PHYSICAL SCIENCES PAPER 1 GRADE 12 MEMORANDUM - NSC EXAMS PAST PAPERS AND MEMOS JUNE 2019
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PAPER 1
GRADE 12
NSC EXAMS
PAST PAPERS AND MEMOS JUNE 2019
GENERAL GUIDELINES
- CALCULATIONS
1.1 Marks will be awarded for: correct formula, correct substitution, correct answer with unit.
1.2 No marks will be awarded if an incorrect or inappropriate formula is used, even though there are many relevant symbols and applicable substitutions.
1.3 When an error is made during substitution into a correct formula, a mark will be awarded for the correct formula and for the correct substitutions, but no further marks will be given.
1.4 If no formula is given, but all substitutions are correct, a candidate will forfeit one mark.
1.5 No penalisation if zero substitutions are omitted in calculations where correct formula/principle is correctly given.
1.6 Mathematical manipulations and change of subject of appropriate formulae carry no marks, but if a candidate starts off with the correct formula and then changes the subject of the formula incorrectly, marks will be awarded for the formula and correct substitutions. The mark for the incorrect numerical answer is forfeited.
1.7 Marks are only awarded for a formula if a calculation has been attempted, i.e. substitutions have been made or a numerical answer given.
1.8 Marks can only be allocated for substitutions when values are substituted into formulae and not when listed before a calculation starts.
1.9 All calculations, when not specified in the question, must be done to a minimum of two decimal places.
1.10 If a final answer to a calculation is correct, full marks will not automatically be awarded. Markers will always ensure that the correct/appropriate formula is used and that workings, including substitutions, are correct.
1.11 Questions where a series of calculations have to be made (e.g. a circuit diagram question) do not necessarily always have to follow the same order. FULL MARKS will be awarded provided it is a valid solution to the problem. However, any calculation that will not bring the candidate closer to the answer than the original data, will no count any marks. - UNITS
2.1 Candidates will only be penalised once for the repeated use of an incorrect unit within a question.
2.2 Units are only required in the final answer to a calculation.
2.3 Marks are only awarded for an answer, and not for a unit per se. Candidates will therefore forfeit the mark allocated for the answer in each of the following situations:- Correct answer + wrong unit
- Wrong answer + correct unit
- Correct answer + no unit
2.4 SI units must be used except in certain cases, e.g. V.m-1 instead of N.C-1, and cm.s-1 or km.h-1 instead of m.s-1 where the question warrants this.
- GENERAL
3.1 If one answer or calculation is required, but two are given by the candidate, only the first one will be marked, irrespective of which one is correct. If two answers are required, only the first two will be marked, etc.
3.2 For marking purposes, alternative symbols (s, u, t etc) will also be accepted.
3.3 Separate compound units with a multiplication dot, no a full stop, for example, m.s-1. For marking purposes, m.s-1 and m/s will also be accepted. - POSITIVE MARKING
Positive marking regarding calculations will be followed in the following cases: Positiewe nasien met betrekking tot berekeninge sal in die volgende gevalle geld:
4.1 Subquestion to subquestion: When a certain variable is calculated in one subquestion (e.g. 3.1) and needs to be substituted in another (3.2 of 3.3), e.g. if the answer for 3.1 is incorrect and is substituted correctly in 3.2 or 3.3, full marks are to be awarded for the subsequent subquestions.
4.2 A multistep question in a subquestion: If the candidate has to calculate, for example, current in die first step and gets it wrong due to a substitution error, the mark for the substitution and the final answer will be forfeited. - NEGATIVE MARKING
Normally an incorrect answer cannot be correctly motivated if based on a conceptual mistake. If the candidate is therefore required to motivate in QUESTION 3.2 the answer given in QUESTION 3.1, and 3.1 is incorrect, no marks can be awarded for QUESTION 3.2. However, if the answer for e.g. 3.1 is based on a calculation, the motivation for the incorrect answer could be considered
MEMORANDUM
QUESTION 1
1.1 C ✔✔ (2)
1.2 A ✔✔ (2)
1.3 B ✔✔ (2)
1.4 D ✔✔ (2)
1.5 C ✔✔ (2)
1.6 A ✔✔ (2)
1.7 B ✔✔ (2)
1.8 D ✔✔ (2)
1.9 A ✔✔ (2)
1.10 A ✔✔ (2) [20]
QUESTION 2
2.1 When a net force/resultant force acts on an object, it produces the acceleration of the object in the direction of the net force/resultant force. This acceleration is directly proportional to the net/resultant force ✔ and inversely proportional to the mass of the object. ✔ (2)
2.2
| OTPION 1 | OPTION 2 |
 | |
Mark awarded for both arrow and label.
