$v=f\lambda$


v 
: speed (m.s^{1}) 
$\lambda$ 
: wavelength (m) 
f 
: frequency (Hz or s^{1}) 
Is this correct? Remember a simple first check is to check the units! On the right hand side we have speed which has units ms^{1}. On the left hand side we have frequency which is measured in s^{1} multiplied by wavelength which is measure in m. On the left hand side we have ms^{1} which is exactly what we want.
Speed of a wave through strings[edit]
The speed of a wave traveling along a vibrating string (v) is directly proportional to the square root of the tension (T) over the linear density (μ):
 $v={\sqrt {\frac {T}{\mu }}}\,$
μ is equal to the mass of the string divided by the length of the string.
$\mu ={\frac {M}{L}}$
Two Types of Waves[edit]
We agreed that a wave was a moving set of peaks and troughs and we used water as an example. Moving peaks and troughs, with all the characteristics we described, in any medium constitute a wave. It is possible to have waves where the peaks and troughs are perpendicular to the direction of motion, like in the case of water waves. These waves are called transverse waves.
There are two additional types of waves. The first is called longitudinal waves and have the peaks and troughs in the same direction as the wave is moving. The question is how do we construct such a wave?
An example of a longitudinal wave is pressure waves moving through a gas. The peaks in this wave are places where the pressure reaches a peak and the troughs are places where the pressure is a minimum.
In the picture below we show the random placement of the gas molecules in a tube. The piston at the end moves into the tube with a repetitive motion. Before the first piston stroke the pressure is the same throughout the tube.
When the piston moves in it compresses the gas molecules together at the end of the tube. If the piston stopped moving the gas molecules would all bang into each other and the pressure would increase in the tube.
When the piston moves out again before the molecules have time to bang around then the increase in pressure moves down the tube like a pulse (single peak and trough, a single wave cycle).
As this repeats we get waves of increased and decreased pressure moving down the tubes. We can describe these pulses of increased pressure (peaks in the pressure) and decreased pressure (troughs of pressure) by a sine or cosine graph.
The second additional type of wave is the torsional wave. The peaks and troughs rotate around the direction of motion. In simpler terms, a "twisting motion" is transmitted through the medium. Of the two wave types, this is the hardest one to describe and visualize.
There are a number of examples of each type of wave. Not all can be seen with the naked eye but all can be detected.
Properties of Waves[edit]
We have discussed some of the simple properties of waves that we need to know. These have just been describing the characteristics that waves have. Now we can progress onto some more interesting and, perhaps, less intuitive properties of waves.
Properties of Waves : Reflection[edit]
When waves strike a barrier they are reflected. This means that waves bounce off things. Sound waves bounce off walls, light waves bounce off mirrors, radar waves bounce off planes and how bats can fly at night and avoid things as small as telephone wires. etc. The property of reflection is a very important and useful one.
(NOTE TO SELF: Get an essay by an air traffic controller on radar) (NOTE TO SELF: Get an essay by on sonar usage for fishing or for submarines)
When waves are reflected, the process of reflection has certain properties. If a wave hits an obstacle at a right angle to the surface (NOTE TO SELF: diagrams needed) then the wave is reflected directly backwards.
If the wave strikes the obstacle at some other angle then it is not reflected directly backwards. The angle that the waves arrives at is the same as the angle that the reflected waves leaves at. The angle that waves arrives at or is incident at equals the angle the waves leaves at or is reflected at. Angle of incidence equals angle of reflection
${\begin{matrix}\theta _{i}=\theta _{r}\end{matrix}}$ 
(2.1)

${\begin{matrix}\theta _{i}=\theta _{r}\end{matrix}}$ 


$\theta _{i}$

: angle of incidence 
$\theta _{r}$

: angle of reflection 
In the optics chapter you will learn that light is a wave. This means that all the properties we have just learnt apply to light as well. Its very easy to demonstrate reflection of light with a mirror. You can also easily show that angle of incidence equals angle of reflection.
If you look directly into you see yourself ....
Need to mention that the incident wave, normal to the surface and the reflected wave all lie in the same plane. The same also holds for refraction at a surface.
Phase shift of reflected wave[edit]
When a wave is reflected from a more dense medium, it undergoes a phase shift. That means that the peaks and troughs are swapped around.
The easiest way to demonstrate this is to tie a piece of string to something. Stretch the string out flat and then flick the string once so a pulse moves down the string. When the pulse (a single peak in a wave) hits the barrier that the string is tied to, it will be reflected. The reflected wave will look like a trough instead of a peak. This is because the pulse had undergone a phase change. The fixed end is like reflection off a more dense medium.
If the end of the string was not fixed, i.e. it could move up and down then the wave would still be reflected but it would not undergo a phase shift.
Properties of Waves : Refraction[edit]
Sometimes waves move from one medium to another. The medium is the substance that is carrying the waves. In our first example this was the water. When the medium properties change it can affect the wave.
Let us start with the simple case of a water wave moving from one depth to another. The speed of the wave depends on the depth. If the wave moves directly from the one medium to the other than we should look closely at the boundary. When a peak arrives at the boundary and moves across it must remain a peak on the other side of the boundary. This means that the peaks pass by at the same time intervals on either side of the boundary. The period and frequency remain the same! But we said the speed of the wave changes, which means that the distance it travels in one time interval is different i.e. the wavelength has changed.
Going from one medium to another the period or frequency does not change only the wavelength can change.
Now if we consider a water wave moving at an angle of incidence not 90 degrees towards a change in medium then we immediately know that not the whole wave will arrive at once. So if a part of the wave arrives and slows down while the rest is still moving faster before it arrives the angle of the wavefront is going to change. This is known as refraction. When a wave bends or changes its direction when it goes from one medium to the next.
If it slows down it turns towards the perpendicular.
If the wave speeds up in the new medium it turns away from the perpendicular to the medium surface.
When you look at a stick that emerges from water it looks like it is bent. This is because the light from below the surface of the water bends when it leaves the water. Your eyes project the light back in a straight line and so the object looks like it is a different place.
Properties of Waves : Interference[edit]
If two waves meet interesting things can happen. Waves are basically collective motion of particles. So when two waves meet they both try to impose their collective motion on the particles. This can have quite different results.
If two identical (same wavelength, amplitude and frequency) waves are both trying to form a peak then they are able to achieve the sum of their efforts. The resulting motion will be a peak which has a height which is the sum of the heights of the two waves. If two waves are both trying to form a trough in the same place then a deeper trough is formed, the depth of which is the sum of the depths of the two waves. Now in this case the two waves have been trying to do the same thing and so add together constructively. This is called constructive interference.
If one wave is trying to form a peak and the other is trying to form a trough then they are competing to do different things. In this case they can cancel out. The height of the peak less the depth of the trough will be the resulting effect. If the depth of the trough is the same as the height of the peak nothing will happen. If the height of the peak is bigger than the depth of the trough a smaller peak will appear and if the trough is deeper then a less deep trough will appear. This is destructive interference.
File:Fhsst waves23.png
Properties of Waves : Standing Waves[edit]
When two waves move in opposite directions, through each other, constructive interference happens. If the two waves have the same frequency and wavelength then a specific type of constructive interference can occur: standing waves can form.
Standing waves are disturbances which don't appear to move, they stand in the same place. Lets demonstrate exactly how this comes about. Imagine a long string with waves being sent down it from either end. The waves from both ends have the same amplitude, wavelength and frequency as you can see in the picture below:
To stop from getting confused between the two waves we'll draw the wave from the left with a dashed line and the one from the right with a solid line. As the waves move closer together when they touch both waves have an amplitude of zero:
If we wait for a short time the ends of the two waves move past each other and the waves overlap. Now we know what happens when two waves overlap, we add them together to get the resulting wave.
Now we know what happens when two waves overlap, we add them together to get the resulting wave. In this picture we show the two waves as dotted lines and the sum of the two in the overlap region is shown as a solid line:
The important thing to note in this case is that there are some points where the two waves always destructively interfere to zero. If we let the two waves move a little further we get the picture below:
Again we have to add the two waves together in the overlap region to see what the sum of the waves looks like.
In this case the two waves have moved half a cycle past each other but because they are out of phase they cancel out completely. The point at 0 will always be zero as the two waves move past each other.
When the waves have moved past each other so that they are overlapping for a large region the situation looks like a wave oscillating in place. If we focus on the range 4, 4 once the waves have moved over the whole region. To make it clearer the arrows at the top of the picture show peaks where maximum positive constructive interference is taking place. The arrows at the bottom of the picture show places where maximum negative interference is taking place.
As time goes by the peaks become smaller and the troughs become shallower but they do not move.
File:Fhsst waves33.png
For an instant the entire region will look completely flat.
File:Fhsst waves34.png
The various points continue their motion in the same manner.
File:Fhsst waves35.png
Eventually the picture looks like the complete reflection through the xaxis of what we started with:
File:Fhsst waves36.png
Then all the points begin to move back. Each point on the line is oscillating up and down with a different amplitude.
If we superimpose the two cases where the peaks where at a maximum and the case where the same waves where at a minimum we can see the lines that the points oscillate between. We call this the envelope of the standing wave as it contains all the oscillations of the individual points. A node is a place where the two waves cancel out completely as two waves destructively interfere in the same place. An antinode is a place where the two waves constructively interfere.
To make the concept of the envelope clearer let us draw arrows describing the motion of points along the line.
Every point in the medium containing a standing wave oscillates up and down and the amplitude of the oscillations depends on the location of the point. It is convenient to draw the envelope for the oscillations to describe the motion. We cannot draw the up and down arrows for every single point!
Reflection from a fixed end[edit]
If waves are reflected from a fixed end, for example tying the end of a rope to a pole and then sending waves down it. The fixed end will always be a node. Remember: Waves reflected from a fixed end undergo a phase shift.
The wavelength, amplitude and speed of the wave cannot affect this, the fixed end is always a node.
Reflection from an open end[edit]
If waves are reflected from end, which is free to move, it is an antinode. For example tying the end of a rope to a ring, which can move up and down, around the pole. Remember: The waves sent down the string are reflected but do not suffer a phase shift.
Wavelengths of standing waves with fixed and open ends[edit]
If the waves that are interfering are not identical then the waves form a modulated pattern with a changing amplitude. The peaks in amplitude are called beats. If you consider two sound waves interfering then you hear sudden beats in loudness or intensity of the sound.
The simplest illustration is to draw two different waves and then add them together. You can do this mathematically and draw them yourself to see the pattern that occurs.
Here is wave 1:
Now we add this to another wave, wave 2:
When the two waves are added (drawn in coloured dashed lines) you can see the resulting wave pattern:
To make things clearer the resulting wave without the dashed lines is drawn below. Notice that the peaks are the same distance apart but the amplitude changes. If you look at the peaks they are modulated i.e. the peak amplitudes seem to oscillate with another wave pattern. This is what we mean by modulation.
The maximum amplitude that the new wave gets to is the sum of the two waves just like for constructive interference. Where the waves reach a maximum it is constructive interference.
The smallest amplitude is just the difference between the amplitudes of the two waves, exactly like in destructive interference.
The beats have a frequency which is the difference between the frequency of the two waves that were added. This means that the beat frequency is given by
$f_{B}=f_{1}f_{2}$

(2.2)

$f_{B}=f_{1}f_{2}$



f_{B} 
: beat frequency (Hz or s^{1}) 
f_{1} 
: frequency of wave 1 (Hz or s^{1}) 
f_{2} 
: frequency of wave 2 (Hz or s^{1}) 
Properties of Waves : Diffraction[edit]
One of the most interesting, and also very useful, properties of waves is diffraction. When a wave strikes a barrier with a hole, only part of the wave can move through the hole. If the hole is similar in size to the wavelength of the wave diffractions occurs. The waves that comes through the hole no longer looks like a straight wave front. It bends around the edges of the hole. If the hole is small enough it acts like a point source of circular waves.
This bending around the edges of the hole is called diffraction. To illustrate this behaviour we start with Huygen's principle.
Huygen's Principle[edit]
Huygen's principle states that each point on a wavefront acts like a point source or circular waves. The waves emitted from each point interfere to form another wavefront on which each point forms a point source. A long straight line of points emitting waves of the same frequency leads to a straight wave front moving away.
To understand what this means lets think about a whole lot of peaks moving in the same direction. Each line represents a peak of a wave.
If we choose three points on the next wave front in the direction of motion and make each of them emit waves isotropically (i.e. the same in all directions) we will get the sketch below:
What we have drawn is the situation if those three points on the wave front were to emit waves of the same frequency as the moving wave fronts. Huygens principle says that every point on the wave front emits waves isotropically and that these waves interfere to form the next wave front.
To see if this is possible we make more points emit waves isotropically to get the sketch below:
You can see that the lines from the circles (the peaks) start to overlap in straight lines. To make this clear we redraw the sketch with dashed lines showing the wavefronts which would form. Our wavefronts are not perfectly straight lines because we didn't draw circles from every point. If we had it would be hard to see clearly what is going on.
Huygen's principle is a method of analysis applied to problems of wave propagation. It recognizes that each point of an advancing wave front is in fact the center of a fresh disturbance and the source of a new train of waves and that the advancing wave as a whole may be regarded as the sum of all the secondary waves arising from points in the medium already traversed. This view of wave propagation helps better understand a variety of wave phenomena, such as diffraction.
Wavefronts Moving Through an Opening[edit]
Now if we allow the wavefront to impinge on a barrier with a hole in it, then only the points on the wavefront that move into the hole can continue emitting forward moving waves  but because a lot of the wavefront have been removed the points on the edges of the hole emit waves that bend round the edges.
The wave front that impinges (strikes) the wall cannot continue moving forward. Only the points moving into the gap can. If you employ Huygens' principle you can see the effect is that the wavefronts are no longer straight lines.
File:Fhsst waves51.png
Riaan Note: still cant find this image, have to get it from the pdf
For example, if two rooms are connected by an open doorway and a sound is produced in a remote corner of one of them, a person in the other room will hear the sound as if it originated at the doorway. As far as the second room is concerned, the vibrating air in the doorway is the source of the sound. The same is true of light passing the edge of an obstacle, but this is not as easily observed because of the short wavelength of visible light.
This means that when waves move through small holes they appear to bend around the sides because there aren't enough points on the wavefront to form another straight wavefront. This is bending round the sides we call diffraction.
Properties of Waves : Dispersion[edit]
Dispersion is a property of waves where the speed of the wave through a medium depends on the wavelength. So if two waves enter the same dispersive medium and have different wavelengths they will have different speeds in that medium even if they both entered with the same speed.
Practical Applications of Waves: Sound Waves[edit]
Doppler Effect[edit]
The Doppler Effect is an interesting phenomenon that occurs when an object producing sound is moved relatively to the listener.
Consider the following: When a car blaring its horn is behind you, the pitch is higher as it is approaching, and becomes lower as it is moving away. This is only noticeable if the object is moving at a fairly high speed, although it is still theoretically present at any speed.
When an object is moving away from the listener, the sound waves are stretched over a further distance meaning they happen less often. The wavelength ends up being greater so the frequency is less and the pitch is lower. When an object is moving towards the listener, the waves are compressed over a small distance making a very small wavelength and therefore a large frequency and high pitch. Since the pitch of the sound depends on the frequency of the waves, the pitch increases when the object is moving towards the listener.
$f'=f({\frac {v\pm v_{0}}{v\mp v_{s}}})$

f' is the observed frequency, f is the actual frequency, v is the speed of sound ($v=336+0.6T$) T is temperature in degrees Celsius, $v_{0}$ is the speed of the observer, and $v_{s}$ is the speed of the source. If the observer is approaching the source, use the top operator (the +) in the numerator, and if the source is approaching the observer, use the top operator (the ) in the denominator. If the observer is moving away from the source, use the bottom operator (the ) in the numerator, and if the source is moving away from the observer, use the bottom operator (the +) in the denominator.
Example problems[edit]
A. An ambulance, which is emitting a 40 Hz siren, is moving at a speed of 30 m/s towards a stationary observer. The speed of sound in this case is 339 m/s.
$f'=40({\frac {339+0}{33930}})$
B. An M551 Sheridan, moving at 10 m/s is following a Renault FT17 which is moving in the same direction at 5 m/s and emitting a 30 Hz tone. The speed of sound in this case is 342 m/s.
$f'=30({\frac {342+10}{342+5}})$
UltraSound[edit]
still to be completed
Ultrasound is sound that has too high a frequency for humans to hear. Some other animals can hear ultrasound though. Dog whistles are an example of ultrasound. We can't hear the sound, but dogs can. Audible sound is in the frequency range between 20 Hz and 20000 Hz. Anything above that is ultrasound, and anything below that is called infrasonic.
Ultrasound also has medical applications. It can be used to generate images with a sonogram. Ultrasound is commonly used to look at fetuses in the womb.

