5 Forces and matter

In this chapter, you will find out:

◆ that forces change the shape and size of a body

◆ how to carry out experiments to produce extension–load graphs
S ◆ how to interpret extension–load graphs
S ◆ about Hooke’s law and how to apply it
◆ what factors affect pressure
◆ how to calculate pressure.

5.1 Forces acting on solids an object. You could imagine holding a cylinder of foam
Forces can change the size and shape of an object. They rubber, which is easy to deform, and changing its shape
can stretch, squash, bend or twist it. Figure 5.1 shows in each of these ways.
the forces needed for these different ways of deforming Foam rubber is good for investigating how things
deform, because, when the forces are removed, it
springs back to its original shape. Here are two more
examples of materials that deform in this way:
◆ When a football is kicked, it is compressed for a
short while (see Figure 5.2). Then it springs back
undeformed
to its original shape as it pushes itself off the foot of
the player who has kicked it. The same is true for a
tennis ball when struck by a racket.

stretched compressed
(tensile forces) (compressive forces)

bent twisted Figure 5.2 This remarkable X-ray image shows how a football
(bending forces) (torsional forces) is compressed when it is kicked. It returns to its original shape
as it leaves the player’s boot. (This is an example of an elastic
Figure 5.1 Forces can change the size and shape of a solid object. deformation.) The boot is also compressed slightly but, because it
These diagrams show four different ways of deforming a solid object. is stiffer than the ball, the effect is less noticeable.

64 Cambridge IGCSE Physics

◆ Bungee jumpers rely on the springiness of the Figure 5.5 shows the pattern observed as the load
rubber rope, which breaks their fall when they is increased in regular steps. The length of the spring
jump from a height. If the rope became increases (also in regular steps). At this stage the
permanently stretched, they would stop spring will return to its original length if the load is
suddenly at the bottom of their fall, rather than removed. However, if the load is increased too far, the
bouncing up and down and gradually coming spring becomes permanently stretched and will not
to a halt.
Some materials are less springy. They become
permanently deformed when forces act on them.
◆ When two cars collide, the metal panels of their
bodywork are bent. In a serious crash, the solid
metal sections of the car’s chassis are also bent.
◆ Gold and silver are metals that can be deformed
by hammering them (see Figure 5.3). People
have known for thousands of years how to
shape rings and other ornaments from these
precious metals.

5.2 Stretching springs

To investigate how objects deform, it is simplest to start
with a spring. Springs are designed to stretch a long way
when a small force is applied, so it is easy to measure
how their length changes.
Figure 5.4 shows how to carry out an investigation
on stretching a spring. The spring is hung from a rigid
clamp, so that its top end is fixed. Weights are hung on
the end of the spring – these are referred to as the load.
As the load is increased, the spring stretches and its Figure 5.4 Investigating the stretching of a spring.
length increases.

Figure 5.3 A Tibetan silversmith making a wrist band. Silver is a Figure 5.5 Stretching a spring. At first, the spring deforms elastically.
relatively soft metal at room temperature, so it can be hammered It will return to its original length when the load is removed.
into shape without the need for heating. Eventually, however, the load is so great that the spring is damaged.

Chapter 5: Forces and matter 65

return to its original length. It has been inelastically 6
deformed.
5
Extension of a spring
As the force stretching the spring increases, it gets 4

Extension / cm
longer. It is important to consider the increase in
length of the spring. This quantity is known as the 3

extension.
2

length of stretched spring 1

= original length + extension
0
Table 5.1 shows how to use a table with three 0 1 2 3 4 5 6 7 8
Load / N
columns to record the results of an experiment to
stretch a spring. The third column is used to record Figure 5.6 An extension–load graph for a spring, based on the
the value of the extension, calculated by subtracting data in Table 5.1.

