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NATIONAL
SENIOR CERTIFICATE
GRADE 12
PHYSICAL SCIENCES: PHYSICS (P1) AND CHEMISTRY (P2)
PRE-TRIAL EXAM
22 AUGUST 2022
MARKS: 150
TIME: 3 hours
This question paper consists of 16 pages and 5 data sheets.
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Physical Sciences Grade 12 Paper 1 Limpopo Mock 2022 Memo
Grade 12 Physical Sciences Paper 1 memorandum from the Limpopo Mock Exam 2022, 34 pages. Read online or download the free PDF.
Physical Sciences · Grade 12 · Limpopo Mock Exam · 2022. Memorandum, 34 pages. Read online or download the PDF.
- Subject
- Physical Sciences
- Grade
- Grade 12
- Document type
- Memorandum
- Year
- 2022
- Exam period
- Limpopo Mock Exam
- Paper
- 1
- Curriculum
- CAPS
- Pages
- 34
- File size
- 1.2 MB
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NSC
INSTRUCTIONS AND INFORMATION
1. Write your NAME in the appropriate space on the ANSWER BOOK.
2. This question paper consists of 7 questions. Answer ALL the questions in the
ANSWER BOOK.
3. Start EACH question on a NEW page in the ANSWER BOOK.
4.
s.co
Number the answers correctly according to the numbering system used in this
question paper.
m
5. si
Leave ONE line between two subquestions, for example between
c
QUESTION 2.1 and QUESTION 2.2.
st
6. an
You may use a non-programmable calculator.
mor
7. You may use appropriate mathematical instruments.
ep
8. hy
You are advised to use the attached DATA SHEETS.
9. om
Show ALL formulae and substitutions in ALL calculations.
10. fr
Round off your FINAL numerical answers to a minimum of TWO decimal
places.
oa
11. de
Give brief motivations, discussions, et cetera where required.
d
12. do
Write neatly and legibly.
wn
l
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QUESTION 1 (Start on a new page)
1.1 ∆n
If X = applies to chemical reactions, where ∆n = change in number of
∆t
moles and ∆t = time interval, then X probably represents …
A equilibrium constant.
B yield of products.
C heat of reaction. s.co
D rate of reaction. m (2)
si c
1.2 Consider the following hypothetical reaction which is at equilibrium:
st
an
2XY(s) ⇌ 2X(l) + Y2(g)
The pressure of the system is INCREASED.
mor
ep
How will the CONCENTRATION of Y2(g) and the value of KC be affected (at
hy
constant temperature) when the new equilibrium is established?
CONCENTRATION of Y2(g) KC VALUE
om
A Unchange
fr Unchange
B oa Decrease Decrease
C
de Increase Increase
d
do
D Decrease Unchange
wn
l (2)
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1.3 Consider the following equilibrium reaction equation:
C(s) + O2(g) ⇌ CO2(g)
Which ONE of the following graphs BEST represents the relationship
between [CO2] and [O2] at CONSTANT TEMPERATURE?
A C
[CO2] [CO2]
[O2] [O2]
B D
[CO2] [CO2]
(2)
[O2] [O2]
1.4 The engine of a car does work, W, to increase the velocity of the car from v to
2v. The work done by the engine of the same car to increase the velocity from
2v to 4v, is …
A W
B 2W
C 3W
D 4W (2)
[8]
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NSC
QUESTION 2 (Start on a new page)
2.1 Two experiments are performed, with everything the same EXCEPT that one
experiment is done at 20°C, and the other is done at 60°C. It is found that the
experiment at 20°C takes 320 s and the experiment at 60°C takes 80 s. For
each question, write down the LETTER from the LIST below which represents
the BEST possible answer.
V The rate of reaction increases when temperature is increased.
W The rate of reaction is higher at 60°C than at 20°C.
X Is the rate of a reaction higher at 60°C than at 20°C?
Y How does increasing temperature affect the rate of a reaction?
Z The rate of reaction is 4 times faster at 60°C than at 20°C.
