unit 1 physics notes[1]

of 69

Unit 1

of 69

Syllabus Topic Pages

Physical Quantities 3-5

Standard Form 6-7

Density 8-9

1-3 Uniform motion 10-19

4 Projectiles 20-24

5-8 Vectors 25-29

9 Newton‟s 1st&2

nd laws 30-35

10 Acceleration due to gravity 36-40

11 Force pairs 41-45

12-17 Energy 46-51

18-21 Fluids 52-56

22-27 Solid materials 57-70

of 69

PHYSICAL QUANTITIES

Physics is a fascinating science. It deals with times that range from less than 10

-20 s,

the half-life of helium 5 years to 1.4 x1010

years, the probable ‟age‟ of our Universe.

Physicists study temperatures from within a millionth of a degree above absolute zero

to almost 200 million degrees, the temperature in the plasma in a fusion reactor. An

investigation of the mass of a quantum of FM radio radiation (2.3x10-42

kg) and the

„size‟ of a proton (1.3x10-15

m) all fall within the World of Physics!

It is vital to realise that all the quantities mentioned above contain a number and then

a unit of measurement.

Without one or other the measurement would be meaningless. Imagine saying that

the world record for the long jump was 8.95 (missing out the metres) or that the mass

of an apple was kilograms (missing out the 0.30)!

All units used in Physics are based on the International System (SI) of units which is

based on the following seven base units.

Mass - measured in kilograms

The kilogram (kg): this is the mass equal to that of the international prototype

kilogram kept at the Bureau International des Poids et Mesures at Sevres, France.

Length - measured in metres

The metre (m): this is the distance travelled by electromagnetic waves in free space in

1/299 792 458 s.

Time - measured in seconds

The second: this is the duration of 9 192 631 770 periods of the radiation

corresponding to the transition between two hyperfine levels of the ground state of

caesium 137 atom.

Electric current - measured in amperes.

The ampere: this is that constant current which, if maintained in two parallel straight

conductors of infinite length and of negligible circular cross section placed 1 metre

apart in a vacuum would produce a force between them of 2 x 10 -7

N.

Temperature - measured in Kelvin

The Kelvin: this is 1/273.16 of the thermodynamic temperature of the triple point of

water.

Angle – measured in radian

The radian: 2 radian = 360

Amount of substance - measured in moles

The mole: this is the amount of substance of a system that contains as any elementary

particles as there are in 0.012 kg of carbon-12.

of 69

It is most important to realise that these units are for separate measurements – you

can‟t add together quantities with different units. For example five kilograms plus

twenty-five metres has no meaning. It is rather similar to having a field in which

there are 30 sheep, twelve cows and 25 pigs and asking how many there are? How

many of what? It's simply a collection of different animals - you cannot add them

together!

This can be summarised by saying

a valid equation must be HOMOGENEOUS

However not all homogeneous equations are necessarily valid.

e.g. The equation time

distance6speed is a homogeneous equation (the units on both

sides are m s-1

) but the equation is wrong (the correct equation is time

distancespeed ).

Units.swf

1 Complete the table to show the missing physical quantity for each unit.

Unit Physical Quantity

m s-1

Velocity

m s-2

kg m-3

N m

N m s-1

2 Listed below are five physical quantities:

energy force power velocity displacement

Select from this list the quantity or quantities fitting each description below. You may

use each quantity once, more than once or not at all.

A quantity that equals the rate of change of another quantity in the list.

A quantity that equals the product of two other quantities in the list.

A quantity with base units kg m2 s

-3.

of 69

3 What are the units of the following quantities?

Quantity Formula Units

Velocity velocity = displacement/time m/s or ms-1

Acceleration

Force

Work

kinetic energy

Pressure

Density

Frequency

wavelength ()

wave speed speed = frequency × wavelength

4 For each of the four concepts listed in the left hand column, place a tick by

the correct example of that concept in the appropriate box.

Concept

A base quantity mole length kilogram

A base unit coulomb ampere volt

A scalar quantity force velocity kinetic energy

A vector quantity mass weight density

of 69

Standard form

Standard form is a way of writing down very large or very small numbers easily and

without using lots of zeros. We sometimes call it scientific notation.

We write 1000 as 103, 10 000 as 10

4 and so on.

Small numbers can also be written in standard form. However, instead of the index of

the power being positive it will be negative.

So 0.001 is written as 10-3

.

5000 would be 5 x 103

0.005 would be 5 x 10-3

5200 would be written as 5.2 x 103

0.0052 would be written as 5.2 x 10-3

Rules for the use of standard form

Addition or subtraction

5000 + 3000 = 5 x 103

+ 3 x 103 = 8 x 10

3

0.001 + 0.002 = 1 x 10-3

+ 2 x 10-3

= 3 x 10-3

Don‟t forget here that 103 is actually 1 x 10

3.

5000 + 400 = 5 x 103

+ 4 x 102 = 5 x 10

3 + 0.4 x 10

3 = 5.4 x 10

3

Notice what we have done to the 400 to make both terms raised to 10 to the power 3.

3000 - 2400 = 3 x 103

- 2.4 x 103 = 0.6 x 10

3 = 6 x 10

2 = 600

Multiplication or division

(add or subtract the indices)

5000 x 3000 = 5 x 103

x 3 x 103 = 15 x 10

6 = 1.5 x 10

7

50000/300 = [5x104]/[3x10

2] = 1.67x10

2 = 167

Standard form is very useful when you are dealing with numbers like the charge on an

electron (1.6x10-19

C) or even worse Planck's constant (6.63x10-34

Js).

Imagine having to write these as:

Electron charge = 0.000 000 000 000 000 000 16 C

Planck's constant = 0.000 000 000 000 000 000 000 000 000 000 000 663 Js

Powers of 10.SWF

We live in a world of about 10-2

m (1 cm) to 10 m (101

m) or „-2 to 1‟.

How big is the world of the following?

The Universe Galaxies Stars the Earth

Continents Cities Cambridge Tutors College

This room a hair a cell an atom

Atomic nucleus proton electron

of 69

pico (p) 10-12

nano (n) 10-9

micro () 10-6

centi (c) 10-2

deci (d) 10-1

kilo (k) 103

mega (M) 106

giga (G) 109

tera (T) 1012

Q1 Write the following out in full:

(a) 10kJ (b) 2.56 kV (c) 10cm (d) 100 g

(e) 5 MJ (f) 1.234 TW (g) 4.56 nm (h) 12 pA

Q2 Convert the following to the nearest power of ten symbols:

(a) 10000 m (b) 0.00004 A (c) 1.2 x 107

C

of 69

DENSITY

The mass of individual atoms and how closely they are packed together can be "felt" on

an every day level - it is called the density of the material.

Volume

MassDensity

and the units are kg m-3

.

Density is important in every day life, for example if this page was made of solid gold it

would have a mass of about 100g! The same applies to power cables, they are made of

aluminium as it is lighter than steel and so the pylons are strong enough to hold them up.

The table below shows the density of some common materials.

In an extreme case we could work out the density of a proton! It has a mass of roughly

1.66 x 10-27

kg and a "radius" of about 10-15

m. This gives it a "volume" of about

4 x 10-45

m-3

and a density of just under 4 x 1017

kgm-3

.

States of matter.SWF

http://www.media.pearson.com.au/schools/cw/au_sch_whalley_sf1_1/int/matter.html

1 Use the data in the table to calculate the mass of a steel ball bearing of radius

0.3mm.

2 3 mg of gas are injected into the vacuum chamber of a fusion reactor. The

volume of the chamber containing the gas is 3.75 m3. What is the density of the

gas under these conditions?

Material Density/ kg m-3

Gold 19300 Water 1000

Uranium 19050 Methyl alcohol 792

Mercury 13546 Chlorine 3.2

Copper 8930 Carbon Dioxide 1.98

Iron 7870 Air 1.3

Steel 7860 Helium 0.18

Marble 2600

Concrete 2400

Brick 2300

Wood (oak) 650

http://www.media.pearson.com.au/schools/cw/au_sch_whalley_sf1_1/int/matter.html

of 69

3 Complete the following table:

material mass/kg volume/m3

density/kg m-3

aluminium 160 0.060

lead 0.032 11x103

steel 60 7.7x103

4 The average radius of the Earth is 6.4x106 m. Its mass is 6.0x10

24 kg. What is

its average density?

5 Calculate the mass of air in a room whose dimensions are 4 m x 5 m x 2.5 m.

6 A neutron star is a dead star consisting entirely of neutrons. Its mass may be

about the same as that of the Sun (2 x 1030

kg). A typical diameter is 18 km.

(a) Calculate the density of the material.

(b) Neutron stars are formed from stars which have masses that are less

than three times the mass of the Sun. Dying stars with more than this

mass collapse to form black holes. Assuming that these have the same

density as neutron stars, what is the diameter of a black hole that has

20 times the mass of the Sun?

of 69

t

v

Area = displacement

Speed and Velocity

Racing balls.SWF

Velocity is defined as the rate of change of displacement with time and if the object

travels a small distance s in a time t then its velocity v is given by the equation:

t

s

takentime

ntdisplacemein changevelocity

and if this is constant the body is said to move with uniform velocity.

Notice that we use the word displacement instead of distance, since displacement is

the distance measured in a particular direction.

