Pearson Edexcel International Advanced Level Mathematics: Newton's laws, pulleys, slopes and friction (YMA01)

This guide covers topics 4.1, 4.2 and 4.4 of Unit M1 (Mechanics 1) in the Pearson Edexcel International Advanced Subsidiary/Advanced Level Mathematics specification (YMA01), Issue 3, April 2019. The specification states that M1 is an optional unit for IAS Mathematics and Further Mathematics and for IAL Mathematics and Further Mathematics. Momentum and impulse (4.3), equilibrium and moments are taught on separate pages, so they are left out here.

Use the Edexcel mathematics hub to find the rest of the course and the printable checklist to tick off what you can do. Once you have read this guide, move on to the revision notes and then the practice questions. The all free 10-minute diagnostics show you where to start.

What this unit covers

Topic	What you must be able to do
4.1	Use the concept of a force and Newton’s laws; solve problems with constant acceleration in scalar form or as a vector ai + bj
4.2 (i)	Two connected particles in a straight line or under gravity; smooth fixed pulleys and pegs
4.2 (ii)	Motion under a force that changes from one fixed value to another, for example a particle hitting the ground
4.2 (iii)	Motion directly up or down a smooth or rough inclined plane
4.4	Use F = μR when a particle is moving

Throughout, g = 9.8 m s⁻². Use the value of g the question gives. Answers are to 3 significant figures unless they are exact.

Force, mass and Newton’s laws (4.1)

A force is a push or pull, measured in newtons (N), where 1 N = 1 kg m s⁻². The three laws you need:

• Law 1: a particle stays at rest or moves at constant velocity unless a resultant force acts on it.
• Law 2: the resultant force F on a particle of mass m gives acceleration a, with F = ma. The resultant and the acceleration point the same way.
• Law 3: if A exerts a force on B, then B exerts an equal and opposite force on A.

Weight is the force of gravity: W = mg, acting vertically downwards. The forces you will meet are weight, normal reaction R (perpendicular to a surface), tension T in a string, thrust in a rod, friction F (along a surface, opposing sliding) and driving or resistive forces.

Models in this unit: a particle has mass but no size; a light string has no mass; an inextensible string does not stretch, so connected particles have the same speed and the same size of acceleration; a smooth surface or smooth pulley has no friction, so tension is the same on both sides of a smooth pulley or peg.

The constant acceleration formulae are listed in the specification among those you are expected to know:

v = u + at
s = ut + ½at²
s = vt − ½at²
v² = u² + 2as
s = ½(u + v)t

Method for every dynamics problem

1. Draw a diagram. Mark every force and the direction of motion.
2. Choose a positive direction, normally the direction of acceleration.
3. Write F = ma for each particle: (forces in the positive direction) − (forces against) = ma.
4. Solve, then use the constant acceleration formulae if the question asks about speed, distance or time.

Vector form

When a force or acceleration is written as ai + bj, add the forces to get the resultant, then divide by the mass.

Example 1. A particle of mass 2 kg is acted on by two forces, (7i − 3j) N and (−i + 11j) N. It has velocity (2i − j) m s⁻¹ at one instant. Find the magnitude and direction of its acceleration, and its speed 3 s later.

Resultant = (7 − 1)i + (−3 + 11)j = (6i + 8j) N
a = (6i + 8j)/2 = (3i + 4j) m s⁻²
|a| = √(3² + 4²) = 5 m s⁻²
angle with i = tan⁻¹(4/3) = 53.1°
v = u + at = (2i − j) + 3(3i + 4j) = (11i + 11j)
speed = √(11² + 11²) = 11√2 = 15.6 m s⁻¹

The acceleration is 5 m s⁻² at 53.1° above the direction of i, and the speed is 15.6 m s⁻¹.

Example 2. A person of mass 65 kg stands on the floor of a lift. Find the normal reaction when the lift accelerates (a) upwards at 0.8 m s⁻² and (b) downwards at 0.8 m s⁻².

(a) positive direction up: R − 65g = 65 × 0.8
    R = 65(9.8 + 0.8) = 689 N
(b) positive direction down: 65g − R = 65 × 0.8
    R = 65(9.8 − 0.8) = 585 N

Connected particles in a line (4.2 i)

Particles joined by a light inextensible string or rod move together. Write F = ma for each particle separately, or for the whole system if the internal tension is not asked for. Tension is internal to the system, so it cancels when you treat the particles as one body.

Example 3. A car of mass 1200 kg tows a caravan of mass 800 kg along a straight horizontal road. The driving force is 3300 N. The resistances are 500 N on the car and 300 N on the caravan. Find the acceleration and the tension in the tow bar.

Whole system: 3300 − 500 − 300 = 2000a
a = 2500/2000 = 1.25 m s⁻²
Caravan: T − 300 = 800 × 1.25
T = 1300 N
Check with the car: 3300 − 500 − T = 1200 × 1.25, so T = 1300 N

Always check with the other particle. If the checks differ, one equation has a sign error.

Smooth pulleys and pegs

If a string passes over a smooth fixed pulley or peg, the tension is the same on both sides. Each particle gets its own equation, and the acceleration has the same size for both. Take “positive” as the direction each particle moves in.

Example 4. Particles of mass 7 kg and 4 kg hang at the ends of a light inextensible string over a smooth fixed pulley. They are released from rest with the 7 kg particle 1.5 m above the ground. Find the acceleration and the tension before the 7 kg particle lands.