2.1.1 Investigative question. (1)
2.1.2 Hypothesis. (1)
2.1.3 Conclusion. (1)
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2.2 A learner conducts a practical investigation in order to test whether the
dissolution of solid ammonium chloride is exothermic or endothermic. The
apparatus used include a beaker, a salt and a certain measuring instrument.
The graph below shows the energy changes that occur when ammonium
chloride dissolves in water.
200
Potential energy (KJ)
170
50
Course of Reaction
2.2.1 Name the measuring instrument that must be used during this
investigation. (1)
2.2.2 Will the reading on the instrument used in QUESTION 2.2.1
INCREASE, DECREASE or REMAIN THE SAME during the course
of the reaction? Explain the answer with reference to the graph.
(2)
For this reaction, calculate:
2.2.3 The heat of reaction (2)
2.2.4 Activation energy (1)
In another experiment, a learner adds a catalyst. On addition of a catalyst,
state whether the following will INCREASE, DECREASE or REMAIN THE
SAME:
2.2.5 Potential energy of the products (1)
2.2.6 Activation energy (1)
[11]
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NSC
QUESTION 3 (Start on a new page)
A group of chemistry learners use the reaction between magnesium and sulphuric
acid to investigate one of the factors that affects reaction rates.
3.1 Define the term reaction rate in words. (2)
They add 2,4 g of magnesium ribbon to EXCESS dilute sulphuric acid and measure
the mass of magnesium used per unit time.
Mg(s) + H2SO4(aq) → MgSO4(aq) + H2(g)
3.2 Suggest a reason as to why the acid must be in excess. (1)
The learners repeat the experiment using EXCESS concentrated sulphuric acid.
3.3 Provide a reason why the learners use the SAME amount (in grams) of
magnesium ribbon in BOTH experiments. (1)
The results obtained for the reaction using DILUTE sulphuric acid are represented in
the graph below.
p
n(Mg)(mol)
0,08
0
160 Time (s)
3.4 Determine the value of p (shown in the graph). (3)
3.5 Using the graph, calculate the mass of magnesium used from t = 0 s to
t = 160 s. (3)
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NSC
3.6 Hence, find the average reaction rate (in gꞏs-1) during the FIRST 160 s. (2)
3.7 Copy the graph above in the ANSWER BOOK (Label it R). On the same set
of axes, use a dotted line to show the curve that will be obtained when
CONCERNTRATED sulphuric acid is used. (Label it S). No numerical values
(2)
needed.
3.8 Use the collision theory to explain the difference in the two graphs. (3)
[17]
QUESTION 4 (Start on a new page)
4.1 Consider the following equilibrium reaction equation:
A2(g) + 3B2(g) ⇌ 2AB3(g), ∆H<0
4.1.1 Write down ONE condition necessary for a chemical system to be
in equilibrium. (1)
A graph showing the concentration of AB3(g) produced versus time is plotted
below.
[AB3(g)]
0 P Q R S Time T
How does the rate of the FORWARD REACTION compare to the rate of the
REVERSE REACTION during the following intervals? (Write down only:
GREATER THAN, EQUAL TO or LESS THAN).
4.1.2 OP (1)
4.1.3 QR (1)
4.1.4 ST (1)
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NSC
Changes are made to the graph from Q to S due to changes in the
temperature.
4.1.5 At which point (Q, R or S) will the temperature be GREATEST?
Justify the answer. (2)
4.1.6 At which point (Q,R or S) will the value of KC be the SMALLEST? (1)
4.1.7 If the changes in the graph from Q to S are due to pressure
changes, at which point (Q,R or S) will the pressure be the
HIGHEST? Explain the answer. (2)
4.2 Initially, 0,086 mol of Br2 is placed in a 1,26 dm3 flask and heated to 1 680 K,
a temperature at which the halogen dissociates to atoms according to the
following equation:
Br2(g) ⇌ 2Br(g), ∆H>0
Br2(g) is 4,8% dissociated at 1 680 K.
