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## Mechanics Introduction

Sections:
• 0, Chapter 1, Introduction,

### Introduction

This Unit introduces us to the broad topic of applied mathematics that is how to use mathematics to solve real life problems, and in particular applying mathemtics to solving problems of simple mechanics.

After completing this unit you should

• Understand the concept of mathematical modelling when applied to problems in mechanics
• Understand the principles of empirical modelling where data is used to model the
practical problem
• Appreciate that force causes a change in motion
• Be aware that there are many types of force, including Weight (gravitational force), Normal Reaction force, Tension and Friction
• Understand the basis for Newton’s Law of Gravitation
• Be able to use dimensional consistency in equations.

You have six sections to work through namely;

1. Mathematical Modelling
2. Modelling from Data
3. Motion and Force
4. Newton’s Law of Gravitation
5. Further Mathematical Modelling
6. Units and Dimensions
• 0, Chapter 2, Mathematical Modelling,

### Mathematical Modelling

Mathematics occurs in many subjects: engineering and physics in particular, but also in economics, biology, epidemiology, and transport, making it vital to industry, commerce, government and health. Applying mathematics to such a wide range of subjects requires not only good mathematical problem solving skills but also the ability of the mathematician to start with a problem in non-mathematical form and to give the results of any mathematical analysis in non-mathematical form. In between these start and end points, the mathematician must:

formulate the problem to allow the use of mathematical analysis,

solve any mathematical problems that have been set up,

interpret the solution in the context of the original setting.

This process is called mathematical modelling and is illustrated by the diagram below.

• 1, Chapter 2.1, Building a mathematical model,

### Building a mathematical model

Central to mathematical modelling is the representation of the real world problem mathematically. In mechanics the real world problem is commonly represented with a formula.  Such a representation is what is meant by a mathematical model. The mathematical model could also come in other forms such as a graph, a spreadsheet, a flow chart, and many others.

For example, the formula $$s = 15t$$ describes distance travelled $$s$$ (in metres) in terms of time $$t$$ (in seconds) when travelling at a speed of $$15 ms^{-1}$$. The formula $$s = 15t$$ is a simple example of a mathematical model.

Broadly speaking, a mathematical model is a relation between two or more variables. The challenge to the applied mathematician is to formulate a good model which accurately describes or represents a given situation. A good model, importantly, can be used to make predictions about future behaviour.

The first step in trying to devise a mathematical model for a given situation is to identify the quantities which can be measured and whose values will describe the real situation.

In mechanics these quantities might be position, velocity, mass, etc. More generally you might be interested in mortality rates, inflation, interest rates, infection rate of a virus, etc.

Such quantities are called variables. In the previous example the variables are distance $$\left(s\right)$$ and time $$\left(t\right)$$.

• 0, Chapter 3, Modelling from Data,

### Modelling from Data

One of the simplest methods for finding a mathematical model for a given situation is to carry out an appropriate experiment to collect relevant data: from this data a formula relating the variables can be found. This is called Empirical Modelling.In the following section we will demonstate this method with two examples.

• 1, Chapter 3, Task 1, Worked Example 1,

### Worked Examples

Use the slider to explore worked examples.

• #### Worked Example

A motorist records the distance they travel along a motorway at various times. The data are recorded in the table below:

Produce a model for this data represented by a formula, using the variables $$s$$ for distance travelled and $$t$$ for time.

• #### Worked Example

A ball is dropped onto a hard floor. A model for the motion of the ball is investigated. Sketch a graph of Height against Time for a ball dropped from $$5$$ metres. You can produce this sketch by doing your own experiment, or you can find videos online which show practical demonstrations of experiments with a bouncing ball.

• 0, Chapter 4, Motion and Force,

### Motion and Force

Mechanics is concerned with the motion of objects and what influences motion. Motion is the change in position and orientation of an object with time. An object’s motion can be described simply by its path: the route it follows on its journey.

In many cases motion is complicated; take for example the motion of a tennis ball from server to receiver.  At first sight it might be thought that the ball moves in a straight line from the server’s racket to the ground, before bouncing to the receiver. However, on closer observation the tennis ball is seen to move in a curved path. In fact, the motion of the ball could also involve a swerve and/or a dipping in its path.

In this section the paths of objects in simple scenarios are considered. To do this an object will be treated as a particle. This means that the size of the object is ignored and the path of the object is modelled by following some point on the object, as shown with the tennis ball below; it also allows the spin/rotation of the object to be ignored.