The difference between displacement and distance is

shown by Figure 1. An object moves from A to B along

the line AXYB. The distance travelled from A to B is

shown by the line AXYB while the displacement is shown

by the vector AB.

Notice that since velocity is a vector, uniform velocity

requires there to be no change in either the magnitude

(size) or direction of the velocity.

For a body moving with uniform velocity a graph of

velocity against time will look like this. It should be

clear that the distance travelled is equal to the area

under the line on the velocity-time graph.

Velocity measurement

The measurement of velocity normally requires the measurement of two quantities

(displacement and time) but it can be found directly as follows:-

The ticker timer A simple method of calculating the velocity of an object

is to use a ticker timer (shown in Figure 4). This makes a

series of dots on a length of tape, usually 50 per second.

This means that there is 1/50 s between one dot and the

next. A section of tape with 5 spaces on it has therefore

passed through the timer in 5/50 or 0.1 s and if this length

is measured the average velocity of the tape may be

found.

average velocity t

s

Figure 1

B

A

Y

X

Figure 4

. . . . . . .

s

of 69

This method has now been superseded by using light gate(s) and computer.

Average speed for a journey

Light gates

Light gate.SWF

t

AB B A to from speed average

Single light gate and an „interrupt card‟

Average speed.SWF

t

card oflength A at speed average

Motion Sensor

Motion sensor.SWF

The Motion Sensor is a device that emits a pulse of ultrasound which travels through

the air at 330 ms-1

before rebounding off a target. The distance of the reflecting object

from the sensor can be found by measuring the time interval between the emitted and

reflected pulses.

How long will a pulse of ultrasound take to travel from the emitter to the trolley if the

trolley is 1.65 m away from the motion sensor?

(Speed of sound 330 m s-1

).

sms

mt 005.0

330

65.11

What is the approximate delay between emitted and reflected pulses when the

trolley is 1.65 m away from the emitter?

st 010.0005.02

1 (a) During a 1500 m race an athlete completes the first 800 m in 2 minutes and

10 seconds. Calculate her average speed for 800 m.

(b) She completes the race in 4 minutes. How long does it take her to run the

remainder of the race?

(c) Calculate her average speed for the remainder of the race.

(d) Decide whether her average speed for the final 700 m is greater or less than

her average speed for the first 800 m.

(e) Calculate her average speed for the entire race.

of 69

2 A student making notes about velocity writes down the following statements.

Decide whether they are true or false.

(a) Velocity is a vector quantity.

(b) To state a velocity value properly you must indicate the direction.

(c) Velocity is the rate of change of distance.

(d) Velocity is the rate of change of displacement.

3 A sports scientist tries to model the motion of a pitch in a game of softball.

She assumes that the pitcher delivers a fast ball at a speed of 30 ms−1

and that

the ball travels horizontally throughout its flight. The batter stands 15 m from

the pitcher.

(a) How long does it take the ball to travel from the pitcher to the batter?

(b) It takes the batter 0.22 s to react to the pitcher's throw. How long does

the batter have to swing the bat to hit the ball?

(c) The swinging bat travels 1.4 m before hitting the ball. Calculate the

speed with which the bat hits the ball.

4 A van driver makes an average of 8 deliveries per day. In a normal working

week of 5 days the driver covers a total distance of 800 km.

(a) Calculate the average distance travelled for each delivery.

(b) The fuel consumption of the van is 7.5 km per litre. Calculate the

volume of fuel used by the van in a week.

5 Which of the following has the highest average speed?

(a) A cyclist who completes a 15 km race in a time of 2 hours and 15

minutes.

(b) A runner completing an 800 m race in 1 minute and 57 seconds.

(c) A swimmer completing a 50 m swim in 48 s.

(d) A yacht sailing 7 km of a race in 1 hour 10 minutes.

(e) A windsurfer travelling 1.2 km in 6 minutes.

6 A robot in the production line at a car factory moves north (000) for 1.2 m. It

turns right and travels a further 1.6 m. Finally, it turns right again and travels

another 1.2 m.

(a) How far has the robot travelled during this operation?

(b) What is the robot's final displacement from its starting position?

of 69

Equations of Motion

Constant velocity

Constant positive velocity.GIF

The gradient shows how quickly the distance travelled is changing with time. In this

example the distance is changing at a rate of about 12 m every 2 seconds.

Changing velocity

Instantaneous velocity.SWF

The gradient of a distance – time graph is the instantaneous velocity.

Velocity-time graphs

A car is travelling at 28 ms-1

and the driver takes 2 s to react to an incident on the road

in front before applying the brakes. The deceleration of the car is -5.6 ms-2

.

How far does the car travel whilst

the driver is reacting?

the brakes are applied?

Distance from velocity time graph.SWF

The area under a velocity-time graph is the distance travelled.

time

distance

time

velocity

time

acceleration

time

velocity

of 69

v

t

v

t

v

t

v

t

v

t

v

t

constant retardation

constant velocity constant acceleration

object thrown

upwards

one bounce

Figure 1

irregular motion

The set of graphs in Figure 1 show how the velocity varies with time for several

different situations.

The area below the line in each graph still represents the distance travelled in a

certain time, whether the acceleration is uniform or not.

The slope of the line at any point

dt

dv gives the instantaneous acceleration.

Interpret graph.SWF

How far north does the object move in the first 2 seconds?

Calculate the average speed of the moving object during the first 3 seconds.

Three seconds after the beginning of the motion of the object is travelling due

…………..

How far is the object away from its starting position six seconds after the start of its

motion?

Calculate the average speed for

(a) the first 2s,

(b) the last 4s,

(c) the whole journey.

Matching Graphs

Match the graphs.SWF

of 69

Equations of motion for uniform acceleration

Motion graphs.SWF

If the acceleration of a particle is uniform the following equations apply to its motion:

axuv

atut

t

uv

uv

2

x distance

aon accelerati

2 velocity average

22

2

21

where u = initial velocity

v = final velocity

a = acceleration

t = time taken

s = displacement

In any problem concerning these equations you would be given three quantities and

asked to work out the fourth

7 A ball is dropped from a high window onto a concrete floor. The velocity-time

graph for part of its motion is shown.

(a) Calculate the gradient from the origin to A.

Comment on the significance of your answer.

(b) What happened to the ball at point A?

(c) Calculate the height of the window above the ground.

of 69

8 An athlete runs a 100 m race. The idealised graph below shows how the

athlete's velocity v changes with time t for a 100 m sprint.

(a) By considering the area under the graph, calculate the maximum velocity

v max of the athlete.

(b) Using the axes below, sketch a graph showing how the acceleration of this

athlete changes with time during this race. Mark any significant values on

the axes.

of 69

x/

m

t/s

9 The diagram below shows a trolley running down a slope.

(a) Complete the diagram to show an experimental arrangement you could use to

determine how the trolley's position varies with time.

The data is used to produce a velocity-time graph for the trolley. Below is the

graph for the motion from point A to point B. Time is taken to be zero as the trolley

passes A, and the trolley passes B 0.70 s later.

(b) The motion shown on the graph can be described by the equation

tauv .

Use information from the graph to determine values for u and a.

u =

a =

(c) Determine the distance AB.

(d) On the axes below sketch a graph to show how the displacement x of the

trolley from point A varies with time t. Add a scale to each axis.

of 69

10 The diagram shows a spade being held above a flat area of soil.

(a) The spade is released and falls vertically. It takes 0.29 s for the blade to

reach the soil.

(i) Show that the speed of the spade at this instant is

approximately 3 m s-1

.

(ii) The spade penetrates 50 mm into the soil. Calculate the

average acceleration of the spade in the soil.

(c) A heavier spade of identical shape is now dropped from the same height

into the same patch of soil. Underline the correct phrase in the brackets to

describe what difference, if any, there would be in the speed at impact and

the acceleration in the soil compared to the lighter spade. Assume the

resistive forces on both spades are the same.

The heavier spade would have {a higher/a lower/the same} speed at impact as the lighter spade. The heavier spade would have { a higher / a lower / the same} acceleration in the soil as the lighter spade.

of 69

11 The graph shows the variation of velocity with time for a body moving in a

straight line.

Calculate

(a) the total distance travelled,

(b) the average speed over the 20 seconds.

of 69

u

vx = u

vy

y

x

Figure 1

Projectiles

BallTossOut.mov

Consider an object projected horizontally from a cliff with a velocity u. The problem

is to find out where this object will be after a time t and how fast it will be travelling.

The important thing to remember is that you can consider the motion in two parts:-

Motion in the horizontal direction - this is uniform velocity since no forces act

in this direction

Motion in the vertical direction - this is uniformly accelerated motion due to

the gravitational pull of the Earth, the vertical acceleration being the strength

of the Earth's field (g = 9.81 ms-2

). Remember that this always acts vertically

downwards.

We will ignore air resistance for the time being.

The time taken for the object to reach the ground along the parabolic path is

the same as if it were dropped vertically.

Notice that if air resistance is ignored the vertical height of the object at given

times after the start is the same no matter what horizontal velocity it had at the

moment of release.

