PHYSICAL SCIENCES: PHYSICS
PAPER 1
GRADE 12 
NSC PAST PAPERS AND MEMOS
FEBRUARY/MARCH 2017

GENERAL MARKING 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 may be relevant symbols and  applicable substitutions. 
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. 
1.3 If no formula is given, but all substitutions are correct, a  candidate will forfeit one mark. 
1.4 No penalisation if zero substitutions are omitted in calculations  where correct formula/principle is given correctly.  
1.5 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 the correct  substitutions. The mark for the incorrect numerical answer is  forfeited. 
1.6 Marks are only awarded for a formula if a calculation has been  attempted, i.e. substitutions have been made or a numerical  answer given. 
1.7 Marks can only be allocated for substitutions when values are  substituted into formulae and not when listed before a calculation  starts. 
1.8 All calculations, when not specified in the question, must be done  to a minimum of TWO decimal places. 
1.9 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. 
1.10 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. Eenhede word slegs in die finale antwoord op 'n vraag verlang. 
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-1instead of  N∙C-1, and cm∙s-1or 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 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-1and m/s will also  be accepted. 

4. POSITIVE MARKING 
Positive marking regarding calculations will be followed in the following cases: 

4.1 Subquestion to subquestion: When a certain variable is  incorrectly calculated in one subquestion (e.g. 3.1) and needs to be  substituted into another subquestion (3.2 or 3.3), full marks are to  be awarded for the subsequent subquestions 
4.2 Multi-step question in a subquestion: If the candidate has to  calculate, for example, current in the first step and gets it wrong  due to a substitution error, the mark for the substitution and the  final answer will be forfeited.

5. 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 to QUESTION 3.1, and 3.1 is incorrect, no  marks can be awarded for QUESTION 3.2. However, if the answer for e.g.  QUESTION 3.1 is based on a calculation, the motivation for the incorrect  answer in QUESTION 3.2 should be considered. 

MEMORANDUM

QUESTION 1
1.1 D✔✔ (2)
1.2 C✔✔ (2)
1.3 A✔✔ (2)
1.4 D✔✔ (2)
1.5 B✔✔ (2)
1.6 C✔✔ (2)
1.7 B✔✔ (2)
1.8 A✔✔ (2)
1.9 D✔✔ (2) 
1.10 B✔✔ (2)

[20]

QUESTION 2
2.1 0 N/zero✔ (1)
2.2 
1M

Accepted labels

w

Fg/Fw/weight/mg/gravitational force/N/19,6 N 

f

Ff riction/Ff/friction/fk 

N

FN/Fnormal/normal force  

 

Deduct 1 mark for any additional force. 

 

Mark is given for both arrow and label 

  • 1 mark if BOTH components of  weight are shown.
  • All other rules in the table apply. (3)  

2.3.1

  • Fnet = ma
    fk - mgsinθ = 0
    fk = mgsinθ 
  • fk - mgsinθ = 0
  • fk = mgsinθ  

✔ 1 mark for any of these

 fk = (2)(9,8) sin7º ✔ 
fk = 2,39 N ✔ (2,389) N (3) 

2.3.2 POSITIVE MARKING FROM QUESTION 2.3.1

  • fk = μkN
    = μkmgcos7º
    ✔ 1 mark for any of these
    2,389 = μk(2)(9,8)cos7º ✔ 
    µk = 0,12 ✔ (3) 

2.3.3 POSITIVE MARKING FROM QUESTION 2.3.2
OPTION 1

  • Fnet = ma
    - fk = ma
    - μkN = ma - μk(mg) = ma
    ✔ 1 mark for any of these
    - (0,12)(2)(9,8) ✔= 2a✔
    a = -1,176 m.s-2      (-1,18)
    vf2 = vi2 + 2aΔx
    0 = (1,5)2+ 2(-1,176)Δx✔
    Δx = 0,96 m
    Distance is 0,96 m✔ 

