Pearson Edexcel International Advanced Level Mathematics: Work, energy and power, and collisions with the coefficient of restitution (YMA01) – Revision Notes

These notes condense Pearson Edexcel International Advanced Level Mathematics (YMA01), Specification Issue 3, April 2019, unit M2 sections 3.1, 4.1, 4.2 and 4.3. The specification states that M2 is an optional unit for IAS Further Mathematics and for IAL Mathematics and Further Mathematics. For full explanations and longer worked examples, read the study guide. Test yourself with the practice set, and tick off topics on the printable checklist. The Edexcel A-level Mathematics hub lists the rest of the course, and the free 10-minute diagnostics show where to focus. Use g = 9.8 m s⁻².

Key results

Topic	Result
3.1 Kinetic energy	½mv² (J)
3.1 Gravitational PE	mgh (J), h measured from a chosen zero level
3.1 Work done	Fd cos θ for a constant force F over distance d, angle θ between force and motion
3.1 Power	Rate of doing work; P = Fv
3.1 Work-energy	Initial energy + work by driving forces − work against resistance = final energy
3.1 Conservation	KE + PE constant if only gravity (and normal reactions at right angles to motion) does work
4.1 Momentum	p = mv, a vector
4.1 Impulse	I = mv − mu = Ft for a constant force
4.1 Conservation	m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (no external impulse along the line)
4.2 Restitution	speed of separation = e × speed of approach, 0 ≤ e ≤ 1
4.2 Energy loss	Loss = KE before − KE after, always ≥ 0

The specification lists kinetic energy = ½mv² and potential energy = mgh as formulae you are expected to know.

3.1 Method boxes

Work-energy problem

1. Draw the motion, mark start and end, set a zero level for PE.
2. Write KE and PE at both ends.
3. Find the work done by the driving force and against the resistance.
4. Initial energy + driving work − resistance work = final energy.

Power problem

1. Find the driving force from F = P/v.
2. Write Newton’s second law along the line of motion: F − R − mg sin α = ma.
3. For a steady speed, a = 0.
4. For maximum speed, set a = 0 and solve for v.

Slope reminder. Moving a distance d up a slope inclined at α raises the particle by d sin α. Weight component along the slope is mg sin α. Normal reaction does no work because it acts at right angles to motion.

Worked reminder: engine power and a maximum speed

A lorry of mass 2000 kg has engine power 30 kW and resistance 1500 N on level ground. Maximum speed: F = R, so v = 30 000 / 1500 = 20 m s⁻¹. At 10 m s⁻¹ the driving force is 3000 N, so 2000a = 3000 − 1500 and a = 0.75 m s⁻².

Worked reminder: braking against a resistance

A car of mass 1200 kg is moving at 20 m s⁻¹ on level ground when the engine is switched off. A constant resistance of 400 N acts. Work-energy: ½ × 1200 × 400 = 400d, so d = 600 m.

Worked reminder: up a smooth slope

A block is projected up a smooth slope at 30° with speed 6 m s⁻¹. At its highest point KE has become PE: ½ × 36 = 9.8 × h, so h = 18/9.8. Distance along the slope = h / sin 30° = 3.67 m (3 s.f.).

4.1 Method boxes

Impulse problem. I = m(v − u). Subtract velocities component by component. Give the result as a vector, or find its magnitude with Pythagoras and its direction if the question asks.

Straight-line momentum. Choose a positive direction. Replace every velocity pointing the other way with a negative value. A particle brought to rest from 6 m s⁻¹ has speed 0, not −6.

Worked reminder. A hockey ball of mass 0.16 kg arrives at 20 m s⁻¹ and leaves at 15 m s⁻¹ the other way. With the arrival direction positive, I = 0.16(−15 − 20) = −5.6, so the impulse is 5.6 N s against the original motion.

4.2 Direct impact

Two equations, always:

(1) Momentum:     m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(2) Restitution:  v₂ − v₁ = e(u₁ − u₂)     (particle 1 behind particle 2)

Value of e	Meaning
e = 1	No loss of mechanical energy
0 < e < 1	Some mechanical energy is lost
e = 0	Particles move on together after impact

Worked reminder. Particle A (1 kg) moves at 6 m s⁻¹ and hits B (2 kg) at rest with e = 0.5. Momentum: vA + 2vB = 6. Restitution: vB − vA = 0.5 × 6 = 3. Then 3vB = 9, so vB = 3 and vA = 0. A is brought to rest.

