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St. Joseph’s Anglo-Chinese School
HKDSE CS – Physics Notes
H eat
M echanics
W aves
E lectricity
St. Joseph’s Anglo-Chinese School
0
Chapter 1 Temperature and Thermometers
1.1 Temperature
Lower fixed point (ice point)
Temperature of pure melting ice at normal atmospheric
pressure
Upper fixed point (steam point)
Temperature of steam over pure boiling water at normal
atmospheric pressure
Celsius temperature scale
Divide 100 equal divisions between the lower and upper
fixed point. Each division is 1oC
1.2 Kinetic theory
(a) All matter is made up of very tiny particles.
(b) These particles are constantly in motion.
(c) Forces between particles:
(i) When particles are close together, they attract/repel each other strongly.
(ii) When particles are far apart, they hardly attract/repel each other.
Solid
Liquid
Gas
Particle arrangement
(1) close together
(2) arranged in
regular pattern
(1) close together
(2) not in fixed
position
(1) very far apart
Particle motion
Vibrate in fixed
positions
Can move freely from
one place to another
Move at random at
very high speed
(~500 ms-1)
1.3 Heat and Internal energy
* Heat is the energy transferred from one body to another due to a temperature difference.
heat
warmer
cooler
internal energy = kinetic energy (K.E.) + potential energy (P.E.) of all particles
K.E (depends on temperature)
T ↑ ⇔ K .E. ↑
P.E. (depends on the state of matter)
solid
liquid
gas
(particles vibrate more rapidly at higher
temperature)
P.E. of particles increases
* Temperature is a measure of average kinetic energy of the particles.
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Chapter 2 Heat Capacity and Specific Heat Capacity
Heat capacity (C)
Definition
Specific heat capacity (c)
Energy required to raise the
Energy required to raise the temperature
o
Formula
temperature of a substance by 1 C
of a 1 kg substance by 1oC
E = C∆T
E = mc∆T
o -1
Unit
J kg-1 oC-1
J C
C = mc
Example 1
After absorbing 1000 J of energy, the temperature of a substance increases by 4oC. If the mass of
the substance is 2 kg, find
(a) the heat capacity, and
(b) the specific heat capacity of the substance.
Solution
(a) For heat capacity, E = C∆T
1000 = C(4)
C = 250 J oC-1
(b) For specific heat capacity, E = mc∆T
1000 = (2)(c)(4)
c = 125 J kg-1 oC-1
Power = rate of energy transferred
P=
E
t
or
E = Pt
Example 2
2 kg of water is heated by a heater of power 1500 W. Find the time it takes for the temperature of
water to increase from 20oC to 98oC.
Given: the specific heat capacity of the water = 4200 J kg-1 oC-1
Solution
∵ E = Pt
∴ Pt = mc∆T
and
E = mc∆T
1500t = (2)(4200)(98 – 20)
t = 436.8 s
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Example 3
The figure below shows the variation of temperature of an object with time.
Temperature / oC
60
20
0
5
time / minute
If the power of the heater is 800 W, find the heat capacity of the object.
Solution
∵ E = Pt
∴ Pt = C∆Τ
and
E = C∆T
(800)(5 × 60) = C(60 – 20)
t = 6000 J oC-1
[5 minutes = 5 × 60 s]
Example 4
A piece of 0.1 kg hot copper is put into a pond of water of 2 kg. If the initial temperatures of the
copper and water are 500oC and 20oC respectively, find the final temperature of the copper.
Given: the specific heat capacity of the water = 4200 J kg-1 oC-1
the specific heat capacity of the copper = 370 J kg-1 oC-1
Solution
Copper:
500 oC → T
Water:
20 oC → T
Assume no heat loss to the surroundings
Energy lost by hot object (copper) = Energy gained by cold object (water)
(0.1)(370)(500 – T) = (2)(4200)(T – 20)
18500 – 37 T = 8400 T – 16800
T = 22.1oC
Remark
Since the specific heat capacity of water is large, water can absorb a large amount of energy with
only a small temperature rise. (∆T = 22.1 – 20 = 2.1 oC)
Uses of high specific heat capacity of water
(1) coolant
(2) body temperature regulation
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Example 5
In an experiment to find the specific heat capacity of aluminium, the following results are obtained.
Mass of aluminium block = 1 kg
Initial joulemeter reading = 98 300 J
Final joulemeter reading = 104 900J
Initial temperature of aluminium block = 28.5 oC
Final temperature of aluminium block = 35.0 oC
(a) Find the specific heat capacity of aluminium.
(b) The standard value of the specific heat capacity of aluminium is 900 J kg-1 oC-1, find the
percentage error of the experiment.
(c) How to improve the accurate of the experiment.
Solution
(a) By E = mc∆T
104900 – 98300 = (1)(c)(35 – 28.5)
c = 1020 J kg-1 oC-1
(b) Percentage error =
1015.4 − 900
× 100% = 12.8%
900
(c) (1) Wrap the aluminium block with cotton to reduce heat loss to the surroundings.
(2) Add a few drop of oil to the holes in the aluminium block to ensure a good thermal
contact between the heater, thermometer and the block.
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Chapter 3 Change of State
3.1 Latent heat
Heating
No change in state
E = mc∆T
Change in state
E = ml
* E = ml, where
l: specific latent heat
* Unit: J kg-1
Gas
Release latent heat (P.E.↓)
∆T = 0 (K.E. remains unchanged)
Absorb latent heat (P.E.↑)
∆T = 0 (K.E. remains unchanged)
Liquid
Release latent heat (P.E.↓)
∆T = 0 (K.E. remains unchanged)
Absorb latent heat (P.E.↑)
∆T = 0 (K.E. remains unchanged)
Solid
Example 1
How much energy is required to melt 2 kg of ice at 0oC and to raise the temperature to 30oC?
Given that:
the latent heat of fusion of ice = 3.34 × 105J kg-1, and
the specific heat capacity of water = 4200 J kg-1oC-1.
Solution
E = mc∆T
E = ml
Ice (0oC)
Water (0oC)
Water (30oC)
= ml + mc∆T
= ( 2)(3.34 × 10 5 ) + ( 2)( 4200)(30 − 0)
Energy required
= 9.2 × 10 5 J
Example 2
A coffee machine injects 0.03 kg of steam at 100oC into a cup of cold coffee of mass 0.17 kg at
20oC. Find the final temperature of the coffee.
Given that:
the latent heat of vaporization of ice = 2.26 × 106 J kg-1, and
the specific heat capacity of coffee = 5800 J kg-1oC-1.
Solution
100oC
Steam (0.03 kg):
E = ml
20 C
coffee
E = mc∆T
water
o
Coffee (0.17 kg):
100oC
water
T (final temperature)
E = ml
coffee
T (final temperature)
Energy lost by steam = Energy gained by coffee
6
(0.03)(2.26 × 10 ) + (0.03)(4200)(100 – T) = (0.17)(5800)(T – 20)
T = 90.0oC
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3.2 Measure the specific latent heat of fusion of ice
Procedures
(1) Set up the apparatus as shown.
(2) Fill both funnels with roughly equal
amounts of crushed melting ice.
(3) Record the initial joulemeter reading (E1).
(4) Switch on the heater for a period of time.
(5) Find the mass of ice (m) melted by the
heater and record the final joulemeter
reading (E2).
(6) Calculate the specific latent heat of fusion
of ice (lf) by E2 – E1 = mlf.
Precautions
Ice should be crushed to increase the contact area with the heater.
Melting ice is used so that it is at 0 °C.
Before switching on the heater, pack the crushed ice in the two funnels so that the drip rates
are steady and about the same.
After switching off the heater, do not remove the beakers; wait until the drip rates have
become steady and about the same.
-
A small piece of wire gauze or steel wool at the neck of the funnels can prevent the crushed ice
from dropping into the beakers directly.
Example 3
The following results are obtained from the above experiment:
Mass of water in experimental cup = 0.050 kg
Mass of water in control cup = 0.014 kg
Initial joulemeter reading = 15 000 J
Final joulemeter reading = 29 200 J
(a) Find the specific latent heat of fusion of ice.
(b) Calculate the experiment percentage error. Account for any difference of the value obtained
from the standard value, 3.34 × 105 J kg-1.
Solution
(a) By E = ml
29200 − 15000 = (0.050 − 0.014)l f
l f = 3.94 × 10 5 J kg-1
3.9444 × 10 5 − 3.34 × 10 5
× 100% = 18.1%
3.34 × 10 5
Possible sources of error include:
(1) Water dripping down the two funnels at different rates.
(b) Percentage error =
(2) Energy is lost to the surroundings.
[Since energy is lost to the surroundings, less amount of ice is melted by the heater.
E
By l f = , the measured lf is greater than the standard value.]
m
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3.3 Measure the specific latent heat of vaporization of water
Procedures
(1) Set up the apparatus as shown.
(2) Take the reading from the electronic balance (m1)
heater
and the kilowatt-hour meter (E1) after the
water boils.
(3) Boil the water for a few minutes and turn off the
heater.
(4) Wait until the water becomes steady and take the
kilowatt-hour
meter
electronic
balance
final reading of the balance (m2) and the kilowatt-hour meter (E2).
(5) Calculate the specific latent heat of vaporization of water by E 2 − E1 = (m1 − m 2 )l v from the
results.
Precaution
Do not switch on the heater unless the heating part is totally immersed in water.
