# Chapter 3- Two dimensional Kinematics

## Vectors and scalars

Vector quantities with number, direction and units:

• Displacement $\vec{r}$ [m]
• Velocity $\vec{v}$ [ms-1]
• Acceleration $\vec{a}$ [ms-2]

Scalar quantities number and units only

• Distance traveled [m]
• Speed [ms-1]

## Graphical representation of vectors and components

It is frequently useful to draw two dimensional vectors as arrows, and to split them in to components that lie along the coordinate axes. The choice of coordinate axes is up to you..but choosing the right ones will make the problem easier or harder.

We can take a look at the acceleration due to gravity as vector using a phone accelerometer, using this tool (click on the link from your phone's browser).

## Vectors - Components

 $v_{1x}=v_{1}\cos\theta_{1}$ $v_{2x}=v_{2}\cos\theta_{2}$ $v_{1y}=v_{1}\sin\theta_{1}$ $v_{2y}=v_{2}\sin\theta_{2}$ $v_{Rx}=v_{1x}+v_{2x}$ $\tan\theta_{R}=\frac{v_{Ry}}{v_{Rx}}$ $v_{Ry}=v_{1y}+v_{2y}$ $v_{R}=\sqrt{v_{Rx}^2+v_{Ry}^2}$

## Vectors - Multiplication by a scalar

Multiplication of a vector by a scalar can change the magnitude, but not the direction of the vector, ie. each component of the vector is multiplied by the scalar in the same way.

## Unit Vectors

It can be useful to express vector quantities in terms of unit vectors. These are dimensionless vectors of length = 1 that point along the coordinate axes. They are usually denoted with carets (hats), i.e. $(\hat{i},\hat{j},\hat{k})$

For example:

$\vec{v}\,\mathrm{ms^{-1}}=v_{x}\,\mathrm{ms^{-1}}\,\hat{i}+v_{y}\,\mathrm{ms^{-1}}\,\hat{j}+v_{z}\,\mathrm{ms^{-1}}\,\hat{k}$

or

$\vec{r}\,\mathrm{m}=x\,\mathrm{m}\,\hat{i}+y\,\mathrm{m}\,\hat{j}+z\,\mathrm{m}\,\hat{k}$

## Vectors and motion

$\vec{r_{1}}=x_{1}\,\hat{i}+y_{1}\,\hat{j}+z_{1}\,\hat{k}$

$\vec{r_{2}}=x_{2}\,\hat{i}+y_{2}\,\hat{j}+z_{2}\,\hat{k}$

$\Delta\vec{r}=\vec{r_{2}}-\vec{r_{1}}=(x_{2}-x_{1})\,\hat{i}+(y_{2}-y_{1})\,\hat{j}+(z_{2}-z_{1})\,\hat{k}$

Average velocity: $\vec{v_{ave}}=\frac{\Delta\vec{r}}{\Delta t}$

Instantaneous velocity: $\vec{v}=\frac{d\vec{r}}{dt}=\frac{dx}{dt}\,\hat{i}+\frac{dy}{dt}\,\hat{j}+\frac{dz}{dt}\,\hat{k}=v_{x}\,\hat{i}+v_{y}\,\hat{j}+v_{z}\,\hat{k}$

Average acceleration: $\vec{a_{ave}}=\frac{\Delta\vec{v}}{\Delta t}$

Instantaneous acceleration: $\vec{a}=\frac{d\vec{v}}{dt}=\frac{dv_x}{dt}\,\hat{i}+\frac{dv_y}{dt}\,\hat{j}+\frac{dv_z}{dt}\,\hat{k}=\frac{d^{2}x}{dt^2}\,\hat{i}+\frac{d^{2}y}{dt^2}\,\hat{j}+\frac{d^{2}z}{dt^2}\,\hat{k}$

## Monkey and Hunter

At the moment the hunter pulls the trigger the monkey can make a decision, should he hang on or drop from the tree in the hope the bullet will miss him?

## Monkey and Hunter solution

In case my monkey shooting fails Physclips has a video you can step through frame by frame and a pretty detailed explanation here

Height of bullet with time:

$y=x\tan\theta-\frac{1}{2}gt^{2}$

Hunter's expectation:

$y=x\tan\theta$

At monkey's $x$ position, $x_{m}$ the bullet will be $-\frac{1}{2}gt^{2}$ below where monkey started…exactly the same distance the monkey will be below his starting point if he lets go.

This is a simple demonstration of the fact that gravity accelerates all objects equally independently of their horizontal velocity.

## Projectile motion

If $g=9.8ms^{-2}$

 $\Large a_{x}=0$ $\Large a_{y}=-g$ $\Large v_{x}=v_{x0}$ $\Large v_{y}=v_{y0}-gt$ $\Large x=x_0+v_{x0} t$ $\Large y=y_{0}+v_{y0} t -\frac{1}{2}gt^{2}$

## Projectile motion for a "cannon"

 $\Large a_{x}=0$ $\Large a_{y}=-g$ $\Large v_{x}=v_{0}\cos\theta$ $\Large v_{y}=v_{0}\sin\theta-gt$ $\Large x=x_0+v_{0}\cos\theta t$ $\Large y=y_{0}+v_{0}\sin\theta t -\frac{1}{2}gt^{2}$

## Motion path and horizontal range

Taking $x_{0}$,$y_{0}$ as $(0,0)$

$t=\frac{x}{v_{0}\cos\theta}$

$y=\frac{\sin\theta}{\cos\theta}x-\frac{g}{(2v_{0}^{2}\cos^2\theta)} x^{2}$

Horizontal Range → Solve for $y=0$

$0=x(\frac{\sin\theta}{\cos\theta}-\frac{g}{(2v_{0}^{2}\cos^2\theta)}x)$

$x=0$ and $x=\frac{2v_{0}^2\sin\theta\cos\theta}{g}$ → $R=\frac{2v_{0}^2\sin\theta\cos\theta}{g}$

## Measurement of initial velocity

0.04/.0095=4.2m/s

## Projectile motion video

Note that in terms of the equations shown previously $\theta=(180^o-angle)$ as the motion is from right to left.

## Angle for which maximum range is acheived

$R=\frac{2v_{0}^2\sin\theta\cos\theta}{g}=\frac{v_{0}^{2}}{g}\sin 2\theta$

$\frac{dR}{d\theta}=\frac{2v_{0}^2}{g}(\frac{d(\sin\theta)}{d\theta}\cos\theta)+\sin\theta\frac{d(\cos\theta)}{d\theta})$

$=\frac{2v_{0}^2}{g}(\cos^2\theta-\sin^2\theta)=\frac{2v_{0}^2}{g}\cos2\theta$

or

$\frac{dR}{d\theta}=\frac{v_{0}^2}{g}\frac{d(\sin{2\theta})}{d\theta}\frac{d(2\theta)}{d\theta}=\frac{2v_{0}^2}{g}\cos2\theta$

$\frac{dR}{d\theta}=0$ at $\theta=45^o,135^o$

$45^o$ and $135^o$ correspond to maximum range.

## Range for a straight shot from height h

From our equations

 $\Large a_{x}=0$ $\Large a_{y}=-g$ $\Large v_{x}=v_{x0}$ $\Large v_{y}=v_{y0}-gt$ $\Large x=x_0+v_{x0} t$ $\Large y=y_{0}+v_{y0} t -\frac{1}{2}gt^{2}$

we can derive that horizontal distance traveled is

$x= v_{x0}\Large \sqrt{\frac{2h}{g}}$