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SECONDARY SCHOOL IMPROVEMENT PROGRAMME (SSIP) 2019 GRADE 12 SUBJECT: PHYSICAL SCIENCE HIGH FLYERS LEARNER GUIDE (Page 1 of 73)
SECONDARY SCHOOL IMPROVEMENT
PROGRAMME (SSIP) 2019
GRADE 12
SUBJECT: PHYSICAL SCIENCE
HIGH FLYERS
LEARNER GUIDE
(Page 1 of 73)
SESSION 1: ELECTROSTATICS
REVISION: GRADE 10 WORK Static electricity is stationary electricity i.e. there is no movement of electrical charges. Electrostatics is the study of static electricity where we try to find out what effect do charges at rest have on one another. TYPES OF CHARGES:
1. Positive charge – A positive charge on an object originates with the removal or
shortage of electrons, i.e. cation
2. Negative charge – A negative charge on an object originates with the addition or
surplus of electrons, i.e.anion
N.B Neutral object - when the number of electrons (negative charges) is equal to the number of protons (positive charges). Charged objects exert forces on each other:
Like charges repel each other – Repulsion
Unlike charges attract each other – Attraction
NB! Use the Van de Graaf generator to demonstrate the effect of charges. Objects can be charged by
Friction (rubbing):
A plastic ruler becomes positively charged when it is rubbed with a dry cloth
because the plastic ruler transfers electrons to the cloth
A glass rod becomes negatively charged when rubbed with a dry cloth because
the dry cloth transfers electrons to the glass rod.
Induction :( polarization of charges) - the act or process by which an electric or magnetic effect is produced in an electrical conductor or magnetizable body when it is exposed to the influence or variationof a field of force.
NB! Use glass and plastic rods to do simple rubbing and induction experiments.
The law of conservation of charge The algebraic sum of the charges remains constant in a closed system.
n
...QQQ 21
Unit of charge
Charge is measured in units called coulombs (C).
A coulomb of charge is a very large charge.Charge is quantized, i.e. found in packages. Principle of quantization of charge- all charges in the universe consists of integer multiple of the charge on an electron, i.e. 1,6x10-19C.
In electrostatics we therefore often work with charge in Micro coulombs (1 × 10−6 C) Nano coulombs (1 × 10−9 C)and Pico coulomb (1 x 10-12C).
Revision Grade 11: Vectors Learners should be able to identify forces and be able to draw free body diagrams.
Coulomb’s Law
The electrostatic force of attraction or repulsion that two charges at rest exert on each other
is directly proportional to the product of the magnitude the two charges and inversely
proportional to the square of the distance between their centres.
Fnet=KQ1Q2
r2
Where:
F is the force in Newton’s (N)
Q1 and Q2are charges in coulombs (C)
r is the distance between the two charges in metres (m)
kis proportionality constant (Coulombs Law constant) with the value of
9x109N∙m2.C-2
NB! Solve problems:
Using the equation 2
21
r
QkQF F for charges in one dimension (1D) – restrict to three
charges.
Using the equation 2
21
r
QkQF for charges in two dimensions (2D) – for three charges
in a right-angled formation (limit to charges at the 'vertices of a right- angled triangle').
Worked Example 1
Two point-like charges carrying charges of +3 nC and −5 nC are 2m apart. Determine the
magnitude of the force between them and state whether it is attractive or repulsive.
Solution 1
Step 1: Data
Step 2: Suitable Equation
No need to substitute the negative
sign on the negative charge
charge.charge
F =? ; Q1= +3 nC = +3 × 10−9C; Q2= -5 nC = −5 × 10−9C;
r = 2 m; k= 9x109N∙m2.C-2
Step 3: Determine the magnitude of the force
Thus the magnitude of the force is 3, 38 × 10−8N. However since both point charges have opposite signs, the force will be attractive. Worked example 2 Two tiny spheres A and B with charges – 4 nC and +4 nC respectively, are placed as in the sketch. They are fixed while a third charge C of + 4 nC has its centre placed at a distance of 40 mm from the centres of both A and B. 1 nC = 1 nano coulomb = 1 x 10 – 9 C. (frictionless)
(a) Calculate the magnitude of the Coulomb force that exists between charge A of – 4 nC and the charge C of +4 nC charge. (b) In which direction will the sphere C move? (c) Calculate the magnitude of the acceleration with which the sphere C will start moving from its position as indicated, if it has a mass of 200 g. Solution 2 Draw a free body diagram showing the forces acting on charge C. For C, take as positive the direction to the left. Let C interact with A first, and then with B
FAC=kQ1Q2
r2 =
(9𝑥109) (4𝑥10−9)(4𝑥10−9)
(40𝑥10−3) = 9𝑥10−5N to the left (Attractive)
(b) FAB = 9x10-5N to the left (Repulsive)
(c) a= Fres
m=
18x10-5
200x10-3=9x10-4m.s-2
ELECTRIC FIELD AROUND CHARGES Electric field An electric field is a region around a charge where any other charge experiences a force. The direction of the electric field at a point is the direction that a positive test charge would move if placed at that point. NB! Revision of Grade 9 and 10 Magnetism. Some important points to remember about electric fields:
They originate and end perpendicularly to the surface of the charged objects.
Field lines never cross.
They are most dense (closer to each other) where the field is the strongest and is least dense (further from each other) where the field is the weakest.
F = kQ1Q2
r2
F = kQ1Q2
r2
= (9, 0× 109)(3 × 10−9)(5 × 10−9)
(2)2
= 3, 38 × 10−8N
They surround the charged object in three dimensions. We only draw a few lines in one plane
There is a uniform field (except at the end points) between two oppositely charged parallel plates.
Electric field lines around a positive point charge:
For a positive point charge, field lines are drawn away from the charge.
Electric field lines around a negative point charge:
For a negative charge, field lines are drawn towards the charge.
Electric field lines of two equal but opposite charges:
Electric field lines of two equal positive charges:
Electric field lines of two equal negative charges:
Electric fields lines between two oppositely charged parallel plates:
Strength of an electric field The test charge placed at a point in electric field will experience a force; the magnitude of the force experienced will depend on the distance of the test charge(q) away from the charge(Q) setting the field. The magnitude of the electric field (electric field strength) at a point in an electric field is the force per positive unit charge (thus + 1C) at that point.
Electric field strength=force
charge
Or in symbols, q
FE
Unit: if the electrostatic force F acts on the charge in Newton (N), the charge q is in coulomb (C), then the electric field strength (E) is in N∙C-1 (Newton per Coulomb). As we will see later, the electric field strength can also be measured in Volt per metre (V∙m-1). Direction: electric field strength is a vector quantity because it has magnitude and direction. A direction of electric field strength E at a specific point in an electric field is the same as the direction of the electrostatic force that a positively charged particle will experience at that point. The positively charged particle will thus move in the direction of the field and a negatively charged particle will move against the field. The force experienced by a test charge when placed in an electric field is given by;
F=qE Worked example 3 If the magnitude of the electric field strength (intensity) is 3 x 10 6 N • C–1 at a point, calculate the magnitude of the force acting on a charge of – 7 nC placed at that point. Electric field at a point due to a number of point charges
F = qE
= 7 × 10-9 × 3 × 106
= 0.021 N
The force between two electric charges is given by:
F = kQq
r2 (If we make the one charge Q and the other q.)
Therefore, the electric field can be written as
𝐸 = 𝑘𝑄
𝑟2 (where Q is the charge setting the field)
N. B. As with Coulomb’s law calculations, do not substitute the sign of the charge into the equation for electric field. Instead, choose a positive direction, and then either add or subtract the contribution to the electric field due to each charge depending upon whether it points in the positive or negative direction, respectively. Worked Example 4 Question: Calculate the electric field strength 30cm from a 5nC charge.
Solution 4 Step 1:Data. Q = +5nc = +5 × 10-9 C
r = 30 cm = 0.30 m k = 9x109N∙m2•C-2
Step 2: Select a suitable equation
E=kQ
r2
Step 3:substitute into equation:
Worked example 5 R and S are two points in the electric field of a small negatively charged sphere Q, Two charges of Q1 = +3nC and Q2 = −4nC are separated by a distance of 40cm. What is the electric field strength at a point that is10cm from Q1 and 30cm from Q2? The point lies between Q1 and Q2.
𝐸 = 𝑘𝑄
𝑟2
= (9 x 10-9) (5 x 10-9
(0, 3)2
=5 × 10-16 N∙C-1
Solution 5 Step 1: Determine what is required: We need to calculate the electric field a distance from two given charges. Step 2: Determine what is given: We are given the magnitude of the charges and the distances from the charges. Step 3: Determine how to approach the problem: We will use the equation:
𝐸 = 𝑘𝑄
𝑟2
We need to work out the electric field for each charge separately and then add them to get the resultant field. Step 4: Substitute into an equation: We first calculate E at x due to Q1: Then for Q2: We need to add the two electric fields because both are in the same direction. The field is away from Q1 and towards Q2. Therefore, Etotal= 2,70 × 103 + 4,00× 102 = 3,10 × 103N∙C−1
𝐸 = 𝑘𝑄
𝑟2
= (9 × 109) (4 × 10−9)
(0.3)2
= 4,00× 102 N∙C−1
𝐸 = 𝑘𝑄
𝑟2
= (9 × 109) (3 × 10−9)
(0, 1)2
= 2, 70 × 103 N∙C−1
POTENTIAL DIFFERENCE
The potential difference between two points in an electric field is defined as the work required to move a unit positive test charge from the point of lower potential to that of higher potential.
OR
The electrical potential difference is the difference in electrical potential energy per unit
charge between two points ( 𝑉 = 𝑊
𝑄).
Unit: The electrical potential difference is measured in volts (V). Using the equation 𝑉 = 𝑊
𝑄
the unit will be joule per coulomb (J∙C-1) which is the same as volt, thus electrical potential difference is also called voltage. Worked example 7: What is the potential difference between two points in an electric field if it takes 600J of energy to move a charge of 2C between these two points? Solution 7 Step 1: Data. Step 2: Suitable equation:
𝑉 = 𝑊
𝑄
Step 3: Substitution:
𝑉 = 𝑊
𝑄
= 600 2 = 300V
PAST EXAM QUESTIONS QUESTION 1 GP 2015 Two identical conducting spheres A and B with charges of Q1and Q2respectively are placed in fixed positions along the same straight line as shown in the diagram below. Spheres A and B are placed 30 cm from each other. Point P is positioned 30 cm to the right of sphere B on the same straight line.
The charge on sphere B is positive. The net electric field Enetat point P as a result of the two charges Q1and Q2is towards the right as shown in the diagram below.
1.1 What is the sign of the charge on sphere A? Give a reason for the answer. (3)
1.2 The net electric field at point P is 1 600 N⋅C-1 to the right and the charge on
sphere B has a magnitude of +12 nC. Calculate the magnitude of the charge on sphere A.