Possible sources of error
(1) Steam condensing on the heater and drips back into the cup ⇒ Larger experimental lv
(2) Energy is lost to the surroundings ⇒ Larger experimental lv
(3) Some water ‘bubbles’ out of the cup ⇒ Smaller experimental lv
Example 4
The following results are obtained from the above experiment:
Mass of water boiled away = 0.10 kg
Energy supplied to the heater = 246 000 J
(a) Find the specific latent heat of fusion of ice.
(b) Account for any difference of the value obtained from the standard value, 2.26 × 106 J kg-1.
Solution
(a) By E = ml
246000 = 0.1l v
l f = 2.46 × 10 6 J kg-1
(b) Possible sources of error include:
(1) Steam condensing on the heater and drips back into the cup.
(2) Energy is lost to the surroundings.
[These will cause a smaller amount of water boiled away.
E
By l v = , the measured lv is greater than the standard value.]
m
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3.4 Evaporation and Boiling
(a) Common:
absorb latent heat (E = ml) to change from liquid state to gas state
(b) Difference:
Evaporation
Boiling
Occurs at any temperature
Occurs at a definite temperature – the boiling point
Occurs at surface
Occurs with liquid
No bubbles formed
Bubbles appear
(c) Evaporation and particle motion
(i) Some of the particles in a liquid have greater K.E. while the
other have smaller K.E.
(ii) Some of the particles at the liquid surface may gain enough
KE to escape into the space above the liquid and become
particles of vapour.
(iii) As fast-moving particles fly away, the average KE of the
remaining particles is lowered; so the liquid becomes colder.
(d) Ways to increase the rate of evaporation
(i) increase temperature of the liquid
(ii) increase the surface area of the liquid
(iii) decrease the humidity of air
(iv) increase the movement of air
Example 5
A person is wearing a wet shirt. There is 0.1 kg of water on the shirt in total.
(a) How much energy is required to evaporate the water?
(b) Where is the energy required taken from?
Specific latent heat of vaporization of water = 2.26 × 106 J kg–1
Solution
(a) Energy required evaporating 0.1 kg of water
E = ml
= (0.1)(2.26 × 106)
= 2.26 × 105 J
(b) The energy required is taken from the surroundings (or the skin of the person).
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Chapter 4 Heat Transfer
Conduction
Convection
Radiation
Description
Hot water
rises
Cold water
falls
– Particles at the hot end
vibrate faster.
– The fast vibrating
– Fluid (gas or liquid)
expands, rises and is
particles bump into the
replaced by the
The solar energy is
slower neighbouring
surrounding cooler fluid.
transferred to the earth by
particles and make
– Such movement of fluid
them vibrate more
is called convection.
radiation
rapidly.
Medium
required
Solid, liquid or gas
Fluid (liquid or gas)
No medium is required
Rate of heat
transfer
Conductor – faster
Dark colour – good
Insulator – slower
absorber and radiator
Daily
examples
A cotton jacket traps air
(Vacuum )
(air is a poor conductor)
– Air conditioners
are
–
installed high on the wall
– Heating element
is fixed
are black in colour
–
near the bottom of an
electric kettle
Most transformers
Car engines are
painted black.
–
Fuel storage tanks
are painted silvery.
Example 1 Solar heater
glass
(a) What is the function of the glass?
(b) The temperature of the 0.3 kg water increases to
70oC after 30 minutes. If the initial temperature of
water is 30oC, find the power of the solar heater.
water
bowl
(c) Give a suggestion to improve the design of the solar
container
heater in order to obtain a greater temperature rise of water. Explain your answer briefly.
Solution
(a) The glass traps warm air to reduce heat loss by convection.
(b) ∵ E = Pt
and
E = mc∆T
∴ Pt = mc∆T
P(30 × 60) = (0.3)(4200)(70 – 30)
P = 28 W
(c) (1) Paint the bowl black in colour since dark colour object is a good radiation absorber.
(2) Stick slivery paper (aluminium paper) onto the inner wall of the container so that more
radiation is reflected to the bowl.
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Vacuum flask
Plastic or cork stopper
reduces heat loss by
conduction and convection.
cap
outer case
Silvery glass or stainless steel
walls reduce heat loss by
radiation.
A vacuum between the double
walls reduces heat loss by
conduction and convection.
insulated support
Vacuum cooker
Inner pot
Outer pot
Inner and outer lids reduce
heat loss by conduction and
Silvery steel walls reduce
convection
heat loss by radiation
Outer vacuum insulated pot (the vacuum
Inner pot
A thin film of air between inner
pot and outer pot. Reduce heat
between the double steel walls of the
outer pot reduces heat loss by
conduction and convection)
loss by conduction and
Food is kept at a high
convection.
temperature for a long time and
is cooked without a fire.
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Chapter 1 – 2
(a) Distance and Displacement
Motion
Speed and Velocity
B
Example 1
A person takes 4 s to walk from A to B along the following path.
Find the distance, displacement, speed and velocity
5m
A
40o
Solution
Distance = length of the path
Displacement = the length of straight line AB
= 10 m (N50oE)
10π
=
= 15.7m
2
Speed
=
Distance
time
=
15.7
= 3.93 ms-1
4
Velocity =
=
Scalar: Distance, Speed
Displacement
time
10
= 2.5 ms-1 (N50oE)
4
Vector: Displacement, velocity
(b) Acceleration
accelerati on =
change in velocity
time
a=
v−u
t
Unit: ms-2
Example 2
It takes 6 s for a racing car to attain a speed of 100 km h-1 from rest. Find the average acceleration
of the car.
Solution
Initial velocity u = 0 ms-1 (start from rest)
Final velocity v = 100 km h-1 =
100 -1
ms = 27.78 ms-1
3 .6
Time t = 6 s
Acceleration a =
=
v−u
t
27.78 − 0
= 4.63 ms-2
6
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(c) Motion graphs (s – t, v – t, a – t graph)
(i)
Uniform velocity
s
v
0
t
a
0
t
Displacement changes at
a constant rate
0
Velocity remains
constant.
t
No acceleration.
a = 0 ms-2
(ii) Uniform acceleration
s
v
0
t
0
t
Displacement:
Velocity increase at a
constant rate.
Beginning: increases a little.
End: increases significantly.
(iii) Slope and area under graph
s
0
slope = velocity
a
s
t
0
t
Acceleration remains
constant.
v
0
t
slope = acceleration
0
t
Area under v-t graph
= displacement
(d) Equations of uniformly accelerated motion
(1) v = u + at
u = initial velocity
(2) v 2 − u 2 = 2as
a = acceleration
(3)
1
s = ut + at 2
2
(4)
s=
u+v
t
2
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v = final velocity
s = displacement
t = time
12
(e) Motion of free fall
In the absence of air resistance, all objects fall with an acceleration of g (9.81 ms-2)
Example 3
A stone is thrown vertically upwards from the ground with a speed of 20 m s-1.
(a) How high does the stone rise?
(b) How long does it take the stone to reach the ground?
(c) Draw a v-t graph for the stone unit it returns to the ground.
Solution
(a) Consider the motion of the stone when it is thrown upward until it reaches the highest point.
u = 20 ms-1
v = 0 ms-1 (highest point)
a = -9.81 ms-2
v = 0 ms
-1
(uniform deceleration)
(momentarily at rest at the highest point)
20 ms-1
s=?
By v 2 − u 2 = 2as
0 2 − 20 2 = 2( −9.81) s
∴
s = 20.4 m
greatest height is 20.4 m
(b)
Method 1
Method 2
Consider the motion of the stone when it is
Consider the motion of the stone when it is
thrown upward until it reaches the highest
thrown upward until it returns to the ground.
point.
u = 20 ms-1
u = 20 ms-1
a = -9.81 ms-2 (uniform deceleration)
a = -9.81 ms-2 (uniform deceleration)
v = 0 ms-1
(momentarily at rest)
t=?
s=0
v = 0 ms
By v = u + at
0 = 20 + ( −9.81)t
20 ms-1
t = 2.04 s
Time required for stone to reach the ground
= 2 × 2.04
(return to ground)
t=?
1
By s = ut + at 2
2
20 ms-1
s=0m
1
( −9.81)t 2
2
t ( 4.905t − 20) = 0
0 = 20t +
t = 0 (rejected) or 4.905t − 20 = 0
∴ t = 4.08 s
= 4.08 s
(c)
-1
v / ms-1
20
0
2.04
4.08
t/s
-20
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Remark: Take upward direction as positive
Stage 1: upward motion
Stage 2: at highest point
decelerate uniformly
Momentarily at rest
a = -9.81 ms-2
a = -9.81 ms-2
Stage 3: downward motion
accelerate downward
uniformly
a = -9.81 ms-2
Misconception
(1) The acceleration of the object at the highest point a = 0 ms-2. (2) We are certain that at the highest point, v = 0 ms-1.
(3) If at the highest point a = 0 ms-2 and v = 0 ms-1, the object will remain at rest forever and will
not fall down.
Example 4
v / ms-1
The v-t graph of a car is shown.
50
(a) Describe the motion of the car from t = 0 s to t = 80 s.
0
(b) Find the deceleration of the car from t = 40 s to t = 70 s.
-25
(c) Find the total displacement of the car.
40
70 80
t/s
Solution
(a) From t = 0 s to 40 s, the car moves with a uniform velocity.
From t = 40 s to 70 s, the car decelerates uniformly and becomes momentarily at rest at t = 70 s.
