Physical Science Paper 1 Grade 12 Memorandum - NSC Past Papers And Memos September 2020 Preparatory Examinations

Friday, 18 February 2022 08:26

Physical Science Paper 1 Grade 12 Memorandum - NSC Past Papers And Memos September 2020 Preparatory Examinations

More in this category: « Physical Science Paper 2 Grade 12 Questions - NSC Past Papers And Memos September 2020 Preparatory Examinations Physical Science Paper 2 Grade 12 Memorandum - NSC Past Papers And Memos September 2020 Preparatory Examinations »

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MEMORANDUM

QUESTION 1:
MULTIPLE-CHOICE QUESTIONS
1.1 C (2)
1.2 C (2)
1.3 A (2)
1.4 D (2)
1.5 A (2)
1.6 B (2)
1.7 D (2)
1.8 B (2)
1.9 D (2)
1.10 C (2) [20]

QUESTION 2
2.1 When a resultant/net force acts on an object, it accelerates in the direction of the force. The acceleration is directly proportional to the force and inversely proportional to the mass of the object. (2)
2.2

OPTION 1	OPTION 2

(4)

Mark awarded for arrow and label.
Do not penalise for length of arrows since drawing is not drawn to scale.
Any other additional force(s)
If force(s) do not make contact with body.

2.3

OPTION 1 (To the right is positive)

OPTION 2 (To the right is negative)

Fnet = ma
Fnet = 0 N Any one
Fg – T = ma
15 x 9,8 - T = 0
T = 147 N

T – Wsin θ – f = ma
T – mgsin θ – f = ma
147– 22 x 9,8sin θ – 43,86 = 0
θ = 28,58^°

Fnet = ma
Fnet = 0 N Any one
T - Fg = ma
T – 15 x 9,8 = 0
T = 147 N

Wsin θ + f – T = ma
mgsin θ + f – T = ma
22 x 9,8 sin θ + 43,86 – 147 = 0
θ = 28,58^°

(4)
2.4
2.4.1The kinetic friction force depends on the angle of the inclined to the horizontal and nature of the surface. (2)
2.4.2

Positive marking from Question 2.3

Marking criteria

For 15 kg block
Fnet = ma 🗸
F - Fg – T = ma
172 – 147 – T 🗸= 15a…(1)

For 22 kg block
T + W sinθ– f = ma
T + mg sinθ– f = ma
[T + 22 x 9,8 sin 28,58^o 🗸– 43,86] 🗸= 22a 🗸
[T + 101,14 – 43,86] = 22a (2)

∴ a = 2,278 m.s^-2 🗸

Fnet = ma
F - Fg – T = ma
T + Wsin θ – f = ma Any one
T + mgsin θ – f = ma
Calculating Fg// 🗸
Substitution for the 15 kg block 🗸
Substitution for the 22 kg block 🗸
Substitution for either 15a or 22a 🗸

Answer🗸

(6) [18]

QUESTION 3
3.1

UPWARD POSITIVE	UPWARD NEGATIVE
vf = vi + a∆t 🗸 vf = -3,28 + (-9,8) x 1,2 🗸 vf = - 15,04 vf = 15,04 m.s^-1 downwards 🗸	vf = vi + a∆t 🗸 vf = 3,28 + (9,8) x 1,2 🗸 vf = 15,04 m.s^-1 downwards 🗸

3.2

OPTION 1
UPWARD POSITIVE	UPWARD NEGATIVE
∆y = vi ∆t + 1 a∆t² 🗸 2 ∆𝑦 = -3,28 x 1,2 🗸+ 1 (-9,8)1,2² 🗸 2 ∆y = -10,99 ∆y = 10,99 m downwards🗸	∆y = vi∆t+ 1 a∆t² 🗸 2 ∆y = 3,28 x 1,2 🗸+ 1 (9,8)1,2² 🗸 2 ∆𝑦 = 10,99 m downwards🗸
OPTION 2
Positive marking from Question 3.1
UPWARD POSITIVE	UPWARD NEGATIVE
vf2= vi2 + 2a∆y 🗸 -15,04² 🗸 = -3,28² + 2 (-9,8) ∆y 🗸 ∆y = 10,99 m downwards 🗸	vf2 = vi2 + 2a∆𝑦 🗸 15,04² 🗸= 3,28² + 2 (9,8) ∆y 🗸 ∆y= 10,99 m downwards 🗸

