Edexcel IAL Maths
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Newton's three laws for a particle, with forces given as magnitudes or in the form ai + bj
Friction on a moving particle, F = μR, including when a pulling force acts at an angle
Motion up and down smooth and rough inclined planes
Lifts, and vehicles towing trailers or caravans (tension and thrust in a tow bar)
Two particles joined by a light inextensible string over a smooth pulley or peg, and the force the string exerts on the pulley
Forces that change from one constant value to another: a string breaking, a particle hitting the ground, a pulling force removed
1. One method solves every dynamics question in M1: draw the forces on each particle, resolve along the direction of motion and write resultant force = mass × acceleration, F = ma (in vector form F = ma).
2. Weight = mg, with g = 9.8 m s⁻². Give final answers to 2 or 3 significant figures.
3. Friction on a moving particle is F = μR and acts against the motion. Find R first by resolving perpendicular to the motion.
4. On a slope at angle α: weight component mgsin α along the slope, and R = mgcos α if nothing else pushes into the slope.
5. Connected particles: a light inextensible string has the same tension throughout (over a smooth pulley too) and both particles have the same size of acceleration. Write one equation per particle.
6. When a force changes (string breaks, particle lands) start a new stage: new forces, new acceleration, and the speed at the change links the stages.
| Law | In words | How you use it in M1 |
|---|---|---|
| First | a particle stays at rest or moves with constant velocity unless a resultant force acts on it | constant velocity ⇒ forces balance in every direction |
| Second | the resultant force on a particle equals its mass × its acceleration, in the direction of the resultant | F = ma along the direction of motion; 0 = m ×0 perpendicular to it |
| Third | if A exerts a force on B, then B exerts an equal and opposite force on A | the floor of a lift and the passenger push on each other equally; a tow bar pulls the car back as hard as it pulls the trailer forward |
The forces you will meet. Weight mg, vertically down (mass in kg, weight in N). Normal reaction R, perpendicular to a surface, pushing away from it. Friction F, along a rough surface, opposing the motion. Tension T in a string, pulling away from the particle; a rod or tow bar can also push, a thrust. Resistance and driving force for vehicles, usually given as constants.
Newton's second law and weight
F = ma W = mg
F = resultant force (N); m = mass (kg); a = acceleration (m s⁻²); g = 9.8 m s⁻². Not in the formulae list: learn both. “Resultant” means you add the forces along your chosen direction, with forces the other way counted negative, before you multiply out ma.
WORKED EXAMPLE · QUESTION
A crate of mass 40 kg is pushed from rest across a horizontal floor by a horizontal force of 150 N. The total resistance to motion is constant and equal to 62 N. The crate is modelled as a particle. Find (a) the acceleration of the crate, (b) the time it takes to reach a speed of 5.5 m s⁻¹, (c) the distance it has moved by then.
SOLUTION
(a) Resultant force in the direction of motion, then F = ma: 150 - 62 = 40a ⇒a = 88 ÷40 = 2.2 m s⁻²
(b) Constant force, so constant acceleration: 5.5 = 0 + 2.2t ⇒ t = 2.5 s
(c) s = ½(u + v)t = ½ ×5.5 ×2.5 = 6.88 m (6.875 m)
Common error
Writing 150 = 40a. The 150 N is only one of the forces; F in F = ma is the resultant. Writing the subtraction out in full shows every force you have used.
Add the forces component by component, (ai + bj) + (ci + dj) = (a + c)i + (b + d)j; then F = ma gives the acceleration vector by dividing each component by the mass. Magnitude: |pi + qj| = √(p² + q²). Direction: tan θ= q/p, then check the quadrant on a sketch. “Parallel to ui + vj” means the components are in the ratio u: v. With constant a, v = u + at and r = ut + ½at² work just like the scalar forms.
WORKED EXAMPLE · QUESTION
Two forces F₁ = (4i - 2j) N and F₂ = (pi + 5j) N act on a particle of mass 2 kg. No other forces act. The acceleration of the particle is parallel to (2i + j). (a) Find the value of p. (b) Find the magnitude of the acceleration and the angle it makes with i. (c) The particle starts from rest. Find its velocity after 4 s.