The examples here will only go as far as describing and sketching the path of an object, but later it will be possible to describe the motion mathematically, even producing a formula, thereby being able to predict where an object will be in the future. The mathematical description of motion considers variables of position, velocity, acceleration and time.

• 1, Chapter 4, Task 1, Worked Example 2,

### Worked Example

• #### Worked Example

In each of the situations below sketch, or describe, the path of the object:

a)     A tennis ball is dropped and falls to the ground.

b)     A golf ball is rolled along a table and falls to the ground from the table’s end.

c)     A tennis ball is thrown away from a spacecraft, into the blackness of space.

d)     A satellite orbits the Earth.

e)     A golf ball is struck into the air and lands a long distance away.

• 2, Chapter 4.1, Force,

### Force

Force causes a change in motion.  If an object which is initially at rest (not moving) suddenly starts to move, then something must have caused the motion to begin: it is force which has this effect.

You are familiar with forces, and know that if you apply a force to an object it will move. If you ‘push’ an object, such as a pen on a table, then let go, you have applied a force to it, so you will see it continue to slide along the table. Note that it can continue to move even after you let go, and have stopped applying the force.

The principles of force and its effect on motion can be seen more easily in space, where an object is free to move without strong effects of gravity, friction and air resistance. If you apply a force to an object in space it will move, and continue to move, in a straight line away from you, unless something else interacts with it. There are some good films which show the effects of forces in space, such as ‘Gravity’, ‘Apollo 13’ and ‘2001: A Space Odyssey’; there are also relevant videos online showing real footage from space missions.

A force can also stop an object from moving. When you slide a pen across a table, it is the force of friction which slows down and eventually stops the pen from moving. On an ice rink there is much less friction and an ice puck will continue to move for longer. You could also apply a force to change its direction, changing its path to the left or the right. Finally, you can apply a force to speed it up a little, or you can apply a force to slow it down a little.

So, in summary, a force can

• start motion
• stop motion
• change the direction of motion
• make an object move faster or more slowly.

These are all examples of forces changing the motion of an object

When there are no forces acting on an object, it will remain at rest or it will continue in a straight line at a constant velocity. This hints at Newton’s laws, to be covered in detail in the another Unit.

Note

It may be counter-intuitive, but force is not required to maintain motion. If you are floating freely in space and you shoot a fire extinguisher, to create a force of thrust, you will begin to move. You will then continue to move, even after you release the fire extinguisher. You will continue to move at a fixed, constant velocity, unless another force is applied to you.

The spacecraft Voyager 2 was launched in 1977: it is travelling at over 9 miles per second, is over 11 billion miles from Earth and counting (you can use the NASA website to see how far from us it is now). But this spacecraft is no longer using jets to propel itself; once it was given an initial force of thrust to start its motion, it did not require further force to continue on its path.

• 1, Chapter 4.1, Task 1, Worked Example 3,

### Worked Example

ADD EXAMPLE ON CHART 24, 25

• 1, Chapter 4.2, Multiple Forces and Net Force,

### Multiple Forces and Net Force

When solving Earth-based problems you will usually be dealing with more than one force. When multiple forces act upon an object, the forces combine and can be treated as a single resultant force, called a net force.

For example consider the situation described in the diagram above. The diagram shows a skydiver who is falling to the ground from a high altitude. Such a skydiver can obtain a terminal velocity, where they fall at a constant speed towards the ground. From what we know already, this tells us that the net force on the skydiver must be zero, as if it was not, the skydiver would be speeding up or slowing down. This is because the skydiver is subject to two forces: gravity and drag, shown in the diagram.

When the skydiver is falling at terminal velocity, the forces of gravity and drag are equal and opposite, cancelling one another out, resulting in a net force of zero. Typically this constant velocity is around 120 miles per hour.

Here is a link to video which shows describes a skydivers path of travel.

• 1, Chapter 4.3, Units of Force,

### Units of Force

The mathematical description of forces leads to a measure for force:

#### Remember

Force is measured in Newtons (N)

The unit is named after Isaac Newton, whose laws will be used in the another of the Mechanics Units to solve mechanics problems in a more mathematical way. When you hold a typical smartphone in your hand you will feel a force of a little over 1 Newton from the weight of the phone. Weight is the first type of force described in the next section.

• 1, Chapter 4.4, Types of Force,

### Types of Force

A number of forces have been mentioned so far: friction, thrust, drag and gravitational force. This section will describe the most common forces that will be encountered in the material that follows. It can be helpful to categorise them as contact forces and non-contact forces.

