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I am currently studying Physics at Durham University. I have always loved physics and maths, and I think that anyone can be successful at them if they are taught properly. I believe that if you learn maths and science well, you will learn to love them and find them much easier as well as more enjoyable.

I have spent time supervising and leading foreign students learning English, as well as having tutored students at secondary school. I think the best way of tutoring students is to guide you through examples and then see if they can apply your new knowledge to all the different sorts of questions you might be asked.

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During our tutorials, what we cover and the pace we cover it is completely up to you. Like most teachers, our initial focus will be on the basic concepts and examples; from there, we can move on to harder questions and the different kinds of exam-style questions you will face.

By using different teaching methods such as diagrams, analogies and examples, you will hopefully be able to explain the topic to me, or anyone else, and make them understand. The only real way to check if you have understood a topic is to teach it to someone else, and in doing so you can see if there are any gaps in your understanding.

Finally, and most importantly, both of us should enjoy the tutorials! My aim is to make you look forward to mastering new topics and taking an active interest in the subject, as well as preparing you as best as possible for the exams.

**What Next?**

If you have any questions, please do not hesitate to send me a "WebMail" or to book a "Meet the Tutor" session!

**About Me: **

I am currently studying Physics at Durham University. I have always loved physics and maths, and I think that anyone can be successful at them if they are taught properly. I believe that if you learn maths and science well, you will learn to love them and find them much easier as well as more enjoyable.

I have spent time supervising and leading foreign students learning English, as well as having tutored students at secondary school. I think the best way of tutoring students is to guide you through examples and then see if they can apply your new knowledge to all the different sorts of questions you might be asked.

**The Sessions:**

During our tutorials, what we cover and the pace we cover it is completely up to you. Like most teachers, our initial focus will be on the basic concepts and examples; from there, we can move on to harder questions and the different kinds of exam-style questions you will face.

By using different teaching methods such as diagrams, analogies and examples, you will hopefully be able to explain the topic to me, or anyone else, and make them understand. The only real way to check if you have understood a topic is to teach it to someone else, and in doing so you can see if there are any gaps in your understanding.

Finally, and most importantly, both of us should enjoy the tutorials! My aim is to make you look forward to mastering new topics and taking an active interest in the subject, as well as preparing you as best as possible for the exams.

**What Next?**

If you have any questions, please do not hesitate to send me a "WebMail" or to book a "Meet the Tutor" session!

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September 19 2016

Emma (Parent from Reading)

October 16 2016

This question involves the principle of the conservation of momentum.

Recall: **Assuming no external forces act, linear momentum is always conserved**. This means the total linear momentum of two objects *before* they collide equals the total linear momentum *after* the collision.

To start, let's work out the momentum of each ball separately. We'll call the moving ball "A" and the stationary ball "B". We will need the formula Momentum = Mass * Velocity, or **p = m * v**.

For ball A:

p = m * v

p = 0.25 * 1.2

p = 0.3 kg m s^{-1}

Let's call this value p_{A}.

For ball B:

p = m *v

p = 0.25 * 0

p = 0.

Let's call this value p_{B}.

So now we can work out the total momentum **before** the collision:

p_{total} = p_{A }+ p_{B} = 0.3 + 0 = 0.3 kg m s^{-1}.

This is where we bring in the principle of conservation of momentum. The question tells us that after the collision, the balls move **together** with the **same velocity**. This means we can treat them as **one object**, and use our formula to work out the velocity after the collision.

We know that after the collision, the total momentum = 0.3 kg m/s, from the principle of conservation.

We also know that the total mass of both balls (remember we're treating them as **one object**, so we can **add the masses**) is 0.5 kg.

We simply plug these numbers into our formula for momentum:

p = m * v

0.3 = 0.5 * v

v = 0.3/0.5

__v = 0.6 m/s__

This question involves the principle of the conservation of momentum.

Recall: **Assuming no external forces act, linear momentum is always conserved**. This means the total linear momentum of two objects *before* they collide equals the total linear momentum *after* the collision.

To start, let's work out the momentum of each ball separately. We'll call the moving ball "A" and the stationary ball "B". We will need the formula Momentum = Mass * Velocity, or **p = m * v**.

For ball A:

p = m * v

p = 0.25 * 1.2

p = 0.3 kg m s^{-1}

Let's call this value p_{A}.

For ball B:

p = m *v

p = 0.25 * 0

p = 0.

Let's call this value p_{B}.

So now we can work out the total momentum **before** the collision:

p_{total} = p_{A }+ p_{B} = 0.3 + 0 = 0.3 kg m s^{-1}.

This is where we bring in the principle of conservation of momentum. The question tells us that after the collision, the balls move **together** with the **same velocity**. This means we can treat them as **one object**, and use our formula to work out the velocity after the collision.

We know that after the collision, the total momentum = 0.3 kg m/s, from the principle of conservation.

We also know that the total mass of both balls (remember we're treating them as **one object**, so we can **add the masses**) is 0.5 kg.

We simply plug these numbers into our formula for momentum:

p = m * v

0.3 = 0.5 * v

v = 0.3/0.5

__v = 0.6 m/s__

We are given the substitution to use, so the first step is to **differentiate "u" with respect to x**.

du/dx = -sin(x)

Now, to replace the "dx" in the original integrand with something in terms of "du", we rearrange the differential:

dx = -1/sin(x) du

We substitute this into the **original expression** we are integrating; this gives:

S -12sin(x)cos^{3}(x) (-1/sin(x)) du

Let's do some simplifying here; the negative signs cancel, and so does sin(x):

S 12cos^{3}(x) du

Now, simplify again using u=cos(x); this gives:

S 12u^{3} du

This is a simple C1-level integration; i**ntegrating with respect to "u" **and adding a constant of integration, we get:

3u^{4} + c

For our final answer, **replace "u" with cos(x)**:

__3cos__^{4}__x) + c__

We are given the substitution to use, so the first step is to **differentiate "u" with respect to x**.

du/dx = -sin(x)

Now, to replace the "dx" in the original integrand with something in terms of "du", we rearrange the differential:

dx = -1/sin(x) du

We substitute this into the **original expression** we are integrating; this gives:

S -12sin(x)cos^{3}(x) (-1/sin(x)) du

Let's do some simplifying here; the negative signs cancel, and so does sin(x):

S 12cos^{3}(x) du

Now, simplify again using u=cos(x); this gives:

S 12u^{3} du

This is a simple C1-level integration; i**ntegrating with respect to "u" **and adding a constant of integration, we get:

3u^{4} + c

For our final answer, **replace "u" with cos(x)**:

__3cos__^{4}__x) + c__