SOLUTION
(a) R = (4 + p)i + 3j and (4 + p): 3 = 2: 1 ⇒4 + p = 6 ⇒ p = 2
(b) a = (6i + 3j) ÷2 = (3i + 1.5j) m s⁻². |a| = √(3² + 1.5²) = 3.35 m s⁻² (3.3541…); tan θ= 1.5 ÷3 ⇒ θ= 26.6° above i
(c) v = u + at = 4(3i + 1.5j) = (12i + 6j) m s⁻¹
Exam tip 1 — sketch every direction
tan⁻¹ of a ratio of components only gives an angle between −90° and 90°; a force such as -3i + 6j points up and to the left, at 116.6° to i, not −63.4°.
Friction on a sliding particle. F = μR, acting along the surface, opposite to the direction of motion. μ is the coefficient of friction between the two surfaces; it has no units. A smooth surface means μ= 0.
For a moving particle the friction always takes its full value μR (the inequality F ≤μR matters only for a particle at rest, on the statics page). The key point is that R is not always mg: find it by resolving perpendicular to the motion, where the acceleration is zero. A pull P at angle θ above the horizontal lifts the particle a little, so R = mg - Psin θ; a push at θ below the horizontal presses it down, so R = mg + Psin θ.
WORKED EXAMPLE · QUESTION
A sledge of mass 25 kg is pulled along rough horizontal snow by a rope inclined at 30° above the horizontal. The tension in the rope is 80 N and the coefficient of friction between the sledge and the snow is 0.15. The sledge is modelled as a particle. Find the acceleration of the sledge.
SOLUTION
Resolve vertically (no vertical acceleration): R + 80sin 30° = 25g ⇒R = 245 - 40 = 205 N.
Friction: F = μR = 0.15 ×205 = 30.75 N.
Resolve horizontally: 80cos 30° - 30.75 = 25a ⇒a = 38.53 ÷25 = 1.54 m s⁻²
Common error
Using R = 25g = 245 N when the rope is at an angle. That gives F = 36.75 N and the wrong acceleration. Resolve perpendicular to the motion every time.
Resolve along the line of greatest slope and perpendicular to it. For a plane at angle α the weight has components mgsin α down the slope and mgcos α into the slope. Perpendicular: R = mgcos α (add or subtract the perpendicular component of any other applied force). Along the slope: F = ma, with friction μR opposing the motion: down the slope while the particle moves up, up the slope while it moves down. If tan α= ¾ you can use sin α= 0.6 and cos α= 0.8 exactly.
WORKED EXAMPLE · QUESTION
A particle of mass 2 kg is projected with speed 8 m s⁻¹ up a line of greatest slope of a rough plane inclined at α to the horizontal, where tan α= ¾. The coefficient of friction is 0.25. (a) Find the distance the particle moves up the plane before coming to rest. (b) The particle then slides back down. Find its speed when it returns to its starting point.
SOLUTION
Perpendicular to the plane: R = 2gcos α= 2 ×9.8 ×0.8 = 15.68 N, so F = 0.25 ×15.68 = 3.92 N.
(a) Moving up, weight component and friction both act down the plane: -(2gsin α+ 3.92) = 2a ⇒a = -(11.76 + 3.92) ÷2 = -7.84 m s⁻². Then 0 = 8² - 2 ×7.84 ×s ⇒ s = 4.08 m (4.0816…)
(b) Moving down, friction now acts up the plane: 2gsin α- 3.92 = 2a ⇒a = (11.76 - 3.92) ÷2 = 3.92 m s⁻² down the plane. v² = 0 + 2 ×3.92 ×4.0816…= 32 ⇒ v = 5.66 m s⁻¹ (4√2)
The particle does slide back because the weight component, 11.76 N, is bigger than the greatest friction available, 3.92 N. Deciding this in general is a statics question (M1.5).
Common error
Keeping the friction in the same direction for the return journey. Friction opposes the current motion: when the velocity reverses, redraw the diagram and reverse the friction.
Lifts. Choose upwards as positive and use the direction of the acceleration, not the velocity: a lift slowing down on its way up has a downward acceleration. For a load of mass m on the floor, R - mg = ma; the floor and the load push on each other with equal force (third law). For the cable, treat lift + contents as one particle: T - (M + m)g = (M + m)a.
WORKED EXAMPLE · QUESTION
A lift of mass 600 kg carries a crate of mass 80 kg on its floor. The lift is raised by a vertical cable. (a) The lift accelerates upwards at 0.5 m s⁻². Find the tension in the cable and the force of the floor on the crate. (b) Later the lift is moving upwards and slowing down at 1.2 m s⁻². Find the force of the floor on the crate.