OPTION 3
Positive marking from Question 3.1
UPWARD POSITIVE	UPWARD NEGATIVE
∆y = vf + vi Δt 2 ∆y = -15,04 + (-3,28)x1,2 2 ∆y = -10,99 m ∆y = 10,99 m downwards	∆y = vf + vi Δt 2 ∆y = 15,04 + 3,28 × 1,2 2 ∆y = 10,99 m downwards

(4)

3.3	OPTION 1
	UPWARD POSITIVE	UPWARD NEGATIVE
	vf²= vi² + 2a∆y vf²= -3,28² + 2 (-9,8) (30) vf²= - 24,47 m.s^-1vf²= 24,47 m.s^-1 downwards	vf²= vi² + 2a∆y vf²= 3,28² + 2 (9,8) (30) vf²= 24,47 m.s^-1 downwards

	OPTION 2		(3)
	UPWARD POSITIVE	UPWARD NEGATIVE
	∆y = vi∆t+ 1 a∆t² 2 -30 = -3,28∆𝑡 + 1 (-9,8)∆𝑡² 2 ∆𝑡 = 2,16 s vf = vi + a∆t vf = -3,28 + (-9,8)(2,16) vf= -24,45 m.s^-1vf = 24,45 m.s^-1 downwards	∆y = vi∆t+ 1 a∆t² 2 30 = 3,28∆𝑡 + 1 (9,8)∆𝑡² 2 ∆t = 2,16 s vf = vi + a∆t vf = 3,28 + (9,8)(2,16) vf = 24,45 m.s^-1 downwards

3.4	Positive marking from Question 3.2		(4)
	UPWARD POSITIVE	UPWARD NEGATIVE
	Fnet.∆t = m(vf - vi) 205 x 0,1 = 0,5 [𝑣_𝑓 – (- 24,47)] vf= 16,53 m.s^-1 upwards	Fnet. ∆t = m(vf - vi) -205 x 0,1 = 0,5 (vf– 24,47)] vf= -16,53 m.s^-1vf = 16,53 m.s^-1 upwards

3.5	OPTION 1
	Positive marking from Question 3.1 and 3.3
	UPWARD POSITIVE	UPWARD NEGATIVE
	vf²= vi² + 2a∆y 0² = 16,53² + 2 (-9,8) ∆y ∆y = 13,94 m width of window = 30 – (10,99+13,94) = 5,07 m	vf²= vi² + 2a∆y 0² = -16,53² + 2 (9,8) ∆y ∆y = - 13,94 m width of window = 30 – (10,99 +13,94) = 5,07 m

	OPTION 2		(5)
	Wnet = ∆E_k Fnet ∆y cos θ = ∆E_k Fg∆y cos θ = 1 mvf² - 1 mvi² Any one 2 2 4,9 ∆ycos 180^o = 0 - 1 x 0,5 x 16,53² 2 ∆y = 13,94 m width of window = 30 – (10,99 + 13,94) = 5,07 m
				[19]

[19]

QUESTION 4
4.1 In an isolated system total linear momentum is conserved. (2)
4.2
4.2.1

MEi = MEf
mgh1 + ½ mvi² = mgh² + ½ mvf²
0 + ½ x 2,005 x vi² = 2,005 x 9,8 x 0,06 + 0
vi = 1,08 m.s^-1
∑pi = ∑pf Any one
m_blockvi_block + m_bulletvi_bullet = (m_block + m_bullet) vf
0 + 0,005 x vi_bullet = 2,005 x 1,08
v_bullet = 433,08 m.s^-1 right (5)

4.2.2

Positive marking from Question 4.2.1
∑pi = ∑pf
(m_gun + m_bullet)vi = m_gunvf_gun + m_bulletvf_bullet = Any one
0 = 5 x vf(gun)+ 0,005 x 433,08
v_f(gun= - 0,43 m.s^-1
v_f(gun = 0,43 m.s^-1 left (3)

[10]

QUESTION 5
5.1 In an isolated system the total mechanical energy is conserved. (2)
5.2

ME_i = ME_f
(E_p + E_k)_A = (E_p + E_k)B
mgh₁ + E_k(A) = mgh₂ + E_k(B)
55 x 9,8 x 15 + 0 = 0 + E_k(B)Any one
E_k(B) = 8085 J (3)

5.3

The net work done on an object is equal to the object’s change in kinetic energy.
OR
The work done by a net force is equal to the object’s change in kinetic energy. (2)

