SENIOR SECONDARY SCHOOL PHYSICS VOLUME I
Quick revision
1. PHYSICAL QUANTITIES AND UNITS
1 What are physical quantities?
 Physical quantities are all measurable properties of a body.
2 write down 0.0000000023 in scientific notation
 0.0000000023 =20.3 × 10 -9
3 write down 230000000000 in scientific notation
 230000000000 = 2.3 × 10 11
4 the speed of radio waves is 3.0 x 108 m /s. Express this speed in ordinary form.
 3.0 x 108 m / s = 300,000000
5 how many significant figure does the number 0.2000 have
 It has 4 significant figures
6 Express 2kg in grams
 2kg = 2 × 1000 =2000g
7 Express 20 g in kilograms
 20 g = 20 ÷ 1000 = 0.02 kg
8 what unit is defined as the force that gives a body of 1kg an acceleration of 1m/s 2.
 Newton
9 what unit is defined as the rate of doing work of 1 joule per second
 Watt (W)
10 how many seconds are in 1 minute
 1 minute = 1 × 60 = 60 s
11 how many seconds are in 1 hour
 1 hour = 1× 60 ×60 = 3600 s
12 (a) What is parallax error ?
 It is the error which arises due to incorrect positioning of the eye.
(a). How do we avoid parallax error ?
 Always place the eye vertically above the mark being read. Or place the eye level in line with the mark being read.
1 2 What gives the most accurate value for the internal diameter of a test tube and pipe a with a single measurement?
 vernier calipers
14 A student has been asked to determine, as accurately as possible, the volume of a piece of wire. The wire is about 80 cm long and
about 0.2 cm in diameter. Which measuring instruments should the student use?
 metre rule for the length of 80 cm and micrometer for the diameter 0.2 cm
15 A car driver takes a total of two hours to make a journey of 75 km. She has a coffee break of half an hour and spends a quarter of
an hour stationary in a traffic jam. At what average speed must she travel during the rest of the time if she wants to complete the
journey in the two hours?
A 38 km/ h
B 50 km/ h
C 60 km/ h
D 75 km/ h
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1.1 MEASUREMENT OF LENGTH, TIME AND VOLUME
1(a) Figure 1.1 shows the micrometer screw gauge, the device used to accurately measure the diameter of a thin wire and thickness of
a razorblade.
D
A
B
E
C
(ii) What physical quantity does the device in figure 1.1 measure?
 length
(iii) Name the parts labeled A, B, C, D, E and F
 A is Anvil
B is spindle C is sleeve (with the main scale) D is Thimble and E is the Ratchet
(iv) What is the function of the ratchet?
 The ratchet knob is used to prevent the user from exerting undue pressure.
(b) What precaution do you need to take before using the micrometer screw gauge?
 check for zero error before using micrometer
2 Find the leading of the micrometer below
(a)
(b)
20
25
mm

45
40
35
35
25
30
Reading = 18 + 0.34 = 18.34
(d)
Reading = 29 + 0.43 = 29.43
(c)
0
15
10
5
25
0
20
15
5
25 Reading = 10.5 + 0.21
10
= 10.71
20
6.5
10
10. 5
0.21
Reading = Reading of main scale + Reading of thimble scale.

Reading = 6.5 + 0.18 = 6.18
3 Describe how you would find the thickness of a sheet of paper used in a magazine using micrometer screw gauge

check for zero error when using micrometer, close instrument on to paper not too tight/use ratchet ,take reading of both
15
scales ,use several sheets, divide reading by no. of sheets
2 Figure 2.1 shows the vernier calipers. An instrument used to measure accurately the internal and external diameter of a pipes with a
single measurement.
Outside jaws
A
B
Inside jaws
(ii) What physical quantity does the device in figure 1.1 measure?
 length
2
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(ii) Name the parts labeled A and B
 vernier scale and B is the main scale
(iii)what are the functions of the inside and outside jaws?
 To measure the inside and outside diameters of pipes
(b) The diagram shows part of a vernier scale. What is the reading on the vernier scale?
4
5
0
cm
0

0
reading = 4.5 + 0.05 = 4.55 cm
10
reading = 2 + 0.04 = 2.04 cm
MULTIPLE CHOICE
1 The diagram shows a vernier V placed against a scale S.
0
5
10
2
3
What is the vernier reading?
A 2.23
B 2.26
C 2.33
D 2.36
2 Which device should be used to accurately measure the diameter of a thin wire?
A measuring cylinder
B metre rule
C micrometer
D vernier calipers
3 A girl uses a rule to measure the length of a metal rod. Because the end of the rule is damaged, she places one end of the rod at the 1
cm mark as shown. How long is the metal rod?
Damaged end
1
A 43 mm
B 46 mm
2
1
Metal rod
5
4
3
6
C 53 mm
cm
7
Length = 5.6 – 1.0 = 4.6 cm
4.6 cm = 4.6 ×10 mm = 46 mm
D 56 mm
4 A rule is used to measure the internal diameter of a pipe.
Pipe
1
What is the internal diameter of the pipe?
A 1.6 cm
1
B 1.8 cm
C 2.0 cm
2
3
1
1
cm
D 2.6 cm
5 write down precautions necessary when using a measuring cylinder to measure the volume of a quantity of water?



Place the measuring cylinder on a horizontal surface (make sure that the measuring cylinder is vertical)
The scale must face you. making sure that your eye is level with the liquid surface
Your eyes must be level with the meniscus. Read the level at the bottom of the meniscus
3 A body of mass 500 g was suspended in 100 cm3 of water by a piece of cotton as shown. What is the density of the body?
A 0.38 g / cm3
B 2.63 g / cm3
C 5.00 g / cm3
3
D 5.56 g / cm3
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PENDULUM
1 A simple pendulum starts with its bob at position X, shown in Fig. 4.1. The bob is pulled aside to Y and then released. It swings
from Y to Z and back to Y.
bob
Y
Z
X
(a) The pendulum takes 1.6 s to swing from position x to position y. Calculate the period of the pendulum
 period = 1.6 × 4 = 6.4 s
(b) If the periodic time is 2.4 s, How long does it for the pendulum to move from Y to X
 It takes 2.4/ 4 = 0.6 s to move from Y to X.
(c) What is the length of a pendulum
 The length of a pendulum is a distance from the point of suspension to the centre of bob at the support.
(d) What is an oscillation
 The maximum displacement of a bob from its rest position is called a swing. The complete swing i.e. from Y to Z and back
to Y again is called an oscillation.
(e) What name is given to the time taken for one complete oscillation?
 It is called a period.
(f) What do we call the number of complete oscillation
 The number of complete oscillation is known as frequency. Frequency is measured in hertz (Hz).
(g) Name two factors on which a period of the pendulum depend
 depends on the length and acceleration due to gravity.
(h) One oscillation of a swinging pendulum occurs when the bob moves from Y to Z and back to Y again. Using a stopwatch, what
would be the most accurate way to measure the time for one oscillation of the pendulum?
 Time 20 oscillations and divide by 20.
(i) A pendulum is set in motion and 20 complete swings are timed. The time measured is 30 s. What is the time for one complete
swing of the pendulum?
𝒕𝒐𝒕𝒂𝒍 𝒕𝒊𝒎𝒆 𝒕𝒂𝒌𝒆𝒏
 Period =
=30/20 =1.5 s
𝟐𝟎
(j) Given that it takes 0.75s for the bob to swing from Y to Z to X, Calculate the period of the pendulum
 period= 0.75 / 3 ×4 = 9 s
(k) State the position, Y, X or Z at which the pendulum has the least potential energy
 X
2(a) How to determine the period of a pendulum accurately
 Pull the bob to one side at a small angle of about 5° to the vertical and release it. Let the pendulum make at least 5
oscillations before timing is started. Using a stop watch, timing is started at the instant that the bob passes the rest position.
One oscillation is counted when the bob moves B to C, and then from C to A and back to B. Time for 20 oscillations and
calculate the average time for one oscillation which is
𝒕𝒐𝒕𝒂𝒍 𝒕𝒊𝒎𝒆 𝒕𝒂𝒌𝒆𝒏
𝟐𝟎
Repeat the experiment several times and then
calculate the average period.
(b) One oscillation of a swinging pendulum occurs when the bob moves from X to Y and back to X again
A student obtained the following values for the time for 20 complete oscillations of the pendulum10.8s, 10.9s, 11s, 10.8s and 10.9s.
Calculate the period of the pendulum.