Calculate the equilibrium constant for this reaction at 1 680 K. (6)
[15]
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QUESTION 5 (Start on a new page)
5.1 Hydrochloric acid (HCl) is added to distilled water.
5.1.1 Write down the balanced equation for the ionisation of hydrochloric
acid in water. (2)
5.1.2 Classify hydrochloric acid as a STRONG acid or WEAK acid. (1)
-
5.2 Calculate the concentration of the hydroxide (OH ) ions in a sulphuric acid
solution with a pH = 3,5 at 25°C. (3)
5.3 A standard solution of oxalic acid ((COOH)2) is prepared by dissolving 3,8 g of
the acid in 250 cm3 of distilled water. The acid is titrated against an impure
solution of sodium hydroxide (NaOH). The impure solution of sodium hydroxide
is prepared by dissolving 5,0 g of the impure mixture in 250 cm 3 of distilled water.
(Assume that the impurities do not react). In a titration, 25,0 cm3 of the
impure sodium hydroxide is neutralized by 30 cm 3 of the acid.
(COOH)2 (aq) + 2NaOH(aq) → (COONa)2(aq) + 2H2O(l)
5.3.1 Is an aqueous solution of (COONa)2 acidic, basic or neutral? Explain
the answer with the aid of a chemical reaction equation. (3)
5.3.2 Calculate the concentration of the oxalic acid solution. (3)
5.3.3 Calculate the mass of the impurities contained in the 5,0 g of the impure
sodium hydroxide. (6)
[18]
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NSC
QUESTION 6 (Start on a new page.)
The diagram below represents a simplified electrolytic cell used to
electroplate a spanner with chromium. The spanner is continuously
rotated during the process of electroplating.
DC power supply
Electrode X
Spanner
Cr(NO3)3(aq)
A constant current passes through the solution and the concentration
of Cr(NO3)3(aq) remains constant during the process. In the process,
a total of 0,03 moles of electrons is transferred in the electrolytic cell.
6.1 Define the term electrolysis. (2)
6.2 State the energy conversion that takes place in the above cell. (1)
6.3 Write down the:
6.3.1 Half-reaction that occurs at the spanner (2)
6.3.2 NAME or FORMULA of the metal of which electrode X is made (1)
6.3.3 NAME or FORMULA of the oxidising agent (1)
6.4 Give a reason why the concentration of the Cr(NO3)3(aq) remains
constant during the process. (2)
6.5. State on reason why the spanner is electroplated. (1)
[10]
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QUESTION 7( Start on a new page)
A standard electrochemical cell is set up, using silver and an
unknown electrode, D, as shown in the sketch below,. The initial
reading on the voltmeter is 2,46 V. After the cell was connected for
some time, the mass of the silver electrode increased.
2 ,46 V
V
Ag D
A B
7.1 What process takes place at the silver electrode? Choose from
OXIDATION or REDUCTION. (1)
7.2 Define an oxidising agent in terms of electron transfer. (2)
7.3 Write down the NAME of a solution which can be used in (1)
beaker A.
7.4 Determine the standard electrode potential for electrode D and
identify D from the standard potential table. (5)
7.5 Write down the cell notation for this cell. (3)
[12]
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NSC
QUESTION 8 (Start on a new page)
During an investigation to verify the speed of sound, Mr Chiv use a siren with a single
frequency of 400 Hz, mounted on a remote-controlled toy car. With the siren
activated, he move the car along a 5 m track at a constant velocity of 7,50 mꞏs-1. One
of the learners stands halfway down the track and records the sound emitted by the
siren for the duration of the motion on the cellphone. They are expecting to record
two different pitches of the sound emitted by the siren.
8.1 For this investigation, write down the:
8.1.1 Independent Variable (1)
8.1.2 Dependent Variable (1)
8.1.3 ONE control Variable (1)
8.2 Write down the NAME of and state in words, the phenomenon on which the
learners depends in order to calculate the speed of sound. (3)
When playing back the sound recorded on the cellphone in the presence of a
frequency meter, two different notes were registered: One with a frequency of
409 Hz and the other with a frequency lower than 400 Hz.
8.3 Using the frequency of one of the recorded notes, calculate the speed of
sound in air. (3)
One of the uses of this phenomenon is in ultrasound.