5.4
5.4.1 Positive marking from Question 5.2

OPTION 1
W_ner = ∆E_k
W_f + W_Fg// = ∆E_k
f x ∆x cosθ + mgsin θ ∆x cos θ = ∆Ek Any one
15 x 10 cos180o + 55 x 9,8 sin 15o x 10 cos 180o = 0 – E_ki
E_ki = 1545,035 J
OPTION 2
W_nc = ∆E_k + ∆E_p
W_f = ∆E_k + ∆E_p
f x ∆x cosθ = ∆E_p + ∆E_kAny one
15 x 10 cos180o = [55 x 9,8 x (10 sin15o)] - 0 + 0 - E_ki
-150 = 1395,035 - E_ki
E_ki = 1545,035 J
OPTION 3
W_ner = ∆E_k
W_f + W_Fg = ∆E_k Any one
f x ∆x cosθ + mg ∆xcos θ = ∆E_k
15 x 10 cos180o + 55 x 9,8 x 10 cos (90 + 15) = 0 – E_ki
E_ki = 1545,035 J (4)

5.4.2

*OPTION 1/OPSIE* 1**	*OPTION 2/OPSIE* 2**
Wnet = ∆E_k Wf = ∆E_k Any one f x ∆x cosθ = ∆E_k f x 10 cos180^o 🗸 = 1545,035 🗸 - 8085 🗸 = 654,00 N 🗸	W_nc = ∆Ek + ∆Ep W_f = ∆Ek + ∆Ep Any one f x ∆x cosθ = ∆Ek + ∆Ep f x 10 cos180^o 🗸= 1545,035 🗸 - 8085 🗸 f = 654,00 N 🗸

Related Items

(5) [16]

QUESTION 6
6.1 Doppler effect (1)
6.2 AWAY. The observed frequency is less than the source frequency. (2)
6.3 As the listener’s velocity increases, the observed frequency decreases. (2)
6.4 The intercept on the vertical axis represents the frequency of the source, fs. (2)
6.5

OPTION 1	OPTION 2
Gradient = 492 - 470 🗸 5 - 20 = - 22 15 = - f_s 🗸 v - 22 🗸= - f_s 🗸 15 340 f_s = 498,67 Hz 🗸	fL = v ±v_L f_s 🗸 v ±v_s492 🗸= 340 - 15 🗸🗸x f_s 340 f_s = 499,34 Hz 🗸

OPTION 3

fL = v ±v_L f_s 🗸
v ±v_s
470 🗸= 340 - 15 🗸🗸x f_s
340
f_s = 499,38 Hz 🗸
Range (498,67 Hz – 499,38 Hz)

(5)
6.6

OPTION 1	OPTION 2	OPTION 3
v = fλ 340 = 498,67 λ 🗸 𝜆 = 0,68 m 🗸 The wavelength produced by the tuning fork is less than the required wavelength. ∴ It is not suitable. 🗸	v = fλ 340 = 499,34 λ 🗸 𝜆 = 0,68 m 🗸 The wavelength produced by the tuning fork is less than the required wavelength. ∴ It is not suitable. 🗸	v = fλ 340 = 499,38 λ 🗸 𝜆 = 0,68 m 🗸 The wavelength produced by the tuning fork is less than the required wavelength. ∴ It is not suitable. 🗸

(3) [15]

QUESTION 7
7.1

n = Q
qe
n = 2 x10^-6
1,6 x10^-19
n = 1,25 x 10¹³(3)

7.2

(3)

Criteria for marking

Correct shape
Direction of electric field
Lines not crossing each other

7.3
7.3.1

vf = vi + a∆t
6,25 x 10³ = 0 + 2 x 10^-3 a Any one
a = 3,125 x 10⁶ m.s^-2
F_net = ma
F_net = 5 x 10^-6 x 3,125 x 10⁶
F_net = 15,625 N
E_net = F_net
q
E_net = 15,625
2 x10^-6
Enet = 7,81 x 10⁶ N.C^-1 (6)