1 ( 10.8  10.9  11  10.8  10.9 ) = 0.544s
20
5
(c) Marlene obtained the following values for the time for 20 complete oscillations of the pendulum14.5s, 14.4s, 14.6s, 14.5s and
14.7s. Calculate the period of the pendulum
4
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VECTORS AND SCALARS QUANTITIES
1 (a) What is a vector quantity
 A vector quantity is a measure that has both a direction and a magnitude (size).
(b) list down at least five Examples of vector quantities
 Force, Velocity, Weight, Acceleration, Displacement and pressure.
2 (a) What is a scalar quantity
 a scalar quantity is a measure that has magnitude only (A scalar has size but no direction.)
(b) list down at least five Examples of scalar quantities
 Mass, Speed, Volume, Temperature: Distance, density. Energy, time
3 Which of the following groups of physical quantities time, acceleration, mass, speed, force and velocity consists only of scalars?
 mass, speed, time
 4 Which of the following groups of physical quantities consists only of scalars?
A acceleration, force, velocity B acceleration, mass, speed C force, time, velocity
D mass, speed, time
SPEED, VELOCITY AND ACCELERATION
1 Explain what is meant by speed.
 distance travelled per unit time or in one second
2 State the difference between speed and velocity.
 velocity has a direction/is a vector or speed does not have a direction/is not a vector or displacement/time and
distance/time
3 Child is standing on the platform of a station, watching the trains.
A train travelling at 30 m / s takes 3 s to pass the child. What is the length of the train?
 Distance = speed × time = 30 m/s × 3s =90 m
4 A tunnel has a length of 50 km. A car takes 20 min to travel between the two ends of the tunnel. What is the average speed of the car
in km/h?
 Speed = distance/ time = 50 ÷ (20/60)=150 km / h
5 A snail crosses a garden path 30 cm wide at a speed of 0.2 cm/s. How long does the snail take?
 Time = distance/speed=30/0.2 = 150 s
1(i) Explain what is meant by velocity?
 Velocity is the rate of change of displacement. or Velocity is speed in a specified direction
(ii) What must change when a body is accelerating?
 the velocity of the body
(iii) What is Acceleration?
 Acceleration is the change of velocity per unit time.
CALCULATIONS
Acceleration is positive when velocity is increasing. When a body changes its velocity from U m/s to V m /s, the acceleration is
𝑽‒𝑼
given by 𝑎 = 𝒕 Where U = initial velocity (m /s), V = final velocity (m /s) and T = time (s) and a = acceleration (m /s 2).
1 A car accelerates from 8 m/s to 20 m/s, and takes 6 seconds to do it
(a) What is its the acceleration?

𝑎=
𝑽‒𝑼
𝒕
=
𝟐𝟎 ‒ 𝟖
𝟔
= 12/ 6 = 2 m/ s2 (that's a pretty impressive car!)
(b) What is its average velocity?
 average velocity = (8 +20) / 2 =14 m/s
(c) What distance does it cover?

Distance = average velocity × time =
𝟐𝟎 + 𝟖
𝟐
× 6 s = 84 m
Acceleration is negative when velocity is decreasing. Negative acceleration is called retardation or deceleration
Retardation = ‒
(𝑽 ‒ 𝑼)
𝒕
Where U = initial velocity (m /s), V = final velocity (m /s) and T = time (s) and a = acceleration (m /s)..
5
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2 The speed of a cyclist reduces uniformly from 2.5 m/s to 1.0 m/s in 12 s.
(a) Calculate the deceleration of the cyclist.

Acceleration = change in velocity/ time ∴ (2.5-1.0 ) / 12 =0.125 m/s2
(b) Calculate the distance travelled by the cyclist in this time.
Distance = average velocity × time =
𝟐.𝟓 + 𝟏.𝟎
𝟐
× 12 s = 21 m
2.2 GRAPHICAL ANALYSIS OF MOTION
1 A car travelling at a uniform speed of 18 m/s. At time t = 0, the driver sees a child run out in front of the car.
At time t = 0.6 s the driver starts to apply the brakes. The car then decelerates uniformly, taking a further 3.0 s to stop.
(i) Draw a graph to show how the speed of the car varies with t.
 sketch
20
15
Speed
(m / s)
10
5
0s
0.3 s
0.6 s
0.9 s
(ii) Calculate the distance travelled in the first 0.6 s of the motion.

Time( s)
Distance = average velocity × time =18 ×0. 6 = 10.8 m
(b) The car has a mass of 920 kg. The maximum forward force produced by the car is 230 N. Calculate the maximum acceleration.
 Use the formula Force = Mass × acceleration.
3 The diagram shows the velocity – time graph of a bus over a period of 90 seconds. The bus reaches a maximum speed of 15 metres
per second.
15
C
B
Velocity (m/s)
D
A
90s
0
(a) Use the graph to describe the motion of a bus between
(i) A and B,
 Constant acceleration
(ii) B and C,
 constant velocity (between B and C, there is no acceleration)
(iii) C and D.
 stopping / deceleration;
(b) What does the shaded part represent?
 The total distance travelled
(a) Express 15 metres per second in kilometres per hour.
 54 km/h (15/1000 ÷ 1/3600)
(b) Given that the acceleration was 0.5 m/s2, calculate the time taken, in seconds, to reach its maximum speed.
 30s
(c) The total distance travelled during the 90 seconds was 750 metres. Calculate the length of time that the bus was travelling at its
maximum speed.
 10 s [ (use ½ (90 +T) 15=750)]
6
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4 Fig. 3.1 shows the speed-time graph of part of a short journey made by a cyclist.
A
20
B
Speed
(m/s)
C
0
0
50
100
Time (s)
a) Which part of Fig. 3.1 shows when the cyclist is travelling at constant speed?)
 AB or 0-50s or the horizontal part
(b) State what is happening during the rest of the journey shown in Fig. 3.1.
 changing speed / speed is reducing/ retardation
(c) (i) Calculate the distance travelled during the first 50 s.
 distance travelled = 20 × 50 = 1000 m
(ii) Calculate the distance travelled between 50 s and 100 s.
 distance travelled = ½ × 20 × 50 = 500 m
(iii) Calculate the total distance travelled.
 total distance travelled =1000 + 500 =1500 m
(iv) Calculate the average speed during the 100 s.
 average speed = 1500/100 = 15 m/s
MULTIPLE CHOICE
1 The diagram shows a speed-time graph. In which region is the acceleration decreasing?
A V to W
B W to X
C X to Y
D Y to Z
X
Speed
Y
W
Z
V
2 On the distance -time graph what does the horizontal line indicate.
 Zero speed as distance is not changing
Time (s)
)
2.3 FREE FALL
1(a) An object is falling under gravity with terminal velocity. What is happening to its speed?
 It is staying constant.
(b) A stone falls freely under gravity. What is meant by the acceleration of the stone?
 The increase in speed of the stone in one second.
2 Fig. 1.1 shows a free-fall parachutist falling vertically downwards. Fig. 1.2 shows how the speed of the parachutist varies with time.
B
speed
downwards
A
time
Fig. 1.
Fig. 1.2
(a) (i) State the name of the downward force acting on the parachutist.
 Weight
(ii) State the name of one upward force acting on the parachutist.
 air resistance
(b) (i) State the initial value of the acceleration of the parachutist. Give the unit of your answer.
 9.8 or 10 m/s2
(ii) Explain why the acceleration decreases from A to B.
 the increase in air resistance as speed increases decreases acceleration
(iii) Explain why the parachutist falls at a constant speed after B.
 constant speed is reached when air resistance balances weight, or when the resultant force is zero
MULTIPLE CHOICE QUESTION
Which graph represents the motion of a body falling vertically that reaches a terminal speed?
B
A
speed
C
speed
time
D
speed
time
7
speed
time
time
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2 A skydiver jumps from an aircraft.
50
velocity
(m / s)
B
40
C
30
20
D
10
E
A
0
0
5
10
15
20
25
30
35
40
45 Time / s
50 travelling with an acceleration which is not uniform?Explain your answer.
(i) Which section of the graph shows the skydiver
 AB
(ii) At which point does she open her parachute? Explain your answer.
 C after point C velocity reduces from 44 m/s to 12/s
(iii) How long after opening her parachute does she reach the ground? Explain your answer.
 15 s ( 35-20)
(iv) When his parachute is fully opened, a parachutist falls towards the ground at a constant speed. Under what condition is this
statement correct?
 The upward force on the parachute is equal to the weight of the parachutist.
4. MASS, WEIGHT AND DENSITY
1 What is mass?
 mass is a measure of the amount of substance in a body
2 what is the density of a material?
 the mass per unit volume of the material
3 what is weight
mass
Density 
 Weight (w) of a body is the force of gravity
volume
4 state the unit of density?
3
3
 g/cm or kg/m
5What are correct units used for mass and for weight?
 Mass is measured in kg while weight in N (Newton)
5 A person measures the length, width, height and mass of a rectangular metal block. Which of these measurements are needed in
order to calculate the density of the metal?
 length, width, height and mass
8 An astronaut has a mass of 80 kg on Earth. He can jump 10 cm high off the surface of the Earth. Explain why when he is on the
Moon he can jump higher than this.
 This is because, on the Moon, his weight is smaller than on Earth.
9 A stone is falling through the air. The acceleration of free fall is 10 m / s2.Ignoring air resistance, what happens to the stone every
second during its fall?
 The speed of the stone increases by 10 m / s.
10 A cube of side length 3 cm is placed in a measuring cylinder. On adding 125 cm3 of water, the cube remains at the bottom of the
cylinder. What is the reading on the measuring cylinder?
 134 cm3
11 A coin is placed on top of a beaker, as shown. If the card is pulled away quickly, the coin does not move sideways but falls into the
beaker
card
beaker
beaker
Which property of the coin makes this possible?
beaker
 inertia
12 A mass resists changes to its motion. Which property of the mass is responsible for this resistance?
 Inertia
13The sides of a cube are 3.00 cm long. The cube is made from a metal of density 7.50 g / cm3.Calculate the mass of the cube.
 use of m = VD (1) use volume of 27 (cm3) as the volume mass= 202.5 (g)
14 A heavy lorry is difficult to stop or start. What property of a heavy truck accounts for this?
 Inertia
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15 A measuring cylinder containing some water stands on a scale pan. A solid ball is lowered into the water. The water level rises
from the 30 cm3 mark to the 40 cm3 mark. The scale reading increases from 100 g to 180 g. What is the density of the material of the
ball?
 8.0 g / cm3
16 Two astronauts are in a rocket being launched to the Moon. One of the astronauts has a mass of 96 kg. The gravitational field
strength on the Moon is about one sixth of that on Earth. State the difference, if any, between
(i) the mass of the astronaut on the Earth and on the Moon,
 no difference
(ii) the weight of the astronaut on the Earth and on the Moon.
 weight is 6 times greater on Earth
(iii) On the Earth, the gravitational field strength is 10 N/kg. On the Moon, the gravitational field strength is 1.6 N/kg. If an object has
a weight of 50 N on Earth, what is its weight on the Moon?
A 1.6N
B 5.0N
C 8.0N
D 80N
SECTION B QUESTIONS
1 The Earth’s gravitational field strength g is lower at Mexico City than at Cape Town. Explain what differences, if any, there would
be for a golf ball if
(i) its mass is measured at both places,
 the mass of an object is the same in all locations
(ii) its weight is measured at both places,
 the weight is determined by the strength of the gravitational field (g).
(iii) the force applied by a golf club to give the ball the same initial acceleration is measured at both places.
 the force required is calculated by finding the product of mass (not weight) and acceleration. So the force would be
the same at both places
2 A measuring cylinder contains 80 cm3 of water and has a total mass of 300 g. A stone is then lowered into the cylinder. The new
reading of the volume is 110 cm3 and the total mass is 390 g. The readings are shown in Fig. 2.1
110 cm3
80 cm3
Mass = 390 g
Mass = 300 g
Fig. 2.1
(a) What is the mass of the stone?
 Mass = 390 g - 300g = 90g;
(b) What is the volume of the stone?
 Volume = 110 cm3 - 80 cm3 = 30 cm3
(c) Use your answers to (a) and (b) to calculate the density of the stone.
 Density =mass/voume =90g/30 cm3 =3g/ cm3
2 An object has a mass of 2.5 kg. On Earth the gravitational field strength, g = 10N/kg.
(a) How much does the object weigh on Earth?

