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St. Joseph’s Anglo-Chinese School
HKDSE Physics Notes
H eat
M echanics
W aves
E lectricity
R adioactivity
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 5 Gases
5.1 The gas laws
(a) Pressure
Pressure =
Force
or
area
P=
F
A
Unit: Nm-2 or Pa
Example 1
Find the pressure exerted on the 5 kg block in each of the following cases.
10 cm
10 cm
(a)
5 cm
5 cm
R = mg cos 30o
(b)
R = mg
30o
Solution
(a) Force acting on the block
(b) Force acting on the block
F = normal reaction = mg cos 30o
F = mg = (5)(9.81) = 49.05 N
Pressure P =
F
49.05
=
= 9810 Pa
A (0.05)(0.1)
Pressure P =
F (5)(9.81) cos 30°
=
= 8500 Pa
A
(0.05)(0.1)
5cm = 0.05 m, 10 cm = 0.1 m ⇒ A = (0.05)(0.1) = 0.005 m2]
[Note:
(b) The gas laws
Boyle’s law
Pressure law
Charles’ law
The pressure of a fixed mass of For a gas with a fixed mass
For a gas with a fixed mass
gas is inversely proportional to
and volume, its pressure is
and pressure, its volume is
its volume at a constant
directly proportional to its
directly proportional to its
temperature.
Kelvin temperature.
Kelvin temperature.
Pressure p
Volume V
Pressure p
0
1/V (m-3)
p∝
1
V
0
Temperature T (K)
p∝T
0
Temperature T (K)
V∝T
General gas law: pV = nRT, where R = 8.31 J mol-1 K-1
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Example 2
A biscuit factory is located at sea level. Its production line is maintained at 20 °C and 100 kPa. Each
pack of biscuit contains 18 cm3 of gas to protect the biscuit from cracking during transportation.
The pack of biscuit is now transported up to a mountain where the temperature and pressure are
12 °C and 75 kPa respectively.
(Take R = 8.31 J mol–1 K–1)
(a) Find the number of moles of gas inside a pack of biscuit.
(b) What is the volume of the gas inside the pack on the mountain?
Solution
(a) By pV = nRT ,
n=
pV (100 × 10 3 )(18 × 10 −6 )
=
= 7.39 × 10 − 4 mol
RT
(8.31)( 20 + 273)
There are 7.39 × 10-4 mol of gas inside the pack.
(b)
Method 1
V=
By pV = nRT
nRT (7.39 × 10 −4 )(8.31)(12 + 273)
=
p
75 × 10 3
Method 2
By
p1V1 p 2V2
=
T1
T2
(100k )(18 × 10 −6 ) (750k )V
=
( 20 + 273)
(12 + 273)
= 2.33 × 10−5 m3
V = 2.33 × 10−5 m3
Remark (1) Always use Kelvin degree for temperature.
i.e. 20 °C = (20 + 273) K,
12 °C = (12 + 273) K
3
-2 3
3
-6
3
(2) 1 cm = (10 ) m = 10 m
5.2 The kinetic theory
(a) Ideal gas
Ideal gas
Obeys the general gas law for all pressures and temperatures.
Real gas
Behaves like an ideal gas at high temperature and low pressure
(b) Assumptions for ideal gas
1.
All the particles are identical and have the same mass.
2.
All the molecules are in constant, random motion.
3.
The size of each particle is negligible compared with the separation between them.
4.
The duration of a collision is negligible compared with the time between collisions.
5.
The collisions of molecules with the container and between the particles are perfectly
elastic.
6.
Intermolecular forces are negligible.
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(c) A three-dimensional kinetic theory model
The ball bearings represent gas molecules.
Physical quantity
Kinetic theory model
weight of the piston
Pressure
(and cardboard discs)
Temperature
Voltage applied to the motor
Volume
Height of the piston
(d) Explanation of the gas laws by the kinetic theory
Boyle’s law
1
p∝
V
Volume ↓
⇒ molecules hitting the
walls more frequently
⇒
pressure ↑
Temperature ↑
Pressure law
⇒
molecules moving
faster
⇒
pressure ↑
p∝T
Temperature ↑
Charles’ law
V∝T
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⇒
molecules moving
faster
⇒
volume ↑ to keep
the same pressure
13
Example 3
A mildly pumped up beach-ball feels hard when pressed between two wooden boards (Fig a). If the
boards are removed, the ball feels soft again (Fig b).
Fig a
Fig b
Explain why the ball feels hard when pressed between the boards using
(a) one of the gas laws,
(b) the kinetic theory.
Solution
(a) When compressed, the volume of the ball decreases.
According to Boyle’s law, pressure increases when the volume decreases.
The ball therefore feels hard.
(b) When compressed, the volume of the ball decreases.
The air molecules can only move in a smaller volume.
As a result, they hit the walls of the ball more frequently and so the pressure increases.
The ball therefore feels hard.
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(e) Pressure and molecular speed
(1)
PV =
1
Nmc r2.m.s
3
(2) c rms =
3RT
mN A
3
(3) KE = nRT
2
Example 4
On a mountain, a sealed metal bottle of volume 800 cm3 contains air at a pressure of 72 kPa and a
temperature of 5 °C. Assume that air behaves like an ideal gas.
(Take R = 8.31 J mol–1 K–1, average molar mass of air = 29.0 g mol–1)
(a) Find the number of moles of air in the bottle.
(b) Find the root-mean-square speed of the air molecules in the bottle.
(c) The bottle is then brought to Hong Kong (at sea level) at 20 °C.
Find the change in the total KE of the gas molecules.