7.3.2	Positive marking from Question 7.3.1
	OPTION 1	OPTION 2
	E = kQ 🗸 r²7,81 x 10⁶ = 9 x10⁹ x 2 x10^-6 🗸 r² ∴ r = 4,80 x 10^-2 m 🗸	v_f² = v_i² + 2a Δx 🗸 (6,25 x 10³)² = 0² + 2(3,125 x 10⁶)Δx 🗸 Δx = 6,25 m 🗸
	Using equations of motion
	*OPTION 3/OPSIE* 3**	*OPTION/OPSIE* 4**
	Δx = ½(vf + vi) Δt 🗸 = ½(6,25 x 10³+ 0)(2 x 10^-3) 🗸 Δx = 6,25 m 🗸	Δx = vi Δt + ½ a Δt² 🗸 = 0 + ½(3,125 x 10⁶)(2 x 10^-3)² 🗸 Δx = 6,25 m 🗸

7.3.3	Positive marking from Question 7.3.1 and 7.3.2		(4)
	OPTION 1	OPTION 2
	FE = kQ₁Q₂🗸 r²🗸 🗸15,625 = 9 x10⁹x Q x 2 x10^-⁶ (4,80 x10^-² )2 ∴ Q = 2,00 x 10^-6 C 🗸	FE = kQ₁Q₂ 🗸 r ² 15,625 🗸= (9 x 10⁹)(Q1)(2 x 10^-6) 6,25² 🗸 ∴ Q = 3,39 x 10^-2 C 🗸
[19]

QUESTION 8
8.1 Resistance (1)
8.2
8.2.1

Gradient = 4,2 - 0,2
24 - 0
= 1
6
= 1
emf
1 = 1
emf 6

8.2.2

Positive marking from Question 8.2.1
OPTION 1	OPTION 2
Intercept on vertical axis = r emf 0,2 = r 🗸 emf 0,2 = r 🗸 6 r = 1,2 Ω 🗸	𝜀 = I (R + r ) 🗸 6 = 1 (24 + r ) 🗸 4,2 r = 1,2 Ω 🗸

8.3
8.3.1

OPTION 1	OPTION 2
P = VI 🗸 3 = I x 1,5 🗸 I = 2 A 🗸	P = V ² R 3 = 1,5² R R = 0,75 Ω P = I² R 🗸 3 = I² x 0,75 🗸 I = 2 A 🗸

8.3.2

Positive marking from Question 8.2.1en 8.2.2

OPTION 1

OPTION 2

P = V² R_{toy car}3 = 1,5² 🗸
R_{toy car}R_{toy car} = 0,75 Ω
emf = I (R_ext + r ) 🗸
6 = 2 (R_ext + 1,2) 🗸
R_ext = 1,8 Ω
R_ext = R_{toy car} + R_p
1,8 = 0,75 + R_p 🗸
Rp = 1,05 Ω
1 = 1 + 1
R_p R₁ R₂
1 = 1 + 1 🗸
1,05 3 R
R = 1,62 Ω 🗸

P = V² R_{toy car}3 = 1,5² 🗸
R_{toy car}

R_{toy car} = 0,75 Ω
emf = I (R_ext + r) 🗸
6 = 2 (R_ext + 1,2) 🗸
R_ext = 1,8 Ω
R_ext = R_{toy car} + R_p
1,8 = 0,75 + R_p 🗸
R_p = 1,05 Ω
R_p = R₁ R₂
R₁ +R₂
1,05 = 3 x R 🗸
3 +R
R = 1,62 Ω 🗸

(6)
8.4

No, the external resistance will increase. Current will decrease.
P = I2R, P 𝛼 I2 since resistance remain constant, power will decrease. (3)

[20]

QUESTION 9
9.1 (Electric) motor (1)
9.2 Electrical energy to mechanical energy (2)
9.3 CLOCKWISE (1)
9.4

Replace the split ring with slip rings.
Replace the cell with a resistor / ammeter / voltmeter. (2)

9.5
9.5.1

P_average = V ²_rms
R
1500= 230²
R
R = 35,27 Ω (3)

9.5.2

OPTION 1

OPTION 2

R = V_rms
I_rms

35,27 = 230² 🗸
I_rmsI_rms = 6,52 A
I_rms = I_max🗸
√2
6,52 = I_max🗸
√2
I_max = 9,22 A 🗸

P_average = V_rmsI_rms1 500 = I_rms x 230 🗸
I_rms = 6,52 A
I_rms = I _max 🗸
√2
6,52 = 𝐼_𝑚𝑎𝑥🗸
√2
𝐼_𝑚𝑎𝑥 = 9,22 A 🗸

(4)

[13]
TOTAL: 150

Last modified on Tuesday, 01 March 2022 06:40

Published in Grade 12 September 2020 Preparatory Examinations