A.M. ONLINE

### Weekly Grinds

A.M. Online has been created with ease of use in mind.

The site can be easily navigated using the menu located on the top left of every page. This menu is split into the five sections described below.

##### 1) Welcome Section

This section (highlighted in red) is where you are currently located and serves as an introduction to the website.

##### 2) Topic Section

This section (highlighted in blue) contains all of the content required for the examination and is therefore where students will spend most of their time.

This section is split into thirteen topics, each of which contains various lessons (in which lesson objectives are explained and discussed) and exercise sets (to test a student’s understanding of the lesson objectives).

An example of an exercise set question is as shown below.

## Exercise Set Question

A smooth sphere $$A$$, of mass $$4\mbox{ kg}$$ and moving with a speed of $$6\mbox{ m/s}$$ along a horizontal surface, collides with a smooth sphere $$B$$ of mass $$2\mbox{ kg}$$ moving with a speed of $$3\mbox{ m/s}$$ in the same direction.

Find the speed and direction of sphere $$B$$ after the collision if:

(a) sphere $$A$$ comes to rest after the collision.

(b) the collision is elastic.

(c) the coefficient of restitution between both spheres is $$0.5$$.

(a) $$15\vec{i}\mbox{ m/s}$$

(b) $$7\vec{i}\mbox{ m/s}$$

(c) $$6\vec{i}\mbox{ m/s}$$

##### Sphere A

$$m_A = 4\mbox{ kg}$$

Initial Velocity

$$\vec{u}_A=6 \vec{i}\mbox{ m/s}$$

Final Velocity

$$\vec{v}_A=v_A\vec{i}$$

##### Sphere B

$$m_B = 2\mbox{ kg}$$

Initial Velocity

$$\vec{u}_B=3 \vec{i}\mbox{ m/s}$$

Final Velocity

$$\vec{v}_B=v_B \vec{i}$$

(a)

$$v_A=0$$

and

\begin{align}m_Au_A+m_Bu_B = m_Av_A+m_Bv_B\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(4)(6)+(2)(3) = (4)(0)+(2)(v_B)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}30 = 2v_B\end{align}

\begin{align}\downarrow\end{align}

\begin{align}v_B=15\end{align}

Therefore, the velocity of sphere $$B$$ after the collision is $$\vec{v}_B=15\vec{i}\mbox{ m/s}$$.

(b)

\begin{align}m_Au_A+m_Bu_B = m_Av_A+m_Bv_B\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(4)(6)+(2)(3) = (4)(v_A)+(2)(v_B)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}30 = 4v_A+2v_B\end{align}

and

\begin{align}E_i&=\frac{1}{2}m_Au_A^2+\frac{1}{2}m_Bu_B^2\\&=\frac{1}{2}(4)(6^2)+\frac{1}{2}(2)(3^2)\\&=72+9=81 \mbox{ J}\end{align}

and

\begin{align}E_f&=\frac{1}{2}m_Av_A^2+\frac{1}{2}v_B^2\\&=\frac{1}{2}(4)(v_A^2)+\frac{1}{2}(2)(v_B^2)\\&=2v_A^2+v_B^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}81=2v_A^2+v_B^2\end{align}

We therefore have the following two equations with two unknowns

\begin{align}30 = 4v_A+2v_B\end{align}

\begin{align}81=2v_A^2+v_B^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}v_A=\frac{30-2v_B}{4}\end{align}

\begin{align}81=2v_A^2+v_B^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}81=2\left(\frac{30-2v_B}{4}\right)^2+v_B^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}81=112.5-15v_B+0.5v_B^2+v_B^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}1.5v_B^2-15v_B+31.5=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}v_B &= \frac{-(-15)\pm \sqrt{(-15)^2-4(1.5)(31.5)}}{2(1.5)}\\&=\frac{15\pm 6}{3}\end{align}

Therefore, either $$v_B=3$$ or $$v_B=7$$. The first result corresponds to no change to the velocity, i.e. both spheres somehow pass through each other, which is ignored.

