IGCSE Physics- year 10 notes

Qatar International SchoolScience Department

IGCSE Physics- year 10 notes

How to use these notes

These notes are not in the final form that you will need for the exam. Throughout the course you will be required to make additions in several forms. Some sections are blank, for you to complete during class, some as home work and other areas are not blank but you will need to make your own notes in your own words to help you remember the class discussions and explanations. Feel free to add comments at any point not only where there are blank spaces. Use the margins or spaces between paragraphs if there is no clear space. Use blank paper and staple in extra sheets.

It is usually followed, in normal type, by an explanation of what you need to know, worked examples, questions for you to finish, etc. Please point out any errors in these notes to Mr. Robinson.

Module 1- General Physics.1.1 length and time.

Units of length;

Units of volume;..

STANDARD unit of time, used for most calculations; There are others such as; ..

To measure the thickness of a sheet of paper, using an ordinary ruler you must;

Measure many sheets and divide to get an average.For example, if 100 pages are 11mm, then one sheet has a thickness of; ..

Show the calculation you used .

Name a more accurate tool we can use to measure small distances..

We can make the measurement more accurate by

Period means the time for one complete swing of a pendulum or the time for one complete wave.

Set up the pendulum as shown in the diagram below, and use a suitable measurement and calculation to find the period. Describe the steps and calculation in the space below..

A simple pendulum

Questions (answer below)

1) A pendulum swings 100 times in 25 seconds. Calculate its period. Would you expect it to swing as far on the last swing as the first? Explain your answer in terms of energy.

2) A sheet of paper is 0.13 mm thick. How thick would a ream of 250 sheets be?

3) If there are 10000 grains of sand in 1 cm3, estimate the volume of a single grain.1.2 Speed, velocity and acceleration

Speed means

Units of speed .

Velocity means.

Units of velocity.

Acceleration means

Units of acceleration..

A very useful point about units and equations.

For any equation that is true, the units must balance, example we can not add kg to m;

3 kg + 2 m = 5 kg, is not a true equation.

But we can add kg to kg as the units balance; 3kg+ 2 kg = 5 kg. (Kg on left equals kg on right)

This applies to all equation, such as; speed = distance / time

So units of speed are equal to units of distance (meters) / units of time (seconds) i.e. m/s

Acceleration = change in speed / time

Units of acceleration are units of speed (m/s) divided by time (s) i.e. m/s/sExamples;1) A car moves 200 km in a time of 5 hours. What is its speed in km/h, in m/s?

Hint; speed in m/s = (distance in metres)/ (time in seconds)

2) A motorbike does the same journey (200km) in half the time. What is its speed?

Always put time on the x-axis.

Learn to recognize and explain a few common shapes, very complex graphs are made up of sections which can be treated separately.

The most common mistake is not to be clear about the difference on the y-axis; check is it speed or time?

at rest means not moving, speed = 0

Moving with constant speed means speed does not change, but distance changes.

Moving with changing speed, if it is getting faster we say it is accelerating. If slowing down we say decelerating (NOT de-accelerating)

At rest at rest means not moving, speed = 0

Speed = 0 means a line along the x-axis on a speed-time graph at rest means not moving, the distance does not change

The line is flat, and may be on the x-axis or may not be on the x-axis for a distance-time graphDraw the line to represent an object at rest on the two graphs below

Constant speed

Moving with constant speed means speed does not change.

Speed does not equal zero so the line is not on the x-axis of a speed-time graph. Moving with constant speed means speed does not change, but distance changes.

A straight line on a distance-time graph means constant speed.Draw lines to show constant speed on the two graphs below.

Draw a scale on the graphs to allow you to plat a line for a car moving with speed of 25km/h and 10m/s respectively.

Changing speed; accelerationDraw line 1 and two on the first axes, line 3 and 4 on the second axes.

Line 1; a straight line going upwards means constant acceleration.

Line 2; Steeper slope up means more acceleration.

Line 3; shallow slope upwards means less acceleration.

Line 4; starting from a speed greater than zero. Small acceleration.

Changing speed; deceleration

Draw these lines on the axes below. Draw two lines on each graph Line 1; if the slope is down, this shows deceleration.