SOLUTION
(a) Lift and crate together, upwards positive: T - 680g = 680 ×0.5 ⇒T = 680 ×10.3 = 7004 = 7000 N (3 s.f.). The crate alone:
R - 80g = 80 ×0.5 ⇒ R = 824 N
(b) Slowing down while moving up means a = -1.2 m s⁻²: R - 80g = 80 ×(-1.2) ⇒R = 80 ×8.6 = 688 N
Check: the cable tension in this stage is 680 ×8.6 = 5848 N, less than the total weight 6664 N, as a downward acceleration needs.
Towed vehicles. The car and trailer move together, so they have the same acceleration; the tow bar is modelled as light, inextensible and parallel to the road. Take the whole system first (the tow-bar force is internal and cancels), then the trailer alone for the tension, and check with the car alone. When the car brakes, the trailer pushes on the car: the tow bar is in thrust (compression). A rope could not do this.
WORKED EXAMPLE · QUESTION
A car of mass 1100 kg tows a caravan of mass 700 kg along a straight horizontal road by a light rigid tow bar. The resistances to motion are constant: 300 N on the car and 450 N on the caravan. (a) The engine produces a driving force of 2100 N. Find the acceleration and the tension in the tow bar. (b) The driver switches off the engine and brakes, and the car and caravan decelerate at 1.5 m s⁻². Find the force in the tow bar, stating whether it is a tension or a thrust, and the braking force on the car.
SOLUTION
(a) Whole system: 2100 - 300 - 450 = 1800a ⇒ a = 0.75 m s⁻². Caravan alone: T - 450 = 700 ×0.75 ⇒ T = 975 N (check, car: 2100 - 300 - 975 = 825 = 1100 ×0.75 ✓).
(b) Caravan alone, with the tow-bar force S pushing backwards on it: S + 450 = 700 ×1.5 ⇒ S = 600 N, a thrust. Car alone, where the thrust pushes the car forwards: B + 300 - 600 = 1100 ×1.5 ⇒ B = 1950 N
Exam tip 2 — a negative tension is a thrust
If you assume a tension and your answer comes out negative, the tow bar is in thrust. Say so in words: “the force in the tow bar is a thrust of 600 N”.
Light string: same tension along its whole length. Inextensible: both particles move with the same speed and the same size of acceleration. Smooth pulley or peg: the tension is the same on both sides.
Draw a separate force diagram for each particle and write F = ma for each in its own direction of motion. Adding the two equations removes T and gives a; substitute back for T.
You may not treat the two particles as one body unless they move along the same straight line: the string turns a corner at the pulley.
Force on the pulley: the two parts of the string each pull on it with tension T. Their resultant bisects the angle between the strings, with magnitude 2Tcos(θ/2), where θ is the angle between the two parts of the string.
WORKED EXAMPLE · QUESTION
A particle A of mass 3 kg lies on a rough horizontal table. The coefficient of friction between A and the table is 0.2. A is attached to a light inextensible string that passes over a small smooth pulley fixed at the edge of the table; the other end carries a particle B of mass 2 kg, which hangs freely. The string between A and the pulley is horizontal. The system is released from rest. Find (a) the acceleration of the particles, (b) the tension in the string, (c) the magnitude and direction of the force exerted by the string on the pulley.
SOLUTION
Friction on A (resolving vertically for A, R = 3g): F = 0.2 ×3g = 5.88 N.
One equation per particle: B (down): 2g - T = 2a; A (towards the pulley): T - 5.88 = 3a.
(a) Add: 2g - 5.88 = 5a ⇒a = 0.28g = 2.74 m s⁻² (2.744)
(b) T = 2(g - a) = 1.44g = 14.1 N (14.112)
(c) Two tensions at 90° to each other: √(T² + T²) = T√2 = 20.0 N, at 45° below the horizontal, towards the table side
Common error
Taking the tension as the weight of the hanging particle (here 19.6 N). That is true only if B is not accelerating. While B accelerates downwards, the tension must be less than its weight.
In M1 a force does not vary smoothly; it switches from one constant value to another. Each stage has constant acceleration, so the suvat formulae apply within each stage, and the link between stages is the velocity at the instant of change.
Hanging particle hits the ground: it stops; the string goes slack, so T = 0 for the other particle, which carries on with its speed at that instant, acted on only by its weight, friction and the normal reaction.
String breaks: T = 0 from that instant for both particles. A pulling force removed: recompute R and the friction too.