Therefore, the velocity of sphere $$B$$ after the collision is instead $$\vec{v}_B=7\vec{i}\mbox{ m/s}$$.

(c)

\begin{align}30 = 4v_A+2v_B\end{align}

and

\begin{align}e=-\frac{v_A-v_B}{u_A-u_B}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}0.5=-\frac{v_A-v_B}{6-3}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}v_A-v_B=-1.5\end{align}

We therefore have the following two equations with two unknowns.

\begin{align}30 = 4v_A+2v_B\end{align}

\begin{align}v_A-v_B=-1.5\end{align}

\begin{align}\downarrow\end{align}

\begin{align}30 = 4(v_B-1.5)+2v_B\end{align}

\begin{align}\downarrow\end{align}

\begin{align}30 = 4v_B-6+2v_B\end{align}

\begin{align}\downarrow\end{align}

\begin{align}6v_B=36\end{align}

\begin{align}\downarrow\end{align}

\begin{align}v_B=6\end{align}

Therefore, the velocity of sphere $$B$$ after the collision is instead $$\vec{v}_B=6\vec{i}\mbox{ m/s}$$.

As the only external force acting on the system (gravity) does not act in the direction of the momentum of either object, we can use conservation of momentum in all three cases.

(a) Sphere $$A$$ has an initial velocity coefficient of $$u_A=6$$.

Sphere $$B$$ has an initial velocity coefficient of $$u_B=3$$.

Sphere $$A$$ has a final velocity coefficient of $$v_A=0$$.

Sphere $$B$$ has an unknown final velocity coefficient of $$v_B$$.

The principle of conservation of momentum states that

\begin{align}m_Au_A+m_Bu_B = m_Av_A+m_Bv_B\end{align}

and therefore

\begin{align}(4)(6)+(2)(3) = (4)(0)+(2)(v_B)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}30 = 2v_B\end{align}

\begin{align}\downarrow\end{align}

\begin{align}v_B=15\end{align}

Therefore, the velocity of sphere $$B$$ after the collision is $$\vec{v}_B=15\vec{i}\mbox{ m/s}$$.

(b) In this case, sphere $$A$$ also has an unknown final velocity coefficient of $$v_A$$. Therefore, conservation of momentum instead gives

\begin{align}(4)(6)+(2)(3) = (4)(v_A)+(2)(v_B)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}30 = 4v_A+2v_B\end{align}

As the collision is elastic, we know that the principle of conservation of energy applies, i.e. $$E_i=E_f$$.

If we set our reference location $$h=0$$ to be the ground, then neither objects have potential energy before and after the collision.

Hence, the total initial energy is given by

\begin{align}E_i&=\frac{1}{2}m_Au_A^2+\frac{1}{2}m_Bu_B^2\\&=\frac{1}{2}(4)(6^2)+\frac{1}{2}(2)(3^2)\\&=72+9\\&=81 \mbox{ J}\end{align}

and the total final energy is instead given by

\begin{align}E_f&=\frac{1}{2}m_Av_A^2+\frac{1}{2}v_B^2\\&=\frac{1}{2}(4)(v_A^2)+\frac{1}{2}(2)(v_B^2)\\&=2v_A^2+v_B^2\end{align}

The principle of conservation of energy therefore gives

\begin{align}81=2v_A^2+v_B^2\end{align}

We therefore have the following two equations with two unknowns

\begin{align}30 = 4v_A+2v_B\end{align}

\begin{align}81=2v_A^2+v_B^2\end{align}

Rearranging the first equation for $$v_A$$

\begin{align}v_A=\frac{30-2v_B}{4}\end{align}

\begin{align}81=2v_A^2+v_B^2\end{align}

and inserting the new first equation into the second equation gives

\begin{align}81=2\left(\frac{30-2v_B}{4}\right)^2+v_B^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}81=112.5-15v_B+0.5v_B^2+v_B^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}1.5v_B^2-15v_B+31.5=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}v_B^2-10v_B+21=0\end{align}

This quadratic equation can be solved by factoring (or by using the quadratic formula)

\begin{align}(v_B-3)(v_B-7)=0\end{align}

Therefore, either $$v_B=3$$ or $$v_B=7$$. The first result corresponds to no change to the velocity, i.e. both spheres passing through each other, which is therefore ignored.