OPTION 2

  • Wnet = ΔK 
    Wnet = ΔE
    Wnc = ∆K + ∆U
    Wnc = ∆EK + ∆EP
    μkN∆x cos θ = ½ mvf2– ½ mvi2 
    1 mark for any of these
    NOTE: substituting into any of the above equations will lead to the  following:
    (0,12) (2)(9,8) ✔∆x cos 180º ✔= 0 – ½ (2)(1,5)2
    ∆x = 0,957 m✔ (5) 

[15] 

QUESTION 3
3.1

  • (Motion of) an object in which the only force acting is the gravitational  force. ✔✔ 
    OR
    (Motion of)an object which has been given an initial velocity and which follows  a path entirely determined by the effects of gravitational  acceleration/force. ✔✔
    OR
    The (motion of )an object that is projected, thrown or shot either upwards or  downwards into the air and on which the only force considered/acting is  gravitational. ✔✔ (2)
    Note: 2 or/of 0 

3.2 No ✔ 

  • The balloon is not accelerating at the rate of 9,8 m∙s-2/moving with constant  velocity/acceleration is 0 m∙s-2✔ (2)
    OR
    There are other forces (e.g.,friction) acting on the balloon besides  gravity.✔
    Net force acting on the balloon is zero

3.3 

OPTION 1
Upward positive
Δy = viΔt + ½ aΔt2✔ 
-22✔ = (-1,2) Δt + ½ (-9,8)Δt2✔  
Δt = 2 s✔ 

Downward positive 
Δy = viΔt + ½ aΔt2✔ 
22✔ = (1,2) Δt + ½ (9,8)Δt2✔  
Δt = 2 s✔

OPTION 2
Upward positive
vf2 = vi2 + 2aΔy✔
vf2 =  =(-1,2)2 + (2)(-9,8)(-22) 
vf = -20,8m.s-1
For both equations ✔

vf = vi + aΔt  ✔
-20,8 = -1,2 + -9,8Δt✔ 
Δt = 2 s✔ 

Downward positive

vf2 = vi2 + 2aΔy
vf2 = (1,,2)2 + (2)(9,8)(22) 
vf = 20,8 m∙s-1 

For both equations ✔

vf = vi + aΔt  
20,8 = 1,2 + 9,8Δt✔ 
Δt = 2 s✔ 

OPTION 3
Upward positive
vf2 = vi2 + 2aΔy
vf2 = [(-1,2)2 +(2)(-9,8)(-22)]
vf = -20,8 m∙s-
Δy = vf + vΔt 
   
         2
-22 = ( -1.2 + -20,8) Δt
                    2
Δt = 2s

For both equations ✔ 

Downward positive
vf2 = vi2 + 2aΔy
vf2 = [(1,2)2 +(2)(9,8)(22)]
vf = -20,8 m∙s-
Δy = vf + vf  Δt 
   
         2
-22 = ( 1.2 +   20,8) Δt
                    2
Δt = 2s
For both equations  ✔ 

OPTION 4
(Emech)Top = (Emech)Ground
(EP +EK)Top = (EP +EK)Bottom
1 mark for any 
(mgh + ½ mv2)Top = (mgh + ½ mv2)Bottom
Wnet = ∆EK 
(9,8)(22) + ½ (1,2)2 = 0 + (½ )(vf2) ✔ 
vf = 20,80 m∙s-1 
vf = vi + aΔt  
20,8 = 1,2 + 9,8Δt✔ 
Δt = 2 s✔ 

NOTES: 

  • Each substitution must include the correct values of 22 m and the velocity  of 1,2 m∙s-1 
  • The values of vf and vi can also be used with Fnet∆t = ∆p = (pf – pi) = (mvf – mvi

 (4) 

3.4 

Upward positive
POSITIVE MARKING FROM QUESTION 3.3 

vf = vi + aΔt ✔ 
0 = 15 + (-9,8)Δt✔ 
Δt = 1,53 s  
For addition
Total time elapsed = 2 + 1,53 + 0,3✔  
= 3,83s

