__About Me:__

Hello, I'm George and I am a second year mathematics undergraduate at Corpus Christi College Cambridge.

I have been enthusiastic about my subjects for a number of years and I started my degree on the one year Maths with Physics course.

Recently I have gained satisfaction from tutoring 3 students prior to their exams, two at A-level and one at GCSE.

Until I reached my A-levels, I was not a strong student in maths. I now use this past experience to relate to students having difficulty with the subject, and determine exactly where the gaps in their understanding are.

__The Session__

I prefer to let the student choose what they wish to do during the session. If you feel like you need part of a course explaining, we can focus on making sure you are confident with it. Alternitively, we can go through past papers before or after you have attempted them.

For STEP tuition, I have a full set of solutions to all three papers from 2004 up to 2014, and I am happy to mark any of your answers. During the Session we can go through some questions and I can help you develop the thought processes needed to tackle them. Alternatively we can focus on a particular type of question, whether it is pure or mechanics.

__About Me:__

Hello, I'm George and I am a second year mathematics undergraduate at Corpus Christi College Cambridge.

I have been enthusiastic about my subjects for a number of years and I started my degree on the one year Maths with Physics course.

Recently I have gained satisfaction from tutoring 3 students prior to their exams, two at A-level and one at GCSE.

Until I reached my A-levels, I was not a strong student in maths. I now use this past experience to relate to students having difficulty with the subject, and determine exactly where the gaps in their understanding are.

__The Session__

I prefer to let the student choose what they wish to do during the session. If you feel like you need part of a course explaining, we can focus on making sure you are confident with it. Alternitively, we can go through past papers before or after you have attempted them.

For STEP tuition, I have a full set of solutions to all three papers from 2004 up to 2014, and I am happy to mark any of your answers. During the Session we can go through some questions and I can help you develop the thought processes needed to tackle them. Alternatively we can focus on a particular type of question, whether it is pure or mechanics.

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4.9from 7 customer reviews

Bilal (Student)

January 1 2016

Really useful and efficient session.

Bilal (Parent)

January 1 2016

An excellent tutor, would recommend to every student looking for a Maths A-Level tutor. George runs through concepts in an efficient way and I really learned a lot from his tutorial!!

Ridwaan (Student)

October 28 2015

Very happy.

Ridwaan (Student)

October 3 2015

A good approach to understanding chapter 6 (C3) mathematics. He gave me a starting point towards this chapter (using graphical forms instead of CAST method), to where i can now explore it in depth. Thank you ever so much :)

**The phenomenon this question relates to is the photoelectric effect, and it can only be explained using Einstein’s photon model of light.**

Electromagnetic waves are emitted in discrete lumps of energy, called photons.

The energy carried by one of these photons is given by E = hf = hc/l where l is the wavelength of the wave ( f=c/l ) , and h is the planks constant.

If a photon collides with a free electron in the surface of the metal, then the electron absorbs all of the energy of the photon. So the electron gains energy equal to hf.

However, energy is required to remove this electron from the metal. This is due to the negative charge of the electron and the positive charge of the nuclei in the metal atoms causing an attraction between them.

The energy required to remove the electron is the 'work function energy', p.

So the maximum kinetic energy that an electron can have once it has absorbed a photon and left the metal is KE = hf – p.

This is because the electron initially gains ‘hf’ of kinetic energy from the photon but then loses ‘p’ energy as it does work against the forces holding it in the metal.

**The phenomenon this question relates to is the photoelectric effect, and it can only be explained using Einstein’s photon model of light.**

Electromagnetic waves are emitted in discrete lumps of energy, called photons.

The energy carried by one of these photons is given by E = hf = hc/l where l is the wavelength of the wave ( f=c/l ) , and h is the planks constant.

If a photon collides with a free electron in the surface of the metal, then the electron absorbs all of the energy of the photon. So the electron gains energy equal to hf.

However, energy is required to remove this electron from the metal. This is due to the negative charge of the electron and the positive charge of the nuclei in the metal atoms causing an attraction between them.

The energy required to remove the electron is the 'work function energy', p.

So the maximum kinetic energy that an electron can have once it has absorbed a photon and left the metal is KE = hf – p.

This is because the electron initially gains ‘hf’ of kinetic energy from the photon but then loses ‘p’ energy as it does work against the forces holding it in the metal.

**How you should approach a question of this type in an exam**

Say you are given two lines L_{1 }and L_{2 }with equations

__r___{1 }= (a_{1},b_{1},c_{1}) + s(d_{1},e_{1},f_{1}) and __r___{2 }= (a_{2},b_{2},c_{2}) + t(d_{2},e_{2},f_{2}) respectively and you are asked to deduce whether they do or do not intersect.

Then we need to either:

Find values of s and t such that both position vectors __r___{1 }and __r___{2 }are equal and thus giving us a point of intersection.

Or show that such a pair of s and t does not exist.

In both cases we try to find the s and t , we either succeed or we reach a contradiction - which shows that they cannot intersect.

**What the best method is for doing this and how to display it to the examiner**

For two lines to intersect, each of the three components of the two position vectors at the point of intersection must be equal.

Therefore we can set up 3 simultaneous equations, one for each component.

However we only have 2 unknowns to find (s and t) so we only need two of these equations, so we pick two of them. (Here I pick the first two)

We write: a_{1 }+ sd_{1 }= a_{2 }+ td_{2}

_{ }b_{1 }+ se_{1 }= b_{2 }+ te_{2}

and we solve these in the usual way to find our s and t, showing our working.

Then we substitute in our values for s and t into our third equation:

If we have a point of intersection, then we should have one side equal to the other. We then may have to find the point of intersection, we do this by plugging our value of s into the equation of the first line, and t into the equation of the second. This gives us our point of intersection as they should be equal, if not , you have made a mistake in solving the two simultaneous equations.

If the lines do not intersect, then both sides will not be equal and we will have something like '5=8'. We now write 'This is a contradiction, so the lines do not intersect.'

**How you should approach a question of this type in an exam**

Say you are given two lines L_{1 }and L_{2 }with equations

__r___{1 }= (a_{1},b_{1},c_{1}) + s(d_{1},e_{1},f_{1}) and __r___{2 }= (a_{2},b_{2},c_{2}) + t(d_{2},e_{2},f_{2}) respectively and you are asked to deduce whether they do or do not intersect.

Then we need to either:

Find values of s and t such that both position vectors __r___{1 }and __r___{2 }are equal and thus giving us a point of intersection.

Or show that such a pair of s and t does not exist.

In both cases we try to find the s and t , we either succeed or we reach a contradiction - which shows that they cannot intersect.

**What the best method is for doing this and how to display it to the examiner**

For two lines to intersect, each of the three components of the two position vectors at the point of intersection must be equal.

Therefore we can set up 3 simultaneous equations, one for each component.

However we only have 2 unknowns to find (s and t) so we only need two of these equations, so we pick two of them. (Here I pick the first two)

We write: a_{1 }+ sd_{1 }= a_{2 }+ td_{2}

_{ }b_{1 }+ se_{1 }= b_{2 }+ te_{2}

and we solve these in the usual way to find our s and t, showing our working.

Then we substitute in our values for s and t into our third equation:

If we have a point of intersection, then we should have one side equal to the other. We then may have to find the point of intersection, we do this by plugging our value of s into the equation of the first line, and t into the equation of the second. This gives us our point of intersection as they should be equal, if not , you have made a mistake in solving the two simultaneous equations.

If the lines do not intersect, then both sides will not be equal and we will have something like '5=8'. We now write 'This is a contradiction, so the lines do not intersect.'