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When I say difficult, I mean difficult in comparison to the usual standard of Higher Level Leaving Cert Applied Maths Connected Particles Questions

## Question (a)

Consider the following:

A rectangular block moves across a stationary horizontal surface with acceleration $g/5$ (the question had $g/5$ m/s${}^2$ but the m/s${}^2$ is repeated as $g=9.8$ m/s${}^2$ and so includes the unit).

There is a serious problem with this question and that is that the asymmetry in the problem means that there is an ambiguity: is the block moving left to right or right to left? I am going to assume the block moves from left to right. One would hope not to see such ambiguity in an official exam paper.

Two particles of mass $M$ placed on the block, are connected by a taut inextensible string. A second string passes over a light, smooth, fixed pulley to a third particle of mass $2M$ which presses against the block as shown in the diagram.

### (i)

If contact between the particles and the block is smooth, find the magnitude and direction of the resultant forces acting on the particles.

#### Solution

Note firstly that there are two accelerations at play. The acceleration of the block relative to the horizontal surface, $g/5$, and the accelerations of the particles relative to the block, say $a$:

We draw all the forces (I lazily didn’t add arrows to the force vectors):

We know that the normal forces for the particles on top of the block because their vertical acceleration is zero and so the sum of the forces in that direction must be zero, and as the down forces are equal for both, necessarily the up forces must be equal too.

On a particular day the velocity of the wind, in terms of $\mathbf{i}$ and $\mathbf{j}$, is $x\mathbf{i}-3\mathbf{j}$, where $x\in\mathbb{N}$.

$\mathbf{i}$ and $\mathbf{j}$ are unit vectors in the directions East and North respectively.

To a man travelling due East the wind appears to come from a direction North $\alpha^\circ$ West where $\tan\alpha=2$.

When he travels due North at the same speed as before, the wind appears to come from a direction North $\beta^\circ$ West where $\tan\beta=3/2$.

Find the actual direction of the wind.

### Solution:

We start by writing

$\overrightarrow{V_{WM}}=\overrightarrow{V_W}-\overrightarrow{V_M}$.

We have two equations.

Firstly, when the man travels due East, $\overrightarrow{V_M}=a\mathbf{i}$ for some constant $a$. We have $\overrightarrow{V_W}=x\mathbf{i}-3\mathbf{j}$ and so

$\overrightarrow{V_{WM}}=(x-a)\mathbf{i}-3\mathbf{j}$.

We know that this, with respect to the man, is coming from North $\alpha^\circ$ West. This means we have:

And furthermore:

$\displaystyle \tan\alpha=\frac{x-a}{3}\overset{!}{=}2$

$\Rightarrow x-a=6\Rightarrow a=x-6$.

Now consider when the man travels due North, $\overrightarrow{V_M}=a\mathbf{j}$. We have $\overrightarrow{V_W}=x\mathbf{i}-3\mathbf{j}$ still and so

$\overrightarrow{V_{WM}}=x\mathbf{i}-(3+a)\mathbf{j}$.

We know that this, with respect to the man, is coming from North $\beta^\circ$ West. This means we have:

And furthermore:

$\displaystyle \tan\beta=\frac{x}{3+a}\overset{!}{=}\frac32$

$\underset{\times_{2(3+a)}}{\Rightarrow} 2x=9+3a$,

but we have $a=x-6$ and so

$2x=9+3(x-6)\Rightarrow x=9$,

so that $\overrightarrow{V_W}=9\mathbf{i}-3\mathbf{j}$.

This means that the velocity of the wind looks like:

That is the wind comes from North $\theta$ West. We have that $\tan\theta=9/3\underset{\tan^{-1}}{\Rightarrow} \theta=\tan^{-1}(3)\approx 71.57^\circ$, so the answer to the question asked is N $71.57^\circ$ W.

Here we present three solutions to the one problem. The vector solution is probably the slickest. The geometry solution here can be simplified by being less rigorous, and the coordinate geometry solution might be made easier by using the $-b\pm\sqrt{\cdots}$ formula.

### LCHL 2007, Q. 2(a)

Ship B is travelling west at 24 km/h. Ship A is travelling north at 32 km/h.
At a certain instant ship B is 8 km north-east of ship A.

(i) Find the velocity of ship A relative to ship B.

(ii) Calculate the length of time, to the nearest minute, for which the ships
are less than or equal to 8 km apart.

Solution to (i): We have that $\overrightarrow{V_A}=32\mathbf{j}$ and $\overrightarrow{V_B}=-24\mathbf{i}$ and so

$\overrightarrow{V_{AB}}=\overrightarrow{V_A}-\overrightarrow{V_B}=32\mathbf{j} -(-24\mathbf{i})=24\mathbf{i}+32\mathbf{j}$.

### Vector Approach to (ii)

First of all we draw a picture. As we are talking relative to ship B we will put B at the origin. If ship B is 8 km north-east of ship A then ship A is 8 km south-west of ship B.