From t = 70 s to 80 s, the car accelerates uniformly in opposite direction.
(b)
by a =
Method 1
v−u
t
From t = 40 s to 70 s
-1
Method 2
a = slope of v-t graph
From t = 40 s to 70 s
-1
u = 50 ms , v = 0 ms , t = 30 s
0 − 50
= −1.67 ms-2
30
∴ Deceleration is 1.67 ms-2
= slope of v-t graph
a
a=
50 − 0
= −1.67 ms-2
70 − 40
Deceleration is 1.67 ms-2
= −
∴
(c) Total displacement of the car
= area under v – t graph
=
( 40 + 70)(50) 1
− (10)( 25)
2
2
= 2625 m
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Chapter 3 Force and Motion
3.1 Newton’s law of motion
Newton’s 1st law
Every object remains in a state of rest or uniform motion (i.e. constant
velocity) unless acted on by a net force, or an unbalanced force.
nd
Newton’s 2 law
The acceleration of an object is directly proportional to, and in the same
direction as, the net force acting on it, and inversely proportional to the mass
of the object. [ a ∝ F , a ∝
Newton’s 3rd law
1
⇒ F = ma]
m
Action and reaction pair
(1) Equal in magnitude
(2) Opposite in direction
(3) Acting on different bodies.
3.2 Forces in daily life
(a) Weight (W)
(b) Normal reaction (R)
(c) Friction (f)
(d) Tension (T)
3.3 Mass m and Weight W
W = mg
Mass m
Weight W
Scalar
Vector
A measure of inertia
Gravitational force acting on the object
Remain unchanged
Vary in different planets (g is different)
3.4 Addition and resolution of forces
Addition: parallelogram of forces method
Resolution: Fx and Fy are components of F
y
FR (resultant force)
F2
F
Fy
θ
F1
Fx = F cos θ
Fx
x
F y = F sin θ
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3.5 Typical exam questions
(a) Inclined plane
Resolve W into two components:
W sin θ
θ
θ
along the plane
W sin θ or mg sin θ
perpendicular to the plane
W cos θ or mg cos θ
W cos θ
W
Example 1
A block is placed on the inclined plane and remains at rest.
(a) (i)
Draw a free diagram of the object.
(ii) If the mass of the block is 5 kg, find the friction
30ο
acting on the block
(iii) Find the normal reaction acting on the block
(b) Now, a force of 120 N is applied on the block and pulls the block upwards along the plane. If
the friction between the block and the plane becomes 60 N, find the acceleration of the block.
Solution
(a) (i)
Normal reaction (R)
Friction (f)
Weight (W = mg)
(ii)
Do not accept W, mg, f, R
R
W sin θ
f = W sin θ = mg sin θ
= (5)(9.81) sin 30°
f
= 24.5 N
θ
W cos θ
W
R = W cos θ = mg cos θ
= (5)(9.81) cos 30°
(iii)
= 42.5 N
(b) Consider the direction along the plane
120 N
120 – W sin θ – f = (5)(a)
W sin θ
θ
By F = ma
f = 60 N
St. Joseph’s Anglo-Chinese School
120 – (5)(9.81)sin θ – 60 = 5a
a = 7.10 ms-2
16
(b) Apparent weight in a lift
Reading of weighing machine = normal reaction = R
When the lift
R
a
at rest or moves with uniform velocity
R=W
Feel normal weight
accelerate upward
R>W
Feel heavier
accelerate downward
R<W
Feel lighter
Weight is always constant (W = mg)
Weighing machine
R – W = ma (Always take upward direction as +ve)
W
Example 2
(a) A boy of mass 55 kg stands on a weighing machine in a lift. Find the reading on the weighing
machine when
(i)
the lift moves upwards with an acceleration of 1 ms-2, and
(ii) the lift moves downward with a deceleration of 1.5 ms-2.
(b) When the lift moves upward and the reading on the weighing machine is 480 N, find the
acceleration of the lift.
Solution
(a) (i) The lift moves upward and accelerate, a = + 1 ms-2
R − W = ma
R − (55)(9.81) = (55)(1)
R = 595 N
[The boy feels heavier]
(ii) The lift moves downward and decelerate, a = + 1.5 ms-2
R − W = ma
R − (55)(9.81) = (55)(1.5)
R = 622 N
(b)
[The boy feels heavier]
R − W = ma
480 − (55)(9.81) = (55)a
a = −1.08 ms-2
∴ The deceleration of the lift is 1.08 ms-2
Remark
(1) When applying R − W = ma
The lift moves
(2)
upward with acceleration
a = +ve
upward with deceleration
a = – ve
downward with acceleration
a = – ve
downward with deceleration
a = +ve
When the cable of the lift break,
R
* W = mg
* a = -9.81 ms-2
* R=0
W
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3.6 The turning effect of a force
(a) Moment = Fs
Unit: Nm
5 sin 30o
5N
2m
30o
2m
pivot
5N
5 cos 30o
pivot
Moment = Fs = (5 sin 30o)(2) = 5 Ns
Moment = Fs = (5)(2) = 10 Ns
(b) Equilibrium of a rigid body
No net force: all forces acting on the body are balanced
No net torque: clockwise moment = anticlockwise moment
Example 3
The mass of the meter rule is 0.5 kg.
W1
School bag
4 kg
R
Block
10 kg
0.2 m
0.3 m
W2
W3
d
(a) Find the distance between the school bag and the pivot to maintain balance.
(b) Find the reaction from the pivot.
Solution
(a) Clockwise moment = anticlockwise moment
(W2)(0.3) + (W3)(d) = (W1)(0.2)
(0.5 × 9.81)(0.3) + (4 × 9.81)(d) = (10 × 9.81)(0.2)
d = 0.463 m
(b) R = W1 + W 2 + W3
= (10)(9.81) + (0.5)(9.81) + (4)(9.81)
= 142 N
Remark
(1) Moment = Force × Distance (distance is usually measured from the pivot)
(2) Must label all forces clearly on the figure.
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Chapter 4 Work, Energy and Power
4.1 Work (W)
Definition:
W = F// s
Unit: J
Example 1
Calculate the work done by F in each of the following case.
(a)
(b)
F=2N
30o
F=2N
5m
5m
Solution
(a) W = (2)(5) = 10 J
(b) W = (2 cos 30o)(5) = 8.66 J
Consider only the component of F along the
direction of displacement. i.e. 2 cos 30o
4.2 Energy
Mechanic energy:
1
mv 2
2
(2) P.E. = mgh
(1) K.E. =
(3) Elastic potential energy
4.3 Energy change
Conservation of energy: Energy can be changed from one form to another, but it cannot
be created or destroyed.
Example 2
Find the speed of the particle at B. Assume that the track is smooth.
A
2 ms-1
10 m
B
3m
Solution
Total mechanical energy at A = Total mechanical energy at B
P.E.A + K.E.A = P.E.B + K.E.B
2
Common mistake
2
m(9.81)(10) + ½ m(2) = m(9.81)(3) + ½ m(v)
98.1 + 2 = 29.43 + ½v2
v = 11.9 ms-1
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cannot use the formula v 2 − u 2 = 2as
reason: not a uniformly accelerated motion
19
Example 3
30o
A block slides down for 8 m along the surface of a 30o inclined plane. It is given that the mass of
the block is 3 kg and the friction acting on the block is 5 N
(a) Find the P.E. loss of the block.
(b) Find the work down against friction.
(c) If the initial speed of the block is 4 ms-1, find its final speed.
Solution
(a) P.E. loss
= mgh
h = 8 sin 30o
= (3)(9.81)(8 sin 30o)
= 117.72 J ≈ 118 J
(b) work done against friction
W = fs
= (5)(8)
= 40 J
(c) K.E. gained = 117.72 – 40 = 77.72 J
8m
30o
5N
8m
30o
½mv2 –½mu2 = 77.72
½(3)(v2 – 42) = 77.72
v = 8.23 ms-1
Remark: Description of energy change
K.E (77.72 J)
P.E. (117.72 J)
Heat (40 J)
P.E. changes into kinetic energy and heat (internal energy)
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Example 4 High diving
A diver of 60 kg dives off a platform 10 m above the water surface.
(a) If she falls directly downwards, what is her speed when she reaches the water?
(b) If the water provides an average water resistance of 2500 N, how deep can she reach under the
water surface?
Solution
(a)
By conservation of energy,
mgh = ½ mv2
v2 = 2gh
v = 2gh = 2(9.81)(10) = 14.0 ms-1
A → B, the diver loses P.E. and K.E.
Total energy loss = work done against water resistance
(b)
K.E. + P.E. = fs
½ mv2 + mgh = fs
½ (60)(14.0)2 + (60)(9.81)(d) = (2500)(d)
d = 3.08 m
Water resistance
(2500 N)
depth
4.4 Power
P = Fv
Unit: W
Conclusion
Work
Energy
Scalar
Power
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Chapter 5
Momentum
5.1 Momentum
Definition: mv
Unit: kg ms-1
5.2 Force and change in momentum
Net force =
change in momentum
time
F=
mv − mu
t
Example 1
20 ms-1
A tennis ball of mass 0.05 kg flies along a horizontal speed of 20 ms-1
towards a player. The player hits the ball, which leaves the racket at
30 ms
-1
30 ms-1 in the opposite direction. If the time of impact is 0.005s,
calculate the average force acting on the ball.