OR
Displacement of the balloon
Δy = viΔt + ½ aΔt
 = -(1,2)(3,83) ✔ 
= - 4,6 m 
Height :
= 22 - 4,6✔ 
= 17,4m✔ 

OR
yf = yi +∆y 
 = [22 – (1,2)(3,83)]✔ ✔ 
 = 17,4 m 
Height = 17,4 m✔ 

Downward Positive
POSITIVE MARKING FROM QUESTION 3.3

vf = vi + aΔt ✔ 
0 = 15 + (-9,8)Δt✔ 
Δt = 1,53 s  
For addition
Total time elapsed = 2 + 1,53 + 0,3✔  
= 3,83s

OR
Displacement of the balloon
Δy = viΔt + ½ aΔt
 = -(1,2)(3,83) ✔ 
= - 4,6 m 
Height :
= 22 - 4,6✔ 
= 17,4m✔ 

OR
yf = yi +∆y 
 = [-22 + (1,2)(3,83)]✔ ✔ 
 = 17,4 m 
Height = 17,4 m✔ 

(6) 
[14] 

4.1 It is the product of the resultant/net force acting on an object✔ and the time  the resultant/net force acts on the object. ✔  (2)
NOTE: ONLY 1 MARK FOR “CHANGE IN MOMENTUM”
4.2.1 (3) 

p = mv✔ 
 = (0,03)(700) ✔ 
 = 21 kg∙m∙s-1✔  
Note: 2/3 if ∆p = (pf – pi) = (mvf – mvi) is used. 

4.2.2 

OPTION 1
POSITIVE MARKING FROM 4.2.1

∆t for a bullet =    60✔    = 0,27 s 
                            220 
Fnet∆t = ∆p = (pf – pi) = (mvf – mvi)  
Fave gun on bullet  Δp     
                               Δt 
✔ 1 mark for any one
= 21- 0
    0,27 
 = 77,01 N ✔ (77,78 N) 
∴average force of bullet on gun
= 77,01 N/ 77,8 N to the west ✔

OR

-77,01 N / - 77,78 N  

OPTION 2
POSITIVE MARKING FROM 4.2.1
Fnet∆t = ∆p = (pf – pi) = (mvf – mvi)  
Fav = Δp    
          Δt 
✔ 1 mark for any one
∆ptot = (21)(220) = 4 620 kg∙m∙s-1 
 Fave gun on bullet = 4 620 - 0✔ 
                                      60 
 = 77,00 N✔ 
∴average force of bullet on gun
= 77,01 N /77,78 N to the west ✔ 

OR

- 77,01 N / -77,78 N 

OPTION 3
vf = vi + a∆t 
a =    700- 0✔   
       (60/ 220) 
a = 2592,59 m∙s-2 
Fnet= ma ✔ 
Fnet = (0,03)(2592,59) ✔ 
Fav = 77,78 N✔ 
∴average force of bullet on gun
= 77,01 N / 77,78 N to the west ✔

OR 

= - 77,01 N/ -77,78 

NOTE: ACCEPT RANGE: 77 N - 77,78 N

(5) 

4.3 

POSITIVE MARKING FROM 4.2.2

77 N/77,78 N✔ to the east✔ 

(2) 
[12] 

QUESTION 5
5.1 (2) 

The rate at which work is done/ Rate at which energy is expended. ✔✔ 

5.2.1 (3) 

OPTION 1
W = F∆xcosθ✔ 
Wgravity  = mg∆ycosθ 
 = (1 200)(9,8)(55)cos180º ✔ 
 = - 646 800 J (6,47 x105J)✔

OPTION 2
W = - ∆Ep✔ 
 = -(1200)(9,8)(55 -0) ✔ 
 = -646800 J✔ 

-1 if either negative is omitted or Ep = mgh is used instead of W  

5.2.2 (2) 