In the following sections we describe the four forces;

• Weight
• Normal Reaction
• Tension and Thrust
• Friction
• 1, Chapter 4.5, Weight (gravitational force),

### Weight (gravitational force)

The gravitational force between two objects is the most familiar non-contact force. If you hold a pen in front of you and let it go, it falls to the ground. The force which produced this motion is the force of gravity. Even while an object is in the air and not in contact with any other object, gravity still acts upon it and it continues to speed up. This is why gravitational force is described as a non-contact force.

The gravitational force an object feels is called its weight. You can roughly work out the weight of an object on the surface of the Earth, in Newtons, by multiplying its mass in kilograms by ten.

All other forces you come across in this section are contact forces in which there must be contact between two surfaces for the force to exist.  Words such as push, pull, tension, hit, knock, load and friction are used to describe contact forces.

• 1, Chapter 4.6, Normal Reaction force,

### Normal Reaction force

Consider the forces acting on you when you sit on a horizontal table. You have your weight acting on you in a downwards direction. But you are not moving, so there must be another force acting upwards to balance your weight. This force comes from the table and is called the Normal Reaction force. These two forces are shown in the diagram on the right.

The normal reaction force is a contact force and its direction is at a right angle to (or is ‘normal’ to) the table.

Note that whenever two objects are in contact there is a contact force between them.  The table exerts a force upwards on you and you exert a force downwards on the table. If the table is not strong enough, it is possible for you to break it.

• 1, Chapter 4.7, Tension and Thrust,

### Tension and Thrust

Consider the forces acting on you when you are hanging from a rope as shown in the diagram below. You have your weight acting on you in a downwards direction. But, again, you are not moving, so there must be another force acting upwards to balance your weight. This force comes from the rope and is called tension. These two forces are shown in the diagram.

Similar to tension is thrust. Whereas tension relates to a material exerting force by pulling you, thrust is a force which pushes you, such as a pogo stick, which pushes you up.

• 1, Chapter 4.8, Friction,

### Frictions

When an object is sliding along a surface, it will tend to slow down. This implies that there is a force acting against it. This force is friction.

It is another contact force. Friction acts parallel to the surface of contact.

If an object is on a slope, friction acts along the line of the slope, as shown in the diagram above. Friction can prevent an object from sliding, or just slow its progress if it is already sliding. The Normal Reaction force and Weight are also shown on the diagram.

• 1, Chapter 4.8, Task 1, Worked Example 4,

### Worked Example

• #### Worked Example

A book is on a table and is not moving.

a)     Which forces are acting on the book?

The table is now lifted at one end so that it is tilted, the book is now on a slope. The book still does not move.

b)     Which forces are now acting on the book?

• 0, Chapter 5, Newton’s Universal Law of Gravitation,

### Newton’s Universal Law of Gravitation

In the last section a rough formula was given to find weight on the Earth’s surface, the force applicable to an object due to its mass: you can multiply the mass of the object in kg by ten to obtain its weight in Newtons.

The weight of an object of mass $$1\;kg$$ is approximately $$10\;Newtons$$.

As a formula this can be written as

#### Formula

$$Weight = mg$$, where $$g \sim 10\;ms^{-2}$$

Experience confirms this result. A bowling ball has a larger mass $$\left(\sim7\;kg\right)$$ than a golf ball $$\left( \sim 0.05\;kg \right)$$. If you hold a golf ball in one hand and a bowling ball in the other hand you will feel the gravitational force of attraction, the weight, of each object, pressing down on your hands. The force from the bowling ball, however, is far stronger. So the force of gravity on an object depends on the mass of the object.

• 1, Chapter 5.1, Mass and Weight,

### Mass and Weight

This section will give a further example of a well-established mathematical model and through it an improved model for gravity. The model will show that a more accurate value for calculating weight on the Earth’s surface is obtained by multiplying the mass by the value $$9.81$$ rather than $$10$$. This section will also introduce a general law for gravity which can be used in problems that are not Earth-based. It will explain why the figure $$9.81$$ is used in more accurate calculations on the surface of the Earth.

You are probably aware that if you were not on the Earth’s surface, on a space station, or on the Moon, the force of gravity would be different. At this point it is important to understand the difference between mass and weight, since the words are often interchanged and confused in everyday language. Your mass, in kg, does not change, whether you are on the Earth or on the Moon, or elsewhere in space, but your weight in Newtons, does change depending where you are.

