PHYSICAL SCIENCES
PHYSICS PAPER 1
GRADE 12 
NSC EXAM PAPERS AND MEMOS
NOVEMBER 2016

MEMORANDUM

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

QUESTION 2
2.1 When a resultant/net force acts on an object, the object will accelerate in the (direction of the net/resultant force). The acceleration is directly proportional  to the net force ✔and inversely proportional to the mass ✔of the object. 
OR
The resultant/net force acting on the object is equal (is directly proportional  to) to the rate of change of momentum of an object (in the direction of the  force). ✔✔  (2) 

2.2 (3) 

fk = μkN✔= μkmg 
= (0,15)(3)(9,8)✔
 = 4,41 N✔ 

2.3   2.3 P

Accepted Labels

w

Fg/Fw/force of Earth on block/weight/14,7 N/mg/gravitational force 

FN/Fnormal/normal force 

Tension/FT 

f

fkinetic friction/ff/w/f//Ff/wkinetic frictiong

25 N 

Fapplied/FA/F 

2.4.1    (3) 

OPTION 1

fk = μkN = μk(25sin 30º + mg) 
= 0,15[(25sin30º)✔ + (1,5)(9,8)✔]
= 4,08 N✔

OPTION 2

fk = μkN = μk(25cos 60º + mg) 
= 0,15[(25cos60º)✔ + (1,5)(9,8)✔] 
= 4,08 N✔ 

2.4.2 

POSITIVE MARKING FROM  QUESTION 2.2 AND QUESTION 2.4.1 
OPTION 1
For the 1,5 kg block
Fnet = ma ✔
Fx + (-T) +(- fk) = ma 
25 cos 30º – T – fk = 1,5a 
(25 cos 30º – T) – 4,08 ✔= 1,5a 
17,571 – T = 1,5a ……….(1) ✔
either AS ABOVE  OR BELOW  
For the 3 kg block 
T – fk = 3a 
T – 4,41✔ = 3a ………….(2) 
13,161 = 4,5 a 
a = 2,925 m∙s-2 
T = 13,19 N ✔ (13,17 N – 13,19 N) 

 

OPTION 2
For the 1,5 kg block
Fnet = ma ✔
Fx + (-T) +(- fk) = ma 
25 cos 30º – T – fk = 1,5a 
(25 cos 30o – T) – 4,08 ✔= 1,5a
17,571 – T = 1,5a ……….(1)✔
either one  
For the 3 kg block 
T – fk = 3a 
T – 4,41✔ = 3a ………….(2) 
35,142 – 2T = T – 4,41  
T = 13,18 N ✔  

OPTION 3
For the 1,5 kg block
Fnet = ma ✔
Fx + (-T) +(- fk) = ma 
25 cos 30º – T – fk = 1,5a 
(25 cos 30º – T) – 4,08 ✔
= 1,5a 17,571 – T = 1,5a ……….(1) ✔
either one 
a = 17,571− T 
          1,5 
For the 3 kg block 
T – fk = 3a 
T – 4,41✔ = 3a ………….(2) 
a = T − 4,41 
          3 
17,571− TT − 4,41 
          1,5           3 
T = 13,18 N✔ 

(5) 
[18] 

QUESTION 3
3.1 The motion of an object under the influence of gravity/weight/gravitational force  only / Motion in which the only force acting is the gravitational force.✔✔  (2) 

OPTION 1
Upwards positive:
vf2 = vi2 + 2aΔy ✔ 
 = 02+ (2)(-9,8)✔(-20)✔
vf = 19,80 m∙s-1 ✔ 
Downwards positive 
vf2 = vi2 + 2aΔy ✔ 
 = 02 + (2)(9,8)✔(20)✔ 
vf = 19,80 m∙s-1