Do not penalise for length of forces since drawing is not drawn to scale.
Any other additional force(s)
If force(s) do not make contact with body. Max. (5)
2.3 Choose East (Right) to be positive
4 kg box
- Fnet = ma
Fx + fk + T = ma
Fx - fk - T = ma
FJaneCos30° + fk + T = ma
80x0,866 + (-8,14) + (-T) = 4a
69,28 - 8,14 - T = 4a
61,14 - T = 4a -------- (1)
5 kg box
- Fnet = ma
T + W = ma
T + (-9,8x5) = 5a
T - 49 = 5a
a = T- 49 ------(2)
5
Subst (2) in (1):(2) in (1):
61,14 - T = 4[T- 49]
[ 5 ]
305,7 - 5 T = 4 T - 196
9 T = 501,7
T = 55,74 N (5) [12]
QUESTION 3
3.1 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. (2)
3.2
| OPTION 1 | OPTION 2 |
| f = Gm1m2 r2 126,30 = 6,67 × 10-11 × 7,35 × 1022m2 (1,74 × 106)2 m2 = 78,00 kg | gm = GMm Rm = 6,67 × 10-11 × 7,35 × 1022 (1,74 × 104)2 = 1,619252874 w = mg 126,30 = m × 1,619252874 m = 78,00 kg |
(4) [6]
QUESTION 4
4.1 Projectile motion is the motion of an object upon which the only force acting is the force of gravity.
OR
Projectile motion is the motion of an object that experiences only gravitational force.(2)
4.2 (3)
| OPTION 1 | |
| (upwards positive) | (downwards positive) |
| vf = vi + g∆t 0 = (13) + (-9.8 ). ∆t ∆t = 1,33 s | vf = vi + g∆t 0 =(-13) + (9.8) .∆t ∴ ∆t = 1,33 s |
| OPTION 2 | |
| (upwards positive) | (downwards positive) |
| vf2 = vi2 + 2a∆y 02 = 132 + 2 (-9,8) ∆y ∆y = 8,62244898 m ∆y = vf + vi ∆t 2 8,62244898 = 0 + 13 ∆t 2 ∴ ∆t = 1,33 s | vf2 = vi2 + 2a∆y 02 = 132 + 2 (9,8) ∆y ∆y = 8,62244898 m ∆y = vf + vi ∆t 2 8,62244898 = 0 - 13 ∆t 2 ∴ ∆t = 1,33 s |
| OPTION 3 | |
| (upwards positive) | (downwards positive) |
| vf2 = vi2 + 2g∆y 02 = 132 + 2 (-9,8) ∆y ∆y = 8,62244898 m ∆y = vi∆t + 1 g∆t2 2 8,62 = 13 ∆t + 1 (- 9,8) ∆t2 2 ∆t = 1,33 s (3) | vf2 = vi2 + 2g∆y 02 = 132 + 2 (9,8) ∆y ∆y = 8,62244898 m ∆y = vi∆t + 1 g∆t2 2 8,62 = 13 ∆t + 1 (9,8) ∆t2 2 ∆t = 1,33 s (3) |
4.3 (4)
| OPTION 1 | |
| (upwards positive) | (downwards positive) |
| vf 2 = vi2 +2 g∆y 02 = (13)2 + 2 (-9,8) ∆y ∆y = 8,62 m | vf 2 = vi2 +2 g∆y 02 = (-13)2 + 2 (-9,8) ∆y ∆y = 8,62 m |
OPTION 2 | |
(upwards positive) | (downwards positive) |
∆y = vi∆t + 1 g∆t2 | ∆y = vi∆t + 1 g∆t2 = -13(1,33) + 1 (9,8)(1,33)2 |
| OPTION 3 | |
| (upwards positive) | (downwards positive) |
| ∆y = vf + vi ∆t 2 = 0 + 13 × 1,33 2 = 8,65 m | ∆y = vf + vi ∆t 2 = 0 - 13 × 1,33 2 = - 8,65 m = 8,65 m |
4.4
4.4.1
| OPTION 1 | |
| Taking top of building as starting point | |
| (upwards positive) | (downwards positive) |
vf = vi + a∆t | vf = vi + a∆t vf = 0 + (-9,8)(3,28 - 1,33) vf = 19,11 m.s-1 |
| OPTION 1 | |
| Taking balcony as a starting point | |
| (upwards positive) | (downwards positive) |