Where $\overrightarrow{R_0}$ is the initial displacement of ship A relative to ship B, the displacement of ship A relative to ship B, as a function of time, is given by

$\overrightarrow{R_{AB}}(t)=\overrightarrow{R_0}+t\cdot \overrightarrow{V_{AB}}$.

Using some trigonometry — vertical and horizontal components of $\overrightarrow{R_0}$ — we have

$\displaystyle\overrightarrow{R_0}=-\frac{8}{\sqrt{2}}\mathbf{i}-\frac{8}{\sqrt{2}}\mathbf{j}$,

and so

$\displaystyle \overrightarrow{R_{AB}(t)}=\left(24t-\frac{8}{\sqrt{2}}\right)\mathbf{i}+\left(32t-\frac{8}{\sqrt{2}}\right)\mathbf{j}$.

For those planning on focusing on questions one to five and ten:

Applied Maths some Notes

Occasionally, it might be useful to do as the title here suggests.

Two examples that spring to mind include:

• solving $a\cdot\cos\theta\pm b\cdot\sin\theta=c$ for $\theta$ (relative velocity example with $-$ below)
• maximising $a\cdot\cos\theta\pm b\cdot\sin\theta$ without the use of calculus

### $a\cdot \cos\theta- b\cdot\sin\theta$

Note first of all the similarity between:

$\displaystyle a\cdot \cos\theta-b\cdot \sin \theta\sim \sin\phi\cos\theta-\cos\phi\sin\theta$.

This identity is in the Department of Education formula booklet.

The only problem is that $a$ and $b$ are not necessarily sines and cosines respectively. Consider them, however, as opposites and adjacents to an angle in a right-angled-triangle as shown:

Using Pythagoras Theorem, the hypotenuse is $\sqrt{a^2+b^2}$ and so if we multiply our expression by $\displaystyle \frac{\sqrt{a^2+b^2}}{\sqrt{a^2+b^2}}$ then we have something:

$\displaystyle \frac{\sqrt{a^2+b^2}}{\sqrt{a^2+b^2}}\cdot \left(a\cdot \cos\theta- b\cdot\sin\theta\right)$

$\displaystyle=\sqrt{a^2+b^2}\cdot \left(\frac{a}{\sqrt{a^2+b^2}}\cos\theta-\frac{b}{\sqrt{a^2+b^2}}\sin\theta\right)$

$=\sqrt{a^2+b^2}\cdot \left(\sin\phi\cos\theta-\cos\phi\sin\theta\right)=\sqrt{a^2+b^2}\sin(\phi-\theta)$.

Similarly, we have

$a\cdot\cos\theta+b\cdot \sin\theta=\sqrt{a^2+b^2}\sin(\phi+\theta)$,

where $\displaystyle\sin\phi=\frac{a}{\sqrt{a^2+b^2}}$.

Consider the following problem:

Two masses of 5 kg and 1 kg hang from a smooth pulley at the ends of a light inextensible string. The system is released from rest. After 2 seconds, the 5 kg mass hits a horizontal table:

i. How much further will the 1 kg mass rise?

ii. The 1 kg mass then falls and the 5 kg mass is jolted off the table. With what speed will the 5 kg mass begin to rise?

[6D Q. 4. Fundamental Applied Maths, 2nd Edition, Oliver Murphy]

It isn’t difficult to answer part i.: the answer is $\displaystyle \frac89 g$ m.

However how to treat part ii.? First of all a picture to help us understand this problem:

The 1 kg mass has dropped under gravity through a distance of $\displaystyle \frac89 g$ m. We can find the speed of the 1 kg mass using $u=0,a=g,s=\frac89 g$. Alternatively, we can use Conservation of (Mechanical) Energy.

Taking the final position as $h=0$, at its maximum height, the 1 kg mass has potential energy and no kinetic energy:

$\text{PE}_0=mgh=1g\frac89 g=\frac{8}{9}g^2$.

When it reaches the point where the string is once again taut, it has not potential energy but the potential energy it had has been transferred into kinetic energy:

$\text{KE}_1=\frac12 mv^2=\frac12 v^2=\frac12 v^2$,

and this must equal the potential energy $\text{PE}_0$:

$\frac{8}{9}g^2=\frac12 v^2\Rightarrow v=\frac43g$.

Now this is where things get trickier. My idea was to use conservation of momentum on the two particles separately. As this clever answer to this question shows, you can treat the 5 kg mass, string, and 1 kg mass as a single particle.

So the prior momentum is the mass of the 1 kg mass by $\frac43 g$:

$p_0=m_0u=1\cdot \frac43 g=\frac43g$.

The ‘after’ momentum is the mass of the 1 kg and 5 kg masses times the new velocity:

$p_1=m_1v=6\cdot v$.

By Conservation of Momentum, these are equal:

$\frac43 g=6v\Rightarrow v=\frac{2}{9} g\text{ m s}^{-1}$.

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:

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.

## 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.

### 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:

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 $\theta$,

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

# 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 $\times$.