Weight = mass × acceleration due to gravity. (W = mg), 25 N
(b) The object has a volume of 1000 cm3. Calculate its density.
 Density =mass/voume =2.4 kg/1000 cm3 =0.0025 kg/cm3
(c) The object is taken to the moon. Will the density increase, decrease or stay the same?
 stay the same
(d)A student is measuring the density of water. Name a piece of apparatus he could use to measure the volume of the water.
 measuring cylinder / graduated beaker
MULTIPLE CHOICES
1 Container is filled with 5 kg of paint. The density of the paint is 2 g / cm3.Which volume of container is needed?
A 10 cm3
B 400 cm3
C 2500 cm3
D 10 000 cm3
2 Ten identical steel balls, each of mass 27 g, are immersed in a measuring cylinder containing20 cm3 of water. The reading of the
water level rises to 50 cm3. What is the density of the steel?
A 0.90 g/cm3
B 8.1 g/cm3
C 9.0 g/cm3
D 13.5 g/cm3
3 Ten identical steel balls, each of mass 27 g, are immersed in a measuring cylinder containing 20 cm3 of water. The reading of the
water level rises to 50 cm3.What is the density of the steel?
A 0.90 g/cm3
B 8.1 g/cm3
C 9.0 g/cm3
D 13.5 g/cm3
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4 A measuring cylinder containing some water stands on a scale pan. A solid ball is lowered into the water.
40 cm3
30 cm3
180 g
100 g
The water level rises from the 30 cm3 mark to the 40 cm3 mark. The scale reading increases from 100 g to 180 g.
What is the density of the material of the ball?
 A 2.0 g / cm3
B 4.5 g / cm3
C 8.0 g / cm3
D 18 g / cm3
5 The diagram shows a rectangular metal block measuring 10 cm × 5.0 cm × 2.0 cm.
Its mass is 250 g. What is the density of the metal?
A 0.20 g / cm3
B 0.40 g / cm3
C 2.5 g / cm3
D 5.0 g / cm3
6(i) A brick is placed on a Newton balance X and then on a beam balance Y. What is measured by each balance?
Balance X
balance Y
A
B
C
D
balance X
mass
mass
weight
weight t
balance Y
mass
weigh t
mass
weigh
(ii) Name the piece of apparatus he could use to measure the mass of the water.
 Balance / scales
7 (i)Describe an experiment how you can Find the density of irregular solid [stone]
 Measure and record the mass (m) of the stone using the beam balance.
 Next fill a measuring cylinder with some water. By looking horizontally at the base of the meniscus and keeping the
cylinder vertical, record the reading of the cylinder scale as V1.
 Place the stone gently into the cylinder so that no water is splashed out of the cylinder and the stone is completely
submerged in the water. Record the reading of the cylinder as V2.
Density of the stone = 𝑽
𝒎
𝟏 ‒ 𝑽𝟐
(ii) 2 Ten identical steel balls, each of mass 27 g, are immersed in a measuring cylinder containing20 cm3 of water. The reading of the
water level rises to 50 cm3. What is the density of the steel?
A 0.90 g/cm3
B 8.1 g/cm3
C 9.0 g/cm3
D 13.5 g/cm3
3.1 FORCE ,MOTION AND FRICTION
1 (a) What is the SI unit for force.
 Newton
(b) What is the a Newton?
 A Newton is defined as the force that gives a body of 1kg an acceleration of 1m/s 2.
(c) In what situations is there no resultant force needed?
 a car moving in a straight line at a steady speed
(d) What is resultant force ?
 The combined effect of all the forces on an object is called the resultant force
(e) Write down the fornula connecting force, mass and acceleration.
 Resultant Force = mass × the acceleration.
2 (a) A girl and a boy are pulling in opposite directions on a rope. The forces acting on the rope are shown in the diagram.
200N
150N
rope
Which single force has the same effect as the two forces shown?
 50 N acting towards the girl
(b) Forces of 30 N and 50 N act on the same body, but in different directions. Which value could not be the resultant force on the
body?
A 10 N
B 30 N
C 50 N
D 70 N
 Any value outside the range (50-30) to (50 +30) or (20 ≤ resultant force ≤ 80). correct option is A 10 N
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(c)When two forces are combined, the size of the resultant depends on the angle between the two forces. Which of the following
cannot be the magnitude of the resultant when forces of magnitude 3N and 4N are combined?
A 1N
B 3N
C 7N
D 8N

Any value outside the range (4-3) to (4 +3) or (1 ≤ resultant force ≤ 7). correct option is D 8 N
(d) what property of a body cannot be changed if a force is applied to it?
 its mass
WORKED EXAMPLES
1 (i) Find the force needed to give a body of 15 kg an acceleration of 3 m / s 2

Force = Mass × the acceleration. F = 15 × 3= 45 N
(ii)A force of 5N was applied to the block of mass 2kg. What was the acceleration of block?