OPTION 2
Upwards positive:
Δy = viΔt + ½ aΔt
-20 = 0 + ½ (-9,8) Δt2
✔either one 
Δt = 2,02 s 
vf = vi + aΔt  
 = 0 + (-9,8)(2,02) ✔ 
 = -19,80 m∙s-1 
 = 19,80 m∙s-1✔ 
Downwards positive 
Δy = viΔt + ½ aΔt
20 = 0 + ½ (9,8) Δt2✔ 
✔either one 
Δt = 2,02 s 
vf = vi + aΔt  
 = 0 + (9,8)(2,02)✔ 
 = 19,80 m∙s-1✔ 

 

OPTION 3
(Emech)Top = (Emech)Ground
✔1 mark for any 
(EP +EK)Top = (EP +EK)Bottom
(mgh + ½ mv2)Top = (mgh + ½ mv2)Bottom 
(9,8)(20) + 0✔ = (0 + ½vf2)✔ 
vf = 19,80 m∙s-1✔ 

OPTION 4
Wnc = ΔEp + ΔEk ✔ 
0 = mgΔh + ½ mΔv2 
0✔ = m(9,8)(0 – 20) + ½ m(vf2 – 0) ✔ 
vf = 19,80 m∙s-1

OPTION 5
Wnet = ΔEk ✔ 
mgΔxcos0º = ½ m(vf2 – 0) 
m(9,8)(20)(1)✔ = ½ mvf2 ✔ 
vf = 19,80 m∙s-1✔ 

(4) 

3.2.2 

POSITIVE MARKING FROM QUESTION 3.2.1
OPTION 1
Downwards positive
vf = vi + aΔt ✔ 
19,80 = 0 + (9,8)Δt ✔ 
Δt = 2,02 s ✔ 

Upwards positive
vf = vi + aΔt ✔ 
-19,80 = 0 + (-9,8)Δt ✔ 
Δt = 2,02 s ✔

 

OPTION 2
Upwards positive:
Δy = viΔt + ½ aΔt2✔ 
-20 = 0 + ½ (-9,8) Δt2✔ 
Δt = 2,02 s✔

Downwards Positive 
Δy = viΔt + ½ aΔt2✔ 
20 = 0 + ½ (9,8) Δt2✔ 
Δt = 2,02 s✔

OPTION 3
Downwards positive: 
Δy = [ v1 + vf] Δt
              2
20 = [ 0 + 19,80] Δt
              2
Δt = 2,02 s✔ 

Upwards positive:

Δy = [ v1 + vf] Δt
              2
-20 = [ 0 - 19,80] Δt
              2
Δt = 2,02 s✔ 

(3)

3.3

Downward positive
downward ve
Upward positive
upward ve

Notes 

✔✔ 

Straight line through the origin. 

(2) 
[11] 

QUESTION 4
4.1 A system on which the resultant/net external force is zero✔ A system which excludes external forces  (1) 
4.2.1 (3) 

OPTION 1
p = mv✔ 
30 000 = (1 500)v ✔ 
v = 20 m∙s-1

OPTION 2
Δp = mvf – mvi ✔ 
0 = (1 500)vf – 30 000 ✔
v = 20 m∙s-1

4.2.2 POSITIVE MARKING FROM QUESTION 4.2.1

OPTION 1

∑pi = ∑pf 
m1 v1i + m2v2i = m1 v1f + m2v2f 
✔ 1 mark for any
30 000 + (900)(-15)✔ = 14 000 + 900vB✔ 
∴vB = 2,78 m∙s-1 ✔east ✔ (Accept: to the right) 

OPTION 2
ΔpA = -ΔpB 
1 mark for any
pf – pi = -(mvf - mvi
14 000 – 30 000 ✔= 900vf – 900(-15) ✔ 
vf = 2,78 m∙s-1✔ east✔ (Accept: to the right) 

(5) 

4.2.3 

OPTION 1

Slope = Δp = Fnet✔ 
             Δt 
 = (14 000 − 30 000) 
        (20,2 - 20,1)  ✔
 = - 160 000  
 Fnet = 160 000 N ✔ 

OPTION 2
FnetΔt = Δp ✔ 
Fnet(0,1)✔ = 14 000 – 30 000✔ 
Fnet = - 160 000 N 
Fnet = 160 000 N ✔

