getting ready to teach online course further mechanics 1 and 2 level/mathem… · al maths...

Getting Ready to Teach – Online Course Further Mechanics 1 and 2

The A level reforms

• All new AS and A levels will be assessed at the same standard as they are currently

• All new AS and A levels will be fully linear

• AS levels will be stand-alone qualifications

• The content of the AS level can be a sub-set of the A level content to allow co-teachability, but marks achieved in the AS will not count towards the A level

A few questions for you!

The A level reforms

A level Further Mathematics

• 50% core (all pure mathematics)

• 50% optional and can include

– pure mathematics

– mechanics

– statistics

– decision mathematics

– any other

The A level reforms - content

AS level Further Mathematics

•20% core (all pure mathematics)

•10% compulsory (selected from the A level core)

•70% optional (same options as A level)

•We have made an additional 20% of the content compulsory (taken from A level core)

The A level reforms

Requirement for the assessment of problem solving, communication, proof, modelling, application of techniques

Requirement that candidates have a calculator with

• the ability to compute summary statistics and access probabilities from standard statistical distributions

• an iterative function

• the ability to perform calculations with matrices up to at least order 3 × 3 (further mathematics only)

Overview of the specification

A Level Further mathematics

Paper 1: Core Pure Mathematics 125%1 hour 30 minutes75 marks Compulsory content – any content

on either paperPaper 2: Core Pure Mathematics 225%1 hour 30 minutes75 marks

Paper 3: Further Mathematics Option 125%1 hour 30 minutes75 marks

Students take two optional papers with options available in

• Further Pure Mathematics

• Further Statistics

• Further Mechanics

• Decision Mathematics

Paper 4: Further Mathematics Option 225%1 hour 30 minutes75 marks

A level Further Mathematics options

For papers 3 and 4 students choose a pair of options, either•any two from column A, or•a matching pair from columns A and B

Overview of the specification

AS Further Mathematics

Interactive Scheme of Work

Schemes of Work

Exam Papers

AS Level

A Level

Existing Resources AS Further Mechanics 1

Unit TitleOld

Module

Chapter

Reference

1 Momentum and impulse M1 3

a Momentum and impulse; derivation of units and formulae

Impulse-momentum principle. Conservation of momentum

applied to collisions and ‘jerking’ string problems

M1 3

2 Work, energy and power M2 3

a Work, kinetic energy; derivation of units and formulae M2 3

b Potential energy, work-energy principle, conservation of

mechanical energy, problem solvingM2 3

c Power; derivation of units and formula M2

3 Elastic collisions in one dimension M2 3

a Direct impact of elastic spheres. Newton’s law of restitution.

Loss of kinetic energy due to impact M2 4

b Problem solving (including ‘successive’ impacts) M2 4

Existing Resources AL Further Mechanics 1

UnitTitle Old Module

Chapter

Reference

4 Momentum and impulse (part 2) M2 4

a Momentum as a vector (i, j problems)

Impulse-momentum principle in vector formM2 4

5 Elastic strings and springs and elastic energy M3 2

a Hooke’s law and definition of modulus of elasticity. Derivation of elastic potential

energy formula.M3 2

bProblem solving: equilibrium and using the work-energy principle M3 2

6 Elastic collisions in two dimensions M4 2

aOblique impact of a smooth sphere with a fixed surface Successive oblique impacts

of a sphere with smooth plane surfaces M4 2

b Oblique impact of two smooth spheres of equal radius M4 2

Existing Resources AS Further Mechanics 2

UnitTitle Old Module

Chapter

Reference

1 Motion in a circle M3 4

aAngular speed, central force, radial acceleration M3 4

b Uniform motion in a horizontal circle M3 4

2 Centres of mass of plane figures M2 2

aMoment of a force

Centre of mass of a discrete mass distribution in one and two dimensions,

framework and uniform lamina (rectilinear shapes)