Projectiles 3.SWF

Projectiles 1.SWF

of 69

Monkey & Hunter.SWF

Truck & ball.GIF

Motor bike jump.SWF

Falling from a helicopter.SWF

Microsoft Office Excel 97-2003 Worksheet

shoot.exe

Bouncing Ball.mdl

Range

(m)

0 100 200 300 400 500 600 700 800

0 5

20

45

80

125

180

245

320

Height fallen (m)

25

m/s

50

m/s

100

m/s

of 69

Projection at an angle

Projectile motion.SWF

Complete the following statements to summarize how the vertical component of the

velocity changes during the flight.

The speed in the vertical direction _________ until the projectile reaches its

_______ _____ .

This is because, if we ignore air resistance, the only force acting on the projectile is

______ downwards.

At the top of its motion its vertical speed is 0 ms-1

. As it starts to fall again, the

downwards speed _______ .

Because there is no force acting on the projectile in the horizontal direction the speed

in the horizontal direction is ______ .

Taking downwards as positive complete the following graphs for

(a) The vertical motion

(b) The horizontal motion

time

distance

time

velocity

time

acceleration

time

distance

time

velocity

time

acceleration

of 69

1 A crate is released from an aircraft that is flying horizontally at a steady speed

of 200 ms-1

at a height 1500 m above the ground:

(a) What is its horizontal velocity?

(i) 2 s after it was released

(ii) 5 s after it was released

(b) What is its vertical velocity:?

(i) 2 s after it was released

(ii) 5 s after it was released

(c) What is its velocity 5s after it was released?

(d) How long will it take to reach the ground?

(e) How far horizontally from the place where it was released will it hit the

ground?

2 If you had taken air resistance into account how would this have affected your

answers to (c) and (e)?

3 A stone is dropped from a balloon which is floating at a height of 50m above

the ground. At the same instant another stone is thrown horizontally from the

balloon at 8 ms-1

.

(a) Which one will hit the ground first?

(b) How fast will each stone be travelling when it hits the ground?

4 Draw an accurate graph to show the parabolic path of a projectile projected

horizontally from a cliff.

5 A coin is flicked off a table so that it initially leaves the table travelling in a

horizontal direction with a speed of 1.5 m s-1

. The diagram shows the coin at

the instant it leaves the table. Air resistance can be assumed to have a

negligible effect throughout this question.

(a) Add to the diagram the path followed by the coin to the floor.

(b) (i) The table is 0.70 m high. Show that the coin takes approximately 0.4 s

to reach the floor.

(ii) Hence calculate the horizontal distance the coin travels in the time it

takes to fall to the floor.

(c) A coin of greater mass is flicked with the same horizontal speed of 1.5 m s-1

.

Compare the path of this coin with that of the coin in the first part of the

question. Explain your answer. You may be awarded a mark for the clarity of

your answer.

of 69

6 A cricketer bowls a ball from a height of 2.3 m. The ball leaves the hand

horizontally with a velocity u. After bouncing once, it passes just over the

stumps at the top of its bounce. The stumps are 0.71 m high and are situated

20 m from where the bowler releases the ball.

(a) Show that from the moment it is released, the ball takes about 0.7 s to

fall 2.3 m.

(b) How long does it take the ball to rise 0.71 m after bouncing?

(c) Use your answers to parts (a) and (b) to calculate the initial horizontal

velocity u of the ball. You may assume that the horizontal velocity has

remained constant.

(d) In reality the horizontal velocity would not be constant. State one

reason why.

of 69

Vectors in equilibrium – general principles for forces

For an object to be in equilibrium it must not have a tendency to:

(a) translate – i.e. move from one place to another

(b) rotate – i.e. turn

In order to satisfy these conditions the following must apply:

ONE FORCE

If there is only one force acting on the body it cannot be in equilibrium. It

may not rotate (if the force passes through the object‟s centre of mass) but

it is bound to move.

TWO FORCES

If two forces act on the body it can only be in equilibrium if the

two forces are equal and opposite and lie in the same line. If they

are equal and opposite but not in the same line they form a couple

and the object will rotate. (See the space station in Figure 2)

If they are in the same line as well as being equal and opposite then the object will be

in equilibrium (See the tug of war or a glass on a table in Figures 3(a) and (b)).

THREE FORCES

If three forces act on a body it can be in

equilibrium if they are parallel or not and

in the same line or not.

The simplest case is where the sum of the

two forces acting in one direction is equal

to the third force acting in the opposite

direction (Figure 4).

The sum of the moments of the forces on the object about any point must be zero

The resultant of the forces on the object in any direction must be zero

Figure 2

Figure 3(a) Figure 3(b)

Figure 4

of 69

A case where the three forces are in parallel but not acting through the same point is

shown by the see saw in Figure 5. The resultant moment must be zero. Notice that

the downward forces (the weights of the two people) are balanced by an upward force

(R) at the pivot.

Finally consider the case of three forces that are not in

parallel.

From the first condition for equilibrium it follows that if

three non-parallel forces act on an object that is in

equilibrium then the lines of action of these three forces

must pass through one point.

If this were not true then taking moments about the point (P) where two of the forces

cross (i.e. their total moment is zero) would still leave a resultant moment due to the

third force.

(See Figure 6).

A simple example of this is a ladder leaning against a wall.

(See Figure 7).

The vectors (in this case the forces) can be represented in magnitude and direction by

the three sides of a triangle – the vectors must be drawn so that their direction follows

round the figure. See the example set of forces for the ladder shown in Figure 8.

forces.exe

Figure 7

P

Figure 6

d

Figure 8

Figure 5 W1 W2

R

of 69

Resolving vectors into COMPONENTS

Unit Vectors.SWF

Vector Maths.SWF

Add components.SWF

Add two vectors.SWF

Equilibrium.HTM

Car on a slope.SWF

How big is the resultant force that acts on the car?

1 A gymnast of weight 720 N is holding himself in the cross position on the high

rings. He is quite still. A free-body force diagram for the gymnast shows the

two upward pulls of the rings on his hands, each of size 380 N.

Calculate the angle between the wires supporting the rings and the vertical.

2 A very heavy sack is hung from a rope and pushed sideways. When the

sideways push is 220 N the rope supporting the sack is inclined at 18° to the

vertical.

(a) Calculate the tension in the rope.

(b) Hence find the mass of the sack.

3 The wireless mast shown in the diagram is

held by two cables and at the bottom by a

frictionless pivot. (This means that the contact

force exerted by the base is at right angles to

the ground).

If the mast has a weight of 5000 N and the

tension in the horizontal cable is 500 N

calculate:

(a) The tension (T) in the left hand cable.

(b) The contact force exerted (R) by the base.

4 A gymnast of mass 65 kg hangs between two parallel bars so that their arms

each make an angle of 15o with the vertical. What is the tension in each arm?

500 N

T 30o

F cos

F

F cos (90 - )

of 69

5 The diagram below shows a drop-down table attached to a wall. The table

is supported horizontally by two side arms attached to the mid-points of the

sides of the table.

(a) The table surface is 80 cm long, 50 cm deep and 1.8 cm thick. It is made from

wood of density 0.70 g cm-3

. Show that its weight is about 50 N.

(b) The free-body force diagram below shows two of the three forces acting on the

table top.

(i) Calculate the horizontal and vertical components of the 83 N force.

Horizontal component:

Vertical component:

(ii) Add appropriately labelled arrows to the free-body force diagram to

show these components.

(iii) Hence find the magnitude of the horizontal force that the hinge applies

to the table top and state its direction.

of 69

6 The diagram shows two forces acting on a body.

On the grid below draw a scale diagram to determine the resultant force acting

on the body. Use a scale of 1 major division to 1 N.

of 69

Newton's 1st Law and Inertia

We can state Newton‟s Laws of motion simply as:

A body remains at rest or in a state of uniform motion unless acted on by a resultant

force.

Also we define inertia as:

The inertia of a body is its reluctance to change its state of motion

The more massive the body is the more inertia it has. The astronauts in the Space

Shuttle had trouble with large pieces of equipment since although they were

"weightless" they still had inertia and were difficult to stop once they were moving.

The astronauts themselves also experienced a problem on the lunar surface. They had

a much smaller weight and therefore lower friction between themselves and the

surface but their mass and therefore inertia was the same as on Earth. Stopping was

difficult due to the lower frictional force.

The period of vibration of a body is also affected by its inertia. Large, heavy objects

will vibrate slowly. One simple way of testing this is to use a ruler loaded with a

lump of plasticene - the greater the load the longer the period of the motion.

Therefore the rate of vibration can be used to compare the masses of two objects since

the rate of vibration depends on inertia (mass) and not weight.

Newton's First Law

Aristotle thought that a force was needed to keep

an object moving and that if this force was

removed the object would naturally come to rest.

Galileo - not being totally happy with this idea

thought about motion in terms of what might

happen to a ball rolling down one side of a U

shaped slope. He reasoned that if there was no

friction the ball would go down one side and then up the other until it reached the

same height as that from which it started (think about this in the design of big

dippers). Now if the other side of the U is steadily lowered the ball will have to go

further before it reaches its original height. Logically if the slope is flattened out the

ball will roll on for every - never getting to its original height. So no force is needed to

keep it going.

Newton expressed this idea in his first law of motion.