Solution
u = -20 ms-1
v = 30 ms-1
F=
(towards the left)
(0.05)(30) − (0.05)( −20)
= 500 N
0.005
5.3 Impulse
Definition: Ft
Impulse = Ft = mv – mu = change in momentum
F/N
Area under F – t graph = Ft = mv – mu
0.02
0.04
t/s
Example 2
If the area under the above F – t graph is 0.52 Ns, find
(a) the change in momentum
(b) The average force of impact.
Solution
(a) Change in momentum = area under F – t graph = 0.52 Ns
(b) Average force of impact
F=
mv − mu
0.52
=
= 26 N
t
0.04 − 0.02
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5.4 Conservation of momentum: m1u1 + m 2 u 2 = m1 v1 + m 2 v 2 (for no external force)
Example 3
Find the final velocity for the following completely inelastic collision.
10 ms-1
1 ms-1
2 kg
4 kg
2 kg
Before collision
4 kg
v
After collision
Solution
By conservation of momentum
m1u1 + m 2 u 2 = m1 v1 + m 2 v 2
( 2)(10) + ( 4)( − 1) = ( 2 + 4)v
(opposite direction)
v = 2.67 ms-1
Example 4 2-dimensional impact (resolution of velocities required)
Find the final velocities of the 5 kg and 1 kg objects.
5 kg
2 ms-1
At rest
10
v1
o
30o
1 kg
5 kg
Before collision
1 kg
After collision
v2
Solution
Along direction of x-axis
(5)( 2) = (5)v1 cos 10° + (1)v 2 cos 30°
5v1 cos 10° + v 2 cos 30° = 10 ---- (1)
Along direction of y-axis
0 = (5)v1 sin 10° − (1)v 2 sin 30°
v 2 sin 30°
---- (2)
5 sin 10°
Sub. (2) into (1)
 v sin 30° 
5 2
 cos 10° + v 2 cos 30° = 10
 5 sin 10° 
v1 =
v 2 = 2.70 ms-1
Sub. v 2 = 2.70 ms-1 into (2):
v1 = 0.576 ms-1
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Example 5 Conservation of energy and energy changes
A bullet of 4 g shoots into a piece of 200 g plasticine. Find the initial velocity of the bullet if the
plasticine rises by 3 cm.
u
200 g
3 cm
4g
Solution
Stage (1)
Stage (2)
impact: m1u1 + m 2 u 2 = m1 v1 + m 2 v 2
u
v
energy change:
1
mv 2 = mgh
2
v
M = 200 g
3 cm
m=4g
Before impact
K.E. → P.E.
After impact
Consider stage (2): K.E. → P.E.
1
( M + m )v 2 = ( M + m ) gh
2
v 2 = 2 gh = 2(9.81)(0.03)
[3 cm = 0.03 m]
v = 0.7672 ms-1
Consider stage (1): m1u1 + m 2 u 2 = m1 v1 + m 2 v 2
mu = ( M + m )v
0.004u = (0.2 + 0.004)(0.7672)
∴
[4 g = 0.004 kg]
u = 39.1 ms-1
the initial velocity of the bullet is 39.1 ms-1
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5.5 Different types of collisions
Total momentum conserved?
Elastic collision
Inelastic collision
Total energy conserved?
Example 6
Identify the type of the following collisions
5 ms-1
4 ms-1
(a)
2 kg
1 kg
1 ms-1
1 kg
2 kg
Before collision
4 ms-1
After collision
Solution
Total K.E. before collision =
1
1
(2)(5) 2 + (1)( 4) 2 = 33 J
2
2
1
1
(2)(1) 2 + (1)( 4) 2 = 9 J
2
2
there is K.E. loss during collision (24 J)
Total K.E. after collision =
∵
∴
the collision is inelastic.
(b)
2 ms-1
At rest
2 kg
1 kg
Before collision
2 kg
2 -1
ms
3
1 kg
8 -1
ms
3
After collision
Solution
Total K.E. before collision =
1
1
(2)( 2) 2 + (1)(0) 2 = 4 J
2
2
1
2
1
8
( 2)( ) 2 + (1)( ) 2 = 4 J
2
3
2
3
there is no K.E. loss during collision
Total K.E. after collision =
∵
∴
the collision is elastic.
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Chapter 6
Projectile motion
Main point: Resolve projectile motion into horizontal motion and vertical motion
y
u
u sin θ
H
θ
u cos θ
x
Range sx
Horizontal motion
Uniform motion
Velocity
Displacement
vx = ux (velocity is constant)
s x = ux t
vy = uy – gt
sy = uyt – ½ gt2
Vertical motion
Free fall motion
(a = -g = -9.81 ms-2)
Other useful formulae
1.
u x = u cos θ
u y = u sin θ
2.
Time of flight T =
3.
u 2 sin 2θ
Range s x =
g
4.
Maximum height H =
5.
Equation of trajectory: s y = (tan θ ) s x −
2
v = vx + v y
2u sin θ
g
u2
Maximum Range s x =
when θ = 45o
g
u 2 sin 2 θ
g
g
2
sx
2
2u cos θ
2
Example 1
A small object is thrown horizontally towards a
vertical wall 1.2 m away. If the initial speed of
the ball is 8 ms-1, find the vertical displacement
of the ball when it hits the wall
Solution:
Given
2
wall
1.2 m
Horizontal motion
ux = 8 ms-1
sx = 1.2 m,
Vertical motion
uy = 0 (thrown horizontally)
sy = ?
For horizontal motion:
a = -9.81 ms-2
s x = ux t
1.2 = 8t ⇒ t = 0.15 s
For vertical motion:
s = ut + ½ at2
= 0 + ½ (-9.81)(0.15)2
= -0.110 m
Vertical displacement sy = 0.110 m (downward)
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Example 2
A ball is projected upward with velocity 10 ms-1 at 30o with the horizontal from a cliff which is
100 m above the sea level as shown. Find
10 ms-1
30 o
100 m
(a) the time of flight of the ball,
(b) the horizontal distance moved when the ball reaches the sea,
(c) the direction of the ball when it reaches the sea, and
(d) the speed of the ball when it reaches the sea.
Solution:
(a) Given
Horizontal motion
ux = 10 cos 30o
Vertical motion
uy = 10 sin 30o
sy = -100
a = -9.81 ms-2
t =?
sy = uyt + ½ at2
For vertical motion:
-100 = 10sin 30ot + ½ (-9.81)t2
4.905t2 – 5t – 100 = 0
t = 5.05 s or t = -4.03 s (rejected)
(b) horizontal distance = uxt = (10 cos 30o)(5.05) = 43.8 m
(c)
10 ms-1
30 o
vx
100 m
vy
θ
v
Horizontal motion : vx = ux = 10 cos 30º = 8.66 ms-1
Vertical motion
tan θ =
vy
vx
=
: vy = uy + at = 10 sin 30º + (-9.81)(5.05) = -44.6 ms-1 (downward)
44.6
8.66
θ = 79.0o
The direction makes 79.0o with the horizontal.
(d) speed v = v x + v y = 8.66 2 + 44.6 2 = 45.4 ms-1
2
2
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Chapter 1 Reflection of Light
1.1 Laws of reflection
(a) angle of incidence i = angle of reflection r
(b) incident ray, reflected ray and the normal all lie on the same plane.
Example 1
(a) Find the angle of incidence.
(b) Find the angle between the incident ray and the reflected ray.
Solution:
(a) i = 90o – 25o
Common mistake:
= 65o
i = 25o
(b) r = i = 65o
∴ angle between the incident ray and the reflected ray
=i+r
= 130o
1.2 Images formed by a plane mirror
* (a) Steps
(i) image (image distance = object distance)
(ii) reflected ray
(iii) incident ray
Note:
(1) appropriate solid lines and dotted lines
(2) arrows for light rays to show direction
(b) Properties of image
(i) Virtual
(ii) Laterally inverted
(iii) Same size as the object
(iv) Image distance = object distance
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Example 2
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Chapter 2 Refraction of Light
2.1 Laws of refraction
(a) sin i and sin r are in direction proportion
(b) incident ray, refracted ray and the normal all lie on the same plane.
Note:
Light bends towards the normal when it travels from a less dense
medium into a denser medium.
(e.g. from air into water)
2.2 Snell’s law n1sin θ1 = n2sin θ2
* Example1:
Find the angle of refraction r.
Glass (ng = 1.5)
Water (nw = 1.33)
Solution:
angle of incidence i = 90o – 30o = 60o
(Common mistake: i = 30o)
By n1sin θ1 = n2sin θ2
1.5 sin 60o = 1.33 sin r
sin r =
1.5 sin 60°
1.33
r ≈ 77.6o
2.3 Refractive index
Glass block
**
Slope of graph of sin i against sin r
=
sin i
= n = refractive index of the glass block
sin r
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**Example 2
Do exercise: page 63 Q 13
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2.4 Refractive index and light speed n =
c
v
Example 3
The speed of light in air is 3 × 108 ms-1. If the refractive index of water is 1.33, find the
speed of light in water.