(7)
A proton is placed at point P without changing the charges and positions of spheres A and B. 1.3 Calculate the net electrostatic force experienced by the proton. (4)
[15]
QUESTION 2 DBE JUNE 2015 Two identical neutral spheres, M and N, are placed on insulating stands. They are brought into contact and a charged rod is brought near sphere M.
When the spheres are separated it is found that 5 x 106 electrons were transferred from sphere M to sphere N. 2.1 What is the net charge on sphere N after separation? (3) 2.2 Write down the net charge on sphere M after separation. (2) The charged spheres, M and N, are now arranged along a straight line, in space, such that the distance between their centres is 15 cm. A point P lies 10 cm to the right of N as shown in the diagram below. 2.3 Define the electric field at a point. (2) Calculate the net electric field
at point P due to M and N. (6) [13]
2.4 Calculate the net electric field at point P due to M and N. (6) [13]
QUESTION 3 DOE FEB/MAR 2016 A sphere Q
1, with a charge of -2,5 μC, is placed 1 m away from a second sphere Q
2, with a
charge +6 μC. The spheres lie along a straight line, as shown in the diagram below. Point P
is located a distance of 0,3 m to the left of sphere Q1, while point X is located between Q
1
and Q2
. The diagram is not drawn to scale.
3.1 Show, with the aid of a VECTOR DIAGRAM, why the net electric field at
point X cannot be zero. (4)
3.2 Calculate the net electric field at point P, due to the two charged spheres Q1
and Q2.
[10]
(6)
QUESTION 4: DOE NOVEMBER 2009 Two metal spheres on insulated stands carry charges of +4 uC and -6 uC respectively. The spheres are arranged with their centres 40 cm apart, as shown below.
4.1 Calculate the magnitude of the force exerted by each sphere on the other. (4) 4.2 By what factor will the magnitude of the force in QUESTION 4.1 change if the distance between the spheres is halved? (Do not calculate the new value of the force.) (1) 4.3 Calculate the net electric field at point P as shown in the diagram above. (6) 4.4 The spheres are now brought into contact with each other and then returned to their original positions. Now calculate the potential energy of the system of two charges. (5)
[16]
SESSION 2 :CURRENT ELECTRICITY PRIOR KNOWLEGDE An atom is the smallest particle of an element.
CHARGE
The unit of charge is the Coulomb. A Coulomb ( C ) consists of 6,25 x 1018 electrons.
A coulomb is the quantity of charge that passes through a conductor when a current of one ampere flows for one second.
Electrical charges may be positive or negative. A positively charged object is caused by a deficiency of electrons while a negatively charged object is caused by an excess of electrons. Charges carry energy to the components of a circuit.
ELECTRICAL SYMBOLS
ENERGY SOURCES
Electrical energy is provided by sources such as a cell, battery or dynamo. A battery is a combination of cells.
Cells can be connected in series or parallel.
ELECTROMOTIVE FORCE (ℰ)
The emf of a cell is the maximum quantity of electrical energy that can be supplied per coulomb of charge. OR
It is the work done per unit charge by the source. It is equal to the potential difference measured across the terminals of the batterywhen no current is flowing through the cell or circuit.
The symbol for emf is ℰ. Cells connected in series:
Cells are connected in series when their terminals are connected in a positive to negative sequence:
The total emf of a battery of cells connected in series is equal to the algebraic sum of the emf’s of the individual cells.
ℰ T = ℰ 1 + ℰ 2 + ℰ 3 + … NB: The advantage of connecting cells in series is that the total emf is increased thus producing a more energy.
A V Ammeter Voltmeter Bulb Resistor
Rheostat (variable
resistor)
cell Switch
Cells connected in parallel: Cells are connected in parallel when their positive terminals are connected to each other and the negative terminals are connected to each other as indicated in the sketch below: The total emf of a battery of cells connected in parallel is equal to the emf of a single cell, provided that all the cells have the same emf. NB:The advantages of connecting cells in parallel are that the cells last longer. The total emf of a battery of cells connected in parallel is equal to theemf of the individual cells.
ℰT = ℰ1 = ℰ2 = ℰ3 = … Examples: Each of the cells below has an emf of 2 V. Calculate the emf of each of the following batteries: 8 V 2 V 6 V SIMPLE CIRCUIT A simple circuit consists of a cell, conducting wires, resistor and a switch.
POTENTIAL DIFFERENCE (V) The potential difference between two points in a circuit is the energy required to move one coulomb of charge between the points. It is measured using a voltmeter ‘v’ which should always be connected in parallel to the component / resistor as illustrated in the diagram below.
From the definition, the formulae for calculating potential difference is
Q
WV , where V = potential difference (SI unit is V),
W = work done (SI unit is J) and Q = charge (SI unit is C).
Example 1: Calculate the potential difference between two points if 20 J of work are required to move a charge of 2 C.
V 22
20
Q
WV
Example 2: Calculate the work done in moving a charge of 5 C through a potential difference of 2V.
W = QV = 5 x 2 = 10 J CURRENT (I) An electric current is the rate of flow of charge (positive or negative) from one point to another in an electrical circuit. It is measured in Amperes (A). An ammeter measures the strength of an electric current. It has a low resistance and is connected in series in a circuit. Conventional current is the flow of positive charge and its direction is from the positive terminal through the circuit to the negative terminal of a cell. Since a current in a metal is conducted by negative electrons, the electron current flows from negative to positive. However, current direction in metals is always based on the direction
As the charges move through a resistor,
from point A to B they release energy.
The voltmeter reads the difference in the
potential energy of the charges at point
A and point B. Current will not flow
through a voltmeter because the
voltmeter has a high resistance.
Current in a series circuit is
the same at all points.
Therefore, the ammeter
readings in A1 andA2 is the
same.
that a positive charge would take and is therefore the direction of a conventional current, i.e. from positive to negative. From the definition, the formula for calculating current is
t
QI , where Iis the current (SI unit is A),
Q is the charge (SI unit is C) and t is time (SI unit is s).
NB: (The use of the abbreviations “amp” and ‘sec’ is incorrect) Example 1: Calculate the current strength when 5 C of charge passes a given point in 2 s.
A 5,22
5
t
QI
Example 2: Calculate the quantity of charge passing a point in a circuit when a current of 5 A flows for 10 s.
Q = It = 5 x 10 = 50 C CONDUCTION IN A METAL In a metal, the atoms are packed closely in a crystal lattice. The outermost electrons of the metal atoms are held loosely and can escape the attractive forces of their nuclei to form positively charged metal ions. These electrons are called free (delocalized) electrons and move around at random. If a potential difference is applied across the ends of the conductor, an electric field is set up
in the conductor. Negatively charged electrons are attracted to the positive terminal while the
metal ions remain stationary because they are held in a crystal lattice. The movement of
electrons is impeded by collisions with the positive metallic ions causing resistance. Every
electron that leaves a conductor at the positive terminal is replaced by another from the
negative terminal. The overall charge of a conductor is therefore neutral.
RESISTANCE (R) Resistance is the ratio of the potential difference across a resistor to the current in the resistor. It is the opposition to the flow of charge. A good conductor has a low resistance and a poor conductor has a high resistance. Resistance is caused by collisions between electrons and metallic ions that interfere with the flow of charge.
+
+
+
+
+
+
+ +
+ +
+
+
+
+
+
+ + +
+ +
+
+
+ - -
- - -
-
-
-
- - -
- +
-
-
- - - -
-
-
-
-
- -
-
Mobile electrons Positive atomic residues (ions)
The following factors affect the resistance of a conductor: (1) type of metal used e.g. copper wire has less resistance that nichrome wire. (2) length of the conductor. The longer the conductor the higher the resistance. (3) thickness (or cross-sectional area). The thicker the conductor the lower the resistance. (4) temperature. The higher the temperature the higher theresistance.
The formula for calculating resistance is
I
V R , where R = resistance (),
V = potential difference (V) and I = current strength (A).
An ohm is equal to one volt per ampere. It is the resistance of a conductor when a potential difference of one volt causes a current of one ampere to flow through it.
Example:
Calculate the resistance of a conductor when a current of 2 A flows when the potential across its ends is 10 V.
52
10
I
VR
Resistors connected in series
Resistors connected in series act as potential dividers.
The total resistance of a combination of resistors connected in series is equal to the algebraic sum of the component resistors.
VT = V1 + V2 + V3 IR = IR1 + IR2 + IR3(Divide by the common factor I.)
RT = R1 + R2 + R3 Example:
RT = R1 + R2 + R3= 2 + 3 + 9 = 14
VT = V1 + V2 + V3
In the accompanying diagrams, a cell with an emf of 10 V and negligible internal resistance is
connected in parallel with a 2 resistor and a rheostat. The resistance of the rheostat is adjusted to vary the total resistance of the circuit thus varying the current and potential difference. Diagram 1 Diagram 2
The
resistance of the rheostat is increased from 3 in Diagram 1 to 8 in Diagram 2 causing a different division in potential difference across the resistors.
In diagram 1: RT = R1 + R2 = 2 + 3 = 5 Ω A 25
10
R
VI
V1 = IR = 2 x 2 = 4 V and V2 = IR = 2 x 3 = 6 V
VT = 4 +6 = 10V
In diagram 2: RT = R1 + R2 = 2 + 8 = 10Ω A 110
10
R
VI
V1 = IR = 1 x 2 = 2 V and V2 = IR = 1 x 8 = 8 V
VT = 2 + 8 = 10V OR
Diagram 1: V4105
21 VR
Rv V 610
5
32 VR
Rv
Diagram 2: V 21010
21 VR
Rv V 810
10
82 VR
Rv
This illustrates that when the resistance of the rheostat increases the potential difference across the rheostat also increases but the total potential difference in the circuit remains the same.
2
8
V1 V2
10 V
0
2
3
V1 V2
10 V
0
Resistors connected in parallel Resistors connected in parallel act as current dividers.
IT = I1 + I2 + I3
The reciprocal of the total resistance is equal to the sum of the reciprocals of the resistances of the component resistors.
IT = I1 + I2 + I3
321
321
R R
V
R
VV
R
V
T
T (Divide by the common factor, V.)
321
11
R
11
RRRT
Example:
0,67 R 6
9
6
621
1
1
3
1
6
11111
321 RRRRT
Consider the following combination of resistors connected in parallel that act as current dividers.
The current flowing through resistors connected in parallel will be inversely proportional to their resistances, i.e. the greater the resistance the smaller the current. The ratio of the resistances is 2: 1 (total 3). Two-thirds of the current will therefore flow down one resistor and the remaining one-third down the other resistor. The larger current will flow down the resistor having the smaller resistance.
6
3
1
1 resistor: A 233
2
2 resistor: A 133
1
OHM’S LAW
Ohm’s Law states that current strength is directly proportional to the potential difference between the ends of a given resistor provided that temperature remains constant.