| vf = vi + a∆t vf = -13 + (-9,8)(3,28 - 1,66) vf = -19,08 m.s-1 vf = 19,08 m.s-1 (3) | vf = vi + a∆t vf = -13 + (-9,8)(3,28 - 1,66) vf = 19,08 m.s-1 (3) |
4.4.2
| OPTION 1 | |
| (upwards positive) | (downwards positive) |
| ∆y = h = vi∆t + 1 a∆t2 2 h = -13(0,62) + 1 (-9,8)(0,62)2 2 h = - 9,94 m Height of building = h + ∆y = 8,62 + 9,94 = 18,56 m | ∆y = h = vi∆t + 1 a∆t2 2 h = 13(0,62) + 1 (9,8)(0,62)2 2 h = - 9,94 m Height of building = h + ∆y = 8,62 + 9,94 = 18,56 m |
OPTION 2 | |
(upwards positive) | (downwards positive) |
vf2 = vi2 + 2a∆y | vf2 = vi2 + 2a∆y |
| OPTION 3 | |
| (upwards positive) | (downwards positive) |
| vf2 = vi2 + 2a∆y (-19,08)2 = (-13)2 + 2(-9,8)∆y h = ∆y = 9,95 m Height of building = ∆y + h = 8,62+ 9,95 = 18,57 m = 18,57 m | vf2 = vi2 + 2a∆y (-19,08)2 = (-13)2 + 2(-9,8)∆y h = ∆y = 9,95 m Height of building = ∆y + h = 8,62+ 9,95 = 18,57 m = 18,57 m |
(6) [18]
QUESTION 5
5.1 Downwards(1)
5.2 Two (times) OR 2 (times) (1)
5.3 Elastic (1)
5.4 4 s and (2)
5.5 Positive marking from QUESTION 5.4 
Criteria for graph
- The height from which the ball was dropped
- The times when the ball was at its maximum height - 4 s and 8 s
- Shape of the graph/Vorm van die grafiek (3) [8]
QUESTION 6
6.1 The total linear momentum of an isolated system is conserved.
OR
In an isolated system the total linear momentum before collision is equal to the total linear momentum after collision. (2)
6.2
- ∑pi = ∑pf
mvViv + mcVic = mvVfv +mcVfc
1500 x 30 + 1200 x 80000 = 1500 Vfv + 1200 x 25
3600
Vfv = 27,78 m.s-1 in the same direction (5)
6.3 (3)
| OPTION 1 | OPTION 2 |
| ∆p = mvf - mvi ∆p = 1200 x 25 - 1200 x 80000 3600 ∆p = 3333,33 kg.m.s-1 In the direction of the motion | ∆p = mvf - mvi ∆p = 1500 x 27,78 - 1500 x 30 ∆p = -3330 kg.m.s-1 ∆p = 3330 kg.m.s-1 In the direction of the motion |
6.4
- ∑Eki = ½ m vic2 + ½mviv2
∑Eki = ½ x 1200 x (80000)2+ ½ x 1500 x 302
3600
∑Eki = 971296,30 J
∑Ekf = ½m vfc2 + ½mvfv2
∑Ekf = ½ x 1200 x (25)2+ ½ x 1500 x 27,782
∑Ekf = 953796,30 J
∑Eki ≠ ∑Ekf
Collision is inelastic (5) [15]
QUESTION 7
7.1 To the right
For momentum to be conserved:
- momentum of trolley B = momentum of trolley A.
- momentum of B must be in an opposite direction to the momentum of trolley A (3)
7.2 Let right/East be positive
- Σpi = Σpf
mAvAi + mBvBi = mAvAf + mBvBf Any one
(mA + mB) vi = mAvAf + mBvBf
(0,5 + 0,75) x 0 = 0,5 x (-2,5) + 0,75vBf
vBf = 1,66 m.s-1 (to the right/East) (4)
7.3.1 21,5 N
- According Newton’s Third law of motion, when the wall exerts a force on the trolley, the trolley simultaneously exerts a force (of 21,5 N) on the wall but in an opposite direction.