8.4 State ONE application of ultrasound in the medical field. (1)
[10]
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QUESTION 9 (Start on a new page)
9.1 A crate of mass 10 kg slides FROM REST down path ABC, as shown in the diagram below.
A
35°
h
C
Along the portion A to B, the coefficient of kinetic friction between the crate and the surface is
0,16. The distance from A to B is 5 m.
9.1.1 Define the term conservative force. (2)
9.1.2 Draw a labelled free-body diagram for the crate as it slides along portion AB.
(3)
9.1.3 Write down the work-(kinetic) energy theorem in words. (2)
9.1.4 Use the work-(kinetic) energy theorem to calculate the speed of the crate as it
reaches point B.
(5)
The crate reaches point C with a speed of 8,5 mꞏs-1.
9.1.5 Assume portion BC is smooth (frictionless) and calculate the vertical height, h, of B
above C, using energy principles. (4)
9.2 Water is pumped from a borehole 100 m deep. A pump with a power output of 2,3 kW (2 300
W) is used.
9.2.1 Define the term power as applied in physics. (2)
9.2.2 Calculate the time (in seconds) taken for the pump to bring the 500 kg of water to
the surface. (3)
[21]
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QUESTION 10 (Start on a new page)
Two charged spheres, S and P, on insulated stands, with charges of 5 µC and -6 µC
respectively, are placed 0,02 m apart, as shown in the diagram below.
0 , 02 m
S P -6 µC
5 µC
10.1 State Coulombs’ Law in words. (2)
10.2 Calculate the magnitude and direction of the electrostatic force that sphere S exerts on
sphere P. (4)
Spheres S and P are brought into contact with each other and then returned to their original
positions. A third, negatively charged, sphere R with an UNKNOWN charge is now placed
0,01 m to the right of sphere P as shown below.
S 0 , 02 m P 0 , 01 m R
10.3 Calculate the magnitude of the NEW charge on S after being in contact with P. (1)
10.4 Draw a free-body diagram showing ALL the electrostatic forces experienced by P. (2)
10.5 The magnitude of the net electrostatic force experienced by P due to S and R is
84,375 N. Calculate the magnitude of the charge on R. (5)
10.6 Calculate the magnitude of the net electric field at a point 0,01 m to the LEFT of R
due to the charges on S and R. (3)
[17]
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QUESTION 11 (Start on a new page.)
11.1 A teacher demonstrates how current can be obtained using a bar magnet, a
coil and a galvanometer. The teacher moves the bar magnet up and down, as shown
by the arrow in the diagram below.
galvanometer
coil
11.1.1. Briefly describe how the magnet must be moved in order to obtain a LARGE
deflection on the galvanometer. (2)
The two devices, A and B, below operate on the principle described in QUESTION 11.1.1
above.
Output
Output
11.1.2 Write down the name of the principle. (1)
11.1.3 Write down the name of part X in device A. (1)
11.2 A 240 V, AC voltage is supplied from a wall socket to an electric kettle of resistance
40 Ω. Wall sockets provide rms voltages and currents.