Mathematical statement: For a given resistor, V I at constant temperature.
Mathematical formula: V = IR
Experiment to verify Ohm’s Law Set up the apparatus as shown in the accompanying diagram. Vary the current flowing through the resistor by adjusting the rheostat and take five readings of potential difference and current. Precaution: Keep the temperature of the resistor ‘R’ constant. Tabulate your results as follows:
Current (A) Potential Difference (V) 𝑽
𝑰 (Ω)
1 0,5 1,0 2
2 1,0 2,0 2
3 1,5 3,0 2
4 2,0 4,0 2
5 2,5 5,0 2
POWER
Power is the rate at which work is done.
t
WP , where P = power in watts (W), W = work in joules (J) and t = time (s).
Example 1:
Calculate the power when 100 J of electrical energy is used in 2 seconds.
W502
100
t
WP
V = RI + 0
y = mx + c
The gradient of the graph is equal to R.
Example 2:
Calculate the electrical energy used when a 100 W electric bulb burns for 2 minutes.
W = Pt = 100 x 2 x 60 = 12 000 J
Other formulae for power and work:
W = QV = VIt (Substituting It for V)
VIt
VIt
t
WP
R
V
R
VVVIP
2
(Substituting forIR
V)
P = VI = IR x I = I2R (Substituting IRfor V)
W = Pt = I2Rt
W = Pt = tR
V2
Electricity Cost
Deduce that the kilowatt hour (kWh) refers to the use of 1 kilowatt of electricity for 1 hour.
Calculate the cost of electricity usage given the power specifications of the appliances used, the duration and the cost of 1 kWh.
ESKOM or local municipality charges a tariff per kWh used. To calculate the cost of electricity, we multiply the kilowatt-hour (energy units) reading on the meter by the unit price per kilowatt-hour.
Energy units (E)= power(P) x time(t) Where: energy units are measured in kWh, power in kW and Time in hours Cost = E x price per kWh
NB: price per kWh must be in rands (R) Examples 1. A heater marked 2000W/ 3000W is switched on for 4 hours. For the first hour, it is on the
highest setting, and for the last 3 hours, it is on lowest setting.
1.1 How much energy does it transfer to the room in kWh? (9kWh) 1.2 What is the cost of heating the room if the electricity costs 70c per kWh? (R6.30)
2. The reading on the electricity meter shows that 30kWh of energy have been supplied to a house during one day. 2.1 What is the average power consumption (per hour) of the house in this time? (1,25 kW) 2.2 Electricity is charged at 70c per kWh. How much will the electricity bill be for that day? (R21,00)
3. It takes 4,05 X 105 J to boil 1250 ml of water. 3.1. How long will it take to boil this water if you use an electric kettle marked 2500 W?
(162 s) 3.2. If the cost of electricity is R1,78 per kWh, calculate how much it will cost to heat this
water. (20 c)
INTERNAL RESISTANCE - LOST VOLTS EFFECT Emfis the maximum energy dissipated by a battery per coulomb / per unit charge passing
through it.
OR
The maximum work done by a battery per coulomb / per unit charge passing through it. It is equal to the potential difference across the terminal of the battery when no charges are flowing in the circuit. Since a cell has resistance, energy is used to overcome the internal resistance when current flows. This results in a decrease in potential difference across the terminals because less energy per coulomb is available to drive a current around the external circuit. The decrease in potential difference is called the lost volts effect. Internal resistance is the opposition to the flow of charges within the cells that make up the battery. Emf = potential difference + energy used per coulomb to overcome internal resistance. ℰ = V + Ir where ℰ is the emf (V), V is the potential difference (V), Iis the current (A) and r is
internal resistance (). PROBLEMS INVOLVING CIRCUIT DIAGRAMS Example 1 In the accompanying circuit diagram, a battery has an emf of 30V. Resistance of the battery and wires is negligible Calculate: 1.1 The total resistance of the circuit. 1.2 The total current in the circuit. 1.3 The current through the 4Ω resistor Example 2: In the accompanying circuit diagram, each cell has an
emf of 2 V and internal resistance of 0,5 . 1. Calculate the readings on A, V1 and V2 when the
switch is open.
6,6 6
4
6
30 V
X Y
A
V1
V2
0,5
6
3
S
2. Calculate the following when switch S is closed: 2.1 the total resistance of the circuit. 2.2 the reading on A 2.3 the reading on V2 2.4 the charge passing through A in 2 minutes.
2.5 the power used by the 0,5 resistor. Solutions 1. V1 = 5 x 2 = 10 V V2 = 0 V and A = 0 A
2.1 Resistance of cells in series: r = r1 + r2 + r3 + r4 + r5 = 5 x 0,5 = 2,5
Resistance of resistors in parallel:
2 R 6
3
6
21
3
1
6
11
R
11//
1// 2RR
Total resistance of circuit: RT = R1 + R2 + R3 = 2,5 + 2 + 0,5 = 5
2.2 A: A 25
10
R
VI
2.3 V2: V = IR = 2 x 2 = 4 V (The resistance between the points to which the voltmeter is connected is equivalent
to the resistance of the parallel combination of resistors. The total current flowing through the two resistors is 2 A.)
2.4 Q = It = 2 x 2 x 60 = 240 C 2.5 P = I2R = 22 x 0,5 = 2 W.
QUESTION 1 Two identical resistors P and Q are connected in the circuit as shown below. The
cell has an emf ‘ℰ’ and negligible internal resistance. Switch ‘s’ is initially CLOSED.
Which ONE of the following combinations of changes will occur in Q and V when switch ‘s’ is OPENED?
CURRENT IN Q READING ON VOLTMETER V
A Decreases Increases
B Remains the same Remains the same
C Decreases Remains the same
D Remains the same Decreases (2)
QUESTION 2
Two identical resistors P and Q are connected in the circuit as shown below. The
cell has an emf ‘ℰ’ and an unknown internal resistance. Switch ‘s’ is initially
CLOSED.
Which ONE of the following combinations of changes will occur in Q and V when
switch ‘s’ is OPENED?
CURRENT IN Q READING ON VOLTMETER V
A Decreases Increases
B Remains the same Remains the same
C Decreases Remains the same(2)
D Remains the same Decreases
PAST EXAM QUESTIONS QUESTION 9 DOE FEB/MAR 2015 A battery of an unknown emf and an internal resistance of 0,5 Ω is connected to three resistors, a high-resistance voltmeter and an ammeter of negligible resistance, as shown below.
The reading on the ammeter is 0,2 A.
9.1 Calculate the: 9.1.1 Reading on the voltmeter (3) 9.1.2 Total current supplied by the battery (4) 9.1.3 Emf of the battery (5) 9.2 How would the voltmeter reading change if the 2 Ω resistor is removed from
the circuit? Write down INCREASE, DECREASE or REMAIN THE SAME. Explain the answer. (3)
Question 6 (ieb Trial 2013 – Hilton College) 6.1 In the circuit in Figure 1, the battery, of emf 15 V and negligible internal
resistance, is connected in series with two lamps and a resistor. The three
components each have a resistance of 12 Ω.
6.1.1 Write down the voltage across each lamp (no working required) (1) 6.1.2 Calculate the current through the lamps. (3) 6.2 A battery of emf∈ and internal resistance r is connected in series to a
variable resistor R and an ammeter of negligible resistance. A voltmeter is connected across R, as shown in the figure below.
6.2.1 State what is meant by the emf of the battery and under what conditions is this achieved by the battery, assuming the battery has internal resistance.
(2)
6.2.2 Explain why the voltmeter must be connected as shown in the diagram above?
(2)
6.3 A student wishes to measure ℰ and r. Using the circuit shown in the figure above the value of R is decreased in steps and at each step the readings V and I on the voltmeter and ammeter respectively are recorded. These are shown in the table.
6.3.3 Write down an expression relating V, I, ℰ and r. (1) 6.3.4 Draw a graph of V (on the y-axis) against I (on the x-axis) (7) 6.3.5 Use your expression in 6.3.3 and the equation for a straight line
(y=mx+c), to determine the values of ℰ and r from the graph. (The graph should be marked appropriately to indicate how you obtained your answer) (4)
Question 7 (ieb Trial 2011 – Hilton College)
The circuit diagram below shows a battery, with internal resistance r, connected to three resistors M, N and Y. The resistance of N is 2 ohms and the reading on the voltmeter V is 14 V. The reading on ammeter A1 is 2A and the reading on ammeter A2 is 1A. (The reading of the ammeter and wires may be ignored).
7.1 State Ohms law in words. (2) 7.2 How does the resistance of M compare with that of N? Explained how you
arrived at the answer. (2) 7.3 If the emf of the battery is 17 V, calculate the internal resistance of battery. (5) 7.4 Calculate the potential difference across resistor N. (3) 7.5 Calculate the resistance of Y. (4) Question 5 (IEB November 2013) Tommy has a toy ambulance which has a light, a siren and a motor. The circuit diagram for the electric circuit of the ambulance is given below. The resistances of the battery, ammeter, switches and connecting wires can be ignored.
When Tommy closes ONLY switch 1 (S
1) and keeps switch 2 (S
2) open, the light
bulb comes on while the ambulance moves at a constant speed. 5.1 Calculate the reading on the ammeter when only switch 1 (S
1) is closed.
Give your answer to 2 decimal places. (4) 5.2 Calculate the rate of energy transfer (power) in the bulb when only switch 1
is closed. (4) When Tommy closes both switches 1 (S
1) and 2 (S
2) the siren sounds and the
reading on the ammeter is 200 mA. 5.3 Calculate the potential difference across the motor. (3) 5.4 Calculate the new current through the bulb. (3) 5.5 Calculate the total resistance of the circuit. (3) 5.6 Calculate the resistance of the siren. (4) 5.7 How will the following change when the siren sounds? (Write only increase,
decrease or no effect). 5.7.1 The brightness of the bulb. (1) 5.7.2 The speed of the ambulance. (1)
5.8 Explain your answer to Question 5.7.1 with reference to one or more suitable formulae. (2)
5.9 Explain your answer to Question 5.7.2 with reference to one or more suitable formulae. (2)
Question 7 (ieb November 2014) 7.1 An electric circuit is set up as shown in the diagram below. The
resistances of the switch, ammeters and connecting wires are negligible. The voltmeters have very high resistance. The battery has an emf of 12 V and has significant internal resistance (r). The switch S1 is CLOSED. The ammeter A2 reads 0,2 A and the voltmeter V2 reads 5,5 V.
7.1.1 Define emf. (2) 7.1.2 Calculate the reading on ammeter 1 A. (4) 7.1.3 Calculate the resistance of resistor X. (3) 7.1.4 Calculate the total external resistance of the circuit. (3) 7.1.5 Calculate the internal resistance (r) of the battery (4) 7.1.6 Resistor X is replaced by a new resistor of greater resistance than that of X.