Hence, in this case, the velocity of sphere $$B$$ after the collision is $$\vec{v}_B=7\vec{i}\mbox{ m/s}$$.

(c) The conservation of momentum equation remains the same, i.e.

\begin{align}30 = 4v_A+2v_B\end{align}

We are also given the coefficient of restitution $$e$$ and can therefore use the law of restitution

\begin{align}e=-\frac{v_A-v_B}{u_A-u_B}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}0.5=-\frac{v_A-v_B}{6-3}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}v_A-v_B=-1.5\end{align}

We therefore have the following two equations with two unknowns.

\begin{align}30 = 4v_A+2v_B\end{align}

\begin{align}v_A-v_B=-1.5\end{align}

Inserting from the second equation that $$v_A=v_B-1.5$$ into the first equation gives

\begin{align}30 = 4(v_B-1.5)+2v_B\end{align}

\begin{align}\downarrow\end{align}

\begin{align}30 = 4v_B-6+2v_B\end{align}

\begin{align}\downarrow\end{align}

\begin{align}6v_B=36\end{align}

\begin{align}\downarrow\end{align}

\begin{align}v_B=6\end{align}

Therefore, in this case, the velocity of sphere $$B$$ after the collision is $$\vec{v}_B=6\vec{i}\mbox{ m/s}$$.

As described on the homepage, each of these questions do not only include the answer but various “stepping stones”.

• First, check to see if you got the right Answer. If you didn’t…
• Read the Solution, i.e. what you should write down when answering the question. If you don’t understand the solution…
• Read the Walkthrough, i.e. what a teacher would say if they were explaining the solution to you. If you don’t understand the walkthrough…
• Ask Mr. Kenny for further guidance on Teacher Chat!

The differences between solutions and walkthroughs will initially be minimal for introductory lessons. The benefit of walkthroughs will be become more apparent as questions get more difficult!

And don’t worry, each of these stepping stones only need to be viewed when necessary! They are provided so that students don’t simply “give up” if they get the wrong answer.

The tabs below show common examples of when each of these stepping stones should be used.

"I got the right answer, nice! I will move on to the next question..."

"I got the wrong answer. Hmm... let me have a read of the solution."

\begin{align}\downarrow\end{align}

"Ah, I made a silly maths mistake. Oops! I fully understand why I was wrong and I will therefore move on to the next question."

"I got the wrong answer. Hmm... let me have a read of the solution."

\begin{align}\downarrow\end{align}

"Hmm... the solution uses a different equation than the one I used, but I don't understand why! I better have a read of the walkthrough...."

\begin{align}\downarrow\end{align}

"Ah of course, I assumed that the object is initially at rest, but it's not! Yes, we can't use the equation that I used in that case, I get it! I fully understand why I was wrong and I will therefore move on to the next question."

"I got the wrong answer. Hmm... let me have a read of the solution."

\begin{align}\downarrow\end{align}

"Hmm... the solution uses a different method to the one I used, but didn't it say in the lesson that we could use either method? I better have a read of the walkthrough...."

\begin{align}\downarrow\end{align}

"Hmm... the walkthrough explains how to get the right answer using the other method, but I still don't understand why my method can't be used as well. I better get some clarity from Mr. Kenny on Teacher Chat."

\begin{align}\downarrow\end{align}

"Ah of course, the acceleration is not uniform! Yes, we can't use the method that I used in that case, I get it! I fully understand why I was wrong and I will therefore move on to the next question."