Calculate the:
9.2.1 Electrical energy consumed by the kettle per second (4)
9.2.2 Maximum (peak) current through the kettle (3)
[11]
Total Marks: 150
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NSC
DATA FOR PHYSICAL SCIENCES GRADE 12
PAPER 1 (PHYSICS)
TABLE 1: PHYSICAL CONSTANTS:
NAME SYMBOL VALUE
Acceleration due to gravity g 9,8 m·s-2
Gravitational constant G 6,67 x 10-11 Nm2kg-2
Radius of Earth RE 6,38 x 106 m
Mass of the Earth ME 5,98 X 1024 kg
Coulomb’s constant k 9,0 x 109 Nm2·C-2
Plank’s contant h 6,63 x 10-34 Jꞏs
Speed of light in a vacuum
c 3,0 x 108 m·s-1
Charge on electron e -1,6 x 10-19 C
Electron mass me 9,11 x 10-31 kg
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TABLE 2: FORMULAE
MOTION
v f v i a t Δx v i Δt 21 at 2 or/of Δy v i Δt 21 at 2
v vf v vf
v f vi 2 ax or/of v f v i 2ay Δx i Δt or/of Δy i Δt
2 2 2 2
2 2
FORCE
Fnet ma p = mv
fsmax = µsN Fk = µkN
Fnet t m v
w mg
p mv f mv i
m 1 m2 m 1 m2 GM
F = G 2 or F = G 2 g
d r r2
WORK ENERGY AND POWER
W Fx cos U mgh or Ep mgh
Wnet K or Wnet Ek
1 1
K mv 2 or Ek mv 2
2 2 K K f K i or E k E kf E ki
W
Wnc K U or Wnc E k E p P
t
Pave Fv ave / pgemid = Fvgemid
WAVES, SOUND AND LIGHT
1
v f T=
f
v vL V ± VL hf
fL f s or FL = V ± V fb E hf or/of E
v vs b
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ELECTROSTATICS
KQ1 KQ2 KQ
F= E=
r2 r2
W F
V= E=
q q
Q Q
n= or n =
e qe
ELECTRIC CIRCUITS
V Emf (ε) = I(R + r)
R=I
Rs = R1 + R2 + … q = I∆t
1 1 1
= + +…
Rp R1 R2
W = Vq W
P=
∆t
W = VI∆t
P = Vi
W = I2R∆t
V2 ∆t P = I 2R
W=
R
V2
P=
R
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ATERNATING CURRENT
1max Pave = Vrms Irms
Irms =
√2
Pave = I2rmsR
Vmax
Vrms = v2
√2 Pave = rms
R
PAPER 2 (CHEMISTRY)
m N
n= n=N
M A
n m V
c= OR c= n=
V MV Vm
Cb Vb n pH = -log[H3O+]
= b
Ca Va na
Kw = [H3O+][OH-] = 1 x 10-14 at 298 K
NAME SYMBOL VALUE
Avogadro’s constant NA 6,02X1023 mol-1
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NATIONAL SENIOR CERTIFICATE
GRADE 12
PHYSICAL SCIENCES: PHYSICS (P1) AND CHEMISTRY (P2)
22 AUGUST 2022
MEMORANDUM
MARKS: 100
These marking guidelines consist of 10 pages.
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Marking Guidelines
QUESTION 1
1.1 D (2)
1.2 A (2)
1.3 B (2)
1.4 D (2)
[8]
QUESTION 2
2.1.1 Y (Ask general question) (1)
2.1.2 V (Predict) (1)
2.1.3 V (State) (1)
2.2.1 Thermometer (1)
2.2.2 Decrease
One mark the first bullet
OR
One mark for the second bullet
(2)
The potential energy for the products is higher than that of the reactants/.
The potential energy for the reactants is lower than that of products.
This shows that the reaction is endothermic/heat absorbing.
2.2.3 ∆H = EP(P) – EP(R)
= 170 – 50
= 120 KJ (2)
2.2.4 Ea = EP(Ca) – EP(R)
= 200 – 50
= 150 KJ (1)
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Marking Guidelines
2.2.5 Remain the same. (1)
2.2.6 Decrease. (1)
[11]
QUESTION 3
3.1 The change in concentration of products or reactants per unit time.
OR:
The change in number of moles/ mass/ volume of reactants/products per
unit time. (2)
3.2 To ensure that all the magnesium gets used up. (1)
3.3 To make it a fair comparison/test. (1)
3.4
m
ni(Mg) = M
2,4
= 24
= 0,10 mol
∴P = 0,10 (3)
3.5 Positive marking from 3.4
n(Mg) used = ni – nf
= 0,10 –0,08 Option 2: 𝒎(𝑴𝒈) = 𝒏. 𝑴
= 0,02 mol = (0,1)(24)
m(Mg)used= nM = 2,4 𝑔 (𝑖𝑡 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑖𝑛 𝑡ℎ𝑒 𝑝𝑟𝑒𝑎𝑚𝑏𝑙𝑒)
= (0,02)(24) 𝒎(𝑴𝒈) = (𝟎, 𝟎𝟖). (𝟐𝟒)
= 0,48 g (3)
= 𝟏, 𝟗𝟐 𝒈
𝒎(𝑴𝒈)𝒖𝒔𝒆𝒅 = 𝟐, 𝟒 − 𝟏, 𝟗𝟐
= 𝟎, 𝟒𝟖 𝒈
3.6 Positive marking from 3.4 & 3.5
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Marking Guidelines
m(Mg)
used
Average rate = Time taken
0,48
(2)
= 160
= 3 x 10-3 gꞏs-1
3.7
n(Mg)(mol)
R
S
0
Time (s)
Steeper gradient
X – intercept for S smaller than for graph R (2)
3.8 Higher concentration for curve S means:
More reactant particles will be present in a given volume /
More particles will have the correct orientation.