Force = Mass × the acceleration. Making a the subject, a = 5/2 = 2.5m / s 2
2 During part of the race, the athlete is travelling at a constant speed. What can be said about the forward and backward forces acting
on the athlete at this time?
 equal and opposite/ balanced
(d) The mass of the athlete is 60 kg. His initial forward acceleration is 2 m/s2. Calculate the force required to give this acceleration.
 force = mass x acceleration;= 120 N
MULTIPLE CHOICES
1 Which property of an object cannot be changed by a force?
A its mass
B its motion
C its shape
D its size
2 Below are four statements about the effects of forces on objects. Three of the statements are correct. Which statement is incorrect?
A A force can change the length of an object
.B A force can change the mass of an object.
C A force can change the shape of an object.
D A force can change the speed of an object.
FORCE OF FRICTION
1 What is friction
 This is the force that opposes motion. Force of friction is always opposite to the direction of force.
2 Name two factors on which the frictional force depend on
 the type of surfaces and how hard the surfaces are pressed together
3 The wheel of a moving car is driven by the engine. The car is accelerating in the direction shown.
A
Driving force
C
D
B
In which direction does the frictional force act on the wheel?
 C (Force of friction is always opposite to the direction of force.)
4 Write down at least four advantages of friction
 It is used to stop vehicles at brakes, It helps nails hold things in position, It prevents objects from sliding down a slope.
 it enables animals or humans to walk or craw without sliding. Without friction between our feet and the ground walking
would be difficult.
 Friction provides the force to accelerate, stop or change the direction of the car. Ice and water on the road reduce this
friction, and make is easier to skid
2 Write down at two disadvantages of friction Disadvantages
 Friction can be a nuisance, because it changes kinetic energy into heat which is usually wasted and reduces the
efficiency machines.
 Friction also tends to wear away at the surfaces, causing damage.
3 write down the ways in which friction may be reduced
 We can reduce friction by oiling ("lubricating") the surfaces. This means that the surfaces no longer rub directly on
each other, but slide past on a layer of oil. It's now much easier to move them
 Hovercraft ride on a cushion of air, which reduces the drag dramatically compared to the drag on the hull of a ship.
Thus hovercraft can easily achieve much higher speeds than ships.
 using "ball bearings" or "roller bearings", where balls or rollers allow the surface to move easily without actually
touching each other using special materials, for example, Teflon, which have a very low coefficient of friction and thus
slide easily (Teflon is used in "non-stick" frying pans for this reason.
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MULTIPLE CHOICE
A wooden block is pushed across a table at constant speed. Which statement is correct?
A The frictional force increases as the block moves at constant speed.
B The frictional force is greater than the pushing force
C. The frictional force is equal and opposite to the pushing force.
D The frictional force is less than the pushing force
BALANCED FORCES
1 (a) Show all the forces that may affect a moving car
Air resistance
Forward driving
force
Force of friction
Weight
(b) How does weight affect an object moving horizontally?
 When an object is moving horizontally, motion is not affected by weight because weight is a vertical force. It will however
affect the motion of a body moving uphill or vertically.
(c) Explain why the car slows down when it climbs a hill, even though the driving force is unchanged.
 gravity or the weight of the car opposes the motion or acting, in some way, backwards
2(a) write down the four forces that govern the flight of an aeroplane.

drag, thrust, weight and lift
(b) show these forces on a moving aeroplane.
Lift
thrust
Drag
Weight
(c) When does the aeroplane move at a constant height
 when the lift and weight are equal because the upward and downward forces are balanced.
3 A car of mass 800 kg was moving along a level road. The forward force produced by its engine was 2500 N and the total resistive
force was 900N.
900 N
2500 N
(a) Find the resultant force
 Resultant force = 2500 N – 900 N = 1600 N
(b) Find the acceleration of the car.
 F = m a. ∴ a = F / m =1600 / 800= 2 m / s 2.
(c) A model car is at rest. A force is applied and it starts to move. State one other way in which a force can affect an object.
 change direction of motion of object; change shape of object; change the speed of an object / speed up / slow down;
4 Fig. 13.1 shows a force of 20 N being used to move a block of mass 5.0 kg across a horizontal frictionless surface. block, mass 5.0
kg direction of motion force 20 N
direction of motion
force of 20 N
Block Mass 5.0 kg
Fig. 13.1
(a) Calculate the acceleration of the block.
 F = m a. ∴ a = F / m = 20 / 5= 4 m / s 2.
(b) On Earth, the gravitational field strength g = 10 N / kg. Calculate the weight of the block.
 Weight = mass × gravitational field strength g= 5×10 = 50N
(c) The force of 20 N moves the block a distance of 40 cm. Calculate the work done by the force.
 work = force x distance or work = weight x distance ; 20N x 0.4 m = 8 J
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3 Fig. 1.1 shows a car travelling at 30 m/s on a level road. At this speed the car has to overcome a total force of 600 N opposing the
car.
Opposing force 600N
Driving force
Fig. 1.1
(a) (i) Calculate the distance travelled by the car in 10 s.
 distance = 300m, [ distance = speed× time]
(ii) State the value of the driving force produced by the engine for a steady speed of 30 m/s.
 600N
(b) Explain why the car slows down when it climbs a hill, even though the driving force is unchanged.
 gravity or weight oppose the motion or weight is acting, in some way, backwards
(c) While on the level road and travelling at 30 m/s, the driving force becomes zero. The mass of the car is 800 kg. Calculate the
deceleration of the car.
 Deceleration = 0.75 m/s2
27 Three horizontal forces act on a car that is moving along a straight, level road.
Air resistance
900 N
Driving force
1200
Friction X
What forces of friction X would result in the car moving at constant speed?
 X + 900 =1200 , x =1200 – 900 =300N
MULTIPLE CHOICES
1 The diagram shows a bird in flight. In which direction does the weight of the bird act?
D
B
A
C
2 A block of mass 6 kg is pulled across a rough surface by a 54 N force, against a friction force F.
F
6 kg
54N
The acceleration of the block is 6 m / s2.What is the value of F ?
A9N
B 18 N
C 36 N
D 54 N
2 Mr Musondela pulls a wooden block of mass 6 kg across the horizontal surface by applying a force of 18 N. Given that the force of
friction acting on the wood is 12 N, calculate the constant acceleration.
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3. The car has mass 800kg. At one point in its motion, when the combined forces of air resistance and friction acting backwards are
400N, its acceleration is 1.4m/s2. Calculate the forward driving force required to accelerate the car.
4 When a block of wood of mass 2 kg is pushed along the horizontal flat surface of a bench, the friction force measured is 4N.When
the block is pushed along the same bench with a force of 10 N, it moves with a constant
A speed of 3m/ s.
B speed of 5m/ s.
C acceleration of 3m/ s2.
D acceleration of 5m/ s2.
6. DEFORMATION
1 Fig. 1.1 shows apparatus that may be used to compare the strengths of two springs of the same size, but made from different
materials.
A
spring
B
masses
ruler
C
(a) (i) Explain how the masses produce a force to stretch the spring.
 force of gravity acts on masses/weight of masses
(ii) Explain why this force, like all forces, is a vector quantity.
 vector has direction/force has direction
(iii) A ruler is used to take the reading of the extensions. Which point A, B and C would give the correct measurement?
 B (place the eye level in line with the mark being read)
(b) Fig. 1.2 shows the graphs obtained when the two springs are stretched.
20
Spring A
Load/N
Spring B
10
10
30
21
40
extension/mm
(i) State which spring is more difficult to extend. Quote values from the graphs to support your answer.
0
 spring A
(ii) On the graph of spring 2, mark a point P at the limit of proportionality. Explain your choice of point P.
 P marked at extension 25 mm to 28 mm
(iii) Use the graphs to find the difference in the extensions of the two springs when a force of 15 N is applied to each one.
 each graph read at 15 N, approx. 25 mm, 19 mm so difference =25- 19 = 6 mm (+/- 1) mm
2 (a) A student adds loads to an elastic cord. He measures the length of the cord for each load. He then plots a graph from the results.
Which length is plotted on the vertical axis?
 (measured length – original length)
(b) A student carries out an experiment to plot the extension-load graph for a spring. The diagrams show the apparatus at the start of
the experiment and with a load added.
With load added
Start
M
N
What is the extension caused by the load?
 extension caused = M – N
(c) Objects with different masses are hung on a 10 cm spring. The diagram shows how much the spring stretches.
10 cm
20 cm
100 g
30 cm
M
100 g → 10
M g → 20
∴ M=
𝟏𝟎𝟎 × 𝟐𝟎
𝟏𝟎
= 200g
The extension of the spring is directly proportional to the mass hung on it. What is the mass of object M?
 200g
[100g gives (20-10 ) and unknown mass gives (30-10)]
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Fig. 8.2 shows a spring. The spring is 10 cm long. A metal nut of 50 g mass is hung on the spring and the length of the spring
increases to 13 cm.
50 g → 3
M g → 12
13cm
10cm
∴ M=
𝟓𝟎 × 𝟏𝟐
𝟑
= 200g
(a) The 50 g mass is replaced by an object of unknown mass. The new length of the spring is 22 cm. Calculate the value of the
unknown mass
 200g
[50g gives (13-10 ) and unknown mass (M)gives (22-10)]
(b) Calculate the length of the spring if 4 more identical nuts are hung on the spring
 25 cm [1nut gives (13-10 ) and 4nuts give (13+ unknown length of the spring )
3 A large spring is repeatedly stretched by an athlete to increase the strength of his arms. Fig. 3.1 is a table showing the force required
to stretch the spring.
Extension of spring/m
0.096
0.192
0.288
0.384
force exerted to produce extension/N
250
500
750
1000
Fig. 3.1
(a) (i) State Hooke’s law.
 Extension proportional to load provided the limit of proportionality is not exceeded
(ii) Use the results in Fig. 3.1 to show that the spring obeys Hooke’s law.
 Any relevant arithmetic to show direct proportion (or straight line graph with values)
(b) Another athlete using a different spring exerts an average force of 400 N to enable her to extend the spring by 0.21m
(i) Calculate the work done by this athlete in extending the spring once.
 work done = Work done = force x distance = 400 x 0.210 = 84.0 J
(ii) She is able to extend the spring by this amount and to release it 24 times in 60 s. Calculate the power used by this athlete while
doing this exercise.
 power = (total) work/time or (24 x) 84/60 = 33.6 W
4 The extension / load graph for a spring is shown. The unloaded length of the spring is 15.0 cm.
Answers (a)
3
1.6
66
66
66
66
6
2
Extension /cm
1
0
4
Load/N
2
5
0
1
3
4
Fig. 4.1
(a) Use Fig. 4.1 to find the extension of the spring for a load of 4.0 N.
 Extension = 1.6 cm
(b) Calculate the length of the spring when the load is 4.0 N
Length = original length + extension =15.0 cm + 1.6 cm = 16.6 cm
(c) When an object of unknown weight is hung on the spring, the length of the spring is 16.4 cm. What is the weight of the object?
 3.5 N (16.4 -15.0 = 1.4 cm, fron the graph 1.4 cm corresponds with 3.5 N)
(d) State the apparatus that may be used in the experiment to measure
(i) the length of the spring
 ruler
(ii) the load on the spring
 spring balance or a Newton meter
DO IT YOUR SELF1
Describe an experiment that would enable you to produce the extension-load graph for a spring. Draw a diagram to show the
arrangement of the apparatus, and state the readings you would take. Explain what is meant by the limit of proportionality.
[7]
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5 In an experiment, forces are applied to a spring as shown in Fig. 2.1a. The results of this experiment are shown in Fig. 2.1b.
16
R
Q
12
ruler
spring
Force (N)
8.0
P
4.0
weight
2.0
4.0
Extension/mm
6.0
(a) What is the name given to the point marked Q on Fig. 2.1b?
 limit of proportionality (elastic limit)
(b) For the part OP of the graph, the spring obeys Hooke’s Law. State what this means.
 force is proportional to extension or in terms of doubling
(c) The spring is stretched until the force and extension are shown by the point R on the graph. Compare how the spring stretches, as
shown by the part of the graph OQ, with that shown by QR.
 (up to Q extension proportional to force applied) Q to R extension/unit force more however expressed
(d) The part OP of the graph shows the spring stretching according to the expression F = kx. Use values from the graph to calculate
the value of k.
 k = force/extension or 8/2 = 4.0 N/mm
MULTIPLE CHOICES
1 A student adds different loads to the end of a spring. She finds the extension in each case and plots a graph of extension against load.
Which is the correct graph?
Extension
Extension
Extension
Extension
load
load
load
D
C
B
A
load
2 The diagrams show a spring having a length of 9 cm when loaded with a 100 g mass and the extension-mass graph for the spring.
9 cm
Extension/mm
3
2
1
100g
100
200
300
Mass (g)
What is the length of the spring after the 100 g mass has been removed?
A 7 cm
B 8 cm
C 9 cm
D 10 cm
3 Which object will experience an elastic deformation?
A a car damaged in a collision
B a football being kicked
C a log hit by an axe
D a target hit by an arrow
30 The table shows the length of a wire as the load on it is increased.
load / N
0
10
20
30
length / cm
50.0 52.1
54.1
56.3
Which subtraction should be made to find the extension caused by the 20 N load?
A 54.1 cm – 0 cm
B 54.1 cm – 50.0 cm
16
C 54.1 cm – 52.1 cm
D 56.3 cm – 54.1 cm
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3 A student investigated the stretching of a spring by hanging various weights from it and measuring the corresponding extensions.
The results are shown below.
weight / N
0
1
2
3
4
5
extension / mm
0
21
40
51
82
103
(a) On Fig. 3.1, plot the points from these results. Do not draw a line through the points yet.
120
100
80
Extension /mm
60
40
20
1
2
3
4
5
6
Weight (N)
 At least 5 points correctly plotted
b) The student appears to have made an error in recording one of the results. Which result is this?
 3 N one, however identified OR 3rd value OR 4th value
(c) Ignoring the incorrect result, draw the best straight line through the remaining points.