A body remains at rest or in a state of uniform motion unless acted on by a resultant

force

Newton's ist law.GIF

of 69

Think about the law in two parts:

at rest

The word resultant really ought to go in because clearly someone sitting on a stool

may be at rest but they are acted on by two forces - their weight and the contact force

of the stool. It is because these two forces are balanced and there is no resultant force

they stay still - i.e. at rest.

uniform motion

This means no change of velocity - and since velocity is a vector this means at a

steady speed in a straight line. Think about a sky diver, as they fall out of the plane

their speed increases - their weight is bigger than the drag - there is a net force and

Newton's first law does not apply. However as the drag increases the two forces on

them become equal and the sky diver falls with a constant velocity - a state of uniform

motion!

Drag.mdl

1 A lift of mass 1200 kg is pulled vertically upwards at a steady speed of

(i) 1.0 m s-1

, (ii) 2.0 m s-1

.

(It will help if you draw a free body force diagram for the lift.)

(a) What is the tension in the cable in each case?

(b) Does it make any difference to your answers if the lift is descending

at a steady speed?

2 An object falls through the air and its motion is affected by two forces. One of

these is constant (weight) and the other increases (viscous drag). The drag

force may be expressed as k v2, where v is the speed of the object and k is a

factor that depends on the viscous properties of the air and also on the shape of

the object.

(a) Draw a free body force diagram of the falling ball when it is moving at

speed v.

(b) Sketch a speed – time graph of its motion.

(c) Derive an expression for the terminal velocity vT in terms of the

object‟s mass m, the acceleration due to gravity g and the drag factor k.

of 69

3 Figure 1 shows a box resting on the floor of a stationary lift. Figure 2 is a free-

body force diagram showing the forces A and B that act on the box.

For each of the following situations, tick the appropriate boxes to show how the

magnitude of the forces A and B change, if at all, compared with when the lift is

stationary.

Situation Force A Force B

increases no

change decreases increases

no

change decreases

Lift

accelerating

upwards

Lift moving

with

constant

speed

upwards

Lift

accelerating

downwards

Lift moving

with

constant

speed

downwards

of 69

Newton's Second Law

This law is all about the situation when there is a resultant force and therefore there is

also a resulting acceleration.

You need a resultant force to change the motion of a body. The bigger the resultant

(net) force the greater the acceleration. (Remember that force is a vector and the

direction of the forces acting on a body need to be considered).

Aristotle thought that the force was in the same direction as the motion (i.e. the

velocity). This is not true - the force is in the same direction as the acceleration if you

don't follow this think of a basket ball being thrown into a net. The ball follows a

curving path in the air but the force - and therefore the acceleration is always in one

direction - vertically downwards.

You can investigate the law practically by using a friction-compensated track and

measuring the acceleration of the truck for a variety of accelerating forces and truck

masses.

It is also important to remember that the mass being accelerated is the total mass of

the truck plus the mass of the accelerating weight although this is not significant if the

truck is very heavy in comparison.

The results can be analysed by plotting two graphs

acceleration against the accelerating force (a v F)

acceleration against the inverse of the mass of the truck (a v 1/m)

This should show you that a F and a m

1

Combining these two results gives

amF

The units are

force N

mass kg

acceleration ms-2

This law also gives us a good definition of the Newton as a unit of force.

1 N is the force that will give a mass of 1 kg an acceleration of 1 m s-2

newton.exe

Although we may think of the second law as “Force equals Mass times Acceleration”

Newton actually stated his famous law as follows:-

The rate of change of momentum of a body is directly proportional to the applied

force and in the direction of that force.

of 69

4 A student is provided with a trolley and a track as shown in the diagram

below. He is required to apply different forces to the trolley, measure the

corresponding accelerations and hence demonstrate the relationship between

the two. Any additional normal school laboratory equipment is available for

him to use.

(a) Describe how he could

(i) apply a constant measurable force;

(ii) measure the velocity of the trolley at a point on the track as the trolley

moves under the action of this applied force. List any additional

apparatus that would be required. You may add to the diagram above

to help your description.

(b) Assuming the velocity has been measured at one point, what additional

measurements are required to determine the acceleration?

(c) How could the student demonstrate the expected relationship between the

force and the acceleration?

(d) In such an experiment, the track is given a slight tilt to compensate for friction.

Why is this necessary if the relationship suggested by Newton's second law is

to be successfully demonstrated?

5 Determine the resultant force on the object below.

What can be deduced about the motion of an object

(i) when the resultant force on it is zero,

(ii) when the resultant force on it is vertically upwards,

(iii) when the resultant force on it is in the opposite direction to its motion?

of 69

6 A lorry is travelling at 25 m s-1

down a mountain road when the driver

discovers that the brakes have failed. She notices that an escape lane covered

with sand is ahead and stops her lorry by steering it on to the sand.

(a) The lorry is brought to a halt in 40 m. Calculate the average deceleration

of the lorry.

(b) Suggest how the depth of the sand affects the stopping distance. Justify

your answer.

7 A student performs an experiment to study the motion of a trolley on a

horizontal bench. The trolley is pulled by a horizontal string which runs over

a pulley to a suspended mass.

Initially the trolley is held at rest at position A. It is then released. When it has

moved some distance, but before the suspended mass hits the floor, a card attached to

the trolley passes through a light gate. A clock controlled by the gate records how

long the card blocks the light beam.

The card, which is 0.130 m long, takes 0.070 s to pass through the beam.

(a) Calculate the average velocity of the trolley as it passes through the light gate.

(b) The light gate is 0.600 m from the start. Show that the acceleration of the

trolley is approximately 3 m s-2

.

(c) The mass of the trolley is 0.950 kg. Calculate the tension in the string pulling

it, stating any assumption which you make.

(d) The tension in the string must be less than the weight of the 0.400 kg mass

Suspended from it. Explain why.

of 69

time

distance

time2

distance

distance

velocity2

Acceleration due to Gravity/Free Fall

Free fall.mdl

0 if2

21

2

21

uat

atuts

Free fall.SWF

Free fall 2.SWF

Diagram of apparatus used to measure g due to freefall / 2

21 ats method

Graph

Gradient = ?

g =

0 if2

222

uas

asuv

v squared.SWF

?

s

of 69

?

s What does the graph look like if the object is

dropped from a great height and we also take

account of air resistance?

Free fall.mdl

Sometimes a falling object can accelerate at an increasing rate

(e.g. a chain sliding off a smooth table)

Microsoft Office PowerPoint 97-2003 Presentation

1 The acceleration of free fall g can be measured by timing an object falling

from rest through a known distance.

(a) Explain one advantage and one disadvantage of making this

distance as large as possible.

(b) In a typical laboratory measurement of g, a steel sphere is

dropped through a distance of the order of one metre. With the

help of a labelled diagram, describe and explain an experimental

method of measuring the time it takes the sphere to fall.

(c) At any given place, the weight of a body is proportional to its

mass. Explain how measurements of g support this statement.

time

distance

time

velocity

time

acceleration

time

distance

time

velocity

time

acceleration

of 69

15

10

5

0

5

–10

–15

–20

–25

–30

–35

time/s

velocity

/m s

0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5

–1

- - - - - - - -

acceleration

/m s

time/s0

1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5

-

0.5

–1

2 A hot-air balloon is rising vertically at a speed of 10 m s–1

. An object is

released from the balloon. The graph shows how the velocity of the object

varies with time from when it leaves the balloon to when it reaches the ground

four seconds later. It is assumed that the air resistance is negligible.

(a) Use the graph to

(i) show that the object continues to rise for a further 5 m after it

is released.

(ii) determine the total distance travelled by the object from when

it is released from the balloon to when it reaches the ground.

(b) Hence determine the object‟s final displacement from its point of

release from the balloon.

(c) Using the axes below, sketch a graph showing how the acceleration

of the object changes during the time from when it leaves the

balloon to when it hits the ground.

Mark any significant values on the axes.

of 69

3 An astronaut on the moon drops a hammer. The gravitational acceleration

is 1.6 m s–2

.

(a) How long does the hammer take to fall 1.0 m from rest?

(b) Calculate the velocity of the hammer just before it hits the ground.

4 A student is working on a spreadsheet to model the fall of a golf ball from

rest from the window of a tall building.

(a) He assumes that the acceleration remains constant at 9.81 m s–2 for

the first two seconds of the fall. Comment on whether this is a

reasonable assumption.

A B C D

1

time from start / s

velocity reached/

m s–1

distance fallen during 0.20 s time

interval / m

total distance from the start

/m

2

3 0.00 0.00 0.00 0.00

4 0.20 1.96 0.20 0.20

5 0.40 3.92 0.59 0.78

6 0.60 5.89 0.98 1.77

7 0.80 7.85 1.37 3.14

8 1.00 9.81 1.77 4.91

9 1.20 11.77 2.16 7.06

10 1.40 13.73 2.55 9.61

11 1.60 15.70 2.94 12.56

12 1.80 17.66 3.34 15.89

13 2.00 19.62 3.73 19.62

(b) Cell B6 is calculated using the formula B6 = 9.81*A6. Explain why

this is appropriate.

(c) Cell C7 is calculated using the formula C7 = ((B6+B7)/2)*0.20.

(i) Explain what (B6+B7)/2 represents.

(ii) Why is this fraction multiplied by 0.20?