Line 2; less slope means less deceleration.

Line 3; steeper means more deceleration.

Line 4; lower initial speed and small deceleration

The area under the graph line for a speed/time graph will give the distance. Because the syllabus states with a constant acceleration this means the graph line will always be a straight line. This means the area you have to calculate will be a rectangle or triangle, and we can use base x height or base x height, reading values from the graph to find the base and height. This is demonstrated below.

Worked exampleFrom the graph calculate the distance travelled between the points a) 0 to A, b) 0 to B, c) B to C, d) B to D

The method used to solve the question.

First break the graph into simple shapes (rectangles and triangles) as shown in dotted lines in step 1 below.

Then read off the axis to find the length of all the sides, as shown with double arrows in step 2 below.

Then work out the areas of the shapes, as shown in boxes in step 3 below.

Finally add the shapes together to get the area required to find the distance, since area under the line between any two points = distance travelled between the two points.

So, the answer is;

0 to A = area of triangle = 5

0 to B = triangle + rectangle = 5 + 20 = 25

B to C = triangle + rectangle = 5 + 10 = 15

B to D = triangle + triangle + rectangle = 5 + 10 + 10 = 25

Example questions.Example1.Describe the motion shown in the graphs below, use the table to help you.Answer the questions below in the space below the graph.

TimeDescription of the motion

0 to 1

1 to 3

3 to 6

6 to 7

7 to 9

How far was the maximum distance between the object and its starting point?How far away was it at the end of the 9 second period?

a) Complete the table to describe the motion (motion = movement).TimeDescription of the motion

0 to 1

1 to 3

3 to 6

6 to 7

7 to 9

b) Calculate the distance travelled in the first 6 seconds.

c) Calculate the total distance travelled in the whole 9 seconds of the motion.

Example 3.A man starts from rest, and then accelerates for 2 seconds until his speed is 5m/s. He keeps a constant speed for 5 seconds then decelerates for 3 seconds until he is at rest again.Draw a speed/time graph below. Calculate from the graph the total distance he moved.

Acceleration when it first starts to fall Acceleration just before landing

A0 m/s/s0 m/s/s

B0 m/s/sg (= 10 m/s/s)

Cg (= 10 m/s/s)0 m/s/s

Dg (= 10 m/s/s)g (= 10 m/s/s)

Example; An apple falls from a tree, which line in the table correctly describes the acceleration of the apple? Explain your answer? Answer .. Because ..

1.3 Mass and Weight

Normally we speak about weight in the wrong way.

If I ask what do you weigh? You may reply 55kg. But since weight is a force it should be measured in Newton.

Your mass is actually what we measure in kg, your mass is 55 kg, but your weight is 550 N.

The equation to connect mass and weight is

Weight = mass x gravity;

In symbols W = mg and on Earth g = 10 m/s/s so 1kg (mass) has a weight of 10 N

On other planets, Gravity is different. Example on the moon gravity is less than Earth.

We can work out the weight of any object if we know its weight and the gravity acting on it, using weight = mass x gravity.

Example complete the following;

A man has a mass of 100kg. On Earth gravity = 10 m/s/s, calculate his weight.

The same man travels to the moon, where gravity is 2m/s/s calculate his weight.

Calculate his weight in space, where gravity is 0m/s/sRepeat the above for a boy with mass of 55 kg.

On Earth weight =

On the moon weight =

In space weight =

Mass is a property which 'resists' change in motion.A large object is hard to move because it has a large mass, not due to its large weight, the object in space still is hard to move as it still has mass even though it is weightless (no weight). This shows that the resistance to being moved depends on the mass, not weight.

Weight is a FORCE caused by gravity acting on a mass.Weight always acts straight downwards (the direction of gravity). An object in space has no weight because there is no gravity.

Experiment on mass and weightYou have several objects and a 1 kg mass.

By feeling the weight of a 1 kg mass (the force is 10 N) and comparing it to the weight of the objects, estimate (guess) the mass and weight of each object. Record the estimates in the table below.Measure the masses using a balance (in kg or g). Record the values in the table.Measure the weights using a Newton-meter. Record the values in the table.Also measure and record the size of the objects (length height width) this will let you calculate the size (or volume) of the objects using length x height x width.

