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The following question gave a little grief:

Two smooth spheres of masses 2m and 3m respectively lie on a smooth horizontal table.

The spheres are projected towards each other with speeds 4u and u respectively.

i. Find the speed of each sphere after the collision in terms of e, the coefficient of restitution

ii. Show that the spheres will move in opposite directions after the collision if e>\frac13.

My contention is that the question erred in not specifying that the answers to part i. were to be in terms of e and u.

Solution: 

i. The following is the situation:

diagram

Let v_1 and v_2 be the velocities of the smaller respectively larger sphere after collision. Note that the initial velocity of the larger sphere is minus u.

Using conservation of momentum,

m_1u_1+m_2u_2=m_1v_1+m_2v_2

\Rightarrow 2m(4u)+3m(-u)=2mv_1+3mv_2

\Rightarrow 5u=2v_1+3v_2.

Using:

\displaystyle e(u_1-u_2)=-(v_1-v_2)\Rightarrow \frac{v_1-v_2}{u_1-u_2}=-e,

Therefore,

\displaystyle \frac{v_1-v_2}{4u-(-u)}=-e

\Rightarrow v_1-v_2=-5ue\Rightarrow v_1=v_2-5ue,

and so

5u=2(v_2-5ue)+3v_2\Rightarrow 5u=2v_2-10ue+3v_2,

\Rightarrow 5v_2=5u+10ue\Rightarrow v_2=u(1+2e).

\Rightarrow v_1=u(1+2e)-5ue=u+2ue-5ue=u(1-3e).

 

ii. v_2>0. If e>\frac13\Rightarrow 3e>1\Rightarrow 1-3e<0 and so

v_1=u(1-3e)<0;

that is the particles move in opposite directions.

 

 

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The First Problem

A car has to travel a distance s on a straight road. The car has maximum acceleration a and maximum deceleration b. It starts and ends at rest.

Show that if there is no speed limit, the time is given by

\displaystyle \sqrt{2s\cdot \frac{(a+b)}{ab}}.

Solution: We are going to use two pieces of information:

  • the area under the velocity-time graph is the distance travelled,
  • the slope of the velocity-time graph is the acceleration.

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Introduction


To successfully analyse and solve the equations of Leaving Cert Applied Maths projectiles, one must be very comfortable with trigonometry.

Projectile trigonometry all takes place in [0^\circ,90^\circ] so we should be able to work exclusively in right-angled-triangles (RATs), however I might revert to the unit circle for proofs (without using the unit circle, the definitions for zero and 90^\circ are found by using continuity).

Recalling that two triangles are similar if they have the same angles, the fundamental principle governing trigonometry might be put something like this:

Similar triangles differ only by a scale factor.

We show this below, but what this means is that the ratio of corresponding sides of similar triangles are the same, and if one of the angles is a right-angle, it means that if you have an angle, say 40^\circ, and calculate the ratio of, say, the length of the opposite to the length of the hypotenuse, that your answer doesn’t depend on how large your triangle is and so it makes sense to talk about this ratio for 40^\circ rather than just a specific triangle:

graph1

These are two similar triangles. The opposite/hypotenuse ratio is the same in both cases.

Suppose the dashed triangle is a k-scaled version of the smaller triangle. Then |A'B'|=k|AB| and |A'C'|=k|AC|. Thus the opposite to hypotenuse ratio for the larger triangle is

\displaystyle \frac{|A'B'|}{|A'C'|}=\frac{k|AB|}{k|AC|}=\frac{|AB|}{|AC|},

which is the same as the corresponding ratio for the smaller triangle.

This allows us to define some special ratios, the so-called trigonometric ratios. If you are studying Leaving Cert Applied Maths you know what these are. You should also be aware of the inverse trigonometric functions. Also you should be able to, given the hypotenuse and angle, find comfortably the other two sides. We should also know that sine is maximised at 90^\circ, where it is equal to one.

In projectiles we use another trigonometric ratio:

\displaystyle \sec(\theta)=\frac{\text{hypotenuse}}{\text{adjacent}}=\frac{1}{\cos\theta}.

Note \cos90^\circ=0, so that \sec90^\circ is not defined. Why? Answer here.

The Pythagoras Identity

For any angle \theta,

\sin^2\theta+\cos^2\theta=1.

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Introduction

In Ireland at least, we first encounter fractions at age 6-8. At this age, because of our maturity, while we might be capable of some conceptional understanding, by and large we are doing things by rote and, for example, multiplying fractions is just something that we do without ever questioning why fractions multiply together like that. This piece is aimed at second and third level students who want to understand why the ‘calculus’ of fractions is like it is.

Mathematicians can in a rigorous way, write down what a fraction is… this piece is pitched somewhere in between these constructions — perhaps seen in an undergraduate mathematics degree — and the presentation of fractions presented in primary school. It is closer in spirit to a rigorous approach but makes no claims at absolute rigour (indeed it will make no attempt at rigour in places). The facts are real number axioms.

Defining Fractions

We will define fractions in terms of integers and multiplication.

To get the integers we first define the natural numbers.

Definition 1: Natural Numbers

The set of natural numbers is the set of counting numbers

\mathbb{N}=0,1,2,3,4,\dots,

together with the operations of addition (+) and multiplication \times.

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Applied maths could be defined as the use of mathematics in studying natural phenomena. The branch of applied maths studied at Leaving Cert level is Newtonian mechanics. Mechanics is the study of systems under the action of forces. Newtonian mechanics is concerned with systems that can be adequately described by Newton’s Laws of Motion. Systems that aren’t adequately described by Newtonian mechanics include systems with speeds approaching the speed of light, systems of extremely small particles and systems with a large number of particles. Hence Leaving Cert Applied Maths is the study of simple macro-systems that have moderate speeds.

Applied maths is essentially a further study of the mathematics of chapters 6 through 13 in Real World Physics; following which more specific and involved questions than those of Leaving Cert Physics may be posed and answered. The emphasis in applied maths is more on problem-solving than anything and reflecting this, the need for rote-learning is almost non-existent. The course content itself could be presented on one A4 sheet. The skills required for the course include:
• Capacity for interpretation and visualisation
• Ability for strategic problem solving
• Competency in mathematics

Anyone who is strong in higher level maths (A or B standard) and is self-motivated can achieve great success in applied maths – especially in conjunction with LC Physics. There is then a three for the price of two and a half effect as applied maths will help your physics, physics will help your applied maths and maths will help your applied maths. For those looking at the bigger picture, if you are intending on pursuing any science or engineering course in college, having done LC Applied Maths will really stand to you in your daunting first year — most 1st year physics and applied maths modules broadly cover the material of LC Applied Maths.

The statistics over the last number of years are that typically over 90% of students taking applied maths do higher level, and of these about a quarter achieve an A, over half achieve an A or B and nearly two-thirds achieve an honour. 90% of students pass. So with strong maths and good, effective application to the task at hand, good grades are there for the taking.

Some LC Applied Math Notes and some Solutions to selected problems. Section 6.2 deals with the Fundamental Theorem of Calculus and Section 6.4 has a little integration. See Chapter 10 of these LC Maths Lecture Notes for some more on integration.

A short note covering integration for Leaving Cert maths.

 

(Please note that the proof of the Fundamental Theorem of Calculus inside isn’t quite correct. We need the Mean Value Theorem to prove it but the one in here is just for illustrative purposes.)

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