MARKS : 150
TIME: 3 hours
This marking guideline consists of 15 pages.
NATIONAL
SENIOR CERTIFICATE
GRADE 12
PHYSICAL SCIENCES P1 (PHYSICS)
PREPARATORY EXAMINATIONS
MARKING GUIDELINE
SEPTEMBER 2021
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MARKS : 150
TIME: 3 hours
This marking guideline consists of 15 pages.
NATIONAL
SENIOR CERTIFICATE
GRADE 12
PHYSICAL SCIENCES P1 (PHYSICS)
PREPARATORY EXAMINATIONS
MARKING GUIDELINE
SEPTEMBER 2021
NSC
QUESTION 1: MULTIPLE CHOICE
1.1 A (2)
1.2 D (2)
1.3 C (2)
1.4 D (2)
1.5 D (2)
1.6 B (2)
1.7 B (2)
1.8 A (2)
1.9 A (2)
1.10 C (2)
[20]
NSC
2.3 For the 4 kg crate
F
net
= ma
T f Fg// = ma
T 1 (9,8)(4)sin
o
T = 30,6 N upwards
For the 6 kg crate
Fnet = ma
F f T Fg// = ma
F 1,5 30,6 (9,8)(6)sin 30
o
F = 76,5 N
2.4.1 INCREASES (1)
2.4.2 The 4 kg block will move upwards/forward (for a brief moment) , stop
and then slide down the plane / backward. (3)
[1 6 ]
QUESTION 3
3.1.1 Zero (1)
3.1.2 9,8 m
.
s
downwards (2)
3.2.1 OPTION 1
UPWARD POSITIVE UPWARD NEGATIVE
y= vi t + ½a t
2
y= vi t + ½a t
2
y= (4,9)( 0,5) + ½(-9,8)(0,5)
2
y= (-4,9)( 0,5) + ½(9,8)(0,5)
2
y = 1,225 m y = -1,225 m
Height after 0,5 s = 80 + 1,
= 81,23 m
Height after 0,5 s = 80 + 1,
= 81,23 m
The ball is 81,23 m above the
ground
The ball is 81,23m above the
ground
Any of the two
NSC
OPTION 2
UPWARD POSITIVE UPWARD NEGATIVE
vf = vi + a t vf = vi + a t
v f
= 4,9 + ( 9,8) (0,5) v f
vf = 0 m
.
s
vf = 0 m
.
s
vf
2
= vi
2
- 2a y vf
2
= vi
2
- 2a y
2
- 2( 9,8) y 0 = - 4,
2
- 2(9,8) y
y = 1,225 m y = -1,225 m
Height after 0,5 s = 80 + 1,
= 81,23 m
Height after 0,5 s = 80 + 1,
= 81,23 m
The ball is 81,23 m above the ground The ball is 81,23 m above the
ground
OPTION 3
UPWARD POSITIVE UPWARD NEGATIVE
v f
= v i
- a t v f
= v i
- a t
v f
= 4,9 + ( 9,8) (0,5) v f
v f
= 0 m
.
s
v f
= 0 m
.
s
t
v
v
y =
f i
t
v
v
y =
f i
[ 0 , 5
y = ] [ 0 , 5
y =
]
y = 1,225 m y = -1,225 m
Height after 0,5 s = 80 + 1,
= 81,23m
Height after 0,5 s = 80 + 1,
= 81,23 m
The ball is 81,23 m above the
ground
The ball is 81,23 m above the
ground
NSC
OPTION 1
UPWARDS POSITIVE UPWARDS NEGATIVE
y= vi t + ½a t
2
y= vi t + ½a t
2
2
2
t = 4,57 s t = 4,57 s
OPTION 2
UPWARDS POSITIVE UPWARDS AS NEGATIVE
vf
2
= vi
2
vf
2
= vi
2
v f
2
2
- 2(-9,8)(-80) v f
2
2
vf = -39,9 m
.
s
vf = 39,9 m
.
s
v f
v f
= v i
t = 4,57 s t = 4,57 s
OPTION 3: POSITIVE MARKING FROM QUESTION 3.2.