Solution
(a) By
pV = nRT
n=
pV (72 × 10 3 )(800 × 10 −6 )
=
= 0.0249 mol
RT
(8.31)(5 + 273)
(b) mNA = 29.0 g =
c rms =
[1 cm3 = 10-6 m3]
29.0
kg = 0.029 kg
1000
3RT
3(8.31)(5 + 273)
=
= 489 ms-1
mN A
0.029
3
(c) Change in total KE = nR (T2 − T1 )
2
=
(f)
3
(0.0249)(8.31)( 20 − 5) = 4.66 J
2
Speed of gas molecules
Xe – 132
Ar – 40
Ne – 20
He – 4
(1) cr.m.s. increases with temperature ( c r .m.s ∝ T ).
(2) cr.m.s. decreases with the mass of the molecule or molar mass of the gas ( c r .m. s ∝
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mN A
).
15
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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
18
(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
a = slope of v-t graph
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
22
(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
St. Joseph’s Anglo-Chinese School
cannot use the formula v 2 − u 2 = 2as
reason: not a uniformly accelerated motion
25
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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St. Joseph’s Anglo-Chinese School
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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 7 Circular Motion
7.1 Velocity and Angular velocity
angular velocity ω
velocity v
velocity =
Unit: ms
distance
time
(v =
-1
s
)
t
Angular velocity ω =
Unit: rad s
θ
t
=
2π
T
-1
v = rω
Example 1
A piece of rod rotates about O. A and B are two particles on the rod. If the rod makes 20
revolutions in 80 s, find
(a) the angular velocity of the rod.
(b) the velocities of A and B.
Solution
(a) 1 revolution = 360o = 2π rad
ω=
θ
t
=
20 × 2π
= 1.57 rad s-1
80
A
O
2m
B
2m
Alternative method:
T = 80 / 20 = 4 s
ω=
2π
= 1.57 rad s-1
4
(b) v = rω
v A = 2 × 1.57 = 3.14 ms-1
v B = 4 × 1.57 = 6.28 ms-1
Remark:
(1) When the rod rotates, both A and B perform circular motion.
(2) Since A and B make one revolution (360o) at the same time (same period T), they have the
same angular velocity.
(3) vB > vA because rB > rA ( v = rω )
Example 2
It takes 23 hours 56 minutes and 4.1 second for the Earth to revolve about its own axis. Find
the angular velocity of the rotation.
Solution
Period T = 23 × 3600 + 56 × 60 + 4.1 = 86164.1 s
ω=
2π
2π
=
= 7.29 × 10 −5 rad s-1
T
86164.1
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7.2 Centripetal Acceleration and Centripetal Force
Note:
v2
= rω 2
r
Centripetal Acceleration
a=
Centripetal force
F = ma = m
v2
= mrω 2
r
Centripetal force is only the net force in a circular motion.
Example 3
A stone of mass 2 kg is tied by a string and moves in a horizontal circular path of radius 0.5 m.
(a) Find the tension in the string if the speed of the stone is 4 ms-1.
(b) The string breaks when its tension exceeds 100 N. Find the maximum speed of the stone
if the string does not break.
Smooth table
Solution
(a)
∵
T is the only force along the radius and it is pointing to the centre
∴
T =m
T
(b) T = m
v2
42
= 2×
= 64 N
r
0 .5
v2
r
100 = 2
v2
0 .5
v =5 ms-1
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Example 4
A 4 kg bob is suspended by a piece of string as shown. If the period of
revolution is 2.5 s and the radius of the circular path is 60 cm, find
θ
(a) angular speed,
(b) θ, and
(c) tension in the string.
Solution
(a) ω =
2π 2π
=
= 2.51 rad s-1
T
2 .5
(b)
T cos θ
Resolve T into its components:
Along the radius:
T sin θ = mrω 2 L (1)
In vertical direction:
T cos θ = mg L ( 2)
θ
T
ω
(1)
:
( 2)
rω 2
g
0.6( 2.51) 2
[60 cm = 0.6 m]
9.81
θ = 21.1o
tan θ =
T sin θ
mg
(c) put θ = 21.1o into (2)
T=
tan θ =
mg
4 × 9.81
=
= 42.0 N
cos θ cos 21.1°
Example 5 Rounding a bend – without banking
The maximum friction between the tires
and the road 0.6 times the weight of a car.
If a car goes around a circular bend of
radius of curvature 50 m, find the maximum
speeds of the car without slipping.
tan θ =
rω 2 v 2
=
g
gr
60 m
f
Solution
Maximum friction f = 0.6 mg
Along the radius:
f =m
v2
r
v2
r
2
v = 0.6 gr = 0.6 × 9.81 × 50
0.6mg = m
v = 17.1 ms-1
Remark: friction provides the centripetal force for a car to perform circular motion.
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37
Example 6
Rounding a bend – with banking
A car moves round a corner with a radius of 80 m on a banked
80 m
road of inclination 20o.
(a) What is the safety speed designed for this bank road?
(b) If the speed of the car is 70 km h-1, is there any
20ο
friction acting on the car by the road? If yes, draw the friction acting on the car.
Solution
(a) The centripetal force does not rely on friction, but provided by the horizontal component of the
normal reaction.
In vertical direction: R cos θ = mg L (1)
In horizontal direction: R sin θ = m
R
v2
L ( 2)
r
R sin θ
(2)
v2
:⇒ tan θ =
(1)
gr
tan 20° =
v2
9.81 × 80
R cos θ
mg
20ο
tan θ =
v2
gr
v = 16.9 ms-1
(b) If the speed of the car = 70 km h-1 =19.4 ms-1 > 16.9 ms-1,
there will be a tendency for the car to move outward.
Friction will act on the car as shown.
St. Joseph’s Anglo-Chinese School
Friction f
38
7.3 Cyclist around a circular track
The centripetal force is provided by friction f
v2
The tilting angle θ of the cyclist is given by tan θ =
gr
7.4 Aeroplane making a turn
Uplifting force
L cos θ
L sin θ = m
v2
L (1)
r
L cos θ = mg L (2)
(1)
v2
:⇒ tan θ =
( 2)
gr
Example 7
Find the angle of inclination of the wings of an aircraft which is traveling in a circular path of
radius 2000 m at a speed of 360 km h-1.