This applies to any object, so if you are standing on the Moon rather than on the Earth and you are holding the golf ball and the bowling ball, you will find that the weights of the objects are far less; little weight is felt from the golf ball and much less weight from the bowling ball.

mass, in kg, is fixed does not change when you are in a different location

weight, in Newton, changes depending where you are

• 1, Chapter 5.2, Newton’s Universal Law of Gravitation,

### Newton’s Universal Law of Gravitation

Every object in the Universe attracts every other object in the Universe with a force, $$F$$, that has magnitude (or size) directly proportional to the masses of the particles, $$m_{1}$$ and $$m_{2}$$, and inversely proportional to the square of their distance apart, $$d$$ (the distance between their centres of mass);

$$F= \frac {Gm_{1}m_{2}}{d^{2}}$$

The proportionality constant $$G$$ is called the gravitational constant and in SI units has the value

$$6.67 \times 10^{-11} m^{3} s^{-2} kg^{-1}$$

This is a general law, which tells you that any two objects anywhere in the universe will attract one another. It tells you that the gravitational force between two objects is bigger if the objects have a larger mass, and is also dependent on how close the objects are to one another, becoming weaker when the objects are further apart.

For you, the largest gravitational force you will feel is towards the ground. This is because the largest object which is close to you is planet Earth, weighing in at $$5.98 \times 10^{24} kg$$ . Note that the distance here is the distance between the objects’ centres of mass; each is a point, which for Earth is at its centre and for a person is inside their body, a little below their navel.

• 1, Chapter 5.2, Task 1, Worked Example 5,

### Worked Example

• #### Worked Example

Apply this formula to an object of mass $$m$$ on the Earth's surface.

$$F=\frac{Gm_{1}m_{2} }{d^{2} }$$

The radius of the Earth is $$6.37 \times 10^{6}$$ metres.

The mass of the Earth is $$5.98 \times 10^{24}$$ kg.

$$G$$ is a constant and the value of $$6.67 \times 10^{-11}$$ , as given previously.

• 0, Chapter 6, Further Mathematical Modelling,

### Further Mathematical Modelling

The formulation of the law of gravitation demonstrates another form of mathematical modelling. In the previous section models were formulated by doing experiments and collecting data. From this data the model in the form of a graph and/or an equation takes shape: this is called empirical modelling.

Newton’s work demonstrates a type of modelling called theoretical modelling. You can formulate a model based on an understanding of the physical situation and the important features that affect the situation. Newton proposed that the force between two objects depends on the masses of the objects and the distance between them.  He then made three important assumptions:

• the force is an attractive force;
• the force is directly proportional to each mass;
• the force is inversely proportional to the square of the distance.

From these assumptions the model follows. Appropriate data or experimental activities are then used to validate the model. Notice how this approach uses the data at the end of the activity to test the model and not at the beginning to formulate the model. The diagram below summarises this type of modelling.

• 0, Chapter 7, Units and Dimensions,

### Units and Dimensions 1

Throughout this unit physical quantities will be measured in an agreed international system of units called SI units. In this system distances, displacements and positions are measured in metres ($$m$$), time in seconds ($$s$$) and mass in kilograms ($$kg$$).

In mechanics, nearly all quantities, somewhat surprisingly, are expressed in terms of basic units of mass, length and time, denoted by the symbols $$M, L$$ and $$T$$.

For example, velocity can be expressed as

$$\frac{L}{T}$$ or $$LT^{-1}$$

Similarly, acceleration $$\left(ms^{-2}\right)$$ can be expressed as

$$\frac {L}{T^{2}}$$ or $$LT^{-2}$$

The dimensions of velocity and acceleration are therefore $$LT^{-1}$$ and  $$LT^{-2}$$ respectively.

Since SI units are used for the measurements of quantities, we use the following units:

M is measured in kilograms

L is measured in metres

T is measured in seconds

For example, a farmer may be able to carry out a particular task at the rate of $$5$$ hectares per hour.  The dimensions of hectares, i.e. an area, are $$L^{2}$$, so

dimensions of hectares per hour are  $$L^{2} T^{-1}$$

Note that numbers like $$2, \pi, 3,$$ etc. have no units: they are said to be dimensionless. Some quantities such as force (in Newtons) are not expressed in terms of the common SI units; however, the dimensions can be established.

• 1, Chapter 7.1, Units of Dimensions 2,

### Units and Dimensions 2

As another example, the Universal Law of Gravitation met in the previous section is given as

$$F=\frac{Gm_{1}m_{2}}{d^{2}}$$

and the proportionality constant $$G$$, in SI units, has the value $$6.67 \times 10^{-11} m^{3} s^{-2} kg^{-1}$$.

The dimensions of $$F$$ are therefore given by combining the dimensions of  $$G, m_{1}, m_{2}$$ and $$d^{2}$$.

