Hellooooo Everybody! Just your friendly neighbourhood ex-physics student passing through!

This thread contains my notes for Unit 3 Physics – I hope they can be a helpful guide for anyone in need of a little support in their physics endeavours

I have also made one of these threads for

Psychology and

Biology – so feel free to check those out and say “Hello”

Hopefully the structure of this thread won’t be too shabby and is easy to follow. I will be using the criteria, set out in the syllabus (which you can access

here), as a guide. If you haven’t already checked out the syllabus, then I highly recommend you do so. Afterall, QCAA can’t ask you any questions on the external exam that don’t relate to the content covered in the syllabus.

Anywho, please feel free to contribute your own notes or tips and tricks to help this thread grow

Also, please feel free to ask any and all questions – I promise I won’t bite

Thread Index – Jump Ahead to Your Desired Topic :)

VectorsProjectile Motion

Inclined Planes

Circular Motion

Gravitational Force and Field

Orbits

Electrostatics

Magnetic Fields

Electromagnetic Induction

Electromagnetic Radiation

So, let’s kick this thread off, shall we!

**Topic 1: Part 1**

**Gravity and Motion**

**Vectors** – **solve** vector problems by resolving vectors into components, adding or subtracting the components and recombining them to determine the resultant vector*Combining vectors in one-dimension*Vectors in the same directionE.g. A boat is heading East down a river at a speed of 10ms^{-1}. The water is also heading East at a speed of 5ms^{-1}. Therefore, the boats actual velocity is 15ms^{-1}. 10ms^{-1} + 5ms^{-1} = 15ms^{-1} to the East

Vectors in the opposite directionsE.g. A boat is heading East down a river at a speed of 10ms^{-1}. The water is heading West at a speed of 5ms^{-1}. Therefore, the boats actual velocity is 5ms^{-1}.10ms^{-1} - 5ms^{-1} = 5ms^{-1} to the East

*Combining vectors in two-dimensions*It is important that you remember the trigonometric ratios SOH, CAH, and TOA for this section. This is because vectors require a magnitude AND direction – thus, you will need to calculate and specify the direction of the resulting vector.

E.g. Imagine you are swimming North across a river at a speed of 1ms^{-1}, however the current is flowing to the East at a speed of 5ms^{-1}. By adding these two vectors in a head-to-tail formation we can calculate the magnitude and direction:**Magnitude:**Resultant = √(1

^{2}+5

^{2})

Resultant = √26

Resultant = 5.1 ms

^{-1} (1 d.p)

**Direction:**Tan(theta) = opp/adj

Tan(theta) = 1/5

theta = tan

^{-1}1/5

theta = 11.3˚

– **use** vector analysis to **resolve** a vector into two perpendicular components**Consider the following scenario: ** you’re playing a game of brandy with your mates, and you happen to be it. Your mate Dylan is running ahead of you, so you launch the tennis ball in his direction. Being a physics student, you take note of the angle with which you release the ball (theta), as well as the initial launch velocity (u). Having just left your physics class on vectors, you now know that you can resolve the launch velocity into two components perpendicular to each other – the horizontal and vertical velocities. Where the horizontal and vertical velocities are given to you by

*u*_{x} = ucos(theta) and

*u*_{y} = usin(theta) respectively.

E.g. Let’s say the angle you release the ball at is 35˚ to the horizontal, and the initial velocity is 10ms^{-1}. **The horizontal component would be:**u

_{x} = ucos(theta)

u

_{x} = (10ms

^{-1})cos(35˚)

u

_{x} = 8.2ms

^{-1} (1 d.p)

**The vertical component would be:**u

_{y} = usin(theta)

u

_{y} = (10ms

^{-1})sin(35˚)

u

_{y} = 5.7ms

^{-1} (1 d.p)

Therefore, we know that Dylan was hit by a tennis ball with a horizontal velocity of 8.2ms

^{-1} and vertical velocity of 5.7ms

^{-1}