Is the largest always the heaviest object?ObjectEstimatedMeasuredSize of the objectvolume

massweightweightmasslengthwidthheight

1.4 Density. The previous experiment on mass and weight showed that a large object is not always the heaviest object.

The mass compared to the size (volume) is called the density of an object.

A small but heavy object means it has a high density. A large but light object has low mass. Objects of different sizes can have the same density. The equation is density = mass / volume.

Can you figure out the units of density? (hint use the equation) Units are units of mass (kg) / units of volume (m3), thats kg/m3 or we can use g/cm3Use the data in the table from the last experiment on mass and weight to calculate the density of each object, and write the values in the last column of the table.The experiment and calculation you have done is listed in the syllabus as;

Can you use the idea of density to explain why a metal coin with a mass of a few grams will sink in water, but a tree with mass of many kg will float in water? Calculate the density of the following objects, and state the units in each case

a) volume = 100 m3 mass = 10000kg

b) volume = 25 cm3 mass = 100g

c) volume = 200 cm3 mass =100g

In the previous examples and the experiment the solid was a regular shape. If the solid is irregular, we measure the volume differently but the rest of the calculation is the same.We use a displacement method to measure the volume. Drop an object into water, and the water level will rise by an amount equal to the volume of the object. This method is shown below.

Experiment. Measuring volume by displacement.

Use a displacement method to find the volume of some objects. Record your results in the space below, including the calculation and units.Name of the objectVolume of waterVolume of Water and object togetherVolume of object

1.5 Forces1.5 (a) Effects of forces

A force in the direction an object is moving will cause acceleration. E.g. pressing the gas when driving at a steady speed makes you get faster.

A force in the opposite direction to that an object is moving will cause deceleration.

E.g. pressing the brake when driving at a steady speed makes you slow down. A force from the side will change the direction an object is moving. E.g. turning the steering wheel causes a sideways force which turns the car in the direction of the force.

Example; pulling a spring causes it to get longer.

Putting a force on a spring is called loading the spring.

The amount of increased length is called the extension.

More load will give more extension.

We can plot a graph of load against extension.

Loading a ruler at one end may cause it to bend.

Hang a spring on a stand and measure its length. This is the original, unloaded length. This can be called l0 Add a load and measure the new length. Record the results in the table in your notes. This can be called l1 l2 l3 etc Convert the mass into a force (in N not g)

Calculate extension by subtracting the new length minus the original length. In symbols, for the first load extension = l1 l0

Repeat this with larger loads. Plot a graph force (x-axis) against extension below.

This means you must be able to use a graph to read off values from one axis, e.g. given the force find the extension.

Make sure you know which axis to begin with, and then read the values just like any other graph question from Maths.Worked Example question.

From the graph below,

a) Find the force to produce an extension of 5 cm.

b) Find the extension produced by a force of 5N.

Answer a)

The extension is 5, so from 5cm on the y-axis you draw across to the graph-line and then down. Read off the x-axis to find the force. Answer 7.5N.Answer b)The force is 5, so from 5N on the x-axis you draw up to the graph-line and then across. Read off the y-axis to find the force. Answer 3cm.

Hookes law states force is proportional to extension.

This means as you increase the force the extension it causes will also increase.

We can write this as an equation using a constant (called a constant of proportionality). Often called the spring constant in a question. So Force = constant x extension or F=kx

Hookes law applies to an elastic material, like a spring. It can be used to calculate extension or load as shown below.

Limit of proportionality means the point where extension stops being proportional to the load.

It can also be called the elastic limit.

Confusing names? Simply look for the point where the line starts to curve. Limit of proportionality is labeled on the graph below.

Hookes Law calculations.Worked example.