"I got the wrong answer. Hmm... let me have a read of the solution."

\begin{align}\downarrow\end{align}

"Hmm... the solution uses a different method to the one I used, but didn't it say in the lesson that we could use either method? I better have a read of the walkthrough...."

\begin{align}\downarrow\end{align}

"Hmm... the walkthrough explains how to get the right answer using the other method, but I still don't understand why my method can't be used as well. I better get some clarity from Mr. Kenny on Teacher Chat."

\begin{align}\downarrow\end{align}

"Ah, I understand now! However, I just tried some more questions in this exercise set and I keep making mistakes. I think it's best that I book a grind with Mr. Kenny as these type of questions are causing me difficulty."

Within the lessons themselves, students will also find the following:

• Guided Examples. These are just like exercise set questions, but students are only expected to read them!

They instead prepare students for the following exercise set, just like examples in a schoolbook.

As students are not expected to try them, Guided Examples don’t include Answers or Solutions as they serve no purpose!

## Guided Example

Using Kruskal’s Algorithm, find the minimum spanning tree(s) of the network below.

We are specifically instructed to use Kruskal’s Algorithm and will therefore perform the corresponding steps.

Step 1: The edge with the smallest weight is the first edge of the MST. Cross it off the network and add it to the MST.

For this network, the edge with the smallest weight is $$BC$$. We therefore cross it off our network and add it to our MST.

Network

MST

Step 2: The edge with the next smallest weight that does not create a cycle is the next edge of the MST. Cross it, and any edges that created cycles, off the network and add it to the MST.

The next edge with the smallest weight is $$DE$$. As our MST would not have a cycle if this edge was added to it, we cross it off our network and add it to our MST.

Network

MST

Step 3: Repeat Step 2 until the MST contains all of the nodes of the network and the MST is connected.

The next edge with the smallest weight is $$CE$$. As our MST would not have a cycle if this edge was added to it, we cross it off our network and add it to our MST.

Network

MST

The next edge with the smallest weight is $$BD$$. However, our MST would have a cycle if it was added. We therefore ignore $$BD$$ and instead look at the next edge with the smallest weight, which is $$AB$$.

As our MST would not have a cycle if this edge was added to it, we cross both $$BD$$ and $$AB$$ off our network and add only $$AB$$ to our MST.

Network

MST

The next edge with the smallest weight is $$EF$$. As our MST would not have a cycle if this edge was added to it, we cross it off our network and add it to our MST.

Network

MST

As our MST now contains all of the nodes of our network and is also connected, it is now complete.

As we see, we get the same MST as we did in the last lesson but without the need to first find each of the spanning trees.

• Knowledge Checks. These are quick-fire questions that give students immediate feedback as to whether a lesson “clicked” or not, thereby indicating whether they are ready to answer the corresponding exercise set or whether they need to do some quick revision first.
/20

$\,$Knowledge Check 8

$\,$A quick check on your understanding of:

Displacement

Velocity

Acceleration

SUVAT equations

$\,$

The rate of change of a vector is itself a vector.

1 / 20

The rate of change of velocity with respect to time is the definition of

The rate of change of a vector is itself a vector.

2 / 20

The rate of change of displacement with respect to time is the definition of

What quantity has that unit?

3 / 20

"A car brakes from $$\mathbf{15\mbox{ m/s}}$$ east to $$0\mbox{ m/s}$$ east at a rate of $$5\mbox{ m/s}^2$$ within $$3$$ seconds". The quantity in bold refers to:

What quantity has this unit?

4 / 20

"A car brakes from $$15\mbox{ m/s}$$ east to $$\mathbf{0}\mbox{ m/s}$$ east at a rate of $$5\mbox{ m/s}^2$$ within $$3$$ seconds". The quantity in bold refers to:

Displacement is a vector quantity.

5 / 20

A pedestrian walks $$25$$ metres west from $$o$$ followed by $$10$$ metres east. Their total displacement from $$o$$ is

Distance is a scalar quantity.

6 / 20

A pedestrian walks $$25$$ metres west from $$o$$ followed by $$10$$ metres east. Their total distance travelled is

Speed is a scalar quantity.