Solution
v = 360 km h-1 = 100 ms-1
[1 km h-1 =
v2
100 2
=
gr 9.81 × 2000
tan θ =
1
ms-1]
3 .6
θ = 27.0o
Example 8 The rotor
A man is riding inside the ‘rotor’ of radius 2 m. The maximum
friction between the man and the wall is 0.4 times the normal
reaction acting on the man. Find the minimum speed v of the
man before the floor is pulled downwards?
f
Solution
Given: f = 0.4R
--- (1)
In vertical direction: f = mg --- (2)
R= m
v2
r
mg
--- (3)
Sub. (2) and (3) into(1): ⇒ mg = 0.4 m
v=
R
v2
r
gr
9.81 × 2
=
= 7.00 ms-1
0.4
0.4
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Chapter 8 Gravitation
Gravitational force
F=
GMm
r2
Gravitational field
g=
F GM
= 2 (g = gravitational force per unit mass)
m
r
Unit: N
Unit: Nm-1 or ms-2
Vector
vector
Gravitational field pattern
Field lines
(i) direction: shows the direction of gravitational force
acting on an object (attraction)
(ii) density: shows the strength of the field
Example 1
Student P of mass 65 kg is 1.5 m away from student Q of mass 60 kg. Find the magnitude of
the gravitational force between them. Assume students P and Q are two spherically
symmetrical objects. Take G = 6.67 × 10–11 N m2 kg–2.
Solution
GMm 6.67 × 10 −11 × 65 × 60
=
= 1.16 × 10 -7 N
2
2
r
1 .5
Example 2
A particle of mass 0.02 kg is located at 100 km above the Earth’s surface.
F=
(a) Find the gravitational force acting on the particle by the Earth.
(b) Find the gravitational field strength of the Earth at the position of the particle.
Given:
G = 6.67 × 10–11 N m2 kg–2
Mass of the Earth ME = 5.98 × 1024 kg
Radius of the Earth RE = 6370 km
Solution
(a) Gravitational force acting on the particle by the Earth =
=
(6.67 × 10 −11 )(5.98 × 10 24 )(0.02)
(100 × 10
3
+ 6370 × 10 3 )
2
(b) gravitational field strength = =
GMm
r2
= 0.191 N
F 0.191
=
= 9.53 N kg-1
m 0.02
Note: since the particle is 100 km above the Earth, the gravitational field strength is
smaller than 9.81 N kg-1 (close to the surface of the Earth)
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Example 3
A spacecraft is orbiting around Venus. The acceleration due to gravity at its orbit is 8 m s–2.
Estimate the height of the orbit above Venus’ surface. Take the acceleration due to gravity at
Venus’ surface to be 8.87 m s–2 and the radius of Venus RV to be 6050 km.
Solution
By g =
GM
r2
For the spacecraft: 8 =
GM
--- (1)
r2
At the surface of Venus: 8.87 =
GM
--- (2)
(6050 × 10 3 ) 2
(1)/(2):
8
(6050 × 10 3 ) 2
=
8.87
r2
8.87(6050 × 10 3 )
r=
= 6070.5 km
8
2
The height of the orbit above Venus’ surface = 6370 km – 6050 km = 320 km
Example 4
A geostationary satellite takes 24 hours to revolve once around the Earth in a circular orbit.
(a) Find the angular velocity ω of the satellite.
(b) Estimate the distance of the satellite h above the Earth’s surface.
Take the acceleration due to gravity at the Earth’s surface to be 9.8 m s–2 and the radius of the Earth
RE to be 6370 km.
(a) ω =
2π
2π
=
= 7.27 × 10 −5 rad s-1
T 24 × 3600
(b) At the surface of the earth: g =
2
GM
⇒ GM = 9.81 × (6370 × 10 3 ) --- (1)
2
r
For the satellite:
satellite
Gravitational force = Centripetal force
2
GMm
= mrω 2 ⇒ GM = r 3ω 2 = r 3 × (7.27 × 10 −5 ) --- (2)
2
r
GMm
r2
Sub. (1) into (2):
GM = r 3 × (7.27 × 10 −5 ) = 9.81 × (6370 × 10 3 )
2
2
9.81 × (6370 × 10 3 )
2
r =
3
∴
(7.27 × 10 )
−5 2
r = 42220 km
The distance of satellite above the Earth = 35900 km = 3.59 × 107 m.
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Example 5
Moon
Assume that the Moon revolves around the Earth in a
circular orbit of radius 3.78 × 108 m. The period is 27.3 days.
(a) Find the linear velocity v of the Moon.
GMm
r2
(b) Estimate the mass of the Earth ME.
Given G = 6.67 × 10–11 N m2 kg–2.
Solution
(a) ω =
2π
2π
=
= 2.66 × 10 − 6 rad s-1
T 27.3 × 24 × 3600
v = rω = 3.78 × 108 × 2.66 × 10-6 = 1010 ms-1
(b) Consider the circular motion of the Moon moving around the Earth.
GMm
= mrω 2
2
r
r 3ω 2 (3.78 × 10 8 ) (2.66 × 10 −6 )
= 5.75 × 10 24 kg
M=
=
−11
G
6.67 × 10
3
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42
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)
47
(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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3.5 The lens formula
1 1 1
= +
f u v
Remark
*
Focal length f
Image distance v
Convex lens
+ ve
Concave lens
– ve
Real image
+ ve
Virtual image
– ve
* 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.
(d)
Using the lens formula to calculate the focal length of the lens.