This page or section is an undeveloped draft or outline.
You can help to develop the work, or you can ask for assistance in the project room. 
Practical Applications of Waves: Electromagnetic Waves[edit]
In physics, waveparticle duality holds that light and matter simultaneously exhibit properties of waves and of particles. This concept is a consequence of quantum mechanics.
In 1905, Einstein reconciled Huygens' view with that of Newton. He explained the photoelectric effect (an effect in which light did not seem to act as a wave) by postulating the existence of photons, quanta of energy with particulate qualities. Einstein postulated that the frequency of light, $f$, is related to the energy, $E$, of its photons:
${\begin{matrix}E=hf\end{matrix}}$ 
(2.3)

where $h$ is Planck's constant ($6.626\times 10^{34}Js$).
In 1924, De Broglie claimed that all matter has a wavelike nature. He related wavelength $\lambda$ and momentum p:
${\begin{matrix}\lambda ={\frac {h}{p}}\end{matrix}}$ 
(2.4)

This is a generalization of Einstein's equation above, since the momentum of a photon is given by
${\begin{matrix}p={\frac {E}{c}},\end{matrix}}$ 
(2.5)

where $c$ is the speed of light in vacuum, and $f={\frac {c}{\lambda }}$.
De Broglie's formula was confirmed three years later by guiding a beam of electrons (which have rest mass) through a crystalline grid and observing the predicted interference patterns. Similar experiments have since been conducted with neutrons and protons. Authors of similar recent experiments with atoms and molecules claim that these larger particles also act like waves. This is still a controversial subject because these experimenters have assumed arguments of waveparticle duality and have assumed the validity of de Broglie's equation in their argument.
The Planck constant h is extremely small and that explains why we don't perceive a wavelike quality of everyday objects: their wavelengths are exceedingly small. The fact that matter can have very short wavelengths is exploited in electron microscopy.
In quantum mechanics, the waveparticle duality is explained as follows: every system and particle is described by state functions which encode the probability distributions of all measurable variables. The position of the particle is one such variable. Before an observation is made the position of the particle is described in terms of probability waves which can interfere with each other.
Important Equations and Quantities[edit]
Frequency:
${\begin{matrix}f={\frac {1}{T}}.\end{matrix}}$ 
(2.6)

Speed:
${\begin{matrix}v&=&f\lambda \\&=&{\frac {\lambda }{T}}\end{matrix}}$
Table 2.1: Units used in Waves
Quantity 
Symbol 
S.I. Units 
Direction 
Amplitude 
A 
m 
 
Period 
T 
s 
 
Wavelength 
$\lambda$ 
m 
 
Frequency 
f 
Hz or s^{1} 
 
Speed 
v 
m.s^{1} 
 
Vectors[edit]
Introduction[edit]
``A vector is `something' that has both magnitude and direction. ```Thing'? What sorts of `thing'?" Any piece of information which contains a magnitude and a related direction can be a vector. A vector should tell you how much and which way.
Consider a man driving his car east along a highway at 100 km/h. What we have given here is a vector — the car's velocity. The car is moving at 100 km/h (this is the magnitude) and we know where it is going — east (this is the direction). Thus, we know the speed and direction of the car. These two quantities, a magnitude and a direction, form a vector we call velocity.
Definition: A vector is a measurement which has both magnitude and direction.
In physics, magnitudes often have directions associated with them. If you push something it is not very useful knowing just how hard you pushed. A direction is needed too. Directions are extremely important, especially when dealing with situations more complicated than simple pushes and pulls.
Different people like to write vectors in different ways. Any way of writing a vector so that it has both magnitude and direction is valid.
Are vectors physics? No, vectors themselves are not physics. Physics is just a description of the world around us. To describe something we need to use a language. The most common language used to describe physics is mathematics. Vectors form a very important part of the mathematical description of physics, so much so that it is absolutely essential to master the use of vectors.
Mathematical representation[edit]
Numerous notations are commonly used to denote vectors. In this text, vectors will be denoted by symbols capped with an arrow. As an example, ${\overrightarrow {s}}$, ${\overrightarrow {v}}$ and ${\overrightarrow {F}}$are all vectors (they have both magnitude and direction). Sometimes just the magnitude of a vector is required. In this case, the arrow is omitted. In other words, F denotes the magnitude of vector ${\overrightarrow {F}}$. ${\overrightarrow {F}}$ is another way of representing the size of a vector.
Graphical representation[edit]
Graphically vectors are drawn as arrows. An arrow has both a magnitude (how long it is) and a direction (the direction in which it points). For this reason, arrows are vectors.
In order to draw a vector accurately we must specify a scale and include a reference direction in the diagram. A scale allows us to translate the length of the arrow into the vector's magnitude. For instance if one chose a scale of 1cm = 2N (1cm represents 2N), a force of magnitude 20N would be represented as an arrow 10cm long. A reference direction may be a line representing a horizontal surface or the points of a compass.
Worked Example 2: Drawing vectors[edit]
Question: Using a scale of $1cm=2m.s^{1}$ represent the following velocities:
a) $6m.s^{1}$ north
b) $16m.s^{1}$ east
Answer:
Some Examples of Vectors[edit]
Displacement[edit]
Imagine you walked from your house to the shops along a winding path through the veld. Your route is shown in blue in Figure 3.1. Your sister also walked from the house to the shops, but she decided to walk along the pavements. Her path is shown in red and consisted of two straight stretches, one after the other.

Figure 3.1: Illustration of Displacement 
Although you took very different routes, both you and your sister walked from the house to the shops. The overall effect was the same! Clearly the shortest path from your house to the shops is along the straight line between these two points. The length of this line and the direction from the start point (the house) to the end point (the shops) forms a very special vector known as displacement. Displacement is assigned the symbol ${\overrightarrow {s}}$
Definition: Displacement is defined as the magnitude and direction of the straight line joining one's starting point to one's final point.
OR
Definition: Displacement is a vector with direction pointing from some initial (starting) point to some final (end) point and whose magnitude is the straightline distance from the starting point to the end point.
(NOTE TO SELF: choose one of the above)
In this example both you and your sister had the same displacement. This is shown as the black arrow in Figure 3.1. Remember displacement is not concerned with the actual path taken. It is only concerned with your start and end points. It tells you the length of the straightline path between your start and end points and the direction from start to finish. The distance travelled is the length of the path followed and is a scalar (just a number). Note that the magnitude of the displacement need not be the same as the distance travelled. In this case the magnitude of your displacement would be considerably less than the actual length of the path you followed through the veld!
Velocity[edit]
Definition: Velocity is the rate of change of displacement with respect to time.
The terms rate of change and with respect to are ones we will use often and it is important that you understand what they mean. Velocity describes how much displacement changes for a certain change in time.
We usually denote a change in something with the symbol $\Delta$ (the Greek letter Delta). You have probably seen this before in maths — the gradient of a straight line is ${\frac {\Delta y}{\Delta x}}$. The gradient is just how much y changes for a certain change in x. In other words it is just the rate of change of y with respect to x. This means that velocity must be
${\begin{matrix}{\overrightarrow {v}}={\frac {\Delta {\overrightarrow {s}}}{\Delta t}}={\frac {{\overrightarrow {s}}_{final}{\overrightarrow {s}}_{initial}}{t_{final}t_{initial}}}\end{matrix}}$
(NOTE TO SELF: This is actually average velocity. For instantaneous $\Delta$'s change to differentials. Explain that if $\Delta$ is large then we have average velocity else for infinitesimal time interval instantaneous!)
What then is speed? Speed is how quickly something is moving. How is it different from velocity? Speed is not a vector. It does not tell you which direction something is moving, only how fast. Speed is the magnitude of the velocity vector (NOTE TO SELF: instantaneous speed is the magnitude of the instantaneous velocity.... not true of averages!).
Consider the following example to test your understanding of the differences between velocity and speed.
Worked Example 3: Speed and Velocity[edit]
Question: A man runs around a circular track of radius 100m. It takes him 120s to complete a revolution of the track. If he runs at constant speed, calculate:
 his speed,
 his instantaneous velocity at point A,
 his instantaneous velocity at point B,
 his average velocity between points A and B,
 his average velocity during a revolution.
Answer:
 1. To determine the man's speed, we need to know the distance he travels and how long it takes. We know it takes $120s$ to complete one revolution of the track. What distance is one revolution of the track? We know the track is a circle and we know its radius, so we can determine the perimeter or distance around the circle. We start with the equation for the circumference of a circle:
${\begin{matrix}C&=&2\pi r\\&=&2\pi (100m)\\&=&628.3\;m.\end{matrix}}$
 2. Now that we have distance and time, we can determine speed. We know that speed is distance covered per unit time. If we divide the distance covered by the time it took, we will know how much distance was covered for every unit of time.
${\begin{matrix}v&=&{\frac {Distance\ travelled}{time\ taken}}\\&=&{\frac {628.3m}{120s}}\\&=&5.23\ m.s^{1}\end{matrix}}$
 3. Consider point A in the diagram:
We know which way the man is running around the track, and we know his speed. His velocity at point A will be his speed (the magnitude of the velocity) plus his direction of motion (the direction of his velocity). He is moving at the instant that he arrives at A, as indicated in the diagram below.
His velocity vector will be $5.23\ m.s^{1}$ West.
 4. Consider point B in the diagram:
 We know which way the man is running around the track, and we know his speed. His velocity at point B will be his speed (the magnitude of the velocity) plus his direction of motion (the direction of his velocity). He is moving at the instant that he arrives at B, as indicated in the diagram below.
 His velocity vector will be $5.23\ m.s^{1}$ South.
 4. So, now, what is the man's average velocity between Point A and Point B?
As he runs around the circle, he changes direction constantly. (Imagine a series of vector arrows pointing out from the circle, one for each step he takes.) If you add up all these directions and find the average it turns out to be ...Right. South west. And, notice that if you just looked for the average between his velocity at Point A and at Point B, that comes out south west, too. So his average velocity between Point A and Point B is $5.23\ m.s^{1}$ south west.
 5. Now we need to calculate his average velocity over a complete revolution. The definition of average velocity is given earlier and requires that you know the total displacement and the total time. The total displacement for a revolution is given by the vector from the initial point to the final point. If the man runs in a circle, then he ends where he started. This means the vector from his initial point to his final point has zero length. A calculation of his average velocity follows:
${\begin{matrix}{\overrightarrow {v}}&=&{\frac {\Delta {\overrightarrow {s}}}{\Delta t}}\\&=&{\frac {0m}{120s}}\\&=&0\ m.s^{1}\end{matrix}}$
Remember: Displacement can be zero even when distance is not!
Acceleration[edit]
Definition: Acceleration is the rate of change of velocity with respect to time.
Acceleration is also a vector. Remember that velocity was the rate of change of displacement with respect to time so we expect the velocity and acceleration equations to look very similar. In fact:
${\begin{matrix}{\overrightarrow {a}}={\frac {\Delta {\overrightarrow {v}}}{\Delta t}}={\frac {{\overrightarrow {v}}_{final}{\overrightarrow {v}}_{initial}}{t_{final}t_{initial}}}\end{matrix}}$ 
(3.2)