POSITIVE MARKING FROM QUESTION 4.2.2
OPTION 3
FnetΔt = Δp ✔ 
Fnet(0,1)✔ = 900[(2,78) – (-15)]✔ 
Fnet = 160 020 N 
FA = - F
Fnet = 160 020 N ✔

 

OPTION 4
p = mv 
14 000 = 1 500vf ✔ 
vf = 9,33 m∙s-1

Fnetm(vf - vi)  ✔  1500(9,33 − 20)✔ 
                ∆t                       0,1
 = -160 050  
 = 160 050 N ✔

vf = vi + aΔt 
9,33 = 20 + a(0,1) 
a = -106.7 m∙s-2 
Fnet = ma ✔ 
 = 1 500(-106,7) ✔ 
Fnet = - 160 050 N 
Fnet = 160 050 N ✔

(4) 
[13] 

QUESTION 5
5.1.1 Ek/K = ½ mv2 ✔ 
 = ½ (2)(4,95)2 ✔ 
 = 24,50 J ✔ (3) 

5.1.2    (4)

POSITIVE MARKING FROM QUESTION 5.1.1
OPTION 1
Emech before = Emech after 
[(Emech)bob + (Emech)block ]before = [(Emech ) Block + (Emech)bob]after (mgh + ½ mv2)before = (mgh + ½ mv2)after
Any one ✔ 
(5)(9,8)h + 0 + 0 ✔= 5(9,8)¼h + 0 + 24,50 ✔ 
h = 0,67 m✔ 

OPTION 2
Wnc = ΔEp + ΔE
0 = ΔEp + ΔEk 
Any one ✔
-ΔEp = ΔEk 
-[(5)(9,8)(¼h) – (5)(9,8)h]✔ = 24,50✔ h = 0,67 m ✔ 

OPTION 3
Loss Ep bob = Gain in Ek of block ✔
mg(¾h) = 24,5 
(5)(9,8)(¾h)✔ = 24,5 ✔ 
h = 0,67 m ✔

OPTION 4 
Before

(mgh + ½ mv2)top = (mgh + ½ mv2)bottom
(5)(9,8)h + 0 = (5)(9,8)ho + ½ (5)v
vi2 = 19,6h - 19,6ho 
After
(mgh + ½ mv2)bottom = (mgh + ½ mv2)top
(5)(9,8)ho + ½(5)vf2 = (5)(9,8)(¼h) + 0 
vf2 = 4,9h – 19,6ho 
Emech before collision = Emech after collision✔
½ mvi2(bob) + 0 = ½ mvf2(bob)+ ½ mv2(block)
½ (5)(19,6h – 19,6ho) ✔ = ½ (5)(4,9h -19,6ho) + 24,5 ✔ 
h = 0,67 m✔

5.2 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 a resultant/net force is equal to the change in the  object's kinetic energy.  (2) 

5.3  

OPTION 1
Wnet = ΔEK✔ 
Wf + mgΔycosθ = ½m(vf2 - vi2
Wf +(2)(9,8)(0,5)cos180o ✔ = ½ (2)(22 – 4,952) ✔
Wf = - 10,7 J✔

OPTION 2
Wnc = ΔEK + ΔU  ✔
Wnc = ΔEK + ΔEP 
Wf = ½ (2)(22 – 4,952) ✔ + (2)(9,8)(0,5-0) ✔ 
 = - 10,7 J✔ 

(4) 
[13]

QUESTION 6
6.1

6.1.1 It is the (apparent) 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
An (apparent) change in (observed/detected) frequency (pitch), (wavelength) ✓as a result of the relative motion between a source and an observer  ✓(listener).  (2) 

6.1.2

v = fλ ✔ 
340 = f(0,28) ✔ 
fs = 1 214,29 Hz ✔ (3)