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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)
(d) u = 10 cm, v = -4 cm (virtual image)
By
1 1 1
= +
f u v
Wrong presentation:
1 1 1
By
= −
f u v
1
1
1
=
+
f 10 − 4
1
1 1
= −
f 10 4
f ≈ -6.67 cm
f ≈ -6.67 cm
The focal length is 6.67 cm
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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
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
(c) d = vt
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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**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
64
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
(iii) all slits should be as narrow as
∆x
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
λ
D
(i) use light of longer wavelength λ
∆x =
a
(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
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) Find the distance between P and Q if the slit separation is 0.5 mm and the distance
between the double-slit and the screen is 1.5 m.
(e) If the yellow source is replaced by a violet light source, what changes would be observed
on the screen?
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65
Solution
(a) v = fλ
λ=
3 × 108
v
=
= 6 × 10 −7 m
f 5 × 1014
(b) Path difference at C = 0 (or 0λ). Constructive interference occurs at C. Hence, a bright fringe is
observed at C.
(c)
P
Q
-6
Path difference
R
-7
λ = 6 × 10 m
2λ = 1.2 × 10 m
S
-7
λ = 6 × 10 m
2λ = 1.2 × 10-6 m
(d) Given a = 0.5 mm = 0.5 × 10-3 m, D = 1.5 m
(6 × 10 −7 )(1.5)
PQ = ∆x =
=
= 0.0018 m
a
0.5 × 10 −3
(e) Violet fringes will be observed.
λD
The fringe separation is smaller because the wavelength of violet light is shorter than that of
yellow light.
7.4 Diffraction Grating
2nd order
θ1
θ2
(b) For nth order maximum,
0th order
d sin θ = nλ
(c) Fringe separation can be increased by
st
1 order
Diffraction grating
(a) bright fringes – constructive interference
1 order
st
2nd order
Screen
(i) decreasing the grating spacing
(ii) using light of longer wavelength
Example 2
A source emits blue light of wavelength 450 nm at a plane transmission grating of 250 lines
per mm.
(a) Find the grating spacing of the grating.
(b) Find the angle of the 1st order bright fringe.
(c) Find the total number of bright fringes on the screen.
Solution
(a) grating spacing
= distance between adjacent lines on
the grating
(c) By d sin θ = nλ⇒ sin θ =
nλ
d
∵
sin θ ≤ 1
=
∴
nλ
≤1
d
d sin θ = nλ
4 × 10 −6
≈ 8.89
λ 450 × 10 −9
Maximum number of order = 8
10 −3
= 4 × 10 −6 m
250
(b) For 1st order bright fringe, n = 1
(1)( 450 × 10 −9 )
4 × 10 −6
θ = 6.46°
sin θ =
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n≤
d
=
Total number of fringes on the screen
= 8 + 1 + 8 = 17
66
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
St. Joseph’s Anglo-Chinese School
Nodal lines and antinode
lines become closer together
67
St. Joseph’s Anglo-Chinese School
68
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
St. Joseph’s Anglo-Chinese School
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
72
(b) Electric field strength and gravitational field strength
Definition
Equation
Unit
Gravitational field strength g
Electric field strength E
Gravitational force per unit mass
Electric force per unit charge
F
E=
q
g=
F
m
GM
g= 2
r
Isolated point charge:
E=
Parallel plates:
E=
N kg-1
N C-1
Vector
Vector
Q
4πε 0 r 2
σ
ε0
Example 3
Two point charges A (3 × 10–8 C) and B (–2 × 10–8 C) are placed 20 cm
apart. Find the magnitude ofthe 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
=
St. Joseph’s Anglo-Chinese School
Eresultant
30o
30o
EC
EB
EC
EC cos 30o
EB cos 30
60°
EB
o
30o 30o
EC sin 30o
EB sin 30o
73
1.4 Electric Potential
Definition
The electric potential V at a point is the electric potential energy of a unit
positive test charge placed at that point.
PE
V=
q
Equation
Isolated point charge V =
Q
4πε 0 r
Parallel plate V = Ed (magnitude only)
Unit
V
Scalar
Example 5
Overall electric potential
Two charges A (3 × 10–8 C) and B (–5 × 10–8 C) are
shown. Find the electric potential at point X.
Solution
By V =
Q
,
4πε 0 r
QA
3 × 10 −8
Electric potential due to A =
=
= 1206 V
4πε 0 r 4π (8.85 × 10-12 ) 0.2 2 + 0.12
Electric potential due to B =
QB
− 5 × 10 −8
=
= −1422 V
4πε 0 r 4π (8.85 × 10-12 ) 0.32 + 0.12
Overall electric potential at X =1206 + (-1422) = -215 V
Remark:
1. A positive charge gives a positive potential while a negative charge gives a negative potential
2. Unlike electric field, resultant electric potential can be obtained by direct addition because
potential is a scalar.
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Example 6 electric potential of parallel plate
An electron is projected from the positive plate to the negative plate with initial speed
1.5 × 106 m s−1. The electric field strength between the two parallel plates is 5000 N C–1.
+
–
10 cm
(a) Find the electric potential difference between the plates.
(b) Find the gain in potential energy by the electron if it reaches the negative plate.
(c) Can the electron reach the negative plate?
(Given the mass of an electron = 9.1 × 10–31 kg, the charge of an electron = –1.6 × 10–19 C)
Solution
(a) V = Ed = (5000)(0.1) = 500 V
PE
(b) V =
q
PE = Vq = (500)(1.6 × 10 −19 ) = 8 × 10 −17 J
1 2 1
mv = (9.1 × 10 −31 )(1.5 × 10 6 ) 2 = 1.02 × 10 −18 J
2
2
initial KE < PE required
(c) Initial KE = =
∵
∴
the electron cannot reach the negative plate.
1.5 Equipotential lines
6V
0V
6V
0V
6V
0V
6V
0V
Note:
(a) No work is due when a charge moves along an equipotential line. Its electric potential
energy remains unchanged.