(a) Will the reading on the voltmeter 1 V connected across the terminals
of the battery increase, decrease or remain the same? (1) (b) Explain your answer to Question 7.1.6 (a), making reference to
relevant formulae. (4) 7.2 An electric kettle is rated 240 V; 1 800 W. 7.2.1 What does 'rated 240 V; 1 800 W' mean in regard to how this
kettle works? (2) 7.2.2 Calculate the current drawn by the kettle when connected to a
240 V source. (3) 7.2.3 Calculate the cost of using the kettle for 15 minutes if electricity
costs R1,40 per kWh. (3) Question 9 (DBE November 2015) A battery with an internal resistance of 1 Ω and an unknown emf (ε) is connected in a circuit, as shown below. A high-resistance voltmeter (V) is connected across the battery. A
1 and A
2 represent ammeters of negligible resistance.
With switch S closed, the current passing through the 8 Ω resistor is 0,5 A. 9.1 State Ohm's law in words. (2)
9.2 Calculate the reading on ammeter A1. (4)
9.3 If device R delivers power of 12 W, calculate the reading on ammeter A2. (5)
9.4 Calculate the reading on the voltmeter when switch S is open. (3) Question 8(Free State Trial 2015) 8.1 The graph below is obtained from an experiment to calculate the internal
resistance of a battery 8.1.1 Calculate Vinternal if the current in the circuit is equal to 0,2 A. (2) 8.1.2 Calculate the internal resistance of the battery. (4) 8.2 A circuit is connected as shown below. When switch S1 is closed, Vexternal is
22,5 V.
8.2.1 Define Ohm’s Law in words. (2) 8.2.2 Calculate the power dissipated by the 16 Ω resistor (7) 8.2.3 Calculate the resistance of R. (5) 8.2.4 Switch S2 is now closed. How will voltmeter reading V1 be influenced?
(Write down only INCREASE, DECREASE or STAYS THE SAME.) Give an explanation to your answer. (4)
SESSIO 3. ORGANIC CHEMISTRY
Representing Chemical Change
Balanced chemical equations
Write and balance chemical equations.
Interpret balanced reaction equations in terms of:
Conservation of atoms
Conservation of mass (use relative atomic masses)
Quantitative Aspects of Chemical Change
Molar volume of gases
1 mole of any gas occupies 22,4 dm3at 0 °C (273 K) and 1 atmosphere (101,3 kPa).
Volume relationships in gaseous reactions
Interpret balanced equations in terms of volume relationships for gases, i.e. under the
same conditions of temperature and pressure, equal number of moles of all gases
occupy the same volume.
Concentration of solutions
Calculate the molar concentration of a solution.
More complex stoichiometric calculations
Determine the empirical formula and molecular formula of compounds.
Determine the percentage yield of a chemical reaction.
Determine percentage purity or percentage composition, e.g. the percentage CaCO3 in
an impure sample of seashells.
Perform stoichiometric calculations based on balanced equations.
Perform stoichiometric calculations based on balanced equations that may include
limiting reagents.
Intermolecular Forces
Intermolecular forces and interatomic forces (chemical bonds)
Name and explain the different intermolecular forces (Van der Waal's forces):
i. Dipole-dipole forces:
Forces between two polar molecules
ii. Induced dipole forces or London forces:
Forces between non-polar molecules
iii. Hydrogen bonding:
Forces between molecules in which hydrogen is covalently bonded to
nitrogen, oxygen or fluorine – a special case of dipole-dipole forces
Describe the difference between intermolecular forces and interatomic forces
(intramolecular forces) using a diagram of a group of small molecules; and in words.
Example:
State the relationship between intermolecular forces and molecular size. For non-polar
molecules, the strength of induced dipole forces increases with molecular size.
Explain the effect of intermolecular forces on boiling point, melting point and vapour
pressure.
Boiling point:
The temperature at which the vapour pressure of a substance equals atmospheric pressure.
The stronger the intermolecular forces, the higher the boiling point.
Melting point:
The temperature at which the solid and liquid phases of a substance are at equilibrium.
The stronger the intermolecular forces, the higher the melting point.
Vapour pressure:
The pressure exerted by a vapour at equilibrium with its liquid in a closed system.
The stronger the intermolecular forces, the lower the vapour pressure.
Organic Molecules
Define organic molecules as molecules containing carbon atoms.
Organic molecular structures – functional groups, saturated and unsaturated
structures, isomers
Write down condensed structural formulae, structural formulae and molecular
formulae (up to 8 carbon atoms, one functional group per molecule) for:
Alkanes (no ring structures)
Alkenes (no ring structures)
Alkynes
Halo-alkanes (primary, secondary and tertiary haloalkanes; no ring structures)
Alcohols (primary, secondary and tertiary alcohols)
Carboxylic acids
Esters
Aldehydes
Ketones
Molecular formula: A chemical formula that indicates the type of atoms and the correct
number of each in a molecule.
Example: C4H8O
Structural formula: A structural formula of a compound shows which atoms are attached to
which within the molecule. Atoms are represented by their chemical symbols and lines are
used to represent ALL the bonds that hold the atoms together.
Example:
C C C C
OH
H
H H
H
H
H
H
Condensed structural formula: This notation shows the way in which atoms are bonded
together in the molecule, but DOES NOT SHOW ALL bond lines.
Example:
CH3CH2COCH3 OR
Hydrocarbon: Organic compounds that consist of hydrogen and carbon only.
Homologous series: A series of organic compounds that can be described by the same
general formula OR in which one member differs from the next with a CH2 group.
Saturated compounds: Compounds in which there are no multiple bonds between C atoms
in their hydrocarbon chains.
Unsaturated compounds: Compounds with one or more multiple bonds between C atoms
in their hydrocarbon chains.
Functional group: A bond or an atom or a group of atoms that determine(s) the physical and
chemical properties of a group of organic compounds.
Homologous Series Structure of functional group
Structure Name/Description
Alkanes
Only C–H and C–C single bonds
Alkenes
Carbon-carbon double bond
Alkynes
Carbon-carbon triple bond
Haloalkanes
Halogen atom bonded to a saturated C atom.
Alcohols
Hydroxyl group bonded to a saturated C atom
Aldehydes
Formyl group
Ketones
Carbonyl group bonded to two C atoms
Carboxylic acids
Carboxyl group
Esters
-
Structural isomer: Organic molecules with the same molecular formula, but different
structural formulae
Identify compounds (up to 8 carbon atoms) that are saturated, unsaturated and are
structural isomers.
Restrict structural isomers to chain isomers, positional isomers and functional isomers.
Chain isomers: Same molecular formula, but different types of chains, e.g.
butane and 2-methylpropane.
C C C C
H
H
H H
H
H
H
H
H
H
C C C
C
H
H
H
H
H
H
H
H
H
H
butane 2-methylpropane
Positional isomers: Same molecular formula, but different positions of the side
chain, substituents or functional groups on the parent chain, e.g. 1-choropropane
and 2-chloropropane or but-2-ene and but-1-ene
C C C
Cl
H
H
H H
H H
H C C C
Cl
H
H
H
H
H
H
H
1-chromopropane 2-chromopropane
C C C C
H
H H H
H
H
H
H
C C C C
H
H
H H
H
H
H
H
but-1-ene but-2-ene
Functional isomers: Same molecular formula, but different functional groups, e.g.
methyl methanoate and ethanoic acid
C O C
H
H
H H
H
H
C C
O
O
H
H
H H
methyl methanoate ethanoic acid
IUPAC naming and formulae
Write down the IUPAC name when given the structural formula or condensed
structural formula for compounds from the homologous series above, restricted
to one functional group per compound, except for haloalkanes. For haloalkanes,
maximum two functional groups per molecule.
Write down the structural formula when given the IUPAC name for the above
homologous series.
Identify alkyl substituents (methyl- and ethyl-) in a chain to a maximum of THREE
alkyl substituents on the parent chain.
When naming haloalkanes, the halogen atoms do not get preference over alkyl
groups – numbering should start from the end nearest to the first substituent,
either the alkyl group or the halogen. In haloalkanes, where e.g. a Br and a
Cℓ have the same number when numbered from different ends of chain,
Br gets alphabetical preference.
When writing IUPAC names, substituents appear as prefixes written alphabetically
(bromo, chloro, ethyl, methyl), ignoring the prefixes di- and tri.
Structure and physical properties (boiling point, melting point, vapour pressure)
relationships
For a given example (from the above functional groups), explain the relationship
between physical properties and:
Strength of intermolecular forces (Van der Waal's forces), i.e. hydrogen bonds,
dipole-dipole forces, induced dipole forces
Type of functional groups
Chain length
Branched chains
Factors that influence the strength of IMF
1. Surface area
Length of the carbon chain and branched molecules.
For compounds that belong to the same homologous series, the larger the surface area the higher the Melting and the Boiling point and the lower the Vapour pressure.
The more branched the organic molecules are, the more compact it becomes.
The surface area is smaller and less Van der Waals forces are available.
The IMF are weaker resulting in lower boiling points and melting points.
The vapour pressure will increase.
2. The type of functional group For compounds with comparable molecular mass (C-chain length) the functional group will be the determining factor regarding the strength of the IMF. The more polar the functional group the stronger the IMF. Carboxylic acid > Alcohol > Ketone, Aldehyde & Ester > Alkyne, Alkane & Alkene
The stronger the intermolecular force the higher the boiling point and melting point, lowervapourpressure
London forces
Dipole-dipole forces
Hydrogen Bond
Alkanes
Alkenes
Alkynes
Aldehydes
Ketones
Halo- alkanes
Esters
Alcohols ( 1 site )
Carboxylic acids ( 2 sites )
The strength of Melting Point and Boiling Point increases and the Vapour pressure decreases
The strength of Melting Point and Boiling Point decreases and the Vapour pressure increases
ORGANIC REACTIONS
Oxidation of alkanes
State the use of alkanes as fuels.
Write down an equation for the combustion of an alkane in excess oxygen.
Esterification
Write down an equation, using structural formulae, for the formation of an ester.
Name the alcohol and carboxylic acid used and the ester formed.
Write down reaction conditions for esterification.
Substitution, addition and elimination reactions
Identify reactions as elimination, substitution or addition.
Write down, using structural formulae, equations and reaction conditions for the
following addition reactions of alkenes:
Hydrohalogenation:
The addition of a hydrogen halide to an alkene
Halogenation:
The reaction of a halogen (Br2, Cℓ2) with a compound
Hydration:
The addition of water to a compound
Hydrogenation:
The addition of hydrogen to an alkene
Write down, using structural formulae, equations and reaction conditions for the
following elimination reactions:
Dehydrohalogenation of haloalkanes:
The elimination of hydrogen and a halogen from a haloalkane
Dehydration of alcohols:
Elimination of water from an alcohol
Cracking of alkanes:
The chemical process in which longer chain hydrocarbon molecules are broken down to
shorter more useful molecules.