A spring has an original, unloaded length of 25 mm. When a load of 2N is applied the length becomes 29mm.

a) Calculate the extension for 2N.

b) Calculate the extension for 5N.

c) Calculate the new length when 5N is added to the spring.

a) Extension = new length original length

Extension = 29 25 = 4mm.

b) F = k e where F=2, e=4 so 2=k4 so k=2/4 = 0.5

(NOTE; we worked out the constant, so we can now use k=0.5 to find the new extension for the force of 5N)

For 5N, F=k e so 5= 0.5 x e so e = 5/0.5 = 10mm.

c) New length = original length + extension

Original length= 25mm (from question), extension=10mm (above)

New length for 5N load = 25+10 = 35mm.Example.A spring has a length of 3cm when unloaded.

The same spring has a length of 3.1 cm when loaded with 1N.

a) Calculate the extension for 1N. Write the value in the table.b) Use the Hookes law equation to find the spring constant.

c) Calculate the extension for 2N. Write the value in the table.d) Calculate the extension for 3N. Write the value in the table.e) Calculate the extension for 4N. Write the value in the table.f) Calculate the length of the spring for a load of 3 N.

g) Given the extension measured for 5N was 8mm plot a graph of force against extension below.

h) Label the elastic limit on the graph.

Force012345

Extension08

This means you need to learn and apply the equation F= Ma

In words; force = mass x acceleration.

A force forwards makes an object increase in speed (accelerate) but a force backwards will slow it down (decelerate).

Example how much force is required to make an object with mass 5kg accelerate at 20m/s/s?

Example how much force is required to make an object with mass 500g accelerate at 20m/s/s?

1.5b) turning effect.

A force can stretch an object, or change the speed an object, or change its direction. Another possible effect is to turn an object, e.g. the force to turn a tap or door handle. We call the turning effect the moment of a force.The moment of a force is given by BOTH the size of the force and the distance from the pivot.

Moment = force x distance (to the pivot)

Example; calculate the moment when a force of 5N is applied 1m away from a point.Answer; moment = force x distance = 5 x 1 = 5

A beam can balance when the masses are not equal, e.g. a see-saw can balance when a large person is closer to the pivot than a small person, as shown below. This is when the moments are equal size but opposite direction. One moment is trying to turn it clockwise, the other trying to turn anti-clockwise which means they will cancel each other, if they are the same size.

The beam means the flat object (e.g. a ruler) which the forces are applied to.

The pivot is the point where the beam will turn (often the middle). The pivot is often drawn as a triangle on a diagram.

Any system is balanced when the moments in opposite directions cancel out.A balanced system is in equilibrium, there is no net moment.This is also called the principal of moments, which states

For any system in equilibrium there is no net moment

We use this in calculations as

Sum of clockwise moments = sum of anticlockwise moments

Where sum of means add them together

We can use this in an experiment, if the moments are equal and opposite there is no net moment. This means equal in size and opposite in direction. Balance a beam without any load, to find its center point (center of mass).

Put a weight on each side and move one weight until the beam is balanced.

Record the size and position of the masses and calculate the moment on each side of the pivot. They should be equal and opposite.

Repeat for several different masses and positions.

A table for results could look like this;

Left hand side (LHS)Right hand side (RHS)Net Moment LHS-RHS

LoaddistanceMoment

(F x d)LoaddistanceMoment

(F x d)

21020120200

41040220400

61060320600

Perform the experiment and record the results below. For each set of results calculate the net moment, it will often not be exactly zero due to experimental errors.

Left hand side (LHS)Right hand side (RHS)Net Moment LHS-RHS

LoaddistanceMoment

(F x d)LoaddistanceMoment

(F x d)

This means you must be able to solve questions similar to the examples below.

Worked examples of calculations using the principal of moments1) Find the force F to balance the system in the diagram. (note this is a simple example since there is only one force at each side)

By the principal of moments, clockwise moments = anticlockwise moments

100 x 40 = F x 50

F = 100 x 40 / 50

F = 80N

2) Find the force F to balance the system in the diagram. (Note there are two forces turning anticlockwise. We add their moments, NOT the forces).

By the principal of moments, sum of clockwise moments = anticlockwise moment

(Moment of 100N force, 100x40) + (moment of 50N force, 50x30) = moment of F

(100 x 40) + (50 x 30) = F x 50

F = (4000 + 1500) / 50

F = 110N

Example Questions on moments (simple systems in equilibrium)1) Partly worked example

a) The diagram shows a crane lifting a load. Calculate the weight of the counterweight (W) to have zero net moment on the crane.

b) What would be the maximum distance between the crane and load so that a load of 9000N could be lifted with zero net moment using the same counter weight as above?