(NOTE TO SELF: average and instantaneous distinction again! expand further — what does it mean?)
Acceleration will become very important later when we consider forces.
Imagine that you and your friend are pushing a cardboard box kept on a smooth floor. Both of you are equally strong. Can you tell me in which direction the box will move ? Probably not. Because I have not told you in which direction each of you are pushing the box. If both of you push it towards north, the box would move northwards. If you push it towards north and you friend pushes it towards east, it would move northeastwards. If you two push it in opposite directions, it wouldn't move at all !
Thus in dealing with force applied on any object, it is equally important to take into account the direction of the force, as the magnitude. This is the case with all vectors.
Mathematical Properties of Vectors[edit]
Vectors are mathematical objects and we will use them to describe physics in the language of mathematics. However, first we need to understand the mathematical properties of vectors (e.g. how they add and subtract).
We will now use arrows representing displacements to illustrate the properties of vectors. Remember that displacement is just one example of a vector. We could just as well have decided to use forces to illustrate the properties of vectors.
Addition of Vectors[edit]
If we define a displacement vector as 2 steps in the forward direction and another as 3 steps in the forward direction then adding them together would mean moving a total of 5 steps in the forward direction. Graphically, this can be seen by first following the first vector two steps forward, and then following the second one three steps forward:
We add the second vector at the end of the first vector, since this is where we now are after the first vector has acted. The vector from the tail of the first vector (the starting point) to the head of the last (the end point) is then the sum of the vectors. This is the tailtohead method of vector addition.
The order in which you add vectors does not matter. In the example above, if you decided to first go 3 steps forward and then another 2 steps forward, the end result would still be 5 steps forward.
The final answer when adding vectors is called the resultant.
Definition: The resultant of a number of vectors is the single vector whose effect is the same as the individual vectors acting together.
In other words, the individual vectors can be replaced by the resultant — the overall effect is the same. If vectors ${\overrightarrow {a}}$ and ${\overrightarrow {b}}$ have a resultant ${\overrightarrow {R}}$, this can be represented mathematically as, ${\begin{matrix}{\overrightarrow {R}}&=&{\overrightarrow {a}}+{\overrightarrow {b}}.\end{matrix}}$
Let us consider some more examples of vector addition using displacements. The arrows tell you how far to move and in what direction. Arrows to the right correspond to steps forward, while arrows to the left correspond to steps backward. Look at all of the examples below and check them.
Let us test the first one. It says one step forward and then another step forward is the same as an arrow twice as long — two steps forward.
It is possible that you end up back where you started. In this case the net result of what you have done is that you have gone nowhere (your start and end points are at the same place). In this case, your resultant displacement is a vector with length zero units. We use the symbol ${\overrightarrow {0}}$ to denote such a vector:
Check the following examples in the same way. Arrows up the page can be seen as steps left and arrows down the page as steps right.
Try a couple to convince yourself!
It is important to realise that the directions aren't special — forward and backwards or left and right are treated in the same way. The same is true of any set of parallel directions:
In the above examples the separate displacements were parallel to one another. However the same tailtohead technique of vector addition can be applied to vectors in any direction.
Now you have discovered one use for vectors; describing resultant displacement — how far and in what direction you have travelled after a series of movements.
Although vector addition here has been demonstrated with displacements, all vectors behave in exactly the same way. Thus, if given a number of forces are acting on a body, you can use the same method to determine the resultant force acting on the body. We will return to vector addition in more detail later.
Subtraction of Vectors[edit]
What does it mean to subtract a vector? Well this is really simple: if we have 5 apples and we subtract 3 apples, we have only 2 apples left. Now lets work in steps — if we take 5 steps forward, and then subtract 3 steps forward, we are left with only two steps forward:
What have we done? You originally took 5 steps forward but then you took 3 steps back. That backward displacement would be represented by an arrow pointing to the left (backwards) with length 3. The net result of adding these two vectors is 2 steps forward:
Thus, subtracting a vector from another is the same as adding a vector in the opposite direction (i.e. subtracting 3 steps forwards is the same as adding 3 steps backwards).
This suggests that in this problem, arrows to the right are positive, and arrows to the left are negative. More generally, vectors in opposite directions differ in sign (i.e. if we define up as positive, then vectors acting down are negative). Thus, changing the sign of a vector simply reverses its direction:
In mathematical form, subtracting ${\overrightarrow {a}}$ from ${\overrightarrow {b}}$ gives a new vector ${\overrightarrow {c}}$
${\begin{matrix}{\overrightarrow {c}}&=&{\overrightarrow {b}}{\overrightarrow {a}}\\&=&{\overrightarrow {b}}+({\overrightarrow {a}})\end{matrix}}$
This clearly shows that subtracting vector ${\overrightarrow {a}}$ from ${\overrightarrow {b}}$ is the same as adding $({\overrightarrow {a}})$ to ${\overrightarrow {b}}$. Look at the following examples of vector subtraction.
Scalar Multiplication[edit]
What happens when you multiply a vector by a scalar (an ordinary number)?
Going back to normal multiplication we know that $2\times 2$ is just 2 groups of 2 added together to give 4. We can adopt a similar approach to understand how vector multiplication works.
Techniques of Vector Addition[edit]
Now that you have been acquainted with the mathematical properties of vectors, we return to vector addition in more detail. There are a number of techniques of vector addition. These techniques fall into two main categories graphical and algebraic techniques.
Graphical Techniques[edit]
Graphical techniques involve drawing accurate scale diagrams to denote individual vectors and their resultants. We next discuss the two primary graphical techniques, the tailtohead technique and the parallelogram method.
The Tailtohead Method[edit]
In describing the mathematical properties of vectors we used displacements and the tailtohead graphical method of vector addition as an illustration. In the tailtohead method of vector addition the following strategy is followed:
 Choose a scale and include a reference direction.
 Choose any of the vectors to be summed and draw it as an arrow in the correct direction and of the correct length remember to put an arrowhead on the end to denote its direction.
 Take the next vector and draw it as an arrow starting from the arrowhead of the first vector in the correct direction and of the correct length.
 Continue until you have drawn each vector each time starting from the head of the previous vector. In this way, the vectors to be added are drawn one after the other tailtohead.
 The resultant is then the vector drawn from the tail of the first vector to the head of the last. Its magnitude can be determined from the length of its arrow using the scale. Its direction too can be determined from the scale diagram.
Worked Example 4 TailtoHead Graphical Addition I[edit]
Question: A ship leaves harbour H and sails 6km north to port A. From here the ship travels 12km east to port B, before sailing 5.5km southwest to port C. Determine the ship's resultant displacement using the tailtohead technique of vector addition.
Answer:
Now, we are faced with a practical issue: in this problem the displacements are too large to draw them their actual length! Drawing a 2km long arrow would require a very big book. Just like cartographers (people who draw maps), we have to choose a scale. The choice of scale depends on the actual question you should choose a scale such that your vector diagram fits the page. Before choosing a scale one should always draw a rough sketch of the problem. In a rough sketch one is interested in the approximate shape of the vector diagram.
Step 1 :
Let us draw a rough sketch of the situation
In a rough sketch one should include all of the information given in the problem. All of the magnitudes of the displacements are shown and a compass has been included as a reference direction.
Step 2 :
Next we choose a scale for our vector diagram. It is clear from the rough sketch that choosing a scale where 1cm represents 1km (scale: 1cm = 1km) would be a good choice in this problem ) the diagram will then take up a good fraction of an A4 page. We now start the accurate construction.
Step 3 :
Construction Step 1: Starting at the harbour H we draw the first vector 6cm long in the direction north (remember in the diagram 1cm represents 1km):
Construction Step 2: Since the ship is now at port A we draw the second vector 12cm long starting from this point in the direction east:
Construction Step 3: Since the ship is now at port B we draw the third vector 5.5cm long starting from this point in the direction southwest. A protractor is required to measure the angle of 45^{o}.
Construction Step 4: As a final step we draw the resultant displacement from the starting point (the harbour H) to the end point (port C). We use a ruler to measure the length of this arrow and a protractor to determine its direction
Step 4 :
We now use the scale to convert the length of the resultant in the scale diagram to the actual displacement in the problem. Since we have chosen a scale of 1cm = 1km in this problem the resultant has a magnitude of 8.38 km. The direction can be specified in terms of the angle measured either as 75.4^{o} east of north or on a bearing of 75.4^{o}.
Step 5 :
Now we can quote the final answer: The resultant displacement of the ship is 8.38 km on a bearing of 75.4^{o}!
Worked Example 5 TailtoHead Graphical Addition II[edit]
Question: A man walks 40 m East, then 30 m North.
a) What was the total distance he walked?
b) What is his resultant displacement?
Answer:
Step 1 :
What distance did the man travel? In the first part of his journey he traveled 40 m and in the second part he traveled 30 m. This gives us a total distance traveled of $40+30=70\ m$
Step 2 :
What is his resultant displacement? The man's resultant displacement is the vector from where he started to where he ended. It is the sum of his two separate displacements. We will use the tailtohead method of accurate construction to find this vector. Firstly, we draw a rough sketch:
File:Fhsst vectors36.png
Step 3 :
Next we choose a scale suitable for the problem. A scale of 1cm represents 5m (1cm = 5m) is a good choice here. Now we can begin the process of construction.
Step 4 :
We draw the first displacement as an arrow 8cm long (according to the scale $8cm=8\times 5m=40m$) in the direction east:
Step 5 :
Starting from the head of the first vector we draw the second displacement as an arrow 6cm long (according to the scale $6cm=6\times 5m=30m$) in the direction north:</math>
Step 6 :
Now we connect the starting point to the end point and measure the length and direction of this arrow (the resultant)
Step 7 :
Finally we use the scale to convert the length of the resultant in the scale diagram to the actual magnitude of the resultant displacement. According to the chosen scale 1cm = 5m. Therefore 10cm represents 50m. The resultant displacement is then 50m 36.9^{o} north of east.
The Parallelogram Method[edit]
When needing to find the resultant of two vectors another graphical technique can be applied the parallelogram method. The following strategy is employed:
 Choose a scale and a reference direction.
 Choose either of the vectors to be added and draw it as an arrow of the correct length in the correct direction.
 Draw the second vector as an arrow of the correct length in the correct direction from the tail of the first vector.
 Complete the parallelogram formed by these two vectors.
 The resultant is then the diagonal of the parallelogram. Its magnitude can be determined from the length of its arrow using the scale. Its direction too can be determined from the scale diagram.
Worked Example 6[edit]
Parallelogram Method of Graphical Addition I
Question: A force of F_{1} = 5 N is applied to a block in a horizontal direction. A second force F_{2} = 4 N is applied to the object at an angle of 30° above the horizontal.
Determine the resultant force acting on the block using the parallelogram method of accurate construction.
Answer:
Step 1 :
Firstly we make a rough sketch of the vector diagram:
Step 2 :
Now we choose a suitable scale. In this problem a scale of 1 cm = 0.5 N would be appropriate, since then the vector diagram would take up a reasonable fraction of the page. We can now begin the accurate scale diagram.
Step 3 :
Let us draw F_{1} first. According to the scale it has length 10 cm:
Step 4 :
Next we draw F_{2}. According to the scale it has length 8 cm. We make use of a protractor to draw this vector at 30° to the horizontal:
Step 5 :
Next we complete the parallelogram and draw the diagonal:
RIAAN NOTE: Image missing img155.png PDF page 51 File:Fhsst vectors43.png
Step 6 :
Finally we use the scale to convert the measured length into the actual magnitude. Since 1 cm = 0.5 N, 17.4 cm represents 8.7 N. Therefore the resultant force is 8.7 N at 13.3° above the horizontal.
The parallelogram method is restricted to the addition of just two vectors. However, it is arguably the most intuitive way of adding two forces acting at a point.
Algebraic Addition and Subtraction of Vectors[edit]
Vectors in a Straight Line[edit]
Whenever you are faced with adding vectors acting in a straight line (i.e. some directed left and some right, or some acting up and others down) you can use a very simple algebraic technique:
 Choose a positive direction. As an example, for situations involving displacements in the directions west and east, you might choose west as your positive direction. In that case, displacements east are negative.
 Next simply add (or subtract) the vectors with the appropriate signs.
 As a final step the direction of the resultant should be included in words (positive answers are in the positive direction, while negative resultants are in the negative direction).
Let us consider a couple of examples.
Worked Example 7[edit]
Adding vectors algebraically I
Question: A tennis ball is rolled towards a wall which is 10m away to the right. If after striking the wall the ball rolls a further 2.5m along the ground to the left, calculate algebraically the ball's resultant displacement.
(NOTE TO SELF: PGCE suggest a `more real looking' diagram, followed by a diagram one would draw to solve the problem (like our existing one with the positive direction shown as an arrow))
Answer:
Step 1 :
Let us draw a picture of the situation:
Step 2 :
We know that the resultant displacement of the ball ( ${\overrightarrow {s}}_{resultant}$) is equal to the sum of the ball's separate displacements (${\overrightarrow {s}}_{1}$ and ${\overrightarrow {s}}_{2}$ ):
${\begin{matrix}{\overrightarrow {s}}_{resultant}&=&{\overrightarrow {s}}_{1}+{\overrightarrow {s}}_{2}\end{matrix}}$
Since the motion of the ball is in a straight line (i.e. the ball moves left and right), we can use the method of algebraic addition just explained.
Step 3 :
First we choose a positive direction. Let's make to the right the positive direction. This means that to the left becomes the negative direction.
Step 4 :
With right positive:
${\begin{matrix}{\overrightarrow {s}}_{1}&=&+10.0m\\&and&\\{\overrightarrow {s}}_{2}&=&2.5m\end{matrix}}$
Step 5 :
Next we simply add the two displacements to give the resultant:
${\begin{matrix}{\overrightarrow {s}}_{resultant}&=&(+10m)+(2.5m)\\&=&(+7.5)m\end{matrix}}$
Step 6 :
Finally, in this case right means positive so:
${\begin{matrix}{\overrightarrow {s}}_{resultant}&=&7.5m{\rm {\ to\ the\ right}}\end{matrix}}$
Let us consider an example of vector subtraction.
Worked Example 8[edit]
Subtracting vectors algebraically I
Question: Suppose that a tennis ball is thrown horizontally towards a wall at 3m.s^{1} to the right. After striking the wall, the ball returns to the thrower at 2m.s^{1}. Determine the change in velocity of the ball.
Answer:
Step 1 :
Remember that velocity is a vector. The change in the velocity of the ball is equal to the difference between the ball's initial and final velocities:
${\begin{matrix}\Delta {\overrightarrow {v}}&=&{\overrightarrow {v}}_{final}{\overrightarrow {v}}_{initial}\end{matrix}}$
Since the ball moves along a straight line (i.e. left and right), we can use the algebraic technique of vector subtraction just discussed.
Step 2 :
Let's make to the right the positive direction. This means that to the left becomes the negative direction.
Step 3 :
With right positive:
${\begin{matrix}{\overrightarrow {v}}_{initial}&=&+3m.s^{1}\\&and&\\{\overrightarrow {v}}_{final}&=&2m.s^{1}\end{matrix}}$
Step 4 :
Thus, the change in velocity of the ball is:
${\begin{matrix}\Delta {\overrightarrow {v}}&=&(2m.s^{1})(+3m.s^{1})\\&=&(5)m.s^{1}\end{matrix}}$
Remember that in this case right means positive so:
${\begin{matrix}\Delta {\overrightarrow {v}}&=&5m.s^{1}{\rm {\textbf {\ to\ the\ {\emph {left}}}}}\end{matrix}}$
Remember that the technique of addition and subtraction just discussed can only be applied to vectors acting along a straight line.
A More General Algebraic technique[edit]
In worked example 3 the tail to head method of accurate construction was used to determine the resultant displacement of a man who travelled first east and then north. However, the man's resultant can be calculated without drawing an accurate scale diagram. Let us revisit this example.
Worked Example 9[edit]
An Algebraic solution to Worked Example 3
Question: A man walks 40 m East, then 30 m North.
 Calculate the man's resultant displacement.
Answer:
Step 1 :
As before, the rough sketch looks as follows:
Step 2 :
Note that the triangle formed by his separate displacement vectors and his resultant displacement vector is a rightangle triangle. We can thus use Pythogoras' theorem to determine the length of the resultant. If the length of the resultant vector is called s then:
${\begin{matrix}s^{2}&=&(40m)^{2}+(30m)^{2}\\s^{2}&=&2500m^{2}\\s&=&50m\\\end{matrix}}$
Step 3 : Now we have the length of the resultant displacement vector but not yet its direction. To determine its direction we calculate the angle $\alpha$ between the resultant displacement vector and East.
We can do this using simple trigonometry:
${\begin{matrix}\tan \alpha &=&{\frac {opposite}{adjacent}}\\\tan \alpha &=&{\frac {30}{40}}\\\alpha &=&\arctan(0.75)\\\alpha &=&36.9^{o}\\\end{matrix}}$
Step 4 :
Our final answer is then:
 Resultant Displacement: 50 m at 36.9^{o} North of East
This is exactly the same answer we arrived at after drawing a scale diagram!
In the previous example we were able to use simple trigonometry to calculate a man's resultant displacement. This was possible since the man's directions of motion were perpendicular (north and east). Algebraic techniques, however, are not limited to cases where the vectors to be combined are along the same straight line or at right angles to one another. The following example illustrates this.
Worked Example 10[edit]
Further example of vector addition by calculation
Question: A man walks from point A to point B which is 12km away on a bearing of 45^{o}. From point B the man walks a further 8km east to point C. Calculate the man's resultant displacement.
Answer:
Step 1 : Let us begin by drawing a rough sketch of the situation
RIAAN NOTE: Image on page 56 is missing File:Fhsst vectors46.png
$B{\hat {A}}F=45^{o}$ since the man walks initially on a bearing of 45^{o}.Then, $A{\hat {B}}G=B{\hat {A}}F=45^{o}$ (alternate angles parallel lines). Both of these angles are included in the rough sketch.
Step 2 :
Now let us calculate the length of the resultant (AC). Since we know both the lengths of $AB$ and $BC$ and the included angle $A{\hat {B}}C$, we can use the cosine rule:
${\begin{matrix}AC^{2}&=&AB^{2}+BC^{2}2\cdot AB\cdot BC\cos(A{\hat {B}}C)\\&=&(12)^{2}+(8)^{2}2\cdot (12)(8)\cos(135^{o})\\&=&343.8\\AC&=&18.5\ km\end{matrix}}$
Step 3 :
Next we use the sine rule to determine the angle $\theta$:
${\begin{matrix}{\frac {\sin \theta }{8}}&=&{\frac {\sin 135^{0}}{18.5}}\\\sin \theta &=&{\frac {8\times \sin 135^{o}}{18.5}}\\\theta &=&\arcsin(0.3058)\\\theta &=&17.8^{o}\end{matrix}}$
Thus, $F{\hat {A}}C=62.8^{o}$
Step 4 :
Our final answer is then:
 Resultant Displacement: 18.5km on a bearing of 62.8^{o}
Components of Vectors[edit]
In the discussion of vector addition we saw that a number of vectors acting together can be combined to give a single vector (the resultant). In much the same way a single vector can be broken down into a number of vectors which when added give that original vector. These vectors which sum to the original are called components of the original vector. The process of breaking a vector into its components is called resolving into components.
While summing a given set of vectors gives just one answer (the resultant), a single vector can be resolved into infinitely many sets of components. In the diagrams below the same black vector is resolved into different pairs of components. These components are shown in red. When added together the red vectors give the original black vector (i.e. the original vector is the resultant of its components).
In practice it is most useful to resolve a vector into components which are at right angles to one another.
Worked Example 11[edit]
Resolving a vector into components
Question: A motorist undergoes a displacement of 250km in a direction 30^{o} north of east. Resolve this displacement into components in the directions north (${\overrightarrow {s}}_{N}$ and east (${\overrightarrow {s}}_{E}$).
Answer:
Step 1 :
Firstly let us draw a rough sketch of the original vector
Step 2 :
Next we resolve the displacement into its components north and east. Since these directions are orthogonal to one another, the components form a rightangled triangle with the original displacement as its hypotenuse:
Notice how the two components acting together give the original vector as their resultant.
Step 3 :
Now we can use trigonometry to calculate the magnitudes of the components of the original displacement:
${\begin{matrix}s_{N}&=&250\sin 30^{o}\\&=&125\ km\end{matrix}}$
and
${\begin{matrix}s_{E}&=&250\cos 30^{o}\\&=&216.5\ km\end{matrix}}$
Remember s_{N} and s_{E} are the magnitudes of the components they are in the directions north and east respectively.
Block on an incline[edit]
As a further example of components let us consider a block of mass m placed on a frictionless surface inclined at some angle $\theta$ to the horizontal. The block will obviously slide down the incline, but what causes this motion?
The forces acting on the block are its weight mg and the normal force N exerted by the surface on the object. These two forces are shown in the diagram below.
Now the object's weight can be resolved into components parallel and perpendicular to the inclined surface. These components are shown as red arrows in the diagram above and are at right angles to each other. The components have been drawn acting from the same point. Applying the parallelogram method, the two components of the block's weight sum to the weight vector.
To find the components in terms of the weight we can use trigonometry:
${\begin{matrix}W_{\}&=&mg\sin \theta \\W_{\perp }&=&mg\cos \theta \end{matrix}}$
The component of the weight perpendicular to the slope W_{$\perp$} exactly balances the normal force N exerted by the surface. The parallel component, however, $W_{\}$ is unbalanced and causes the block to slide down the slope.
Vector addition using components[edit]
In Figure 3.3 two vectors are added in a slightly different way to the methods discussed so far. It might look a little like we are making more work for ourselves, but in the long run things will be easier and we will be less likely to go wrong.
In Figure 3.3 the primary vectors we are adding are represented by solid lines and are the same vectors as those added in Figure 3.2 using the less complicated looking method.
Figure 3.2:An example of two vectors being added to give a resultant
Each vector can be broken down into a component in the xdirection and one in the ydirection. These components are two vectors which when added give you the original vector as the resultant. Look at the red vector in figure 3.3. If you add up the two red dotted ones in the xdirection and ydirection you get the same vector. For all three vectors we have shown their respective components as dotted lines in the same colour.
But if we look carefully, addition of the x components of the two original vectors gives the x component of the resultant. The same applies to the y components. So if we just added all the components together we would get the same answer! This is another important property of vectors.
Worked Example 12[edit]
Adding Vectors Using Components
Question: Lets work through the example shown in Figure 3.3 to determine the resultant.
Answer:
Step 1 :
The first thing we must realise is that the order that we add the vectors does not matter. Therefore, we can work through the vectors to be added in any order.
Step 2 :
Let us start with the bottom vector. If you are told that this vector has a length of 5.385 units and an angle of 21.8^{o} to the horizontal then we can find its components. We do this by using known trigonometric ratios. First we find the vertical or y component:
${\begin{matrix}\sin \theta &=&{\frac {y}{\mbox{hypotenuse}}}\\\sin(21.8)&=&{\frac {y}{5.385}}\\y&=&5.385\sin(21.8)\\y&=&2\end{matrix}}$
Secondly we find the horizontal or x component:
${\begin{matrix}\cos \theta &=&{\frac {x}{\mbox{hypotenuse}}}\\\cos(21.8)&=&{\frac {x}{5.385}}\\x&=&5.385\cos(21.8)\\x&=&5\end{matrix}}$
We now know the lengths of the sides of the triangle for which our vector is the hypotenuse. If you look at these sides we can assign them directions given by the dotted arrows. Then our original red vector is just the sum of the two dotted vectors (its components). When we try to find the final answer we can just add all the dotted vectors because they would add up to the two vectors we want to add.
Step 3 :
Now we move on to considering the second vector. The green vector has a length of 5 units and a direction of 53.13 degrees to the horizontal so we can find its components.
${\begin{matrix}\sin \theta &=&{\frac {y}{\mbox{hypotenuse}}}\\\sin(53.13)&=&{\frac {y}{5}}\\y&=&5\sin(53.13)\\y&=&4\end{matrix}}$
${\begin{matrix}\cos \theta &=&{\frac {x}{\mbox{hypotenuse}}}\\\cos(53.13)&=&{\frac {x}{5}}\\x&=&5\cos(53.13)\\x&=&3\end{matrix}}$
Step 4 :
Now we have all the components. If we add all the xcomponents then we will have the xcomponent of the resultant vector. Similarly if we add all the ycomponents then we will have the ycomponent of the resultant vector.
The xcomponents of the two vectors are 5 units right and then 3 units right. This gives us a final xcomponent of 8 units right.
The ycomponents of the two vectors are 2 units up and then 4 units up. This gives us a final ycomponent of 6 units up.
Step 5 :
Now that we have the components of the resultant, we can use Pythagoras' theorem to determine the length of the resultant. Let us call the length of the hypotenuse l and we can calculate its value
${\begin{matrix}l^{2}&=&(6)^{2}+(8)^{2}\\l^{2}&=&100\\l&=&10.\\\end{matrix}}$
The resultant has length of 10 units so all we have to do is calculate its direction. We can specify the direction as the angle the vectors makes with a known direction. To do this you only need to visualize the vector as starting at the origin of a coordinate system. We have drawn this explicitly below and the angle we will calculate is labeled $\alpha$.
Using our known trigonometric ratios we can calculate the value of $\alpha$
${\begin{matrix}\tan \alpha &=&{\frac {6}{8}}\\\alpha &=&\arctan {\frac {6}{8}}\\\alpha &=&36.8^{o}.\end{matrix}}$
Step 6 :
Our final answer is a resultant of 10 units at 36.8^{o} to the positive xaxis.
Do I really need to learn about vectors? Are they really useful?[edit]
Vectors are essential to do physics. Absolutely essential. This is an important warning. If something is essential we had better stop for a moment and make sure we understand it properly.
Summary of Important Quantities, Equations and Concepts[edit]
Table 3.1:Summary of the symbols and units of the quantities used in Vectors

Quantity 
Symbol 
S.I. Units 
Direction 
Displacement 
${\overrightarrow {s}}$ 
m 
yes 
Velocity 
${\overrightarrow {u}}$ ${\overrightarrow {v}}$ 
m.s^{1} 
yes 
Distance 
d 
m 
 