the original length from the value in the second

column. Activity 5.1
To see how the extension depends on the load, we Investigating springs
draw an extension–load graph (Figure 5.6). You can see
that the graph is in two parts.
Skills
◆ At first, the graph slopes up steadily. This shows that AO3.1 Demonstrate knowledge of how to safely use
the extension increases in equal steps as the load techniques, apparatus and materials (including
increases. following a sequence of instructions where appropriate)
◆ Then the graph bends. This happens when AO3.3 Make and record observations, measurements and
estimates
the load is so great that the spring has become
AO3.4 Interpret and evaluate experimental observations
permanently damaged. It will not return to its
and data
original length.
(You can see the same features in Table 5.1. Look at the
Use weights to stretch a spring, and then plot a
third column. At first, the numbers go up in equal steps.
graph to show the pattern of your results.
The last two steps are bigger.)
1 Select a spring.
2 Fix the upper end of the spring rigidly in a clamp.
Load / N Length / cm Extension / cm
3 Position a ruler next to the spring so that you
0.0 24.0 0.0 can measure the complete length of the spring,
1.0 24.6 0.6 as shown in Figure 5.4.
4 Measure the unextended length of the spring.
2.0 25.2 1.2
5 Prepare a table for your results, similar to Table 5.1.
3.0 25.8 1.8 Record your results in your table as you go along.
4.0 26.4 2.4 6 Attach a weight hanger to the lower end of the
spring. Measure its new length.
5.0 27.0 3.0
7 Carefully add weights to the hanger, one at a time,
6.0 27.6 3.6 measuring the length of the spring each time.
7.0 28.6 4.6 8 Once you have a complete set of results, calculate
the values of the extension of the spring.
8.0 29.5 5.6
9 Plot a graph of extension (y-axis) against load
Table 5.1 Results from an experiment to find out how (x-axis) and comment on its shape.
a spring stretches as the load on it is increased.

66 Cambridge IGCSE Physics

 Questions a
S

5.1 A piece of elastic cord is 80 cm long. When it is

stretched, its length increases to 102 cm. What is

Extension
limit of
its extension? proportionality
5.2 The table shows the results of an experiment
to stretch an elastic cord. Copy and complete
the table, and draw a graph to represent this
data.
Load
Load / N Length / mm Extension / mm
0.0 50 0
b
1.0 54
2.0 58
3.0 62

Extension
4.0 66
5.0 70
6.0 73
7.0 75 Load
8.0 76
Figure 5.7 a An extension–load graph for a spring. Beyond the
limit of proportionality, the graph is no longer a straight line, and
the spring is permanently deformed. b This graph shows what
happens when the load is removed. The extension does not
return to zero, showing that the spring is now longer than at the
start of the experiment.
S 5.3 Hooke’s law
The mathematical pattern of the stretching spring was
first described by the English scientist Robert Hooke. The behaviour of the spring is represented by the
He realised that, when the load on the spring was graph of Figure 5.7a and is summed up by Hooke’s law:
doubled, the extension also doubled. Three times the
load gave three times the extension, and so on. This The extension of a spring is proportional to the load
shows up in the graph in Figure 5.7. The graph shows applied to it, provided the limit of proportionality is
how the extension depends on the load. At first, the not exceeded.
graph is a straight line, leading up from the origin.
This shows that the extension is proportional to the
load. We can also write Hooke’s law as an equation:
At a certain point, the graph bends and the line
slopes up more steeply. This point is called the limit of
proportionality. (This point is also known as the elastic F = kx
limit.) If the spring is stretched beyond this point, it
will be permanently damaged. If the load is removed, In this equation, F is the load (force) stretching the
the spring will not return all the way to its original, spring, k is the spring constant of the spring, (a measure
undeformed length. of its stiffness) and x is the extension of the spring.

Chapter 5: Forces and matter 67

S S
Studyy tip
p
Activity 5.2
If you double the load that is stretching a spring, Investigating rubber
the spring will not become twice as long. It is the
extension that is doubled. Skills
AO3.1 Demonstrate knowledge of how to safely use
techniques, apparatus and materials (including
following a sequence of instructions where appropriate)
AO3.3 Make and record observations, measurements and
Worked example 5.1 estimates
AO3.4 Interpret and evaluate experimental observations
A spring has a spring constant k = 20 N/cm. and data
What load is needed to produce an extension
of 2.5 cm? Carry out an investigation into the stretching
of a rubber band. This is a good test of your
Step 1: Write down what you know and what you
experimental skills. You will need to work carefully
want to find out.
if you are to see the effects described above.
load F = ?
1 Hang a rubber band from a clamp. Attach a
spring constant k = 20 N/cm weight hanger at the lower end so that the band
extension = 2.5 cm hangs straight down.
2 Clamp a ruler next to the band so that you can
Step 2: Write down the equation linking these measure the length of the rubber band.
quantities, substitute values and calculate 3 Prepare a table for your results.
the result. 4 One by one, add weights to the hanger. Record
F = kx the length of the band each time. Add the
F = 20 × 2.5 = 50 N weights carefully so that you do not allow the
band to contract as you add them.
So a load of 50 N will stretch the spring by 2.5 cm.
5 Next, remove the weights one by one. Record
the length of the band each time. Remove the
weights carefully so that you do not stretch the
How rubber behaves band or allow it to contract too much.
A rubber band can be stretched in a similar way to a 6 Calculate the extension corresponding to each
spring. As with a spring, the bigger the load, the bigger weight.
the extension. However, if the weights are added with 7 Plot your results on a single graph. Can you see
great care, and then removed one by one without the effect shown in Figure 5.7b?
releasing the tension in the rubber, the following can be
observed: Hooke and springs
◆ The graph obtained is not a straight line. Rather, it Why was Robert Hooke so interested in springs? Hooke
has a slightly S-shaped curve. This shows that the was a scientist, but he was also a great inventor. He was
extension is not exactly proportional to the load. interested in springs for two reasons:
Rubber does not obey Hooke’s law. ◆ Springs are useful in making weighing machines,
◆ Eventually, increasing the load no longer produces and Hooke wanted to make a weighing machine
any extension. The rubber feels very stiff. When that was both very sensitive (to weigh very light
the load is removed, the graph does not come back objects) and very accurate (to measure very precise
exactly to zero. quantities).