(b) Equipotential lines and electric field lines are perpendicular to each other.
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1.6 Relationship between electric field strength and electric potential
Consider the E-field of parallel plates.
V = –Ex
+
The negative sign expresses that V decreases
–
along the direction of E-field
E
Slope of a V – x graph is -E
x
Example 7
A voltage of 1000 V is applied across a pair of parallel plates as shown below.
+
–
(a) Draw the electric field between the plates.
(b) If the distance between the two plates is 4 cm, sketch a V – x graph of the plates. Take the
potential of the negative plate as zero and take the direction to the right as positive.
(c) Find the E-field between the plates.
Solution
(a)
+
–
E
Electric field lines go from +ve plate to –ve plate
V/V
(b) E-field is uniform
1000
V – x graph is linear (straight line)
When x = 0 (at +ve plate), V = 1000 V
When x = 4 cm = 0.04 m (at –ve plate), V = 0 V
(c)
0
Method 1
Method 2
By V = Ed
Slope of V – x graph = –E
1000 = E(0.04)
-1
E = 25000 Vm (to the right)
St. Joseph’s Anglo-Chinese School
0.04
x/m
–1000/0.04 = –E
E = 25000 Vm-1 (to the right)
76
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
St. Joseph’s Anglo-Chinese School
Conductor Y
(Lower resistance, smaller slope)
78
Example 3
Remark
1. Ammeters must be connected in series.
2. Voltmeters must be connected in parallel.
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79
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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80
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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81
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
N
x
N
uniform
field
4.2 Magnetic field due to current carrying conductors
Current-carrying
conductor
Magnetic flux
density (B)
B=
Long straight wire
Magnetic field pattern
current direction
up out of page
µ0 I
2π r
wire (top view)
B=
Circular coil
µ 0 NI
2r
solenoid
N
Solenoid
B=
µ 0 NI
l
S
= µ 0 nI
current direction
St. Joseph’s Anglo-Chinese School
Inside the solenoid:
Uniform field
Outside the solenoid:
Similar to that around a bar magnet
85
Example 1
Magnetic field due to a long straight wire
A straight wire X carries a current of 5 A (Fig a).
(a) Find the magnitude of the magnetic field at the point P.
(b) Another straight wire is put near wire X such that P is the mid-point between two wires (Fig b).
Find the magnitude of the resultant magnetic field at P if the currents flow
(i)
in opposite direction.
(ii) in the same direction.
I=5A
8 cm
I=5A
8 cm
8 cm
Fig a
Fig b
I=2A
Given the permeability of free space µ0 = 4π × 10–7 T m A–1
Solution
(a)
B=
µ0 I
2πr
=
(4π × 10−7 )(5)
= 1.25 × 10− 5 T
2π (0.08)
(b) BX = 1.25 × 10–5 T
BY =
(i)
µ0 I
2πr
=
(4π × 10−7 )(2)
= 5 × 10− 6 T
2π (0.08)
BX and BY have the same direction.
The magnitude of the resultant magnetic field is
BX + BY = 1.25 × 10–5 + 5 × 10–6
= 1.75 × 10–5 T
(ii) BX and BY have opposite direction.
The magnitude of the resultant magnetic field is
BX – BY = 1.25 × 10–5 – 5 × 10–6
= 7.5 × 10–6 T
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86
Example 2
Magnetic field of a solenoid
A solenoid of 30 cm has 180 turns of coil. A current of 5 A passing through the solenoid.
(a) Identify the magnetic poles at the end A and B.
(b) Find the magnitude of the magnetic field at the centre of the solenoid.
(c) If the size of the current is halved, what happens to the answer in (b)?
B
A
(d) Give three ways to increase magnetic field strength of a solenoid.
Solution
(a) A: south pole
B: north pole
(b)
B=
µ 0 NI
l
=
(4π × 10 −7 )(180)(5)
= 3.77 × 10 − 3 T
0.3
(c) The answer in (b) will be halved.
(d) (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
4.3 Magnetic force acting on a current-carrying conductor
Magnetic force F = BIl
F = BIl
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 )
When the current I is at an angle θ to the magnetic field B , the magnetic force F = BIl sin θ.
Note:
when θ = 90o (conductor ⊥ field), F = BIl.
l
θ
I
St. Joseph’s Anglo-Chinese School
B
when θ = 0o
(conductor // field), F = 0.
87
Example 3
Two straight parallel wires X and Y are 20 cm apart. Wire X carries current 3Aand wire Y carries
current 6A in the opposite direction.
(a) Draw the magnetic field produced by Y at the position of wire X and calculate the magnitude.
(b) Find the magnetic force per unit length acting on wire 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) magnetic field produced by wire Y
BY =
µ0 I
2πr
=
(4π × 10 −7 )(6)
= 6 × 10 − 6 T
2π (0.2)
BY
IX
(b) By F = BIl,
Magnetic force per unit length
F
= BI = (6 × 10 − 6 )(3) = 1.8 × 10 −5 Nm-1
l
(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.
Remark:
The magnetic force acting on X is due to the magnetic field produced by Y and the current
passing through X. Hence, F=BYIXlX.
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88
4.4 Magnetic force on a moving charge in a magnetic field
uniform
magnetic
field B
Circular path of
motion of +q
F = Bqv sin θ
uniform
+q
+q
magnetic
field B
Example 4
An electron moves in a circular path in a uniform magnetic field of field strength 0.01T.
Given that the charge and mass of an electron are -1.6 × 10-19 C and 9.1 × 10-31 kg respectively.
If the speed of the electron is 6 × 107 ms-1, find
(a) the radius of the circular path, and
(b) the period of circular motion.
