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

These notes condense topics 4.1, 4.2 and 4.4 of Unit M1 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. For full explanations and longer worked examples, read the study guide first.

Related pages: the practice questions, the Edexcel mathematics hub, the printable checklist and the all free 10-minute diagnostics. Take g = 9.8 m s⁻² unless a question says otherwise.

Key results

Result	Meaning
F = ma	Resultant force = mass × acceleration, same direction
W = mg	Weight acts vertically downwards
F = μR	Friction on a moving particle, opposing sliding
a = F/m with F = ai + bj	Add vector forces, then divide by the mass
mg sin α	Weight component down a slope at angle α
mg cos α	Weight component into the slope, so R = mg cos α when nothing else pushes perpendicular to the plane

Constant acceleration formulae, which the specification lists as formulae you are expected to know:

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

Models and assumptions (4.2)

Term	What you may assume
Particle	Mass but no size, so no turning effects
Light string	No mass, so tension is the same along it
Inextensible string	Does not stretch, so both particles have the same speed and acceleration size
Smooth pulley or peg	No friction, so tension is the same on both sides
Smooth surface	Friction is zero
Rough surface	Friction F = μR while sliding

Method boxes

Vector dynamics (4.1)

1. Add the forces component by component.
2. Divide by the mass to get a = (ai + bj).
3. Use v = u + at and s = ut + ½at² with vectors, or find a magnitude with √(a² + b²).
4. Give a direction as an angle to i or j, with a diagram if it helps.

Connected particles (4.2 i)

1. Draw each particle with its own forces. Mark the same tension on both sides of a smooth pulley.
2. Choose the direction of motion as positive for each particle.
3. Write one F = ma equation per particle.
4. Add the equations to remove T and find a, then substitute back for T.
5. Check T with the second particle.

For two masses M and m with M the heavier:

Set-up	Acceleration	Tension
Both hang over a smooth pulley	(M − m)g/(M + m)	2Mmg/(M + m)
Smooth table, M on the table, m hanging	mg/(M + m)	Mmg/(M + m)

Use these only to check a worked answer. Show the equations in the exam.

Force changes (4.2 ii)

1. Stage 1: find a, then the speed when the particle lands or the string goes slack.
2. Stage 2: remove the tension. Recalculate the acceleration from the forces that are left.
3. Use the speed at the end of stage 1 as u for stage 2.

Inclined planes (4.2 iii) and friction (4.4)

1. Draw weight, R, friction and any other forces.
2. Resolve perpendicular to the plane: R = mg cos α (add or subtract other perpendicular components).
3. Find F = μR.
4. Resolve along the plane: sliding down, friction acts up; moving up, friction acts down.

Motion	Equation along the slope
Down a smooth plane	ma = mg sin α
Down a rough plane	ma = mg sin α − μmg cos α
Up a smooth plane (deceleration)	ma = −mg sin α
Up a rough plane (deceleration)	ma = −mg sin α − μmg cos α

Worked reminders

Reminder 1. A 2 kg block is sliding at 5 m s⁻¹ on a rough horizontal surface, μ = 0.2, with no other horizontal force. Find the distance and the time to stop.

R = 2g, F = 0.2 × 2g, so 2a = −0.2 × 2g
a = −1.96 m s⁻²
0 = 25 − 2 × 1.96 × s, so s = 6.38 m
0 = 5 − 1.96t, so t = 2.55 s

Reminder 2. Masses 5 kg and 3 kg hang over a smooth pulley. Equations: 5g − T = 5a and T − 3g = 3a. Adding gives 2g = 8a, so a = 2.45 m s⁻², and T = 3(g + a) = 36.75 N.

Now suppose the 5 kg particle starts 0.8 m above the ground and lands. Its speed is given by v² = 2 × 2.45 × 0.8 = 3.92. The string then goes slack, and the 3 kg particle decelerates at g, so it rises a further 3.92/(2 × 9.8) = 0.2 m before stopping.

Reminder 3. A particle is released from rest on a rough plane at 25° to the horizontal, μ = 0.15, and slides 2 m down the line of greatest slope.