68 Cambridge IGCSE Physics

S proportional to the force producing it. You can see S
Hooke’s straight-line graph in Figure 5.8.

Questions
5.3 A spring requires a load of 2.5 N to increase its
length by 4.0 cm. The spring obeys Hooke’s law.
What load will give it an extension of 12 cm?
5.4 A spring has an unstretched length of 12.0 cm. Its
spring constant k is 8.0 N/cm. What load is needed
to stretch the spring to a length of 15.0 cm?
5.5 The results of an experiment to stretch a spring
are shown in table. Use the results to plot an
extension–load graph. On your graph, mark the
limit of proportionality and state the value of
the load at that point.

Load / N Length / m
0.0 0.800
2.0 0.815
4.0 0.830
6.0 0.845
Figure 5.8 Robert Hooke’s diagrams of springs. 8.0 0.860
10.0 0.880
◆ He also realised that a spiral spring could be used to 12.0 0.905
control a clock or even a wristwatch.
Figure 5.8 shows a set of diagrams drawn by Hooke,
including a long spring and a spiral spring, complete
with pans for carrying weights. You can also see some of 5.4 Pressure
his graphs. If you dive into a swimming pool, you will experience
For scientists, it is important to publish results so the pressure of the water on you. It provides the
that other scientists can make use of them. Hooke was upthrust on you, which pushes you back to the surface.
very secretive about some of his findings, because he The deeper you go, the greater the pressure acting
did not want other people to use them in their own on you. Deep-sea divers have to take account of this.
inventions. For this reason, he published some of They wear protective suits, which will stop them being
his findings in code. For example, instead of writing crushed by the pressure. Submarines and marine
his law of springs as given above, he wrote this: exploring vehicles (Figure 5.9) must be designed to
ceiiinosssttuv. Later, when he felt that it was safe to withstand very great pressures. They have curved
publish his ideas, he revealed that this was an anagram surfaces, which are less likely to buckle under pressure,
of a sentence in Latin. Decoded, it said: Ut tensio, sic and they are made of thick metal.
vis. In English, this is: ‘As the extension increases, This pressure comes about because any object
so does the force.’ In other words, the extension is under water is being pressed down on by the

Chapter 5: Forces and matter 69

 a

weight
of water

area

Figure 5.10 a Pressure is caused by the weight of water (or

other fluid) above an object, pressing down on it. b This dam
is thickest near its base, because that is where the pressure is
greatest.