Solution:
n=
c
v
1.33 =
v=
3 × 108
v
3 × 10 8
= 2.26 × 10 8 ms −1
1.33
2.5 Image formed by refraction
* (a) Steps:
(i) image (nearer to water surface)
(ii) refracted ray
(iii) incident ray
Note:
(1) appropriate solid lines and dotted lines
(2) arrows for light rays to show direction
(b) image property – virtual
2.6 Total internal reflection
(a) Critical angle C:
(i) when angle of refraction r = 90o, i = C
n
1
(ii) C = sin −1
or C = sin −1 2 (n1 > n2)
n
n1
water
[n1: denser medium]
(b) Conditions:
* (i)
light enters from a denser medium
(ii) angle of incidence i > critical angle C
**Example 4
Determine whether total internal reflection would occur in each of the following cases:
(a)
(b)
60o
Glass (ng = 1.7)
water (nw = 1.33)
water (na = 1.33)
diamond (ng = 2.42)
Solution:
(a) ∵ Light enter from a denser medium,
1.33
= 51.5° < 60°
1 .7
total internal reflection would occur.
C = sin −1
∴
60o
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(b) ∵
∴
Light enters from a less dense medium
total internal reflection would not occur.
(No need to calculate the critical angle)
33
(c) Natural phenomena (自然現象)
(i)
mirage (海市蜃樓)
(ii) sparkling of diamonds
(d) Applications
(i)
optical fibres (transmit telephone signals)
(ii) cat’s eyes found on highways
(iii) using prisms as mirrors in cameras
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Chapter 3 Lenses
** 3.1
Images formed by a convex lens
The nature of image depends on the distance between the object and the lens.
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** 3.2
Images formed by a concave lens
The images formed by a concave lens are always virtual, erect and diminished.
3.3 Magnification
image size image distance
m=
=
object size object distance
m>1
m=1
m=
hi v
=
ho u
m<1
Magnified Same size
Diminished
*3.4 Real image and Virtual image
Real image
Virtual image
Can be seen by
eyes?
Can be projected
on screens?
Remark
We can see virtual images
formed by plane mirrors
Virtual images cannot be
formed on screens
– Plane mirrors
Examples
Convex lens (in
– Convex lens
projectors)
(magnifying glasses)
– Convex lens
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* Example 1
Some letters on a single-lined paper are observed by using a lens as shown below.
J J J J
J J J J
J J J J
(a)
What kind of lens is used? Explain briefly.
(b)
If the distance between the lens and the paper is 10 cm and the magnification is 0.4, find
the image distance.
(c)
Draw a ray diagram to find the focal length of the lens.
Solution:
(a) Concave lens is used because the image formed is erect and diminished.
(b) Object distance = 10 cm, u = 10 cm
By m =
v
u
0.4 =
v
10
v = 4 cm
(c)
1.2
1.1 5
1.1
1.0 5
F
C
1
2 cm
0.9 5
0
1
2
3
4
5
6
7
8
9
10
Focal length f ≈ 6.8 cm (measure the distance between F and the optical centre)
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Chapter 4 The Electromagnetic Spectrum
4.1 The EM spectrum
Frequency f, Energy Ε
Radio
Waves
Infra-red
Visible
Microwaves
light
Ultra-violet
X – ray
γ – ray
Wavelength λ
4.2 Properties of EM waves
(a) show reflection, refraction, diffraction and interference
(b) are transverse waves
(c) can travel through vacuum
(d) travel at speed of light in air / vacuum (3 × 108 ms-1)
(e) can apply the equation v = fλ
* 4.3 Values commonly used in calculations
f
1kHz = 103 Hz
1MHz = 106 Hz
1GHz = 109 Hz
λ
1mm = 10-3 m
1µm = 10-6 m
1nm = 10-9 m
t
1ms = 10-3 s
1µs = 10-6 s
1ns = 10-9 s
Example 1
The wavelengths of visible light lie between 4 × 10-7 m to 7 × 10-7 m.
Find the frequency range of light.
Solution:
∵ v = fλ
∴ f=v/λ
Minimum frequency =
3 × 10 8
= 4.29 × 1014 Hz
−7
7 × 10
3 × 108
= 7.5 × 1014 Hz
−7
4 × 10
∴ frequency range of visible light is 4.29 × 1014 Hz to 7.5 × 1014 Hz.
Maximum frequency =
* Example 2:
Many auto-focus cameras use infra-red radiation to judge the distance of an object from the
camera. If the time between emitting an infra-red pulse and receiving the pulse is 25 ns, find the
distance of the object from the camera.
Solution:
Distance
= speed × time
= (3 × 108) × (25 × 10-9 ÷ 2)
[25 ns = 25 × 10-9 s]
= 3.75 m
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4.4 Useful information
EM wave
Detector
Application
Radio waves
Radio & TV receiver
Radio communication
Microwaves
Microwave receiver
Microwave oven
Satellite telecommunication
Radar
Infra-red
Thermometer
Auto-focus camera
IR photographic film
IR telescopes
Skin
Visible light
Eyes
-7
λ: red (4 × 10 m)
violet (7 × 10-7 m)
Photographic film
Ultra-violet
Fluorescent paper
Solar cells
Sterilizing (消毒) drinking water
Checking fake banknotes
X – ray
Photographic film
Medical diagnosis
γ – ray
Photographic film
Radiothery
Sterilization (消毒)
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Chapter 5 Nature of Waves
5.1 Transverse and longitudinal waves
Transverse Wave
Waveform
crest
trough
Longitudinal wave
crest
compression
trough
rarefaction
* Definition
Vibrations are at right angle to the
direction of travel of the wave
Vibrations are parallel to the direction of
travel of the wave
Wavelength
Distance between 2 adjacent crests
Distance between 2 adjacent compressions
or troughs
or rarefactions
EM waves, water waves
Sound waves
Examples
5.2 Equations:
(a) v = fλ
1
(b) f =
T
(c) d = vt
frequency f:
no. of waves produced in 1 s.
If f = 5 Hz, 5 waves are generated in 1s.
* period T:
time required to produce 1 wave,
OR time for a particle to make a complete vibration
If T = 0.2 s, it takes 0.2 s to generate 1 wave,
OR it takes 0.2 s for a particle to make 1 complete vibration.
Example 1:
It takes 10s for a slinky to produce 20 waves.
(a) Find the frequency of the wave.
(b) If the wavelength of the wave is 0.2 m, find the speed of the wave.
Solution:
(a) frequency f = no. of waves produced in 1 s
f =
20
= 2 Hz (2 waves are produced in 1 s.)
10
(b) By v = fλ
v = (2)(0.2) = 0.4 ms-1
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5.3 Vibrations of particles
* Example 2:
The following figure shows a transverse wave travelling to the right.
direction of travel of the wave
a
e
bI
d
f
c
(a) Which particles are
(i)
moving upwards,
(ii) moving downwards,
(iii) momentarily at rest?
(b) Which particles are
(i)
vibrating in-phase,
(ii) vibrating anti-phase?
** (c) If the frequency of the wave is 2 Hz, draw the shape of the waves after 0.125 s.
Solution:
(a)
a
e
ib
d
f
c
(i)
moving upards: b, f
(ii) moving downwards: d
(iii) momentarily at rest: a, c, e
(b)
(i)
a,e or b,f are in-phase
(ii) a,c or b,d or c,e or d,f are anti-phase
Remark
In phase
Separation of 2 particles is λ, 2λ, 3λ, …
Anti-phase Separation of 2 particles is 0.5λ, 1.5λ, 2.5λ, …
a
(c)
Steps:
(1) f = 2 Hz ⇒ T = 1/2 = 0.5 s
(2) 0.125s = ¼ T
e
b
d
¼λ
f
(3) In ¼ T, the wave travels by ¼ λ.
c
Note: particles only move up and down while the wave travels forward.
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5.4 Displacement-time (s – t) graph and displacement-distance (s – d) graph
s – t graph
s – d graph
Shows the displacement of ONE particle at Shows the displacement of ALL particle at a
different times
certain time
s
s
Time for one complete
vibration
t
d
Τ
λ
* Example 3:
An s – t graph and s – d graph of a wave are shown below.
s / cm
s / cm
5
5
0
0
0.1
0.2
0.3
0.4
t/s
-5
-5
(a)
P
0.1
0.2
d/m
Q
Find the speed of the wave.
** (b) Draw an s – t graph for particle Q (see the s – d graph.)
Solution:
(a) From s – t graph, we have T = 0.2 s ⇒ f = 1/0.2 = 5 Hz.
From s – d graph, we have l = 0.2 m
∴
wave speed v = fλ = (5)(0.2) = 1 ms-1
(b) s / cm
5
0
0.1
0.2
0.3
0.4
t/s
-5
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Chapter 6 Wave Phenomena
6.1 Ripple tank
*
Crest: bright fringe
Trough: dark fringe
λ: distance between two adjacent bright fringes or dark fringes
6.2 Reflection
i=r
Frequency f
No change
Wavelength λ
No change
Wave speed v
No change
6.3 Refraction
(a) Deep region → Shallow region
Deep region
Shallow region
(Put a glass block)
Frequency f
No change
Wavelength λ
Decrease
Wave speed v
Decrease
Direction
Bend towards normal
Frequency f
No change
Wavelength λ
Increase
Wave speed v
Increase
Direction
Bend away from normal
(b) Shallow region → Deep region
Shallow region
Deep region
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6.4 Diffraction
(a) Diffraction is bending or spreading of waves around the edge of an obstacle.
(b)
Degree of diffraction depends on the relative size of the gap.