Solution
(a) For the circular motion
magnetic force = centripetal force
v2
r
mv (9.1 × 10 −31 )(6 × 10 7 )
r=
=
= 0.0341 m
Bq
(0.01)(1.6 × 10 −19 )
Bqv = m
(b) v = rω ⇒ ω =
T=
2π
ω
=
v
6 × 10 7
=
= 1.76 × 10 9 rads-1
r 0.034125
2π
= 3.57 × 10 − 9 s
9
1.758 × 10
Notes:
(1) The magnetic force is always perpendicular to the motion of the charged particle; no work is
done by the magnetic field.
(2) Kinetic energy of the charged particle remains constant.
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89
4.5 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
4.6 Hall effect
(a) Drift velocity
The movement of an electron in a metal (i) without and (ii) with an electric field.
When a voltage is applied across a conductor, the electrons accelerate and collide with +ve
ions constantly. Finally, they attain a steady velocity (called drift velocity) of about 10–5 m s–1
and produce a current.
I = nAvq
Example 5
Find the drift velocity of electrons in a copper wire if the diameter of the wire is 1 mm and the
current flowing through the wire is 2 A.
Given: charge density of copper = 6 × 1028 m-3
Solution
By I = nAvq , A = πr2 = π (0.5 ×10-3)2
I
2
v=
=
= 2.65 × 10 − 4 ms-1
28
-3 2
-19
nAq (6 × 10 ) π × (0.5 × 10 ) × (1.6 × 10 )
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(b) Hall voltage
(i)
If a piece of current-carrying metal is placed in a perpendicular B-field, magnetic
force will act on the charges.
Electrons will accumulate on one side of the conductor and a potential difference
called the hall voltage is developed.
This phenomenon is known as the Hall Effect.
– – –
– – –
–
FB = Bqv
v
–
Hall voltage
I
B
+
(ii) If a semi-conductor with positive charge carries is used, positive charges will be
accumulated on the top surface. The Hall voltage will be reversed.
+ + +
+ + +
+
FB = Bqv
+
Hall voltage
v
I
B
(iii) Hall voltage VH =
–
BI
. To obtain a greater Hall voltage, the charge density (n)
nQt
should be smaller, such as in a semi-conductor and the thickness (t) should be
smaller.
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Example 6
A semiconductor strip carrying a current of 3 A and is put in a uniform magnetic field. The Hall
voltage is 8 µV. The top surface of the strip has a higher potential than the bottom surface. (Given
the charge of each charge carrier = 1.6 × 10–19 C)
B = 0.5 T
I
2 cm
5 mm
I=3A
top surface
(a) What is the sign of the charge carriers?
(b) Find the Hall electric field.
(c) What is the drift velocity of the charge carriers in the strip?
(d) Find the number of charge carriers per unit volume.
Solution
(a) By Fleming’s left-hand rule, the magnetic force is acting upwards. Therefore the charge
carriers accumulate on the top surface. As the top surface has a higher potential, the sign of the
charge carriers is positive.
V H 8 × 10 −6
(b) Hall E-field =
=
= 1.6 × 10 − 3 Vm-1
−3
d
5 × 10
(c) Electric force = magnetic force
E H q = Bqv
E H 1.6 × 10 −3
=
= 8 × 10 − 4 ms-1
v=
B
0 .5
BI
(d) By VH =
,
nQt
⇒
n=
BI
(0.5)(3)
=
= 2.34 × 10 26 m-3
−6
V H Qt (8 × 10 )(1.6 × 10 −19 )(5 × 10 − 3 )
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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 Magnetic flux and flux density
area A
If B-field ⊥ coil,
magnetic flux Φ = BA
magnetic field B
normal to A
B cos θ
B sin θ
If B-field is at an angle θ to the normal of the coil
θ
magnetic flux Φ = BA cos θ
magnetic
field B
If a coil of N turns is used, the total
magnetic flux through the coil, called Magnetic flux linkage = NΦ
magnetic flux linkage is equal to NΦ.
Unit
weber (Wb)
5.3 Faraday’s law
The induced e.m.f (e) is equal to the rate of change of magnetic flux, or the rate of flux cutting.
i.e
ε =−
∆Φ
∆Φ
(for a single-turn coil) ε = − N
(for a coil of N turns)
∆t
∆t
– ve sign denotes that induced e.m.f. oppose the change that starts it. (Lenz’s law)
Example 1
A 1-turn circular coil of radius 10 cm is placed perpendicularly
to a uniform magnetic field. The magnetic flux density decreases
at a rate of 0.02 T s–1.
(a) Draw the induced current on the loop.
(b) Find the magnitude of the induced e.m.f. in the loop.
Solution
(a)
I
(b) magnitude of induced e.m.f.
ε=
∆Φ
∆BA
∆B
=
=A
= π (0.1) 2 (0.02) = 6.28 × 10 − 4 V
∆t
∆t
∆t
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]
94
5.4 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
(3) Increase the area of coil within the magnetic field
(4) winding the coil on a soft-iron core
(5) rotate the coil at a higher speed
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Example 2
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Chapter 6
Transmission of Electricity
6.1 Alternating current a.c.
Formulae
I r . m. s. = I 2
(1)
2
V r . m. s.
= V r .m. s . I r .m. s .
R
I
V
(3) For sinusoidal a.c. I r .m. s. = 0 , V r . m. s. = 0
2
2
2
P = I r .m. s. R =
(2)
Example 1
A resistor of 3 Ω is connected to the following alternating currents. Find the average power for each
type of alternating current.
I/A
I/A
6
5
0
0
t/s
-2
t/s
-5
Type 1
Type 2
Solution
For type 1 a.c.:
6 2 + ( −2 ) 2
= 20 = 4.47 A
2
I r . m. s. =
2
Power = I r .m. s. R = ( 20 ) 2 (3) = 60 W
For type 2 a.c.:
I r .m. s . =
I0
2
=
5
2
= 3.54 A
2
Power = I r .m. s .