good straight line through origin and candidate’s remaining points
(d) State and explain whether this spring is obeying Hooke’s Law.
 straight line / constant gradient so it does obey Hooke’s Law because force is directly proportion to extension
(e) Describe how the graph might be shaped if the student continued to add several moreweights to the spring.
 graph becomes non-linear / curves / bends
(f) The student estimates that if he hangs a 45 N load on the spring, the extension will be 920 mm. Explain why this estimate may be
unrealistic.
 will have exceeded / reached proportional / elastic limit OR permanently deformed OR staightened OR will have broken
OR no longer elastic
4 A spring is suspended from a stand, and a mass of 500 g is hung from its free end. The extension is 1.6 cm. The mass is then
removed, and a second identical spring is hung from the free end of the first. State and explain what will be the total extension when
the mass of 500 g is hung from the lower spring.
 Total extension = 1.6 × 2 = 3.2 cm (each spring had a load of 500 g, so each was extended by 1.6 cm.)
EXERCISE
1 (i) Describe an experiment to investigate how the extension of a spring varies with the load applied. Sketch a graph of the result you
would expect.
[6]
(ii) Explain how you would know if the limit of proportionality had been exceeded.
[1]
2 A student uses a spring to find the weight of an object X. She measures the length of the spring with no load and then with a 6.0 N
weight attached. She then measures the length of the spring with object X attached. The spring returns to its unstretched length when
each load is removed. The pupil’s results are shown below. Length of spring with no load = 21.0 cm, length of spring with the 6.0 N
weight = 33.0 cm and length of spring with the object X = 28.0 cm .Calculate the weight of object X
[3]
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5. TURNING EFFECT OF FORCES (MOMENTS)
1 (a)What is the correct Physics term for the turning effect of a force?
 moment
(b) State the two factors on which the turning effect of a force depends.
 (magnitude of) force and distance (from fulcrum
(c) State the formula used to calculate the moment of a force and state its s I unit
 Moment = perpendicular force × distance and The S.I of moment is a Newton metre (Nm).
2 Fig. 16.1 shows a spanner being used to tighten a nut.
40 N
0.15 m
Fig. 16.1
(a) Calculate the moment of the 40 N force about the centre of the nut.
Moment = perpendicular force × distance = 40 N × 0.15 m = 6 Nm
(b) Explain the reason why if a nut and bolt are difficult to undo, it may be easier to turn the nut by using a longer spanner.
 This is because the longer spanner gives a larger turning moment.
 because a longer spanner requires a smaller force to produce the same moment as a shorter spanner
(b)The nut and the bolt are so rust that the nut cannot be turned. What effect may be produced on the nut and bolt by the application of
this force?
 Both the nut and bolt turn and eventually the bolt may break
4 A door requires a minimum moment of 32.5 N m in order to open it.
hinges
Distance
handle
What is the minimum distance of the handle from the hinges, if the door is to be pulled open with a force at the handle of 50 N?
A 0.33 m B 0.65 m C 0.77 m D 1.54 m
(c)The child has to apply a perpendicular force of 2.5 N at 3 cm from the propeller pivot in order to turn the propeller. Calculate the
moment of this force about the pivot
 Moment = perpendicular force × distance = 2.5N × 3 cm =7.5 N cm or 0.075 Nm
(d)A uniform rule with a weight of 6N is pivoted at the 35cm mark and a weight W is placed at the 5cm mark to balance the rule.
0
5 cm
35 cm
50 cm
100 cm
W
6N
(i) Calculate the moment of the weight of the ruler about the 35cm mark.
 Moment = force  distance. Therefore = 6  0.15 =0.9Nm
(50-35 =15)
(ii) Calculate the moment of the weight W about the 25cm mark.
 When balanced: Clockwise moment anti-clockwise moment. Hence moment =0.9Nm
(iii) Calculate the value of W
 clockwise moment = anti clockwise moment. 30W = 90.hence W = 3N
7. A uniform half –metre rule AB is balanced horizontally across a knife –edge placed 20cm from A. Amass of 22.5g is hang from end
A as shown below
A
20 cm
B
22.5 kg
(a) What is the mass of the ruler?
 (25 -20) M = 22.5 x 20. ∴ 5 M = 450 ⟹ M = 90 g
(b) What force is exerted on the rule by the knife-edge? Take g to be 10 N/kg.
 Weight = mass x acceleration due to gravity. Hence w =0.9 x 10 = 9 N
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5. The diagram below shows a metre ruler pivoted off centre but kept in equilibrium by a suspended mass of 240g
0
5 cm
20 cm
50 cm
100 cm
240 kg
The centre of mass of the ruler is at the 50 cm mark. Calculate the mass of the ruler
 Let M be the required mass
(50 – 20) M = (20 - 5) (240) ∴ 30 M = 3600 ⟹ M= 120 g
2 A metal cube has a mass of 0.05 kg. On Earth, the gravitational field strength g = 10 N/ kg.
(a) Calculate the weight of the metal cube.
 0.5N
(b) Fig. 5.1 shows a stone and the metal cube on a balanced lever
0.70m
stone
0.35m
metal cube
Fulcrum
Fig. 5.1
The distance of the stone from the fulcrum (pivot) is 0.70 m. The distance of the metal cube from the fulcrum is 0.35 m.
(i) State the principle of moments.
 clockwise and anticlockwise moments are equal on a balanced
 1.5 cm × F = 11cm × 30N
(ii) Calculate the weight of the stone.
 0.25N
⟹ 1.5 F = 330
3 Fig. 3.1 shows a bottle opener.
𝟏.𝟓
𝟑𝟑𝟎
⟹ 𝟏.𝟓𝑭 = 𝟏.𝟓
11cm
pivot P
1.5 cm
F
⟹ Force, F = 220
bottle opener
Bottle cap
effort = 30N
Fig. 3.1
A force of 30 N is applied at a distance of 11 cm from the pivot P. The force F on the bottle cap is 1.5 cm from the pivot P. Calculate
the force F on the edge of the cap.
 moment calculated as 330 (Ncm) .equating moments 1.5F = 330 , force F = 220 (N)
3 A uniform metre rule is balanced by a 4 N weight as shown in the diagram. What is the weight W of the metre rule?
10cm
50cm
0
100cm
Pivot
4N
W
 A 1 N [ ( 50-10)× weight = 4 × 10]
4 A load is to be moved using a wheelbarrow. The total mass of the load and wheelbarrow is 60 kg. The gravitational field strength is
10 N / kg.
F
load
weight
Pivot
600N
70 cm
(a) To which type of simple machines does a wheelbarrow belong?