So dimensions of $$F$$ are:

$$F=\frac{\frac{L^{3}}{T^{2}M} \times M \times M }{L^{2}} =\frac{L^{3}M}{T^{2}L^{2}} = MLT^{-2}$$

To back up the previous finding, another way of expressing force, $$F$$, is by using Newton's law of motion, $$F=ma$$, where m is the mass and a the acceleration, this is discussed in the next unit but can be seen to  confirm the same result, with Force having dimensions of $$MLT^{-2}$$.

As force is in such common use, the unit of the Newton, $$N$$, is used for efficiency rather than the unit $$kg\;m\;s^{-2}$$, based around $$MLT^{-2}$$, which would be a bit cumbersome. Many measures in common use have their own units, e.g. Pascals, Pa, for pressure rather than the unit $$kg\; m^{-1}\; s^{-2}$$ as it would be based on its dimensions, again, a bit too long-winded.

• 1, Chapter 7.2, Dimensional Consistency,

### Dimensional Consistency

Equations can be checked for dimensional consistency. In other words the dimensions of both sides of the equations should be the same

The Worked Examples in the next section show how this is done

• 1, Chapter 7.2, Task 1, Worked Example 6,

### Worked Example

ADD EXAMPLES ON CHARTS 55 and 56 AND 57

• 1, Chapter 7.3, Using Dimensions to Produce Formulae,

### Using Dimensions to Produce Formulae

Experimenters sometimes use dimensions to evolve formulae. They create a formula by beginning with an educated guess at it, using measures which would be expected to be related, then they use the principles of balancing dimensions to produce the final formula.

For example, it is known that the frequency, $$f$$, of a stretched wire depends on the mass per unit length, $$m$$, the length of the vibrating wire, $$l$$, and the force, $$F$$, used to stretch the wire.  The problem is, therefore, to find the relationship between these quantities.

An initial formula is proposed,

$$f = k m^{a} t^{b} F^{c}$$

where $$k$$ is a constant to be found later and $$a, b$$ and $$c$$ are powers to now be determined.

The dimensions of f must be the same as the dimensions of $$m^{a} l^{b} F^{c}$$

The dimensions of each quantity are identified:

The frequency means the number of vibrations per second, so its dimensions are $$T^{-1}$$.

Mass per unit length $$m$$ has dimensions $$ML^{-1}$$.

Length $$l$$ has dimension $$L$$.

Force $$F$$  has dimensions $$MLT^{-2}$$.

$$f= k m^{a} l^{b} F^{c}$$

So the dimensions of the proposed formula can be substituted in:

$$T^{-1} = \left(ML^{-1}\right)^{a} L^{b} \left(MLT^{-2}\right)^{c}$$

Remove brackets

$$T^{-1} = M^{a}L^{-a}L^{b}M^{c}L^{c}T^{-2c}$$

Collect powers of each dimension

$$T^{-1} = M ^{a+c} \times L^{-a+b+c} \times T^{-2c}$$

Equating the indices for $$M, L$$ and $$T$$ on both sides of the equation gives

$$a+c=0$$

$$-a + b + c = 0$$

$$-2c = -1$$

Solving these equations gives $$a= -\frac{1}{2}$$, $$b= -1$$ and $$c=\frac{1}{2}$$

So the formula becomes

$$f= km^{-\frac{1}{2}}$$ $$l^{-1}$$ $$F^{\frac{1}{2}}$$

or $$f=\frac{k}{l} \sqrt{\frac{F}{m}}$$

where $$k$$  is a constant whose value could be found from experimentation.

• 0, Chapter 8, Summary,

### Summary

• The process of considering a real world problem, formulating a mathematical problem, solving the mathematical problem
and interpreting the solution in the context of the original problem is called Mathematical Modelling and is represented
in the diagram below:
• A force on a particle can start motion, stop motion, change the direction of motion and make an object move faster or slower.
• The weight of an object of mass $$1\;kg$$ is approximately $$10\;Newtons$$, that is weight $$= mg$$, where $$g\;10\;m$$
• Note that mass, in kg, is fixed does not change when you are in a different location but weight, in Newtons, changes depending
where you are.
• Newton’s Universal Law of Gravitation states that “Every object in the Universe attracts every other object in the Universe with a force, $$F$$,
that has magnitude (or size) directly proportional to the masses of the particles, and , and inversely proportional to the square of their distance apart, d (the distance between their centres of mass); as given by the formula: where the proportionality constant $$G$$ is called the gravitational constant and in $$SI$$ units has the value $$G = “$$
• Dimensions, given in terms of $$M$$ (mass), $$L$$ (length) and $$T$$ (Time) must balance across an equation.
Interactive Exercises:
• Mechanics Text Exercises, https://www.cimt.org.uk/sif/mechanics/m1/text.pdf
• Introduction to Mechanics, https://www.cimt.org.uk/sif/mechanics/m1/index.htm
File Attachments:

## Projectiles

Sections:
• 0, Chapter 1, Introduction,

### Introduction

This unit explores the motion of projecticles. The suvat equations which we describe in the unit One Dimnesional Motion can be applied to this scenario, which becomes a problem based in two dimensions.

