# Modelling

## p Terms:

• A particle is an object with mass but no size
• A rigid body has size but won't deform when forces are applied
• Some factors may be simplified or ignored as they are usually negligible:
- Smooth - no friction
- Rough - there is friction
- Light - no mass
- Inelastic/inextensible - does not stretch
- Elastic - does stretch
- Thin - can be assumed to have no width
- Rigid - does not stretch or bend
• If something is uniform, it is always the same for that object/environment. For example, if velocity is uniform, it remains constant

## SI units:

• There are seven SI base units. The three relevant to this module are:
- Mass, $m$$m$ ($\mathrm{kg}$$\mathrm{kg}$)
- Time, $t$$t$ ($\mathrm{s}$$\mathrm{s}$)
- Length, $l$$l$ (scalar), $d$$d$ (scalar), or $s$$s$ (vector) ($\mathrm{m}$$\mathrm{m}$)
• Every other measurement (e.g. speed) can be made from these
• Speed is measured in metres per second, $\mathrm{m/s}$$\mathrm{m/s}$ or $\mathrm{m}\ \mathrm{s}^{-1}$$\mathrm{m}\ \mathrm{s}^{-1}$
• Acceleration is measured in $\mathrm{m/s/s}$$\mathrm{m/s/s}$ or $\mathrm{m}\ \mathrm{s}^{-2}$$\mathrm{m}\ \mathrm{s}^{-2}$
• Force is measured in newtons ($\mathrm{N}$$\mathrm{N}$)

## Constants:

• $g$$g$ is the constant for the acceleration due to gravity. On Earth this is $9.8$$9.8$ $\mathrm{m}\ \mathrm{s}^{-2}$$\mathrm{m}\ \mathrm{s}^{-2}$ to 2 significant figures. Unless told otherwise by the question, you should always use $g = 9.8$$g = 9.8$ $\mathrm{m}\ \mathrm{s}^{-2}$$\mathrm{m}\ \mathrm{s}^{-2}$ in this module

## Significant figures:

• Always give answers in the exam to 3 s.f. unless the question says otherwise
• Don't do this for A Level sciences, though!

# Vectors

## v Notation:

• Vectors represent a magnitude and direction
- e.g. displacement, velocity, acceleration, weight
• Scalars represent only a magnitude
- e.g. time, distance, speed, mass
• Vectors are written as $\mathbf{v}$$\mathbf{v}$ (bold) or $\vec{v}$$\vec{v}$ (with an arrow above)
• The magnitude component is written italicised (e.g. $v$$v$), or as $|\mathbf{v}|$$|\mathbf{v}|$. It is calculated with the formula $|\mathbf{v}| = \sqrt{x^2 + y^2}$$|\mathbf{v}| = \sqrt{x^2 + y^2}$
• In a diagram, an arrow is often used - the length represents the magnitude and it points in the direction of the vector

## Unit vectors:

• A unit vector is a vector with a magnitude of 1
• $\mathbf{i}$$\mathbf{i}$ represents the $x$$x$ direction, and $\mathbf{j}$$\mathbf{j}$ the $y$$y$ direction
• For example, $5\mathbf{i} + 4\mathbf{j}$$5\mathbf{i} + 4\mathbf{j}$ represents the vector $\begin{pmatrix}5 \\ 4\end{pmatrix}$$\begin{pmatrix}5 \\ 4\end{pmatrix}$, 5 right and 4 up. This doesn't have to start at the origin, it can be a translation from one point to another
• For 3D vectors, $\mathbf{k}$$\mathbf{k}$ is used to represent the unit vector for the third dimension

## Position vectors:

• A position vector is the distance and direction of a point from the origin, $O$$O$
• The notation $\vec{OP}$$\vec{OP}$ is used to represent the position of point $P$$P$
• The letter $\mathbf{r}$$\mathbf{r}$ can also be used to represent the position vector of a body