NOTE: If learners just state Newton’s third law without using it to explain, no marks will be awarded (3)
7.3.2
- Fnet ∆t = m ∆v Any one
Fnet ∆t = m ∆(vAf - vAi)
-21,5 x 0,1 = 0,5(vAf - 2,5)
-2,15 = 0,5 vAf - 1,25
vAf = -1,8 m.s-1
vAf = 1,8 m.s-1 (to the left/west) (5)
7.3.3 Remains the same.
- When the contact time increases, the net force decreases. (3) [18]
QUESTION 8
8.1 The (total) mechanical energy of an isolated system is conserved. (2)
8.2 (4)
| OPTION 1 | OPTION 2 |
| MEB = MEC (EP + EK)B = (EP + EK)C (mgh + ½m v2)B = (mgh + ½m v2)B 0+ ½(0,2)vi2 = 0,2(9,8)(1,2)+ 0 Vi = 4,85 m.s-1 | Wnc = ∆Ek + ∆Ep 0 = ½m vf2 - ½m vi2 + mgh2 - mgh1 0 =0 - ½(0,2)vi2 + 0,2(9,8)(1,2) - 0 Vi = 4,85 m.s-1 |
8.3
| OPTION 1 | OPTION 2 | OPTION 3 |
| ∆x = vf + vi ∆t 2 = 6 + 4,85 x 0,82 2 = 4,45m | vf = vi + a∆t 4,85 = 6 + a x 0,82 a = - 1,40 m.s-2 ∆x = vi∆t + ½ a∆t2 = 6x0,82 + 0,5xax0,822 = 4,45 m | vf = vi + a∆t 4,85 = 6 + a x 0,82 a = - 1,40 m.s-2 vf2 = vi2 + 2a∆x 4,852 = 62 + 2(-1,4) ∆x ∆x = 4,46 m |
(3) [9]
QUESTION 9
9.1The net work done on an object is equal to the change in the kinetic energyof the object.
OR
The amount of work done by a net force on object is equal to the change in the object’s kinetic energy. (2)
9.2 (4)
| OPTION 1 | OPTION 2 |
 | |
- Mark awarded for both the arrow and label
- Do not penalise for length of forces since drawing is not drawn to scale.
- If force(s) do not make contact with body
9.3 OPTION 1
- Wnet = ∆EK
Wnet = WApp + W// + Wf
WApp + W// + Wf = ∆EK Any one
260(8)Cos0°) + 0,16(25)(9,8)(Cos40°)(8)(Cos180°) + 25(9,8)(sin40°)(8)(Cos180°) = ½(25)vf2 - 0
2080 - 240,23 -1259,86 = 12,5 vf²
579,91 = 12,5 vf2
vf = 6,81 m.s-1
OPTION 2
- Wnc = ∆EK+ ∆Ep
Wapp + Wf = ∆EK+ WFg Any one
260(8)(cos0°) + 0,16(25)(9,8)(Cos40°)(8)(cos180°) = 25(9,8)(Sin40°)(8)(cos180°) + ½ (25)vf2 - 0
vf = 6,81 m.s-1 (6)
9.4
| OPTION 1 | OPTION 2 |
| ∆x = (vi + vf) ∆t 2 8 = (0 + 6,81) ∆t 2 ∆t = 2,35s P = W ∆t P = 2080 2,35 = 885,11 W | Vav = vf + vi 2 Vav = 0 + 6,81 2 Vav = 3,405 Pav = F.Vav Pav = 260 × 3,405 = 885,11 W |
(4) [16]
QUESTION 10
10.1.1 Doppler effect is 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
OR
Doppler effect is the apparent change in frequency of a wave when thereis relative motion between the source and an observer
OR
Doppler Effect is an (apparent) change in observed/detected frequency(pitch), (wavelength) as a result of the relative motion between a source and an observer (listener) (2)
10.1.2 750 Hz
10.1.3
- Waves are compressed as Khosi moves towards the source.
- Wavelength become shorter.
- Wavelength is inversely proportional to frequency, hence the detected frequency is higher. (3)
10.1.4
- FL = V ± VL FS
V ± VS
When Khosi approaches - 750 = 340 +vL fs
340
fs = 750 × 340 ................ (1)
340 + vL
When the Khosi moves away - 700 = 340 - vL
340
fs = 750 × 340 ................ (2)
340 + vL
(1)=(2): - 750 × 340 = 750 × 340
340 + vL 340 + vL
750(340 + vL) = 700(340 + vL)
1450vL = 17000
vL =11,72 m.s-1 (6)
(3)
(3)10.2.1
- Away
- It shows a red shift.
- Frequency is decreasing.
- Wavelength becomes longer. (4)
10.2.2The universe is expanding. (1) [20]
QUESTION 11
11.1 The force per unit positive charge. (2)
11.2
- E = kQ
r2 Any one
Enet = E1 + E2
(6) [8]
(6) [8]TOTAL: 150