A sketch of the velocity–time graph keeps the stages straight: straight segments, with a change of gradient at each change of force.
WORKED EXAMPLE · QUESTION
A particle A of mass 2 kg rests on a rough plane inclined at 30° to the horizontal. The coefficient of friction between A and the plane is 0.25. A is attached to a light inextensible string that runs parallel to a line of greatest slope, over a small smooth pulley fixed at the top of the plane, to a particle B of mass 3 kg. B hangs freely 1.2 m above the ground. The system is released from rest with the string taut, and A does not reach the pulley. (a) Find the acceleration of the system and the tension in the string. (b) Find the magnitude of the force exerted by the string on the pulley. (c) Find the speed of B as it hits the ground. (d) B does not rebound. Find the further distance A moves up the plane before it first comes to rest.
SOLUTION
Normal reaction and friction on A: R = 2gcos 30° = 16.97 N, F = 0.25R = 4.244 N.
(a) B: 3g - T = 3a; A: T - 2gsin 30° - 4.2435 = 2a. Adding: 29.4 - 9.8 - 4.2435 = 5a ⇒ a = 3.07 m s⁻² (3.0713); T = 3(g - a) = 20.2 N
(b) The two parts of the string make 60° with each other (one along the plane at 30° above the horizontal, one vertical): 2Tcos 30° = T√3 = 35.0 N (34.96)
(c) B falls 1.2 m from rest with a = 3.0713…: v² = 2 ×3.0713 ×1.2 = 7.371 ⇒ v = 2.71 m s⁻¹
(d) The string is slack, so for A only its weight component and friction act along the plane, both down it: -(gsin 30° + 0.25gcos 30°) = a ⇒a = -7.022 m s⁻². Then 0 = 7.371 - 2 ×7.022 ×s ⇒ s = 0.525 m
Use the unrounded v² = 7.3711 here; rounding v to 2.71 first gives 0.523 m, which shows how early rounding drifts. The mass of A cancels in (d): once the string is slack, A's motion does not depend on B at all.
Exam tip 3 — label each stage
Write “string taut”, “string slack” before you start, and list u, v, a, s, t for each stage. The reasoning at the change of stage, such as stating that the tension becomes zero, is part of the answer, so write it down.
Exam tip 4 — one diagram per particle
Draw a separate force diagram for each particle, with every force labelled and the direction of the acceleration marked. A missing force, or one pointing the wrong way, spoils a whole equation and everything that follows from it.
Exam tip 5 — resolve in two directions, every time
Perpendicular to the motion gives R (acceleration zero); along the motion gives F = ma. Write each equation as “resolving parallel to the plane: …” so that each one stands on its own. Keep at least four significant figures in intermediate values and give final answers to 2 or 3, since g is only given to 2.
Exam tip 6 — “show that” answers
When a value is given (“show that the acceleration is 0.28g”), every step must appear, with the forces substituted before simplifying. Then use the given value in the following parts, even if your own answer differed.
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| The normal reaction always equals mg. | Slopes and angled pulls change the perpendicular balance. | Resolve perpendicular to the motion: R = mgcos α on a slope, R = mg - Psin θ for an upward pull. |
| Friction acts in a fixed direction on a given slope. | Friction opposes the relative motion. | Down the plane while moving up; up the plane while moving down. |
| The tension equals the weight of the hanging particle. | That would make the resultant force on it zero. | While it accelerates down, T <mg; find T from mg - T = ma. |
| A lift going up always has R >mg. | The sign depends on the acceleration, not the velocity. | Slowing on the way up: acceleration downwards, so R <mg. |
| After a string breaks, the particle stops at once. | No force can change the velocity instantly. | It keeps its speed at that instant and starts a new stage with new forces. |
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| The force on a pulley is 2T whatever the set-up. | The two tensions add as vectors. | 2T only when both strings are parallel; otherwise 2Tcos(θ/2). |
M1.2 Kinematics and M1.1 Vectors
F = ma supplies the constant acceleration that the suvat formulae need, and the i, j methods carry straight over to forces.
M1.4 Momentum and M1.5 Statics
When a slack string becomes taut again, the sudden jerk is a momentum problem. A particle at rest on a rough surface, where friction can take any value up to μR, belongs to statics.