By the principal of moments

Moment of W = moment of load

(x...) = (x ..)

Distance =

1.5 c) Conditions for equilibrium.

Short summary: If no resultant force and no net moment are applied, nothing changes and we call this equilibrium.Long explanation: We now have studied various effects of forces. When a force is applied to an object, that force can change the shape an object (e.g. stretching a spring), or change the speed an object (e.g. brakes on a car), or change its direction, or turn the object, called the moment of a force.If the object does not change its speed, direction or shape, nor does it turn, this must be because there is no force applied, since these effects are due to a force.

More precisely, there is no RESULTANT force acting on the object, there could be equal and opposite forces acting, which cancel each other out. We call this situation equilibrium. This is when an object has no resultant force and no net moment acting, so it does not change in any of the ways we have studied.

1.5 d) Centre of mass.

Meanings used in this phrase are;

Centre of mass of an object The balance point. The middle, where it has as much mass on one side as the other. A scientific definition is a point at which the mass of an object can be considered to act.Plane lamina This simply means any thin flat shape e.g. a piece of card. We do not need to worry about 3D shapes.Center of mass experimentHang any shape where it is free to swing, and it will settle with the center of mass hanging straight down. If the center of mass is not at the lowest point then the object swings downwards due to its weight.

Use a vertical line (e.g. a plumb line is a mass on a string to mark vertical line) to mark the object. The center of mass is somewhere on the line.

Rotate the object, mark a second vertical line. Since the center of mass is somewhere on the second line too, the center of mass is where the lines cross.

This is shown below in the diagram.

Stability describes if an object will stand up or fall over. A stable object will stand, an unstable object will fall. A less stable object will fall over easier than a more stable one.

There are two factors which are involved here, the position of the centre of mass, and the width of the base.

In general, the LOWER the centre of mass and the WIDER the base then the more stable an object will be.

You will need to answer two types of question in IGCSE as shown below.

ExamplesType 1 questions) which object is the most stable? Which is least stable? (Hint you want the widest base AND lowest centre of mass for most stable)

Answer most stable D, least stable C

Type 2 questions) A student has designed a cup, shown below, but it falls over too easily. How would you advise him to change his design?

Can you draw some of these design ideas below, and explain briefly why they are more stable than the students design?1.7 Pressure

Think of high pressure as concentrating the force on a small area, low pressure means the force is spread out over a larger area.

E.g. Explain why a nail can easily be hammered into a wooden block point first, but cannot be hammered in head first.

Answer; the force of the hammer on the nail is the same in both cases, but the point concentrates the force into a smaller area (high pressure) where as the head of the nail applies the force over a larger area (low pressure)

Pressure = force/area

This allows us to calculate values of pressure.

The units of pressure are (units of area)/ (units of area) i.e. N/m2We can also use Pa (Pascal) note 1Pa = 1 N/m2Worked exampleA point of a drawing pin has an area 0.1of mm2 the head has an area 100mm2If a force of 10N is applied on the pin, find the pressure at both ends.

Answer

Point; Force = 10 N, area = 0.1mm2P=F/A

P=10/0.1 =100N/ mm2(Note units are units of force N divided by units of area mm2 i.e. N/ mm2)Head; Force = 10 N, area = 100 mm2P=F/A

P=10/100 = 0.1N/ mm2(Note, as above, units are units of force N divided by units of area mm2)

In both cases we could work out pressure in N/m2 if we wish, the easiest way is to convert the area into m2 before working out pressure. This is often a good idea if we need to use the answers of pressure to calculate in later parts of a question. This is shown below. Note 1m=100cm=1000mm. 1m2= (1000x1000) mm2 = 1,000,000mm2.Point; Force = 10 N, area = 0.1mm2 = 0.1/1000000m2 =0.0000001 m2P=F/A

P=10/0.0000001 = 100,000,000N/ m2(Note units are units of force N divided by units of area m2 i.e. N/ m2)Head; Force = 10 N, area = 100 mm2 = 100/1000000 m2 =0.0001 m2P=F/A

P=10/0.0001 = 100,000N/ m2(Note, as above, units are units of force N divided by units of area mm2)

As you swim deeper underwater you feel more pressure on your ears. This shows pressure increases with depth.