7 / 20

A cyclist travels $$60$$ metres east in $$2$$ seconds at a constant acceleration. The cyclist's speed is

Velocity is a vector quantity.

8 / 20

A cyclist travels $$60$$ metres east in $$2$$ seconds at a constant acceleration. The cyclist's velocity is

Speed is equivalent to the magnitude of velocity.

9 / 20

Speed and velocity have the same units.

Acceleration is the rate of change of velocity.

10 / 20

A cyclist increases her velocity from $$10\mbox{ m/s}$$ east to $$20\mbox{ m/s}$$ east before reducing it again to $$10\mbox{ m/s}$$ east. As her starting and final velocities are the same, she did not accelerate during this motion.

A vector has both a magnitude and a direction.

11 / 20

When a car uses a roundabout, it must accelerate.

Acceleration is the rate of change of velocity.

12 / 20

If an object is moving, it must be accelerating.

If an object is not moving, its velocity is zero.

13 / 20

In order for a moving car to come to a stop, it must decelerate.

Displacement is a vector quantity.

14 / 20

In the below image, the car is parked and is not moving. What is its displacement from $$o$$ after 10 seconds?

$\,$

$\,$

If there is no acceleration, then there is no change in velocity.

15 / 20

In the SUVAT vector equations, if $$\vec{a}=0$$, then $$\vec{u}=\vec{v}$$.

What SUVAT variable is not included in this triangle?

16 / 20

"Dad's Silly Triangle" can be used:

What quantities do we know and what quantity are we looking for?

17 / 20

A cyclist increases his velocity from $$15\mbox{ m/s}$$ east to $$25\mbox{ m/s}$$ east over a period of $$5$$ seconds. Which of the following SUVAT equations should we use to find the cyclist's acceleration in that time?

What quantities do we know and what quantity are we looking for?

18 / 20

A cyclist increases his velocity from $$15\mbox{ m/s}$$ east to $$25\mbox{ m/s}$$ east over a period of $$5$$ seconds. Which of the following SUVAT equations should we use to find the cyclist's displacement in that time?

What happens when we multiply a scalar by a vector?

19 / 20

$$5t^3\vec{a}$$ is a vector

What assumption was made when deriving the SUVAT equations?

20 / 20

If the acceleration of an object is gradually changing over time, we cannot use the SUVAT equations.

0%

Finally, note that, unlike the official Leaving Cert Maths syllabus, the corresponding Applied Maths syllabus does not go into great detail about the differences between Higher Level and Ordinary Level learning objectives.

As such, rather than only highlighting those few differences, we instead make an additional distinction so that students are more aware of what the likely differences will be between the Higher Level and Ordinary Level exam.

Lesson content that is definitely Higher Level only (according to the official curriculum) will be surrounded by a solid red box.

Lesson content that is likely Higher Level only (based on the difficulty of the content) will be surrounded by a dashed red box.

##### 3) Project Section

In 6th year, students will create a project associated with Applied Maths. This project is worth 20% of a student’s final grade, with the Leaving Cert examination worth the remaining 80%.

Don’t worry! This project requires submitting a report of at most 900 words and is considered to be an easy way to collect marks!

This section (highlighted in green) offers various tips and strategies on how to tackle this project, including several examples of completed projects.

The topic that the report should be about changes each year, with those details being released near the beginning of 6th year. We will notify our 6th years about those details when they are released.

Until then, we recommend that students ignore this section and focus purely on the topic section described above!

##### 4) Revision Section

This section (highlighted in orange) contains all of the material that students need to revise the subject in the second half of 6th year, including revision sheets and sample papers (with solutions!).

##### 5) Weekly Grinds Section

Each week, Mr. Kenny streams a live grind that is free to watch for all members.

In this section, you will find a link to join the upcoming grind as well as the recordings of all previous grinds.

The material that will be discussed in each grind is posted on social media a day or two prior. Therefore, be sure to follow us on Facebook and Instagram if you are yet to do so!