NB : 60 kg = 600N on earth
50 cm
levers of the 2nd order [class ii levers
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
(b) What is the size of force F needed just to lift the loaded wheelbarrow?

(70+ 50) cm × F = 70 × 600N
⟹ 120 F = 42000
F = 350 N [(70+50) cm × F = 70 cm × 600N]
(ii) What is the M.A. of this machine?

𝟏𝟐𝟎
𝟒𝟐𝟎𝟎𝟎
𝑭 = 𝟏𝟐𝟎
𝟏𝟐𝟎
⟹
𝐥𝐨𝐚𝐝
𝟔𝟎𝟎
⟹ Force, F = 350
Mechanical advantage M.A = 𝐞𝐟𝐟𝐨𝐫𝐭 =𝟑𝟓𝟎 = 1.71 (2dp)
2 The diagram shows a wheelbarrow and its load, which have a total weight of 150 N. This is supported by a vertical force F at the
ends of the handles.

F
(0.75+0.75) m × F = 0.75 × 150N
⟹ 1.5 F = 112.5
150 N
⟹
𝟏.𝟓
𝟏𝟏𝟐.𝟓
𝑭 = 𝟏.𝟓
𝟏.𝟓
⟹ Force, F = 75
0.75 m
What is the value of F?
 F = 75 [(0.75+0.75) m × F = 0.75 × 150N]
0.75 m
b) Fig. 6.1 shows a method of measuring the mass of a uniform loaded ruler. The ruler is pivoted at the 18 cm mark.
0 cm
8cm
18cm
80 cm
50 g
Fig. 6.1
(i) The ruler is uniform. What does this tell you about the position of its centre of mass?
 centre of mass at its centre ;
(ii) The total length of the ruler is 80 cm. The 50 g mass is hung from the 8 cm mark on the ruler. Calculate the mass of the ruler.
attempted use of moments ; use of 10 cm and 22 cm ; 50 x 10 = m x 22 ; mass = 22.7(3)
2 Fig. 2.1 shows apparatus for investigating moments of forces. The uniform metre rule shown in Fig. 2.1 is in equilibrium.
Spring
Balance
0
10
20
30
40
Horizontally balanced ruler
50
6.0N
Weight
60
70
80
90 100
Horizontal pivot
(a) Write down two conditions for the metre rule to be in equilibrium.

condition 1 : upwards force = downwards force or no resultant force

condition 2 : opposing moments equal or A.C.M. = C. M.
(b) Show that the value of the reading on the spring balance is 8.0 N. [2]

30 x spring balance reading = 40 x 6.0, spring balance reading= 240/30 = 8.0 N
(c) The weight of the uniform metre rule is 1.5 N. Calculate the force exerted by the pivot on the metre rule.

magnitude of force = 0.5 N downwards
6 Fig. 6.1 shows a balanced uniform metre rule. The knife edge is at the 50.0 cm mark and the 0.1 N weight is at the 20.0 cm mark.
0.0 cm
mark
50.0 cm
mark
20.0 cm
mark
0.1N
0.2N
Knife edge
Fig. 6.1
(a) Calculate the anticlockwise moment of the 0.1 N weight about the knife edge.
 Moment = force × distance = 0.1N × 30 cm = 3.0 Ncm or = 0.1N × 30 cm/100 = 0.03 Nm
(b) Calculate the distance of the 0.2 N weight from the knife edge.
 clockwise moment = anticlockwise moment ∴ 0.2 × d = 0.1N × 30 cm ⟹ distance = 3.0 Ncm /0.2N = 15 cm
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4 (a) A metal bar, PQ, has a weight of 5 N and is pivoted at P.It is held horizontal by a Newton meter acting at Q.
Newton meter
20N
20N
P
5N
Q
What is the reading on the newton meter?
Newton meter reading × 40 = 20 × 5. ∴ Newton meter reading = 100/ 40 = 2.5 N
(b) Fig. 1.2 shows the same spring being used in a device for weighing objects. The spring pulls down on one side of a wooden strip
with a force of 8.0 N. The wooden strip is horizontal.
12.0 cm
6.0 cm
stretched
spring
pivot
object
Wooden strip
8.0N
(i) Calculate the anticlockwise moment, about the pivot, of the force in the spring.
 Moment = perpendicular force × distance = 8 N× 6 cm = 48Ncm or 0.4 Nm
(ii) State the clockwise moment of the weight of the object. The weight of the wooden strip can be ignored.
 The wooden strip is horizontal so clockwise moment = anticlockwise moment = 48Ncm
(iii) Calculate the weight of the object.
 so clockwise moment = anticlockwise moment , 12 × weight = 48 ∴ weight = 48/12 =4N
(c) Identical apparatus is used to weigh the same object on the Moon. The wooden strip is horizontal but the pivot is not in the same
position as it is on Earth. Explain why.
 gravity and/or the weight of the object on the Moon is less than that on Earth
5 (a) A uniform metre rule is pivoted at its centre, which is also the position of its centre of mass. Three loads, 2.0 N, F and 3.0 N are
positioned on the rule at the 20 cm, 30 cm and 90 cm marks respectively, as shown in Fig. 3.1.
20 cm 30 cm
50 cm
90 cm 100 cm
0 cm
2.0 N
F
3.0 N
(i) Calculate the moment of the 3.0 N load about the pivot.
 moment = perpendicular force × distance = (90 - 50 ) cm × 3.0 N = 120 Ncm OR 1.2 Nm
(ii) Calculate the moment of the 2.0 N load about the pivot.
 60 Ncm OR 0.6 Nm
(iii) The force F maintains the metre rule in equilibrium on the pivot. Calculate the value of F.
 CW moments = ACW moments so 60 + 20F = 120 OR 0.6 + 0.2F = 1.2 ∴ F = 3.0 N OR 3 N
(b) The weight of the metre rule is 1.2 N and can be considered to act at the 50 cm mark. All the weights in (a) are removed. The pivot
is positioned under the 30 cm mark and the 2.0 N load is placed on the rule as shown in Fig. 3.2.
30 cm
50 cm
Fig. 3.2
2.0 N
1.2 N
The position of the 2.0 N load is adjusted until the metre rule is again in equilibrium. Determine the position of the 2.0 N load
 1.2 × 20 = 2.0 × d = 12 cm OR 1.2 × 0.2 = 2.0 × d = 0.12m
6 A driver’s foot presses with a steady force of 20 N on a pedal in a car as shown.
F
5 cm
20 N
piston
Pivot
40 cm
(a) What is the force F pulling on the piston?
 160 N [ 5 cm × F = 20 N × 40 cm so F = 800Ncm / 5 cm = 160 N]
(b) Force is a vector quantity. State which two of the following are also vector quantities?
acceleration, distance, mass, speed, velocity
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5.2 CENTRE OF MASS
1 A hole is drilled in a square tile. The diagram shows the tile hanging freely on a nail. Where is the centre of gravity of the tile?
Nail
D
Tile
C
B
A
3 The diagram shows a step in an experiment to determine the position of the centre of mass of a thin card. A pencil line is drawn
between P and Q.
Card
Thin string
P
R
Q
Q
bob
What is the next step in this experiment?
A Find the mid-point of PQ. B Hang the card from point R. C Measure the mass of the card. D Measure the thickness of the card.
2 A flat metal plate that may be hung from a nail so that it can rotate about any of four holes. What is the smallest number of holes
from which the flat metal plate should be hung in order to find its centre of gravity?
5.3 STABILITY
1 Explain what is meant by an object being in equilibrium?
 Equilibrium is the exact balance of equal and opposite forces
1 The front views of two cars are shown in Fig. 5.1, to the same scale.
Family car
racing car
Suggest which car has the greater stability, and give two reasons.
 Racing car has the greater stability because it has wider base and low centre of mass
2(a) A heavy lorry is difficult to start and stop .Why is a heavy lorry both difficult to start and stop
 It has a high inertia
(b) What is the best position for its centre of mass and why is it placed there?
 as low as possible for greater stability
(c) What affects the stability of an object?
 its base area and the location of its centre of mass
(d) Passengers are not allowed to stand on the upper deck of double-decker buses. Why is this?
 They would cause the bus to become unstable as standing rises the centre of mass.
(e) State two conditions which can help to prevent the lorry toppling over when tilted
 Centre of gravity should not be too high and its base should not be too narrow
3 Fig. 3.1 and Fig. 3.2 show a thick piece of wood with one corner on a table. Fig. 3.3 shows the same piece of wood balanced on the
table. B is the centre of mass.
Fig. 3.1
Fig. 3.2
Fig. 3.3
(i) Explain why in Fig. 3.1 the piece of wood falls to the right and in Fig. 3.2 it falls to the left.
 Wood falls to the right or left because of moment or turning effect of weight . or anticlockwise and clockwise moment or
weight to right and left of corner
(ii) Explain why the piece of wood in Fig. 3.3 does not fall over. [1]
 moments balance/cancel or weight inside base
(iii) Suggest how the thickness of the wood in Fig. 3.3 affects its stability. [1]
 thicker wood is more stable and the thinner wood is less stable
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(a) Describe in detail how you would experimentally determine the position of the centre of mass of the sheet of metal.
 Suspend the lamina from a pin and let it freely swing. When it stops swinging. A plumb line is then suspended from the
support (hole). Draw a vertical line below the point of suspension along a plumb line. having drawn the line, hang the
lamina at another point and again draw another vertical line using the plumb line. The centre of mass is the point of
intersection of the lines drawn. To limit errors, repeat the experiment at a third point.
P
Sheet of metal
Centre of mass
R
Q
Q
Plumb line
bob
2 Fig. 4.1 shows a torch standing on a table. M is the position of the centre of mass of the torch.
B
M
A
Fig. 4.1
(i) What is meant by the term centre of mass?
 A point at which the whole mass may be considered to act
(ii) Explain why the torch is more stable if it stands on end A rather than on end B. Use diagrams in your answer.
B
M
A
M
A