Write down, using structural formulae, equations and reaction conditions for the
following substitution reactions:
Hydrolysis of haloalkanes
Hydrolysis: The reaction of a compound with water
Reactions of HX (X = Cℓ, Br) with alcohols to produce haloalkanes
Halogenation of alkanes
The reaction of a halogen (Br2, Cℓ2) with a compound
Distinguish between saturated and unsaturated hydrocarbons using bromine water.
REACTIONS OF ALKANES
1. OXIDATION (COMBUSTION) 2C6H14 + 19O2 12CO2 + 14H2O ∆ H < 0
Alkane + oxygen carbon dioxide + water + energy
Highly exothermic thus used as fuel.
2. SUBSTITUTION Alkane → haloalkane
C H
H
H
H Br Br+ C Br
H
H
H + H Br
Substitution reaction will only occur when the compound is saturated.
3. ELIMINATION Alkane → alkene(s) + alkane with shorter chain or
Alkane → alkene(s) + hydrogen
C C C C CH
H
H H
H H
H
H
H H
H
H C C CH
H
H H
H H
H
H + C C
H
H H
H
C C CH
H
H H
H H
H
H
Type of elimination:Thermal cracking
Conditions: High temperature (80 0 oC) and high pressure (in absence of oxygen)
Type of elimination:Catalytic cracking
Conditions: Mixture of SiO2 and Al2O3 as catalyst at lower temperature (500 oC) and lower
pressure.
Products: alkane + alkene(s) OR alkene + Hydrogen
Eliminationreaction will only occur when the compound is saturated.
REACTIONS OF ALKENES
ADDITION REACTION 1 (Halogenation)
C C
H
H H
H
+ C CH
Cl
H
Cl
H
HCl Cl
Bromine testto distinguish between alkane and alkene
Add bromine water (orange-brown) to unknown substances.
If bromine water discolours the substance is an alkene.
CH2CH2 + Br2 → CH2BrCH2Br
ADDITION REACTION 2 (Hydrogenation)
C C
H
H H
H
+ H H C CH
H
H
H
H
H
Application: Hydrogenation of unsaturated vegetable oils is used to manufacture
margarine.
Conditions: Heat OR
sunlight(uv)
Reactants: alkane + X2 (F, Cl,
Br, I)
Type of substitution:
halogenation(bromination)
Products: haloalkane + HX
Conditions: Unreactive solvent
Type of addition: halogenation
Reactants: alkene + X2 (X = Cl,
Br)
Product: haloalkane
Conditions: Pt, Pd or Ni
as catalyst
Type of addition:
hydrogenation
Reactants: alkene + H2
Product: Alkane
ADDITION REACTION 3 (Hydrohalogenation)
C C C
C
H
H
H H
H H
H
H+
C C C C
Br
H
H
H H
H H
H
H
H
H Brmajor product
C C C C
H
H
H
H H
H H
Br
H
H
minor product
Conditions:No water; Unreactive solvent Type of addition: hydrohalogenation Reactants: alkene + HX (X = I, Br, Cl) Product(s): haloalkane(s) Major product: The H-atom attaches to the C-atom already having the greater number of H-
atoms. (Markovnikov’s rule) ADDITION REACTION 4 (Hydration)
C C C
C
H
H
H H
H H
H
H
+
C C C C
OH
H
H
H H
H H
H
H
H
C C C C
H
H
H
H H
H H
OH
H
H
major product
minor product
H OH
Conditions:Excess H2O; Acid (H2SO4 / H3PO4) as catalyst. Type of addition: hydration Reactants: alkene + H2O Product: Alcohol(s) Major product: The H-atom attaches to the C-atom already having the greater number of H-
atoms. (Markovnikov’s rule)
Addition reaction will only occur when the compound is unsaturated.
REACTIONS OF HALOALKANES
1. ELIMINATION
Haloalkane → alkene
C C C CH
H
H H
Br H
H
H
H
H + NaOH
C C C C
H
H
H H
H
H
H
H
+heat NaBr + H2O
C C C C
H
H
H
H
H
H
H
H
major product
minor product
Conditions:concentrated strong basein ethanol as solvent (NaOH, KOH, LiOH) + heat Type of elimination: dehydrohalogenation Reactants: Haloalkane + concentrated strong base. Products: Alkene + NaBr + H2O Major product: The H-atom is removed from the C-atom with the least number of H-atoms
(most substituted double bond forms i.e. double bond with most alkyl groups)
2. SUBSTITUTION :
Haloalkane → alcohol
2.1
C C C C
BrH
H
H
H
H
H H
H
H + C C C C
OHH
H
H
H
H
H H
H
H +Na OH Na Br
Conditions:Diluted strong base (NaOH/KOH/LiOH) + mild heat
Type of substitution: hydrolysis/ hydration
Reactants:Haloalkane in ethanol + diluted strong base
Products: Alcohol + NaBr/KBr/LiBr
2.2
C C C C
BrH
H
H
H
H
H H
H
H +OH
HC C C C
OHH
H
H
H
H
H H
H
H + BrH
Conditions:Excess H2O + mild heat Type of substitution: hydrolysis Reactants: Haloalkane + H2O Products: Alcohol + HBr
REACTIONS OF ALCOHOL
1. ELIMINATION
Alcohol → alkene
C C C C
OH
H
H
H
H
H
H
H
H
H
C C C C
H
H
H H H H
H
H
C C C C
H
H H H
H
H
H
H
major product H OH+
minor product
Conditions: Dehydrating agent (conc H2SO4 + heat) Type of elimination: dehydration Reactants: Alcohol + conc H2SO4 Products: Alkene(s) + H2O Major product: The H-atom is removed from the C-atom with the least number of H-atoms
2. SUBSTITUTION
Alcohol → haloalkane
C C
H
H
H OH
H
H
BrH+ C C
H
H
H Br
H
H
+ OH2
Conditions: Heat Reactants needed: Alcohol + HX For primary & secondary alcohols: NaBr + conc H2SO4 is used to prepare HBr in reaction flask. For tertiary alcohols: HBr (or HCl) are directly applied.
Products: Haloalkane + H2O
ESTERIFICATION
C C C CH
H
H H
H H
H
H
H
O
H
+ C
O
O
C
H
H
H
H C C C C O
C
O
CH
H
H
H
H H
H H
H
H
H
H
+ O
H
H
Conditions:Concentrated H2SO4 as catalyst + heat
Reactants: Alcohol + carboxylic acid + H2SO4
Products: Ester + water
The process whereby esters is formed is called esterification.
Esters form when an alcohol reacts with a carboxylic acid when heated (condensation reaction).
The catalyst is concentrated sulphuric acid (H2SO4), a dehydrating agent that extracts the water.
The homologous series is an esters and the name, therefore, ends in “-oate”. The name of the example above is therefore, butylethanoate.
Plastics and polymers (ONLY BASIC POLYMERISATION as application of organic
chemistry)
Describe the following terms:
Macromolecule: A molecule that consists of a large number of atoms
Polymer: A large molecule composed of smaller monomer units covalently bonded to
each other in a repeating pattern
Monomer: Small organic molecules that can be covalently bonded to each other in a
repeating pattern
Polymerisation: A chemical reaction in which monomer molecules join to form a polymer
Distinguish between addition polymerisation and condensation polymerisation:
Addition polymerisation: A reaction in which small molecules join to form very large
molecules by adding on double bonds
Addition polymer: A polymer formed when monomers (usually containing a double bond)
combine through an addition reaction
Condensation polymerisation: Molecules of two monomers with different functional
groups undergo condensation reactions with the loss of small molecules, usually water.
Condensation polymer: A polymer formed by two monomers with different functional
groups that are linked together in a condensation reaction in which a small molecule,
usually water, is lost
Identify monomers from given addition polymers.
Write down an equation for the polymerisation of ethene to produce polythene.
State the industrial uses of polythene.
Type of reaction
Polymer Structure
Monomer Name
Monomer Structure
Addition
Polyethene (polythene or polyethylene)
-[-CH2-CH2-]n-
ethene (ethylene)
CH2=CH2
Condensation Polylactic acid
(PLA)
-O-C(CH3)- OC-
Question 1 Multiple-choice Questions
Four options are given as possible answers to the following questions.
Each question has only ONE correct answer.
1.1 Which ONE of the compounds below is an aldehyde?
A CH3CHO
B CH3COCH3
C CH3COOH
D CH3OH (2)
1.2 The reaction represented by the equation below takes place in the presence
of a catalyst.
𝐶13𝐻28(ℓ) ⟶ 𝐶2𝐻4(𝑔) + 𝐶3𝐻6(𝑔) + 𝐶8𝐻18(ℓ)
This reaction is an example of …
A addition.
B cracking.
C substitution.
D polymerisation. (2)
1.3 Consider the structural formula of an organic compound below.
C C C C C
C C
C
H
H
H
H
H
H
H
H
H
H H
H
H
H
H
H
Which ONE of the following is the correct IUPAC name of this compound?
A 2,2,4-trimethylpent-2-ene
B 2,2,4-trimethylpent-3-ene
C 2,4,4-trimethylpent-2-ene
D 2,4,4-trimethylpent-3-ene (2)
1.4 Which ONE of the following statements is CORRECT?
Alkenes …
A have the general formula CnH2n+2.
B are unsaturated hydrocarbons.
C readily undergo substitution reactions.
D have one triple bond between two carbon atoms. (2)
1.5 The following equation represents the cracking of a hydrocarbon at high
temperature and pressure:
𝐶11𝐻24 ⟶ 2𝐶2𝐻4 + 𝐘 + 𝐶4𝐻10
Which ONE of the following is the IUPAC name of product Y?
A Prop-1-ene.
B Propane.
C Ethene.
D Ethane. (2)
1.6 When 2-chlorobutane is strongly heated in the presence of concentrated
sodium hydroxide, the major product formed is …
A but-1-ene.
B but-2-ene.
C butan-1-ol.
D butan-2-ol. (2)
1.7 Which ONE of the following compounds is an aldehyde?
A Pentanal
B Pentan-2-ol
C Pentan-2-one
D Ethyl propanoate (2)
1.8 Consider the reaction represented by the equation below:
𝐶𝐻3𝐶𝐻𝐶𝐻2 + 𝐻2 ⟶ 𝐶𝐻3𝐶𝐻2𝐶𝐻3
This reaction is an example of …
A hydration.
B dehydration.
C substitution.
D hydrogenation. (2)
1.9 Consider the structural formula of a compound below.
C C C O C C
OH
H
H H
H
H
H
H
H
H
Which ONE of the following pairs of reactants can be used to prepare this
compound in the laboratory?
