M8 Forces and Newton's laws — notes on one page

Every Edexcel A-level Maths 9MA0 specification point in M8, read straight through: 6 points with definitions, explanations, mark-scheme wording and worked examples. Practice questions are in the printable question pack.

Checked against Edexcel 9MA0 section M8. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0) specification; registry verification recorded 11 July 2026.

How this checking works

M8.1 · Understand the concept of a force; understand and use Newton's first law.

A force is an interaction with magnitude and direction; common modelled forces include weight, normal reaction, tension, thrust, compression and resistance.
Draw a force diagram for the chosen body, including only forces acting on that body, then resolve forces along useful perpendicular directions.
Newton's first law says that a particle remains at rest or moves with constant velocity when the resultant force on it is zero.
A common error is to infer that a moving particle must have a forward resultant force; constant non-zero velocity also means zero resultant force.

Worked example

A powered trolley moves in a straight line with constant velocity. Its motor exerts a forward force of $8\,\text{N}$ . Find the resistance force and explain your answer.

1.Constant velocity means zero acceleration and hence, by Newton's first law, zero resultant force.
2.The resistance must therefore balance the motor force, so it is $8\,\text{N}$ backwards.

Answer: Resistance $=8\,\text{N}$ opposite to the motion; Constant velocity means the resultant force is zero

Common mistakes

Don't balance only the horizontal forces and declare equilibrium while a vertical resultant remains.
Don't assume motion implies a resultant force even when the velocity is constant.

Exam tip. For constant velocity, invoke zero acceleration and balance the forces with equal magnitudes in opposite directions.

M8.2 · Understand and use Newton's second law for motion in a straight line (forces in two perpendicular directions or simple 2-D vectors); extend to situations where forces need to be resolved (restricted to 2 dimensions).

Newton's second law is $\sum\mathbf{F}=m\mathbf{a}$ ; in two dimensions it gives one scalar equation in each resolved direction.
Choose axes parallel and perpendicular to the motion where possible, resolve every force onto those axes, then apply $\sum F=ma$ separately.
For motion along a fixed smooth plane, perpendicular acceleration is zero, so the perpendicular force equation determines the normal reaction while the parallel equation determines acceleration.
A common error is to write $F=ma$ for one force instead of the resultant, or to reverse signs for only some components after choosing a positive direction.

Resolve forces in perpendicular directions, then apply

\sum\mathbf F=m\mathbf a

component by component.

Worked example

A $5\,\text{kg}$ particle is on a smooth plane inclined at $30^\circ$ to the horizontal. A force of $40\,\text{N}$ pulls it up the line of greatest slope. Using $g=9.8\,\text{m s}^{-2}$ , find its acceleration.

1.Resolve up the plane.
2.The component of weight down the plane is $5(9.8)\sin30^\circ=24.5\,\text{N}$ .
3.Hence the resultant up the plane is $40-24.5=15.5\,\text{N}$ .
4.From $15.5=5a$ , $a=3.1\,\text{m s}^{-2}$ up the plane.

Answer: $3.1\,\text{m s}^{-2}$ up the plane

Common mistakes

Don't include the normal reaction in the equation parallel to an inclined plane.
Don't resolve weight using the wrong trigonometric component relative to the inclined plane.

Exam tip. Draw a force diagram, choose axes parallel and perpendicular to the plane, then resolve every force consistently.

M8.3 · Understand and use weight and motion in a straight line under gravity; gravitational acceleration, g, and its value in S.I. units to varying degrees of accuracy.

Weight is the gravitational force $W=mg$ , directed vertically downwards; mass is measured in kilograms and does not depend on location.
The value of $g$ is not universal and depends on location; use the stated value, or the Edexcel default $9.8\,\text{m s}^{-2}$ when none is supplied.
For free motion near Earth's surface in the constant- $g$ model, acceleration is $g$ downwards, but a support or tension changes the resultant acceleration.
A common error is to call mass a force or to assume that the normal reaction always equals weight when the body has vertical acceleration.

Worked example

A particle is released from rest and falls freely through $19.6\,\text{m}$ . Using $g=9.8\,\text{m s}^{-2}$ , find the time taken and its speed after falling this distance.

1.Taking downwards as positive, $s=ut+\tfrac12gt^2$ gives $19.6=0+4.9t^2$ , so $t=2.0\,\text{s}$ .
2.Then $v=u+gt=0+9.8(2)=19.6\,\text{m s}^{-1}$ downwards.

Answer: Time $=2.0\,\text{s}$ ; Speed $=19.6\,\text{m s}^{-1}$

Common mistakes

Don't use $g=9.8$ with distances in centimetres and times in seconds without converting units.
Don't use positive gravitational acceleration with an upward-positive displacement equation, producing inconsistent signs.

Exam tip. Declare the positive direction before substituting gravitational acceleration into a constant-acceleration equation.