B
(ii) diagram showing torch on end B and position of M; centre of mass is closer to base; base has larger area;
Do it alone
Describe in detail how you would experimentally determine the position of the centre of mass of the sheet of metal.
7. ENERGY, WORK AND POWER
1 What is work?
 Work is the product of force and distance in the direction of force.
2 Write down the formula for calculating work and state its S.I unit.
 Work = force  displacement. The S.I unit of work is a joule (j).
 The work done lifting an object up will be: Work Done = Force × Distance .The force that moves when we lift an object is
its weight and the distance is the Height. In that case, Work e = Energy Transferred = Weight × Height.
3 What is a joule?
 A joule is work done when a force of one Newton moves a body through a displacement of one metre.
4 When is work done?
 Work is done when the force applied moves a body. A woman carrying a bucket of water on her head along a level
ground does no work.
5. A boy of mass 30kg runs up a fight of stairs to a floor which is at s height of 5.5m.taking the height of 1kg to be 10N, calculate the
work done by the boy against gravity.
 Work Done = Energy Transferred = Weight × Height = 30 × 10 ×5.5 = 1650 J.
6. A bricklayer lifts 12 bricks each weighing 20N through a vertical of 1.2m in 30s and place them at rest on a wall. Calculate the
work done.
 Work Done = Energy Transferred = total weight × height = 12 × 20 × 1.2 = 288J.
MULTIPLE CHOICE
1 The diagram shows a girl lifting a box of weight 100 N from a low shelf to a high shelf.
1.5m
100N
0.5m
How much work is done by the girl?
A 50 J
B 100 J
C 150 J
D 200 J
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7 Fig. 7.1 shows two boys applying a force to an object.
Alex holds a box in a steady position.
John pushes the box along the bench.
Fig. 7.1.
(a) State and explain which boy is doing useful work on the box.
 John because work is done when force moves the box
(b) The box has a mass of 1.8 kg. Calculate the weight of the box. (g = 10 N/kg)
 Weight = mg, w =1.8 x 10 = 18 N
(c) Alex drops the box. Describe the motion of the box as it falls to the ground
2 What quantity is not measured in joules
A power
B work
C gravitational potential
D heat
POWER
1 Define power
 Power is the rate of doing work (or converting energy).
2 Write down the formula for calculating power and state its S.I unit.
 Power = work/ time t or P = f d / t or p = mgh / t, where p =power (w), w = work (j), t = time (s).The SI unit of power is a
Watt.
3 what is a watt
 A Watt is the rate of doing work of 1 joule per second.
1(a) A mass of 5000g is pushed a distance of 400 cm across a floor in 8 seconds by applying the force of 80 N
 Power = work/ time = 4 × 80/ 8 = 40 w
(b) A bricklayer lifts 12 bricks each weighing 20N through a vertical of 1.2m in 30s and place them at rest on a wall. Calculate the
power developed by the bricklayer.
 P = mgh / t = (12 × 20 × 1.2) / 30 = 288 / 30 = 9.6 W
(c)The water from the reservoir falls to the turbine through a height of 0.75 m. The mass of water that flows through the turbine in
5.0 s is 2.0 kg. The gravitational field strength is 10 N / kg. Calculate the power delivered to the turbine.

Power =
𝒎𝒈𝒉
𝒕
=
𝟐 × 𝟏𝟎 × 𝟎.𝟕𝟓
𝟓
= 3.0 w
(Power = work done per second = mgh ÷ t )
2(b) A boy, who weighs 50 N, runs up a flight of stairs 6.5m high in 7 seconds. How much power does he develop?
(c) The mass of the athlete is 60 kg. The athlete does 3000 J of work in 5 seconds. Calculate the power developed by the athlete
 power = work/time; = 600 W;
5 When a 300 N force is applied to a box weighing 600 N, the box moves 3.0 m horizontally in 20 s.
300N
600 N
3.0 m
What is the average power?
A 45 W
B 90 W
C 900 W
D 1800 W
4 A crane lifts a concrete block, whose weight is 60 000 N, to a height of 20 m in 30 s. What power is achieved by the crane?
A 100W
B 4000
W C 40 000W
D 90 000 W
(b) Mr. Hamweemba lifted a brick of mass 20000g through the height of 2.5 m in 25 seconds. Calculate the rate of doing work.
 P = mgh / t = (20 kg × 2 0 × 2.5) / 25 = 20W
ENERGY
1What is energy
 Energy is capacity to do work. Its SI unit is a joule. Examples of energy are potential, kinetic, nuclear, solar etc
2 A body moving with a speed of 30 m / s has a kinetic energy of 1800 J. What is its mass?
 A 120 kg
B 60 kg
C 4 kg 
D 2 kg
3 Which property of a body is affected by the gravitational field strength?
 weight
4 State the principle of conservation of energy
 energy cannot be created energy cannot be destroyed or created (i.e. the other one as well) or (merely) transformed
or total energy in an isolated system is constant
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5 A child pushes a toy car along a level floor and then lets it go. As the car slows down, what is the main energy change?
 from kinetic to heat
6 A ball of mass 100 g is balanced on the edge of a ledge 10 m above the ground. It rolls off the ledge and falls. How much
gravitational potential energy is lost when the ball falls to the ground? (Gravitational field strength = 10N/ kg.)

10j (convert mass into kilograms and then use the formula P e = m g h)
7 A block of mass 2 kg slides from rest through a distance of 20 m down a frictionless slope, as shown
20 m
16 m
What is the kinetic energy of the block at the bottom of the slope? [The acceleration of free fall is 10 m/s2.]

Kinetic energy at the bottom = potential energy at the highest point = mgh =2 kg×10 m/s2 × 16 m= 320 J
HYDROELECTRIC POWER
Dam
lake
Turbine
house
to river
1 (a) In a hydroelectric power station, water flows from a high reservoir to turn turbines to generate electricity. Which energy
conversions take place?
 gravitational potential → kinetic → electrical
(b) Describe the energy conversions involved in a hydro-electric power station.
 Potential energy of the water (at a high position) is converted into kinetic energy as the water falls. This then produces
kinetic energy of the turbines that produce electrical energy by rotating the generator.
(c) State one harmful effect that the hydroelectric power station may have on the environment.
 flooding or fish unable to pass or turbines kill fish or destroy habitats or less land or uses up large space or fells trees or
unsightly/destroys scenery or lake/river silt up or more rain/evaporation or no need to transport coal or renewable or
rapid response to power demand or less heat produced/more efficient
 The dams are very expensive to build.
However, many dams are also used for flood control or irrigation, so building costs can be shared.
 Building a large dam will flood a very large area upstream, causing problems for animals that used to live there.
 Finding a suitable site can be difficult - the impact on residents and the environment may be unacceptable.
 Water quality and quantity downstream can be affected, which can have an impact on plant life
(c) Some power stations burn coal to produce the same electrical power output. State one advantage of the hydroelectric power station.
 no costs for water/energy supply or less pollution

Once the dam is built, the energy is virtually free.

No waste or pollution produced.

Much more reliable than wind, solar or wave power.