(d) Give an appropriate spreadsheet formula that uses cell D9 to

calculate cell D10.

of 69

5 Some people think that all raindrops fall at the same speed; others think

that their speed depends on their size.

(a) Calculate the speed of a raindrop after it has fallen freely from

rest for 0.2 s.

(b) The raindrop falls for longer than 0.2 s. Explain why its

acceleration does not remain uniform for the whole of its fall.

(c) Sketch a graph to show how the raindrop‟s velocity increases

from rest to terminal velocity.

of 69

Forces

Forces always come in pairs called a force pair.

(This is Newton's 3rd law of motion)

Moonlander 2.SWF

Load Lab.SWF

Gravitational forces exist between bodies on account of their mass.

When you are standing on the Earth, the Earth pulls you towards it and provides your

weight.

You pull the Earth with an equal and opposite force.

Gravity & star collapse.SWF Fred Hoyle Jocelyn Burnell

Microsoft Office Word 97 - 2003 Document

Supernova

Electrostatic forces act between objects which are charged.

They can be forces of attraction or repulsion, depending on the sign of the charges.

Electromagnetic forces act between

(a) a magnet and another magnet

(b) a magnet and a magnetic material,

(c) magnets and current-carrying conductors.

In (a) the forces may be attraction or repulsion. It depends on the polarity of the

magnets.

In (b) the forces are always of attraction.

In (c) the force is at right angles to both the current and the magnetic field.

Nuclear Forces act inside the nucleus of an atom which are neither gravitational or

electromagnetic.

Nuclear forces hold the nucleus together (electrostatic forces alone would cause the

positively charged protons to fly apart).

Gluons & the strong force.SWF

forces ticelectrostaboth are F and

F of magnitude F of magnitude

21

21

F

directions oppositein act F and F

Bobject on acts F A,object on acts

21

21F

Similarities

Differences

F2 + +

F1

A B

http://scienceblogs.com/startswithabang/upload/2009/06/the_last_100_years_1950s_the_t/lecturing.jpg

http://www.google.co.uk/imgres?imgurl=http://www.centauri-dreams.org/wp-content/uploads/2006/11/cosmic_ray_supernova.jpg&imgrefurl=http://www.centauri-dreams.org/?p=906&h=349&w=400&sz=23&tbnid=OA-pVbjcAIQoAM:&tbnh=108&tbnw=124&prev=/images?q=supernova+image&hl=en&usg=__2SX5NHW7mEMX7UdBhGFrbKDoAKY=&ei=pt12SoPxLJe5jAfkl-SnBg&sa=X&oi=image_result&resnum=4&ct=image

of 69

sphere S

table

Q

P

P and Q are forces

acting on the sphere S

Contact forces act between bodies when they are very close together.

They are, in fact, electrostatic forces acting over very short distances. They are caused

by forces between the outer layers of electrons of the two bodies in contact.

Frictional forces on solid bodies moving over each other or on bodies moving through

fluids (Viscous Drag) are all contact forces.

Viscous drag forces act on an object which is moving through a fluid.

Viscous drag forces always oppose the motion of the object.

Viscous drag increases with the speed of the object.

Parachute game.SWF

Tension forces exist when a body has a pair of equal and opposite forces acting on it.

The top of the spring is pulled up by the support.

The bottom of the spring is pulled down by the hanging object.

The spring is in a state of tension.

Tension forces.SWF

1

(a) On which body does the force which pairs with force P act?

Give its direction.

(b) On which body does the force which pairs with force Q act?

Give its direction.

2 A child sits at rest on a swing. The figure shows free-body force diagrams for

(i) the child, (ii) the swing seat.

of 69

(a) For each of the forces P, W, T, w and P', identify the body that is producing

the force. [ P' is not produced by the Earth.]

Body that exerts the force Body that the force acts on

P swing seat child

W

T

w

P‟

(b) Write a phrase describing each force as

the push or pull exerted on the ………….. by the ……………

P is the upward push exerted by the swing seat on the child

W

T

w

P‟

(c) In this situation P = W; T = w + P' and P = P'.

Explain each equation in terms of Newton's laws.

3 A man checks the weight of a bag of potatoes with a newtonmeter. Two of the

forces acting are shown in the diagram.

The table below gives these forces. For each force there is a corresponding force, the

„Newton‟s third law pair force‟. In each case state

the body that the Newton‟s third law pair force acts upon

the type of force (one has been done for you)

the direction of the Newton‟s third law pair force.

of 69

4 The diagram shows two magnets, M1 and M2, on a wooden stand. Their faces

are magnetised as shown so that the magnets repel each other.

(a) Draw a fully labelled free-body force diagram for the magnet M1.

(b) The table gives the three forces acting on the magnet M2. For each force on

M2 there is a corresponding force known as its „Newton‟s third law pair‟.

In each case state

(i) the body on which this corresponding force acts,

(ii) the direction of this corresponding force.

of 69

Energy

There are about 20 million households in Britain alone and on average each one of

these uses some 21 000 kW h of energy annually! The study of energy is therefore of

vital importance in our lives.

World energy use

An account of the growth in world energy consumption makes sobering reading. If

we define a quantity of energy (W) as 1021

J then in the 2000 years up to 1850 the

world is thought to have used between 6 and 9 Q. In the following hundred years up

to 1950 it had burned a further 6 Q and from then on we have been using at least 1Q

every ten years. Fortunately in recent years this rate of increase has slowed down

considerably, partly because of the high cost of energy.

Energy conservation is also important. Some 5 per cent of the total energy bill of the

western world can be saved by conservation measures such as the proper use of

insulation!

To give you some idea of the value of energy, the table below shows the energy used

in a number of situations.

Typical energy values (J)

Moonlight on face for 1 s 10-3

Pressing down a typewriter key 1

House brick raised to shoulder height 30

Burning a march 1000

Potential energy of a person at the top of the stairs 1500

Kinetic energy of a car travelling at 70 mph 500000

Electrical energy in a fully charged car battery 2 000 000

Chemical energy in a day's food intake 11000000

Chemical energy in one litre of petrol 35 000 000

First atomic bomb 1013

Very severe earthquake 1020

World energy consumption (1964) 1.4 x 1021

Earth's annual share of the Sun's heat 1025

Rotational kinetic energy of the Earth 1029

Present rate of total energy use

USA 10 000W per person 24 hours a day

Japan and Europe 4000W

Less developed countries 100-1000W

World average 2000W

Annual energy consumption

Total = 5 x 109 people x 2000W x 86400 x 365 = 315 x 10

18 = 3.15 x 10

20 J

Renewable energy explained.SWF

of 69

Conservation of energy

You should notice that we talk about the transformation or conversion of energy from

one form to another and not its use. This is because although we may use up energy

in one form it always reappears as another. This is a most important principle of

Physics: that of the conservation of energy. The principle states that:

Energy is never created or destroyed but only changed from one form to

another.

Bungee jump.mov

http://jersey.uoregon.edu/Work/index.html (Cost of Energy)

Force, work energy and power

Work is done when a body exerts a force and moves a distance in the direction of the

force.

Work = force x displacement in the direction of the force

(This is an example of two vectors being multiplied together to give a scalar)

Car braking to a

stop

sFmv 2

21



Braking distances Toy car experiment

s

½ mv2 0 F

http://jersey.uoregon.edu/Work/index.html

of 69

Making use of kinetic energy

Launching a plane.SWF

http://lectureonline.cl.msu.edu/~mmp/kap5/work/work.htm

(Force and work animation)

Making use of potential energy

Traction Trebuchet

Used people power to haul down the shorter end of the beam, which in turn flipped up a sling

that was connected to the longer end. As the longer end reached its apex, the sling opened

releasing a large stone or other object.

Ballista

The oldest form of catapult used in siege warfare. The Ballista is

best described as a giant crossbow which fired spears instead of

arrows. There were many types of Ballista that were popular with

the Greeks and Romans, but the one thing they have in common is

that they were all powered by twisted sinew ropes.

The ballista was accurate and could fire spears a great distance but

they were difficult to build and they were limited to hitting only

what they could see.

http://www.physicsclassroom.com/mmedia/qt/energy/coastwin.html

(Roller coaster)

Electromagnetic potential energy depends on the arrangement of charges.

It is responsible for the energy stored in

batteries

capacitors

magnets which are pushed together

compressed springs

energy released in chemical reactions.

Conservation of energy

In any system isolated from its surroundings the total amount of energy in that system

remains constant.

http://lectureonline.cl.msu.edu/~mmp/kap5/work/work.htm

http://www.physicsclassroom.com/mmedia/qt/energy/coastwin.html

of 69

1 (a) A car is travelling along a horizontal road. The driver applies the

brakes and the car comes to rest. Describe the principal energy

transformation which occurs as the car comes to rest.

(b) On another occasion, the same car is travelling with the same speed, but

down a hill. The driver applies the brakes, which produce the same average

braking force as before. With reference to the energy transformations which

occur, explain why the braking distance will be greater on the hill than on the

horizontal road.

2 The diagram shows a small vehicle which is free to move in a vertical plane along a

curved track.

The vehicle of mass m is released from rest from point A. It runs down to point B, a

distance h vertically below A. Its speed at point B is v.