After completing this unit you should be able to

• Understand the basis of the motion of an object under the effect of gravity
• Develop the model for a particle moving in two dimensions under gravity, finding the maximum height reached and the horizontal range
• Use vectors to describe two-dimentional motion under gravity.

You have two sections to work through namely;

1. Projectiles and 2D problems
2. Further Projectile Problems
• 0, Chapter 2, Projectiles,

### Projectiles

A projectile is an object which has been given an initial velocity and travels under the influence of gravity along a curved path; examples are a cannonball, or a stone that has been thrown into the air. A number of assumptions will be used to model projectile motion:

#### Assumptions

• The projectile is treated as a particle: its size and spin are negligible.
• The effect of gravity can be treated as a constant downwards acceleration on the object. An acceleration of $$a=10\;ms^{-2}$$ will be used for convenience, which is sufficiently accurate. Note that in reality the effect of gravity varies depending on where you are on the Earth’s surface and how far you are from the centre of the Earth, as discussed in the Mechanics Introduction unit.
• Air resistance is negligible; the only force acting on the projectile is gravity.
• 1, Chapter 2.1, Motion Under Gravity,

### Motion under Gravity

To begin, the motion of an object under gravity which has been thrown into the air directly upwards can be considered. This motion is addressed in the unit on One Dimensional Motion. If an object is given an initial velocity upwards, it travels up, with gravity acting downwards, directly against it.

For acceleration, $$a$$, a positive sign indicates acceleration upwards. A negative sign indicates deceleration, or a ‘pull’ downwards. This will be gravity, so in this scenario,

$$a = g = - 10\;^{-2}$$ will be used.

For $$s$$, a position of zero is set as the starting position, then positive $$s$$ is above this position (height) and negative $$s$$ is below it.

For $$v$$, a positive sign indicates that the direction of travel is upwards, a negative sign indicates that the direction of travel is downwards.

For this scenario the suvat equations can be easily applied, with gravity as the constant acceleration acting upon the motion of the object.

This is illustrated in the next worked examples

• 1, Chapter 2.1, Task 1, Worked Example 1,

### Worked Example

Use the slider to explore worked examples.

• #### Worked Example

A ball is thrown vertically upwards with an initial speed of $$15 ms^{-1}$$.

Calculate the height reached after $$2$$ seconds.

• #### Worked Example

A ball is thrown vertically upwards with a velocity of $$12 ms^{-1}$$.

Calculate the times on its path when the ball is at a height of $$4$$ metres.

• 1, Chapter 2.2, Working in 2 Dimensions,

### Working in 2 Dimensions

Following the recap on motion of an object under the influence of gravity in one dimension, the extension can be made to two dimensions, which allows for the general motion of a projectile to be modelled.

It will be demonstrated that projectiles, under the assumptions used here, will travel along a parabolic path as shown below.

To show this the progression needs to be made to two dimensions. The position in the upwards direction, the vertical position described earlier, will be labelled as $$y$$ and the horizontal position will be labelled as $$x$$. This is shown in the diagram below.

The horizontal position $$x$$ and the vertical position $$y$$ can be handled independently from one another. So for a given time the horizontal position and the vertical position can be calculated using the suvat equations.

For a given initial velocity u of magnitude u at an angle of $$\theta$$ to the horizontal, the horizontal and vertical components can be calculated using some basic trigonometry. This is illustrated in the diagram on the right.

The horizontal component of initial velocity will be called $$u_{x}$$ and the vertical component of initial velocity will be called $$u_{y}$$. They can be calculated as follows:

Initial horizontal velocity $$u_{x} = u \cos \theta$$

Initial vertical velocity $$u_{y} = u \sin \theta$$

• 1, Chapter 2.3, Applying the Suvat Equations,

### Applying the Suvat Equations

The suvat equations can now be applied to horizontal and vertical directions for a projectile which has initial speed u at an angle of $$\theta$$ to the horizontal and the position of the projectile can be found at a given time.

