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Cambridge International AS & A Level Mathematics 9709 · Mechanics

Newton's laws of motion

Newton's second law for a particle under constant forces, mass and weight, vertical motion, motion on smooth and rough inclined planes, and connected particles: pulleys, strings, tow-bars and thrust.

4.4Paper 4: MechanicsFree lesson

Take $g = 10 {m s}^{- 2}$ wherever the acceleration due to gravity is needed, as in the Paper 4 exam.

What the syllabus says

4.4 (Paper 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
solve simple problems which may be modelled as the motion of a particle moving vertically or on an inclined plane with constant acceleration
solve simple problems which may be modelled as the motion of connected particles.

Notes and examples

If any other forces resisting motion are to be considered (e.g. air resistance) this will be indicated in the question.
$W = m g$ . In this component, questions are mainly numerical, and use of the approximate numerical value $10 (m s^{- 2})$ for $g$ is expected.
Including, for example, motion of a particle on a rough plane where the acceleration while moving up the plane is different from the acceleration while moving down the plane.
e.g. particles connected by a light inextensible string passing over a smooth pulley, or a car towing a trailer by means of either a light rope or a light rigid tow-bar.

Quoted from the Cambridge International AS and A Level Mathematics 9709 syllabus (2026-2027 and 2028-2030 editions, which share this content).

Newton's laws of motion Paper 4

Kinematics describes motion; Newton's laws explain it.

First law: a particle stays at rest, or moves with constant velocity, unless a resultant force acts on it. So constant velocity means the forces balance, exactly as in equilibrium.
Second law: the resultant force on a particle equals its mass times its acceleration,

$F = m a$

with $F$ in newtons, $m$ in kilograms and $a$ in ${m s}^{- 2}$ . One newton is the force that gives a mass of $1 kg$ an acceleration of $1 {m s}^{- 2}$ . The acceleration is in the direction of the resultant force.

Third law: the forces two bodies exert on each other are equal and opposite.

In Paper 4 the particle has constant mass and the forces are constant, so the acceleration is constant and the formulae from kinematics apply. The forces can include friction, the tension in an inextensible string and the thrust in a connecting rod.

Method: draw a force diagram and mark the acceleration with a double-headed arrow beside it. Resolve perpendicular to the motion, where there is no acceleration, so the forces balance (this usually gives $R$ ). Resolve along the motion and write resultant force $= m a$ , taking the direction of the acceleration as positive. Only forces mentioned in the question act: if air resistance or any other resistance is to be included, the question will say so.

Mass and weight Paper 4

Mass and weight are different quantities. Mass ( $m$ , in kilograms) is the amount of matter and does not change. Weight ( $W$ , in newtons) is the gravitational force on the mass:

$W = m g$

In Paper 4, questions are mainly numerical and you are expected to use $g = 10 {m s}^{- 2}$ , so a $65 kg$ student has weight $650 N$ . Writing " $65 N$ " for the weight, or putting $m = 650$ into $F = m a$ , are the two commonest slips in the whole paper.

Check yourself Paper 4

A car of mass $900 kg$ moves along a straight horizontal road. The driving force is $2500 N$ and the resistance to motion is $700 N$ . Find the acceleration of the car.

Answer it in the portal, where your answer is checked as you type →

Motion in a vertical line Paper 4

For a particle moving vertically, the weight acts along the line of motion. Take the direction of the acceleration as positive. A person standing in a lift is a classic case: the forces on the person are the weight $m g$ down and the normal reaction $R$ of the floor up.

Lift accelerating upwards at $a$ : $R - m g = m a$ , so $R = m (g + a)$ and the person feels heavier.
Lift accelerating downwards at $a$ (or moving up and slowing down): $m g - R = m a$ , so $R = m (g - a)$ .
Lift at constant velocity: $R = m g$ , whichever way it moves.

The same idea covers a crate raised by a rope (tension instead of reaction) and a stone falling against air resistance.