Write down expressions for

(a) the gravitational potential energy lost by the vehicle as it runs from A to B,

(b) the kinetic energy of the vehicle at B.

(c) Hence derive an expression for the speed v.

(d) State one assumption you have made in your derivation.

(e) Would you expect the vehicle to pass point C? Explain your answer.

3 A lift has a mass of 400 kg. A man of mass 70 kg stands on a weighing

machine fixed to the floor of the lift. Four seconds after starting from rest the

lift has reached its maximum speed and has risen 5 m.

(a) What will be the reading on the weighing machine during the period of

acceleration?

(b) How may it be decided whether the acceleration was uniform?

(c) How much energy will be used by the lift motor in

(i) the first four seconds,

(ii) the next four seconds?

of 69

4 (a) A car of mass m is travelling in a straight line along a horizontal road

at a speed u when the driver applies the brakes. They exert a constant force F

on the car to bring the car to rest after a distance d.

(i) Write down expressions for the initial kinetic energy of the car and

the work done by the brakes in bringing the car to rest.

Kinetic energy

Work done

(ii) Show that the base units for your expressions for kinetic energy and

work done are the same.

(b) A car is travelling at 13.4 m s-1

. The driver applies the brakes to decelerate

the car at 6.5 m s-2

. Show that the car travels about 14 m before coming to

rest.

(c) On another occasion, the same car is travelling at twice the speed. The driver

again applies the brakes and the car decelerates at 6.5 m s-2

. The car travels

just over 55 m before coming to rest. Explain why the braking distance has

more than doubled

5 A dummy is used in a test crash to test the suitability of a seat belt. If the

dummy had a mass of 65 kg and it was brought to rest in a distance of 65 cm

from a velocity of 12 ms-1

calculate

(a) The mean deceleration during the crash

(b) The average force exerted on the dummy during the crash.

(c) The loss of k.e. of the dummy.

Power

t

sF

takentime

doneWork Power

e.g. A car travelling at a constant speed v has to overcome a resistance force F and the power

of the engine is

vFPower .

6 (a) A car travelling at 30 ms-1

along a level road is brought to rest in a

distance of 35 m by its brakes. If the car has a mass of 900 kg

calculate the average force exerted by the brakes.

(b) If the same car travels up a slope of 1 in 15 at a constant speed of

25 ms-1

what power does the engine develop if the total frictional

resistance is 120 N?

7 A weightlifter raised a bar of mass of 110 kg through a height of 2.22 m. The

bar was then dropped and fell freely to the floor.

(i) Show that the work done in raising the bar was about 2400 J.

(ii) It took 3.0 s to raise the bar. Calculate the average power used.

of 69

8 A certain power station generates electricity from falling water. The diagram

shows a simplified sketch of the system.

(a) (i) In what form is the energy of the water initially stored?

(ii) What energy form is this transformed into in order to drive the

turbine?

(b) State the principle of conservation of energy.

(c) The force of the water at the turbine is 3.5 x 108 N and the output power

generated is 1.7 x 109 W. Use this data to calculate the minimum speed at

which the water must enter the turbine.

(d) Explain why, in practice, the speed at which the water enters the turbine is

much greater than this.

(e) When working at this output power, 390 m3 of water flows through the

turbine each second. The top reservoir holds 7.0 x 106 m

3 of water. For how

long will electricity be generated?

(f) This power station is used at peak periods, after which the water is pumped

back to the top reservoir. The water has to be raised by

500 m. How much work is done to return all the water to the top reservoir?

(The density of water is 1000 kg m-3.)

of 69

Fluids

The rate at which fluid flows through a tube is likely to depend on

(a) the viscosity of the fluid

(Oil pipelines in Siberia have to be shut down in winter because the oil

viscosity increases and the oil will not flow)

(b) the dimensions of the tube, and

(c) the pressure difference between its ends.

This flow rate is of great importance in our lives since it governs things like the flow

of blood round our bodies and the transmission of gas, water or oil through long

distances in pipelines.

Laminar and turbulent flow

In the nineteenth century Reynolds investigated the conditions that would give

turbulence in the flow of a fluid. The streamlining of bodies is most important in the

design of cars, submarines and the nose cones of aircraft and rockets since a reduction

in drag can reduce vibration and also save large amounts of fuel.

In laminar flow the particles in the fluid follow streamlines, and the motion of

particles in the fluid is predictable.

If the flow rate is very large, or if objects obstruct the flow, the fluid starts to

swirl in an erratic motion. No longer can one predict the exact path a particle

on the fluid will follow. This region of constantly changing flow lines is said

to consist of turbulent flow.

Laminar flow.GIF

Turbulent flow 2.GIF

Turbulent flow in a pipe.flv

Turbulent flow.flv

of 69

Upthrust

The buoyancy (upthrust) force that you experience in a swimming pool is a

consequence of the greater pressure below an immersed object than above it.

The upthrust is due to the difference between the force due to water pressure at the

bottom of the cylinder, F2, and that at the top F1.

gmgVAghhFFU

AghApF

AghApF

1212

212

111

The upthrust equals the weight of the displaced fluid.

This is Archimedes’ Principle.

An object will float in a fluid if the upthrust the weight of the object.

The apparent weight of an object = weight – upthrust.

h1

h2

p1

P2

Liquid of density

of 69

Data:

Density of brine = 1 150 kgm-3

Density of meths = 800 kgm-3

Density of water = 1000 kgm-3

Density of sea water = 1020 kgm-3

Density of mercury = 13600 kgm-3

Density of ice = 950 kgm-3

Density of platinum = 21000 kgm-3

Density of gold = 19000 kgm-3

Density of iron = 7800 kgm-3

Density of copper = 8930 kgm-3

Density of air = 1.2 kgm-3

1 Calculate the apparent weight in the following cases:

(a) 12 kg of iron immersed in water

(b) 3 kg of gold immersed in meths

(c) 5 kg of copper immersed in brine

(d) 8 kg of platinum immersed in mercury

2 A wooden rod, density 650 kgm-3

and one metre long floats upright in a liquid.

If the cross sectional area of the rod is 4 cm2 calculate the depth to which it sinks

in:

(a) water (b) meths (c) brine

3 A pure water ice cube with sides 2 cm long floats in a liquid in a glass beaker.

What volume will be below the liquid surface if the liquid is:

(a) pure water

(b) brine

(c) What happens to the level of the water in the beaker if the ice cube

melts and why?

4 A hot air balloon with a volume of 250 m3 hangs in the air. If the density of

the hot air is 0.8 kgm-3

and that of the cool air outside the balloon is 1.2 kgm-3

what is the biggest load it can support if the fabric of the balloon and the

basket have a total mass of 50 kg.

5 Calculate the total pressure:

(a) 2.5 m below the surface of water

(b) 0.5 m below the surface of mercury

(c) 3.5 m below the surface of brine

(Take the atmospheric pressure to be 105 Pa)

6 Calculate the difference if pressure between the top and bottom of a diving

bell 2.5 m long immersed in water.

7 If the atmosphere was uniform and with the same density as that at sea level

how high would it be? (Height of a mercury barometer = 0.76 m)

of 69

Velocity of

falling sphere

Terminal velocity

Figure 2

Stokes’ law (Viscous Drag Force vrF 6 )

Consider a sphere falling through a viscous fluid. As the sphere falls so its velocity

increases until it reaches a velocity known as the terminal velocity. At this velocity

the frictional drag due to viscous forces is just balanced by the gravitational force and

the velocity is constant (shown by Figure 2).

At this speed:

Viscous drag = 6πrv = mg - upthrust

If the density of the material of the sphere is and

that of the liquid , then

weight – upthrust = 4

3𝜋𝑟3𝑔 𝜌 − 𝜍

Therefore we have for the viscosity

𝜂 =2𝑟2𝑔 𝜌 − 𝜍

9𝑣

where v is the terminal velocity of the sphere.

From the formula it can be seen that the frictional drag is smaller for large spheres

than for small ones, and therefore the terminal velocity of a large sphere is greater

than that for a small sphere of the same material.

9

2 2

grv

Stokes' law is important in Millikan's experiment for the measurement of the charge

on an electron, and it also explains why large raindrops hurt much more than small

ones when they fall on you - it's not just that they are heavier, they are actually falling

faster.

Gravitational pull (weight) mg

Viscous drag (6rv)

m

Figure 2

Velocity v

of 69

8 Calculate the terminal velocities of the following raindrops falling through air:

(a) one with a diameter of 0.3 cm,

(b) one with a diameter of 0.01 mm.

(Take the density of water to be 1000 kg m-3

and the viscosity of air to be

1.8x10-5

Pas. The buoyancy effect of the air may be ignored.)

9 Calculate the viscous drag on a drop of oil of 0.1 mm radius falling through air

at its terminal velocity. (Viscosity of air = 1.8 x 10-5

Pas; density of oil = 850

kg m-3

)

10 Discuss the following statements:

(a) Large ball-bearings fall through glycerol faster than do smaller ones of

the same material.

(b) The acceleration of a sphere falling through a fluid varies with the

distance that it has fallen in the fluid.

(c) The viscosity of a liquid could be used to measure temperature.

11 Compare the speed at which a steel ball (density 7800 kg m-3

) of radius 2 mm

will fall through treacle, with that at which an air bubble (density 1.3 kg m-3

)

of radius 1 mm will rise through the same liquid. (Take the density of treacle

to be 1600 kg m-3

.)