## Components:

• A component of a vector is its magnitude in a given direction - e.g. the horizontal component of $\begin{pmatrix}a \\ b\end{pmatrix}$$\begin{pmatrix}a \\ b\end{pmatrix}$ is $a$$a$
• If you only have a vector's magnitude and direction, the horizontal and vertical components can be calculated with right-angle trigonometry using $\sin$$\sin$ or $\cos$$\cos$. Splitting a vector into the two components (or 3 if it's 3D) is called resolving

## Finding the direction of a vector:

• If you know the two components of a vector, you can use $\tan$$\tan$ to calculate its direction
• $\theta = \tan^{-1}$$\theta = \tan^{-1}$$(\frac{y}{x})$$(\frac{y}{x})$
• This will produce the angle of elevation from the positive $x$$x$ axis. If $x < 0$$x < 0$, add 180° to $\theta$$\theta$. If the point is in the bottom-right quadrant, add 360° to make the angle positive
• It's best to draw a diagram/sketch to help with this
• You will need to adjust this if the question asks for a bearing so that it's the distance clockwise from the positive $y$$y$ axis. Also, remember that a bearing is always three digits - e.g. 40° from positive $y$$y$ is 040° as a bearing

## Operations on vectors:

• Add vectors by adding their individual terms. The same applies for subtraction
• Multiplying a vector by a scalar will increase its magnitude by the scalar (e.g. multiplying a vector $\mathbf{v}$$\mathbf{v}$ by 2 will double its magnitude). In vector notation, simply multiply the scalar by each element (e.g. $s \times \begin{pmatrix}a \\ b\end{pmatrix} = \begin{pmatrix}a \times s \\ b \times s\end{pmatrix}$$s \times \begin{pmatrix}a \\ b\end{pmatrix} = \begin{pmatrix}a \times s \\ b \times s\end{pmatrix}$)

# Kinematics

## k Terms:

• The distance is the amount of space between two points
• The position of an object is its location relative to the origin. It's a vector so it has a distance and direction
• The displacement of an object is the direction and distance relative to the starting point
• Speed is a scalar which measures how fast something is moving
• Velocity measures how fast something is moving in a specified direction - it's a vector
• Acceleration is a measure of how much something is increasing in velocity per unit of time
• Acceleration is a vector, so the magnitude of acceleration is the magnitude only - not the direction
• Relative velocity is the velocity of one object from the perspective of another. For example, if you are moving at $5\ \mathrm{m}\ \mathrm{s}^{-1}$$5\ \mathrm{m}\ \mathrm{s}^{-1}$ and another object is moving at $5\ \mathrm{m}\ \mathrm{s}^{-1}$$5\ \mathrm{m}\ \mathrm{s}^{-1}$ towards you, it has a relative velocity of $10\ \mathrm{m}\ \mathrm{s}^{-1}$$10\ \mathrm{m}\ \mathrm{s}^{-1}$

## Equations for uniform acceleration (aka 'suvat'/'uvast'):

• Where $s$$s$ = displacement, $u$$u$ = initial velocity, $v$$v$ = final velocity, $t$$t$ = time and $a$$a$ = acceleration:
$s =$$s =$ $\frac{(u+v)t}{2}$$\frac{(u+v)t}{2}$
$v = u + at$$v = u + at$
$s = ut + \frac{1}{2}at^2$$s = ut + \frac{1}{2}at^2$
$v^2 = u^2 + 2as$$v^2 = u^2 + 2as$
• If an object is uniformly accelerating, average velocity = $\frac{1}{2}(u + v)$$\frac{1}{2}(u + v)$
• These equations need to be memorised
• They only work if the acceleration is uniform/constant (i.e. it's not changing)
• If working with two dimensions, deal with each one separately
• Remember that displacement, velocity and acceleration are all vectors and one direction must be positive and the other negative, so you may need to make some negative

