George R. A Level Further Mathematics  tutor, A Level Maths tutor, A ...

George R.

Currently unavailable: for regular students

Degree: Mathematics (Bachelors) - Cambridge University

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About me

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.

Subjects offered

SubjectLevelMy prices
Further Mathematics A Level £20 /hr
Maths A Level £20 /hr
Physics A Level £20 /hr
Maths GCSE £18 /hr
Physics GCSE £18 /hr
.STEP. Uni Admissions Test £25 /hr

Qualifications

QualificationLevelGrade
MathsA-LevelA*
Further MathsA-LevelA*
PhysicsA-LevelA*
ChemistryA-LevelA
Step IIIUni Admissions TestS
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

Currently unavailable: for regular students

Ratings and reviews

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 :)
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Questions George has answered

Derive the formula for the maximum kinetic energy of an electron emitted from a metal with work function energy p , that is illuminated by light of frequency f.

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...

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.

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1 year ago

535 views

How do you show that two lines do, or do not intersect?

How you should approach a question of this type in an exam Say you are given two lines L1 and L2 with equations  r1 = (a1,b1,c1) + s(d1,e1,f1) and r2 = (a2,b2,c2) + t(d2,e2,f2) respectively and you are asked to deduce whether they do or do not intersect. Then we need to either: Find values o...

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

Say you are given two lines Land Lwith equations

 r= (a1,b1,c1) + s(d1,e1,f1) and r= (a2,b2,c2) + t(d2,e2,f2) 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 rand rare 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+ sd= a+ td2

                      b+ se= b+ te2

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.'

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1 year ago

582 views
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