6.1.3 POSITIVE MARKING FROM QUESTION 6.1.2

fL = v ± vLfs  OR    fLv ± vL ×  OR   fL   v      fs  OR   FL     fs      
      v ± vs                     v ± vs     λs                v - v                       1 - vs/v
 fL =    340   1214,29      OR    fL =   340        ×     340    OR   FL = 1214,29 
      (340 - 30)                               (340 - 30)           0,28                  1 - 30/340 
= 1 331,80 Hz✔ (1 331,80 Hz – 1 335,72 Hz) (5)
6.1.4 Decreases ✔ (1) 

6.2 The spectral lines of the star are/should be shifted towards the lower frequency ✔ end, which is the red end (red shift) of the spectrum. ✔ 

 (2)
[13] 

QUESTION 7

7.1.1 The (magnitude of the) electrostatic force exerted by one (point) charge on  another is directly proportional to the product of the charges ✓ and inversely  proportional to the square of the distance between their (centres) them. ✓  (2) 
7.1.2 FE/Electrostatic force✓ (1) 
7.1.3 The electrostatic force is inversely proportional to the square of the distance  between the charges ✔ 
OR
The electrostatic force is directly proportional to the inverse of the square of  the distance between the charged spheres (charges). ✔ 
OR
F α 1  ✔ 
      r2
OR 
They are inversely proportional to each other  (1) 

7.1.4   (6) 

OPTION 1 
1 mark for using slope
Slope = ∆FE ✔ = (O,027 - 0)
            ∆1/ r2 ✔     (5,6 - 0) 
 = 4,82 x 10-3 N∙m2 (4,76 x 10-3 – 5 x 10-3
Slope = FEr2 = kQ1Q2 = kQ2 ✔ 
4,82 x 10-3 ✓= 9 x 109 Q2 ✓ 
∴ Q = 7,32 x 10-7C ✔ 
OPTION 2
Accept any pair of points on the line
F = kQ1Q2✔ 
          r2
( ) ✔ = (9 × 109) Q2 
                 (  ) ✔✔ 
Q = 7,32 x 10-7C ✔
(7,32 x 10-7 – 7,45 x 10-7 C) 

Examples 

( 0,005) ✔ = (9 × 109) Q2 
                             ( 1 ) ✔✔
Q = 7,45 x 10-7 C ✔ 

(0,027 ) ✔ = (9 × 109) Q2 
                     (1/5,6) ✔✔

Q = 7,32 x 10-7 C ✔ 

7.2.1

Criteria for drawing electric field: 

Marks

Direction 

Field lines radially inward

7.2.2 
E =  kQ  
        r2
Take right as positive
EPA = (9 × 109) (0,75 ×10-6
                    (0,09)
= 8,33 x 105 N∙C-1 to the left
EPB = (9 × 109) (0,8 ×10-6
                        (0,03)
= 8 x 106 N∙C-1 to the left
Enet = EPA + EPC 
 = [-8,33 x 105 + (- 8 x 106)] ✔ 
= -8,83 x 10 
 = 8,83 x 106 N∙C-1✔  
1 mark for the addition of same signs
Take left as positive
EPA = (9 × 109) (0,75 ×10-6
                    (0,09)
= 8,33 x 105 N∙C-1 to the left
EPB = (9 × 109) (0,8 ×10-6
                        (0,03)
= 8 x 106 N∙C-1 to the left
Enet = EPA + EPC 
 = (8,33 x 105 + 8 x 106) ✔ 
1 mark for the addition of same signs
= 8,83 x 106 N∙C-1

(5)
[17] 

QUESTION 8
8.1.1 (Maximum) energy provided (work done) by a battery per coulomb/unit charge passing through it ✔✔  (2) 
8.1.2 12 (V)✔ (1)
8.1.3 0 (V) / Zero✔ (1) 
8.1.4

ε = I(R + r)
ε = Vext + Vint 
12 = 11,7 +Ir 
0,3 = Itot(0,2) ✔ 
Itot = 1,5 A ✔ 

OR
V = IR ✔ (Accept: V”lost” = Ir) 
0,3 = Itot(0,2) ✔ 
Itot = 1,5 A✔ (3)  