Speed 
v 
m.s^{1} 
 
Acceleration 
${\overrightarrow {a}}$ 
m.s^{2} 
yes 
Vector: A vector is a measurement which has both magnitude and direction.
Displacement: Displacement is a vector with direction pointing from some initial (starting) point to some final (end) point and whose magnitude is the straightline distance from the starting point to the final point.
Distance: The distance traveled is the length of your actual path.
Velocity: Velocity is the rate of change of displacement with respect to time.
Acceleration: Acceleration is the rate of change of velocity with respect to time.
Resultant: The resultant of a number of vectors is the single vector whose effect is the same as the individual vectors acting together
What is a force?[edit]
The simplest way to describe force is to say that it is a `push' or a `pull'. The push or pull on an object may cause either deformation or may change the state of motion of the object under consideration. The harder you 'push' or 'pull', the more force you are applying.
If we leave aside the deformation aspects, then force can be considered to produce change in the state of the motion of the object i.e. velocity. We have, though, experienced in real life that a 'push' or 'pull' does not always manifest in the change of motion. The reason is simple. A change in the state of motion requires a net force. For example, if the force is great enough to overcome friction the object being pushed or pulled will move. So long as the forces on an object are balanced (i.e. net force is zero), the state of motion described by "velocity" will remain same.
In fact, the acceleration of a body is directly proportional to the net force acting on it. The word net is important forces are vectors and what matters in any situation is the vector sum of all the forces acting on an object.
The unit of force is the newton (symbol N)

Force was first described by Archimedes. Archimedes of Syracuse (circa 287 BC  212 BC), was a Greek mathematician, astronomer, philosopher, physicist and engineer. He was killed by a Roman soldier during the sack of the city, despite orders from the Roman general, Marcellus, that he was not to be harmed.
EQUILIBRIUM OF FORCES.
DEFINITIONThe Equilibrium of forces is the single fore required to produce an equilibrium When only three forces act on an object this closed figure is a triangle. This leads to the triangles law for three forces in equilibrium.
Force diagrams[edit]
The resultant force acting on an object is the vector sum of the set of forces acting on that one object. It is very important to remember that we consider all the forces that act on the object under consideration  not the forces that the object might, in turn, apply on other objects.
The easiest way to determine this resultant force is to construct what we call a force diagram. In a force diagram we represent the object by a point and draw all the force vectors connected to that point as arrows. Remember from the Vectors chapter that we use the length of the arrow to indicate the vector's magnitude and the direction of the arrow to show which direction it acts in.
The second step is to rearrange the force vectors so that it is easy to add them together and find the resultant force.
Let us consider an example to get started:
Two people push on a box from opposite sides with a force of 5 N.
When we draw the force diagram we represent the box by a dot. The two forces are represented by arrows, with their tails on the dot.
See how the arrows point in opposite directions and have the same magnitude (length). This means that they cancel out and there is no net force acting on the object.
This result can be obtained algebraically too, since the two forces act along the same line. Firstly we choose a positive direction and then add the two vectors taking their directions into account.
Considering direction towards right as the positive direction
${\begin{matrix}F_{res}&=&(+5{\mbox{ N}})+(5{\mbox{ N}})\\&=&0N\end{matrix}}$
As you work with more complex force diagrams, in which the forces do not exactly balance, you may notice that sometimes you get a negative answer (e.g. 2 N). What does this mean? Does it mean that we have something which is opposite of the force? No, all it means is that the force acts in the opposite direction to the one that you chose to be positive. You can choose the positive direction to be any way you want, but once you have chosen it you must stick with it.
Once a force diagram has been drawn the techniques of vector addition introduced in the previous chapter can be implemented. Depending on the situation you might choose to use a graphical technique such as the tailtohead method or the parallelogram method, or else an algebraic approach to determine the resultant. Since force is a vector, all of these methods apply!
Always remember to check your signs
Worked Example 13 Single Force on a block[edit]
Question: A block on a frictionless flat surface weighs 100 N. A 75 N force is applied to the block towards the right. What is the net force (or resultant force) on the block?
Answer:
Step 1 : Firstly let us draw a force diagram for the block:
File:Fhsst forces4.png
RIAAN Note image on page 68 is missing
Be careful not to forget the two forces perpendicular to the surface. Every object with mass is attracted to the centre of the earth with a force (the object's weight). However, if this were the only force acting on the block in the vertical direction then the block would fall through the table to the ground. This does not happen because the table exerts an upward force (the normal force) which exactly balances the object's weight.
Step 2 :
Thus, the only unbalanced force is the applied force. This applied force is then the resultant force acting on the block.
Equilibrium of Forces[edit]
At the beginning of this chapter it was mentioned that resultant forces cause objects to accelerate. If an object is stationary or moving at constant velocity then either:
 no forces are acting on the object, or
 the forces acting on that object are exactly balanced.
A resultant force would cause a stationary object to start moving or an object moving with a given velocity to speed up or slow down or change direction such that the velocity of the object changes.
In other words, for stationary objects or objects moving with constant velocity, the resultant force acting on the object is zero. The object is said to be in equilibrium.
If a resultant force acts on an object then that object can be brought into equilibrium by applying an additional force that exactly balances this resultant. Such a force is called the equilibrant and is equal in magnitude but opposite in direction to the original resultant force acting on the object.
Definition: The equilibrant of any number of forces is the single force required to produce equilibrium.
As an example of an object in equilibrium, consider an object held stationary by two ropes in the arrangement below:
Let us draw a force diagram for the object. In the force diagram the object is drawn as a dot and all forces acting on the object are drawn in the correct directions starting from that dot. In this case, three forces are acting on the object.
Each rope exerts a force on the object in the direction of the rope away from the object.
Since the object has mass, it is attracted towards the centre of the earth. This weight is represented in the force diagram as ${\overrightarrow {W}}$.
Since the object is stationary, the resultant force acting on the object is zero. In other words the three force vectors drawn tailtohead form a closed triangle:
[In general, when drawn tailtohead the forces acting on an object in equilibrium form a closed figure with the head of the last vector joining up with the tail of the first vector. When only three forces act on an object this closed figure is a triangle. This leads to the triangle law for three forces in equilibrium:
Triangle Law for Three Forces in Equilibrium: Three forces in equilibrium can be represented in magnitude and direction by the three sides of a triangle taken in order.
Newton's Laws of Motion[edit]
Our current laws of motion were discovered by Sir Isaac Newton. It is said that Sir Isaac Newton started to think about the nature of motion and gravitation after being struck on the head by a falling apple.
Newton discovered 3 laws describing motion:
First Law[edit]
Newton's first law basically says that a force has to be applied to an object to make it move or to make it stop. The first part of that statement definitely makes sense. The only way I can make something move is to have something give it a push. The second part of that statement might not be quite as easy to just take as fact. We've all witnessed objects slow down when nobody is pushing them. How then can we say that the only way to stop an object's motion is with a force? The answer is that there are forces that we don't always see. Most of the time, the force that we don't see is the force of friction.
Friction is the force that resists motion when two things are sliding past one another. To understand what friction is, think about sandpaper. If you try to rub 2 pieces of sandpaper together, it will be hard to get them to slide. This same phenomenon happens between all objects to some degree. This frictional force is what slows objects down or stops their motion.
Second Law[edit]
Definition: The time rate of change in momentum is proportional to the applied force and takes place in the direction of the force.
The law is represented in the following basic form (the system of measurement is chosen such that constant of proportionality is 1) :
 $\mathbf {F} ={d(m\mathbf {v} ) \over dt}$
The product of mass and velocity i.e. mv is called the momentum. The net force on a particle is, thus, equal to rate change of momentum of the particle with time. Generally mass of the object under consideration is constant and thus can be taken out of the derivative :
 $\mathbf {F} =m{d(\mathbf {v} ) \over dt}$
 $\mathbf {F} =m\mathbf {a}$
For constant mass,
$\mathbf {F} =m\mathbf {a}$
Force is equal to mass times acceleration. This version of Newton's Second Law of Motion assumes that the mass of the body does not change with time, and as such, does not represent a general mathematical form of the Law. Consequently, this equation cannot, for example, be applied to the motion of a rocket, which loses its mass (the lost mass is ejected at the rear of the rocket) with the passage of time.
It makes sense that the direction of the acceleration is in the direction of the resultant force. If you push something away from you it doesn't move toward you unless of course there is another force acting on the object towards you!
Worked Example 16 Newton's Second Law[edit]
Question: A block of mass 10 kg is accelerating at 2 m·s^{−2}. What is the magnitude of the net force acting on the block?
Answer:
Step 1 :
We are given
 the block's mass
 the block's acceleration
all in the correct units.
Step 2 :
We are asked to find the magnitude of the force applied to the block. Newton's Second Law tells us the relationship between acceleration and force for an object. Since we are only asked for the magnitude we do not need to worry about the directions of the vectors:
${\begin{matrix}F_{Net}&=&ma\\&=&10\ {\mbox{kg}}\times 2{\mbox{ m}}\cdot {\mbox{s}}^{2}\\&=&20\ {\mbox{N}}\end{matrix}}$
Thus, there must be a net force of 20 N acting on the box.
Worked Example 17 Newton's Second Law 2[edit]
Question: A 12 N force is applied in the positive xdirection to a block of mass 100 mg resting on a frictionless flat surface. What is the resulting acceleration of the block?
Answer:
Step 1 :
We are given
 the block's mass
 the applied force
but the mass is not in the correct units.
Step 2 :
Let us begin by converting the mass:
${\begin{matrix}100{\mbox{ mg}}&=&100\times 10^{3}{\mbox{ g}}=0.1{\mbox{ g}}\\1000{\mbox{ g}}&=&1\ {\mbox{kg}}\\1&=&1kg\times {\frac {1}{1000g}}\\&=&{\frac {1kg}{1000g}}\\0.1g&=&0.1g\times 1\\&=&0.1g\times {\frac {1kg}{1000g}}\\&=&0.0001\ kg\\\end{matrix}}$
Step 3 :
We know that net force results in acceleration. Since there is no friction the applied force is the resultant or net force on the block (refer to the earlier example of the block pushed on the surface of the table). The block will then accelerate in the direction of this force according to Newton's Second Law.
Step 4 :
To determine the magnitude of the acceleration:
${\begin{matrix}F_{Res}&=&ma\\12{\mbox{ N}}&=&(0.0001{\mbox{ kg}})a\\a&=&{\frac {12{\mbox{ N}}}{0.0001{\mbox{kg}}}}\\&=&120000{\frac {\mbox{N}}{\mbox{kg}}}\\&=&120000{\frac {{\mbox{kg}}\cdot {\mbox{m}}}{{\mbox{s}}^{2}\cdot {\mbox{kg}}}}\\&=&120000{\frac {\mbox{m}}{{\mbox{s}}^{2}}}\\&=&1.2\times 10^{5}\ {\mbox{ m}}\cdot {\mbox{s}}^{2}\end{matrix}}$
From Newton's Second Law the direction of the acceleration is the same as that of the resultant force. The final result is then that the block accelerates at $1.2\times 10^{5}\ {\mbox{ m}}\cdot {\mbox{s}}^{2}$ in the positive xdirection.
Weight and Mass[edit]
You must have heard people saying ``My weight is 60 kg. This is actually correct, though physicists often have difficulty understanding it because it is mass that is measured in kilograms. This is a different meaning of the word weight from the one used in mechanics, where weight is the force of gravity exerted by the earth on an object with mass:
${\begin{matrix}F_{weight}=mg\end{matrix}}$
As such, weight is measured in newtons.
If you compare this equation to Newton's Second Law you will see that it looks exactly the same with the a replaced by g. Thus, when weight is the only force acting on an object (i.e. when $F_{weight}$
F_{weight} is the resultant force acting on the object) the object has an acceleration g. Such an object is said to be in free fall. The value for g is the same for all objects (i.e. it is independent of the objects mass):
${\begin{matrix}g=9.8{\mbox{m}}\cdot {\mbox{s}}^{2}\approx 10{\mbox{m}}\cdot {\mbox{s}}^{2}\end{matrix}}$
${\begin{matrix}g=9.8{\mbox{m}}\cdot {\mbox{s}}^{2}\approx 10{\mbox{m}}\cdot {\mbox{s}}^{2}\end{matrix}}$ 
(4.3)

You will learn how to calculate this value from the mass and radius of the earth in Chapter ???. Actually the value of g varies slightly from place to place on the Earth's surface.
The reason that we often get confused between weight and mass, is that scales measure your weight (in newtons) and then display your mass using the equation above.
Worked Example 18 Calculating the resultant and then the acceleration[edit]
Question: A block (mass 20 kg) on a frictionless flat surface has a 45 N force applied to it in the positive xdirection. In addition a 25 N force is applied in the negative xdirection. What is the resultant force acting on the block and the acceleration of the block?
Answer:
Step 1 :
We are given
 the block's mass
 Force F_{1} = 45 N in the positive xdirection
 Force F_{2} = 25 N in the negative xdirection
all in the correct units.
Step 2 :
We are asked to determine what happens to the block. We know that net force results in an acceleration. We need to determine the net force acting on the block.
Step 3 :
Since F_{1} and F_{2} act along the same straight line, we can apply the algebraic technique of vector addition discussed in the Vectors chapter to determine the resultant force. Choosing the positive xdirection as our positive direction:
Positive xdirection is the positive direction:
${\begin{matrix}F_{Res}&=&(+45\ {\mbox{N}})+(25{\mbox{ N}})\\&=&+20\ {\mbox{N}}\\&=&20{\mbox{N}}\ in\ the\ positive\ xdirection\end{matrix}}$
where we remembered in the last step to include the direction of the resultant force in words. By Newton's Second Law the block will accelerate in the direction of this resultant force.
Step 4 :
Next we determine the magnitude of the acceleration:
${\begin{matrix}F_{Res}&=&ma\\20{\mbox{N}}&=&(20{\mbox{kg}})a\\a&=&{\frac {20{\mbox{N}}}{20{\mbox{kg}}}}\\&=&1{\frac {\mbox{N}}{\mbox{kg}}}\\&=&1{\frac {{\mbox{kg}}\cdot {\mbox{m}}}{{\mbox{s}}^{2}\cdot {\mbox{kg}}}}\\&=&1\ {\mbox{ m}}\cdot {\mbox{s}}^{2}\\\end{matrix}}$
The final result is then that the block accelerates at $1\ {\mbox{ m}}\cdot {\mbox{s}}^{2}$ in the positive xdirection (the same direction as the resultant force).
Worked Example 19 Block on incline[edit]
Question: A block (mass 10 kg) is released on an inclined plane. What is the resulting rate of acceleration of the block if the angle theta is 25 degrees and the coefficient of friction between the block and the plane is 0.25?
Solution:
Step 1: The acceleration of the block will be parallel to the plane, so we break the problem into two parts  forces perpendicular to the plane (which must cancel out since there is no acceleration perpendicular to the plane) and parallel to the plane.
Step 2: We know that there is no acceleration in the direction perpendicular to the surface of the inclined plane, or in other words $a_{\perp }=0$. Therefore we know that:
 ${\begin{matrix}\\\sum {F_{\perp }}&=&0\\NW_{\perp }&=&0\\Nm\ g\ {\cos \theta }&=&0\\N&=&m\ g\ {\cos \theta }\\&=&(10\ kg)(9.8\ {m \over {s^{2}}})\ {\cos 25^{\circ }}\\&=&88.82\ N\end{matrix}}$
Step 3: Summing forces parallel to the plane yields:
 ${\begin{matrix}\\\sum {F_{\}}&=&m\ g\ {\sin \theta }\mu \ N\\ma_{\}&=&m\ g\ {\sin \theta }\mu \ N\\(10kg)a_{\}&=&(10kg)(9.8\ {m \over s^{2}})\sin 25^{\circ }(0.25)\ (88.82\ N)\\a_{\}&=&(9.8\ {m \over s^{2}})\ \sin 25^{\circ }(0.25)\ {{88.82\ N} \over {10\ kg}}\\&=&5.5\ {m \over {s^{2}}}\end{matrix}}$
The final result is that the block accelerates down the inclined plane with an acceleration of $5.5\ m\cdot s^{2}$.
Third Law[edit]
Definition: For every force one body applies to another (action) there is always an equal but opposite in direction force another body applies back to the first one (reaction.)
This law is a direct consequence of the Principle of Conservation of Linear Momentum.
Newton's Third Law is easy to understand but it can get quite difficult to apply it. An important thing to realise is that the action and reaction forces can never act on the same object and hence cannot contribute to the same resultant.
Worked Example 20 Identifying actionreaction Pairs[edit]
Question: Consider pushing a box on the surface of a rough table.
 Draw a force diagram indicating all of the forces acting on the box.
 Identify the reaction force for each of the forces acting on the box.
Answer:
 The following force diagram shows all of the forces acting on the box
There is an important thing to realise which is related to Newton's Third Law. Think about dropping a stone off a cliff. It falls because the earth exerts a force on it (see Chapter ???) and it doesn't seem like there are any other forces acting. So is Newton's Third Law wrong? No, the reactionary force to the weight of the stone is the force exerted by the stone on the earth. This is illustrated in detail in the next worked example.
Worked Example 21 Newton's Third Law[edit]
Question: A stone of mass 0.5 kg is accelerating at 10 m·s^{−2} towards the earth.
 What is the force exerted by the earth on the stone?
 What is the force exerted by the stone on the earth?
 What is the acceleration of the earth, given that its mass is 5.97 × 10^{27} kg?
Answer:
 Step 1 : We are given

 the stone's mass
 the stone's acceleration (g)
 By Newton's Third Law the stone must exert an equal but opposite force on the earth. Hence the stone exerts a force of 5 N towards the stone on the earth.
 We have

 the force acting on the earth
 the Earth's mass
File:Fhsst forces13.png

Newton first published these laws in Philosophiae Naturalis Principia Mathematica (1687) and used them to prove many results concerning the motion of physical objects. Only in 1916 were
Newton's Laws superseded by Einstein's theory of relativity.