7 kg (moving down):  7g − T = 7a
4 kg (moving up):    T − 4g = 4a
Add:                 3g = 11a, so a = 3g/11 = 2.67 m s⁻²
T = 7(g − a) = 56g/11 = 49.9 N

A force that changes (4.2 ii)

When a particle hits the ground, or a string goes slack, the forces on the particles change from one fixed set to another. Split the motion into stages. The speed at the end of stage 1 is the starting speed of stage 2.

Continue Example 4. The 7 kg particle stops when it hits the ground. How high above its starting position does the 4 kg particle rise?

Stage 1: v² = 0 + 2 × (3g/11) × 1.5 = 9g/11 = 8.02
After the landing the string is slack. The 4 kg particle moves
freely under gravity: a = −g.
Stage 2: 0 = 8.02 − 2g s, so s = 9/22 = 0.409 m
Total rise = 1.5 + 0.409 = 1.91 m

The particle that lands does not bounce in this model. The other particle rises as a projectile until it stops.

Friction (4.4)

When a particle is sliding on a rough surface, friction acts along the surface against the motion and has the value

F = μR

where μ is the coefficient of friction and R the normal reaction. In this unit you use F = μR for a moving particle. Find R first by resolving perpendicular to the surface, because R is not always equal to mg.

Example 5. A particle A of mass 6 kg lies on a rough horizontal table, μ = 0.3. It is joined by a light inextensible string over a smooth pulley at the edge of the table to a particle B of mass 4 kg hanging 1.2 m above the floor. The system is released from rest. Find (a) the acceleration and the tension, (b) the distance A moves after B lands, assuming A does not reach the pulley.

(a) R = 6g = 58.8 N, F = 0.3 × 58.8 = 17.64 N
    A: T − 17.64 = 6a
    B: 4g − T = 4a
    Add: 39.2 − 17.64 = 10a, so a = 2.156 = 2.16 m s⁻²
    T = 4g − 4a = 30.6 N
(b) B lands: v² = 2 × 2.156 × 1.2 = 5.174
    A now has no tension. Only friction acts: −17.64 = 6a, so a = −2.94
    0 = 5.174 − 2 × 2.94 × s, so s = 0.88 m

Motion on an inclined plane (4.2 iii)

For a plane at angle α to the horizontal, resolve the weight mg along and perpendicular to the plane.

Component down the plane      = mg sin α
Component into the plane      = mg cos α
Normal reaction (no other perpendicular force): R = mg cos α
Friction when moving: F = μR = μmg cos α, acting up the plane if the particle slides down

Smooth plane: the only force along the slope is mg sin α, so a = g sin α down the slope. The mass cancels.

Example 6. A particle is released from rest on a smooth plane at 30° to the horizontal. Find its speed and time after it has moved 4 m.

a = g sin 30° = 4.9 m s⁻²
v² = 2 × 4.9 × 4 = 39.2, so v = 6.26 m s⁻¹
4 = ½ × 4.9 × t², so t = 1.28 s

Rough plane, sliding down: friction acts up the slope, so m a = mg sin α − μmg cos α.

Example 7. A block of mass 5 kg slides down a rough plane at 35° to the horizontal, μ = 0.4, starting from rest. Find its acceleration and its speed after 3 m.

R = 5g cos 35° = 40.14 N
F = 0.4 × 40.14 = 16.06 N
5a = 5g sin 35° − 16.06 = 28.10 − 16.06 = 12.04
a = 2.41 m s⁻²
v² = 2 × 2.41 × 3 = 14.46, so v = 3.80 m s⁻¹

Moving up a rough plane, friction and the weight component both act down the slope, so the deceleration is g sin α + μg cos α.

Connected particles with a plane

Example 8. A particle A of mass 4 kg lies on a rough plane inclined at α, where tan α = 3/4, with μ = 0.25. It is joined by a string over a smooth pulley at the top of the plane to a particle B of mass 6 kg hanging freely. The system is released from rest with B moving down. Find the acceleration and the tension.

sin α = 3/5, cos α = 4/5
R = 4g × 4/5 = 31.36 N, F = 0.25 × 31.36 = 7.84 N
A (up the slope): T − 4g × 3/5 − 7.84 = 4a
B (down):         6g − T = 6a
Add: 58.8 − 23.52 − 7.84 = 10a
a = 2.744 = 2.74 m s⁻²
T = 58.8 − 6 × 2.744 = 42.3 N

Common errors

• Writing R = mg on a slope. The normal reaction is mg cos α unless another force has a perpendicular component.
• Using different signs for the same direction in the two equations. Pick the direction of motion for each particle and keep to it.
• Using the same acceleration but a different tension, or the reverse: one string over a smooth pulley has one tension and one acceleration.
• Forgetting that after a particle lands the string is slack, so the tension disappears.
• Leaving friction in after the particle has stopped. F = μR applies when the particle is moving.
• Using sin and cos the wrong way round. The component down the slope is mg sin α, the component into the plane is mg cos α. A quick check at α = 0: the slope component should be zero.

Check your understanding

Try the practice questions after the revision notes.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics, Specification, Issue 3, April 2019, Pearson Education Limited. Unit M1: Mechanics 1, topics 4.1, 4.2 and 4.4.