OPTION 1 
1  1   +   1  
R//   R1      R2
1  1   +   1  
R//   10     15
R = 6 Ω ✔ 

OPTION 2 

R|| =  R1R2   
       R1 + R2
R|| =  (10)(15)     
       10 + 15
= 6 Ω ✔ 

(2) 

 

POSITIVE MARKING FROM QUESTIONS 8.1.4 AND 8.1.5
OPTION 1
V = IR ✔ 
11,7✔ = 1,5(6 + R) ✔ 
R = 1,8 Ω ✔ 
OR
V = IR ✔ 
11,7 = 1,5R ✔ 
R = 7,8 Ω  
RR = 7,8 – 6 ✔  
 = 1,8 Ω ✔

 

OPTION 2 
ε = I(R + r) ✔ 
12 = 1,5(R + 0,2) ✔ 
R = 7,8 Ω 
RR = 7,8 – 6 ✔ 
 = 1,8 Ω ✔

OPTION 3 
V||= IR|| 
 = (6)(1,5) ✔ 
 = 9 V 
VR = IR ✔ 
(11,7 - 9) = (1,5)R✔
R = 1,8 Ω✔ 

(4) 

8.2.1  (3) 

Pave= Fvave✔= mg(vave)  
 = (0,35)(9,8)(0,4)✔ 
 = 1,37 W✔ 
OR 
P = Wnc  = ∆Ek  +  ∆E =0 + (0,35)(9,8)(0,4 − 0)✔= 1,37 W ✔
       ∆t                ∆t                             1   
OR
P = W=    = (0,35)(9,8)(0,4 )✔= 1,37 W ✔
      ∆t     ∆t                1 

8.2.2 

POSITIVE MARKING FROM QUESTION 8.2.1

OPTION 1
P = VI  
1,37 = (3)I ✔ 
I = 0,46 A 
✔Any one 
ε = Vext + Vint 
 = VT + VX + Vint 
12 = VT + 3 + (0,2)(0,46) ✔
VT = 8,91 V 
VT = IR
8,91 = (0,46)RT ✔ 
RT = 19,37 Ω✔ 

OPTION 2
P = V
       R
1,37 = 32  ✔ 
            R
R = 6,57 Ω 
✔Any one
P = VI  
1,37 = (3)I ✔ 
I = 0,46 A 
ε = I(R + r) 
12 = 0,46(6,57 + RT + 0,2) ✔
RT = 19,38 Ω ✔ 

 

OPTION 3
P = VI ✔ 
1,37 = (3)I ✔ 
I = 0,46 A 

Ptot = Pr + Pmotor + P
(12)(0,46)✔ = (0,46)2(0,2) + 1,37 + (0,46)2RT
RT = 19,41 Ω ✔ 
OR 
P = VI ✔ 
1,37 = (3)I ✔ 
I = 0,46 A 
Ptot = Pr + Pmotor + P
(12)(0,46) = (0,46)2(0,2) + 1,37 + PT ✔ 
PT = 4,07 W 
P=I2
4,07 = (0,46)2RT ✔ 
RT = 19,49 Ω ✔

 

OPTION 4
P = VI 
1,37 = (3)I ✔ 
I = 0,46 A 

✔Any one 

ε = I(R + r) 
12 = (0,46)(R + 0,2) ✔ 
R = 25,87 Ω 

V = IR 
3 = (0,46)R✔ 
R = 6,52 Ω 
RT = 25,87 – 6,52  
= 19,35 Ω ✔

P = I2
1,37 = (0,46)2R✔ 
R = 6,47 Ω 
RT = 25,87 – 6,47
 = 19,4 Ω ✔

Pmotor V   
                  R
1,37 = 3
           R
R = 6,56 Ω 
RT = 25,87 – 6,56

 = 19,31 Ω ✔

(5) 
[21]