Wall or plane. Speed after = e × speed before, direction reversed. A ball dropped from height h rebounds to height e²h.

4.3 Successive impacts

1. Solve the first impact with both equations.
2. Update every particle’s velocity. Mark which way each is moving.
3. Check whether two particles are now closing. If the particle behind is faster, they collide.
4. Repeat. After a wall rebound the particle reverses with speed multiplied by e.
5. Stop when no pair is closing: every particle is either moving away from the one behind it, or moving at least as fast as it.

Worked reminder. P (2 kg) moves at 3 m s⁻¹ towards Q (1 kg) at rest, with a wall beyond Q. Take e = 0.5 between P and Q and e = 1/3 between Q and the wall.

P and Q:   2p + q = 6 and q − p = 0.5 × 3, so p = 1.5, q = 3
Q and wall: Q returns at (1/3) × 3 = 1 m s⁻¹
Q is now coming back at 1 m s⁻¹ and P is still going forward at 1.5 m s⁻¹, so they collide again
Second impact (towards the wall positive): 2a + b = 2 and b − a = 0.5 × 2.5
So a = 0.25, b = 1.5

Both now move towards the wall, with Q ahead and faster, so for now they separate. Q then reaches the wall again and returns at 0.5 m s⁻¹, while P is still moving towards the wall at 0.25 m s⁻¹, so a third impact between P and Q follows.

Must-know distinctions

• Work done by a force is Fd cos θ. The work done by weight on a slope is mg multiplied by the vertical drop, not mg multiplied by distance along the slope.
• Momentum is conserved in a particle-particle collision, but not in a collision with a fixed wall.
• Kinetic energy is conserved only when e = 1. Momentum is conserved whatever e is.
• Impulse is a change of momentum, so its units are N s. Work is a change of energy, so its units are J.

Quick self-test

1. Find the KE of a 1.5 kg ball moving at 4 m s⁻¹.
2. A 2 kg object is raised 3 m. Find the gain in PE.
3. A 40 N force acts at 60° to the direction of motion and moves a body 5 m. Find the work done.
4. An engine of power 18 kW drives a vehicle at 24 m s⁻¹. Find the driving force.
5. A 0.3 kg ball changes velocity from 6i to (−2i + 4j) m s⁻¹. Find the impulse.
6. A 2 kg particle at 4 m s⁻¹ hits a 1 kg particle at rest and they coalesce. Find the speed afterwards.
7. A ball is dropped from 5 m onto a floor. e = 0.6. How high does it rebound?
8. Two particles of equal mass: A moves at 6 m s⁻¹, B is at rest, e = 0.5. Find both speeds afterwards.
9. In question 8 each particle has mass 2 kg. Find the loss of KE.
10. A particle hits a smooth wall at 8 m s⁻¹ and rebounds at 2 m s⁻¹. Find e.
11. A car with engine power 30 kW has a constant resistance of 1000 N. Find the maximum speed on level ground.
12. A 3 kg block slides at 5 m s⁻¹ on rough ground against a constant 6 N resistance. How far does it slide?

Answers

1. 12 J
2. 58.8 J
3. 40 × 5 × cos 60° = 100 J
4. 18 000 / 24 = 750 N
5. 0.3[(−2i + 4j) − 6i] = (−2.4i + 1.2j) N s
6. 8 / 3 = 2.67 m s⁻¹ (3 s.f.)
7. e²h = 0.36 × 5 = 1.8 m
8. A has 1.5 m s⁻¹ and B has 4.5 m s⁻¹, both in the original direction
9. Before: ½ × 2 × 36 = 36 J. After: ½ × 2 × (2.25 + 20.25) = 22.5 J. Loss = 13.5 J
10. e = 2 / 8 = 0.25
11. 30 000 / 1000 = 30 m s⁻¹
12. ½ × 3 × 25 = 37.5 J = 6d, so d = 6.25 m

Where marks are usually lost

• Putting the slope length into mgh. The height gained is d sin α.
• Leaving the weight component out of the force equation in a power problem on a hill.
• Writing the restitution equation with the wrong sign of approach or separation. Sketch before and after each time.
• Stopping after solving a single collision when the question asks whether there is a further collision.
• Using momentum conservation for a collision with a fixed wall.
• Forgetting that a negative final velocity means reversal, then giving a negative speed.
• Treating i and j components as one number. Handle each component separately.
• Rounding the g = 9.8 values early. Keep full precision until the final line.

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 M2: Mechanics 2, sections 3.1, 4.1, 4.2 and 4.3.