Worked example 1 Paper 4

A woman of mass $70 kg$ stands in a lift. Find the normal reaction of the floor on her when (a) the lift accelerates upwards at $0.8 {m s}^{- 2}$ , (b) the lift is moving upwards and decelerating at $1.2 {m s}^{- 2}$ .

Her weight is $70 \times 10 = 700 N$ .
(a) Up is positive, $a = 0.8$ : $R - 700 = 70 \times 0.8 = 56$ , so $R = 756 N$ .
(b) Decelerating while moving up means the acceleration is $1.2 {m s}^{- 2}$ downwards. Down positive: $700 - R = 70 \times 1.2 = 84$ , so $R = 616 N$ .
By Newton's third law she pushes on the floor with the same force; a set of scales in the lift would read these values.

Answer: (a) $756 N$ (b) $616 N$

Motion on an inclined plane Paper 4

On a plane inclined at $α$ , resolve along and perpendicular to the plane. The weight has component $m g \sin α$ down the plane and $m g \cos α$ into it, so on a plane with no other perpendicular forces $R = m g \cos α$ .

On a rough plane, sliding friction is $μ R$ , directed against the motion. This means the acceleration moving up the plane is different from the acceleration moving down it:

moving up: weight component and friction both act down the plane, deceleration $g (\sin α + μ \cos α)$ ;
moving down: weight component down, friction up, acceleration $g (\sin α - μ \cos α)$ , and only if this is positive; if $μ ⩾ \tan α$ the particle stays at rest once it stops.

Treat the journey up and the journey down as separate stages, each with its own constant acceleration.

Worked example 2 Paper 4

A particle is projected at $8 {m s}^{- 1}$ up a line of greatest slope of a rough plane inclined at $30^{°}$ . The coefficient of friction is $0.2$ . Find (a) the distance it travels up the plane, (b) its speed when it returns to its starting point.

Let the mass be $m$ . $R = m g \cos 30^{°}$ , so friction $= 0.2 m g \cos 30^{°}$ .
(a) Up the plane, both friction and $m g \sin 30^{°}$ act down the plane: $m \times deceleration = m g \sin 30^{°} + 0.2 m g \cos 30^{°}$ , so the deceleration is $5 + 1.732 = 6.732 {m s}^{- 2}$ . The mass cancels.
$0 = 8^{2} - 2 \times 6.732 \times s$ , so $s = 4.75 m$ (3 s.f.).
(b) Coming down, friction acts up the plane: acceleration $= 5 - 1.732 = 3.268 {m s}^{- 2}$ . Since this is positive, the particle does slide back.
$v^{2} = 0 + 2 \times 3.268 \times 4.753 = 31.07$ , so $v = 5.57 {m s}^{- 1}$ (3 s.f.). It returns slower than it left, because friction has done work against it both ways.

Answer: (a) $4.75 m$ (b) $5.57 {m s}^{- 1}$

Check yourself Paper 4

A particle is released from rest on a smooth plane inclined at $20^{°}$ to the horizontal. Find its acceleration, correct to 3 significant figures.

Answer it in the portal, where your answer is checked as you type →

Connected particles Paper 4

Particles joined by a light inextensible string move together: while the string is taut they have the same speed and the same size of acceleration. If the string passes over a smooth pulley, or a smooth peg, the tension is the same all along it.

Typical set-ups: two particles hanging either side of a fixed pulley; a block on a table pulled by a particle hanging over the edge; a particle on an inclined plane joined to a hanging one; and a car towing a trailer by a light rope or a light rigid tow-bar.

Method: draw a separate force diagram for each particle, write $F = m a$ for each in its own direction of motion, then solve the equations together. Adding the equations eliminates the tension and gives the acceleration.

A string can only pull, so its tension is never negative. When a hanging particle hits the floor, the string goes slack and each particle then moves on its own.
A rod or tow-bar can push as well as pull. When a car brakes, the trailer tends to run on, and the tow-bar pushes back on it: the force in the tow-bar is then a thrust.
For a car and trailer moving in one straight line you may treat the whole system as one body to find the acceleration; the tension is an internal force and cancels. You still need a separate equation for one part to find the tension.