M2 2

bCentre of mass of triangular, circular-based and composite laminas and

centre of mass of a uniform circular arcM2 2

c Modelling equilibrium: hanging bodies and systems free to rotate (about a

fixed horizontal axis)M2 2

3 Further kinematics M3 1

aMotion in a straight line when the acceleration is a function of time (t);

Setting up and solving differential equationsM3 1

bMotion in a straight line when the acceleration is a function of the velocity

(v); Setting up and solving differential equationsNew

Existing Resources AL Further Mechanics 2

UnitTitle Old Module

Chapter

Reference

4 Motion in a circle (part 2)

aMotion in a vertical circle: radial and tangential acceleration; Conservation

of energy in this contextM3 4

5 Further centres of mass M3 5

aCentre of mass of uniform/non-uniform rod, lamina, 3-D rigid body using

integration (and symmetry); Deriving formulae in formula bookM3 5

bCentre of mass of composite bodies; Simple cases of equilibrium of rigid

bodies.M3 5

c Conditions for toppling/sliding M3 5

6 Further kinematics (part 2)

aMotion in a straight line when the acceleration is a function of time (t);

Setting up and solving differential equationsM3 1

bMotion in a straight line when the acceleration is a function of the velocity

(v); Setting up and solving differential equationsNew

7 Further dynamics

aParticle moving in straight line with variable applied force; Using F = ma

to set up differential equations and solvingM3 1

b Newton’s law of gravitation M3 3

c Simple harmonic motion (SHM) M3 3

Key PrerequisitesAS Further Mechanics 1

Unit Title Prerequisite Knowledge

1 Momentum and impulse SUVAT Formulae (introduction)

a Momentum and impulse; derivation of units and formulae

Impulse-momentum principle. Conservation of momentum applied

to collisions and ‘jerking’ string problems

2Work, energy and power

Resolving Forces

Frictional Forces

a Work, kinetic energy; derivation of units and formulae

b Potential energy, work-energy principle, conservation of

mechanical energy, problem solving

c Power; derivation of units and formula

3Elastic collisions in one dimension

SUVAT Formulae

Geometric Progression Summation

a Direct impact of elastic spheres. Newton’s law of restitution. Loss of

kinetic energy due to impact

b Problem solving (including ‘successive’ impacts)

Key Prerequisites AL Further Mechanics 1

UnitTitle Prerequisite Knowledge

4 Momentum and impulse (part 2) SUVAT in 2D

a Momentum as a vector (i, j problems)

Impulse-momentum principle in vector form

5Elastic strings and springs and elastic energy

Resolving Forces, Basic

Integration, Friction

a Hooke’s law and definition of modulus of elasticity. Derivation of elastic potential

energy formula.

b Problem solving: equilibrium and using the work-energy principle

6Elastic collisions in two dimensions

AS Maths Trig Identities, AS

FM Scalar product

a Oblique impact of a smooth sphere with a fixed surface Successive oblique impacts

of a sphere with smooth plane surfaces

b Oblique impact of two smooth spheres of equal radius

Key Prerequisites AS Further Mechanics 2

Unit Title Prerequisite Knowledge

1Motion in a circle

A Level Maths Radians

A Level Maths Resolving Components

aAngular speed, central force, radial acceleration

b Uniform motion in a horizontal circle

2 Centres of mass of plane figures A Level Maths Moments

aMoment of a force

Centre of mass of a discrete mass distribution in one and two

dimensions, framework and uniform lamina (rectilinear shapes)

bCentre of mass of triangular, circular-based and composite laminas and

centre of mass of a uniform circular arc

c Modelling equilibrium: hanging bodies and systems free to rotate

(about a fixed horizontal axis)