M8.4 · Understand and use Newton's third law; equilibrium of forces on a particle and motion in a straight line; apply to smooth pulleys and connected particles; resolve forces in 2 dimensions; equilibrium of a particle under coplanar forces.

Definition. Specification M8.4: understand and use Newton's third law; equilibrium of forces on a particle and motion in a straight line; application to problems involving smooth pulleys and connected particles; resolving forces in two dimensions; equilibrium of a particle under coplanar forces.

Draw a separate force diagram for each body, then resolve: in equilibrium the components in two perpendicular directions each sum to zero; in motion the resultant equals $ma$ along the line of motion, with one sign convention throughout.
For connected particles joined by a light inextensible string over a smooth light pulley, the tension is the same on both sides and both particles have the same magnitude of acceleration, so write $F=ma$ for each particle separately and solve the pair simultaneously;
the string, pulley and modelling assumptions are what justify those two facts and are the answer when the question asks why.
On an inclined plane resolve along and perpendicular to the plane, with $R=mg\cos\alpha$ , never $mg$ , and the weight component $mg\sin\alpha$ down the slope.
Newton's third law pairs act on different bodies, so the force on the pulley is $2T$ downwards. Give forces in newtons with $g=9.8$ , to two or three significant figures.

Two particles connected by a light string over a smooth pulley share one tension magnitude while the string is taut.

Worked example

Two particles

A

and

B

, of masses

5

kg and

3

kg respectively, are connected by a light inextensible string passing over a small smooth fixed pulley. The particles hang freely and the system is released from rest with the string taut. Find the acceleration of the particles and the tension in the string.

(6)

(Total for Question 1 is 6 marks)

Working

1.For $A$ , taking downwards as positive: $5g-T=5a$ .
2.For $B$ , taking upwards as positive: $T-3g=3a$ .
3.Add the equations to eliminate $T$ : $2g=8a$ , so $a=\dfrac{g}{4}=2.45$ m s $^{-2}$ .
4.Substitute back: $T=3g+3a=3(9.8)+3(2.45)=36.75$ N, that is $37$ N to 2 s.f.

Answer: $a=2.45$ m s $^{-2}$ ; $T=36.75$ N ( $37$ N)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1.	Equation of motion for $A$ , with the correct terms: $5g-T=5a$	M1
	$5g-T=5a$	A1
	Equation of motion for $B$ : $T-3g=3a$	M1
	$T-3g=3a$	A1
	$a=2.45$ m s $^{-2}$	A1
	$T=36.75$ N	A1
Notes
M1: Condone sign errors
A1: Or the whole-system equation $2g=8a$ for M1A1
A1: Accept $\frac{g}{4}$ or $2.5$
A1: Accept $37$ N; $\frac{15g}{4}$

(6 marks)

Mark-scheme wording to use

“0.5g, (1/2)g or 4.9 (N) seen … B1 cao. (must be positive) … This answer must appear in (a) to earn this mark.” — Pearson 9MA0, June 2024 Paper 32, Q1(a)
“M1 for resolving horizontally or an equivalent complete method. A1 for X = Tcos(60). M1 for resolving vertically or an equivalent complete method. A1 for Y = Tcos(30) - 20g as the downward component magnitude” — Pearson 9MA0, Specimen Paper 32, Q4(b)
“B1 for R = mg cos(alpha). M1 for resolving parallel to the plane. A1 for the correct equation -F - mg sin(alpha) = -0.8mg, or a sign-equivalent form” — Pearson 9MA0, Sample Paper 32, Q2(a)

What the examiner wants. Each stage named and the diagram annotated. The June 2024 Paper 32 report says the most successful candidates indicated what they were doing at each stage, resolving horizontally or taking moments about $A$ , and that on the inclined plane those who used $\tan$ , or divided by $\sin$ or $\cos$ instead of writing $R=mg\cos\alpha$ , lost both marks.

Common mistakes

Don't assign different tension magnitudes to the two ends of one light inextensible string over a smooth pulley.
Don't treat a Newton third-law pair as two forces on the same particle or assume connected-particle tensions without a force diagram.

Exam tip. Draw a separate force diagram for each particle and distinguish interaction pairs from forces that can balance.

M8.5 · Understand and use addition of forces; resultant forces; dynamics for motion in a plane.

Forces add as vectors, so a resultant may be written in component form and then converted to magnitude-direction form.
Resolve each force along two perpendicular axes, add corresponding components, and apply $\sum\mathbf{F}=m\mathbf{a}$ to obtain the acceleration vector.
If the resultant is $X\mathbf{i}+Y\mathbf{j}$ , its magnitude is $\sqrt{X^2+Y^2}$ and its direction must be placed in the correct quadrant from the component signs.
A common error is to add force magnitudes without accounting for their directions, or to quote an inverse-tangent angle in the wrong quadrant.

The resultant force is the vector sum of the component forces.

