Ross W. GCSE Physics tutor, A Level Physics tutor, GCSE Maths tutor, ...

Ross W.

Currently unavailable: for regular students

Degree: MPhys (Masters) - Durham University

MyTutor guarantee

Contact Ross
Send a message

All contact details will be kept confidential.

To give you a few options, we can ask three similar tutors to get in touch. More info.

Contact Ross

About me

I am a second year student studying Physics at Durham University (very pleased with my first year results - got a 1st - yay!) I have always been fascinated by science and maths and their ability to explain the world around us.

What Do I know?

I've done lots of teaching; instructing sailing (ages 8-18) for two years, supporting in GCSE science classes and peer support at A-level throughout sixth form. I was also a volunteer team leader on a camp for inner-city children (ages 7-11) and an assistant on several Bronze Duke of Edinburgh expedition weekends, having achieved Gold myself. These experiences have taught me to be able to respond to individual needs in different learning environments. I come from a family of teachers (sadly), so I have a good understanding of listening to what you need and discovering your unique learning style so you can achieve optimum results. We will develop your confidence and, most importantly, enjoyment of the subject (I hope...)  - a happy student learns far more!

Sessions:

I will prepare material before the online tutorials and so am happy to be in touch before tutorials to get an idea of the areas in which you would like support. The work I prepare will be based on the level at which you are working, your exam board and any extension work you require or feel inspired to try! We will use a variety of learning techniques, for example, diagrams, pictures, equations, and my teaching style will adapt to what you find most effective. 

Applying to University?

Having been through the process very recently, I am keen to help any prospective students with their personal statements (such a nightmare!) and with the application as a whole. This can seem daunting at first, but with a little help is no problem at all.

Getting in Touch

If you have any questions, send me a 'WebMail' or book a 'Meet the Tutor Session' (both accessible through this website). 

Looking forward to working with you!

Subjects offered

SubjectLevelMy prices
Maths A Level £20 /hr
Physics A Level £20 /hr
Biology GCSE £18 /hr
Chemistry GCSE £18 /hr
Further Mathematics GCSE £18 /hr
Maths GCSE £18 /hr
Physics GCSE £18 /hr
-Personal Statements- Mentoring £20 /hr

Qualifications

QualificationLevelGrade
PhysicsA-LevelA*
MathsA-LevelA*
ChemistryA-LevelA*
Further Maths (AS)A-LevelA
Biology (AS)A-LevelA
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

Currently unavailable: for regular students

General Availability

Weeks availability
MonTueWedThuFriSatSun
Weeks availability
Before 12pm12pm - 5pmAfter 5pm
MONDAYMONDAY
TUESDAYTUESDAY
WEDNESDAYWEDNESDAY
THURSDAYTHURSDAY
FRIDAYFRIDAY
SATURDAYSATURDAY
SUNDAYSUNDAY

Please get in touch for more detailed availability

Questions Ross has answered

A rollercoaster carriage of mass 100kg has 45kJ of Kinetic Energy at the lowest point of its ride. Ignoring air resistance and friction between the wheels and the tracks, what is the maximum height above this point it could reach? [Take g as 10m/s/s)

[A useful tip: always start by drawing a diagram!!] This question is asking you to apply conservation of energy, i.e. at the highest point it can reach above the lowest point, all of this Kinetic Energy will have been transferred to Gravitational Potential Energy. This requires use of the equ...

[A useful tip: always start by drawing a diagram!!]

This question is asking you to apply conservation of energy, i.e. at the highest point it can reach above the lowest point, all of this Kinetic Energy will have been transferred to Gravitational Potential Energy. This requires use of the equations KE=GPE and GPE=mgh, where m is the mass (in kg), g is the acceleration due to gravity (in m/s/s) and h is the maximum height above the lowest point (in m). All units of energy must be in J for these equations to give the right answers.

To get the height, rearrange the equations to give h=KE/mg, or rather h=45000/(100x10)=45m

(Note: the question gave the energy in kJ, so the number had to be multiplied by 1000 to give it in J, which gave the height in m) 

see more

4 months ago

125 views

Two electrons are a distance r apart, the first electron exerts a force F on the second electron. a) What force does the second electron exert on the first? b) In terms of r, at what distance is the force that the first electron exerts on the second F/9?

This question is on electric forces between charged particles. A useful equation to consider is Coulomb's law: F=k(Q1Q2)/R2 Where k is the Coulomb's law constant: k~9.0x109Nm2/C2 Q is the charge on each particle in Coulombs, R is the distance in metres and F is the force in Newtons. a) Thi...

This question is on electric forces between charged particles. A useful equation to consider is Coulomb's law:

F=k(Q1Q2)/R2

Where k is the Coulomb's law constant:

k~9.0x109Nm2/C2

Q is the charge on each particle in Coulombs, R is the distance in metres and F is the force in Newtons.

a) This part is a simple application of Newton's third law, as the first electron is exerting a repulsive force F on the second, the second must also be exerting a repulsive force F on the first. (Every force has an equal and opposite reaction force!)

b) This section requires you to look at Coulomb's law. It is what is known as an inverse square law, this effectively means the force decreases proportionally to the square of the distance, so for the force to have decreased by a factor of 9, the distance must have increased by a factor of the square root of nine, this equals 3, so the new distance is 3r. Nothing else in the equation changes, so they all other terms can be treated as constants and ignored.

This can be seen more explicitly by mathematically manipulating Coulomb's law, however I find it easier and more useful to instead find the answer by just thinking about the underlying link between force and distance in this equation, this means you develop a proper understanding of the inverse square relationship.

see more

4 months ago

137 views
Send a message

All contact details will be kept confidential.

To give you a few options, we can ask three similar tutors to get in touch. More info.

Contact Ross

Still comparing tutors?

How do we connect with a tutor?

Where are they based?

How much does tuition cost?

How do tutorials work?

Cookies:

We use cookies to improve our service. By continuing to use this website, we'll assume that you're OK with this. Dismiss

mtw:mercury1:status:ok