More effective collisions per unit time/Higher frequency of effective
collisions.
Increased reaction rate.
OR: Lower concentration for graph R means:
Less reactant particles will be present in a given volume /
Less particles will have the correct orientation.
Less effective collisions per unit time/ Lower frequency of effective
collisions.
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Decreased reaction rate.
(3)
[17]
QUESTION 4
4.1.1 ANY ONE:
Closed system (1)
Reversible reaction
Accept isolated system
4.1.2 Greater than (1)
4.1.3 Less than (1)
4.1.4 Equal to (1)
4.1.5 R
The reverse reaction is endothermic and so the temperature is greatest
where [AB3 (g)] is lowest. (2)
4.1.6 R (1)
4.1.7 Q
An increase in pressure will the favour the forward reaction, resulting in
the [AB2] increasing.
Therefore, high pressure corresponds to a high [AB3]. (2)
4.2 Equation Br2(g) ⇌ 2Br(g)
Ratio 1 2
Initial amount (mol) 0,086 0
Change in amount (mol) -4,128x10-3 +8,256x103
Equilibrium amount (mol) 0,081872 8,256x10-3
n
C = v (molꞏdm-3) 0,06498 6,5524x10-3
𝑘𝑐 =
4,8
n(Br2)decomposed = 100(0,086)
= 4,128x10-3 mol
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[𝐵𝑟]2
𝑘𝑐 = [𝐵𝑟 ]
2
(6,5524×10−3 )2
= (6)
(0,06498)
−4
=6,61 × 10
[15]
QUESTION 5
5.1.1 HCl(aq) + H2O(l) ⇌ H3O+(aq) + Cl-(aq) Accept single arrow (2)
5.1.2 Strong acid (1)
5.2 OPTION 1: OPTION 2:
pH = -log[H3O+] pH + pOH = 14
3,5 = -log[H3O+] pOH= 14 - pH = 14 - 3,5 = 10,5
-3,5 = log[H3O+] pOH = -log[OH ] -
[H3O+]= 10-3,5 10,5 = -log[OH-]
[H3O+]= 3,1622 X 10-4 molꞏdm-3 [OH-] = 10-10,5
[OH-] = 3,16228 x 10-11 molꞏdm-3
Kw = [H3O+][OH-]
1 X 10-14 = (3,16228 X 10-4) [OH-]
[OH-]= 3,16228 X 10-11 molꞏdm-3 (3)
5.3.1 Basic
(COO-)2(aq) + 2H2O(l) ⇌ (COOH)2(aq) + 2OH-(aq)
Excess OH- ions are formed
Therefore the solution is basic
Hence the pH > 7 (3)
m m
5.3.2 c = MV OR: n = M
n
3,8 3,8 ∴c = V
=(90)(0,25) = 90
0,0422222
=0,169 molꞏdm-3 = 0,0422222 mol = 0,25
= 0,169 molꞏdm-3 (3)
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5.3.3 OPTION 1: OPTION 2:
PacaVa = PbcbVb 1 mol (COOH)2 reacts with 2 mol NaOH
cb Vb nb na= cV = (0,169)(0,03)
OR c V = n
a a a
= 5,07x10-3 mol
(2)(0,169)(30) =(1)cb(25)
Pbnb = Pana
Cb = 0,4056 molꞏdm-3
(1)nb = (2)( 5,07x10-3)
∴ nb = 0,01014 mol in 25 cm3
n(NaOH)=cV
∴nb =(10)(0,01014)
=(0,4056)(0,25)
= 0,1014 mol in 250 cm3
=0.1014 mol
∴m(NaOH) = Nm
= (0,1014)(40) = 4,056 g
m(NaOH) = nM
∴m(impurities) = 5,0 –4,056 = 0,944 g
=(0,1014)(40)
= 4,056 g
m(impurities) = 5,0
– 4,056
(6)
= 0,944 g
[18]
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QUESTION 8
8.1.1 The direction of the car with the siren relative to the observer/listener. (1)
8.1.2 The observed/detected frequency of the siren. (1)
8.1.3 ANY ONE:
The actual frequency of the siren. (1)
The velocity of the car.