Related Items

Wcounterw eight = mg∆ycosθ 
 = (950)(9,8)(55)cos0º ✔ 
 = 512 050 J (5,12 x105J)✔ 

5.3 

OPTION 1
POSITIVE MARKING FROM QUESTIONS 5.2.1 AND 5.2.2 
Wnet = ∆EK ✔ 
Wgravity + Wcountw eight + Wmotor = 0 
Wmotor = - (Wgravity + Wcountweight
Wnc = ∆EK + ∆E
1 mark for any one
NOTE: Substituting into any of the above equations will lead to: 

-646800✔ + 512050 ✔+ Wmotor = 0 
∴Wmotor = 134 750 J 
P ave motor W  
                       Δt✔ 
134750 ✔ 
       180 
 = 748,61 W✔ 

OPTION 2
Fnet = 0 
Fgcage + Fgcount + Fmotor = Fnet 
✔ 1 mark for any one
-117600✔ + 9310✔ + Fmotor = 0
Fmotor = 2450 N 
Pave = Fvave✔ 
 = 2450  55  ✔ 
            180 
 = 748,61 W 

OPTION 3
P ave = Fvave ✔✔ 
                   ✔              ✔
 = [1200 (9,8) – 950(9,8)]   55    ✔     
                                          180 
 = 748,61 W✔ 

(6) 
[13] 

QUESTION 6
6.1.1 The Doppler effect.✔ (1)
6.1.2 Measuring the rate of blood flow
OR
Ultrasound (scanning)✔ (1) 

6.1.3 (6) 

FL = V   ±  VL  F  OR       FL    V     FS     OR  FL =      V    FS
        V  ±   VS                              V - VS                          V + VS
2600 =     340    FS
         ( 340  - VS)  
1750  =    340      FS
            ( 340  - VS)  
 2600(340-vs) = 1750(340 + vs
vs = 66,44 m∙s-1 ✔ 

6.1.4

  1. Increase✔ (1) 
  2. Decrease/✔ (1) 

6.2.1 The spectral lines (light) from the star are shifted towards longer  wavelengths. ✔✔ (2) 
6.2.2 Decrease ✔ (1)

[13]

QUESTION 7
7.1.1 Removed✔  (1) 
7.1.2 (3) 

n = 
      e 

Do not penalise for negative sign of charge used in calculation

  6 × 10-6
     1,6 ×10-19 
 = 3,75 x 1013 electrons

7.2.1 Negative ✔ (1) 
7.2.2 WHHHAT(2) 

NOTE: 
Vectors not drawn to scale
Learners forfeit 1 mark for:

  1. Wrong directions/verkeerde rigtings 
    OR
    Arrows not shown/Pyltjies nie aangedui nie 

Give credit to the required forces even if a triangle of forces is drawn.
ACCEPT: two separate diagrams 
ACCEPT:correctly drawn vector but no labels

7.2.3 (3)

F = KQ1Q2 
         r2
F1,3X = (9 × 109)(2 × 10-6) (6 × 10-6  (cos 45º)  = (0,0764)
                               r2                                                   r2
ACCEPT:
F = KQ1Q2 
         r2
F1,3XKQ1Q2   (cos 45º) 
                  r2                                                (3)

7.2.4 (4) 

POSITIVE MARKING FROM QUESTION 7.2.3
OPTION 1
F = KQ1Q2 
         r2
F2,3X = (9 × 109)(2 × 10-6) (6 × 10-6  (cos 45º)  = (0,0764)
                               r2                                                   r2

Fx = F1,3x + F2,3x 
Fx = 0,0764     0,0764    = 2 0,0764  
          r2                   r2                   r2  

1 mark for the addition 

NOTE: Fy net = 0
(0,12) ✔ =  0,1528
                      r2   
r = 1,128 m ✔ 

OPTION 2
Fnet2 = (F1,3 )2 + (F2,3)2 

= (K  Q1Q3 )2   + (K  Q2Q3)2
           r2                       r2
= 2 (K  Q1Q3 )2  
           r
✔1 mark for any of the three 
= 2 [(9 × 109)(2 × 10-6) (6 × 10-6)]2
                                    r2

