5\text{ kg}$ is pulled along rough horizontal ground by a horizontal rope with tension $30\text{ N}$ . The friction force on the block is $10\text{ N}$ . Find the acceleration of the block, and the normal reaction on it.

Show full working

1
Force diagram. Horizontally: the tension $30\text{ N}$ forwards and friction $10\text{ N}$ backwards. Vertically: the weight $5 \times 10 = 50\text{ N}$ down and the normal reaction $R$ up.
Friction always acts against the motion, so it points backwards.
2
Choose the positive direction: the direction the block moves, forwards.
The acceleration will be forwards too, because the forward pull is bigger than the friction.
3
Vertically there is no acceleration, so the vertical forces balance: $R = 50\text{ N}$
The block slides along the ground; it does not accelerate up or down.
4
Resultant force along the motion: forwards minus backwards: $30 - 10 = 20\text{ N}$
This 20 N is the F in F = ma. Using 30 N on its own is the classic mistake.
5
Apply $F = ma$ : $20 = 5a$
Mass 5 kg on the right, not the weight 50 N.
6
Solve: $a = \frac{20}{5} = 4\text{ m s}^{-2}$
Positive, so the block accelerates in the direction we chose.

Answer

$a = 4\text{ m s}^{-2}$ in the direction of the pull; $R = 50\text{ N}$ .

Working backwards: finding a force from the acceleration

A box of mass $8\text{ kg}$ is pushed across a rough floor by a horizontal force $P\text{ N}$ . The friction force on the box is $6\text{ N}$ , and the box accelerates at $1.5\text{ m s}^{-2}$ . Find $P$ .

Show full working

1
Horizontal forces: $P$ forwards, friction $6\text{ N}$ backwards. Positive direction: forwards.
Same set-up as before; only the unknown has moved.
2
Resultant force: $P - 6$
Written with P still unknown. The equation will find it.
3
Apply $F = ma$ with $m = 8$ and $a = 1.5$ : $P - 6 = 8 \times 1.5$
Every N2L question is this one equation; the unknown can be the force, the mass or the acceleration.
4
Evaluate the right-hand side: $P - 6 = 12$
8 × 1.5 = 12 N is the resultant force needed for this acceleration.
5
Solve: $P = 18$
P has to beat the friction (6 N) and still leave 12 N over to accelerate the box.

Answer

$P = 18$ .

The first law: constant speed means the forces balance

A car moves along a straight horizontal road at a constant speed of $20\text{ m s}^{-1}$ . The driving force of its engine is $600\text{ N}$ . Find the total resistance to its motion.

Show full working

1
Read "constant speed" as "no acceleration": $a = 0$
Newton's first law: constant velocity happens only when the resultant force is zero.
2
Resultant force along the road: driving force minus resistance, $600 - R_{\text{es}}$ where $R_{\text{es}}$ is the resistance.
The 20 m s⁻¹ does not appear anywhere in the force equation. Speed alone tells you nothing about the force; only a change of speed does.
3
Apply $F = ma$ with $a = 0$ : $600 - R_{\text{es}} = 0$
The same equation as before, with zero on the right.
4
Solve: $R_{\text{es}} = 600\text{ N}$
The engine is working, but only hard enough to cancel the resistance.

Answer

The resistance is $600\text{ N}$ .

Students often think a moving car must have a forward resultant force. It does not: a car cruising at constant speed has zero resultant force, exactly like a parked one.

Newton's second law on a real paper

9709/41 O/N 2025 Q1(a)2 marks

A car of mass $900\text{ kg}$ is moving along a straight horizontal road against a constant resistance to motion of $350\text{ N}$ . At an instant when the car is moving at $15\text{ m s}^{-1}$ its acceleration is $0.25\text{ m s}^{-2}$ .

Find the driving force of the car's engine at this instant.

Show full working

1
Horizontal forces on the car: the driving force $D$ forwards and the resistance $350\text{ N}$ backwards. Positive direction: forwards.
The speed of 15 m s⁻¹ is for part (b) (the power); it plays no part in the force equation.
2
Resultant force: $D - 350$
Forwards minus backwards.
3
Apply $F = ma$ with $m = 900$ and $a = 0.25$ : $D - 350 = 900 \times 0.25$
This is the M1 line: N2L with the correct number of terms.
4
Evaluate the right-hand side: $D - 350 = 225$
900 × 0.25 = 225 N of resultant force.
5
Solve: $D = 575\text{ N}$
Driving force = resistance + resultant needed for the acceleration.