A Propanoic acid and ethanol
B Propanoic acid and methanol
C Ethanoic acid and propan-1-ol
D Methanoic acid and propan-1-ol (2)
1.10. Which ONE of the following compounds has dipole-dipole forces between
its molecules?
A Ethanal
B Ethane
C Ethene
D Ethyne (2)
1.11. Which ONE of the following is a product formed during the hydrolysis of
bromoethane?
A Water
B Ethene
C Ethanol
D Bromine (2)
1.12. Which ONE of the following is the EMPIRICAL FORMULA of
1,2-dichloroethane?
A 𝐶𝐻𝐶ℓ
B 𝐶𝐻2𝐶ℓ
C 𝐶𝐻𝐶ℓ2
D 𝐶2𝐻4𝐶ℓ2 (2)
1.13. Which ONE of the following compounds is an aldehyde?
A 𝐶𝐻3𝐶𝑂𝐶𝐻3
B 𝐶𝐻3𝐶𝐻2𝐶𝐻𝑂
C 𝐶𝐻3𝐶𝐻2𝐶𝑂𝑂𝐻
D 𝐶𝐻3𝐶𝐻2𝐶𝐻2𝑂𝐻 (2)
1.14. Which ONE of the following pairs of compounds are FUNCTIONAL
isomers?
A Methanol and methanal
B Butane and 2-methylpropane
C Propan-1-ol and pronan-2-ol
D Propanoic acid and methyl ethanoate (2)
Questions 2 to 6 are Examples Question on the basics of Organic Chemistry.
Question 2
The letters A to G in the table below represent seven organic compounds.
A C C C C C
C
O
H
H
H H
H
H H
H
H
H
H
H
B
C
C C C
C
O
H
H
H
H
H
H
H
H
H
H
D C C C C
C
O
H
H
H H
H
H
H
H
H
H
H
H
E Butane F C C C C
H
H
H H
H
H
G Ethyl propanoate
2.1. Write down the:
2.1.1. Name of the homologous series to which compound F belongs.
(1)
2.1.2. Name of the functional group of compound D. (1)
2.1.3. Letter that represents a primary alcohol. (1)
2.1.4. IUPAC name of compound A. (2)
2.1.5. Structural formula of the monomer of compound B. (2)
2.1.6. Balanced equation, using molecular formulae, for the combustion of
compound E in excess oxygen. (3)
2.2. Briefly explain why compounds C and D are classified as POSITIONL
ISOMERS. (2)
2.3. Compound G is prepared using an alcohol as one of the reactants.
Write down the balanced equation for the reaction using structural
formulae for all the organic reagents. (7)
[19]
Question 3
Consider the organic compounds represented by the letters A to F in the table below.
A 2,2,4-trimethylhexane B CH3CH2CH2CH2CHO
C
C C C
C
C C
Cl BrH
H
H
H
H
H
H
H
H H
H
H
D
E
F Pentan-2-one
3.1. Write down the LETTER that represents the following:
3.1.1. An aldehyde. (1)
3.1.2. A condensation polymer (1)
3.1.3. A compound which has a carbonyl group bonded to two carbon atoms
as its functional group. (1)
3.2. Write down the IUPAC name of:
3.2.1. Compound C (3)
3.2.2. The monomer of compound D (1)
3.3. Write down the structural formula of:
3.3.1. Compound A (2)
3.3.2. Compound F (2)
3.4. The table contains compounds which are functional isomers.
3.4.1. Define the term functional isomer. (2)
3.4.2. Write down the LETTERS that represent two compounds that are
functional isomers. (1)
[14]
Question 4
The letters A to F in the table below represent six organic compounds.
A C C C
C
H
H
H H
H
H
HH
B 2-methylbutanoic acid
C C C C
CH2
C C
CH3
OH
H
H
H H
H H
H
H
D C C C C
CH3
CH3
CH2 CH3
CH2 CH3
H
H
H H
H
H
E But-2-ene F
4.1. Write down the:
4.1.1. NAME of the functional group of compound B (1)
4.1.2. Homologous series to which compound C belongs. (1)
4.1.3. Type of polymerisation reaction that produces compound F (1)
4.2. Write down the IUPAC name of:
4.2.1. The monomer used to prepare compound F (1)
4.2.2. Compound C (2)
4.2.3. Compound D (2)
4.3. Write down the NAME or FORMULA of each product formed during the
complete combustion of compound D. (2)
4.4. Write down the structural formula of:
4.4.1. Compound B (2)
4.4.2. A CHAIN ISOMER of compound A (2)
4.5. A laboratory assistant uses bromine water to distinguish between compounds
D and E. She adds bromine water to a sample of each in two different test
tubes. She observes that the one compound decolourises the
bromine water immediately, whilst the other one only reacts after placing the
test tube in direct sunlight.
Write down the:
4.5.1. Letter (D or E) of the compound that will immediately decolourise the
bromine water. (1)
4.5.2. Name of the type of reaction that takes place in the test tube containing
compound D (1)
4.5.3. Structural formula of the organic product formed in the test tube
containing compound E (2)
[18]
Question 5
The letters A to D in the table below represents four organic compounds.
A C C C C C C
CH3
CH3
CH2CH3
H
H
H
H
H
H
H
B C C C C
OH
H
H H
H
H
H
H
C CH3CH2CHO D Butane
Use the information in the table to answer the questions that follow.
5.1. Write down the:
5.1.1. Letter that represents a ketone. (1)
5.1.2. Structural formula of the functional group of compound C. (1)
5.1.3. General formula of the homologous series to which compound A
belongs. (1)
5.1.4. IUPAC name of compound A. (3)
5.1.5. IUPAC name of compound B. (2)
5.2. Compound D is a gas used in cigarette lighters.
5.2.1. To which homologous series does compound D belong? (1)
5.2.2. Write down the STRUCTURAL FORMULA and IUPAC NAME of a
structural isomer of compound D. (4)
5.3. Compound D reacts with bromine (Br2) to form 2-bromobutane.
Write down the name of the:
5.3.1. Homologous series to which 2-bromobutane belongs. (1)
5.3.2. Type of the reaction that takes place. (1)
[16]
Questions 6 to 11 are Examples of Questions on the Properties of Organic
Compounds and Intermolecular Forces.
Question 6
The table below shows the results obtained from experiments to determine the boiling
point of some alkanes and alcohols of comparable molecular masses.
Compound Relative molecular mass Boiling point (ºC)
CH3CH3 30 -89
CH3OH 32 65
CH3CH2CH3 44 -42
CH3CH2OH 46 78
CH3CH2CH2CH3 58 0
CH3CH2CH2OH 60 97
CH3CH2CH2CH2CH3 72 36
CH3CH2CH2CH2OH 74 117
6.1 Define the term boiling point. (2)
6.2 Consider the boiling points of the four alkanes in the above table.
6.2.1 Describe the trend in their boiling points. (1)
6.2.2 Fully explain the trend in QUESTION 6.2.1. (3)
6.3 The boiling point of each alcohol is much higher than that of the alkane
of comparable relative molecular mass. Explain this observation by referring
to the type and strength of the intermolecular forces in alkanes and alcohols. (2)
[8]
Question 7 7.1 Give a reason why alkanes are saturated hydrocarbons. (1)
7.2 Write down the structural formula of:
7.2.1 The functional group of alcohols. (1)
7.2.2 A tertiary alcohol that is a structural isomer of butan-1-ol. (2)
7.3 Learners investigate factors that influence the boiling points of alkanes
and alcohols. In one of the investigations they determine boiling points
of the first three alkanes.
7.3.1 Write down an investigative question for this investigation. (2)
7.3.2 Fully explain why the boiling point increases from methane to propane. (3)
7.4 The learners find that the boiling point of propan-1-ol is higher than that
of propane.Explain this observation by referring to the TYPE of
INTERMOLECULAR FORCES present in each of these compounds. (3)
[12]
Questions 8 to 16 are Examples of Questions on Types of Organic Reactions.
Question 8
The flow diagram below shows the preparation of the organic compounds using
CH3CH=CH2 as starting material. X, Y, Z and P represent different organic reactions.
8.1 To which homologous series does CH3CH=CH2 belong? (1)
8.2 Write down the:
8.2.1 Type of reaction of which X is an example. (1)
8.2.2 Structural formula and IUPAC name of the alcohol produced during
reaction P. (3)
8.2.3 The type of reaction of which Y is an example. (1)
8.2.4 Function of the acid in reaction Y. (1)
8.3 For reaction Z, write down:
8.3.1 The NAME of the inorganic reagent needed. (1)
8.3.2 TWO reaction conditions needed. (2)
8.3.3 A balanced equation for the production of the alkene, using structural
formulae. (5)
[15]
Question 9
The diagram below shows the preparation of an ester using prop-1-ene as a starting
reagent. P, Q, R and S represent different organic reactions.
CH3CH=CH2
CH3CHCCH3
An alcohol
X
P
An Alkene
Y
H2SO4
Z
Prop-1-ene Propane Haloalkane
Propan-1-ol
Ester
P Q
R
S
C2
9.1 Write down the type of reaction represented by:
9.1.1 Q (1)
9.1.2 R (1)
9.2 For reaction P write down the:
9.2.1 Type of addition reaction (1)
9.2.2 Balanced equation using structural formulae (3)
9.3 Write down the structural formula of the haloalkane formed in reaction Q.
(2)
9.4 In reaction S propan-1-ol reacts with ethanoic acid to form the ester.
For this reaction write down the:
9.4.1 Name of the reaction that takes place (1)
9.4.2 FORMULA or NAME of the catalyst needed (1)
9.4.3 Structural formula of the ester formed (2)
9.4.4 IUPAC name of the ester formed (2)
9.5 The propan-1-ol formed in reaction R can be converted to prop-1-ene.
Write down the FORMULA or NAME of the inorganic reagent needed. (1)
[15]
SESSION 4 : Momentum and Impulse
Momentum Define momentum as the product of an object's mass and its velocity. Describe the linear momentum of an object as a vector quantity with the
same direction as the velocity of the object. Calculate the momentum of a moving object using p = mv. Describe the vector nature of momentum and illustrate it with some simple
examples. Draw vector diagrams to illustrate the relationship between the initial
momentum, the final momentum and the change in momentum for each of the above examples.
Newton's second law of motion in terms of momentum State Newton's second law of motion in terms of momentum: The
resultant/net force acting on an object is equal to the rate of change of momentum of the object in the direction of the resultant/net force.
Express Newton's second law of motion in symbols: 𝐹𝑛𝑒𝑡 = ∆𝑝
∆𝑡
Calculate the change in momentum when a resultant/net force acts on an
object and its velocity:
Increases in the direction of motion, e.g. 2nd
stage rocket engine fires Decreases, e.g. brakes are applied Reverses its direction of motion, e.g. a soccer ball kicked back in the
direction it came from
Impulse
Define impulse as the product of the resultant/net force acting on an object and the time the resultant/net force acts on the object.