= 2 (0,108)   
             r4
(0,12)2 = 2 (0,108)2   
                     r4
∴r = 1,128 m✔ 
NOTE:  Fnet = Fnet(x) since Fnet (y) = 0

7.3.1 The electric field at a point is the (electrostatic) force experienced ✔per unit positive charge ✔placed at that point   (2)
7.3.2  (5) 

OPTION 1
E =  kQ  
         r2

100 =  (9×109 )Q 
               (0,6)2 
Q = 4 x 10-9

When the electric field strength 50 is N∙C-1

E =  kQ  
         r2

50 =  (9×109 )  (4 x 10-9)          
                     (0,6)2 
✔ For the equation

 r = 0,85 m (0,845) m✔ 

OPTION 2

E =  kQ  
         r2

E1 r22
   E    r12
  100   =    r2
   50        (0,6)2

∴r = 0,85 m (0,849 m)✔ 

[21]

QUESTION 8
NEGATIVE MARKING FOR 8.1.1,8.1.2 AND 8.1.3
8.1.1

  • P and Q burn with the same brightness ✔ same potential difference/same  current✔  (2) 

8.1.2

  • P is dimmer (less bright) than R
    OR
  • R is brighter than P✔
    R is connected across the battery alone therefore the voltage (terminal pd) is  the same as the emf source (energy delivered by the source). ✔
    OR
    The potential difference across R is twice (larger/greater than) that of P./The  current through R is twice (larger/greater than) that of P.
    OR
    P and Q are in series and are both connected across the same battery, ✔ hence the voltage (terminal pd) is shared equally ✔(P and Q are potential  dividers) Therefore R is brighter.
    OR
    Potential difference across P is half that across R(2)

8.1.3 T does not light up at all✔ 
ACCEPT

  • T is dimmer (less bright) than R✔
  • R is brighter than T✔
    Reason
    • The wire acts as a short circuit. ✔
      OR
    • The potential difference across T / current in T is zero.✔  (2) 

8.2.1  opt 2
opt 2 2(6) 

8.2.2 

OPTION 1 
V = IR 
V5 = ℇ – (V8 + V1
Any one✔
= 20✔ – [ 1,62(8 + 1)]✔ 
 = 5,42 V✔ 

OPTION 2
POSITIVE MARKING FROM 8.2.1
R// =    ( 5)( 10)   = 3,33Ω                                   V|| = IR|| ✔ 
             5  + 10                                                    = (1,62)(3.33) ✔✔  = 5,39 V ✔

VR// =  R//     × Vtot 
           Rtot 
VR// =  (3,33)  (20)✔✔ 
          (12,33 )
 = 5,41 V✔ 

OPTION 3
POSITIVE MARKING FROM 8.2.1 ✔ 
I5R5 = I10R10
5I5    = 10(1,62 - I5) ✔ 
I5 = 1,08 A 
V5 = (1,08 )(5) ✔ 
 = 5,4 V✔ 

(4)

POSITIVE MARKING FROM 8.2.1
OPTION 4

I5 10   ×   Itot
        15
= 2 (1,62)
   3
= 1,08 A 

V5 = I 5 R5 ✔ 
V5 = (1,08 )(5) ✔ 
 = 5,4 V✔ 

(4) 
8.2.3 

POSITIVE MARKING FROM 8.2.1
OPTION 1 
P = IV = Iℇ ✔ 
 = (1,62)(20)✔ 
= 32,4 W✔

POSITIVE MARKING FROM 8.2.1
OPTION 2
P = IV✔ 
Ptot = P + P// + P 
 = IV8 + IV// + IV1 
 = I2(R8 +R//+ R1
 = (1,62)2[8 +3,33 + 1)] ✔ 
 = 32,36 W✔