The vertical component was effectively discussed at the beginning of this section: the suvat equations are applied with the upwards direction as positive; gravity acts against the motion so $$a = - g$$, and the initial speed is $$u_{y}$$. The vertical position $$y$$ can be found using the suvat equations $$s=ut+\frac{1}{2}at^{2}$$:

$$y=u_{y}t+\frac{1}{2}at^{2}$$

With $$a$$ set as $$-10$$, this equation becomes:

$$vertical\: position\; y=u_{y}t-5t^{2}$$

If you consider the horizontal component of motion, the situation is actually much simpler. Gravity acts in a vertical direction only, so horizontally there are no forces acting on the projectile, hence no acceleration. The initial speed horizontally is $$u_{x}$$ and this does not change! Any calculations for position, or time, need only use the relationship distance $$= speed \times time$$. Therefore the following equation for horizontal position holds:

$$horizontal\;position\;x=u_{x}t$$

So given the initial velocity in terms of speed $$u$$ and angle to the horizontal $$\theta$$, problems involving the position both vertically and horizontally are easily tackled. The suvat equations can also provide information on the final velocity of the projectile at a given time.

These are just applications of the suvat equations to the projectile scenario. $$g$$ is often given the more accurate value of $$9.81$$, or can be left in solutions as ‘$$g$$’, and other resources will vary on this matter.

• 1, Chapter 2.3, Task 1, Worked Example 2,

### Worked Example

• #### Worked Example

A stone is thrown at a speed of $$15\;ms^{-1}$$ at an angle of $$60^{\circ}$$ to the horizontal. Find its position horizontally and vertically at the following times:
a)     $$0.5\;seconds$$
b)     $$1\;second$$
c)     $$2\;seconds$$

• 1, Chapter 2.3, Task 2, Worked Example 3,

### Worked Example

• #### Worked Example

A ball is thrown with a speed of $$8ms^{-1}$$ at an angle of $$30^{\circ}$$ to the horizontal. How high above its point of projection is it when it has travelled $$2\;m$$ horizontally?

• 1, Chapter 2.4, Final Velocity and Speed,

### Final Velocity and Speed

The suvat equations can also be applied to find the horizontal speed and vertical speed at a given time, for a projectile which has initial speed u at an angle of $$\theta$$ to the horizontal.

In keeping with the components of initial velocity, the following notation can be used for the components of the final velocity:

Final horizontal velocity     $$v_{x}$$

Final vertical velocity     $$v_{y}$$

Once again, for the horizontal component, the situation is very simple. Gravity acts in a vertical direction only, so horizontally there are no forces acting on the projectile, hence no acceleration. The initial speed horizontally is $$u_{x}$$ and this does not change.

So    $$v_{x}=u_{x}$$

The vertical component of final velocity $$v_{y}$$ is calculated by applying the suvat equations, again with the upwards direction as positive; gravity acts against the motion so $$a=-g$$, and the initial speed is $$u_{y}$$. The final vertical velocity is found using the equation $$v=u+at$$:

$$v_{y} = u_{y} + (-g)t$$

So $$v_{y} = u_{y} - gt$$

With $$a$$ set as $$-10$$, this equation becomes:

$$v_{y}=u_{y}-10t$$

So given the initial velocity in terms of speed $$u$$ and angle to the horizontal $$\theta$$, the horizontal and vertical components of final velocity are easily found.

The speed of the projectile and its angle to the horizontal can be found from these two components. The diagram below shows the final horizontal and vertical components of velocity as a vector $$v$$.

From the diagram some basic trigonometry tells you that the final speed $$v$$ is calculated by finding the magnitude of the vector $$v$$:

$$v=\sqrt{v^{2}_{x}+v^{2}_{y} }$$

The angle with the horizontal is also found using basic trigonometry:

$$\theta =\tan ^{-1}(\frac{v_{y}}{v_{x}} )$$

Now the scenario from the last Worked Example is returned to and will demonstrate these results.

• 1, Chapter 2.4, Task 1, Worked Example 4,

### Worked Example

• #### Worked Example

A stone is thrown at a speed of $$15ms^{-1}$$ at an angle of $$60^{\circ}$$ to the horizontal. Find the horizontal and vertical components of velocity at the following times, along with the speed and the angle to the horizontal:

a) $$0.5\;seconds$$

b) $$1\;second$$

c)  $$2\;seconds$$

• 0, Chapter 3, Further Projectile Problems - Maximum Height,

### Further Projectile Problems - Maximum Height

The first section introduced a technique for solving problems based around projectiles, by simply applying the suvat equations. This section continues to use this approach to find significant attributes in the motion of a projectile such as the maximum height and the range, then addresses the possibility of starting the motion from a different height.

