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

This guide covers Pearson Edexcel International Advanced Level Mathematics (YMA01), following the Pearson Edexcel International Advanced Subsidiary/Advanced Level specification, Issue 3, April 2019. It teaches unit M2 sections 3 and 4: work and energy (3.1) and collisions (4.1 to 4.3). The specification states that M2 is an optional unit for IAS Further Mathematics and for IAL Mathematics and Further Mathematics. Use the revision notes for recall and the practice set to test yourself. The Edexcel A-level Mathematics hub, the printable checklist and the free 10-minute diagnostics are there to help you find gaps.

Throughout, g = 9.8 m s⁻². Units: newtons (N) for force, joules (J) for energy and work, watts (W) for power, N s for impulse and momentum.

What this unit covers

Spec topic	What you must be able to do
3.1	Use kinetic and potential energy, work and power; apply the work-energy principle and conservation of mechanical energy; solve problems with constant resistance and inclined planes
4.1	Treat momentum as a vector; use the impulse-momentum principle in vector form; apply conservation of linear momentum
4.2	Analyse direct impact using Newton’s law of restitution, with 0 ≤ e ≤ 1; find the loss of mechanical energy
4.3	Handle successive impacts of up to three particles, or two particles and a smooth plane (no oblique impact)

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

3.1 Kinetic energy, potential energy, work and power

Kinetic energy of a particle of mass m and speed v is ½mv². Gravitational potential energy is mgh, where h is the height above a chosen zero level. Only changes in potential energy matter, so choose the zero level that makes the working easiest.

Work done by a force is the force multiplied by the distance moved in the direction of the force. For a constant force F moved through a distance d along its own line of action, work = Fd. If the force makes an angle θ with the direction of motion, work = Fd cos θ.

Power is the rate of doing work. For a force F acting on an object moving at speed v in the direction of F, power = Fv. Engine power P gives a driving force of P/v.

The work-energy principle

Work done by driving forces, minus work done against resistances, equals the change in total mechanical energy (KE plus PE):

Work done by the driving force − work done against resistance
   = (change in KE) + (change in PE)

If there are no driving forces other than gravity and no resistance, mechanical energy is conserved: KE + PE stays constant. This is the principle of conservation of mechanical energy.

Method

1. Sketch the motion, mark the start and end, and choose a zero level for PE.
2. Write KE and PE at the start and end.
3. Find the work done by each driving force and against each resistance.
4. Substitute into: initial energy + driving work − resistance work = final energy.

Worked example 1: constant force and resistance on a slope

A crate of mass 12 kg is pulled from rest up a rough plane inclined at 20° to the horizontal, along the steepest direction. A constant force of 90 N acts parallel to the slope. The constant resistance to motion is 25 N. Find the speed of the crate after it has moved 15 m.

Work done by the 90 N force = 90 × 15 = 1350 J
Work done against resistance = 25 × 15 = 375 J
Gain in PE = 12 × 9.8 × 15 sin 20° = 603.3 J
Work-energy: 1350 − 375 = ½ × 12 × v² + 603.3
½ × 12 × v² = 371.7
v² = 61.95
v = 7.87 m s⁻¹ (3 s.f.)

Worked example 2: power on a hill

A car of mass 900 kg has engine power 24 kW. The resistance to motion is a constant 600 N. The road is inclined at α to the horizontal where sin α = 1/20. Find (a) the steady speed up the hill, (b) the acceleration up the hill when the speed is 15 m s⁻¹.

(a) At a steady speed the driving force equals the sum of resistance and the weight component down the slope.

Driving force F = 600 + 900 × 9.8 × (1/20) = 600 + 441 = 1041 N
P = Fv, so v = 24 000 / 1041 = 23.05
Steady speed = 23.1 m s⁻¹ (3 s.f.)

(b) At 15 m s⁻¹ the driving force is 24 000 / 15 = 1600 N.

F − 600 − 441 = 900a
1600 − 1041 = 900a
a = 559/900 = 0.621 m s⁻² (3 s.f.)

Worked example 3: conservation of energy, then resistance

A particle of mass 0.8 kg slides from rest down a smooth curved slide, dropping through a vertical height of 2.5 m. It then moves on rough horizontal ground against a constant resistance of 4 N. Find its speed at the bottom of the slide and the distance it travels on the ground before it stops.

Smooth slide: mechanical energy conserved
½ × 0.8 × v² = 0.8 × 9.8 × 2.5 = 19.6
v² = 49, so v = 7 m s⁻¹
On the ground: work against resistance = loss of KE
4 × d = 19.6
d = 4.9 m

4.1 Momentum, impulse and conservation of momentum

Momentum is a vector: p = mv. For velocities written as i and j components, multiply each component by the mass.

The impulse-momentum principle states that impulse equals final momentum minus initial momentum:

I = mv − mu

with u the starting velocity and v the final velocity. Impulse is a vector in the direction of the net force. For a constant force F acting for time t, I = Ft.

Worked example 4: impulse as a vector

A ball of mass 0.4 kg moving with velocity (5i − 2j) m s⁻¹ is struck so that its velocity becomes (−i + 4j) m s⁻¹. Find the impulse and its magnitude. If the contact lasted 0.05 s, find the average force.