Considering ball from the maximum height
UPWARDS POSITIVE
UPWARDS AS NEGATIVE
y= vi t + ½a t
2
y= vi t + ½a t
2
2
t = 4,07 s t = 4,07 s
Time to reach ground = 0,5 + 4,07 Time to reach ground = 0,5 + 4,
= 4,57 s = 4,57 s
NSC
3.3 POSITIVE MARKING FROM Q3.2.
OR
CRITERIA FOR MARKING OF GRAPH
Correct shape
Indication of initial velocity
Indication of the time for the entire motion
[14]
UPWARD NEGATIVE
Time (s)
UPWARD POSITIVE
Time (s)
NSC
OPTION 3 OPTION 4
W
net k
f
2
- ½mv i
2
t = 4 s
2
2
f = 7500 N
F
net
= f =
f =
= -7500 N
Magnitude of f is 7500 N
[16]
QUESTION 5
5.1.1 OPTION 1
W
gravity
= (1 300)(9,8)(60)cos
o
= - 764 400 J (-7,64 x
5
J)
OPTION 2
W = -
= - 764 400 J (-7,64 x
5
J)
-1 mark if either negative is omitted
5.1.2 Wcounterweight
= (900)(9,8)(60)cos
o
= 529 200 J (5,29 x 10
5
J) (2)
NSC
5.2 OPTION 1 OPTION 2
POSITIVE MARKING FROM 5.1.
AND 5.1.
Wnet K
Wgravity + Wcountweight + Wmotor = 0
Wmotor = - (Wgravity + Wcountweight)
-764 400 + 529 200 + Wmotor = 0
Wmotor = 235 200 J
F
motor
= T
right
- T
left
= 3920 N
Wmotor = Fmotor
= (3920)(60) cos
= 235 200 J
Pave motor =
Pave motor =
= 1960 W (5)
[1 0 ]
QUESTION 6
6.1 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.
(2)
v =
t
d
v =
v L
= 15 m
.
s
f
v± v
v± v
f S
S
L
L f
s
v
f L
fL = 721,69 Hz (5)
6.3 Any two
Ultrasound waves (to measure the heartbeat of a foetus in the womb).
Doppler flowmeter (to measure the rate of blood flow)
Traffic management systems, (especially speed control)
Radar, (allowing for the tracking of weather systems)
Astronomy, (where the application of the red-shift and blue-shift of
light from the stars has revolutionised our understanding of the
universe) (2)
6.4.1 The spectral lines (light) from the star are shifted towards longer
wavelengths. (2)
6.4.2 Decrease (1)
[12]
NSC
7.2.1 The electric field at a point is the (electrostatic) force experienced per
unit positive charge placed at that point. (2)
CRITERIA FOR MARKING THE ABOVE ELECTRIC FIELD PATTERN
Correct direction of field lines
Shape of the electric field lines (At least 4 lines on each sphere)
No field lines crossing each other/No field lines inside the spheres
Q =
+ Q
Q
1 2
Q =
= - 2,5 x 10
C (-2,
r
Q
E = k
2
EAP = 9 × 10
9
2 , 5 × 10
2
EQP = 1,11x 10
7
N
.
C
to the left
EBP = 9 × 10
9
2 , 5 × 10
2
E
BP
= 1,00x 10
8
N
.
C
to the left
Enet = EAP + EBP
E
net
= 1,11 x 10
7
- 1,00 x 10
8
Enet = 1,11 x 10
8
N
.
C
[21]
A
B
NSC
QUESTION 8
8.1 (Maximum) energy provided (work done) by a battery per coulomb/unit
charge passing through it (2)
8.2.1 OPTION 1 OPTION 2
P = I
2
R P
s
= V
3
I
2
R 3 40 = V 3 (2)
V 3 = 20 V
R
3
R 3 =
I
v 3
R
P
R
P
R
P
R
EXT
= R
P
+ R
s
R
EXT
= R
P
+ R
s
R
EXT
= R
P
+ R
s
REXT = 5 + 10 REXT = 5 + 10 REXT = 5 + 10
R
EXT
= R
EXT
= R
EXT
POSITIVE MARKING FROM QUESTION 8.2.
= I(R + r)
= 30,2 V
OPTION 1 OPTION 2 (Positive marking
from Q8.2.1)
W = P x t W = I
2
Rt
W = 40 x 20 x 60 W = (2)
2
(10)(20 x 60)
W = 48000 J W = 48000 J (3)
[13]
NSC
QUESTION 10
10.1 Threshold frequency, f
o
, is the minimum frequency of light needed to
emit electrons from a certain metal surface. (2 or 0)
10.2 Q = nqe
Q= 2,01 x 10
9
X 1,6 x 10
Q = 3,22 x 10
C
I = 3,22 x 10
A (4)
10.3 Decreases
When the intensity of the light is decreased, the number of photons
per second will decrease
10.4 E = W
o
+ K
max
E = hf o
+ K
max
2,12x
=(6,63x
)(2,21x
15
) + K
max
K
max
= 6,55 x 10
J
10.5 Decreases
More energy is used to release the electrons.
OR
Work function is greater. (2)
[14]
TOTAL: 150
NSC
CONVERSION OF MARKS FOR QUESTION 7
MARK
OBTAINED
OUT OF 13
CONVERTED
MARK OUT
OF 21
0 0
1 2
2 3
3 5
4 6
5 8
6 10
7 11
8 13
9 15
10 16
11 18
12 19
13 21