Swimming in salt water causes even more pressure, as the water in more dense. This shows that pressure also depends on the density of the liquid.

You need to know salt water is more dense than fresh water, and oil is less dense than water (it floats on water). You should also know that pressure under a liquid acts equally in all directions, but increases as you get deeper.Worked example questions1a) At which position is the pressure largest?b) At which position is pressure lowest?

Answer

a) Most pressure at C (largest depth at C and D, but salt water is denser than water)b) Lowest pressure at B (least depth at A and B, but oil is less dense than water)

2) Which arrow shows the direction of the least pressure acting at the point underwater?

Answer;

All are the same pressure, because pressure in a liquid is equal in all directions.

The equation is used to calculate pressure under a liquid, where we know depth and density.

In the equation

h = depth below the surface of the liquid = density of the liquid

g = acceleration due to gravity = 9.8m/s/s (approx. 10 m/s/s)Note we can still use the equation P=F/A under a liquid to find the force applied to an area, as shown in part b) of the worked example below.Worked example pressure questionWater has a density of 1000kg/m3 and a human body has an area of 1 m2.(g = 10m/s/s)a) Find the pressure at a depth of 10m.

b) Find the force applied by this pressure on a human swimming at a depth of 10m.

c) How many kg of mass is this force equivalent to?

Answer

a) P = hg Where h = 10m, = 1000kg/m3 and g = 10m/s/s

We are using P = hg because we know depth density and gravity.P = 10 x 1000 x 10

P = 100000 N/m2b) P = F/AWhere P = 100000 N/m2 (Use the value of pressure from above calculation, because both are at the same depth and so have the same pressure) A = 1 m2 from the question.

We are using P=F/A, because we know the area and need to find the force. Force does not appear in the equation P = hg100000 = F/1

F=100000N

c) weight = mass x gravity

100000 = M x 10

M = 10000kg

Note that this shows there is a pressure causing 100000N which is the force due to 10000kg on you, if you swim at a depth of 10m.

A barometer is used to measure atmospheric pressure.

You need to know that the pressure is represented by the height of the mercury column shown as h on the diagram, this comes up a lot in paper one.

You need to know that there is a vacuum in the tube above the mercury, so the pressure above the mercury in the tube is zero.

Strictly speaking you need to be able to draw this diagram but it has never come up in IGCSE questions in recent years.

The Mercury Barometer

diagram for questions

SHAPE \* MERGEFORMAT

Which letter shows the measurement taken to determine atmospheric pressure?

Name liquid X and state the value of pressure P

Answers;

B shows the pressure

X is mercury

P = 0 (vacuum)

A manometer is also known as a U-tube. It is a U-shaped tube used to compare gas pressures. When water (or mercury) is put in a U-tube, if the pressure at each end is equal the liquid stays at the bottom and the two sides are level, as shown in Fig 2.1. But if the pressure is not equal, then the highest pressure will push water downwards. The bigger the difference, the more the liquid moves.Fig 2.2 shows a pressure higher than atmospheric pressure being applied on the left side.

Fig 2.3 shows a pressure lower than atmospheric pressure (e.g. a vacuum) being applied on the left side.

New Work

New Work

New Work

Year 10

75% of

Syllabus

Study

Leave

And

IGSCE

Revise

And

Mock

Exams

New Work

Year 11

25% of

Syllabus

One swing is from A to B and back to A

Fixed point

B

A

Distance

Speed

Time

Time

Time

Time

Distance

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5

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Volume 100 cm3

Volume 150 cm3

Volume of water = 100

Volume of water + star = 150

Volume of star = 150 100

Volume of star = 50 cm3

Force

N

Extension

cm

Force

N

Extension

cm.

10

10

Limit of proportionality; where the line first begins to curve.