Deduce the impulse-momentum theorem: 𝐹𝑛𝑒𝑡∆𝑡 = 𝑚∆𝑣. Use the impulse-momentum theorem to calculate the force exerted, the time
for which the force is applied and the change in momentum for a variety of situations involving the motion of an object in one dimension.
Explain how the concept of impulse applies to safety considerations in everyday life, e.g. airbags, seatbelts and arrestor beds.
Conservation of momentum and elastic and inelastic collisions
Explain what is meant by a closed/an isolated system (in Physics), i.e. a system on which the resultant/net external force is zero.
A closed/an isolated system exclude external forces that originate outside
the colliding bodies, e.g. friction. Only internal forces, e.g. contact forces between the colliding objects, are considered.
State the principle of conservation of linear momentum: The total linear
momentum of a closed system remains constant (is conserved). Apply the conservation of momentum to the collision of two objects moving
in one dimension (along a straight line) with the aid of an appropriate sign convention.
Distinguish between elastic collisions and inelastic collisions by calculation.
Question 1 Multiple choice questions
Four options are provided as possible answers to the following questions.
Each question has only ONE correct answer.
1.1. Two trolleys, P and Q, of mass m and 2m respectively are at rest on a
frictionless horizontal surface. The trolleys have a compressed spring
between them.
The spring is released and the trolleys move apart. Which ONE of the
following statements is TRUE?
A P and Q have equal kinetic energies.
B The speed of P is less than the speed of Q.
C The sum of the kinetic energies of P and Q is zero.
D The sum of the final momentum of P and Q ia zero.
1.2. An object of mass m moving at constant velocity v collides head-on with
an object of mass 2m moving in the opposite direction at velocity v in the
opposite direction and the larger mass is brought to rest. Refer to the
diagram below.
Ignore the effects of friction.
Which ONE of the following is CORRECT?
MOMENTUM MECHANICAL ENERGY
A Conserved Conserved
B Not Conserved Conserved
C Conserved Not Conserved
D Not Conserved Not Conserved
1.3. Two bodies undergo an INELASTIC collision in the absence of friction.
Which ONE of the following combinations of momentum and kinetic
energy of the system is CORRECT?
MOMENTUM KINETIC ENERGY
A Not Conserved Conserved
B Conserved Not Conserved
C Not Conserved Not Conserved
D Conserved Conserved
1.4. Airbags in modern cars provide more safety during an accident.
The statements below are made by a learner to explain how airbags can
ensure better safety in a collision.
(i) The time of impact increases.
(ii) The impact force decreases.
(iii) The Impulse increases.
A (i) only
B (ii) only
C (ii) and (iii) only
D (i) and (ii) only
[4 2 = 8]
Questions 2-6 is possible examples of question 4 from final exam papers on Momentum and Impulse
Question 2
Two boys, each of mass m, are standing at the back of a flatbed trolley of mass
4m. The trolley is as at rest on a frictionless horizontal surface.
The boys jump of simultaneously at one end of the trolley with a horizontal
velocity of 2m·s-1. The trolley moves in the opposite direction.
2.1. Write down the principle of conservation of linear momentum in words.
(2)
2.2. Calculate the final velocity of the trolley. (5)
2.3. The two boys jump of the trolley one at a time. How will the velocity of
the trolley compare to that calculated in QUESTION 2.2? Write down
only GREATER THAN, SMALLER THAN or EQUAL TO. (1)
[8]
Question 3
Dancers have to learn many skills, including how to land correctly. A dancer of
mass 50 kg leaps into the air and lands feet first on the ground. She lands on
the ground with a velocity of 5 m·s-1. As she lands, she bends her knees and
comes to a complete stop in 0,2 seconds.
3.1. Calculate the momentum with which the dancer reaches the ground.
(3)
3.2. Define the term Impulse of a force. (2)
3.3. Calculate the magnitude of the net force acting on the dancer as she
lands. (3)
Assume that the dancer performs the same jump as before but lands
without bending her knees.
3.4. Will the force now be GREATER THAN, SMALLER THAN or EQUAL TO
the force calculated in QUESTION 3.3? (1)
3.5. Give a reason for the answer to QUESTION 3.4. (3)
[12]
Question 4
A bullet of mass 20 g is fired from a stationary rifle of mass 3 kg. Assume that
the bullet moves horizontally. Immediately after firing, the rifle recoils (moves
back) with a velocity of 1,4 m·s-1.
4.1. Calculate the speed at which the bullet leaves the rifle. (4)
The bullet strikes a stationary 5 kg wooden block fixed to a flat,
horizontal table. The bullet is brought to rest after travelling a distance of
0,4 m into the block. Refer to the diagram below.
4.2. Calculate the magnitude of the average force exerted by the block on the
bullet. (5)
4.3. How does the magnitude of the force calculated in QUESTION 4.2
compare to the magnitude of the force exerted by the bullet on the block?
Write down only LARGER THAN, SMALLER THAN or THE SAME.
(1)
[10]
Question 5
The diagram below shows two trolleys, P and Q, held together by means of a
compressed spring on a flat, frictionless horizontal track. The masses of P and Q
are 400 𝑔 and 600 𝑔 respectively.
When the trolleys are released, it takes 0,3 𝑠 for for the spring to unwind to its
natural length. Trolley Q then moves to the right at 4 𝑚 ∙ 𝑠−1.
5.1. State the principle of conservation of linear momentum in words. (2)
5.2. Calculate the:
5.2.1. Velocity of trolley P after the trolleys are released. (4)
5.2.2. Magnitude of average force exerted by the spring on trolley Q. (4)
5.3. Is this an elastic collision? Only answer YES or NO. (1)
[11]
SESSION 5: Newton's laws and application of Newton's laws
Different kinds of forces: weight, normal force, frictional force, applied force (push, pull), tension (strings or cables) Define normal force, N, as the force or the component of a force which a
surface exerts on an object with which it is in contact, and which is perpendicular to the surface.
Define frictional force, f, as the force that opposes the motion of an object and which acts parallel to the surface.
Define static frictional force, fs, as the force that opposes the tendency of
motion of a stationary object relative to a surface. Define kinetic frictional force, f
k, as the force that opposes the motion of a
moving object relative to a surface. Know that a frictional force:
Is proportional to the normal force Is independent of the area of contact Is independent of the velocity of motion
Solve problems using 𝑓𝑠
𝑚𝑎𝑥 = 𝜇𝑠𝑁 where 𝑓𝑠𝑚𝑎𝑥 is the maximum static
frictional force and μs is the coefficient of static friction.
NOTE: If a force, F, applied to a body parallel to the surface does not cause the
object to move, F is equal in magnitude to the static frictional force. The static frictional force is a maximum (𝑓𝑠
𝑚𝑎𝑥) just before the object starts to move across the surface.
If the applied force exceeds 𝑓𝑠𝑚𝑎𝑥, a resultant (net) force accelerates the
object.
Solve problems using fk = μ
kN, where f
k is the kinetic frictional force and μ
k
the coefficient of kinetic friction.
Force diagrams, free-body diagrams Draw force diagrams.
Draw free-body diagrams. (This is a diagram that shows the relative magnitudes and directions of forces acting on a body/particle that has been isolated from its surroundings)
Resolve a two-dimensional force (such as the weight of an object on an inclined plane) into its parallel (x) and perpendicular (y) components.
Determine the resultant/net force of two or more forces.
Newton's first, second and third laws of motion
State Newton's first law of motion: A body will remain in its state of rest or motion at constant velocity unless a non-zero resultant/net force acts on it.
Discuss why it is important to wear seatbelts using Newton's first law of motion.
State Newton's second law of motion: When a resultant/net force acts on an object, the object will accelerate in the direction of the force at acceleration directly proportional to the force and inversely proportional to the mass of the object.
Draw force diagrams and free-body diagrams for objects that are in equilibrium or accelerating.
Apply Newton's laws of motion to a variety of equilibrium and non-equilibrium problems including:
A single object:
Moving on a horizontal plane with or without friction Moving on an inclined plane with or without friction Moving in the vertical plane (lifts, rockets, etc.)
Two-body systems (joined by a light inextensible string): Both on a flat horizontal plane with or without friction One on a horizontal plane with or without friction, and a second
hanging vertically from a string over a frictionless pulley Both on an inclined plane with or without friction Both hanging vertically from a string over a frictionless pulley
State Newton's third law of motion: When one body exerts a force on a
second body, the second body exerts a force of equal magnitude in the opposite direction on the first body.
Identify action-reaction pairs. List the properties of action-reaction pairs.
Newton's Law of Universal Gravitation State Newton's Law of Universal Gravitation: Each body in the universe
attracts every other body with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.
Solve problems using: 𝐹 = 𝐺𝑚1𝑚2
𝑟2
Calculate acceleration due to gravity on a planet using: 𝑔 = 𝐺𝑚
𝑟2
Describe weight as the gravitational force the Earth exerts on any object on
or near its surface. Calculate weight using the expression w = mg. Calculate the weight of an object on other planets with different values of
gravitational acceleration. Distinguish between mass and weight. Explain weightlessness.
Question 1 Multiple choice questions
Four options are provided as possible answers to the following questions. Each question has only ONE correct answer. 1.1. Two identical metal spheres, each of mass m and separated by a
distance r, exert a gravitational force of magnitude F on each other. The distance between the spheres is now HALVED. The magnitude of the force the spheres now exerts on each other is: A ½ F B F C 2 F D 4 F
1.2. Which one of the following physical quantities is a measure of the inertia of a body? A Mass B Energy C Velocity D Acceleration
1.3. The magnitude of the gravitational force exerted by one body on another body is F. When the distance between the centres of the two bodies is doubled, the magnitude of the gravitational force, in terms of F, will now be … A ¼ F B ½ F C 2 F D 4 F
1.4. Which ONE of the following forces always acts perpendicular to the surface on which a body is placed? A Normal force B Frictional force C Gravitational force D Tension force
1.5. Two isolated bodies, A and B, having masses m and 2m respectively, are placed a distance r apart.
Consider the following statements regarding the gravitational force exerted by the bodies on each other. (i) The force exerted by B on body A is half that exerted by A on body
B. (ii) The force exerted on the bodies is independent of the masses of the
bodies. (iii) The force exerted on body A by B is equal but opposite to that
exerted on body B on A. (iv) The forces will always be attractive. Which of the statements is/are TRUE? A (i), (ii) and (iv) only B (ii), (iii) and (iv) only C (iii) and (iv) only D (iv) only
1.6. Two forces, F1 and F2, are applied on a crate lying on a frictionless, horizontal surface, as shown in the diagram below. The magnitude of force F1 is greater than that of force F2.
The crate will … A accelerate towards the east. B accelerate towards the west. C move at a constant speed towards the east. D move at a constant speed towards the west.