Significant diffraction
Insignificant diffraction
Gap size is relatively small (d ≈ λ)
Gap size is relatively large (d >> λ)
(c)
Ways to increase the degree of diffraction
(i)
increase wavelength
(lower the frequency or increase the depth of water)
(ii) decrease the size of gap
6.5 Interference
(a) Condition: two sets of waves of the same frequency meet each other
** (b) Stable interference pattern: two sources MUST be coherent.
Coherent sources are sources with
(i) same frequency
(ii) constant phase difference
(e.g. constantly in-phase or antiphase)
– amplitudes are not necessarily the same.
** (c) Constructive interference, destructive interference and path difference
Path difference
Type of interference
A
4λ – 3λ = λ
Constructive
B
4λ – 3.5λ = ½ λ
Destructive
C
3.5λ – 2.5λ = λ
Constructive
** Constructive interference: path difference = 0λ, λ, 2λ, 3λ, …
** Destructive interference: path difference = ½λ, 1½λ, 2½λ, 3½λ, …
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*Example 1
If λ = 0.25 cm, determine what kind of interference occurs at P.
Solution:
P
Path difference = S1P – S2P
3.2 cm
= 3.2 – 2.7
2.7 cm
= 0.5 cm
S1
S2
∴
= 2λ
Constructive interference occurs at P.
Common mistake 1:
Common mistake 2:
Did not express the path difference in terms of Wrong presentation
λ.
Path difference = 3.2 - 2.7
∴
Path difference = 3.2 - 2.7
= 0.5
Destructive interference occurs at P.
= 0.5
=
[Note:
0 .5
= 2λ 0.25
0 .5
= 2 ≠ 2 λ]
0.25
**(d) Factors affecting interference pattern
Increase source
separation
Result:
Nodal lines and antinodal lines
become closer together
Decrease
wavelength
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47
**Example 2:
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6.6 Stationary Waves
A stationary wave is formed by the superposition of two
waves with the same frequency and amplitude travelling in
opposite directions.
N: Node (amplitude = 0) A: Antinode (maximum amplitude)
(a) Waveforms of a stationary wave
(i)
a and b (c and d) are in-phase but with
different amplitudes
(ii) a and c, a and d, b and c, b and d are
anti-phase
(iii) when t = 0 s, all particles (a, b, c and d) are
momentarily at rest.
*(b) Comparing travelling waves and stationary waves
1
Traveling waves
Stationary waves
Energy is transmitted from one place to
Energy is localized.
another
2
3
All particles vibrate with the same
Different particles vibrate with different
amplitude.
amplitudes.
Neighbouring particles always vibrate
Particles in the same loop vibrate in phase.
out of phase.
Particles in two adjacent loops vibrate in
anti-phase.
4
Different particles reach their
All particles reach their maximum
maximum displacement at different
displacement at the same times.
times.
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* Example 3:
The figure below shows a stationary. At the instant shown, all particles reach their maximum
displacement. It is given that the frequency of the vibrator is 25 Hz.
(a) Find particles which are
(i)
vibrator
in-phase
(ii) anti-phase
(iii) momentarily at rest
(b) Find the wave speed.
* (c) Draw the shape of the wave after 0.01 s.
(d) What should be the frequency of the vibrator in order to produce 4 vibrating loops?
Solution:
(a) (i) in-phase: c and d
(ii) anti-phase: b and c, b and d
(iii) b, c and d (a is always at rest)
(b) λ = 1.2 × 2 / 3 = 0.8 m
Wave speed
(c) Steps:
v = fλ
= (25)(0.8) = 20 ms-1
(1) f = 25 Hz ⇒ T = 1/25 = 0.04 s
(2) 0.01 s = ¼ T
After 0.01 s (¼ T), the waveform is:
(d) for 1 vibrating loop: f1 =
25
= 8.33 Hz
3
For 4 vibrating loops: f4 = 4f1 = 4 ×
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25
= 33.3 Hz
3
50
Chapter 7 Light Waves
7.1 Diffraction of light
(a) Laser passes through a narrow slit and spreads into the
shadow of the slit.
Red
Green
Blue
(b) λred > λblue ⇔ diffraction of red light > diffraction of blue
light
(c) λred = 7 ×10-7 m. To show the diffraction of light, the slit
must be very narrow.
7.2 Interference of light
Young’s double-slit experiment
*(a) Precautions:
(i)
use strong light source (or black
out the laboratory)
(ii) use monochromatic light (單色光)
a
∆x
(iii) all slits should be as narrow as
possible
(iv) slit separation should be very small
(~ 0.5 mm)
D
(v) screen should be placed 1 – 2 m
(b)
Ways to increase fringe separation ∆x
behind the double-slit
(i) use light of longer wavelength λ
(ii) decrease slit separation a
(iii) increase the distance between the double-slit and the screen D.
7.3 Values commonly used in calculations
f
1kHz = 103 Hz
1MHz = 106 Hz
1GHz = 109 Hz
λ
1mm = 10-3 m
1µm = 10-6 m
1nm = 10-9 m
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Example 1
A yellow light of frequency 5 × 1014 Hz is used in Young’s double-slit experiment
(a) Find the wavelength of light.
(b) What is the path difference at C? What can
be observed at C?
(c) If successive bright fringes are formed at P,
Q, R and S, what are the path differences at P,
Q, R and S respectively?
(d) If the yellow source is replaced by a violet light source, what changes would be observed
on the screen?
Solution
(a) v = fλ
λ=
3 × 108
v
=
= 6 × 10 −7 m
14
f 5 × 10
(b) Path difference at C = 0 (or 0λ). Constructive interference occurs at C. Hence, a bright fringe is
observed at C.
(c)
P
Path difference
Q
-6
2λ = 1.2 × 10 m
R
-7
λ = 6 × 10 m
S
-7
λ = 6 × 10 m
2λ = 1.2 × 10-6 m
(d) Violet fringes will be observed.
The fringe separation is smaller because the wavelength of violet light is shorter than that of
yellow light.
7.4 Diffraction Grating
2nd order
1st order
0th order
1st order
Diffraction grating
2nd order
Screen
(a) bright fringes – constructive interference
(b) Fringe separation can be increased by
(i)
decreasing the grating spacing (using fine grating)
(ii) using light of longer wavelength
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7.5 Electromagnetic Waves
(a) show reflection, refraction, diffraction and interference
(b) are transverse waves
(c) can travel through vacuum
(d) travel at speed of light in air / vacuum (3 × 108 ms-1)
(e) can apply the equation v = fλ
Example 3
Microwaves are used to study interference as shown below. The amplitude of signal collected
by the received is showed in the following graph.
amplitude
A
P
(a) Explain why the amplitude varies as the receiver moves from A to B?
(b) If XP = 23 cm, YP = 30 cm, find
(i)
the path difference at P,
(ii) the wavelength of the microwave
(c) Now the separation between the slits is increased. Sketch the graph of amplitude of
microwave received along AB.
Solution:
(a) Interference occurs. Maximum amplitude is obtained because of constructive interference
and minimum signal is due to destructive interference.
(b) (i) Path difference at P = YP – XP = 30 – 23 = 7 cm
(ii) Path difference at P = 2λ
∴ 2λ = 7 cm
(constructive interference)
λ = 3.5 cm
(c)
amplitude
Separation between positions of
maximum amplitudes is reduced.
Increase source
separation
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Nodal lines and antinode
lines become closer together
53
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Chapter 8 Sound
8.1 Wave nature of sound
(a) Reflection of sound – echoes heard from an obstacle.
(b) Refraction
microphone
speaker
CO2
Refraction of light
Sound waves are focused by the balloon filled with carbon dioxide. A large sound is received
by the microphone.
(c) Diffraction
frequency of human speech f ~ 100 Hz – 300 Hz
wavelength λ ~ 1.1 m – 3.3 m [λ = v / f, v = 330 ms-1]
∵
λ > width of doorway
∴
significant diffraction of sound
Door
(d) Interference
* Experiment procedures:
Signal generator
Speaker
Speaker
1.
Connect two speakers to a signal generator.
2.
Set the frequency of the signal generator at 2000 Hz.
3.
λ = v / f = 330 / 2000 = 0.165 m.
4.
Place the two speakers 3λ apart. i.e. 3 × 0.165 ≈ 0.5 m apart.
Walk across in front of the loud speakers. Detect any change in the loudness of the sound.
5.
Connect a microphone to a CRO and move it across in front of the two loudspeakers.
Detect any change in the amplitude on the waveform on the CRO.
CRO amplitude
O
**
A
B
C
mic. position
–
Constructive interference: O, B
(maximum amplitude)
–
Destructive interference: A, C
(minimum amplitude)
–
Due to background noise, the amplitude at destructive interference ≠ 0.
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8.2 Properties of sound
(a) Longitudinal waves
(b) Need a medium to transmit sound.
(sound cannot travel through vacuum)
(c) vsolid > vliquid > vgas
(sound travels fastest in solid)
(d) Sounds are produced by vibrations
(e) Audible frequency range: 20 Hz – 20 kHz
8.3 Musical notes and Noise
Musical note
Noise
Waveform on CRO (regular in shape)
Waveform on CRO (irregular)
Measured in decibel (dB) (分貝)
Waveform of a tuning fork
*
Pitch
-------------
frequency
Loudness
-------------
amplitude
Sound quality -------------
waveform on CRO
8.4 Ultrasound
(a) sound waves of frequency greater than 20 kHz
(b) properties are the same as sound waves
(c) applications (i) ultrasonic scan of the foetus (胎兒)
(ii) detecting shoal of fish (魚群)
(iii) ultrasound flaw (裂紋) detector
Example 1:
30 kHz ultrasound is used to detect shoal of fish. It is given that the speed of ultrasound in sea
water
is 1500 ms-1.