I = 0.4[(−i + 4j) − (5i − 2j)] = 0.4(−6i + 6j) = (−2.4i + 2.4j) N s
|I| = 2.4√2 = 3.39 N s (3 s.f.)
F = I / t = (−2.4i + 2.4j) / 0.05 = (−48i + 48j) N

Conservation of linear momentum

When two particles collide and no external horizontal impulse acts, total momentum is unchanged:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

This holds as a vector equation. In straight-line problems pick one direction as positive and give every velocity a sign.

Worked example 5: coalescing particles

Particle A (mass 2 kg, velocity (3i + 2j) m s⁻¹) collides with particle B (mass 3 kg, velocity (−i + 4j) m s⁻¹). They join and move together. Find the common velocity, the speed, and the loss of kinetic energy.

Total momentum = 2(3i + 2j) + 3(−i + 4j) = (3i + 16j)
5v = 3i + 16j, so v = (0.6i + 3.2j) m s⁻¹
Speed = √(0.36 + 10.24) = √10.6 = 3.26 m s⁻¹ (3 s.f.)
KE before = ½(2)(13) + ½(3)(17) = 13 + 25.5 = 38.5 J
KE after = ½(5)(10.6) = 26.5 J
Loss = 12 J

4.2 Direct impact and Newton’s law of restitution

In a direct impact the particles move along the line joining their centres before and after. Newton’s law of restitution relates speeds of separation and approach:

(speed of separation) = e × (speed of approach)

The coefficient of restitution e satisfies 0 ≤ e ≤ 1. If e = 1 the impact is perfectly elastic and no mechanical energy is lost. If e = 0 the particles do not separate, so they move on together. Always use two equations: conservation of momentum and the law of restitution.

For velocities v₁ and v₂ along the same positive direction, with particle 1 behind particle 2, the law reads v₂ − v₁ = e(u₁ − u₂).

Worked example 6: direct impact

Particle A of mass 2 kg moves at 5 m s⁻¹ and collides directly with particle B of mass 3 kg moving at 1 m s⁻¹ in the same direction. The coefficient of restitution is 0.6. Find the speeds after impact and the loss of kinetic energy.

Momentum: 2(5) + 3(1) = 2vA + 3vB, so 2vA + 3vB = 13
Restitution: vB − vA = 0.6(5 − 1) = 2.4
Substitute vB = vA + 2.4: 5vA + 7.2 = 13, so vA = 1.16
vB = 3.56
KE before = 25 + 1.5 = 26.5 J
KE after = ½(2)(1.16)² + ½(3)(3.56)² = 1.3456 + 19.0104 = 20.356 J
Loss = 6.144 J

Both velocities are positive, so both particles continue forwards. A negative velocity would mean that particle reversed.

4.3 Successive impacts

Take each collision in turn. After every impact write the new velocities of all particles, then decide whether any two will meet. Two particles moving the same way collide if the one behind is faster.

For a collision with a smooth plane or wall, treat the wall as an immovable object: the speed after impact is e times the speed before, and the direction reverses. No momentum equation is used for the wall because the wall absorbs momentum.

Worked example 7: two particles and a wall

A smooth horizontal surface has a fixed vertical wall on the right. Particle B (mass 3 kg) is at rest between particle A (mass 1 kg) and the wall. A is projected towards B at 4 m s⁻¹. The coefficient of restitution between A and B is 0.5 and between B and the wall is 0.6. Show that A and B collide a second time, and find their speeds afterwards.

Take towards the wall as positive.

First collision
Momentum: vA + 3vB = 4
Restitution: vB − vA = 0.5 × 4 = 2
Solve: vB = 1.5, vA = −0.5
A moves away from the wall at 0.5 m s⁻¹; B moves towards it at 1.5 m s⁻¹
B hits the wall: rebound speed = 0.6 × 1.5 = 0.9 m s⁻¹ away from the wall

Both now move away from the wall. B (speed 0.9) is behind A (speed 0.5) and faster, so they collide again. Take away from the wall as positive.

Second collision
Momentum: 1(0.5) + 3(0.9) = a + 3b, so a + 3b = 3.2
Restitution: a − b = 0.5(0.9 − 0.5) = 0.2
Solve: a = 0.95, b = 0.75

A moves away at 0.95 m s⁻¹ and B at 0.75 m s⁻¹. A is ahead and faster, so there are no more collisions.

A ball and a floor

A ball dropped onto a smooth horizontal floor is the simplest plane problem. If it falls through height h, it hits at speed √(2gh) and rebounds at e√(2gh). It then rises to e²h. Dropped from 2 m with e = 0.5, it arrives at 6.26 m s⁻¹, leaves at 3.13 m s⁻¹ and rises to 0.5 m.

Common errors

• Using the law of restitution with a sign slip. Draw a diagram of before and after with arrows, and label the positive direction.
• Using only one equation in a collision. You need both momentum and restitution.
• Forgetting the PE term on a slope. The weight component along the slope is mg sin α, and the height gained is distance × sin α.
• Mixing up driving force and power. If a speed is given, find F from P = Fv before using F = ma.
• Assuming a collision with a wall conserves momentum. The wall changes the particle’s momentum, so only use the restitution law.
• Treating momentum as a scalar when the question gives i and j components. Give vectors as vectors, or magnitude and direction if asked.

Where next

Try the practice set next.

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.