Worked example

Two forces of magnitudes $12\,\text{N}$ and $10\,\text{N}$ act at an angle of $120^\circ$ to each other. Find the magnitude of their resultant and the angle the resultant makes with the $12\,\text{N}$ force.

1.Take the $12\,\text{N}$ force along the positive horizontal direction.
2.The other force has components $10\cos120^\circ=-5$ and $10\sin120^\circ=5\sqrt3$ .
3.The resultant is $7\mathbf{i}+5\sqrt3\mathbf{j}$ , with magnitude $\sqrt{49+75}=2\sqrt{31}\,\text{N}$ .
4.Its angle is $\tan^{-1}(5\sqrt3/7)=51.1^\circ$ .

Answer: Magnitude $=2\sqrt{31}\,\text{N}$ ; Angle $=51.1^\circ$ towards the $10\,\text{N}$ force

Common mistakes

Don't find a resultant magnitude using Pythagoras even though the force components are not perpendicular.
Don't add force magnitudes arithmetically despite a non-zero angle between them.

Exam tip. Resolve forces into perpendicular components or use the cosine rule, then find the resultant direction from its components.

M8.6 · Understand and use the F ≤ μR model for friction; coefficient of friction; motion of a body on a rough surface; limiting friction and statics.

Definition. Specification M8.6: understand and use the $F\leq\mu R$ model for friction; coefficient of friction; motion of a body on a rough surface; limiting friction and statics.

Friction acts along the surface, opposing the motion or the tendency to move, and satisfies $F\leq\mu R$ where $R$ is the normal reaction. Find $R$ first by resolving perpendicular to the surface; on a slope $R=mg\cos\alpha$ , and an applied force with a vertical component changes $R$ .
Friction equals $\mu R$ only when the body is moving or is in limiting equilibrium, on the point of moving; in ordinary equilibrium friction takes whatever value the other forces need, and the question is whether that value exceeds $\mu R$ .
To decide whether a body on a slope stays at rest, compare the component of weight down the slope, $mg\sin\alpha$ , with the maximum friction $\mu mg\cos\alpha$ and state the conclusion.
For a body in motion write $F=\mu R$ and $F=ma$ along the surface, with friction opposite to the velocity, which reverses when the body slides back down.
A range of values for a force comes from the two limiting cases.

On a rough horizontal surface, friction opposes impending or actual motion and satisfies

F\leq\mu R

Worked example

A block of mass

6

kg is moving on a rough horizontal surface under the action of a horizontal force of magnitude

20

N acting in the direction of motion. The coefficient of friction between the block and the surface is

0.25

. Find the acceleration of the block.

(4)

(Total for Question 1 is 4 marks)

Working

1.Resolve vertically: $R=6g=58.8$ N.
2.The block is moving, so friction is limiting: $F=\mu R=0.25\times58.8=14.7$ N, acting against the motion.
3.Horizontally, $F=ma$ : $20-14.7=6a$ , so $a=\dfrac{5.3}{6}=0.883$ m s $^{-2}$ .

Answer: $0.88$ m s $^{-2}$ (2 s.f.)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1.	$R=6g=58.8$ N	B1
	$F=\mu R$ used: $0.25\times$ their $R$	M1
	Equation of motion with the correct terms: $20-F=6a$	M1
	$a=0.88$ m s $^{-2}$ or $0.883$	A1
Notes
M1: Condone sign errors
A1: awrt 0.88

(4 marks)

Mark-scheme wording to use

“(2/7) × 4.9 oe seen … M1 (2/7) × their (a) (must be a numerical value) … 1.4, 1.40 or (1/7)g … A1” — Pearson 9MA0, June 2024 Paper 32, Q1(b)
“Equation of motion down the plane … M1 … mg sin α − F = ma or mg sin α − F = −ma … A1 … (F =) μ × their R … M1 … A1* Given answer correctly obtained, with at least one further line of working with both trig ratios substituted as fractions.” — Pearson 9MA0, June 2024 Paper 32, Q3(b)
“B1 for R = 3gcos(20) and hence maximum friction 0.9gcos(20). M1 for comparing or subtracting 3gsin(20) and 0.9gcos(20). A1* for the justified conclusion that the box starts to move down the plane” — Pearson 9MA0, Specimen Paper 32, Q3(c)

What the examiner wants. $R$ before $F$ , and $\mu R$ only when friction is limiting. The June 2024 Paper 32 report says most found the limiting friction by multiplying their reaction by $\frac27$ and then realised the applied force had the same value, though some, seeing the word “magnitude”, squared or square-rooted a correct answer and lost the accuracy mark.

Common mistakes

Don't reverse the direction of friction so that it assists the impending relative motion.
Don't set friction equal to the limiting value in every situation, even when equilibrium requires a smaller force.

Exam tip. Decide whether friction is limiting; otherwise use the force balance to find its actual value within the inequality.