(a) Find the wavelength of ultrasound.
* (b) Explain ultrasound is used instead of audible sound.
Solution:
(a) λ = v / f
= 1500 / (30 × 103) = 0.05 m
* (b) Since wavelength of ultrasound is shorter, the degree of diffraction of ultrasound is
smaller.
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Chapter 1 Electrostatics
1.1 Electric charge
(a) Like charges repel; unlike charges attract.
(b) Charging methods
(i)
Rub an acetate strip with a duster. The rubbed acetate strip becomes positively
charged.
(Electrons are transferred from the acetate strip to the duster)
(ii) Rub a polystyrene strip with a duster. The rubbed ac polystyrene strips become
negatively charged.
(Electrons are transferred from the duster to the polystyrene strip)
(iii) Charging by Van de Graaff generator
(c) Charge Q
unit: C
Example 1
The charge of an electron is e = -1.6 ×10-19 C. How many extra electrons are gained by a
negatively charged rod of 2 ×10-8 C?
Solution
No. of extra electrons gained
=
2 × 10 −8
= 1.25 × 1011
−19
1.6 × 10
(d) Attraction of a neutral object by a charged object
(i)
When a positively charged rod is put near to a neutral object, negative charges are
induced on the side near the rod and positive charges are induced on the other side.
(ii) The attraction between the rod and the negative charges is greater than the repulsion
between the rod and the positive charges.
(iii) Therefore, the neutral object is attracted by the charged rod.
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1.2 Coulomb’s law
Coulomb’s law states that the force between two point charges is directly proportional to the
product of the charges and inversely proportional to the square of the distance between them.
F=
Q1Q2
4πε 0 r 2
ε0 = 8.85 x 10-12 C2 N-1 m-2
Example 2
There are three charges Q1 (–2 µC), Q2 (1 µC) and Q3 (–3 µC) as shown below. Find the
resultant force acting on Q2.
Solution
Q2 is attracted by Q1 and Q3.
By F =
Q1Q2
4πε 0 r 2
(2 × 10 )(1 × 10 ) = 1.798 N
=
4π (8.85 × 10 )(0.1 )
(3 × 10 )(1 × 10 ) = 2.698 N
=
4π (8.85 × 10 )(0.1 )
−6
F1
−6
−12
−6
F3
2
[Note: 10 cm = 0.1 m]
−6
−12
2
Resultant force = F3 − F1 = 0.899 N (toward the right)
1.3 Electric Field
(a) Electric field pattern
Direction of
field lines
Meaning
Density of
field lines
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Go from a positive charge to a
negative charge
Show the direction of electric force
acting on a positive test charge
Strength of electric field
58
(b) Electric field strength and gravitational field strength
Electric field strength E
Definition
Electric force per unit charge
F
E=
q
Isolated point charge:
E=
Parallel plates:
E=
Equation
Unit
Q
4πε 0 r 2
σ
ε0
N C-1
Vector
Example 3
Two point charges A (3 × 10–8 C) and B (–2 × 10–8 C) are placed 20 cm apart. Find the magnitude of
the electric field strength at B due to A.
Solution
Electric field strength due to A:
3 × 10 −8
E=
=
= 6740 NC-1
2
−12
2
4πε 0 r
4π (8.85 × 10 )(0.2)
QA
Remark: Electric field strength due to A is calculated from the charge of A.
Example 4
Resultant electric field
Three point charges A (–3 × 10–8 C), B (2 × 10–8 C) and
C (2 × 10–8 C) are shown. Find the electric field
strength due to B and C at the position of A.
Solution
Electric field strength due to B
2 × 10 −8
EB =
=
= 4496 NC-1
4πε 0 r 2 4π (8.85 × 10 −12 )(0.2) 2
QB
Electric field strength due to C
2 × 10 −8
= 4496 NC-1
4πε 0 r 2 4π (8.85 × 10 −12 )(0.2) 2
Eresultant = E B cos 30° + EC cos 30° = 7790 NC-1
EC =
QC
=
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Eresultant
30o
30o
EC
EB
EC
EC cos 30o
EB cos 30
60°
EB
o
30o 30o
EC sin 30o
EB sin 30o
59
Chapter 2
Electric Circuits
2.1 Simple circuit
electron flow
battery
conventional
current
bulb
bulb
The conventional current goes round the circuit from
the positive terminal of the battery to the negative
Current is the same at every part of a
simple circuit
terminal.
2.2 Current
Definition:
current is the rate of flow of charge
current =
charge
time
Q

I = 
t 

or
Q = It
Unit: A
2.3 Potential difference, electromotive force, voltage
Electromotive force (e.m.f)
Potential difference (p.d.)
Energy gained by a unit charge when the charge is the amount of electrical energy which
passes through a battery
changes into other forms of energy when a unit
charge passes between two points in a circuit.
e.m.f = 4V ⇔ 4J of electrical energy is gained p.d. = 1V ⇔ 1 J of electrical energy is
by 1C of charge
changed into heat and light energy when 1 C of
charge passes through the bulb
e.m.f = 4V
Suppose 6C of charge passes through the circuit.
Total amount of energy gained by the charge from the
battery = (4)(6) = 24 J
For 1st bulb,
Total amount of charge converted into heat and light
energy = (1)(6) = 6 J
p.d. = 1V
e.m.f
p.d. = 3V
ε = V1 + V2
V =
For 2nd bulb,
Total amount of charge converted into heat and light
energy = (3)(6) = 18 J
E
Q
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2.4 Resistance
Definition: Resistance =
voltage
current
R=
V
or
I
V= IR
Unit: Ω
Example 1
A hairdryer has a resistance of 24 Ω. Find the current passing through it when it is connected to
a 220-V power supply.
Solution
By V= IR
220 = I(24)
I = 9.17 A
2.5 Ohm’s law
The voltage across a conductor is directly proportional to the current flowing through it,
provided the temperature and other physical conditions are unchanged.
R=
[V ∝ I]
V
V
= slope of V – I graph
I
Higher R
Lower R
I
Example 2
The V–I graph of conductor X is shown.
(a) Find the resistance of conductor X.
(b) Conductor Y obeys Ohm’s law and its resistance is half
that of conductor X. Sketch the V–I graph of conductor Y.
Solution
(a) Resistance = slope of V – I graph
=
5
= 0.5Ω
10
(b) Resistance of Y = ½ (0.5) = 0.25 Ω
i.e. when V = 5V, I = 20 A
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Conductor Y
(Lower resistance, smaller slope)
61
Example 3
Remark:
1. Ammeters must be connected in series.
2. Voltmeters must be connected in parallel.
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Factors affecting resistance
Temperature T
Length l and thickness A
For metals: T↑ ⇒ R↑
For semi-conductors: T↑ ⇒ R↓
R∝
l
A
R=ρ
Conductor
Resistivity (Ωm)
Copper: 1.7 × 10
-8
longer wire ⇒ R↑
l
A
Thicker wire ⇒ R↓
ρ = resistivity
Semi-conductor
Insulator
Silicon: 2300
Polystyrene 1015
Example 4
Find the resistance of a copper wire if its length and diameter are 70 cm and 1 mm
respectively.
Given that the resistivity of copper is 1.7 × 10-8 Ωm.
Solution
By R = ρ
l
0.7
= (1.7 × 10 −8 )
= 0.0152 Ω
2
A
 10 − 3 

 π
 2 
[Α = πr2 = π(d/2)2]
2.6 Resistors in Series and Parallel
Resistors in Series
Resistors in parallel
ε
ε (e.m.f of a cell)
I1
R1
I2
V1
R2
I
V2
I1
R1
I2
IV1 1
R2
V2
I1 = I2 (Common)
I = I1 + I2
ε = V1 + V2
ε = V1 = V2 (Same)
1
1 −1
R eq = ( +
)
R1 R 2
Req = R1 + R2
If R1 = R2, then Req = ½ R1 = ½R2
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Example 5 Circuit analysis 1
Find I6, I8, I12; V6, V8, V12.
Example 6
Circuit analysis 2
Figure 1
Find the current passing through each resistor and
voltage across each resistor.
Solution: simplify the above circuit
Solution
1.
2.
1 1
R eq = 8 + ( + ) −1 = 8 + 4 = 12 Ω
6 12
12 12
I =
=
=1A
Req 12
3.
∵
∴
4.
V6= V12 = 12 – V8 = 4 V
5.
I6 =
I8 = I = 1A
V8 = I8(8) = (1)(8) = 8V
Figure 2
V6 4
= = 0.667 A
6 6
V12
4
=
= 0.333 A
12 12
(OR I12 = I8 – I6 = 1 – 0.667 = 0.333 A)
I 12 =
Figure 3
In Figure 2
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2.8 Internal Resistance in Ammeters, Voltmeters and Cells
* Addition of an ammeter will increase the total
resistance of the circuit ⇒ I ≠ I’
* An ideal ammeter should have small resistance.
⇒ I ≈ I’
* Addition of a voltmeter will affect the voltage
across 4Ω resistor.
⇒ V4 ≠ Voltmeter reading
* An ideal voltmeter should have large resistance.
⇒ V4 ≈ Voltmeter reading
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Chapter 3
Domestic Electricity
3.1 Electric power and energy
E
V2
= VI = I 2 R =
t
R
Electric power:
P=
Electrical energy:
E = Pt = VIt = I 2 Rt =
V2
t
R
Example 1
Find the total amount of energy stored by the rechargeable battery when it is fully charged.