Answer

$575\text{ N}$ .

The first law on a real paper: a coupling with no tension

9709/42 F/M 2023 Q4(a)1 mark

A toy railway locomotive of mass $0.8\text{ kg}$ is towing a truck of mass $0.4\text{ kg}$ on a straight horizontal track at a constant speed of $2\text{ m s}^{-1}$ . There is a constant resistance force of magnitude $0.2\text{ N}$ on the locomotive, but no resistance force on the truck. There is a light rigid horizontal coupling connecting the locomotive and the truck.

State the tension in the coupling.

Show full working

1
Look at the truck on its own. The only horizontal force that could act on it is the force from the coupling, $T$ . There is no resistance on the truck.
Choosing the right body to look at is the whole question. The locomotive has three horizontal forces; the truck has one.
2
Constant speed, so no acceleration: $a = 0$ for the truck.
First law: constant velocity means zero resultant force.
3
Apply $F = ma$ to the truck: $T = 0.4 \times 0 = 0$
The coupling is not needed to keep the truck moving; with nothing slowing it down, it carries on by itself.

Answer

The tension is $0\text{ N}$ .

A body moving at constant speed with no resistance needs no force at all. The mark scheme accepts the answer simply stated, but the reason is the first law.

Common mistakes

Using one force (for example the pull) as $F$ in $F = ma$
$F$ is the resultant: add the forces along the motion with signs
A 30 N pull against 10 N of friction gives a resultant of 20 N, not 30 N.
Writing the weight on the right-hand side: $50a$ for a $5\text{ kg}$ block
The right-hand side is always mass × acceleration: $5a$
Mark schemes check that 'masses must be correct' in the N2L equation.
Assuming a moving object must have a resultant force in its direction of motion
Constant velocity means zero resultant force (first law)
Only a change in velocity needs a resultant force.
Putting both forces of a Newton's-third-law pair on the same force diagram
The two forces act on different bodies; each body's diagram shows only the forces acting on it
If both were on one diagram they would always cancel, and nothing could ever accelerate.

Your turn

Draw the forces, choose a positive direction, then write resultant = ma.

1
A crate of mass $6\text{ kg}$ is pulled across a rough horizontal floor by a horizontal force of $40\text{ N}$ . The friction force is $16\text{ N}$ . Find the acceleration of the crate.
Stuck? Show hint
The resultant is the pull minus the friction.
Show solution
1. 1
  Resultant force, forwards positive: $40 - 16 = 24\text{ N}$
  Friction acts backwards, against the motion.
2. 2
  $F = ma$ : $24 = 6a$
  Mass 6 kg on the right.
3. 3
  Solve: $a = 4\text{ m s}^{-2}$
  Divide the resultant force by the mass.
Answer
$4\text{ m s}^{-2}$ .
2
A trolley is pushed with a horizontal force of $45\text{ N}$ against a resistance of $15\text{ N}$ , and accelerates at $2.5\text{ m s}^{-2}$ . Find the mass of the trolley.
Stuck? Show hint
This time the unknown is $m$ .
Show solution
1. 1
  Resultant force: $45 - 15 = 30\text{ N}$
  Forwards minus backwards.
2. 2
  $F = ma$ : $30 = m \times 2.5$
  The mass is the unknown.
3. 3
  Solve: $m = \frac{30}{2.5} = 12\text{ kg}$
  Divide the resultant by the acceleration.
Answer
$12\text{ kg}$ .
3
(a) Find the weight of a particle of mass $0.35\text{ kg}$ . (b) A particle has weight $12\text{ N}$ . A resultant force of $3\text{ N}$ acts on it. Find its acceleration.
Stuck? Show hint
In (b), convert the weight to a mass first.
Show solution
1. 1
  (a) $W = mg = 0.35 \times 10 = 3.5\text{ N}$
  Weight is a force, so the answer is in newtons.
2. 2
  (b) Mass from the weight: $m = \frac{W}{g} = \frac{12}{10} = 1.2\text{ kg}$
  F = ma needs the mass.
3. 3
  $F = ma$ : $3 = 1.2a$
  The 3 N is already the resultant.
4. 4
  (b) Solve: $a = 2.5\text{ m s}^{-2}$
  Using 12 instead of 1.2 would give 0.25, ten times too small.
Answer
(a) $3.5\text{ N}$ . (b) $2.5\text{ m s}^{-2}$ .
4
A sledge is pulled across horizontal snow at a constant speed of $3\text{ m s}^{-1}$ by a horizontal rope. The tension in the rope is $40\text{ N}$ . Find the friction force on the sledge. What would the friction be if the speed were a constant $6\text{ m s}^{-1}$ with the same tension?
Stuck? Show hint
Constant speed means zero resultant force.
Show solution
1. 1
  Constant speed: $a = 0$ , so the resultant force is zero.
  Newton's first law.
2. 2
  Horizontally: $40 - F = 0$
  The tension exactly balances the friction.
3. 3
  Solve: $F = 40\text{ N}$
  Add F to both sides.
4. 4
  At a constant $6\text{ m s}^{-1}$ the reasoning is identical: $a = 0$ again, so $F = 40\text{ N}$ again.
  The value of a constant speed never enters the force equation.
Answer
$40\text{ N}$ in both cases.

