Gr 10 Physical Sciences P1 (English) November 2022 Possible Answers_hlayiso.com_.pdf

Downloaded from hlayiso.com PROVINCIAL EXAMINATION NOVEMBER 2022 GRADE 10 MARKING GUIDELINES PHYSICAL SCIENCES (PAPER 1) 8 pages

Downloaded from hlayiso.com PHYSICAL SCIENCES MARKING GUIDELINES (PAPER 1) GRADE 10 QUESTION 1 1.1 A ✓✓ (2) 1.2 D ✓✓ (2) 1.3 B ✓✓ (2) 1.4 C ✓✓ (2) 1.5 B ✓✓ (2) 1.6 B ✓✓ (2) 1.7 C ✓✓ (2) 1.8 B ✓✓ (2) [16] QUESTION 2 2.1 Physical quantity with both magnitude and direction. ✓✓ (2) 2.2 2.2.1 50 m ✓✓ (2) 2.2.2 0 m ✓✓ (2) 2.2.3 Vector ✓ – It has magnitude, unit and direction. ✓ (2) [8] 2

Downloaded from hlayiso.com PHYSICAL SCIENCES MARKING GUIDELINES (PAPER 1) GRADE 10 QUESTION 3 3.1 135 km.h −1 ✓ 3,6 37,5 m·s–1 ✓ (2) 3.2 vf = vi+ at ✓ = 0 ✓ + (1,5)(11,5) ✓ = 17,25 m·s–1 ✓ (4) 3.3 x = vit + 12 at2 ✓ = 0 ✓ + 12 (1,5)(11,5)2 ✓ = 99,19 m ✓ (4) 3.4 x = vit + 12 at2 ✓ 40 = (37,5)(4) ✓ + 12 (a)(4)2 ✓ a = –13,75 m·s–2 ✓ (4) [14] 3

Downloaded from hlayiso.com PHYSICAL SCIENCES MARKING GUIDELINES (PAPER 1) GRADE 10 QUESTION 4 4.1 The rate of change in velocity. ✓✓ OR The change in velocity per unit time. ✓✓ (2) 4.2 4.2.1 – The car starts from rest and velocity increases to 10 m·s–1 in 20 seconds. ✓ – Constant positive acceleration or uniformly accelerated motion. ✓ (2) 4.2.2 – Velocity is constant (uniform). ✓ – Acceleration is zero. ✓ (2) 4.2.3 Car has stopped. ✓ Acceleration is zero. ✓ (2) 4.3 y 2 − y1 a=m= ✓ x2 − x1 0 − 40 = ✓ 60 − 50 =–4✓ = 4 m·s–2, in the opposite direction or west (deceleration in opposite direction) ✓ OR vf = vi + a. t ✓ 0 = 40 ✓ a  10 ✓ a =–4 = 4 m·s–2, in the opposite direction or west (deceleration) ✓ (4) 4.4 BC ✓, Steeper slope. ✓ (2) [14] 4

Downloaded from hlayiso.com PHYSICAL SCIENCES MARKING GUIDELINES (PAPER 1) GRADE 10 QUESTION 5 5.1 Energy of an object as a result of its position/height above the surface of the Earth. ✓✓ (2) 5.2 EMA = mgh + mv2 ✓ = (2 x 9,8 x 30) + 21  2  02 ✓ = 588 + 0 = 588 J ✓ (3) 5.3 Total mechanical energy is conserved in an isolated system. ✓✓ OR Mechanical energy at the top equals mechanical energy at the bottom in the absence of friction. ✓✓ (2) 5.4 POSITIVE MARKING FROM QUESTION 5.2 EMA = EMB ✓ 588 ✓ = mgh + 21 + mv2 588 = 2  9,8  10 + 21  2 v2 ✓ 588 – 196 = v2 v= 392 = 19,80 m·s–1 ✓ (4) 5.5 EQUAL TO ✓. Mechanical energy is conserved. ✓ (2) [13] 5

Downloaded from hlayiso.com PHYSICAL SCIENCES MARKING GUIDELINES (PAPER 1) GRADE 10 QUESTION 6 6.1 A wave where the movement of particles of the medium is perpendicular to the direction of propagation of the wave. ✓✓ (2) 6.2 A – Trough ✓ B – Wavelength ✓ C – Crest ✓ (3) 6.3 T = 1f T = 301 ✓ = 0,03s ✓ (2) 6.4 No, ✓ two points in phase are separated by a complete number of wavelengths. or They are not separated by wavelength. ✓ (2) 6.5 v=f✓ v = 30  4 ✓ = 120 m·s–1 ✓ (3) [12] 6

Downloaded from hlayiso.com PHYSICAL SCIENCES MARKING GUIDELINES (PAPER 1) GRADE 10 QUESTION 7 7.1 Neutral charge – an atom that has equal number of electrons and protons. ✓✓ (2) 7.2 The net charge of an isolated system remains constant during any physical process. ✓✓ (2) 7.3 Due to polarisation, a negative charge is developed on the side of sphere B near sphere A and a positive charge is developed on the side of sphere B that is away from sphere A ✓. Sphere B moves towards sphere A (attraction) as opposite charges attract. ✓ (2) 7.4 Q = n.e ✓ Q = 20  (–1,6  10–19) ✓ Q = –3,2  10–18 C ✓ (3) 7.5 Q1 + Q 2 Qnet = 2 ✓  2  10−9  +  −3,2  10−18  Qnet =    2  ✓ Qnet = 9,99  10–10 C ✓ (3) [12] 7

Downloaded from hlayiso.com PHYSICAL SCIENCES MARKING GUIDELINES (PAPER 1) GRADE 10 QUESTION 8 8.1 7,5 v ✓ (1) 8.2 8.2.1 1 = 1 + R1 + R1 Rp R1 2 3 1 = 1 + 1 +1 ✓ Rp 6 4 3 Rp = 1,3  RT = R s + Rp RT = 2,67 + 1,3 ✓ RT = 3,97  (3) 8.2.2 I = VR 7 ,5 I= ✓ 3,97 I = 1,88 A ✓ (2) 8.2.3 I = VR 1,88 = 2,V67 ✓ V = 5,01 V ✓ (2) 8.3 Q I= t Q 1,88 = 360 ✓ Q = 676,8 C ✓ (3) [11] TOTAL: 100 8