Solution
2000 mAh:
(1) I = 2000 mA = 2A
(2) t = 1 h = 3600 s
Energy stored E = VIt
= (1.2)(2)(3600) = 8640 J
3.2 Pay for electricity
Kilowatt-hour meter measures the electrical energy consumed.
Unit: kWh
(1 度電)
1 kWh is the amount of electrical energy consumed by an appliance of 1000 W for 1 hour
Example 2
Find the cost of electricity to operate five lamps of power 60 W for 8 hours. It is given that
electricity costs $1.02 per kWh.
Solution
 60 
E = Pt = 
(8) × 5 = 2.4 kWh
 1000 
Cost of electricity = $1.02 × 2.4 = $2.448
3.3 Mains electricity and household wiring
(a) Electric socket and plug
Earth wire
Earth hole
Switch
Live hole
Fuse
Neutral wire
Live wire
Neutral hole
Remark:
The earth pin is designed longer to open ‘shutters’ on the live and the neutral holes.
This ensures the earth wire is connected before the live wire.
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(b) Cable
Live wire
Neutral wire
Earth wire
+220 V , – 220 V
0V
To ground (0V)
Brown
Blue
Yellow and green
*
Connects the metal body of an electrical
appliance to the Earth.
*
In case of a fault, a large current will flow
through the earth wire to earth. This
prevents the user from getting an electric
shock.
(c) Switch
(i) Must be fitted in the live wire.
(ii) This makes sure that no part of the electrical appliance is at high voltage when the
switch is turned off.
(d) Fuse
(i) Must be fitted in the live wire.
(ii) If an excess current flows through the circuit or the circuit overloads, the fuse blows
and breaks the circuit before the cable overheats and causes a fire.
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Chapter 4
Electromagnetism
4.1 Magnetic field due to permanent magnets
Magnetic field lines are used to show the strength and direction of a magnetic field.
(a) Field lines run from the N-pole round to the S-pole.
(b) When field lines are closely-spaced, field is strong and vice-versa.
neutral
point
N
S
x
N
N
uniform
field
4.2 Magnetic field due to current carrying conductors
Current-carrying
conductor
Magnetic field pattern
current direction
up out of page
Long straight wire
wire (top view)
solenoid
N
S
current direction
Solenoid
Inside the solenoid:
Uniform field
Outside the solenoid:
Similar to that around a bar magnet
Ways to increase magnetic field strength of a solenoid
(1) increase the current
(2) increase the number of turns of solenoid (for the same
length of solenoid)
(3) insert a soft-iron core through the solenoid
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4.3 Magnetic force acting on a current-carrying conductor
Magnetic force F = BIl
Fleming’s left hand rule
The magnetic force is increased if
(1) the current is increased, ( F ∝ I )
(2) the magnetic field is increased, ( F ∝ B )
(3) there is a greater length of wire inside the magnetic field. ( F ∝ l )
Example 1
The figure below shows two straight parallel wires X and Y . Wire X carries current 3A and wire Y
carries current 6 A in the opposite direction.
(a) Draw the magnetic field produced by Y at the position of wire X.
(b) Determine the direction of magnetic force acting on X.
(c) Mary says, ‘Since the current through Y is two times that through X, the magnetic force acting
on Y is two times that acting on X.’ Comment on Mary’s statement.
Solution
(a)
BY
IX
(b) By Fleming’s left-hand rule, the direction of magnetic force acting on X is towards the right.
BY
F
(c) Her statement is incorrect.
According to Newton’s third law, the forces acting on both wires are an action and reaction
pair. The magnetic force acting on X is equal to that acting on Y.
4.4 Motor
Structure
(1) magnets
(3) carbon brushes
(2) a coil
(4) commutator
Function of a commutator:
The commutator reverses the current every
half turn to keep the coil rotating
continuously in one direction.
Ways to increase the turning speed of the coil
(1) increase the current
(2) increase the magnetic field
(3) increase the number of turns of the coil
(4) increase the area of the coil within the
magnetic field
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Chapter 5
Electromagnetic Induction
5.1 Induced e.m.f and induced current
S
N
N
An e.m.f. is induced because
S
the coil An e.m.f is induced because the conductor
experiences a changing magnetic field.
cuts across magnetic field lines.
A current is induced because the circuit is A current is NOT induced because the circuit is
complete.
not complete.
Ways to increase induced e.m.f or induced current:
(1) move the magnet or the wire faster
(2) use a stronger magnet
(3) use a coil of more turns or increase the length of wire within the magnetic field
(4) insert a soft iron core in the coil
Working out the direction of induced current
(a) Lenz’s law
An induced current always flows to oppose the change which started it in a magnetic field.
S
N
N
S
induced current
S
N
S
N
induced current
(b) Fleming’s right-hand rule
(i) motion or force F
(ii) magnetic field B
(iii) induced current I
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5.2 Generator
An a.c. generator
A d.c. generator
coil rotated
carbon brushes
carbon brushes
commutator
slip rings
(i)
The coil cuts through the
field lines most rapidly.
(max. induced current I)
(ii)
The coil does not cut through
any field line momentarily.
(no induced current I)
An a.c. generator becomes a d.c. generator if the slip rings are replaced by a commutator.
*
The commutator reverses the connections of the coil to the outside circuit every half turn.
Therefore, the current in the outside circuit always flows in the same direction.
Ways to increase induced e.m.f or induced current:
(1) increase the number of turns of the coil
(2) use a stronger magnet
(1) Increase the area of coil within the magnetic field
(2) winding the coil on a soft-iron core
(3) rotate the coil at a higher speed
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Example 1
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Chapter 6
6.1 Transformers
Vp N p
(a)
=
Vs
Ns
Transmission of Electricity
Soft iron core
Primary coil
Secondary coil
step-up transformer
step-down transformer
N p < N s ⇔ Vp < Vs
N s < N p ⇔ Vs < Vp
(b) Efficiency η
Ip
Is
Vp
Vs
Power input
= VpIp
Power output
= VsIs
Efficiency η
Power output
=
× 100%
Power input
=
Vs I s
× 100%
Vp I p
If efficiency η = 100% (ideal transformer / no power loss in transformer),
then VpIp = VsIs
(c) Power loss in transformers
Cause
Improvement
Resistance of coils
Use thick wire for the coil
Magnetization and demagnetization of the core
Use soft-iron core
Induced currents in the core
Use laminated core
Example 1
If the transformer has 40 turns in its secondary coil and
both bulbs can operate at their rated values, assume the
transformer is ideal, find
(a) the number of turns in the primary coil.
(b) the currents in the primary coil and secondary coil.
220 V
a.c.
10 V, 20 W
Ns = 40
Solution
Vp N p
N p 220
(a)
=
⇒
=
⇒ N p = 880
Vs
Ns
40
10
(b) Power output = 40 W (two light bulbs)
V s I s = 40 ⇒ 20 I s = 40 ⇒ I s = 2 A
Power input = Power output (ideal transformer)
V p I p = 40 ⇒ 220 I p = 40 ⇒ I p = 0.182 A
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6.2 High voltage power transmission
*
Power loss in transmitting electricity
*
Reason: resistance in power transmission cable
Analysis: consider the following power transmission system.
I
Cable 10 Ω
Power station
Consumer unit
200 V, 1200 W
Cable 10 Ω
Procedures:
(1) Current in cable
I cable =
P 1200
=
=6A
V
200
(2) Power loss (in cable)
2
Ploss = I cable R = (6) 2 (20) = 720 W
(3) Efficiency of the power transmission system
useful power output
η=
× 100%
power input
=
1200 − 720
× 100% = 40%
1200
(4) Voltage drop (due to cable)
V drop = I cable R = (6)( 20) = 120 V
(5) Voltage available to consumer unit
Vo = 1200 − Vdrop = 200 − 120 = 80 V
Conclusion
(a) low efficiency (η = 40 %) large amount of power loss
(b) insufficient terminal voltage (200V → 80V)
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Solution: use high voltage to transmit electrical power
⇒ reduce current in transmission cable
⇒ reduce power loss in cable
C Cable 10 Ω E
A
Cable 10 W
G
200 V
Consumer
1200 W
Unit
power station B
1:20
Cable 10 Ω
20:1
Cable
10 W F
D
H
Assumption: ideal transformers are used
(1) ∵ VAB = 200 V
∴
VCD = 200 × 20 = 4000 V
(step-up transformer)
(2) Current in transmission cable
IcableVCD = 1200
(no power loss in transformer)
Icable(4000) = 1200
Icable = 0.3 A
(3) Power loss (in cable)
(high voltage ⇒ low current)
2
Ploss = I cable R = (0.3) 2 ( 20) = 1.8 W
(4) Efficiency of the power transmission system
useful power output
η=
× 100%
power input
=
1200 − 1.8
× 100% = 99.85%
1200
(5) Voltage drop (due to cable)
V drop = I cable R = (0.3)( 20) = 6 V
VEF = VCD − Vdrop = 4000 − 6 = 3994 V
(6) Voltage available to consumer unit
Vo = VEF ×
1 3994
=
= 199.7 V
20
20
Conclusion
(a) high efficiency η = 99.85% (nearly perfect!)
(b) terminal voltage (Vo = 199.7 V ≈ 200V)
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