POSITIVE MARKING FROM 8.2.1 AND 8.2.2
OPTION 3
P = I2R
I5 = V5 = 5,4  = 1,08 A 
       R5     5 
∴ I10 = 0,5A
Ptot = I28R8  +  I25R5   + I210R10
= (1,62)2[8 + 1] + (1,08)2(5) + (0,54)2(10) ✔ = 32,37 W✔

OPTION 4

P = V2
      R
P =        202           
         (8 +1+ 3,33) 

 = 32,44 W✔ 

P = I2Rtot ✔ 
 = (1,62)2(12,33) ✔ 
= 32,36 W✔

(3) 

NOTE: Range 32,35- 32,45 

[19]

QUESTION 9
9.1 Slip rings ✔ (1)
9.2 Q9GHSD    (2) 

Marking criteria

 

Sine graph starts from 0. 

Two complete waves (between t0 and t2) 

9.3 Any TWO

  • Increase the speed of rotation✔
  • Increase the number of coils (turns)✔
  • Use stronger magnets
    ACCEPT: Increase surface area(2) 

9.4

  • The rms value of an AC voltage it that value of the AC voltage which will  dissipate the same amount of energy as DC. 
    OR
  • The rms value of an AC voltage it that value of the AC voltage which will  produce the same joule heating effect as DC.  (2)

9.5 (3) 

OPTION 1
Pave = Irms/wgkVrms/w gk ✔ 

1500 = Irms/w gk(240) ✔ 

Irms/w gk = 1500 
                 240 
 = 6,25 A ✔

OPTION 2 
Pave =2✔ 
           R

1500 =  2402    
                R 

R = 38,4 Ω 
Irms = V 
          R
=    240✔ 
     38,4 

 = 6,25 A✔ 

[10]

QUESTION 10
10.1

  • The minimum frequency of light ✔needed to emit electrons from a certain  metal surface.✔
    OR
  • The minimum frequency of light✔ below which electrons will not be emitted  from the surface of a certain metal. ✔ . (2) 

10.2 The speed remains unchanged. ✔  (1) 
10.3 (5)

OPTION 1
c = fλ✔, 
3 x 108 = f(6 x 10-7) ✔ 
∴ f = 5 x 1014 Hz✔ 

  • The value of f is less than the threshold frequency of the metal, ✔ therefore  photoelectric effect is not observed.✔ 

OPTION 2 
For the given metal
W0 = hf0✔ 
 = (6,63 x 10 -34)(6,8 x 1014 )✔ 
 = 4,51 x 10-19
For the given wavelength
Ephoton = hc  
                 λ
(6,63 ×10-34)(3×108 )✔ 
               6×10 -7
= 3,32 x 10-19 J  
Ephoton = hf 
 =( 6,63 x 10-34)(5 x 1014) ✔ 
 = 3,32 x 10-19 J  

  •  This energy is less than the work function✔ of the metal,therefore  photoelectric effect is not observed.✔ 

OPTION 3

c = f0λ0✔ 
3 x 108 = 6,8 x 10140) ✔ 
λ0 = 4,41 x 10-7 m✔ 

  • The threshold wavelength (λ0) is smaller than 6 x 10-7 m ✔ therefore  photoelectric effect is not observed.✔ 

10.4 (5)  

E = Wº   +   EK(MAX)

  • E =  Wº   +   ½mv2max
  • hc/λ = hfº   +    ½mv2max
  • hf = hfº   +    ½mv2max

Any one of the three ✔

(6,63 x 10-34)(7,8 x 1014) ✔ = (6,63 x 10-34)(6,8 x 1014) + ½mv2max
 ½mv2max = 6,63 x 10-20
½ (9,11 x 10-31) v2max ✔= 6,63 x 10-20 
                         vmax = 3,82 x 105 m∙s-1✔ 

[13] 
TOTAL: 150

Last modified on Monday, 12 July 2021 07:18