M2 and IAL Physics Unit 1
The same pulley and slope systems are solved with work and energy in M2. Free-body diagrams, lifts, towing and Newton's third law appear in Physics too, with g = 9.81 N kg⁻¹ instead of 9.8 m s⁻².
| Formula | Use | In the formulae list? |
|---|---|---|
| F = ma · F = ma | Resultant force and acceleration, scalar or vector | No — learn it |
| W = mg | Weight, g = 9.8 m s⁻² | No — learn it |
| F = μR | Friction on a moving particle | No — learn it |
| R = mgcos α; mgsin α along the slope | Particle on a plane at angle α | No — derive it |
| 2Tcos(θ/2) | Force of a string on a smooth pulley, θ = angle between the two parts of the string | No — derive it |
| v = u + at, s = ut + ½at², v² = u² + 2as, … | Each stage of constant acceleration | No — learn them |
1. The only force on a particle of mass 3 kg is (6i - 9j) N. Find the magnitude of its acceleration.
2. A box slides across a rough horizontal floor with no other horizontal force acting. The coefficient of friction is 0.4. Find its deceleration.
3. A particle slides down a smooth plane inclined at 20° to the horizontal. Find its acceleration.
4. Particles of mass 4 kg and 6 kg hang at the ends of a light inextensible string over a smooth fixed pulley and are released from rest. Find the acceleration and the tension.
5. Why can the tension be taken as the same on both sides of a smooth pulley?
1. a = (2i - 3j) m s⁻², so |a| = √13 = 3.61 m s⁻².
2. R = mg, so friction = 0.4mg and the deceleration is 0.4g = 3.92 m s⁻². The mass cancels.
3. a = gsin 20° = 3.35 m s⁻² down the plane.
4. 6g - T = 6a and T - 4g = 4a, so a = 2g/10 = 1.96 m s⁻² and T = 4(g + a) = 47.0 N.
5. The string is light (no mass, so no resultant force is needed to accelerate it) and the pulley is smooth (no friction between string and pulley), so nothing changes the tension as the string passes over.
Check understanding
The answer and each step appear one click at a time.
Two boxes A and B, of masses 4 kg and 6 kg, are at rest in contact on a rough horizontal floor. The coefficient of friction between each box and the floor is 0.1. A horizontal force of magnitude P newtons is applied to A, pushing it against B as shown in Figure 1, which also shows how P varies with the time t seconds. The boxes are modelled as particles moving along the same straight line. Take g = 9.8 m s⁻².
(a) Find the acceleration of the boxes for 0 < t < 4. [2]
(b) Find the magnitude of the force exerted by A on B for 0 < t < 4. [2]
(c) Find the speed of the boxes at t = 10. [3]
(d) For t > 10, explain why the force between the boxes is zero, and find the time taken for the boxes to come to rest after t = 10. [2]
Mark scheme click to reveal each step
both boxes: 30 − 0.1 × 10g = 10a
a = 2.02 m s⁻²
B: C − 0.1 × 6g = 6 × 2.02
C = 18.0 N
at t = 4: v = 2.02 × 4 = 8.08 m s⁻¹
for 4 < t < 10: 12 − 9.8 = 10a, a = 0.22 m s⁻²
v = 8.08 + 0.22 × 6 = 9.40 m s⁻¹
each box on its own would decelerate at μg = 0.98 m s⁻², the same for both, so neither needs a push from the other
A small box slides in a straight line across a rough horizontal warehouse floor. It passes a point A with speed 6 m s⁻¹ and comes to rest 9 m beyond A. No force other than friction acts on the box horizontally. The box is modelled as a particle. Take g = 9.8 m s⁻².
(a) Find the deceleration of the box. [2]
(b) Find the coefficient of friction between the box and the floor. [3]
(c) A second box of twice the mass slides on the same floor with the same coefficient of friction. It passes A with speed 6 m s⁻¹. State how far beyond A it comes to rest, giving a reason. [1]
Mark scheme click to reveal each step
0 = 6² − 2 × a × 9
a = 2 m s⁻²
resolving vertically: R = mg
μmg = m × 2
μ = 2 ÷ 9.8 = 0.204
9 m, because the mass cancels: the deceleration μg does not depend on the mass
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ExaminerPrep is an independent tutoring service. It has been developed independently from and is not endorsed by the International Baccalaureate Organization. "International Baccalaureate", "IB" and "IB Diploma Programme" are registered trademarks of the IBO, used here for descriptive purposes only.
Pearson Edexcel and Cambridge are trademarks of their owners; ExaminerPrep is not affiliated with any exam board.