R = mg cos 25°, so F = 0.15mg cos 25°
a = g(sin 25° − 0.15 cos 25°) = 2.81 m s⁻²
v² = 2 × 2.809 × 2, so v = 3.35 m s⁻¹
2 = ½ × 2.809 × t², so t = 1.19 s

Mass cancels on a slope when only gravity, R and friction act. If a question gives the mass, you still do not need it for a.

Must-know distinctions

• Mass is in kg and does not change. Weight is in N and equals mg.
• Normal reaction is perpendicular to the surface. Friction is along it.
• On a horizontal surface with no vertical forces other than weight, R = mg. On a slope, R = mg cos α.
• Tension pulls on both particles it joins. Thrust in a rod pushes. If an answer for T comes out negative in a rod, it is a thrust.
• Speed is a magnitude. Velocity and acceleration in ai + bj form have direction. Find the speed as √(a² + b²).
• A smooth surface has no friction. A rough surface has F = μR only while the particle is moving. Once it stops, there is no sliding friction to use in this unit.

Quick self-test

1. A particle of mass 5 kg is acted on by a single force (10i − 15j) N. Find its acceleration as a vector and its magnitude.
2. Find the weight of a 12 kg object.
3. Masses 8 kg and 2 kg hang from the ends of a string over a smooth pulley. Find the acceleration and the tension.
4. Find the acceleration of a particle released on a smooth plane at 40° to the horizontal.
5. A 10 kg box is pushed along rough horizontal ground by a horizontal force of 30 N. The coefficient of friction is 0.15. Find the acceleration.
6. A particle slides down a rough plane at 20° to the horizontal, with μ = 0.3. Find its acceleration.
7. The particle in question 6 is now projected up the same plane. Find its deceleration.
8. A particle slides on rough horizontal ground, μ = 0.25, at 7 m s⁻¹ with no other horizontal force. Find the distance to rest.
9. A particle of mass 3 kg lies on a smooth horizontal table. It is joined by a string over a smooth pulley to a 1 kg particle hanging freely. Find the acceleration and the tension.
10. A particle of mass 3 kg has acceleration (2i + j) m s⁻². One force is (4i − j) N and the only other force is F. Find F.

Answers

1. a = (10i − 15j)/5 = (2i − 3j) m s⁻²; magnitude √13 = 3.61 m s⁻².
2. 12 × 9.8 = 117.6 N.
3. 8g − T = 8a and T − 2g = 2a, so 6g = 10a, a = 5.88 m s⁻². T = 2(9.8 + 5.88) = 31.4 N.
4. a = g sin 40° = 6.30 m s⁻².
5. R = 98 N, F = 14.7 N, 10a = 30 − 14.7, so a = 1.53 m s⁻².
6. a = g(sin 20° − 0.3 cos 20°) = 0.589 m s⁻².
7. Deceleration = g(sin 20° + 0.3 cos 20°) = 6.11 m s⁻².
8. a = −0.25g = −2.45, so 0 = 49 − 2 × 2.45 × s, s = 10 m.
9. 1g − T = 1a and T = 3a, so a = g/4 = 2.45 m s⁻² and T = 7.35 N.
10. Total force = 3(2i + j) = (6i + 3j) N, so F = (6i + 3j) − (4i − j) = (2i + 4j) N.

Where marks are usually lost

• Writing R = mg on an inclined plane.
• Putting friction in the wrong direction. It opposes the motion, not the weight component.
• Using one equation for two particles that have different forces, instead of one equation per particle.
• Writing 2T on the left of a pulley equation. A smooth pulley gives the same tension on each side, not double.
• Keeping the tension in stage 2 after a particle has landed.
• Mixing signs: a particle moving up should have T − mg = ma, and a particle moving down should have mg − T = ma.
• Rounding too early. Carry full values through the working and round the final answer, using the value of g the question gives.
• Stating an answer as a magnitude when the question asks for a vector, or the reverse.
• Forgetting units, or leaving an acceleration negative without saying what the sign means.

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.