## Displacement/velocity/acceleration - time graphs:

• Time is always on the $x$$x$-axis
• On a displacement-time graph, the gradient is equal to the velocity
• On a velocity-time graph, the gradient is equal to the acceleration and the area under the graph is the displacement. If this graph is curved, it should be split into small chunks (boxes and triangles, or trapezia). Calculate the area of each chunk and then sum these areas to find the approximate distance travelled
• On an acceleration-time graph, the area underneath is equal to the change in velocity. If the area is 0 square units, the object is moving with constant average velocity
• The above gradients and areas can be derived; the gradient is the $y$$y$ units divided by the $x$$x$ units and the area is $xy$$xy$
- For example, on a velocity-time graph, the units are $\mathrm{m}\ \mathrm{s}^{-1}$$\mathrm{m}\ \mathrm{s}^{-1}$ and $\mathrm{s}$$\mathrm{s}$
- So the gradient is $y$$y$ units ($\mathrm{m}\ \mathrm{s}^{-1}$$\mathrm{m}\ \mathrm{s}^{-1}$) divided by $x$$x$ units ($\mathrm{s}$$\mathrm{s}$) which is $\mathrm{m}\ \mathrm{s}^{-2}$$\mathrm{m}\ \mathrm{s}^{-2}$; acceleration
- The area is $\mathrm{m}\ \mathrm{s}^{-1}$$\mathrm{m}\ \mathrm{s}^{-1}$ $\times$$\times$ $\mathrm{s}$$\mathrm{s}$ $=$$=$ $\mathrm{m}$$\mathrm{m}$; displacement

## Differentiation and integration:

• Velocity is the rate of change of displacement
• Acceleration is the rate of change of velocity
• Therefore, differentiate displacement with respect to time to get velocity and differentiate velocity with respect to time to get acceleration
• This also works with integration in the opposite direction
• A quick way to remember this is to remember this pattern: $s \rightarrow v \rightarrow a$$s \rightarrow v \rightarrow a$ (or $r \rightarrow v \rightarrow a$$r \rightarrow v \rightarrow a$ for vectors). Differentiating is moving right and integration is moving left

# Forces

## d Types of force:

• Weight is a downwards force, the product of an object's mass and the acceleration due to gravity (more specifically, gravitational field strength but this isn't covered in M1)
• The normal reaction force is the reaction from a surface due to Newton's third law. It is always at 90° to the surface
• Tension is a force stretching something
• Thrust is a force compressing something
• Friction is caused by the roughness between an object and a surface. It acts in the opposite direction to motion
• Resistance is a force which opposes motion (e.g. air resistance)

## Resolving:

• Draw a right-angled triangle with the hypotenuse being the force you want to resolve
• Now use right-angle trigonometry to calculate the horizontal and vertical components

## Equilibrium:

• Particles in equilibrium have no resultant force - the components of their forces add up to zero resulting in zero acceleration
• If you draw the forces of an object in equilibrium tip-to-tail, a closed polygon will be formed
• For equilibria involving three forces, the specification states that you can optionally use Lami's Theorem (this external website explains it)

# Newton's laws of motion

## n The laws:

• Newton's first law is that an object continues at a constant velocity/remains at rest, provided that there is no external force acting upon it (i.e. a net force is needed to change velocity)
• Newton's second law states that acceleration is produced when force acts on a mass and this acceleration is directly proportional to mass
• Newton's third law is that forces always occur in equal, opposite pairs. The pairs are also always of the same type (e.g. both kinetic). In other words, if object A exerts a force on object B, object B will exert an equal and opposite force on object A

## Newton's second law:

• Newton's second law is represented by the formula $F = ma$$F = ma$ where $F$$F$ is the resultant force, $m$$m$ is the object mass, and $a$$a$ is the object's acceleration
• It also works for vectors, where it is written as $\mathbf{F} = m\mathbf{a}$$\mathbf{F} = m\mathbf{a}$ (or $\vec{F} = m\,\vec{a}$$\vec{F} = m\,\vec{a}$)

# Projectiles

## y Projectiles:

• Use the equations for uniform acceleration ('suvat'/'uvast') for these problems
• You may need to solve simultaneous equations

## Range:

• The range of a projectile is how far it travels along the ground
• To get the distance travelled by a projectile at time $t$$t$, multiply the initial horizontal velocity component by $t$$t$