Figure 5.9 This underwater exploring vehicle is used to carry withstand the pressure of the water behind it. Because
tourists to depths of 600 m, where the pressure is 60 times that
at the surface. The design makes use of the fact that spherical
the pressure is greatest at the greatest depth, the dam
and cylindrical surfaces stand up well to pressure. The viewing must be made thickest at its base.
window is made of acrylic plastic and is 9.5 cm thick. In a fluid such as water or air, pressure does not
simply act downwards – it acts equally in all directions.
This is because the molecules of the fluid move around
weight of water above it. The deeper you go, in all directions, causing pressure on every surface they
the greater the amount of water pressing down collide with.
on you (see Figure 5.10a). In a similar way, the
atmosphere exerts pressure on us, although we are Pressure measurements
not normally conscious of this. The Earth’s gravity A manometer is a simple instrument for showing the
pulls it downwards, so that the atmosphere presses difference in pressure between two gases or liquids.
downwards on our heads. Mountaineers climbing to Figure 5.11 shows how a manometer is used to measure
the top of Mount Everest rise through two-thirds of the pressure of the laboratory gas supply. This pressure
the atmosphere, so the pressure is only about one- must be higher than atmospheric pressure, or gas would
third of the pressure down at sea-level. There is much not flow out of the pipe.
less air above them, pressing down. ◆ A manometer is a U-shaped tube, holding a small
The pressure caused by water is much greater than amount of liquid.
that caused by air because water is much denser than ◆ When both ends are open, the levels of the liquid in
air. Figure 5.10b shows how a dam is designed to the two sides are the same.

70 Cambridge IGCSE Physics

 a

vacuum

atmospheric atmospheric
pressure pressure mercury

b
glass tube

trough

higher lower
pressure pressure

Figure 5.11 Using a manometer to measure the pressure Figure 5.12 A mercury barometer is used to measure
difference between two gases. a With atmospheric pressure on atmospheric pressure.
both sides of the U-tube, the liquid is at the same level in both
sides. b With higher pressure on one side, the liquid is pushed
round. The greater the pressure difference, the greater is the
difference in levels, h. pressure falls, the force on the mercury decreases, and
the level in the tube decreases.
Mercury is used in barometers like this because it
◆ If one side is connected to the gas supply, the gas
has a high density (more than 13 times the density of
pushes down on the liquid and forces it round the
water). A barometer made using water would require a
bend. The levels are now unequal, showing that
much taller tube, over 10 m in height!
there is a difference in pressure.
A barometer can be used to measure atmospheric
pressure. One simple type, the mercury barometer, is Studyy tip
p
shown in Figure 5.12. It consists of a long glass tube, at
least 80 cm in length. The tube is filled with mercury and You may see atmospheric pressure given as
then carefully inverted into a trough containing mercury. 760 mm Hg. The units ‘mm Hg’ mean millimetres
This must be done carefully, so that no air enters the tube. of mercury, the height of the mercury column.
Once the tube is safely inverted, the level of mercury These are not SI units.
in the tube drops. The length l of the mercury column,
measured from the surface of the mercury in the trough,
is about 76 cm. The space above the mercury column is a
vacuum (with a small amount of mercury vapour). Activity 5.3
The column length l depends on the atmospheric Pressure experiments
pressure. On a day when the atmospheric pressure is Try out some simple experiments to explore the
high, the air presses more strongly on the mercury in idea of pressure.
the trough, so that it rises further in the tube. If the

Chapter 5: Forces and matter 71

 (N/m2). In the SI system of units, this is given the name
Questions
pascal (Pa).
5.6 Name an instrument that is used to measure:
a atmospheric pressure Key definition
b differences in pressure.
5.7 The diagram shows two closed tanks, A and pascal (Pa) – the SI unit of pressure, equivalent to
B. Each tank contains gas and is fitted with a one newton per square metre (1 N/m2).
manometer to show how the pressure compares
N
with atmospheric pressure outside the tank. Pa =
m2

A B
Worked example 5.2
Shoes with stiletto heels go in and out of fashion.
(‘Stiletto’ is an Italian word meaning a small
a In which tank is the gas pressure greater than
and murderous dagger.) Such very narrow heels
atmospheric pressure? Explain how you can
can damage floors, and dance halls often have
tell.
notices requiring shoes with such heels to
b What can you say about the pressure of the
be removed.
gas in the other tank?
Calculate the pressure exerted by a woman
dancer weighing 600 N standing on a single heel
of area 1 cm2. If the surface of the dance floor is
broken by pressures over five million pascals
5.5 Calculating pressure (5.0 MPa), will it be damaged?
A large force pressing on a small area gives a high
Step 1: To calculate the pressure, we need to know
pressure. We can think of pressure as the force per unit
the force, and the area on which the force
area acting on a surface, and we can write an equation
acts, in m2.
for pressure, as shown.
force F = 600 N
area A = 1 cm2 = 0.0001 m2 = 10−4 m2
Key definition
Step 2: Now we can calculate the pressure p.
pressure – the force acting per unit area at right
F
angles to a surface. p=
A
force
pressure = 600 N
area =
0.0001 m2
F
P= = 6000000 Pa = 6 .0 MPa
A
The pressure is thus 6.0 × 106 Pa, or 6.0 MPa.
This is more than the minimum pressure needed
Now let us consider the unit of pressure. If force F is to break the surface of the floor, so it will be
measured in newtons (N) and area A is in square metres damaged.
(m2), then pressure p is in newtons per square metre