3Further kinematics

A Level Maths Integration separation of

variables

aMotion in a straight line when the acceleration is a function of time

(t); Setting up and solving differential equations

bMotion in a straight line when the acceleration is a function of the

velocity (v); Setting up and solving differential equations

Key Prerequisites AL Further Mechanics 2

Unit Title

4 Motion in a circle (part 2) Radians, Projectile motion, resolving forces

aMotion in a vertical circle: radial and tangential

acceleration; Conservation of energy in this context

5 Further centres of mass Integration

a

Centre of mass of uniform/non-uniform rod, lamina, 3-D

rigid body using integration (and symmetry); Deriving

formulae in formula book

bCentre of mass of composite bodies; Simple cases of

equilibrium of rigid bodies.

c Conditions for toppling/sliding

6Further kinematics (part 2)

Integration

A Level Maths and Further Maths

a Motion in a straight line when the acceleration is a

function of time (t); Setting up and solving differential

equations

b Motion in a straight line when the acceleration is a

function of the velocity (v); Setting up and solving

differential equations

7

Further dynamics

AL Maths Trigonometry, Differentiation

AL FM1 Hooke’s Law

AL FM Integration, Differentiation, Hyperbolics

a

Particle moving in straight line with variable applied

force; Using F = ma to set up differential equations

and solving

b Newton’s law of gravitation

c Simple harmonic motion (SHM)

Getting Ready to Teach – Online Course Elastic Collisions–Further Mechanics 1 (AS)

Objectives

•

Objectives

• be able to solve problems of the following types involving elastic impacts:

– successive collisions between pairs of spheres (horizontal motion);

– bouncing ball (off a horizontal elastic plane);

– successive collisions including two spheres and sphere against a wall;

– determination of number of collisions or deriving the possible range of values of e.

Newton’s (Experimental) Law of Restitution

•

Key Formulae

•

Example

•

151

AS FM 1 Qu 1

Getting Ready to Teach – Online Course Elastic Collisions in 2D–Further Mechanics 1 (A Level)

Elastic collisions in two dimensions: Prior Knowledge

Covered so far

•Momentum and impulse (Unit 1)

•Direct collisions, Newton’s (experimental) law of restitution (Unit 3)

•Change in kinetic energy due to impact (Unit 3)

•Momentum and Impulse-momentum principle in vector form (Unit 4)

•Change in kinetic energy due to impact (Unit 5)

•AS Mathematics – Pure content

– Work with i, j vectors, magnitude & direction (See SoW Year 1 Unit 5)

– Trigonometric identities (See SoW Year 1 Unit 7)

•A level Mathematics – Pure content

– Trigonometric identities (See SoW Year 2 Unit 8)

•AS Further Mathematics – Core Pure content

– Scalar product of vectors in 2-D (See SoW Unit 7b)*

•* Note that, although an understanding of scalar product is useful in problem solving, only vectors in two dimensions are being considered and so other techniques are possible

Elastic collisions in two dimensions

• 5.1 Oblique impact of smooth elastic spheres and a smooth sphere with a fixed surface. Loss of kinetic energy due to impact.

• 5.2 Successive oblique impacts of a sphere with smooth plane surfaces.

Elastic collisions in two dimensions: Objectives

By the end of the sub-unit, students should:

•understand that during an impact the impulse acts perpendicularly to the surface through the centre of the sphere;

•be able to apply Newton’s (experimental) law of restitution in the direction of the impulse;

•appreciate that perpendicular to the impulse, the velocity component does not change;

•understand and be able to calculate an angle of deflection;

•be able to calculate the kinetic energy ‘lost’ in an impact;

•be able to work in speeds and angles or in velocity vectors (i, j).

Impulse in 2 DimensionsA Level

Teaching Points

•

Example

•

Example

•

Situations

• Oblique impact of a smooth sphere with a fixed surface

• Successive oblique impacts of a sphere with smooth plane surfaces

• Oblique impact of two smooth spheres of equal radius (Spin is ignored)

Two Spheres

Two Spheres

Angle of Impact?

Two Spheres

Angle of Impact?