10

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Extension/cm.

F/N

Force

N

Extension

cm

Large force

(at a)

small distance.

Small force

(at a)

large distance.

BALLANCE

40 cm

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100 N

F

40 cm

50 cm

100N

F

50N

30 cm

Hang shape from one end, COM is on the line shown.

Hang shape from another end, COM is on both the lines

COM where lines cross

Centre of mass (COM) experiment

A

B

D

C

Students design

You could answer either

make the cup wider

make the cup shorter

put a wider base on the cup

Other ways to lower the COM are

make the base thicker

put the handle lower

use a shape which is wider at the bottom of the cup

water

oil

Salt water

water

A

B

C

D

D

A

B

mercury

Vacuum

h

Liquid X

Pressure P

C

Use and describe the use of a mechanical method for the measurement of a small distance

4m

1m

10000 N

W

By the principal of moments

..

Moment of W = moment of load

(.) = ()

W =

Use and describe the use of clocks and devices for measuring an interval of time.

Use and describe the use of rules and measuring cylinders to determine a length or a volume.

The text in these text boxes is taken directly from the syllabus. Typing in bold is from the extended syllabus, normal type is from core syllabus.

Measure and describe how to measure a short interval of time (including period of a pendulum).

Define speed and calculate speed from (total distance) / (total time)

Plot and interpret a speed/time graph or a distance/time graph.

Recognise from the shape of a speed/time graph when a body is:at rest, moving with constant speed, or moving with changing speed.

Calculate the area under a speed/time graph to determine the distance travelled for motion with a constant acceleration.

10

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Speed

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Step 2; work out the length of all the sides

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Step 1; break the graph into simple shapes

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Step 3 work out areas of the triangles and rectangles

Example2

5

2

4x5 = 20

1/2x2x10 = 10

1/2x2x5 = 5

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2

5

5

10

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B

A

2x5 = 10

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Speed

10

5

4

5

2

1/2x2x5 = 5

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10

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Speed

10

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5

10

State that the acceleration of free fall for a body near to the Earth is constant.

This means acceleration caused by gravity pulling any falling object is constant, as long as the object is close to the Earth (not in space). We call this acceleration gand its value is g= 9.8 m/s/s (approx 10 m/s/s)

Show familiarity with the idea of the mass of a body.

State that weight is a force.

Demonstrate an understanding that mass is a property which 'resists' change in motion.Describe, and use the concept of, weight as the effect of a gravitational field on a mass.

Describe an experiment to determine the density of a liquid and of a regularly shaped solid and make the necessary calculation.

Describe the determination of the density of an irregularly shaped solid by the method of displacement and make the necessary calculation.

Describe the ways in which a force may change the motion of a body

State that a force may produce a change in size and shape of a body.

Plot extension/load graphs and describe the associated experimental procedure

Interpret extension/load graphs.

State Hookes Law and recall and use the expression F = k x

Recognise the significance of the term 'limit of proportionality' for an extension/load graph.

Describe the moment of a force as a measure of its turning effect and give everyday examples.

Describe, qualitatively, the balancing of a beam about a pivot.

Perform and describe an experiment (involving vertical forces) to verify that there is no net moment on a body in equilibrium.

Apply the idea of opposing moments to simple systems in equilibrium.

State that, when there is no resultant force and no resultant turning effect, a system is in equilibrium.

Perform and describe an experiment to determine the position of the centre of mass of a plane lamina.

Describe qualitatively the effect of the position of the centre of mass on the stability of simple objects

Relate, without calculation, pressure to force and area, using appropriate examples.

Recall and use the equation p = F/A

Relate, without calculation, the pressure beneath a liquid surface to depth and to density, using appropriate examples.

Recall and use the equation p = hg (pressure = depth x density x gravity)

Describe the simple mercury barometer and its use in measuring atmospheric pressure.

Use and describe the use of a manometer.

A

D

C

B

A

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Recall and use the relation between force, mass and acceleration (including the direction).

Created by Mr Phillips

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Triangle area = base x height

Rectangle area = base x height

Triangle area = base x height

Total Area = sum of the three areasTotal Area = distance traveled

Speed

Time