72 Cambridge IGCSE Physics

 Questions Studyy tip
p S

5.8 Write down an equation that defines pressure. You may prefer to use the symbol D to represent
5.9 What is the SI unit of pressure? density, so p = hDg.
5.10 Which exerts a greater pressure, a force of
100 N acting on 1.0 cm2, or the same force
acting on 2.0 cm2?
5.11 What pressure is exerted by a force of 40 000 N Worked example 5.3
acting on 2.0 m2?
5.12 A swimming pool has a level, horizontal, bottom Calculate the pressure on the bottom of a
of area 10.0 m by 4.0 m. If the pressure of the swimming pool 2.5 m deep. How does the
water on the bottom is 15 000 Pa, what total force pressure compare with atmospheric pressure,
does the water exert on the bottom of the pool? 105 Pa? (Density of water = 1000 kg/m3.)
Step 1: Write down what you know, and what you
want to know.
S Pressure, depth and density h = 2.5 m
We have seen that the deeper one dives into water, the ρ = 1000 kg/m3
greater the pressure. Pressure p is proportional to depth
g = 10 m/s2
h (we use the letter h, for height). Twice the depth means
twice the pressure. Pressure also depends on the density ρ p=?
of the material (here ρ is the Greek letter ‘rho’). If you dive Step 2: Write down the equation for pressure,
into mercury, which is more than ten times as dense as substitute values and calculate the answer.
water, the pressure will be more than ten times as great.
p = hρg = 2.5 m × 1000 kg/m3 × 10 m/s2
We can write an equation for the pressure at a depth
= 2.5 × 104 Pa
h in a fluid of density ρ:
This is one-quarter of atmospheric pressure. We
live at the bottom of the atmosphere. There is about
pressure = depth × density × acceleration due to gravity 10 km of air above us, pressing downwards on us –
p = hρg that is the origin of atmospheric pressure.

Questions
5.13 A water tank holds water to a depth of 80 cm. a Calculate the volume of the tank from the
What is the pressure on the bottom of the tank? dimensions shown in the diagram.
(Density of water = 1000 kg/m3.) b Calculate the weight of the oil in the tank.
5.14 The diagram shows a tank that is filled with oil. c The pressure on the bottom of the tank is
The density of the oil is 920 kg/m3. caused by the weight of the oil. Calculate the
pressure using:
F
p=
A
1.0 m d Now calculate the pressure using p = hρg.
Do you find the same answer?
2.0 m
1.5 m

Chapter 5: Forces and matter 73

 Summary

You should know:

◆ the effects of forces, including stretching
S ◆ Hooke’s law
S ◆ how to interpret extension–load graphs
◆ about the idea of pressure
◆ how to calculate pressure from force and area
S ◆ how to calculate pressure in fluids.

End-of-chapter questions

1 A student measures an unstretched spring. He adds weights to the spring and measures its new length each time.
a Copy the correct equation that shows how to calculate the extension of the spring.
extension = length of spring − load
extension = original length − length when stretched
extension = length when stretched − original length
b Copy the correct graph to show how the extension of a spring changes as the load on it is increased.

a b c
Extension

Extension

Extension

Load Load Load

S 2 Hooke’s law describes how the extension of a spring relates to the load on the spring.
a State Hooke’s law in words.
b Hooke’s law can be written as F = kx. Rewrite this as a word equation.
c Sketch an extension–load graph for a spring that obeys Hooke’s law. Mark the part of the graph that shows
that the spring obeys Hooke’s law. Mark also the limit of proportionality.

3 Copy and complete the following sentences, by writing either increases or decreases in each gap.
a Pressure …… when the force acting increases.
b Pressure …… when the force acts on a greater area.
c Pressure …… when you go deeper in a liquid.
d Pressure …… when you go higher in the atmosphere.

74 Cambridge IGCSE Physics

 4 a Write a word equation that relates pressure, force and area.
b Write the same equation in symbols.
c Copy and complete the table to show the unit for each quantity.