Water can be stored above the dam ready to cope with peaks in demand.

hydro-electric power stations can increase to full power very quickly, unlike other power stations

Electricity can be generated constantly
WIND ENERGY
1Wind farms are areas of land containing many wind turbines. Four thousand wind turbines can produce the same power as one coalfired power station.
(a) (i) State the main energy change that takes place in a wind turbine.
 kinetic/movement → electrical
(b) Wind power is said to be a renewable source of energy. Explain what the term renewable means.
 can be used again/replaced/will not run out/ replicated if qualified e.g. wood;
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2 One disadvantage of wind turbines is the high cost of manufacture and installation. State one other disadvantage of using wind
turbines to generate electricity.
 noise / eyesore / only effective over a certain range of wind speeds;
 Suitable areas for wind farms are often near the coast, where land is expensive.
 Some people feel that covering the landscape with these towers is unsightly.
 Can kill birds - migrating flocks tend to like strong winds.
However, this is rare, and we tend not to build wind farms on migratory routes anyway.
 Can affect television reception if you live nearby.
3State one other disadvantage of using wind turbines to generate electricity.
 Wind is free, wind farms need no fuel.
 Produces no waste or greenhouse gases.
 The land beneath can usually still be used for farming.
 A good method of supplying energy to remote areas.
SOLAR ENEYGY
1 Electricity can be generated by many methods, including the use of solar energy.
(i) State one non-renewable fuel that is used to generate electricity.
 coal/oil/gas/nuclear;
(ii) Name the process that produces energy within the Sun.
 nuclear fusion
2 A bank of solar cells is used to supply electricity to a house. What form of energy is converted into electrical energy by the solar
cells?
 light energy
(c) Nuclear fission is used to produce electricity in nuclear power stations. The Sun’s energy is produced by nuclear fusion. Explain
the difference between nuclear fission and nuclear fusion.
 fission means nuclei break (into smaller pieces); fusion means nuclei join together
CHEMICAL
1 Name the chemical element present in all fossil fuels.
 carbon / hydrogen
2 Which type of energy is converted to thermal energy when atoms combine?
 Chemical
3 Name one other fossil fuel, apart from coal, that can be burned in a power station.
 oil / gas ( not crude oil )
PAPER TWO QUESTIONS
1 As the bicycle moves along the road at 4 m/s, the brakes are suddenly applied. The bicycle comes to a stop after 10 m. The average
frictional force stopping the bicycle is 250 N. As the bicycle slows down, work is done
.
(i) Calculate the work done as the bicycle slows down from 4 m/s to a stop.
 work done = force × distance ; = 250 × 10 = 2500 J
(ii) A cyclist travels down a hill from rest at point X without pedalling. The cyclist applies his brakes and the cycle stops at point Y.
x
y
Which energy changes have taken place between X and Y?
 potential → kinetic → heat
(iii) Brakes are used to stop the bicycle moves. What is most of the kinetic energy converted into?
 heat energy
(iv) Which energy changes take place when a pedalling cyclist uses a generator (dynamo) to light his bicycle lamp?
 Chemical → kinetic → electrical → light
(v) When a bicycle accelerates up hill, state what happens to the potential and kinetic energy
 Potential energy increases as height is gained and kinetic energy increases as it accelerates ( speed increases)
MULTIPLE CHOICE
A rock of mass 20 kg is travelling in space at a speed of 6m/ s. What is its kinetic energy?
A 60 J
B 120 J
C 360 J
D 720 J
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2 A man exerts a horizontal force of 200 N on a 60 kg box to move it across a room. The box moves with a constant speed of 0.5m/ s.
(a) How big is the force of friction on the box? Explain your answer.
 200 N , The resultant force must be zero if there is no acceleration. Therefore the friction force must be the same size
as the force exerted by the man
(b) Calculate the kinetic energy of the box.
 7.5 J , ( use the formula ke= 1/2mv 2)
(c) (i) How much work does the man do on the box every second?
 100 J ( work = Force multiplied by distance )
(ii) How much power is the man producing to move the box?
 100 W ( The work done per second gives the power. )
(iii) What is the main form of energy into which his work is converted?
 Heat is the main form of energy into which the work is converted.
3 Some pupils tested a motorcyclist’s crash helmet. They dropped a steel ball of mass 5.0 kg so that it hit the helmet with a speed of
7.8m/ s.
(a) Calculate the kinetic energy of the steel ball just before it hit the helmet.
 ke= 1/2mv 2 = ½ ×5× 7.8×7.8 =152 J
(b) State the form of energy that the ball had before it was dropped.
 Gravitational potential energy
(c) The ball bounces off the helmet at a lower speed. State two energy transfers that occur when the ball hits the helmet.
 Sound and heat
4 Fig. 4.1 shows a weightlifter.
bar
(a) The total weight lifted by the weightlifter is 1600 newtons, which he lifts through 2 metres in 0.5 seconds.
Wheel
(i) Calculate the work done on the bar and weights.
 work = force x distance or work = weight x distance ; 1600 x 2 = 3200 J
(ii) Calculate the power developed by the weightlifter while lifting the bar and weights.
 power = work ÷ time or power = energy ÷ time ;= 3200 ÷ 0.5 = 6400 W
(b) (i) What form of energy has been gained by the bar and weights as a result of lifting
4 Fig. 4.1
 (gravitational) potential (energy) ;
(ii) The weightlifter suddenly drops the bar and weights, and the energy gained in (i) is now transferred to another form. Name this
form of energy.
 kinetic ;
5 An object of mass 0.4 kg is set into motion so that it slides up a friction-free slope with an initial speed of 6m/ s. the object comes to
rest after 2 s. calculate, for this object,
(a) its initial kinetic energy,
 7.2 j; ( use the formula ½ mv 2 )
(b) the potential energy it gains in moving up the slope,
 7.2 j (principle of conservation of energy is used here. The formula mgh is of no value in this question )
(c) its acceleration.
 3, (or –3), m/s2.
6 A spider climbs vertically upwards along a thread.
(i) The spider weighs 0.02N.Calculate the work done when it climbs 21 cm up the thread.

work = force x distance; w = 0.02 x 21/100 = 0.0042 J; ( 21 cm =21/100 m )
(ii) Calculate the power generated by the spider as it climbs up the thread. It climbs 21 cm in 7 seconds.

power = work/time; power = 0.0042/7 = 0.0006 W;
(iii) The mass of the spider is 2g. It begins to move up the thread with an acceleration of 2cm /s 2. Calculate the resultant force
causing this acceleration.

force = mass x acceleration; = 0.002 x 0.02;= 0.00004 N;
5 (a) A hotel has a lift (elevator). It moves through a vertical height of 3 m between each. Floor
(i) A passenger travels in the lift. The passenger has a mass of 80 kg and weighs 800 N. The mass of the empty lift is 1200 kg.
Calculate the total weight of the passenger and lift.
 weight of empty lift = 12000 N ;
(ii) Calculate the work done when the lift and passenger move up three floors, from Floor 1 to Floor 4.
State the formula that you use and show your working
 W = F × D ; (work done =) height × (total) weight]= 12 800 x 9 =115 200 J
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MULTIPLE CHOICE
1Which quantity is calculated by multiplying the magnitude of a force by the distance moved in the direction of the force?
A acceleration
B power
C pressure
D work
8 A student wishes to work out how much power she uses to lift her body when climbing a flight of stairs. Her body mass is 60 kg and
the vertical height of the stairs is 3.0 m. She takes 12 s to walk up the stairs.
(a) A student uses chemical energy to run up some stairs .She stops at the top of the stairs. What has the chemical energy been
converted to when she is at the top of the stairs?
 potential energy and heat energy
(b) Calculate
(i) the work done in raising her body mass as she climbs the stairs,
 work done = force x dist = 600 x 3 = 1800 J
(ii) the output power she develops when raising her body mass.
 power = work/time or 1800/12 = 150 W
(c) At the top of the stairs she has gravitational potential energy. Describe the energy transformations taking place as she walks back
down the stairsand stops at the bottom.
 P.E. decreases/transformed the decrease becomes heat
(d) Another girl of weight 500 N runs up a flight of stairs in 10 seconds. The vertical height of the stairs is 5 m. What is the average
power developed by the girl?
A 50 W
B 100 W
C 250 W
D 1000 W
9 A car of total mass 800 kg has a constant acceleration. It starts from rest. After 8.0 s it has a speed of 20 m / s. Over the 8 second
interval, calculate for this car, showing your working,
(a) the acceleration,
𝐯‒𝐮
 2.5 m/s2 (acceleration = ( 𝐭 )
(b) the resultant force,
 2000 N ( force = mass × acceleration)
(c) the distance travelled,
𝟏
 80 m (distance = average speed × time = 𝟐(𝟐𝟎 + 𝟎 ) × 𝟖 or drew a sketch to find the area under the velocity-time graph
(d) the kinetic energy gained.

𝟏
160000 J. ( KE = 𝟐 mv 2 )
THANKS
WAIT FOR VOLUME II
Four solids have the same mass.
Which solid has the greatest density?
The solid with the smallest volume
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