1.7. A person stands on a bathroom scale that is calibrated in newton, in a stationary elevator. The reading on the bathroom scale is W. The elevator now moves with a constant upward acceleration of ¼ g, where g is the gravitational acceleration.
What will the reading on the bathroom scale be now?
A 1
4 W
B 3
4 W
C W
D 5
4 W
1.8. The simplified diagram below shows a rocket that has been fired
horizontally, accelerating to the west.
Which ONE of the statements below best explains why the rocket accelerates? A The speed of the exhaust gases is smaller than the speed of the rocket. B The pressure of the atmosphere at the back of the rocket is less than at
the front. C The air outside the rocket exerts a greater force on the back of the rocket
than at the front. D The rocket pushes the exhaust gases to the east and the exhaust gases
push the rocket to the west.
1.9. A net force 𝐹 which acts on a body of mass 𝑚 causes an acceleration 𝑎. If the same net force 𝐹 is applied to a body of mass 2 𝑚, the acceleration of the body will be …
A 1
4𝑎
B 1
2𝑎
C 2𝑎
D 4𝑎
1.10. Two objects of masses 2𝑚 and 𝑚 are arranged as shown in the diagram below.
Which ONE of the changes below will produce the GREATEST increase in the gravitational force exerted by the one mass on the other? A Double the larger mass. B Halve the smaller mass. C Double the distance between the masses. D Halve the distance between the masses.
Questions 2 - 6 is possible examples of question 2 from final exam papers on Newton’s laws
Question 2
A light inelastic string connects two objects of mass 6 kg and 3 kg respectively. They are pulled up an inclined plane that makes an angle of 30º with the horizontal, with a force of magnitude F. Ignore the mass of the string.
The coefficient of kinetic friction for the 3 kg object and the 6 kg object is 0,1 and 0,2 respectively. 2.1 State Newton’s Second Law of Motion in words. (2)
2.2 How will the coefficient of kinetic friction be affected if the angle between
the incline and the horizontal increases? Write down only INCREASES, DECREASES or REMAINS THE SAME. (1)
2.3 Draw a labelled free-body diagram indicating all the forces acting on the 6 kg object as it moves up the inclined plane. (4)
2.4 Calculate the:
2.4.1. Tension in the string if the system accelerates up the inclined plane at 4 m·s-2. (5)
2.4.2. Magnitude of F if the system moves up the inclined plane at CONSTANT VELOCITY. (6)
2.5 How would the tension in the string, calculated in QUESTION 2.4.1, be
affected if the system accelerates up a FRICTIONLESS inclined plane at 4 m·s-2? Write down only INCREASES, DECREASES or REMAINS THE SAME. (1)
[19]
Question 3 Two blocks of masses 20 kg and 5 kg respectively are connected by a light inextensible string, P. A second light inextensible string, Q, attached to the 5 5 kg block, runs over a light frictionless pulley. A constant force of 250 N pulls the second string as shown in the diagram below. The magnitudes of the tensions in P and Q are T1 and T2 respectively. Ignore the effects of air friction.
3.1. State Newton’s Second Law of Motion in words. (2)
3.2. Draw a labelled free-body diagram indicating ALL the forces acting on the 5 kg block. (3)
3.3. Calculate the magnitude of the tension T1 in string P. (6)
3.4. When the 250 N force is replaced by a sharp pull on the string, one of the two strings break. Which ONE of the two strings, P or Q, will break? (1) [12]
Question 4
A block of mass 1 kg is connected to another block of mass 4 kg by a light inextensible string. The system is pulled up a rough plane inclined at 30º to the horizontal, by means of a constant 40 n force parallel to the plane as shown in the diagram below.
The magnitude of the kinetic frictional force between the surface and the 4 kg block is 10 N. The coefficient of kinetic friction between the 1 kg block and the surface is 0,29. 4.1. State Newton’s third law in words. (2)
4.2. Draw a labelled free-body diagram showing ALL forces acting on the 1 kg
block as it moves up the incline. (5)
4.3. Calculate the magnitude of the:
4.3.1. Kinetic frictional force between the 1 kg block and the surface. (3)
4.3.2. Tension in the string connecting the two blocks. (6) [16]
Question 5
5.1. Two blocks of mass M kg and 2,5 kg respectively are connected by a light, inextensible string. The string runs over a light, frictionless pulley, as shown in the diagram below. The blocks are stationary.
5.1.1. State Newton’s THIRD law in words. (2)
5.1.2. Calculate the tension in the string. (3)
The coefficient of static friction (μs) between the unknown mass M and the surface of the table is 0,2.
5.1.3. Calculate the minimum value of M that will prevent the blocks from moving. (5) The block of unknown mass M is now replaced with a block of mass 5 kg. The 2,5 kg block now accelerates downwards. The coefficient of kinetic friction (μk) between the 5 kg block and the surface of the table is 0,15.
5.1.4. Calculate the magnitude of the acceleration of the 5 kg block. (5)
5.2. A small hypothetical planet X has a mass of 6,5 × 1020 kg and a radius of 550 km. Calculate the gravitational force (weight) that planet X exerts on a 90 kg rock on this planet’s surface. (4) [19]
SESSION 6: Doppler Effect (relative motion between source and observer)
With sound and ultrasound State the Doppler effect as the change in frequency (or pitch) of the
sound detected by a listener because the sound source and the listener have different velocities relative to the medium of sound propagation.
Explain (using appropriate illustrations) the change in pitch observed when a source moves toward or away from a listener.
Solve problems using the equation 𝑓𝐿 = 𝑣 ± 𝑣𝐿
𝑣 ± 𝑣𝑠 𝑓𝑠 , when EITHER the
source or the listener is moving. State applications of the Doppler effect. With light – red shifts in the universe (evidence for the expanding universe) Explain red shifts and blue shifts using the Doppler Effect. Use the Doppler effect to explain why we conclude that the universe is
expanding.
Question 1 Multiple choice questions
Four options are provided as possible answers to the following questions. Each question has only ONE correct answer.
1.1. An astronomer, viewing light from distant galaxies, observes a shift of spectral lines toward the red end of the visible spectrum. This shift provides evidence that … A the universe is expanding. B the galaxies are moving closer towards Earth. C Earth is moving towards the distant galaxies. D the temperature of Earth's atmosphere is increasing.
1.2. The diagram below shows the electron transitions P, Q, R and S between different energy levels in an atom. Which ONE of the transitions will result in an emission of a radiation with the longest wavelength?
A P B Q C R D S
1.3. Which ONE of the following CANNOT be explained using the Doppler effect?
A Emission of electrons from a metal surface
B 'Flow meters' used in hospitals
C Red spectral lines from distant stars being shifted D Observed frequency of light from moving bodies being higher than
expected
1.4. A line emission spectrum is formed when an excited atom moves from a … A higher to a lower energy level and releases energy B higher to a lower energy level and absorbs energy C lower to a higher energy level and releases energy D lower to a higher energy level and absorbs energy
1.5. Light reaching the Earth from a galaxy moving away is shifted towards … A greater velocities B higher frequencies C longer wavelengths D shorter wavelengths
Questions 2-5 is possible examples of question 6 from final exam papers on Doppler effect.
Question 2 The siren of a stationary police car emits sound waves of wavelength 0,55 m. With its siren on, the police car now approaches a stationary listener at constant
velocity on a straight road. Assume that the speed of sound in air is 345 m·s-1
. 2.1. Will the wavelength of the sound waves observed by the listener be
GREATER THAN, SMALLER THAN or EQUAL TO 0,55 m? (1)
2.2. Name the phenomenon observed in QUESTION 2.1. (1) 2.3. Calculate the frequency of the sound waves observed by the listener if
the car approaches him at a speed of 120 km·h-1
. (7) 2.4. How will the answer in QUESTION 2.3 change if the police car moves
away from the listener at 120 km·h-1
? Write down only INCREASES, DECREASES or REMAINS THE SAME. (1)
[10]
Question 3
3.1. The siren of a stationary ambulance emits a note of frequency 1 130 Hz. When the ambulance moves at a constant speed, a stationary observer detects a frequency that is 70 Hz higher than that emitted by the siren.
3.1.1. State the Doppler effect in words. (2) 3.1.2. Is the ambulance moving towards or away from the observer? Give a
reason for the answer. (2) 3.1.3. Calculate the speed at which the ambulance is travelling. Take the
speed of sound in air as 343 m∙s-1
. (5) 3.2. A study of spectral lines obtained from various stars can provide
valuable information about the movement of the stars. The two diagrams below represent different spectral lines of an element. Diagram 1 represents the spectrum of the element in a laboratory on Earth. Diagram 2 represents the spectrum of the same element from a distant star.
Is the star moving towards or away from the Earth? Explain the answer by referring to the shifts in the spectral lines in the two diagrams above. (2) [11]
Question 4
The Doppler effect is applicable to both sound and light waves. It also has very important applications in our everyday lives. 4.1. A hooter on a stationary train emits sound with a frequency of 520 Hz,
as detected by a person standing on the platform. Assume that the
speed of sound is 340 m∙s-1
in still air. Calculate the:
4.1.1. Wavelength of the sound detected by the person. (2) 4.1.2. Wavelength of the sound detected by the person when the train moves
towards him/her at a constant speed of 15 m∙s-1
with the hooter still emitting sound. (6)
4.2. Explain why the wavelength calculated in QUESTION 4.1.1 differs
from that obtained in QUESTION 4.1.2. (2) 4.3. Use your knowledge of the Doppler effect to explain red shifts.
(2) [12]
Question 5
5.1. The data below was obtained during an investigation into the relationship between the different velocities of a moving sound source and the frequencies detected by a stationary listener for each velocity. The effect of wind was ignored in this investigation.
Experiment number 1 2 3 4
Velocity of the sound source (m·s-1) 0 10 20 30
Frequency (Hz) of the sound detected by the stationary listener
900 874 850 827
5.1.1. Write down the dependant variable for this investigation. (1)
5.1.2. State the doppler effect in words. (2)
5.1.3. Was the sound source moving TOWARDS or AWAY FROM the listener? Give a reason for the answer. (2)
5.1.4. Use the information in the table to calculate the speed of sound during the investigation. (5)
5.2. The spectral lines of a distant star are shifted towards the longer wavelengths of light. Is the star mowing TOWARDS or AWAY from the Earth. (1)
[11]
Question 6
Reflection of sound waves enables bats to hunt for moths. The sound wave
produced by a bat has a frequency of 222 kHz and a wavelength of 1,5 x 10-3
m.
6.1. Calculate the speed of this sound wave through the air. (3)
6.2. A stationary bat sends out a sound signal and receives the same signal reflected from a moving moth at a frequency of 230,3 kHz.
6.2.1. Is the moth moving TOWARDS or AWAY FROM the bat? (1) .
6.2.2. Calculate the magnitude of the velocity of the moth, assuming that the velocity is constant. (6)
[10]