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2
 5 
R=
 (3) = 37.5 W
 2
97
6.2 Transformers
Vp N p
(a)
=
Vs
Ns
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 2
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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6.3 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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Chapter 1 Radiation and Radioactivity
1.1 Properties of alpha, beta and gamma radiation
α particles
Helium nuclei
β particles
Fast-moving electrons
waves
Nature
+2
–1
No charge
Up to 10% of speed of
Up to 90% of speed of
Speed of light
Charge
Speed
light
α or
4
2
Symbol
Range in air
γ rays
Electromagnetic
4
2
light
He
0
−1
β or
0
−1
e
γ
Several centimeter
Several metres
Over 100 m
Strong
Weak
Very weak
Thick and straight
Thin and twisted track
Tracks can hardly be
Ionizing power
Cloud chamber track
track
seen
Slight deflection
Large deflection
No deflection
Slight deflection
Large deflection
No defection
Electric deflection
Magnetic deflection
(Apply Fleming’s left
hand rule to determine
direction of
deflection)
××××
××××
××××
× × γ× ×
Weak
Penetrating power
Medium
Strong
25 mm lead block to
Paper
5 mm aluminium sheet
absorb half of
radiation
Photographic film,
Photographic film,
Photographic film,
spark counter, cloud
cloud chamber, GM
cloud chamber, GM
chamber, GM tube
tube
tube
Absorber
Detectors
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Example 1
To investigate the kind(s) of radiation emitted by a radioactive source, a Geiger-Muller counter is
placed close in front of the source and sheets of different absorbers are placed in turn between the
source and the counter. Three readings are taken at one-minute intervals for each absorber. The
following results are obtained:
Absorber
Recorded count rate/counts per minute
1st reading
2nd reading
3rd reading
Air
600
610
593
Paper
602
603
601
5 mm Aluminium
100
102
98
25 mm Lead
98
101
100
Table 1
(a) Explain why the three readings for each absorber are not identical.
(b) Explain how the above results show that the source emits radiation β only and it does not emit
α andγradiation.
(c) Account for the count rates for lead as absorber.
Solution
(a) Radiation is emitted at random.
(b) The count rates with air and paper as absorbers are about the same. This shows that α radiation
is not emitted.
The count rate decreases significantly when aluminium is used as an absorber. Hence, β
radiation is emitted.
There is no decrease in count rate when lead is used instead of aluminium. Hence, γ radiation
is not emitted.
In conclusion, the source emits β radiation only.
Remark: about presentation
α particles β particles γ particles α radiation β radiation γ radiation (c) The count rate for lead as absorber is due to the background radiation.
1.2 Background radiation
Everyone is exposed to a small amount of radiation from the environment all the time. Such
background radiation comes from either natural or man-made sources.
Cosmic rays
12%
Radioactive
Radioactive
Living bodies, and
materials in rocks
and soil 15%
gases
40%
food and drinks
15%
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Medical practice
17%
Nuclear discharge
and others
1%
102
Chapter 2 The Atomic Model
2.1 Atomic number, mass number and isotopes
Atomic number
Number of protons in the nucleus. Also called proton number.
Mass number
Total number of protons and neutrons in the nucleus. Also called nucleon
number.
Isotopes
Nuclides of an element with the same atomic number but different mass
number.
e.g. C-12 and C-14 are isotopes
2.2 Radioactive decay
Alpha decay
A
Z
Beta decay
X → ZA−−42Y + 42 He
e.g.
A
Z
U→ 23490Th + 42 He
238
92
X → Z +A1Y + −10 e
234
90
e.g.
0
Th → 234
91 Pa + −1 e
Gamma emission
A
Z
X * → ZA X + γ
e.g.
234
90
Th *→ 23490Th + γ
Example 1
The chart in the figure below shows part of a radioactive series.
(a) Write down equations representing the radioactive decay D1 and D2.
(b) What are the values of A and Z in ZA X ?
(c) Which of the above nuclides are isotopes?
Solution
(a)
234
4
D1 : 238
92 U → 90 Th + 2 He
234
0
D 2 : 234
91 Pa → 92 Th + −1 e
(b) A = mass number = 226
Z = atomic number = 88
(c)
238
90
U and
238
92
U,
230
90
Th and
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234
90
Th
103
Example 2
A decay series
229
Plutonium-241 ( 241
94 Pu ) changes to thorium-229 ( 90Th ) by emitting α and β particles.
Pu → L → 22990Th
How many α and β particles are emitted in the decay series?
Solution
241
94
Method 1
Method 2
241 − 229
No. of α particles emitted =
=3
4
241
94
[decrease in mass number is only due to the
emission of α particles]
No. of β particles emitted
= 90 – (94 – 3 × 2) = 2
Pu →
Th + a 42 He + b
229
90
0
−1
e
Consider mass number:
241 = 229 + 4a + 0b ⇒ a = 3
Consider atomic number:
94 = 90 + 2a – b ⇒ b = 2
No. of α and β particles emitted are 3 and 2
respectively.
2.3 The decay curve
Activity A
A0
½A0
¼A0
⅛A0
N = N0e–kt
Half-life t 1
2
A = A0e–kt
Time taken for half of the nuclei in a sample to decay.
Time taken for the activity of a sample to fall to half.
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Example 3
The decay curve of a radioactive source is shown below.
activity / s
–1
time / day
(a) Find the activity of the radioactive source at t = 15 days.
(b) Find the half-life of the radioactive source.
Solution
(a) Activity of background radiation = 25 s-1
Activity of the source at t = 15 days = 75 – 25 = 50 s-1
(b) Initial activity of the source = 225 – 25 = 200 s-1
200
+ 25 = 125 s-1
2
[read from the graph]
Expected activity recorded after 1 half-life =
∴
Half-life ≈ 7.5 days
Formulae
(1) Half-life t 1 =
2
ln 2
, k = decay constant
k
(2) N = N0e–kt
A = A0e–kt
Example 4
The initial activity of a radioactive sample is 1.8 × 107 Bq and its half-life is 1.5 hours.