SAM Questions

Getting Ready to Teach – Online Course Elastic Springs and EnergyFurther Mechanics 1 (A Level)

Elastic Springs and Elastic Energy

• Covered so far– KE, GPE and the work-energy principle (Unit 2)

• AS Mathematics – Pure content

– 8.3 Basic integration (See SoW Year 1 Unit 4)

• AS Mathematics – Mechanics content – 8.1, 8.2, 8.4 Types of forces (in particular tension) and Newton’s

laws

– (See SoW Year 1 Unit 9)

• A level Mathematics – Mechanics content – 8.2, 8.4, 8.5, 8.6 Resolving forces, friction, equilibrium and dynamics

– (See SoW Year 2 Units 4, 7 and 8)

Elastic Springs and Elastic Energy: Objectives

•

Hooke’s Law

•

Example

•

Further Problems

•

Further Problems

•

Work Energy PrincipleObjectives

By the end of the sub-unit, students should:

•be able to calculate the tension in a string or spring when a system is held in equilibrium;

•be able to include EPE when using the work-energy principle;

•know the conditions for conservation of mechanical energy;

•be able to solve string/spring problems involving work and energy (i.e. KE, GPE and EPE).

Energy

•

Example

•

Example

•

Getting Ready to Teach Online course

Hookes Law A Level

Getting Ready to Teach – Online Course Horizontal Circular Motion– Further Mechanics 2 (AS)

Horizontal Circular MotionPRIOR KNOWLEDGE

AS level Mathematics – Pure content

Calculus: Rates of Change, Integration (unit 6 & 7)

Vectors (Unit 5)

A level Mathematics – Pure content

Trigonometry: Radians (Unit 6)

Differentiation: sin & cos, function of a function (Unit 8)

AS Mathematics – Mechanics content

Kinematics 2 : Variable acceleration (Unit 4)

A level Mathematics – Mechanics content

Resolving using Newton’s Second Law: F = ma (Units 4 & 8b)

Friction (Unit 7)

Modelling assumptions made throughout this course

(e.g. particle, light, rod, inextensible, smooth)

Horizontal Circular Motion

Horizontal Circular Motion

OBJECTIVES 2

• understand that a particle moving in a horizontal circle with constant speed has an acceleration of magnitude rω2 directed towards the centre of the circle

• know that the resultant force acting on the particle has magnitude mrω2 and is directed towards the centre of the circle

• be able to model a variety of physical situations which give rise to uniform circular motion, including use of the period, T

• understand that a banked surface is a plane that is at an angle to the horizontal

• be able to consider how conditions affect the described circular motion, for example the greatest and least possible angular speeds

Horizontal Circular MotionAS SAMs Question

Horizontal Circular MotionDERIVATION

If the particle P moves in the circle shown then the angular speed

is defined as rate of change of angle θ w.r.t. time t or dθ/dt .

We can formalise this relationship by writing down the arc length

formula: s = rθ (where r is the radius)

Differentiating both sides with respect to t gives: ds/dt = r dθ/dtas r is constant for a circle.

Denoting angular speed dθ/dt by ω we have the important

relationship

The angular velocity, ω will be measured an rad s–1 and rev min–1 v = rω

When a particle moves in a circle with a constant speed the direction of its linear

velocity (which is tangential to the circle) is constantly changing; this means there

is an acceleration (even though the speed is constant). This acceleration is always

directed towards the centre of the circle (radial acceleration)

The next approach shows why the force is acting in this direction:

If dθ/dt = ω, then if we integrate both sides we get θ = ∫ ω dt

So, θ = ωt + C (Taking θ = 0 when t = 0 gives C = 0) Therefore, θ = ωt

Horizontal Circular Motion

Horizontal Circular Motion

Horizontal Circular Motion

Horizontal Circular MotionMATHEMATICAL MODELLING

For each of the following examples it is a good idea to draw a

diagram and label the forces on the particle. Discuss where the

centre of the circle is and which force, or combination of forces,

provides the force towards the centre.

