PHYSICAL SCIENCES PAPER 1
GRADE 12
NATIONAL SENIOR CERTIFICATE
MEMORANDUM
JUNE 2018

GENERAL GUIDELINES

1. 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 not count any marks.
2. 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.
3. 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, not a full stop, for example, m∙s^-1.
   For marking purposes, m∙s^-1 and m/s will also be accepted.
   Positive marking regarding calculations will be followed in the following cases:
   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.
   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.

QUESTION 1
1.1 B ✓✓ (2)
1.2 A ✓✓ (2)
1.3 C ✓✓ (2)
1.4 A ✓✓ (2)
1.5 B ✓✓ (2)
1.6 B ✓✓ (2)
1.7 A ✓✓ (2)
1.8 C ✓✓ (2)
1.9 A ✓✓ (2)
1.10 A ✓✓ (2)
[20]

QUESTION 2
2.1 If the resultant/net force acts on an object, the object will accelerate in the direction of the resultant/net force with an acceleration that is directly proportional to the resultant/net force ✓and inversely proportional to the mass ✓of the object.(2)
2.2 ✓ (4)
2.3
2.3.1
f_k = μkN ✓
f_k = μkmg    any one
f_k = 0,2 X 2 x 9.8 ✓
f_k = 3.92 N ✓ (3)
2.3.2 2 kg block
Fnet = ma
T + (-f ) = ma ✓any one
T + ( - 3,92) = 2 x 4 ✓
T = 8 + 3,92
T = 11,92 N
(NB: If right is positive f - T = - ma)

X kg block
Fnet = ma
w + (-T ) = ma (any mass substitution)
mg + (-T) = ma
m x 9,8 ✓ -11,92 = m x 4 ✓
5,8 m = 11,92
m = 2,06 kg ✓ (5)
[14]

QUESTION 3
3.1 Free-fall is the motion of an object when the only force acting on it is gravitational force ✓✓ (2)
3.2.1

3.2.2

3.2.3

3.3

OPTION 1 Downwards is positive

OPTION 2 Upwards is positive

(4)
[19]

QUESTION 4
4.1 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)✓✓
4.2F =  Gm₁m₂ 
               r²
34.9 = (6.67 x 10^-11) (5.98 x 10²⁴) (m)
                             (2 x 10⁸)²
m = 3500kg(4)✓✓✓✓
4.3 F_net = ma
F_earth = F_moon
Gm_sm_E = Gm_sm_m
   r²_E             r²_m
5.98 x 10²⁴ =   7.35 x 10 ²²
        d²           (3.8 x 10⁸ -d)²
(3.8 x 10⁸ -d)² =    7.35 x 10 ²²
     d²                    5.98 x 10²⁴
3.8 x 10⁸ - d = 0.11
        d
3,8 ×10⁸ – d = 0,11 d
1,11 d = 3,8 × 10⁸
d = 3,42 × 10⁸
Distance(PQ) = 3,42 × 10⁸ - 2 × 10⁸
= 1,42 × 10⁸ m 
(5)✓✓✓✓✓
[11]

QUESTION 5
5.1 The total linear momentum in a closed system remains constant (is conserved)✓✓
OR
The total momentum before a collision is equal to the total momentum after the collision in a closed system(2)
5.2 Left/ Backwards/ West ✓(1)
5.3 p= mv ✓
p = 1 x 10 ✓
p = 10 kg. m·s^-1 right v(3)
5.4 Fnet. Δt = Δp ✓
Fnet . 0,1 = 1 ( 10 – 0) ✓
Fnet = 1000 N right ✓✓(4)
5.5 Take motion to the right as positive
Σpi = Σpf
(m₁ + m₂) vi = m₁vf₁ + m₂vf₂ (Any one) ✓
(1 +100) (0) = (1) (10) + (100) v_f2 ✓
vf 2 = - 0,1m∙s¹
Speed = 0,1 m∙s^-1
Δx = v Δt ✓
60 = (0,1) Δt ✓✓
Δt = 600 s
Yes, it takes 10 min ✓(5)
[15]

QUESTION 6
6.1 (3)
6.2 Force that is independent on the path taken. ✓✓(2)
6.4 Gravitational force / weight ✓ (1)
6.4 Fnet = ma = 0
FN + (- mg cos 30) = 0
FN = mg cos 30 (any 1 ✓)

FN - 20 x 9.8 x cos 30º = 0 ✓
FN = 169.74 N ✓ (3)
6.5 From B TO A✓(1)
6.6 Option 1
Wnet = Δ EK
WFA + Wg// =½m(v²f – v²i) ✓ any one 1
FA. Δx ∙ cosθ + mgsin 30. Δx .cosθ = ½x 20( 10.8² – 12²) ✓
FA x 4 x 1 ✓ + 20 x9.8 sin 30 x 4 x -1✓ = - 273.6
4 FA - 392 = - 273.6
FA = 29.6 N ✓ (5)