12 Two spherical raindrops of equal size are falling through air at a velocity of

0.08 ms-1

. If the drops join together forming a large spherical drop, what will

the new terminal velocity be?

of 69

Solid Materials

Structure

In a liquid or a gas the atoms and/or molecules are relatively free and can move round

at will. In solids, however, they are almost completely lacking mobility; all they can

do is to vibrate about a mean position when energy is added to the solid. This lack of

mobility gives a solid it‟s most characteristic property - that of retaining whatever

shape it is given.

The way in which a solid behaves depends on its internal structure and there are three

main types of solid:

crystalline solids such as sugar,

amorphous solids such as glass,

polymeric solids such as rubber.

Crystalline solids

A crystalline solid is one where the internal structure is regular in nature, the atoms

within it being set in well-defined patterns. Some crystals are isotropic - that is, their

physical properties are the same in whichever direction they are measured, while

others are anisotropic - their properties are different in different directions.

Metals are generally polycrystalline materials, being composed of a large number of

small crystals or grains aligned in a variety of different directions. In metals the atoms

are held together by a cloud of free electrons that move between the atoms.

Alloys are solid mixtures of two or more metals. An alloy will usually have properties

that are very different from those of the constituent metals. For example the melting

point of solder (50% lead and 50% tin) is 490 K while that of lead is 600 K and that of

tin is 505 K.

Figure 1 shows the structures of the four most common types of crystal (for the

moment we will assume that they are perfect and contain no impurities or

dislocations).

The four types are listed below together with one example of each:

(i) face-centred cube - sodium chloride (FCC)

(ii) hexagonal close-packed – zinc (HCP)

(iii) body-centred cube – potassium (BCC)

(iv) tetrahedral – silicon (T)

of 69

Note: The tetrahedral structure is shown in plan view and then as a solid. The individual

molecules have been shown different colours to make it easier to distinguish between

molecular plains.

The characteristics of each structure can be investigated using the techniques of X-ray

diffraction. In general, if the solid exhibits a regular structure as shown by the first three

crystalline states mentioned the X-ray diffraction pattern will show a series of dots. If it

is irregular, however, as in graphite in which the layers are free to slide one over the

other, a series of rings will result. (This can also be seen by the electron diffraction

through graphite).

The face-centred cube and the hexagonal close-packed crystals are the most closely

packed structures, and 60 per cent of all metals exist in one or other of these forms.

Face centred cubic Body centred cubic

Hexagonal close-packed

Tetrahedral

Figure 1

of 69

Amorphous solids

An amorphous solid exhibits no regular internal structure; glass, plastic and soot are

examples. In a way such a solid is like an instantaneous snapshot of a liquid, although

one with an enormously great viscosity. An amorphous material has the density of a

solid but the internal structure of a liquid. They are considered to be super-cooled

liquids in which the molecules are arranged in a random manner similar to that of the

liquid state. Amorphous solids are also unlike crystalline solids in that they do not

have definite melting points.

Polymeric solids

In these solids the molecules form long chains which may contain anything between

1000 and 10 000 molecules. Many are natural organic materials such as plant

constituents but there are many synthetic polymers, one example being polythene. The

precise properties of the polymer depend on just how tightly these chains of molecules

are bound together. They may be tangled together in a haphazard way or lie side by

side.

Any cross-chain linking will enormously increase the strength of the material and the

vulcanising of rubber, for example, eventually produces the hard material ebonite: this

is the result of cross-chain linking by sulphur atoms. Below a critical temperature

polymers behave much like glass (but with a greater degree of ductility) but above it

they are more rubber-like. Cooling rubber in liquid nitrogen strikingly illustrates this

change of properties.

Deforming a Solid - Hooke’s Law

The simplest form of variation of the extension of

an object when a force is applied is known as

Hooke‟s Law, proposed by Robert Hooke, the

founder of the Royal Society, in 1676. He showed

that:

xkF

xF

extension Force

Figure 2

Figure 1

Extension

Fo

rce

of 69

This shows that the extension is directly proportional to the applied force – doubling

the force will double the extension. If a graph of force is plotted against extension a

straight line will be obtained (Figure 1).

This is the kind of graph that you would get if you loaded a helical spring or a copper

wire as long as you kept the loads fairly small. Where xkF a molecule displaced

from its original position (by squashing or stretching the material) will try and return

to its original position.

The constant k is known as the elastic constant for the material and is defined asx

F .

The units for k are N m-1

.

1 Two tug-of-war teams each pull on a rope with a force of 5000 N. The rope is

horizontal. What is the tension in the rope at its mid-point?

2 State Hooke‟s Law. If a spring extends by 0.02 m when a force of 25 N is

applied to it what is the spring constant?

3 A spring has an unstretched length of 12 cm and a stiffness of 50 N m-1

. What

is the force needed to

(a) double its length

(b) treble its length?

Rubber

Rubber is a polymer and so consists of many long chain molecules. When the rubber

is in an unstretched (relaxed) state these molecules are tangled up as shown in Figure

1(a). As a steadily increasing force is applied these molecules begin to straighten out

(Figure 1(b)) – the bonds between adjacent chains are broken. These bonds are

relatively weak compared with the bonds along each chain molecule. During this

untangling and while the bonds are breaking the rubber warms.

Eventually all the molecules have been straightened (Figure.1(c). Up to this point it is

quite easy to extend the rubber because all that was being done was to untangle the

chains and break the weak bonds between them. However when the molecules are

straightened it becomes much more difficult. This state can be noticed easily by using

a rubber band since when this point is reached the surface of the rubber becomes

whiter and rougher.

Figure 1(a) Figure 1(b) Figure 1(c)

Chain molecules

of 69

As the rubber is allowed to relax by

slowly removing the force the chains

of molecules intertwine again but the

cross links do not completely reform.

The heat energy produced during

stretching is not recovered. Plotting a

graph of force against extension the

amount of energy converted to heat

within the specimen can be found.

(Figure 2).

Area:

OABDO energy given to band during

stretching

PCBD energy released from band

during contraction

OABCPO energy converted to heat

within the band

OP permanent extension of band

Using a rubber band about 2mm wide and 20cm long and a maximum load of 15N the

energy converted to heat is of the order of 0.5 - 1 J.

4 In an experiment to measure the extension of a rubber band the following

graph was obtained. The line represents the extension during loading and

unloading.

(a) Label the two lines to indicate which represents loading and which

represents unloading.

12

10

8

6

4

2

00 100 200 300 400 500

Extension / mm

Force / N

Figure 2

extension

A

B

C D

Force

O P

of 69

(b) What is the name for the characteristic behaviour shown by the

shape of this graph?

(c) If the rubber band has a cross-sectional area of 6.0 × 10–6 m2

calculate the stress produced in the elastic band when it is fully

loaded.

(d) Estimate how much work is done on the rubber band as it is fully

loaded.

(e) Hence show that the energy dissipated during the loading and

unloading process is approximately 1 J.

(f) When the rubber band has a load placed on it a new reading is taken.

Over the next minute this reading increases by a few millimetres. If

a material deforms plastically in this way when stress is applied,

what is the name of this mechanism?

(g) Draw a labelled diagram of the apparatus that could be used to

produce a force-extension graph for a rubber band, for loads up to

12 N.

5 (a) The sap from a rubber tree may flow like thick treacle or thick

oil. State one word that describes this flow behaviour.

(b) The sap is treated to produce a lump of rubber. Choose two words

from the list below and explain the meaning of each as it applies to

rubber.

Elastic, brittle, hard, durable, stiff

(c) The solid line on the following force-extension graph is obtained

when a rubber band is stretched.

Use the graph to estimate the work done in stretching the rubber band to a

tension of 1.0 N.

Curve obtainedunloading

Curve obtained onloading

1.4

1.2

1.0

0.8

0.6

0.4

0.2

00 5 10 15 20 25 30

Extension / m × 10

Force / N

–2

of 69

(d) When the force is reduced gradually, the force-extension graph follows

the dotted line. What does the graph tell you about the work done by the

rubber band when it returns to its original length?

(e) Rubber tyres are constantly being compressed and released as a car travels

along a road. Explain why the tyres become quite hot.

It is useful to have a property of a material that is independent of the size of the

sample and can be use to compare its elastic properties with another - this is called the

Modulus of Elasticity for the material. The modulus of elasticity is defined as:

Strain

Stress Elasticity of Modulus

There are three types of moduli

Young Modulus - tensile and compression longitudinal stress

Shear modulus - a shearing stress

Bulk modulus - volume changes of the specimen

Young Modulus

This is defined as the ratio of longitudinal stress to longitudinal strain. This is the

modulus we need if we want to investigate the change of length of an object - more

accurately any linear dimension (width, length or height).

Strain

StressModulusYoung

)ength (Original l

)(ΔExtension al Strain Longitudin

a (A)tional AreCross-

Force (F) al Stress Longitudin

sec

F

L

A

e

Figure 1

of 69

The values for the Young modulus for some common materials are given in the

following table.