Quantity Unit
force

area

pressure

S 5 The pressure in a fluid of density ρ depends on the depth h. It also depends on the acceleration due to
gravity g.
a What is a fluid?
b What is the value of g on the Earth’s surface?
c Write an equation in symbols to show how to calculate the pressure in a fluid.

6 When a spring is stretched, its length increases from 58.0 cm to 66.0 cm. Calculate its extension. [3]

7 A student has a short spring. He is required to investigate how the length of the spring changes as the load
stretching it increases. Describe the experimental procedure he should follow, stating the equipment he
should use and the measurements he should make. [6]

8 The table shows the results of an experiment in which a long piece of plastic foam was stretched by hanging
weights from one end.

Load / N Length / cm Extension / cm

0.0 83.0 0.0
5.0 87.0
10.0 91.0
15.0 95.0
20.0 99.0

a Copy the table and complete the third column to show the value of the extension produced by each load. [4]
b Use your completed table to plot an extension–load graph. [3]

9 a Draw a labelled diagram to show a simple mercury barometer. [3]

b Describe how such a barometer shows changes in atmospheric pressure. [1]

10 Your friend has fallen through the thin ice on a frozen pond. You come to the rescue by laying a ladder
across the ice and crawling along the ladder to reach your friend. Use the idea of pressure to explain
why it is safer to use the ladder than to walk on the ice. [3]

Chapter 5: Forces and matter 75

S 11 An unstretched spring is 12.0 cm long. A load of 5.0 N stretches it to 15.0 cm. How long will it be under a load
of 15 N? (Assume that the spring obeys Hooke’s law.) [3]

12 A group of students carried out an experiment in which they stretched a length of wire by hanging weights on
the end. For each value of the load, they measured the length of the wire. The table shows their results.

Load / N 0 10.0 20.0 30.0 40.0 50.0 60.0 70.0

Length / m 3.200 3.207 3.215 3.222 3.230 3.242 3.255 3.270

a Copy the table and add a row showing the extension for each load. [4]
b Use the data in your table to draw an extension–load graph for the wire. [4]
c From your graph, determine the extension produced by a load of 25.0 N. [2]
d Determine the value of the load at the limit of proportionality. [2]

13 The pressure of the atmosphere is 100 000 Pa.

a Calculate the force with which the atmosphere presses on the outside of a large window 2.0 m
high and 1.25 m wide. [3]
b Explain why this force does not break the window. [1]

S 14 On a particular day, the height of the mercury column in a simple barometer is 760 mm. Calculate
the atmospheric pressure on this day. (Density of mercury = 13 600 kg/m3, g = 10 m/s2.) [3]

15 An unstretched spring of overall length 50.0 mm is hung from a support, as shown in the diagram.

50.0 mm

load

Different loads are placed on the spring and the extension is measured each time.
a Copy the diagram, and mark clearly on it the extension caused by the load. [1]

76 Cambridge IGCSE Physics

b The extensions for different loads are given in the table.

Load / N Extension / mm
0 0
1.0 10.0
2.0 20.5
3.0 31.0
4.0 41.5

i Copy the graph axes shown below onto graph paper. Plot these values in the table shown, using dots in
small circles (~), and draw the best straight line for the points. [3]
60

50

40
Extension / mm

30

20

10

0 1 2 3 4 5
Load / N

ii Copy and complete the following sentence by inserting the appropriate word. [1]
Within the limits of experimental accuracy, the load and the extension of the spring are ……
to each other.
iii A load of 2.5 N is hung on the spring.
1. What does the letter N stand for? [1]
2. Use the graph to estimate the overall length in mm of the spring when 2.5 N is hanging from it. [2]
[Cambridge IGCSE® Physics 0625/22, Question 2, October/November, 2010]

Chapter 5: Forces and matter 77

16 a The diagram shows end views of the walls built by two bricklayers.

A B

soil

reinforced reinforced
concrete concrete
foundation foundation

Which wall is the least likely to sink into the soil, and why? [2]
b The diagram shows two horizontal squares P and Q.

P Q

The atmosphere is pressing down on both P and Q.

i Name two quantities that would need to be known in order to calculate the atmospheric pressure on
square P. [2]
ii The area of P is four times that of Q.
Copy and complete the following sentences. [3]
1. The atmospheric pressure on P is …… the atmospheric pressure on Q.
2. The force of the atmosphere on P is …… the force of the atmosphere on Q.
[Cambridge IGCSE® Physics 0625/22, Question 4,Paper May/June, 2010]

78 Cambridge IGCSE Physics