QUESTION 9
9.1.1 DC/GS-generator✔ 
Uses split ring/commutator✔ (2)
9.1.2 
q9 p

9.2.1 

OPTION 1
Vrms Vmax
              √2
Pave = VrmsIrms  ✓ 
800 =  340    (Irms )
              √2
Irms = 3,33 A ✓ 

OR
Vrms Vmax
              √2
 340   = 240,416
     √2
Pave = VrmsIrms  ✓ 
800 = Irms  (240,416)
Irms  = 3,33 A ✓ 

9.2.2 

OPTION 2
Pave =( V2rms) = ✓  ( V2rms
               R                (2)(R)
800 =       3402        (Irms )
              (√2)2(R)
R  = 72,25 Ω

EITHER:
Pave = Irms2R
800 = Irms2(72,25)
Irms2 = 3,33A
OR:
Vrms = IrmsR
Irms = 240,416 ✓ 
           72, 25 
 = 3,33 A✓ 

(3) 

 

POSITIVE MARKING FROM QUESTION 9.2.1 
OPTION 1
Pave =  VrmsIrms= I ✓ 
for the kettle: 
2000 = 340 (Irms) rms✓
             √2 
Irms = 8,32 A  
Itot = (8,32 + 3,33) ✓ 
 = 11,65 A ✓ 

 

OPTION 2
Pave (V2rms) = (V2max) ✓ 
                R          (2)(R)
800 =       3402        
              (√2)2(R)
R  = 72,25 Ω

2000 =         3402            
               (√2)2(R2000)
R =  28,9 Ω
1  1   +   1          ⇒ ⇒ ⇒  ⇒       R =  (28,9)(72,25)  = 20,64 Ω
R     R1      R2                                            (28,9 + 72,25) 
Vrms = IrmsR  
240,42 = Irms20,64) ✓ 
Irms = 11,65 A✓ 

 

OPTION 3
Pave = Vrms Irms ✓ = VmaxImax
                                       2 
2800 = (380) Irms✓ 
                 2
Imax = 16,47 A 
Irms ✓ = Imax   = 16,47
              √2         √2 
Irms = 11,65 A✓

OPTION 4
Pave = VrmsIrms✓ 
2 800 ✔=  340 Irms✔ 
                    √2 
Irms = 11,65 A ✔

OPTION 5
PT : P
800 : 2 000 ✔ 
1 : 2,5 

IT : I
3,33 : 8,325 ✔ 
Irms = 3,33 + 8,325 ✔ 
 = 11,66 A ✔ 

(4) 
[11] 

QUESTION 10
10.1

10.1.1 The minimum frequency (of a photon/light) needed✓ to emit electrons from  (the surface of) a metal. (substance) ✓ 
OR
The frequency (of a photon/light) needed✓ to emit electrons from (the surface  of) a metal. (substance) with zero kinetic energy✓ (2)
10.1.2 Silver/Silwer✔ 
Threshold/cutoff frequency (of Ag) is higher✔ 
Wo α fo / Wo = hfo ✔ 
OR
To eject electrons with the same kinetic energy from each metal, light of a  higher frequency/energy is required for silver. ✓ Since E = Wo + Ek(max) (and  Ek is constant), the higher the frequency/energy of the photon/light required,  the greater is the work function/Wo.✓  (3) 
10.1.3 Planck’s constant  ✔ (1)
10.1.4 Sodium✔ (1) 

10.2

10.2.1 Energy radiated per second by the blue light

= (5/100)(60 x 10-3) ✔ = 3 x 10-3 J∙s-1 
Ephoton = hc ✔ 
                λ
= (6,63 × 10-34 )(3 × 108
          470  ×  10-9
 = 4,232 x 10-19J ✔
Total number of photons incident per second
     3  × 10-3      
    4,232  ×  1015
 = 7,09 x 1015 ✔ (5)  
10.2.2 POSITIVE MARKING FROM QUESTION 10.2.1 
7,09 x 1015 (electrons per second) ✔ 
OR 
Same number as that calculated in Question 10.2.1 above (1)

[13] 
TOTAL: 150 

Last modified on Wednesday, 16 June 2021 08:11