File:Fhsst forces15.png
The next two worked examples are quite long and involved but it is very important that you understand the discussion as they illustrate the importance of Newton's Laws.
Worked Example 22 Rockets[edit]
Question: How do rockets accelerate in space?
Answer:
 Gas explodes inside the rocket.
 This exploding gas exerts a force on each side of the rocket (as shown in the picture below of the explosion chamber inside the rocket).
File:Fhsst forces17.png
Note that the forces shown in this picture are representative. With an explosion there will be forces in all directions.
 Due to the symmetry of the situation, all the forces exerted on the rocket are balanced by forces on the opposite side, except for the force opposite the open side. This force on the upper surface is unbalanced.
 This is therefore the resultant force acting on the rocket and it makes the rocket accelerate forwards.
Systems and External Forces[edit]
The concepts of a system and an external forces are very important in physics. A system is any collection of objects. If one draws an imaginary box around such a system then an external force is one that is applied by an object or person outside the box. Imagine for example a car pulling a trailer.
Concept of system is extremely important. Consider an example of a person holding a load (box) on his head and standing on a platform.
Now count the forces :
(i) Box presses the person down (ii) The person pushes the box up (reaction to box's weight) (iii) Box is pulled down by the earth (iv) Earth is pulled up by the box (v) Person presses the floor (vi) Floor pushes the person up (vii) Person is pulled down by earth and (viii) Earth is pulled up by the person.
Quite a mess. Which ones are internal forces and which ones are external forces? Identification of system comes handy here. We can select the box or the person as a system for analyzing forces or we may even consider box and person together as a single system. Only condition to watch while selecting a system is that all parts of the system should have same acceleration. If two parts are at different accelerations then each of them should be treated as a separate system.
Newton's Law of Universal Gravitation[edit]
Why does the Earth stay in orbit around the Sun? Shouldn't it fly off tangentially into outer space?
These questions intrigued Newton and inspired his study of gravitation.
Newton realized that a force must be constantly pulling on the Earth, redirecting its motion and preventing it from being flung off. Newton reasoned that this force, which he termed 'gravity', acted between all bodies with mass and varied inversely to the square of the distance between the two bodies.
 $F={\frac {Gm_{1}m_{2}}{r^{2}}}$
where $G$ is a universal gravitational constant, $m_{1}$ and $m_{2}$ are the 2 masses, and $r$ is the distance between the centers of mass.
Newton also realized that this same force which redirects the path of the Earth around the Sun, was also responsible for an apple falling to the ground. In this case, the two masses, the Earth, and the apple, are attracted each other and this exerts a force which pulls the apple towards the center of the Earth. While we can use the Universal Law of Gravitation formula to solve this problem it is often more convenient to realize two facts:
 The Earths mass, $M_{E}$ is constant
 The distance between the apple (or other object) can usually be approximated with just $R_{E}$, the radius of the Earth, because this is the dominating term.
Thus we can rewrite the equations such that
 $F=mg$
where, $g={\frac {GM_{E}}{R_{E}^{2}}}$. g is the acceleration on Earth and is $9.8m/s^{2}$.
Examples of Forces Studied Later[edit]
Most of physics revolves around forces. Although there are many different forces we deal with them all in the same way. The methods to find resultants and acceleration do not depend on the type of force we are considering.
At first glance, the number of different forces may seem overwhelming  gravity, drag, electrical forces, friction and many others. However, physicists have found that all these forces can be classified into four groups. These are gravitational forces, electromagnetic forces, strong nuclear force and weak nuclear force. Even better, all the forces that you will come across at school are either gravitational or electromagnetic. Doesn't that make life easy?
Newtonian Gravity[edit]
Gravity is the attractive force between two objects due to the mass of the objects. When you throw a ball in the air, its mass and the Earth's mass attract each other, which leads to a force between them. The ball falls back towards the Earth, and the Earth accelerates towards the ball. The movement of the Earth toward the ball is, however, so small that you couldn't possibly measure it.
Electromagnetic Forces[edit]
Almost all of the forces that we experience in everyday life are electromagnetic in origin. They have this unusual name because long ago people thought that electric forces and magnetic forces were different things. After much work and experimentation, it has been realised that they are actually different manifestations of the same underlying theory.
The Electric Force[edit]
If we have objects carrying electrical charge, which are not moving, then we are dealing with electrostatic forces (Coulomb's Law). This force is actually much stronger than gravity. This may seem strange, since gravity is obviously very powerful, and holding a balloon to the wall seems to be the most impressive thing electrostatic forces have done, but think about it: for gravity to be detectable, we need to have a very large mass nearby. But a balloon rubbed in someone's hair can stick to a wall with a force so strong that it overcomes the force of gravity  with just the charges in the balloon and the wall!
Magnetic force[edit]
The magnetic force is a different manifestation of the electromagnetic force. It stems from the interaction between moving charges as opposed to the fixed charges involved in Coulomb's Law.
Examples of the magnetic force in action include magnets, compasses, car engines, computer data storage and your hair standing on end. Magnets are also used in the wrecking industry to pick up cars and move them around sites.
= Friction[edit]
Newton's First Law states that an object moving without a force acting on it will keep moving. Then why does a box sliding on a table stop? The answer is friction. Friction arises from the interaction between the molecules on the bottom of a box with the molecules on a table. This interaction is electromagnetic in origin, hence friction is just another view of the electromagnetic force. The great part about school physics is that most of the time we are told to neglect friction but it is good to be aware that there is friction in the real world.
Friction is also useful . If there was no friction and you tried to prop a ladder up against a wall, it would simply slide to the ground. Rock climbers use friction to maintain their grip on cliffs.
Friction is a force that impedes motion. It is in parallel to the contact surface and acts against the motion of the body.
Drag Force[edit]
This is the force an object experiences while travelling through a medium. When something travels through the air it needs to displace air as it travels and because of this the air exerts a force on the object. This becomes an important force when you move fast and a lot of thought is taken to try and reduce the amount of drag force a sports car experiences.
The drag force is very useful for parachutists. They jump from high altitudes and if there was no drag force, then they would continue accelerating all the way to the ground. Parachutes are wide because the more surface area you show, the greater the drag force and hence the slower you hit the ground.
Summary of Important Quantities, Equations and Concepts[edit]
Table 4.1: Summary of the symbols and units of the quantities used in Force
Units 
Quantity 
Symbol 
S.I. Unit 
Fundamental Units 
Direction 
Mass 
m 
kg 
kg 
no 
Acceleration 
${\overrightarrow {a}}$ 
m·s^{2} 
m·s^{2} 
yes 
Force 
${\overrightarrow {F}}$ 
N 
kg·m·s^{−2} 
yes 
Equilibrium: Objects at rest or moving with constant velocity are in equilibrium and have a zero resultant force.
Equilibrant: The equilibrant of any number of forces is the single force required to produce equilibrium.
Triangle Law for Forces in Equilibrium: Three forces in equilibrium can be represented in magnitude and direction by the three sides of a triangle taken in order.
Newton's First Law: Every object will remain at rest or in uniform motion in a straight line unless it is made to change its state by the action of an external force.
Newton's Second Law: The resultant force acting on a body results in an acceleration which is in the same direction as the resultant force and is directly proportional to the magnitude of this force and inversely proportional to the mass of the object.
Newton's Third Law: For every force or action there is an equal but opposite force or reaction.
Rectilinear Motion[edit]
What is rectilinear motion?[edit]
Rectilinear motion means motion along a straight line. This is a useful topic to study for learning how to describe the movement of cars along a straight road or of trains along straight railway tracks. In this section you have only two directions to worry about: (1) along the direction of motion, and (2) opposite to the direction of motion.
To illustrate this imagine a train heading east.
If it is accelerating away from the station platform (P), the direction of acceleration is the same as the direction of the train's velocity — east. If it is braking the direction of acceleration is opposite to the direction of its motion, i.e. west.
Speed and Velocity[edit]
Let's take a moment to review our definitions of velocity and speed by looking at the worked example below:
Worked Example 23 Speed and Velocity[edit]
Question: A cyclist moves from A through B to C in 10 seconds. Calculate both his speed and his velocity.
Answer:
Step 1 :
Analyse the question to determine what is given. The question explicitly gives
 the distance between A and B
 the distance between B and C
 the total time for the cyclist to go from A through B to C
all in the correct units!
Step 2 :
What is being asked? We are asked to calculate the average speed and the average velocity of the cyclist.
His speed  a scalar  will be
${\begin{matrix}v&=&{\frac {s}{t}}\\&=&{\frac {30m+40m}{10s}}\\&=&7{\frac {m}{s}}\end{matrix}}$
Since velocity is a vector we will first need to find the resultant displacement of the cyclist. His velocity will be
${\begin{matrix}{\overrightarrow {v}}={\frac {\overrightarrow {s}}{t}}\end{matrix}}$
The total displacement is the vector from A to C, and this is just the resultant of the two displacement vectors, i.e.
${\begin{matrix}{\overrightarrow {s}}={\overrightarrow {AC}}={\overrightarrow {AB}}+{\overrightarrow {BC}}\end{matrix}}$
Using the rule of Pythagoras:
${\begin{matrix}{\overrightarrow {s}}&=&{\sqrt {{(30m)}^{2}+{(40m)}^{2}}}\\&=&50m\ in\ the\ direction\ from\ A\ to\ C\end{matrix}}$
∴
${\begin{matrix}{\overrightarrow {v}}&=&{\frac {50\ {\mbox{m}}}{10\ {\mbox{s}}}}\\&=&5\ {\frac {\mbox{m}}{\mbox{s}}}\ in\ the\ direction\ from\ A\ to\ C\end{matrix}}$
For this cyclist, his velocity is not the same as his speed because there has been a change in the direction of his motion. If the cyclist traveled directly from A to C without passing through B his speed would be
${\begin{matrix}v&=&{\frac {50m}{10s}}\\&=&5{\frac {m}{s}}\end{matrix}}$
and his velocity would be
${\begin{matrix}{\overrightarrow {v}}&=&{\frac {50m}{10s}}\\&=&5{\frac {m}{s}}\ in\ the\ direction\ from\ A\ to\ C\end{matrix}}$
In this case where the cyclist is not undergoing any change of direction (i.e. he is traveling in a straight line) the magnitudes of the speed and the velocity are the same. This is the defining principle of rectilinear motion.
Important: 
For motion along a straight line the magnitudes of speed and velocity are the same, and the magnitudes of the distance and displacement are the same. 
In physics we often use graphs as important tools for picturing certain concepts. Below are some graphs that help us picture the concepts of displacement, velocity and acceleration.
DisplacementTime Graphs[edit]
Below is a graph showing the displacement of the cyclist from A to C:
This graphs shows us how, in 10 seconds time, the cyclist has moved from A to C. We know the gradient (slope) of a graph is defined as the change in y divided by the change in x, i.e. ${\frac {\Delta y}{\Delta x}}$. In this graph the gradient of the graph is just ${\frac {\Delta {\overrightarrow {s}}}{\Delta t}}$  and this is just the expression for velocity.
Important: 
The area between a velocitytime graph and the `time' axis gives the displacement of the object. 
The slope is the same all the way from A to C, so the cyclist's velocity is constant over the entire displacement he travels. In figure 5.1 are examples of the displacementtime graphs you will encounter.
a) shows the graph for an object stationary over a period of time. The gradient is zero, so the object has zero velocity.
b) shows the graph for an object moving at a constant velocity. You can see that the displacement is increasing as time goes on. The gradient, however, stays constant (remember: its the slope of a straight line), so the velocity is constant. Here the gradient is positive, so the object is moving in the direction we have defined as positive.
c) shows the graph for an object moving at a constant acceleration. You can see that both the displacement and the velocity (gradient of the graph) increase with time. The gradient is increasing with time, thus the velocity is increasing with time and the object is accelerating.
VelocityTime Graphs[edit]
Uniform acceleration and gradient (slope)[edit]
Look at the velocitytime graph below:
This is the velocitytime graph of a cyclist traveling from A to B at a constant acceleration, i.e. with steadily increasing velocity. The gradient (slope) of this graph is just ${\frac {\Delta {\overrightarrow {v}}}{\Delta t}}$  and this is just the expression for acceleration. Because the slope is the same at all points on this graph, the acceleration of the cyclist is constant. And the constant acceleration is $2/m/s^{2}$, or 2 meters per second per second (or 2 meters per second squared)
${\begin{matrix}\mathrm {slope\ of\ line} &=&\Delta v/\Delta t\\&=&{\frac {10m/s}{5s}}\\&=&2{\frac {m}{s^{2}}}\\\end{matrix}}$
Important: 
The gradient (slope) on a velocitytime graph equals the acceleration. 
Distance travelled[edit]
Not only can we get the acceleration of an object from its velocitytime graph, but we can also get some idea of the displacement traveled. Look at the graph below:
This graph shows an object moving at a constant velocity of 10m/s for a duration of 5s. The area between the graph and the time axis (the shaded area) of the above plot will give us the displacement of the object during this time. In this case we just need to calculate the area of a rectangle with width 5s and height 10m/s
${\begin{matrix}\mathrm {area\ of\ rectangle} &=&\mathrm {height} \times \mathrm {width} \\&=&{\overrightarrow {v}}\times t\\&=&10{\frac {m}{s}}\times 5s\\&=&50m\\&=&{\overrightarrow {s}}=\mathrm {displacement} \\\end{matrix}}$
So, here we've shown that an object traveling at 10m/s for 5s has undergone a displacement of 50m.
Important: 
The area between a velocitytime graph and the `time' axis gives the displacement of the object. 
Here are a couple more velocitytime graphs to get used to:

Figure 5.2: Some common velocitytime graphs: 
In figure 5.2 are examples of the displacementtime graphs you may encounter.
a) shows the graph for an object moving at a constant velocity over a period of time. The gradient is zero, so the object is not accelerating.
b) shows the graph for an object which is decelerating. You can see that the velocity is decreasing with time. The gradient, however, stays constant (remember: its the slope of a straight line), so the acceleration is constant. Here the gradient is negative, so the object is accelerating in the opposite direction to its motion, hence it is decelerating.
AccelerationTime Graphs[edit]
In this chapter on rectilinear motion we will only deal with objects moving at a constant acceleration, thus all accelerationtime graphs will look like these two:
Here is a description of the graphs below:
a) shows the graph for an object which is either stationary or traveling at a constant velocity. Either way, the acceleration is zero over time.
b) shows the graph for an object moving at a constant acceleration. In this case the acceleration is positive  remember that it can also be negative.
We can obtain the velocity of a particle at some given time from an acceleration time graph  it is just given by the area between the graph and the timeaxis. In the graph below, showing an object at a constant positive acceleration, the increase in velocity of the object after 2 seconds corresponds to the shaded portion.
${\begin{matrix}\mathrm {area\ of\ rectangle} &=&{\overrightarrow {a}}\times t\\&=&5{\frac {m}{s^{2}}}\times 2s\\&=&10{\frac {m}{s}}\\&=&{\overrightarrow {v}}\\\end{matrix}}$
Its useful to remember the set of graphs below when working on problems. Figure 5.3 shows how displacement, velocity and time relate to each other. Given a displacementtime graph like the one on the left, we can plot the corresponding velocitytime graph by remembering that the slope of a displacementtime graph gives the velocity. Similarly, we can plot an accelerationtime graph from the gradient of the velocitytime graph.

Figure 5.3: A Relationship Between Displacement, Velocity and Acceleration 
Worked Examples[edit]
Worked Example 24 Relating displacement, velocity, and accelerationtime graphs[edit]
Question: Given the displacementtime graph below, draw the corresponding velocitytime and accelerationtime graphs, and then describe the motion of the object.
Answer:
Step 1 : Analyse the question to determine what is given. The question explicitly gives a displacementtime graph.
Step 2 : What is asked?
3 things:
 Draw a velocitytime graph
 Draw an accelerationtime graph
 Describe the behaviour of the object
For the first 2 seconds we can see that the displacement remains constant  so the object is not moving, thus it has zero velocity during this time. We can reach this conclusion by another path too: remember that the gradient of a displacementtime graph is the velocity. For the first 2 seconds we can see that the displacementtime graph is a horizontal line, i.e. it has a gradient of zero. Thus the velocity during this time is zero and the object is stationary.
For the next 2 seconds, displacement is increasing with time so the object is moving. Looking at the gradient of the displacement graph we can see that it is not constant. In fact, the slope is getting steeper (the gradient is increasing) as time goes on. Thus, remembering that the gradient of a displacementtime graph is the velocity, the velocity must be increasing with time during this phase.
For the final 2 seconds we see that displacement is still increasing with time, but this time the gradient is constant, so we know that the object is now travelling at a constant velocity, thus the velocitytime graph will be a horizontal line during this stage.
So our velocitytime graph looks like this one below. Because we haven't been given any values on the vertical axis of the displacementtime graph, we cannot figure out what the exact gradients are and hence what the values of the velocity are. In this type of question it is just important to show whether velocities are positive or negative, increasing, decreasing or constant.
Once we have the velocitytime graph its much easier to get the accelerationtime graph as we know that the gradient of a velocitytime graph is the just the acceleration.
For the first 2 seconds the velocitytime graph is horizontal at zero, thus it has a gradient of zero and there is no acceleration during this time. (This makes sense because we know from the displacement time graph that the object is stationary during this time, so it can't be accelerating).
For the next 2 seconds the velocitytime graph has a positive gradient. This gradient is not changing (i.e. its constant) throughout these 2 seconds so there must be a constant positive acceleration.
For the final 2 seconds the object is traveling with a constant velocity. During this time the gradient of the velocitytime graph is once again zero, and thus the object is not accelerating.
The accelerationtime graph looks like this:
A brief description of the motion of the object could read something like this: At t = 0s and object is stationary at some position and remains stationary until t = 2s when it begins accelerating. It accelerates in a positive direction for 2 seconds until t = 4s and then travels at a constant velocity for a further 2 seconds.
Worked Example 25 Calculating distance from a velocitytime graph[edit]
Question: The velocitytime graph of a car is plotted below. Calculate the displacement of the car has after 15 seconds.
Answer: We are asked to calculate the displacement of the car. All we need to remember here is that the area between the velocitytime graph and the time axis gives us the displacement.
For t = 0s to t = 5s this is the triangle on the left:
${\begin{matrix}Area\triangle &=&{\frac {1}{2}}b\times h\\&=&{\frac {1}{2}}5s\times 4m/s\\&=&10m\end{matrix}}$
For t = 5s to t = 12s the displacement is equal to the area of the rectangle
${\begin{matrix}Area\Box &=&w\times h\\&=&7s\times 4m/s\\&=&28m\end{matrix}}$
For t = 12s to t = 14s the displacement is equal to the area of the triangle above the time axis on the right
${\begin{matrix}Area\triangle &=&{\frac {1}{2}}b\times h\\&=&{\frac {1}{2}}2s\times 4m/s\\&=&4m\end{matrix}}$
For t = 14s to t = 15s the displacement is equal to the area of the triangle below the time axis
${\begin{matrix}Area\triangle &=&{\frac {1}{2}}b\times h\\&=&{\frac {1}{2}}1s\times 2m/s\\&=&1m\end{matrix}}$
Now the total displacement of the car is just the sum of all of these areas. HOWEVER, because in the last second (from t = 14s to t = 15s) the velocity of the car is negative, it means that the car was going in the opposite direction, i.e. back where it came from! So, to get the total displacement, we have to add the first 3 areas (those with positive displacements) and subtract the last one (because it signifies a displacement in the opposite direction).
${\begin{matrix}{\overrightarrow {s}}&=&10+28+41\\&=&41m\ in\ the\ positive\ direction\end{matrix}}$
Worked Example 26 Velocity from a displacementtime graph[edit]
Question: Given the diplacementtime graph below,
 what is the velocity of the object during the first 4 seconds?
 what is the velocity of the object from t = 4s to t = 7s?
Answer:
 for the first 4sec velocity is slope of the curve i.e. 2/4=0.5m/s
 For the last 3 seconds we can see that the displacement stays constant, and that the gradient is zero. Thus ${\overrightarrow {v}}=0m/s$.
=Worked Example 27 From an acceleration to a velocitytime graph[edit]
Question: Given the accelerationtime graph below, assume that the object starts from rest and draw its velocitytime graph in simple launge
Equations of Motion[edit]
This section is about solving problems relating to uniformly accelerated motion. We'll first introduce the variables and the equations, then we'll show you how to derive them, and after that we'll do a couple of examples.
 u = starting velocity (m/s) at t = 0
 v = final velocity (m/s) at time t
 s = displacement (m)
 t = time (s)
 a = acceleration (m/s²}
$v=u+at\,$ 
(5.1) 
$s={\frac {(u+v)}{2}}t$ 
(5.2) 
$s=ut+{\frac {1}{2}}at^{2}$ 
(5.3) 
$v^{2}=u^{2}+2as\,$ 
(5.4) 
$\Delta x=1/2*a*(\Delta t)^{2}+u(\Delta t)$
Make sure you can rhyme these off, they are very important! There are so many different types of questions for these equations. Basically when you are answering a question like this:
 Find out what values you have and write them down.
 Figure out which equation you need.
 Write it down!!!
 Fill in all the values you have and get the answer.
(1.5,0)

Galileo Galilei of Pisa, Italy, was the first to determine the correct mathematical law for acceleration: the total distance covered, starting from rest, is proportional to the square of the time. He also concluded that
objects retain their velocity unless a force  often friction  acts upon them, refuting the accepted Aristotelian hypothesis that objects "naturally" slow down and stop unless a force acts upon them. This principle was incorporated into Newton's laws of motion (1st law).