The maximum height of a projectile can easily be found by considering its vertical motion. At the maximum height the vertical component of the velocity is momentarily zero, as the projectile slows to a halt before changing direction and travelling back down.

The maximum height and the time taken to reach maximum height can each be calculated by letting $$v_{y}=0$$.

• 1, Chapter 3, Task 1, Worked Example 5,

### Worked Example

• #### Worked Example

A ball is thrown with initial speed $$20\;ms^{-1}$$ at an angle of $$60^{}\circ$$ to the horizontal.

a)     How high does it rise?

b)     How long does it take to reach this maximum height?

c)     How far has it travelled horizontally at its maximum height?

• 1, Chapter 3.1, Range of a projectile,

### Range of a projectile

The range of a projectile is the total distance it travels horizontally, as shown below.

It is found by first calculating the total time the projectile spends in flight. Given this time the horizontal distance travelled can easily be calculated by substituting this time into the formula for horizontal component of velocity, $$x=u_{x}t$$.

The total flight time can be found by calculating the time to maximum height and doubling this time; this takes advantage of the fact that the path of a projectile is symmetric: it takes the same amount of time to reach a given height as it does to return from that height back to the original position.

• 1, Chapter 3.1, Task 1, Worked Example 6,

### Worked Example

• #### Worked Example

A stunt motorcyclist takes off at a speed of $$35ms^{-1}$$ up a ramp of $$30^{\circ}$$ to the horizontal to clear a river $$50\;m$$ wide. Does the cyclist succeed in doing this?

• 1, Chapter 3.2, Initial Height,

### Initial Height

Some problems require more consideration of the initial height of the projectile. The same equations used so far are still applicable, with the understanding that they are based around the position of the projectile as relative to its starting position.

We illustrate how to handle this factor in the next Worked Example

• 1, Chapter 3.2, Task 1, Worked Example 7,

### Worked Example

• #### Worked Example

A tennis player makes a return at a speed of $$15ms^{-1}$$ and at a height of 3 m to land in the opposite court. If she hits the ball at an angle of $$1^{\circ}$$ above the horizontal, how far does the ball travel horizontally?

• 1, Chapter 3.3, Vector Representation of Projectiles,

### Vector Representation of Projectiles

The two dimensional scenario of a projectile given initial speed 𝑢 at an angle of 𝜃 to the horizontal naturally calls upon the use of vectors, as a means to express the independent horizontal and vertical motions.

The components of initial velocity are:

$$u=\begin{pmatrix} u_x \\ \\ u_y \end{pmatrix} =\begin{pmatrix} ucos\theta \\ \\ usin\theta \end{pmatrix}$$

The velocity at a later time $$t$$ is given by:

$$v(t)=\begin{pmatrix} u_x \\ \\ u_y -gt \end{pmatrix}$$

The acceleration is simply

$$a(t)=\begin{pmatrix} 0 \\ \\ -g \end{pmatrix}$$

Notice that this acceleration can be confirmed by differentiating $$v(t)$$. Similarly, differentiating the position $$s(t)$$ gives you $$v(t)$$. This is in keeping with the material in the previous Unit on the relationships between $$s$$, $$v$$ and $$a$$.

The techniques used here can be applied to other scenarios in mechanics, other than projectiles. They extend easily to three dimensions, providing a modelling technique which is extremely powerful and has allowed a deep understanding of the world around us.

• 0, Chapter 4, Summary,

### Summary

Acceleration due to gravity is $$9.31\;ms^{-2}$$ acting vertically downwards and often approximated to $$10 \;ms^{-2}$$

For a given initial velocity $$u$$ of magnitude u and at an angle $$\theta$$ to the horizontal,

• Initial horizontal velocity is given by: $$u_{x} = u \cos \theta$$
• Initial vertical velocity is given by: $$u_{y} = u \sin \theta$$

This is shown in the diagram below:

For two-dimensional motion, taking gravity as $$10\; ms^{-2}$$,

• vertical position is given by $$y = u_{y} − 5t^{2}$$
• horizontal position $$x = u_{x}t$$

The vector velocity $$v$$ has magnitude $$v =\sqrt{v ^{z}_{y}+v^{z}_{y}}$$ and at angle to the horizontal given by $$\theta = \tan^{-1}\left(\frac{v_{y}}{v_{x}} \right)$$

Interactive Exercises:
• Projectiles Questions, https://www.cimt.org.uk/sif/mechanics/m3/text.pdf