Material Young

modulus/GPa

Material Young

modulus/GPa

Diamond 1200

Mild steel 210 Bone 18

Copper 120 Concrete 16.5

Cast iron 110 Beech wood 15

Bronze 96-120 Oak 11-12

Slate 110 Pine 11-14

Aluminium 70 Sandstone 6.3

Granite 40-70 Plastic 2.0

Lead 18 Nylon 2.0

Titanium 116 Rubber 0.02

Nylon fishing line extension

Fishing line behaves in an odd way - the initial extension is not static - it increases as

the creep continues! Fishing lines are designed to not only support a static load they

must cope with the sharp shock of the rod being flipped back or the fish jerking the

line. 6 Express these stresses in N m

-2, giving the number in standard form:

(a) 101 kPa (b) 0.27 MPa (c) 35 MPa (d) 2.8 GPa (e) 235 GPa. 7 Complete the following table:

Force Cross-sectional

area Stress

6.0 N 0.10 mm2

12 kN 2.0 mm

2

3.4 mm2

6.0 x 106 N m

-2

(0.50 mm)2

6.4 MPa 0.11 kN 0.22 GPa

8 A steel wire has a diameter of 0.36 mm.

(a) What is its radius? (b) What is its cross-sectional area in m

2?

(c) It is pulled by a force of 3.5 N. What is the tensile stress in the wire? 9 What is the tensile strain when

(a) A copper wire of length 2.0 m has an extension of 0.10 mm (b) A rubber band of length 50 mm is stretched to a length of 150 mm?

of 69

10 Complete the following table:

stress strain Young Modulus E

50 MPa 6.0 x 10-4

0.10 GPa 5.0 x 10

-4

0.054 0.22 GPa 0.30 GPa 300 GPa 1.8 GPa 180 GPa

11 A copper wire of length 1.2 m and cross-sectional area 0.10 mm

2 is hung

vertically; for copper E = 130 GPa. A steadily increasing force is applied to its lower end to stretch it. When the force has reached a value of 10 N (a) What is the stress in the wire (b) What is the strain in the wire (c) What is the extension in the wire?

Elastic Energy

When a person jumps up and down on a trampoline it is clear that the bed of the

trampoline stores energy when it is in a state of tension. This energy is converted to

kinetic and potential energy of the jumper when the tension is removed.

Similarly, when a piece of elastic in a catapult is stretched energy is stored in it, and

when the catapult is fired this energy is convened into the kinetic energy of the

projectile.

What actually happens within some of the materials mentioned in the examples may

be quite complex, but we can calculate the energy stored in a stretched metal wire

where Hooke‟s law is obeyed as follows.

Let the wire be of unstretched length and let a force F produce an extension x.

(Assume that the elastic limit of the wire has not been exceeded and that no energy is

lost as heat.)

Consider Figure 1(a). The work done by the force is Fs but in this case the force

varies from 0 at the start to F at the end when the wire is stretched by an x.

Force

extension

l

Figure 1(a)

F

extension e1 e2

Figure 1(b)

F1

F2

Force

of 69

Therefore:

xFstretchingduringwiretheondoneWork2

1

Remember

EAxFhence

Ax

F

xA

F

E ,

Therefore

2

2

1

2

1 EAxxFstretchingduringwiretheondoneWork

And this energy is the shaded area of the graph.

If the extension is increased from e1 to e2 then the extra energy stored is given by

2

1

2

2

21storedenergyextra

eeEA

This is the shaded area on the graph in Figure 1(b).

If the wire has been extended beyond the elastic

limit and then the force removed the extension is

only partially recoverable. Energy is therefore lost

due to heat and this phenomenon is known as

hysteresis. The force-extension curve for the wire

will follow the line OAB on the graph in Figure 2,

where the area OABDO is the energy input, OCBD

the recoverable energy and the shaded area

OABCO represents the energy converted to heat

within the specimen. The larger this area is the

bigger the energy loss due to hysteresis.

The effect of hysteresis is usually very small for

metals, but is noticeable for polythene, glass and

rubber. You can easily investigate this using a rubber band. By simply stretching it

and then holding it against your lips you can detect a rise in temperature.

12 (a) Sketch a graph of force F (on the y-axis) against extension

(on the x-axis) for spring of stiffness 50 N m-1

for values of x from 0 to

1.0 m. Assume that the spring obeys Hooke‟s law.

(b) Use your graph to find the increase in energy stored in the spring

when its extension increases from

(i) 0 to 0.20 m (ii) 0.20 m to 0.40 m(iii) 0.80 to 1.00 m

Figure 2

extension

A

B

C D

Force

O

of 69

10

8

6

4

2

0

0 10 20

Strain/ 10–3

Stress/GPa

13 Show that the units on both sides of the equation 2

21 kxW are the same.

14 A copper wire is stretched with a steadily increasing force: Hooke‟s law

ceases to be obeyed when the force reaches 80 N. The extension is 5.0 mm

when the force is 20 N. How much elastic potential energy is stored in the

wire when the force is (a) 20 N (b) 40 N (c) 60 N?

15 The table below gives the corresponding values of load and extension when

masses were hung on a wire of length 2.0 m and diameter 0.40 mm.

Load/kg 0 0.20 0.40 0.60 0.80 1.00 1.10 1.12

Extension/mm 0 1.0 2.1 3.1 4.2 5.4 7.3 9.0

(a) Plot a graph of load (on the y-axis) against extension (on the x-axis).

(b) From the straight part of the graph, where Hooke‟s law is obeyed, calculate the

gradient of the graph (express the answer in units N m-1

).

(c) Estimate the work done in stretching the wire up to the point where the load is

1.12 kg.

(d) Estimate how much of this energy has become internal energy, and how much

is stored as elastic potential energy (energy that is recoverable).

16 This question is about the design of a car seat belt. The seat belt has to

restrain a passenger when the car is involved in an accident.

(a) Use definitions of stress and strain to show that stress × strain

has the same units as energy stored per unit volume of seat belt.

The graph shows how stress varies with strain for the seat belt material.

of 69

(b) Use the graph to show that the energy stored per unit volume of seat

belt material when the strain is 20 × 10–3 is about 1 × l08 Jm–3.

The car is travelling at 20 m s–1 carrying a 60 kg passenger who is

wearing a seat belt.

(c) (i) Show that the kinetic energy of the passenger is 12 000 J.

(ii) Calculate the volume of seat belt material which would be

required to stop the passenger when the car stops suddenly.

Assume that the maximum strain in the seat belt is 20 × 10–3.

(iii) Use your answer to part (ii) to suggest a suitable width

and thickness of seat belt for this situation, assuming its

length is 2.0 m.

of 69

Ductile Brittle, Tough and Hard Materials

A ductile material is one such as copper which shows plastic deformation, e.g.

it may be drawn out into a wire.

A brittle material (e.g. glass) is one that will break or crack with little

deformation. The opposite of brittle is tough. Tough materials (e.g.

polythene) are able to withstand impact forces without breaking. A lot of

energy is required to break a tough material.

Hard materials resist plastic deformation usually by denting, scratching or

cutting.

If a ductile material such as copper is stretched until it breaks and its stress and strain

measured and plotted, a graph like that in Figure 1 may be obtained.

There are a number of important points about such a graph:

OP is a straight line - in this region

Hooke's law is obeyed.

P is the limit of proportionality – up

to P strain is proportional to stress.

E is the elastic limit - up to E, if the

load is removed the material will return

to its original length (although the stress

may not be proportional to the strain up

to this point).

Y is the yield point - between E and Y

the material becomes plastic, that is, if

the load is removed the material will

contract but all the extension is not recoverable. The material follows the

dotted line YS on the graph during contraction and the remaining extension is

known as a permanent set.

Z - after this point none of the extension is recoverable.

B - this is the breaking stress beyond which the material will break.

A material like copper is known as ductile -

that is, it will flow, and can be drawn out into

a wire without fracture.

Materials such as glass that can be extended

but do not show plastic deformation and will

easily fracture are known as brittle materials.

The repeated bending, heating and beating

known as work hardening increases the

strength of metals and is used in the

manufacture of swords.

If the steel is heated and then rapidly cooled it will become more brittle but by heating

a sample of steel and then slow cooling it the effect can be reversed.

E

Stress

Strain S

P

O

Y Z

B

Figure 1

Stress

Strain

breaks

Figure 2

of 69

17 Control of high volume manufacturing production, such as in the steel

industry, is achieved through regular sampling and testing of the product.

The picture below shows a machine called a tensile tester. It is stretching a

sample at a constant rate. The test sample is a rod of steel approximately

the size of a pencil.

The results below were from a test on a sample of steel of 1.3 × 10–4 m2

cross-sectional area and 6.5 × 10–2 m length. The tension T applied to the

sample and its resulting extension x were measured until the sample

failed.

T/103 N 0 5 10 15 20 25 30 35

x/10–6 m 0 12 24 36 48 60 74 100

(a) Plot these values and draw the graph on the grid below.

(b) Indicate on the graph with the letter P the limit of proportionality.

(c) Calculate the stress applied to the specimen at this point.

(d) Calculate the strain in the sample at point P.

(e) Calculate the Young modulus for this steel.

(f) A second sample of exactly the same size is stiffer, weaker and brittle.

Sketch a line on your graph predicting the results for the sample.

Label this line X.

unit 1 physics notes[1]

Documents