Equation 5.1[edit]
By the definition of acceleration $a={\frac {\Delta v}{t}}$ where $\Delta$v is the change in velocity, i.e. $\Delta v=vu$. Thus we have
${\begin{matrix}a&=&{\frac {vu}{t}}\\v&=&u+at\end{matrix}}$
Equation 5.2[edit]
In the previous section we saw that displacement can be calculated from the area between a velocitytime graph and the timeaxis. For uniformly accelerated motion the most complicated velocitytime graph we can have is a straight line. Look at the graph below — it represents an object with a starting velocity of u, accelerating to a final velocity v over a total time t.
To calculate the final displacement we must calculate the area under the graph — this is just the area of the rectangle added to the area of the triangle.
${\begin{matrix}Area\triangle &=&{\frac {1}{2}}b\times h\\&=&{\frac {1}{2}}t\times (vu)\\&=&{\frac {1}{2}}vt{\frac {1}{2}}ut\end{matrix}}$
${\begin{matrix}Area\Box &=&w\times h\\&=&t\times u\\&=&ut\end{matrix}}$
${\begin{matrix}Displacement&=&Area\Box +Area\triangle \\s&=&ut+{\frac {1}{2}}vt{\frac {1}{2}}ut\\&=&{\frac {(u+v)}{2}}t\end{matrix}}$
Equation 5.3[edit]
This equation is simply derived by eliminating the final velocity v in equation 5.2. Remembering from equation 5.1 that
$v=u+at$
then equation 5.2 becomes
${\begin{matrix}s&=&{\frac {u+u+at}{2}}t\\&=&{\frac {2ut+at^{2}}{2}}\\&=&ut+{\frac {1}{2}}at^{2}\end{matrix}}$
Equation 5.4[edit]
This equation is just derived by eliminating the time variable in the above equation. From Equation 5.1 we know
$t={\frac {vu}{a}}$
Substituting this into Equation 5.3 gives
${\begin{matrix}s&=&u({\frac {vu}{a}})+{\frac {1}{2}}a({\frac {vu}{a}})^{2}\\&=&{\frac {uv}{a}}{\frac {u^{2}}{a}}+{\frac {1}{2}}a({\frac {v^{2}2uv+u^{2}}{a^{2}}})\\&=&{\frac {uv}{a}}{\frac {u^{2}}{a}}+{\frac {v^{2}}{2a}}{\frac {uv}{a}}+{\frac {u^{2}}{2a}}\\2as&=&2u^{2}+v^{2}+u^{2}\\v^{2}&=&u^{2}+2as\end{matrix}}$

(5.5) 
This gives us the final velocity in terms of the initial velocity, acceleration and displacement and is independent of the time variable.
Worked Example 28[edit]
Question: A racing car has an initial velocity of 100 m/s and it covers a displacement of 725 m in 10 s. Find its acceleration.
Answer:
Step 1 :
We are given the quantities u, s and t — all in the correct units. We need to find a.
Step 2 :
We can use equation 5.3 $s=ut+{\frac {1}{2}}at^{2}$
Step 3 :
Rearranging equation 5.3 we have $a={\frac {2(sut)}{t^{2}}}$
Substituting in the values of the known quantities this becomes
${\begin{matrix}a&=&{\frac {2(725{\mbox{ m}}100{\frac {\mbox{m}}{\mbox{s}}}\cdot 10{\mbox{ s}})}{10^{2}{\mbox{ s}}^{2}}}\\&=&{\frac {2(275{\mbox{ m}})}{100{\mbox{ s}}^{2}}}\\&=&5.5{\frac {\mbox{m}}{{\mbox{s}}^{2}}}\end{matrix}}$
The racing car is accelerating at 5.5 m/s², or we could say it is decelerating at 5.5 m/s². this ans is wrong
its better to use a=vu/t then a=(725100)/10s you will get a=625/10 which will be equal to 62.5m/s2
Worked Example 29[edit]
Question: An object starts from rest, moves in a straight line with a constant acceleration and covers a distance of 64 m in 4 s. Calculate
 its acceleration
 its final velocity
 at what time the object had covered half the total distance
 what distance the object had covered in half the total time.
Answer:
Step 1 : We are given the quantities u, s and t in the correct units.
Step 2 : To calculate the acceleration we can use equation [#eq:eq3 5.3]. Rearranging it we have:
$a={\frac {2(sut)}{t^{2}}}$
Substituting in the values of the known quantities this becomes
${\begin{matrix}a&=&{\frac {2(64m0{\frac {m}{s}}4s}{4^{2}s^{2}}}\\&=&{\frac {128m}{16s^{2}}}\\&=&8{\frac {m}{s^{2}}}\end{matrix}}$
Step 3 : To calculate its final velocity we can use equation 5.1  remember we now also know the acceleration of the object.
${\begin{matrix}v&=&u+at\\&=&0{\frac {m}{s}}+(8{\frac {m}{s^{2}}})(4s)\\&=&32{\frac {m}{s}}\end{matrix}}$
Step 4 : The time at which the object had covered half the total distance. Half the distance is 32m. Here we have the quantities s, u and a so we first use equation 5.4 to calculate the velocity at this distance:
${\begin{matrix}v^{2}&=&u^{2}+2as\\&=&(0m)^{2}+2(8m/s^{2})(32m)\\&=&512m^{2}/s^{2}\\v&=&22.6m/s\end{matrix}}$
Now we can use equation 5.2 to calculate the time:
${\begin{matrix}t&=&{\frac {2s}{u+v}}\\&=&{\frac {(2)(32m)}{0m/s+22.6m/s}}\\&=&2.8s\end{matrix}}$
Step 5 : The distance the object had covered in half the time. Half the time is 2s. Thus we have u, a and t  all in the correct units. We can use equation 5.3 to get the distance:
${\begin{matrix}s&=&ut+{\frac {1}{2}}at^{2}\\&=&(0m/s)(2s)+{\frac {1}{2}}(8{\frac {m}{s^{2}}})(2s)^{2}\\&=&16m\end{matrix}}$
Worked Example 30[edit]
Question: A ball is thrown vertically upwards with a velocity of 10 m/s from the balcony of a tall building. The balcony is 15 m above the ground and gravitational accleration is 10 m/s^{2}. Find a) the time required for the ball to hit the ground, and b) the velocity with which it hits the ground.
Answer:1[edit]
Step 1 : In this case it often helps to make the problem easier to understand if we draw ourselves a picture like the one below:
First the ball goes upwards with gravitational acceleration slowing it until it reaches its highest point — here its speed is 0 m/s — then it begins descending with gravitational acceleration causing it to increase its speed on the way down. We can separate the motion into 2 stages:
Stage 1  the upward motion of the ball
Stage 2  the downward motion of the ball.
We'll choose the upward direction as positive  this means that gravitation acceleraton is negative  and we'll begin by solving for all the variables of Stage 1. So far we have these quantities:
${\begin{matrix}u_{1}&=&10m/s\\v_{1}&=&0m/s\\a_{1}&=&10m/s^{2}\\t_{1}&=&?\\s_{1}&=&?\end{matrix}}$
Using equation 5.1 to find t_{1}:
${\begin{matrix}v_{1}&=&u_{1}+a_{1}t_{1}\\t_{1}&=&{\frac {v_{1}u_{1}}{a_{1}}}\\&=&{\frac {0m/s10m/s}{10m/s^{2}}}\\&=&1s\end{matrix}}$
We can find s_{1} by using equation 5.4
${\begin{matrix}v_{1}^{2}&=&u_{1}^{2}+2a_{1}s_{1}\\s_{1}&=&{\frac {v_{1}^{2}u_{1}^{2}}{2a}}\\&=&{\frac {(0m/s)^{2}(10m/s)^{2}}{2(10m/s^{2})}}\\&=&5m\end{matrix}}$
For Stage 2 we have the following quantities:
${\begin{matrix}u_{2}&=&0m/s\\v_{2}&=&?\\a_{2}&=&10m/s^{2}\\t_{2}&=&?\\s_{2}&=&15m5m=20m\end{matrix}}$
We can determine the final velocity v_{2} using equation 5.4:
${\begin{matrix}v_{2}^{2}&=&u_{2}^{2}+2a_{2}s_{2}\\&=&(0m/s)^{2}+2(10m/s^{2})(20m)\\&=&400(m/s)^{2}\\v_{2}&=&20m/s{\mbox{ downwards}}\end{matrix}}$
Now we can determine the time for Stage 2, t_{2}, from equation 5.1:
${\begin{matrix}v_{2}&=&u_{2}+a_{2}t_{2}\\t_{2}&=&{\frac {v_{2}u_{2}}{a_{2}}}\\&=&{\frac {20m/s0m/s}{10m/s^{2}}}\\&=&2s\end{matrix}}$
Finally,
a) the time required for the stone to hit the ground is
$t=t_{1}+t_{2}=1{\mbox{ s}}+2{\mbox{ s}}=3{\mbox{ s}}$
b) the velocity with which it hits the ground is just
$v_{2}=20\ {\mbox{m}}/{\mbox{s}}$
Answer:2[edit]
applying <br\> ${\vec {s}}(t)={\vec {S}}_{i}+{\vec {U}}_{i}\ (tT_{i})+{\frac {\vec {A}}{2}}(tT_{i})^{2}$<br\> $0=15{\hat {j}}+10{\hat {j}}(T_{total}0)+{\frac {10{\hat {j}}}{2}}(T_{total}0)^{2}$<br\> $T_{total}^{2}2\ T_{total}3=0$<br\> solving we will get <br\> $T_{total}=3,1$<br\> so time to reach is 3 second , now applying <br\> ${\vec {v}}(t)={\vec {U}}_{i}+{\vec {A}}(tT_{i})$<br\> ${\vec {v}}(3)=10{\hat {j}}+(10{\hat {j}})(30)$<br\> ${\vec {v}}(3)=20{\hat {j}}$<br\>
These questions do not have the working out in them, but they are all done in the manner described on the previous page.
Question: A car starts off at 10 m/s and accelerates at 1 m/s^{2} for 10 seconds. What is its final velocity?
Answer: 20 m/s
Question: A car starts from rest, and accelerates at 1 m/s^{2} for 10 seconds. How far does it move?
Answer: 50 m
Question: A car is going 30 m/s and stops in 2 seconds. What is its stopping distance for this speed?
Answer: 30 m
Question: A car going at 20 m/s stops in a distance of 20 m.
 What is its deceleration?
 If the car is 1 tonne (1000 kg, or 1 Mg) how much force do the brakes exert?
Important Equations and Quantities[edit]
Table 5.1: Units used in Rectilinear Motion
Units 
Quantity 
Symbol 
Unit 
S.I. Units 
Direction 
Displacement 
${\overrightarrow {s}}$ 
 
m 
yes 
Velocity 
${\overrightarrow {u}}$, ${\overrightarrow {v}}$ 
 
m·s^{1} 
yes 
Distance 
s 
 
m 
 
Speed 
v 
 
m·s^{1} 
 
Acceleration 
${\overrightarrow {a}}$ 
 
m·s^{2} 
yes 
Momentum[edit]
What is Momentum?[edit]
Momentum is a physical quantity which is closely related to forces. We will learn about this connection a little later. Remarkably momentum is a conserved quantity. This makes momentum extremely useful in solving a great variety of realworld problems. Firstly we must consider the definition of momentum.
Definition: 
The momentum of an object is defined as its mass multiplied by its velocity. 
Mathematically,
${\overrightarrow {p}}=m{\overrightarrow {v}}$ 