8.2 The Doppler Effect.
The apparent change in frequency (or pitch/wavelength) of the sound
detected by the listener because the sound source and the listener have
different velocities relative to the medium of sound propagation. (3)
8.3 v±v
fL = ( v ± vL )fs
s
v+0
409 = (v - v )(400)
s
(409)(v-7,5) = (v)(400)
V = 340,83 mꞏs-1 (3)
8.4 ANY ONE:
To monitor blood flow. (1)
To detect the heartbeat of a foetus.
[10]
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QUESTION 9
9.1.1 A force for which the work done (in moving an object between two points) is
independent of the path taken.
OR: A force for which the work done on a particle moving through any closed path
is zero. (2)
9.1.2 N/FN
fk / Ff/f
f: Frictional force/friction
N: Normal force
Fg/w: Gravitational force/weight
w/Fg
(3)
9.1.3 The net/ total work done on an object is equal to the change in the object’s kinetic
energy.
OR: The work done on an object by the net force is equal to the change in the
object’s kinetic energy. (2)
9.1.4 Wnet = ∆Ek
WN + Wfk + W w = ∆Ek
1
0 + fkꞏ∆xꞏcosꞵ+ mgꞏ∆xꞏcosα = 2m(v2f -v2i ) Any one
1
µk ꞏNꞏ∆xꞏcosꞵ+mgꞏ∆xꞏ cosα = 2m(v2f -v2i )
1
µk ꞏmgcosθꞏ∆xꞏcosꞵ+mgꞏ∆xꞏ cosα = 2m(v2f -v2i )
1
(0,16)(10)(9,8)(cos 35°)(5)(cos180°)+(10)(9,8)(5)(cos 55°/305°)= (10) (v2f -02 )
2
-64,22152027 + 281,0524538 = 5v2f
216,8309335 = 5v2f
v2f = 43,36618671
vf = 6,5853 mꞏs-1
∴The speed is 6,5853 mꞏs-1 (5)
𝑊𝑤 𝑐𝑎𝑛 𝑎𝑙𝑠𝑜 𝑏𝑒 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑏𝑦 𝑢𝑠𝑖𝑛𝑔 1) 𝑊𝑤 = 𝐹𝑔‖ × ∆𝑥 × 𝑐𝑜𝑠𝜃
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2) 𝑊𝑤 = −∆𝐸𝑝 = −𝑚𝑔(ℎ𝑓 − ℎ𝑖 )
9.1.5 (EP + EK)B = (EP + EK)C
1 1
Any one
(mgh + 2mv2)B = (mgh + 2mv2)C
1 1
(10)(9,8)h + 2(10)(6,5853)2 = (10)(9,8)(0) + 2(10)(8,5)2
(98)h + 216,8308805 = 0 + 361,25
(98)h = 144,4191195
∴h = 1,4737 m
∴The height is 1,4737 m
(4)
9.2.1 The rate at which work is done.
OR: The rate at which energy is expended/transferred. (2)
9.2.2 W = EP
= mgh
= (4500)(9,8)(100)
= 4410000 J
W
P = ∆t
4410000
2300 =
∆t
4410000
∆t = 2300
= 1917,39 s (3)
[21]
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GRAND TOTAL: 150 MARKS
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