Worked example 3 (exam style) Paper 4

A block $A$ of mass $3 kg$ lies on a rough horizontal table, where the coefficient of friction is $0.4$ . A light inextensible string attached to $A$ passes over a smooth pulley at the edge of the table and carries a particle $B$ of mass $2 kg$ , which hangs $0.5 m$ above the floor. The system is released from rest.

(a) Find the acceleration of the system and the tension in the string. [5]

(b) Find the speed of $B$ when it reaches the floor. [2]

(a) For $A$ : $R = 30 N$ , friction $= 0.4 \times 30 = 12 N$ . Along the table: $T - 12 = 3 a$ .
For $B$ , down positive: $20 - T = 2 a$ .
Add: $8 = 5 a$ , so $a = 1.6 {m s}^{- 2}$ . Then $T = 20 - 2 \times 1.6 = 16.8 N$ .
(b) $v^{2} = 0 + 2 \times 1.6 \times 0.5 = 1.6$ , so $v = 1.26 {m s}^{- 1}$ (3 s.f.).
(c) The string goes slack, so only friction acts on $A$ : deceleration $= \frac{12}{3} = 4 {m s}^{- 2}$ .
$0 = 1.6 - 2 \times 4 \times s$ , so $s = 0.2 m$ .

Answer: (a) $1.6 {m s}^{- 2}$ , $16.8 N$ (b) $1.26 {m s}^{- 1}$ (c) $0.2 m$

Worked example 4 Paper 4

A car of mass $1000 kg$ tows a trailer of mass $500 kg$ along a straight horizontal road by a light rigid tow-bar. The driver brakes, and the braking force on the car is $3000 N$ . Ignoring other resistances, find the deceleration and the force in the tow-bar.

Whole system: $3000 = 1500 \times deceleration$ , so the deceleration is $2 {m s}^{- 2}$ .
Trailer alone: the only horizontal force on it is the force from the tow-bar, which must provide a deceleration of $2 {m s}^{- 2}$ : force $= 500 \times 2 = 1000 N$ , acting backwards on the trailer.
So the tow-bar is in compression: it exerts a thrust of $1000 N$ . A rope could not do this; the trailer would run into the car.

Answer: $2 {m s}^{- 2}$ ; thrust $1000 N$

Check yourself Paper 4

Particles of masses $4 kg$ and $6 kg$ are connected by a light inextensible string passing over a smooth fixed pulley, and released from rest with both parts of the string vertical. Find the tension in the string.

Answer it in the portal, where your answer is checked as you type →

Key points

Resultant force $= m a$ , in the direction of the acceleration; perpendicular to the motion the forces balance.
$W = m g$ with $g = 10 {m s}^{- 2}$ ; mass in kg, weight in N.
In a lift: $R = m (g + a)$ accelerating up, $R = m (g - a)$ accelerating down.
On a rough plane the acceleration up and the acceleration down are different: friction always opposes the motion.
Connected by a light inextensible string over a smooth pulley: same tension, same size of acceleration.
Write $F = m a$ for each particle; adding the equations removes the tension.
A string only pulls; a rod or tow-bar can push (thrust). A slack string exerts no force.

Where marks are lost

Using the weight in newtons where the mass is needed in $F = m a$ , or the other way round.
Writing $F = m a$ with $F$ as one force rather than the resultant force.
Taking the tension equal to the hanging weight. If the system accelerates, it is not.
Using the same acceleration for motion up and down a rough plane.
Forgetting that a string goes slack once the hanging particle lands, so the other particle then moves under friction or gravity alone.
Applying a whole-system equation to particles that move in different directions, such as either side of a pulley, without care over signs.

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An original lesson written to the Cambridge International AS & A Level Mathematics 9709 syllabus (2026 to 2027 and 2028 to 2030). The worked examples and questions are our own, not Cambridge questions. Back to every AS & A Level Mathematics 9709 lesson.