${\overrightarrow {p}}$ 
: momentum ($kg.m.s^{1}$ + direction) 
m 
: mass (kg) 
${\overrightarrow {v}}$ 
: velocity (m.s^{1} + direction) 
Thus, momentum is a property of a moving object and is determined by its velocity and mass. A large truck travelling slowly can have the same momentum as a much smaller car travelling relatively fast.
Note the arrows in the equation defining momentum momentum is a vector with the same direction as the velocity of the object.
Since the direction of an object's momentum is given by the direction of its motion, one can calculate an object's momentum in two steps:
 include in the final answer the direction of the object's motion
Worked Example 31 Calculating Momentum 1[edit]
Question: A ball of mass 3kg moves at 2m.s^{1} to the right. Calculate the ball's momentum.
Answer:
Step 1 :
Analyse the question to determine what is given. The question explicitly gives
 the ball's mass, and
 the ball's velocity
in the correct units!
Step 2 :
What is being asked? We are asked to calculate the ball's momentum. From the definition of momentum,
${\begin{matrix}{\overrightarrow {p}}=m{\overrightarrow {v}},\end{matrix}}$
we see that we need the mass and velocity of the ball, which we are given.
Step 3 :
Firstly we calculate the magnitude of the ball's momentum,
${\begin{matrix}p&=&mv\\&=&(3kg)(2m.s^{1})=6\ kg.m.s^{1}.\end{matrix}}$
Finally we quote the answer with the direction of the ball's motion included,
${\begin{matrix}{\overrightarrow {p}}=6\ kg.m.s^{1}{\textbf {\ to\ the\ right}}\end{matrix}}$
Worked Example 32 Calculating Momentum 2[edit]
Question: A ball of mass 500g is thrown at 2m.s^{1}. Calculate the ball's momentum.
Answer:
Step 1 :
Analyse the question to determine what is given. The question explicitly gives
 the ball's mass, and
 the magnitude of the ball's velocity
but with the ball's mass in the incorrect units!
Step 2 :
What is being asked? We are asked to calculate the momentum which is defined as
${\begin{matrix}{\overrightarrow {p}}=m{\overrightarrow {v}}.\end{matrix}}$
Thus, we need the mass and velocity of the ball but we have only its mass and the magnitude of its velocity.
Step 3 :
In order to determine the velocity of the ball we need the direction of the ball's motion. If the problem does not give an explicit direction we are forced to be general. In a case like this we could say that the direction of the velocity is in the direction of motion of the ball. This might sound silly but the lack of information in the question has forced us to be, and we are certainly not wrong! The ball's velocity is then m.s^{1} in the direction of motion.
Step 4 :
Next we convert the mass to the correct units,
${\begin{matrix}1000g&=&1kg\\1&=&{\frac {1kg}{1000g}}\\500g\times 1&=&500g\times {\frac {1kg}{1000g}}\\&=&0.500kg\end{matrix}}$
Step 5 :
Now, let us find the magnitude of the ball's momentum,
${\begin{matrix}p&=&mv\\&=&(0.500kg)(2m.s^{1})=1\ kg.m.s^{1}\end{matrix}}$
Step 6 :
Finally, we quote the answer with the direction of the momentum included,
${\begin{matrix}{\overrightarrow {p}}&=&1\ kg.m.s^{1}{\textbf {\ in\ the\ direction\ of\ motion\ of\ the\ ball}}\end{matrix}}$
Worked Example 33 Calculating the Momentum of the Moon[edit]
Question: The moon is $384\ 400km$ away from the earth and orbits the earth in 27.3 days. If the moon has a mass of $7.35\times 10^{22}kg$ [footnode.html#foot11743 ^{6.1}] what is the magnitude of its momentum if we assume a circular orbit?
Answer:
Step 1 :
Analyse the question to determine what is given. The question explicitly gives
 the moon's mass,
 the distance to the moon, and
 the time for one orbit of the moon
with mass in the correct units but all other quantities in the incorrect units.
The units we require are
 seconds (s) for time, and
 metres (m) for distance pesteng physics→→→→→→
Step 2 :
What is being asked? We are asked to calculate only the magnitude of the moon's momentum (i.e. we do not need to specify a direction). In order to do this we require the moon's mass and the magnitude of its velocity, since ${\begin{matrix}p=mv.\end{matrix}}$
Step 3 :
How do we find the speed or magnitude of the moon's velocity? Speed is defined as,
${\begin{matrix}speed&=&{\frac {Distance}{time}}\end{matrix}}$
We are given the time the moon takes for one orbit but not how far it travels in that time. However, we can work this out from the distance to the moon and the fact that the moon's orbit is circular. Firstly let us convert the distance to the moon to the correct units,
${\begin{matrix}1km&=&1000m\\1&=&{\frac {1000m}{1km}}\\384\ 400km\times 1&=&384\ 400km\times {\frac {1000m}{1km}}\\&=&384\ 400\ 000m\\&=&3.844\times 10^{8}\ m\end{matrix}}$
Using the equation for the circumference, C, of a circle in terms of its radius, we can determine the distance travelled by the moon in one orbit:
${\begin{matrix}C&=&2\pi r\\&=&2\pi (3.844\times 10^{8}\ m)\\&=&2.42\times 10^{9}\ m.\\\end{matrix}}$
Next we must convert the orbit time, T, into the correct units. Using the fact that a day contains 24 hours, an hour consists of 60 minutes, and a minute is 60 seconds long,
${\begin{matrix}1day&=&(24)(60)(60)seconds\\1&=&{\frac {(24)(60)(60)s}{1day}}\\27.3days\times 1&=&27.3days{\frac {(24)(60)(60)s}{1day}}\\&=&2.36\times 10^{6}s\end{matrix}}$
Therefore, ${\begin{matrix}T=2.36\times 10^{6}s.\end{matrix}}$
Combining the distance travelled by the moon in an orbit and the time taken by the moon to complete one orbit, we can determine the magnitude of the moon's velocity or speed,
${\begin{matrix}v&=&{\frac {Distance}{time}}\\&=&{\frac {C}{T}}\\&=&1.02\times 10^{3}\ m.s^{1}.\end{matrix}}$
Step 4 :
Finally we can calculate the magnitude of the moon's momentum,
${\begin{matrix}p&=&mv\\&=&(7.35\times 10^{22}kg)(1.02\times 10^{3}\ m.s^{1})\\&=&7.50\times 10^{25}\ kg.m.s^{1}.\end{matrix}}$
→33→
The Momentum of a System[edit]
In Chapter 4 Forces the concept of a system was introduced. The bodies that make up a system can have different masses and can be moving with different velocities. In other words they can have different momenta.
Definition: 
The total momentum of a system is the sum of the momenta of each of the objects in the system. 
Since momentum is a vector, the techniques of vector addition discussed in This chapter must be used to calculate the total momentum of a system. Let us consider an example.
Worked Example 34 Calculating the Total Momentum of a System[edit]
Question: Two billiard balls roll towards each other. They each have a mass of 0.3kg. Ball 1 is moving at $v_{1}=1\ m.s^{1}$ to the right, while ball 2 is moving at $v_{2}=0.8\ m.s^{1}$ to the left.
Calculate the total momentum of the system.
Answer:
Step 1 :
Analyse the question to determine what is given. The question explicitly gives
 the mass of each ball,
 the velocity of ball 1, ${\overrightarrow {v_{1}}}$ and
 the velocity of ball 2, ${\overrightarrow {v_{2}}}$,
all in the correct units!
Step 2 :
What is being asked? We are asked to calculate the total momentum of the system. In this example our system consists of two balls. To find the total momentum we must sum the momenta of the balls,
${\begin{matrix}{\overrightarrow {p}}_{total}&=&{\overrightarrow {p_{1}}}+{\overrightarrow {p_{2}}}\end{matrix}}$
Since ball 1 is moving to the right, its momentum is in this direction, while the second ball's momentum is directed towards the left.
Thus, we are required to find the sum of two vectors acting along the same straight line. The algebraic method of vector addition introduced in this chapter can thus be used.
Step 3 :
Firstly we choose a positive direction. Let us choose right as the positive direction, then obviously left is negative.
Step 4 :
The total momentum of the system is then the sum of the two momenta taking the directions of the velocities into account. Ball 1 is travelling at $1\ m.s^{1}\ to\ the\ right$ or $+1\ m.s^{1}$. Ball 2 is travelling at $0.8\ m.s^{1}\ to\ the\ left$ or $0.8\ m.s^{1}$. Thus,
Right is the positive direction
${\begin{matrix}{\overrightarrow {p}}_{total}&=&m_{1}{\overrightarrow {v_{1}}}+m_{2}{\overrightarrow {v_{2}}}\\&=&(0.3kg)(+1\ m.s^{1})+(0.3kg)(0.8\ m.s^{1})\\&=&(+0.3\ kg.m.s^{1})+(0.24\ kg.m.s^{1})\\&=&+0.06\ kg.m.s^{1}\\&=&0.06\ kg.m.s^{1}{\textbf {\ to\ the\ right}}\end{matrix}}$
In the last step the direction was added in words. Since the result in the second last line is positive, the total momentum of the system is in the positive direction (i.e. to the right).
Change in Momentum[edit]
If either an object's mass or velocity changes then its momentum too will change. If an object has an initial velocity ${\overrightarrow {u}}$ and a final velocity ${\overrightarrow {v}}$, then its change in momentum, $\Delta {\overrightarrow {p}}$, is
$\Delta {\overrightarrow {p}}$ 
= ${\overrightarrow {p}}_{final}{\overrightarrow {p}}_{initial}$ $m{\overrightarrow {v}}m{\overrightarrow {u}}$ 
Worked Example 35 Change in Momentum[edit]
Question: A rubber ball of mass 0.8kg is dropped and strikes the floor at a velocity of $6\ m.s^{1}$. It bounces back with an initial velocity of $4\ m.s^{1}$. Calculate the change in momentum of the rubber ball caused by the floor.
Answer:
Step 1 :
Analyse the question to determine what is given. The question explicitly gives
 the ball's mass,
 the ball's initial velocity, and
 the ball's final velocity
all in the correct units.
Do not be confused by the question referring to the ball bouncing back with an ``initial velocity of $4\ m.s^{1}$. The word initial is included here since the ball will obviously slow down with time and $4\ m.s^{1}$ is the speed immediately after bouncing from the floor.
Step 2 :
What is being asked? We are asked to calculate the change in momentum of the ball,
${\begin{matrix}\Delta {\overrightarrow {p}}&=&m{\overrightarrow {v}}m{\overrightarrow {u}}.\end{matrix}}$
We have everything we need to find $\Delta {\overrightarrow {p}}$. Since the initial momentum is directed downwards and the final momentum is in the upward direction, we can use the algebraic method of subtraction discussed in the vectors chapter.
Step 3 : Firstly, we choose a positive direction. Let us choose down as the positive direction. Then substituting,
Down is the positive direction
${\begin{matrix}\Delta {\overrightarrow {p}}&=&m{\overrightarrow {v}}m{\overrightarrow {u}}\\&=&(0.8kg)(4\ m.s^{1})(0.8kg)(+6\ m.s^{1})\\&=&(0.8kg)(10\ m.s^{1})\\&=&8\ kg.m.s^{1}\\&=&8\ kg.m.s^{1}{\textbf {\ up}}\end{matrix}}$
where we remembered in the last step to include the direction of the change in momentum in words.
What properties does momentum have?[edit]
You may at this stage be wondering why there is a need for introducing momentum. Remarkably momentum is a conserved quantity. Within an isolated system the total momentum is constant. No matter what happens to the individual bodies within an isolated system, the total momentum of the system never changes! Since momentum is a vector, its conservation implies that both its magnitude and its direction remains the same.
Momentum is conserved in isolated systems!
This Principle of Conservation of Linear Momentum is one of the most fundamental principles of physics and it alone justifies the definition of momentum. Since momentum is related to the motion of objects, we can use its conservation to make predictions about what happens in collisions and explosions. If we bang two objects together, by conservation of momentum, the total momentum of the objects before the collision is equal to their total momentum after the collision.
Principle of Conservation of Linear Momentum: 
The total linear momentum of an isolated system is constant.

or

In an isolated system the total momentum before a collision

(or explosion) is equal to the total momentum after the

collision (or explosion).

Let us consider a simple collision of two pool or billiard balls. Consider the first ball (mass m_{1}) to have an initial velocity (${\overrightarrow {u_{1}}}$). The second ball (mass m_{2}) moves towards the first ball with an initial velocity ${\overrightarrow {u_{2}}}$. This situation is shown in Figure 6.1. If we add the momenta of each ball we get a total momentum for the system. This total momentum is then
${\begin{matrix}{\overrightarrow {p}}_{total\ before}=m_{1}{\overrightarrow {u_{1}}}+m_{2}{\overrightarrow {u_{2}}},\end{matrix}}$

Figure 6.2: After the collision. 
After the two balls collide and move away they each have a different momentum. If we call the final velocity of ball 1 ${\overrightarrow {v_{1}}}$ and the final velocity of ball 2 ${\overrightarrow {v_{2}}}$ (see Figure 6.2), then the total momentum of the system after the collision is
${\begin{matrix}{\overrightarrow {p}}_{total\ after}=m_{1}{\overrightarrow {v_{1}}}+m_{2}{\overrightarrow {v_{2}}},\end{matrix}}$
This system of two balls is isolated since there are no external forces acting on the balls. Therefore, by the principle of conservation of linear momentum, the total momentum before the collision is equal to the total momentum after the collision. This gives the equation for the conservation of momentum in a collision of two objects,


$m_{1}{\overrightarrow {u_{1}}}+m_{2}{\overrightarrow {u_{2}}}=m_{1}{\overrightarrow {v_{1}}}+m_{2}{\overrightarrow {v_{2}}}$ 


m_{1} 
: mass of object 1 (kg) 
m_{2} 
: mass of object 2 (kg) 


${\overrightarrow {u_{1}}}$ 
: initial velocity of object 1 (m.s^{1} + direction) 
${\overrightarrow {u_{2}}}$ 
: initial velocity of object 2 (m.s^{1} + direction) 


${\overrightarrow {v_{1}}}$ 
: final velocity of object 1 (m.s^{1} + direction) 
${\overrightarrow {v_{2}}}$ 
: final velocity of object 2 (m.s^{1} + direction) 
This equation is always true  momentum is always conserved in collisions.
The chapter `Collisions and Explosions' deals with applications of momentum conservation.
At the beginning of this chapter it was mentioned that momentum is closely related to force. We will now explain the nature of this connection.
Consider an object of mass m moving with constant acceleration ${\overrightarrow {a}}$. During a time $\Delta$t the object's velocity changes from an initial velocity ${\overrightarrow {u}}$ to a final velocity ${\overrightarrow {v}}$ (refer to Figure 6.3). We know from Newton's First Law that there must be a resultant force ${\overrightarrow {F}}_{Res}$ acting on the object.

Figure 6.3: An object under the action of a resultant force. 
Starting from Newton's Second Law,
${\begin{matrix}{\overrightarrow {F}}_{Res}&=&m{\overrightarrow {a}}\\&=&m({\frac {{\overrightarrow {v}}{\overrightarrow {u}}}{\Delta t}})\qquad \qquad {\rm {{since}\qquad {\overrightarrow {a}}={\frac {{\overrightarrow {v}}{\overrightarrow {u}}}{\Delta t}}}}\\&=&{\frac {m{\overrightarrow {v}}m{\overrightarrow {u}}}{\Delta t}}\\&=&{\frac {{\overrightarrow {p}}_{final}{\overrightarrow {p}}_{initial}}{\Delta t}}\\&=&{\frac {\Delta {\overrightarrow {p}}}{\Delta t}}\end{matrix}}$
This alternative form of Newton's Second Law is called the Law of Momentum.
Mathematically,
${\overrightarrow {F}}_{Res}={\frac {\Delta {\overrightarrow {p}}}{\Delta t}}$ 


${\overrightarrow {F}}_{Res}$ 
: resultant force (N + direction) 
$\Delta {\overrightarrow {p}}$ 
: change in momentum ($kg.m.s^{1}$ + direction) 
$\Delta$t 
: time over which ${\overrightarrow {F}}_{Res}$ acts (s) 
Rearranging the Law of Momentum,
${\begin{matrix}{\overrightarrow {F}}_{Res}\Delta t&=&\Delta {\overrightarrow {p}}.\end{matrix}}$
The product ${\overrightarrow {F}}_{Res}\Delta t$ is called impulse,
$\mathrm {Impulse} \equiv {\overrightarrow {F}}_{Res}\Delta t=\Delta {\overrightarrow {p}}$ 
From this equation we see, that for a given change in momentum,${\overrightarrow {F}}_{Res}\Delta t$ is fixed. Thus, if F_{Res} is reduced, $\Delta t$ must be increased (i.e. the resultant force must be applied for longer). Alternatively if $\Delta t$ is reduced (i.e. the resultant force is applied for a shorter period) then the resultant force must be increased to bring about the same change in momentum.
Worked Example 36 Impulse and Change in momentum[edit]
Question: A 150 N resultant force acts on a 300 kg object. Calculate how long it takes this force to change the object's velocity from $2\ m.s^{1}\ to\ the\ right$ to $6\ m.s^{1}\ to\ the\ right$
Answer:
Step 1 :
Analyse the question to determine what is given. The question explicitly gives
 the object's mass,
 the object's initial velocity,
 the object's final velocity, and
 the resultant force acting on the object
all in the correct units!
Step 2 :
What is being asked? We are asked to calculate the time taken $\Delta t$ to accelerate the object from the given initial velocity to final velocity. From the Law of Momentum,
${\begin{matrix}{\overrightarrow {F}}_{Res}\Delta t&=&\Delta {\overrightarrow {p}}\\&=&m{\overrightarrow {v}}m{\overrightarrow {u}}\\&=&m({\overrightarrow {v}}{\overrightarrow {u}}).\end{matrix}}$
Thus we have everything we need to find $\Delta t$!
Step 3 :
First we choose a positive direction. Although not explicitly stated, the resultant force acts to the right. This follows from the fact that the object's velocity increases in this direction. Let us then choose right as the positive direction.
Step 4 :
Substituting,
Right is the positive direction
${\begin{matrix}{\overrightarrow {F}}_{Res}\Delta t&=&m({\overrightarrow {v}}{\overrightarrow {u}})\\(+150N)\Delta t&=&(300kg)((+6{\frac {m}{s}})(+2{\frac {m}{s}}))\\(+150N)\Delta t&=&(300kg)(+4{\frac {m}{s}})\\\Delta t&=&{\frac {(300kg)(+4{\frac {m}{s}})}{+150N}}\\\Delta t&=&8s\end{matrix}}$
Worked Example 37 Calculating Impulse[edit]
Question: A cricket ball weighing 156 g is moving at 54 km/h towards a batsman. It is hit by the batsman back towards the bowler at 36\ km/h. Calculate i) the ball's impulse, and ii) the average force exerted by the bat if the ball is in contact with the bat for 0.13 s.
Answer:
Step 1 :
Analyse the question to determine what is given. The question explicitly gives
 the ball's mass,
 the ball's initial velocity,
 the ball's final velocity, and
 the time of contact between bat and ball
all except the time in the wrong units!
Answer to (i):
Step 2 :
What is being asked? We are asked to calculate the impulse
${\begin{matrix}\mathrm {Impulse} =\Delta {\overrightarrow {p}}={\overrightarrow {F}}_{Res}\Delta t.\end{matrix}}$
Since we do not have the force exerted by the bat on the ball (${\overrightarrow {F}}_{Res}$), we have to calculate the impulse from the change in momentum of the ball. Now, since
${\begin{matrix}\Delta {\overrightarrow {p}}&=&{\overrightarrow {p}}_{final}{\overrightarrow {p}}_{initial}\\&=&m{\overrightarrow {v}}m{\overrightarrow {u}},\end{matrix}}$
we need the ball's mass, initial velocity and final velocity, which we are given.
Step 3 : Firstly let us change units for the mass
${\begin{matrix}1000g&=&1kg\\1&=&{\frac {1kg}{1000g}}\\156g\times 1&=&156g\times {\frac {1kg}{1000g}}\\&=&0.156kg\end{matrix}}$
Step 4 :
Next we change units for the velocity
${\begin{matrix}1km&=&1000m\\1&=&{\frac {1000m}{1km}}\end{matrix}}$
${\begin{matrix}3600s&=&1hr\\1&=&{\frac {1hr}{3600s}}\end{matrix}}$
${\begin{matrix}54{\frac {km}{hr}}\times 1\times 1&=&54{\frac {km}{hr}}\times {\frac {1000m}{1km}}\times {\frac {1hr}{3600s}}\\&=&15{\frac {m}{s}}\end{matrix}}$
${\begin{matrix}36{\frac {km}{hr}}\times 1\times 1&=&36{\frac {km}{hr}}\times {\frac {1000m}{1km}}\times {\frac {1hr}{3600s}}\\&=&10{\frac {m}{s}}\end{matrix}}$
Step 5 :
Next we must choose a positive direction. Let us choose the direction from the batsman to the bowler as the positive direction. Then the initial velocity of the ball is ${\overrightarrow {u}}=15\ m.s^{1}$, while the final velocity of the ball is ${\overrightarrow {v}}=+10\ m.s^{1}$
Step 6 :
Now we calculate the change in momentum,
Direction from batsman to bowler is the positive direction
${\begin{matrix}\Delta {\overrightarrow {p}}&=&{\overrightarrow {p}}_{final}{\overrightarrow {p}}_{initial}\\&=&m{\overrightarrow {v}}m{\overrightarrow {u}}\\&=&m({\overrightarrow {v}}{\overrightarrow {u}})\\&=&(0.156kg)((+10\ m.s^{1})(15\ m.s^{1}))\\&=&+3.9\ kg.m.s^{1}\\&=&3.9\ kg.m.s^{1}\ {\textbf {in\ the\ direction\ from\ batsman\ to\ bowler}}\end{matrix}}$
where we remembered in the last step to include the direction of the change in momentum in words.
Step 7 :
Finally since impulse is just the change in momentum of the ball,
${\begin{matrix}\mathrm {Impulse} &=&\Delta {\overrightarrow {p}}\\&=&3.9\ {\mbox{kg}}\cdot {\mbox{m}}\cdot {\mbox{s}}^{1}\\&=&3.9\ {\mbox{N}}\cdot {\mbox{s}}\ {\textbf {in\ the\ direction\ from\ batsman\ to\ bowler}}\end{matrix}}$
Answer to (ii):
Step 8 :
What is being asked? We are asked to calculate the average force exerted by the bat on the ball, ${\overrightarrow {F}}_{Res}$. Now,
${\begin{matrix}\mathrm {Impulse} ={\overrightarrow {F}}_{Res}\Delta t=\Delta {\overrightarrow {p}}.\end{matrix}}$
We are given $\Delta t$ and we have calculated the change in momentum or impulse of the ball in part (i)!
Step 9 :
Next we choose a positive direction. Let us choose the direction from the batsman to the bowler as the positive direction. Then substituting,
Direction from batsman to bowler is the positive direction
${\begin{matrix}{\overrightarrow {F}}_{Res}\Delta t&=&\mathrm {Impulse} \\{\overrightarrow {F}}_{Res}(0.13s)&=&+3.9{\frac {kg.m}{s}}\\{\overrightarrow {F}}_{Res}&=&{\frac {+3.9{\frac {{\mbox{kg}}\cdot {\mbox{m}}}{\mbox{s}}}}{0.13{\mbox{ s}}}}\\&=&30{\mbox{ N}}\ {\textbf {in\ the\ direction\ from\ batsman\ to\ bowler}}\end{matrix}}$
where we remembered in the final step to include the direction of the force in words.
Summary of Important Quantities, Equations and Concepts[edit]
Table 6.1: Summary of the symbols and units of the quantities used in Momentum
Units 
Quantity 
Symbol 
Unit 
SI base units 
Direction 
Momentum 
${\overrightarrow {p}}$ 
 