Option 2
Wnc = Δ E_p + Δ E_K ✓
Wf = mghf – mghi + ½mv²f – v²i
F x 4.cos 0⁰ ✓= 20 x 9.8 x 4sin 30⁰ ✓– 0 + ½ . 20 . ( 10.8² – 12²) ✓
F = 29,6 N ✓
[15]

QUESTION 7
7.1 Air Friction ✓and Tension ✓✓(2)
7.2 (3)
7.3 The work done by the (net) force is equal to the change in the kinetic energy of an object ✓✓
OR
Net work done by the force is equal to the change in kinetic energy of the object. ✓✓(2)
7.4
Wnet = Δ Ek ✓
Fnet. Δx. cos θ= ½mvf² - ½mvi²
Fnet (30) cos180⁰ ✓= ½ (65) (0)² - ½ (65) (2,2)² ✓
Fnet x (- 30) = -15,73 N
ma = + 5,243333333
(65) a = 5,243333333 ✓
a = 0,08 m·s^-2 ✓
(5)
[12]

QUESTION 8
8.1 The Doppler effect is the change in the observed 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
The apparent change in the (observed) frequency when there is relative motion between the sound source and the observer. ✓✓(2)
8.2 The relative velocity between Ruby and the source of the sound is zero. ✓✓
The is no relative velocity between Ruby and the source of sound. ✓✓ (2)
8.3 INCREASES ✓ (1)
8.4 Wavelength of the sound source received by listener per second is inversely proportional to the frequency produced and hence the longer wavelength will produce lower frequency. ✓✓ (2)
8.5 f_L = v ± v_Lfs
             v ± v_s
(any one / enige een) ✓
f_L = v + v_L fs
         v
188 ✓ =  340 + v_L ✓ x 180 ✓
                  340
v_L = 15,11 m·s^-1 ✓ (5)
8.6 f_L = v ± v_Lfs
             v ± v_s
f_L = v + v_L fs (any one)
         v
f_L =  340 - 5 x 180 ✓
          340
fL = 177,35 m·s^-1 ✓
(4)
[16]

QUESTION 9
9.1 Option 1 
Emech at A = Emech at B ✓
(mgh + ½mv²) at A = (mgh + ½mv²) at B (any 1)
(0,4 x 9,8 x 1,2) + ½ x 0,4 x 02 ✓= (0,4x 9,8 x 0) + ½ x 0,4 x v² ✓
v = 4,85 m.s^-1 ✓
Option 2
Wnc = Δ E_p + Δ E_K ✓
0 = mghf – mghi + ½ mv²f – ½ v²i (any one 1)
0 = 0,4 x 9.8 x 0 - 0,4 x 9.8 x 1,2 ✓+ ½ . 0,4 vf² – 0 ✓
vf = 4,85 m.s-1 ✓ (4)
9.2.1 Σpi = Σpf
mv_iM + mv_iN =(m_c + m_m)v_f ✓ (any one 1)
0,4 x 4,85 + 0,3 x 0 ✓= ( 0,4 +0,3 ) v ✓
1,94 + 0 = 0,7 v
v = 2,77 m.s^-1 (right) ✓ (4)
9.2.2 ΣEk_i = ( ½ mv² ) M+( ½ mv²)N ✓
= ( 2½ x 0,4 x 4,85²) ✓ + ( ½ x 0,3 x 0²) ✓
= 4,7045 J
ΣKf = ( ½ mv ² ) M+( ½ mv²)N
= ( ½ x 0,4 x 2,77 ²) + ( ½ x 0,3 x 2,77 ²) ✓
= 1,53458 + 1,150935
= 2,69 J
Energy lost = ΣEk_f - ΣEk_i ✓
= 2,01 J ✓ (6)
9.2.3 Inelastic ✓ (1)
[15]

QUESTION 10
10.1 The electric field at a point is a force per unit positive charge ✓✓(2)
10.2

(2)
10.3.1
F = KQ₁Q₂✓
          r²
F =  (9 x 10⁹)(2 x 10^-6)(3 x 10^-6)
                        (0.1²)
F = 2,11x10⁶ N left ✓
(4)
10.3.2 E = k_Q
                  r²
E_M = 9 x 10⁹ x 2x 10^-6
                (0.1)²
= 1,8 x106 N∙C-1 left
E_N = 9 x 10⁹ x 3x 10^-6
                (0.06)²
= 7,50 x106 N∙C-1 left
E_net = E_M + E_N
E_net = 1,8 x10⁶ +7,50 x10⁶ ✓
= 9,3 x10⁶ N∙C^-1 left ✓
(5)
[13]

TOTAL:150