Also note how the vertical forces balance.

• Particle on the rough surface of a rotating disc – the friction

force acts towards centre. (e.g. Roundabout in a playground)

• Conical pendulum (including elastic string) – a light string has

one end attached to a fixed point and the other attached to a

particle which is moving in a horizontal circle. The horizontal

component of tension provides the central force.

• Smooth bead on a rotating wire – component of reaction acts

towards centre of circular path.

Horizontal Circular MotionMATHEMATICAL MODELLING 1 (Friction)

The person is standing 5m from the axis of the roundabout and is rotating at 1 rads-1

Solution

Resolving vertically: R – 40g = 0 (Equilibrium) So, R = 40gN

For circular motion: X = mrω2 (These two forces identical) So, X = 40 x 5 x (12) = 200N

Now, Friction force X = µR (at point of sliding) Therefore, 40g µ = 200N

Giving µ = 200/40g = 0.51

So, to prevent sliding at this speed µ ≥ 0.51

Horizontal Circular MotionMATHEMATICAL MODELLING 2 (Conical Pendulum)

The chord is 1.2m long and the 0.2kg stone is spun in a horizontal circle at 5 rads-1

Solution

Resolving vertically: T cos θ – 0.2g = 0 (Equilibrium) So, T cos θ = 0.2gN (1)

For circular motion: T sin θ = mrω2 (2)

By Trigonometry the radius r of the circle is 1.2 sin θ (since the hypotenuse is 1.2m)

Substitute r and given ω into equation (2) gives T sin θ = 0.2 x 1.2 sin θ x 52

Cancel sin θ to give T = 0.2 x 1.2 x 52 = 6N The tension in the chord is 6N

Subst. T into equation (1) gives cos θ = g/30, so θ = 70.90 (1 d.p.)

Horizontal Circular MotionMATHEMATICAL MODELLING 3 (Banked Tracks)

The coefficient of friction between the tyres and the road is 0.6, find the max speed of the car in km h-1.

Source:Mr Barton Maths

Horizontal Circular MotionSUMMARY

Source:MEI

Horizontal Circular Motion

PROBLEM SOLVING 1

Particle supported by two light strings (See diagram) – Discuss why the

tensions in this case cannot be equal and which one is greater. What is

a condition for circular motion to be possible in this configuration with

both strings taut? The fact that the tension in the lower string must be

greater than zero leads to a minimum possible value for the angular

speed (and consequently a maximum value for the period of revolution).

T

This is like the SAMs question we saw earlier

You could also discuss what happens

to the tension in AP as the angular

speed increases leading to the

possibility that the string breaks so that

there is a minimum value for the period

of revolution.

Horizontal Circular MotionPROBLEM SOLVING 2

Particle moving on a smooth fixed surface – for example a particle

moving on the inside of a smooth fixed hollow cone.

Source: Mr Barton Maths

Horizontal Circular MotionPROBLEM SOLVING 3

Particle moving on a smooth fixed surface – for example a particle

moving on the inside of a smooth fixed hemispherical shell as shown

in the diagram below: Here it is important to note that:

the reaction is directed towards the

centre of the hemisphere and not

towards the centre of the circular path of

the particle.

The radius of the circular path is not the

same as the radius of the hemisphere

(called r here, so care must be taken

when quoting standard formula ‘rω2’)

The radius of circle can be found by

using Pythagoras theorem and the angle

of the reaction with horizontal is found by

trigonometry.

Horizontal Circular MotionWHAT THE EXAMINER SAID

3. A rough disc rotates about its centre in a horizontal plane with constant angular speed 80 revolutions per minute. A particle P lies on the disc at a distance 8 cm from the centre of the disc. The coefficient of friction between P and the disc is μ. Given that P remains at rest relative to the disc, find the least possible value of μ.