kg·m·s^{1} 
yes 
Mass 
m 
kg 
kg 
 
Velocity 
${\overrightarrow {u}}$, ${\overrightarrow {v}}$ 
 
m·s^{1} 
yes 
Change in momentum 
$\Delta {\overrightarrow {p}}$ 
 
kg·m·s^{1} 
yes 
Force 
${\overrightarrow {F}}$ 
N 
kg·m·s^{2} 
yes 
Impulse 
J 
N·s 
kg·m·s^{1} 
yes 
Momentum: The momentum of an object is defined as its mass multiplied by its velocity.
Momentum of a System: The total momentum of a system is the sum of the momenta of each of the objects in the system.
Principle of Conservation of Linear Momentum:: `The total linear momentum of an isolated system is constant' or `In an isolated system the total momentum before a collision (or explosion) is equal to the total momentum after the collision (or explosion)'.
Law of Momentum:: The applied resultant force acting on an object is equal to the rate of change of the object's momentum and this force is in the direction of the change in momentum.
Work and Energy[edit]
What are Work and Energy?[edit]
The energy is the ability to do a work.
Energy is measured by the result of applied force/power over a period of time to make changes/work.
Work is defined by a force applied to an object to change this object’s physical properties.
Power = Energy / Time No force used no change and none work was done.
Power and energy are closely related, although, they are not the same.
Power is the rate at which energy is delivered, not an amount of energy itself.
Electrical unit of power is Watt (named after the scientist James Watt):
1 Watt = 1 Joule / Second.
Energy = Power × Time.
Energy exist in various forms: kinetic [Ek = ½ m, v2 ], potential [Ep = m g h], thermal (heat), chemical, electrical, electrochemical, magnetic, sound, light, and nuclear.
Energy can be converted from one form into another in many ways:
Through the gravitation forces When gravity accelerates a falling object, it converts its potential energy to kinetic energy, or when an object is lifted, the gravitational field stores the energy exerted by the lifter as potential energy in the earthobject system.
Through the electric and magnetic fields forces Electrically charged particles in the presence of an electric field possess potential energy. The fields’ forces can accelerate particles and convert this particle's potential energy into kinetic energy. Charged particles can interact via electric and magnetic fields to transfer energy between them, i.e.: an electrical current in a conductor transforms electrical energy into heat.
Frictional Forces A mass object with its potential and kinetic energy associated with the position, orientation, and the object’s motion can be converted into thermal energy (heat), whenever the object slides against another object. The sliding causes the molecules on the surfaces of contact to interact via electromagnetic fields with one another and start vibrating.
Through emitting or absorbing photons of light When photons of lights fall on an object, a photon may pass through the object, be reflected by the object, or be absorbed by the atoms making up the object. Depending on the smoothness of the surface, the scale of the photon's wavelength, the reflection may be either diffuse (rough surface) or coherent (smooth surface). If the photon is absorbed, the photon's energy may also be split and converted in one of these ways: • Photothermal effect The energy absorbed may simply produce thermal energy, or heat in the object. In this case, the photon's energy is converted into vibrations of the molecules called phonons, which is actually heat energy. • Photoelectric effect The energy absorbed may be converted into kinetic energy of conduction electrons, and hence electrical energy. • Photochemical effect The energy may bring about chemical changes that effectively store the energy. Nuclear reactions occur when the nuclei of particles combine [fusion reaction] or when nuclei split apart [fission reaction].
In the International System of Units [SI system], the electrical energy unit is 1 Joule (named after the English physicist James Prescott Joule).
A one Joule is the amount of energy we expend as work if we exert a force of a one Newton of force over a distance of one meter.
It takes a one Joule of energy to lift 1 lb. about 9 inches.
The unit of force in the International System of Units is 1 Newton (named after the English physicist Isaac Newton).
One Newton of force is the force that can accelerate a mass of 1 kilogram (about 2.205 lbs), such that it picks up 1 meter per second of velocity during each second that the force is exerted.
Appliances rating is how much energy per unit time these appliances draw. This quantity is called the "power":
Energy = Power x Time = (100 Joules/Second) × (3600 Seconds) = 360,000 Joules
To do work on an object, one must move the object by applying a force with at least a component in the direction of motion. The work done is given by
$W=F_{\}s$ 


W 
: work done (N.m or J) 
F_{} 
: component of applied force parallel to motion (N) 
s 
: displacement of the object (m) 
It is very important to note that for work to be done there must be a component of the applied force in the direction of motion. Forces perpendicular to the direction of motion do no work.
As with all physical quantities, work must have units. As follows from the definition, work is measured in N.m. The name given to this combination of S.I. units is the joule (J).
Definition: 
1 joule is the work done when an object is moved 1m under the application of a force of 1N in the direction of motion.

The work done by an object can be positive or negative. Since force (F_{}) and displacement (s) are both vectors, the result of the above equation depends on their directions:
 If F_{} acts in the same direction as the motion then positive work is being done. In this case the object on which the force is applied gains energy.
 If the direction of motion and F_{} are opposite, then negative work is being done. This means that energy is transferred in the opposite direction. For example, if you try to push a car uphill by applying a force up the slope and instead the car rolls down the hill you are doing negative work on the car. Alternatively, the car is doing positive work on you!
Worked Example 38 Calculating Work Done I[edit]
Question: If you push a box 20m forward by applying a force of 15N in the forward direction, what is the work you have done on the box?
Answer:
Step 1 : Analyse the question to determine what information is provided
 The force applied is F = 15N.
 The distance moved is s = 20m.
 The applied force and distance moved are in the same direction. Therefore, $F_{\}=15N$
These quantities are all in the correct units, so no unit conversions are required.
Step 2 : Analyse the question to determine what is being asked
 We are asked to find the work done on the box. We know from the definition that work done is W = F_{}s
Step 3 : Next we substitute the values and calculate the work done
${\begin{matrix}W&=&F_{\}s\\&=&(15N)(20m)\\&=&300\ N\cdot m\\&=&300\ J\end{matrix}}$
Remember that the answer must be positive as the applied force and the motion are in the same direction (forwards). In this case, you (the pusher) lose energy, while the box gains energy.
Worked Example 39 Calculating Work Done II[edit]
Question: What is the work done by you on a car, if you try to push the car up a hill by applying a force of 40N directed up the slope, but it slides downhill 30cm?
Answer:
Step 1 : Analyse the question to determine what information is provided
 The force applied is F = 40N
 The applied force and distance moved are in opposite directions. Therefore, if we take s = 0.3m, then $F_{\}=40N$.
Step 2 : Analyse the question to determine what is being asked
 We are asked to find the work done on the car by you. We know that work done is W = F_{}s
Step 3 : Substitute the values and calculate the work done
Again we have the applied force and the distance moved so we can proceed with calculating the work done:
${\begin{matrix}W&=&F_{\}s\\&=&(40N)(0.3m)\\&=&12N\cdot m\\&=&12\ J\end{matrix}}$
Note that the answer must be negative as the applied force and the motion are in opposite directions. In this case the car does work on the person trying to push.
What happens when the applied force and the motion are not parallel? If there is an angle between the direction of motion and the applied force then to determine the work done we have to calculate the component of the applied force parallel to the direction of motion. Note that this means a force perpendicular to the direction of motion can do no work.
Worked Example 40 Calculating Work Done III[edit]
Question: Calculate the work done on a box, if it is pulled 5m along the ground by applying a force of F = 10N at an angle of 60^{o} to the horizontal.
Answer:
Step 1 :
Analyse the question to determine what information is provided
 The force applied is F = 10N
 The distance moved is s = 5m along the ground
 The angle between the applied force and the motion is 60^{o}
These quantities are in the correct units so we do not need to perform any unit conversions.
Step 2 : Analyse the question to determine what is being asked
 We are asked to find the work done on the box.
Step 3 : Calculate the component of the applied force in the direction of motion
Since the force and the motion are not in the same direction, we must first calculate the component of the force in the direction of the motion.
From the force diagram we see that the component of the applied force parallel to the ground is
${\begin{matrix}F_{}&=&F\cdot \cos(60^{o})\\&=&10N\cdot \cos(60^{o})\\&=&5\ N\end{matrix}}$
Step 4 : Substitute and calculate the work done
Now we can calculate the work done on the box:
${\begin{matrix}W&=&F_{\}s\\&=&(5N)(5m)\\&=&25\ J\end{matrix}}$
Note that the answer is positive as the component of the force F_{} is in the same direction as the motion.
We will now discuss energy in greater detail.
As we mentioned earlier, energy is the capacity to do work. When positive work is done on an object, the system doing the work loses energy. In fact, the energy lost by a system is exactly equal to the work done by the system.
Like work (W) the unit of energy (E) is the joule (J). This follows as work is just the transfer of energy.
A very important property of our universe which was discovered around 1890 is that energy is conserved.
Energy is never created nor destroyed, but merely

transformed from one form to another.

Energy conservation and the conservation of matter are the principles on which classical mechanics is built.
IN THE ABSENCE OF FRICTION 
When work is done on an object by a system:

the object gains energy equal to the work done by the system

Work Done = Energy Transferred 
Thermal energy (heat) is the disorganized movement of microscopic particles. Once energy is converted to this form, it has limited usefulness for doing further work in the system. Friction is the general name for forces that converts energy to heat.
IN THE PRESENCE OF FRICTION 
When work is done by a system: 
only some of the energy lost by the system is transferred into useful energy 
the rest of the energy transferred is lost to heat by friction 
Total Work Done = Useful Work Done + Work Done Against Friction 
Types of Energy[edit]
So what different types of energy exist? Kinetic, mechanical, thermal, chemical, electrical, radiant, and atomic energy are just some of the types that exist. By the principle of conservation of energy, whenwork is done energy is merely transferred from one object to another and from one type of energy to another.
Kinetic Energy[edit]
Kinetic energy is the energy of motion that an object has. Objects moving in straight lines possess translational kinetic energy, which we often abbreviate as E_{k}.
The translational kinetic energy of an object is given by
$E_{k}={\frac {1}{2}}mv^{2}$ 


E_{k} 
: kinetic energy (J) 
m 
: mass of object (kg) 
v 
: speed of the object (m.s^{1}) 
Note the dependence of the kinetic energy on the speed of the object kinetic energy is related to motion. The faster an object is moving the greater its kinetic energy.
Worked Example 41 Calculation of Kinetic Energy[edit]
Question: If a rock has a mass of 1kg and is thrown at 5m/s, what is its kinetic energy?
Answer:
Step 1 : Analyse the question to determine what information is provided
 The mass of the rock m = 1kg
 The speed of the rock v = 5m/s
These are both in the correct units so we do not have to worry about unit conversions.
Step 2 : Analyse the question to determine what is being asked
 We are asked to find the kinetic energy. From the definition we know that to work out E_{k}, we need to know the mass and the velocity of the object and we are given both of these values.
Step 3 : Substitute and calculate the kinetic energy
${\begin{matrix}E_{k}&=&{\frac {1}{2}}mv^{2}\\&=&{\frac {1}{2}}(1kg)(5{\frac {m}{s}})^{2}\\&=&12.5{\frac {kg\cdot m^{2}}{s^{2}}}\\&=&12.5\ J\end{matrix}}$
To check that the units in the above example are in fact correct:
${\begin{matrix}{\frac {kg\cdot m^{2}}{s^{2}}}&=&\left({\frac {kg\cdot m}{s^{2}}}\right)\cdot m=N\cdot m\\&=&J\end{matrix}}$
The units are indeed correct!
Study hint: Checking units is an important crosscheck
and

you should get into a habit of doing this. If you, for example, 
finish an exam early then checking the units in your calculations is a very good idea. 
Worked Example 42 Mixing Units and Kinetic Energy Calculations 1[edit]
Question: If a car has a mass of 900kg and is driving at 60km/hr, what is its kinetic energy?
Answer:
Step 1 : Analyse the question to determine what information is provided
 The mass of the car m = 900kg
Step 2 : Analyse the question to determine what is being asked
 We are asked to find the kinetic energy.
Step 3 : Substitute and calculate
We know we need the mass and the speed to work out E_{k} and we are given both of these quantities. We thus simply substitute them into the equation for E_{k}:
${\begin{matrix}E_{k}&=&{\frac {1}{2}}mv^{2}\\&=&{\frac {1}{2}}(900kg)(16.67{\frac {m}{s}})^{2}\\&=&125\ 000{\frac {kgm^{2}}{s^{2}}}\\&=&125\ 000\ J\end{matrix}}$
Worked Example 43 Mixing Units and Kinetic Energy Calculations 2[edit]
Question: If a bullet has a mass of 150kg and is shot at a muzzle velocity of 960m/s, what is its kinetic energy?
Answer:
Step 1 : Analyse the question to determine what information is provided
 We are given the muzzle velocity which is just how fast the bullet leaves the barrel and it is v = 960m/s.
Step 2 : Analyse the question to determine what is being asked
 We are asked to find the kinetic energy.
Step 3 : Substitute and calculate
We just substitute the mass and velocity (which are known) into the equation for E_{k}:
${\begin{matrix}E_{k}&=&{\frac {1}{2}}mv^{2}\\&=&{\frac {1}{2}}(150kg)(960{\frac {m}{s}})^{2}\\&=&69\ 120{\frac {kgm^{2}}{s^{2}}}\\&=&69\ 120\ J\end{matrix}}$
Potential Energy[edit]
If you lift an object you have to do work on it. This means that energy is transferred to the object. But where is this energy? This energy is stored in the object and is called potential energy. The reason it is called potential energy is because if we let go of the object it would move.
Definition: 
Potential energy is the energy an object has due to its position or state.

As an object raised above the ground falls, its potential energy is released and transformed into kinetic energy. The further it falls the faster it moves as more of the stored potential energy is transferred into kinetic energy. Remember, energy is never created nor destroyed, but merely transformed from one type to another. In this case potential energy is lost but an equal amount of kinetic energy is gained.
In the example of a falling mass the potential energy is known as gravitational potential energy as it is the gravitational force exerted by the earth which causes the mass to accelerate towards the ground. The gravitational field of the earth is what does the work in this case.
Another example is a rubberband. In order to stretch a rubberband we have to do work on it. This means we transfer energy to the rubberband and it gains potential energy. This potential energy is called elastic potential energy. Once released, the rubberband begins to move and elastic potential energy is transferred into kinetic energy.
Gravitational Potential Energy[edit]
As we have mentioned, when lifting an object it gains gravitational potential energy. One is free to define any level as corresponding to zero gravitational potential energy. Objects above this level then possess positive potential energy, while those below it have negative potential energy. To avoid negative numbers in a problem, always choose the lowest level as the zero potential mark. The change in gravitational potential energy of an object is given by:
$\Delta E_{P}=mg\Delta h$ 


$\Delta E_{P}$ 
: Change in gravitational potential energy (J) 
m 
: mass of object (kg) 
g 
: acceleration due to gravity (m.s^{2}) 
$\Delta h$ 
: change in height (m) 
When an object is lifted it gains gravitational potential energy, while it loses gravitational potential energy as it falls.
Worked Example 44 Gravitational potential energy[edit]
Question: How much potential energy does a brick with a mass of 1kg gain if it is lifted 4m.
Answer:
Step 1 : Analyse the question to determine what information is provided
 The mass of the brick is m = 1kg
 The height lifted is $\Delta h=4m$
These are in the correct units so we do not have to worry about unit conversions.
Step 2 : Analyse the question to determine what is being asked
 We are asked to find the gain in potential energy of the object.
Step 3 : Identify the type of potential energy involved
Since the block is being lifted we are dealing with gravitational potential energy. To work out $\Delta E_{P}$, we need to know the mass of the object and the height lifted. As both of these are given, we just substitute them into the equation for $\Delta E_{P}$.
Step 4 : Substitute and calculate
${\begin{matrix}\Delta E_{P}&=&mg\Delta h\\&=&(1kg)\left(10{\frac {m}{s^{2}}}\right)(4m)\\&=&40{\frac {kg\cdot m^{2}}{s^{2}}}\\&=&40\ J\end{matrix}}$
Mechanical Energy and Energy Conservation[edit]
Kinetic energy and potential energy are together referred to as mechanical energy. The total mechanical energy (U) of an object is then the sum of its kinetic and potential energies:
${\begin{matrix}U&=&E_{P}+E_{K}\\U&=&mgh+{\frac {1}{2}}mv^{2}\end{matrix}}$ 
(7.1)

Now,
IN THE ABSENCE OF FRICTION 
Mechanical energy is conserved 
$U}_{before}={U}_{$     