(a) Find the decay constant of the sample.
(b) Find the activity of the sample after 10 hours.
Solution
(a) 1.5 hours = 1.5 × 3600 s = 5400 s
t1 =
2
ln 2
ln 2 ln 2
⇒k =
=
= 1.28 × 10 − 4 s-1
k
t1
5400
2
(b)
Method 1
(Apply A = A0e–kt)
Method 2
(Concept of half-life)
10 hours = 10 × 3600 s = 36000 s
10 hours = 6.667 half-life
By A = A0e–kt
Activity required
= (1.8 × 107 )e − (1.284×10
= 1.77 × 105 Bq
St. Joseph’s Anglo-Chinese School
−4
)(36000)
= (1.8 × 107) ×
1
2
6.667
5
= 1.77 × 10 Bq
105
2.4 Uses of radioisotopes
Uses
Radiation
Radiotherapy (treatment of γ radiation
cancer cells)
Reasons
Can penetrate deeply into the
body
β radiation or γ radiation with Ionizing power of β radiation or
short half-life
γ radiation is weak
e.g. Iodine-131
Tracers
(β source, t½ = 8 days)
Technetium-99
(γ source, t½ = 6 hrs)
Sterilization of syringes and γ radiation
other equipment
Thickness gauge
β radiation with long half-life
Smoke detector
α radiation with long half-life
e.g. americium-241 (t½ = 432 yrs)
Strong ionizing power of α
radiation
Air molecules are ionized easily
Archaeology (C – 14 dating) β radiation
2.5
(a)
(b)
(c)
Handling precautions
Store and transport all sources in a suitable lead container.
Always handle the sources with forceps.
Keep the sources at arm's length, and point it away from the human body, especially the
eyes.
(d) Carefully plan the experiments to minimize the time the source is used.
Handle radioactive sources with care!
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Warning sign of radioactivity
106
Chapter 3
Nuclear Energy
3.1 Nuclear fission and nuclear fusion
Fission
Fusion
A heavy nucleus splits up into smaller nuclei
Two nuclei join together to form a heavier
nucleus
e.g.
235
92
U + 01 n →
142
56
Ba +
91
36
Kr + 3 01 n
Nuclear reactor to generate electricity
Atomic bomb
e.g.
2
1
H + 31 H → 42 He + 01 n
The Sun undergoes nuclear fusion to give out
(Controlled reaction)
heat and light
(Uncontrolled reaction)
(Uncontrolled reaction)
Hydrogen bomb
(Uncontrolled reaction)
3.2 Chain reaction and controlled fission – nuclear reactor
Chain reaction (鏈式反應
鏈式反應)
鏈式反應
When a uranium nucleus splits, two or three
neutrons are emitted. These neutrons can carry on
splitting other uranium nuclei, which results in a
chain reaction.
Controlled fission – nuclear reactor
The energy released from controlled nuclear fission can be used to generate a huge amount of
electricity.
Component
Description
Nuclear fuel elements
Uranium oxide enriched with extra U-235.
Moderator
–
Pressurized water
–
To slow down neutrons. If the neutrons move too
fast, the chain reaction will stop.
Control rods
St. Joseph’s Anglo-Chinese School
–
Boron-steel
–
Can be raised and lowered
–
To absorbs neutrons to control the rate of fission.
107
Advantages and disadvantages of nuclear energy
Advantages
Disadvantages
Nuclear energy helps to solve the world’s future The consequence of an accident is extremely
energy shortage crisis.
serious.
There is no serious fuel transportation problem.
Nuclear energy will not be cheap if large sums
of money have to spend on maintaining and
upgrading the safety standards of the reactor.
Nuclear energy is in many cases cheaper than
Nuclear energy is not necessary. Future energy
coal or oil for generating electricity.
needs can be met by using alternative energy
sources.
Nuclear energy is clean and causes little
Any country which operates a nuclear reactor
environmental pollution.
can produce nuclear weapons.
3.3 Mass-energy relationship
Einstein’s theory of relativity points out that mass can change into energy
E = mc2
Example 1
Uranium-235 is used to generate electricity in a nuclear reactor of a nuclear power station. The
following shows the fission of U-235.
235
1
142
91
1
92 U + 0 n → 56 Ba + 36 Kr + 3 0 n
(a) Find the loss in mass after reaction and hence, find the amount of nuclear energy released.
(b) If 5 × 10–6 kg of U-235 undergoes fission every second, what is the rate of energy produced?
Given:
1u = 1.661 × 10–27 kg
mass of U-235 = 235.043 923 u,
mass of neutron = 1.008 665 u
mass of Ba-142 = 141.916 453 u, mass of Kr-91 = 90.923 442 u
Solution
(a) Total mass before reaction
= 235.043 923 u + 1.008 665 u = 236.052 588 u
Total mass after reaction
= 141.916 453 u + 90.923 442 u + 3 × 1.008 665 u = 235.865 890 u
Loss in mass = 236.052 588 u – 235.865 890 u
= 0.186 698 u = (0.186 698)(1.661 × 10–27) kg
= 3.10 × 10–28 kg
Energy released E = mc2
= (3.101 × 10–28)(3× 108)2 = 2.79 × 10–11 J
(b) no. of U-235 consumed in 1 s =
5 × 10 -6
= 1.281 × 1019
- 27
(235.043923)(1.661 × 10 )
Amount of energy released in 1 s = (1.281 × 1019 )( 2.791 × 10 −11 ) = 3.57 × 10 8 J
Rate of energy produced P =
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E 3.57 × 10 8
=
= 3.57 × 10 8 W
t
1
108
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