Practise applying Newton's second lawReal past-paper questions · Applying Newton's second law to linear motion under constant forces Practise mass and weightReal past-paper questions · Relationship between mass and weight (W = mg)

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Can you do all of these?

Draw a force diagram for each particle and mark the direction of the acceleration before writing any equation
F in F = ma is the resultant: forces with the acceleration minus forces against it
The right-hand side is mass × acceleration: convert a weight to a mass by dividing by 10
Constant speed (or steady speed) means a = 0: resolve as in equilibrium
Perpendicular to the motion there is no acceleration: use it to find R
A force at an angle changes R: R = mg − P sin θ for a pull above the horizontal, R = mg + P sin θ for a push below it
Before writing F = μR for a body starting from rest, check that the pull beats μR
Friction opposes the motion, not the applied force: decide the direction of motion first
Vertically, the weight is on the line of motion: T − mg = ma for a lift, mg − resistance = ma for a pile-driver
Rising and slowing down means a < 0 (upwards positive); falling and slowing down means a > 0
On a slope: mg sin α along, mg cos α perpendicular; turn ratios like tan α = 3/4 into exact sin and cos
Keep the slope angle (with the weight) apart from a rope's angle to the slope (with T)
A rough slope gives one acceleration going up and another coming down; it slides back only if tan α > μ
Light string over a smooth pulley: one tension; inextensible: one size of acceleration
Write one equation per particle; the system equation gives a but never T
Two strings means two tensions
'Does it move?': compare the driving force with the greatest friction, state both, then conclude
Towing: the tow-bar pulls the trailer forward and the car back; the trailer's equation is the quickest route to T
A negative T is a thrust in a tow-bar, or a slack rope; the least braking force for a slack rope comes from T = 0
Driving force = power ÷ speed, with the power in watts; at steady speed it equals the total resistance (plus mg sin α uphill); the other power questions are in §4.5
When a string goes slack or breaks, the particles part company: find the speed at that instant, then a new F = ma for each
Add every stage when asked for a total distance or a greatest height
Each new section of a track, or a force starting or stopping, needs a new acceleration
Two particles on one slope: one origin, one clock (t and t − 1), then set the displacements equal
Use the unrounded acceleration in the suvat that follows, and give answers to 3 significant figures unless the question asks otherwise

Now do the questions

117 real Paper 4 parts from 2021–2025, sorted by difficulty, with mark schemes

Momentum

Energy, Work and Power

Newton's Laws of Motion

Newton's three laws, F = ma, and W = mg

Horizontal motion: friction, resistance, and the link to suvat

A force at an angle: when the normal reaction changes

Can you do all of these?

Newton's three laws, F = ma, and W = mg

The rest of this note

Can you do all of these?