(7 marks)

Candidates seemed to have great difficulty changing from revolutions perminute to radians per second. This seemed to be wrong at least as often as itwas correct. A surprising number of solutions involved inequalities, not alwaysthe correct way round. Unfortunately, some gave their final answer as aninequality and so failed to answer the question as set. Many did not notice thatthe distance was given in centimetres and so used 8 instead of 0.08 in theircalculation. Those who knew how to tackle this question produced succinctsolutions.

Horizontal Circular MotionReturn to AS SAMs Question

Horizontal Circular MotionAS SAMs Question

Horizontal Circular MotionAS SAMs Question

Horizontal Circular MotionMark Scheme

Horizontal Circular MotionMark Scheme

Horizontal Circular Motion

This topic is extended in

AL Further Mechanics 2:

Vertical Circular Motion

Thank you for your attention

Getting Ready to Teach – Online Course Centres of Mass – Further Mechanics 2 (AS/AL)

Centres of Mass - Moments

PRIOR KNOWLEDGE

A level Mathematics – Pure content

Sigma notation (Unit 4b)

AS Mathematics – Mechanics content

Modelling assumptions made throughout this course

(e.g. particle, rigid, light, lamina, etc.)

Types of forces and force diagrams (Unit 8a)

A level Mathematics – Mechanics content

Equilibrium and statics (Unit 7a)

Moments (Unit 4)

Centres of Mass - MomentsOBJECTIVES 1

• be able to extend the principle of moments to develop an

approach and formula for finding the centre of mass of a

one dimensional discrete mass distribution;

• be able to find the centre of mass of a two dimensional

discrete mass distribution and a framework;

• be able to find the centre of mass of a plane uniform

lamina made up of rectilinear shapes e.g. L shape.

• know and be able to quote the position of the centre of

mass for a uniform triangular lamina;

• know and be able to quote the position of the centre of

mass for a uniform circular arc or sector of a circle;

• be able to adapt the given formulae for semi-circular

laminas and arcs;

Centres of Mass - Moments

OBJECTIVES 2

• be able to find the centre of mass of composite 2-D shapes.

• be able to use the position of the centre of mass for a 2-D lamina or framework to determine the orientation of the shape relative to either the vertical or horizontal when suspended freely from a point;

• be able to use the position of the centres of mass for a 2-D lamina or framework to establish a condition for equilibrium to exist when the body is free to rotate about a fixed horizontal axis.

Centres of Mass - MomentsAS SAMs Question

Centres of Mass - MomentsAS SAMs Question

Centres of Mass - Moments

TEACHING POINTS 1

Centres of Mass - Moments Centre of mass of a discrete mass distribution in two dimensions, framework and lamina. (rectilinear shapes)

1) 2)

3a) 3b)

Centres of Mass - MomentsCentre of mass of uniform bodies

Triangular lamina, arc & sector of a circle (Special case:- Semi-circular lamina)

Centres of Mass - Moments

MATHEMATICAL MODELLING (Hanging Body)

In the example below the lamina is freely

suspended from A. It is required to find

the size of the angle that the side AB

makes with the vertical.

The diagram illustrates that the creation

of a right-angled triangle is the most

effective way to find the size of this angle,

reading off the distances to the centre of

mass to find the lengths and using ‘arc

tan’ to find the size of the required angle

(usually in degrees to 1 decimal place).

Note that, although when suspended the

centre of mass lies vertically below A, it is

not necessary to draw a new diagram.

Centres of Mass - MomentsPROBLEM SOLVING

The mass of the lamina is M. A particle of mass kM is attached to

the lamina at D to form the system S. The system S is freely

suspended from A and hangs in equilibrium with AO horizontal.

A horizontal force of magnitude P is applied at C in the direction CD.

The loaded plate L remains suspended from A and rests in

equilibrium with AB horizontal and C vertically below B.

This topic is extended in

AL Further Mechanics 2:

Further Centres of Mass

Thank you for your attention

Centres of Mass - Moments

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