Physical Sciences Grade 10 KZN September Test 2025 QP & Memo

J KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE GRADE 10 | —————— lll SSCS SE AG SESS = 6 See, a PHYSICAL SCIENCES : t SEPTEMBER TEST ‘ " 2025 ‘ ee eerrrenhteies cor MARKS: 100 DURATION: 2 hours This question paper consists of 9 pages and 2 data sheets. Copyright reserved Please turn over es

Physicapeinnoaded from Stanmerephysics.com September 2025 Test INSTRUCTIONS AND INFORMATION 1. 10. 11. This question paper consists of SEVEN questions. Answer ALL the questions in the ANSWER BOOK. Start EACH question on a NEW page in the ANSWER BOOK. Number the answers correctly according to the numbering system used in this question paper. Leave ONE line between two sub-questions, for example between QUESTION 2.1 and QUESTION 2.2. You may use a non-programmable calculator. You may use appropriate mathematical instruments. You are advised to use the attached"DATA SHEETS. ’ Show ALL formulae and in ALL calculations. i Round off your final num: s to a minimum of TWO decimal places. Give brief motivations, dis¢ dns et cetera where required. Write neatly and legibly. Copyright reserved

PhysicdDanatretoaded from Stanmoraphysits.com September 2025 Test QUESTION 1: MULTIPLE-CHOICE QUESTIONS Four options are provided as possible answers to the following questions. Each question has only ONE correct answer. Choose the answer and write only the letter (A-D) next to the question number (1.1—1.8) in the ANSWER BOOK, for example 1.11 = 1.1 Which ONE of the following pairs of physical quantities are BOTH vectors? A __ Displacement and speed B Distance and velocity C_ Acceleration and force D-Mass and time (2) 12 The velocity — time graph for a car travelling on a straight horizontal surface is shown below: velocity Which ONE of the following graphs is the correct position — time graph for the motion of the car? downloaded from stanmorephysics.com (2) Copyright reserved

PhysicBhSaiepttoaded from Stanmerephysias.com September 2025 Test 1.3 1.4 1.5 1.6 Which ONE of the following statements is TRUE? An object experiences negative acceleration when it is moving in the... negative direction at a constant speed. negative direction, experiencing a constant decrease in speed. positive direction, experiencing a constant decrease in speed. positive direction, experiencing a constant increase in speed. (2) downloaded from stanmorephysics.com Which ONE of the following gases occupies the largest volume at STP? 0OW> A 40g of Ne(g) B 6 gof Hag) C 44g of CO2(g) D 17 g of NHsa(g) (2) Consider the balance for the reaction below: What is the minimum len (Oz) required to form 22,4 dm° of sulphur trioxide (SOs) a’ A 22,4 dm? B 16g Cc 1mol D 44,8 dm? (2) Which ONE of the following is the correct number of hydrogen (H) ATOMS in 2 moles of ammonium phosphate ( (NH4)3POs )? 6,02 x 1023. (2) [12] Copyright reserved

Physienatoaded from Stanmore physias.com September 2025 Test SECTION A: PHYSICS QUESTION 2 (Start on a new page.) aa | A cheetah travelling eastwards at a constant speed of 30 m:s* is in pursuit of a buck also moving eastwards at a constant speed of 22 m-s". At the instant the cheetah reaches point A, the buck is at point B, as illustrated in the diagram below: downloaded from stanmorephysics.com 22 m:s" N —> Ss { . 132m ‘ The distance between point A and point B is 60 m. 2441 Calculate the time taken for the cheetah to reach point B. (4) 2.1.2 Determine the position of the buck relative to the cheetah, when the cheetah reaches point B. (3) The cheetah speeds up at point B to eventually catch the buck at point C. From points B to C, the buck maintains its constant velocity of 22 m-s™. Point C is located 132 m due east of point B. 213 Define the term average velocity in words. (2) 2.1.4 Determine the average velocity of the cheetah from point B to point C. (6) Copyright reserved

Physical[Seiqnaaloaded from Stanntsrephysics.com September 2025 Test 2:2 Three forces, F1, F2 and F3 acts on a block on a frictionless horizontal surface as shown in the diagram below. Fi and F2 have forces of 120 N right and 80 N right respectively. The magnitude of Fs, acting to the left, is unknown. Fs <—__—__ The RESULTANT FORCE acting on the box is = Using the tail-to-head method and a scale of 1 cm represents 20 N, draw an accurate vector diagram to determine the magnitude of the force, Fs, acting on the box. (5) [20] QUESTION 3 (Start ona ne The POSITION-TIME GRAPH (not S for a cyclist initially moving west on a straight horizontal road is shown | a N tN E ' Ss o 7 Time (s) 3.1 e after 60 seconds? (1) 3.2 Calculate the velocity of the cyclist for the first 30 seconds. (4) 3.3 Write down the acceleration of the cyclist for the first 30 seconds. (1) 3.4 Explain the answer to QUESTION 3.3 referring to the shape of the graph. (2) 3.5 The MAGNITUDE of the velocity from 60 to 70 seconds is 2,5 m-s*. Using to graph only (no equations of motion), determine the total distance covered by the cyclist for the entire 70 second period. (4) [12] Copyright reserved

Physic etatoaded from Stanmecaphysic¢s.com September 2025 Test 7 QUESTION 4 (Start on a new page.) In the diagram below, a ship is travelling due east towards an iceberg on a foggy night, where the visibility is low. The ship travels at a constant speed of 32 m-s"' and is 500 m away from the iceberg. downloaded from stanmorephysics.com N E Ss Iceberg gees Due to the fog, the captain takes 3 is to react before applying brakes. The ship thereafter slows down at a uniform rate of 2,5 m:s? . 4.1 Define the term distance in words. (2) 4.2 Determine the distance the ship travels during the reaction time. (2) 4.3 Calculate the time taken for the ship to stop after applying brakes. (4) The iceberg moves toward the ship at a constant speed of 5 m-s*, during the entire motion. After the brakes are applied, the ship covers a distance of 204.8 metres. 4.4 Determine the distance between the ship and iceberg once the ship stops. (4) [12] Copyright reserved

Physicd welHtadoaded from Stanmeocephysirs -com September 2025 Test 8 SECTION B: CHEMISTRY QUESTION 5 (Start on a new page.) When potassium chlorate (KCIO3) decomposes upon heating, it produces potassium chloride (KCI) and oxygen (Oz) gas as represented in the balanced equation below: 2KCIO3(s) — 2KCI(s) + 302(g) In one such reaction, 30,3 g of potassium chlorate (KCIOs) is heated and decomposes completely at standard temperature and standard pressure (STP) 651 Define the term one mole of a substance in words. (2) 52 Calculate the: 5.2.1 Number of moles of KCIO3 decomposed. (4) 5.2.2 Number of KCl formula units formed. (4) §.23. Volume of oxygen (Oz) formed (in dm*) at standard temperature (4) and standard pressure (STP). [14] QUESTION 6 (Start on a new page) 6.1 A compound is made up of potassium (K), chromium (Cr) and oxygen (O) only. A 1 kg sample of this compound has the following composition by mass: Element Mass(g) Potassium (K) 265,31 g Chromium (Cr) 353,74 g Oxygen (O) Remainder 6.1.1 Calculate the mass of oxygen(O) in the 1 kg sample of the compound. (2) 6.2 Determine the empirical formula of this compound. (6) 6.2 An unknown hydrated salt with the formula X2COs-10H20 contains 62,937% water (H20) by mass. downloaded from stanmorephysics.com Determine the FORMULA of the element represented by the letter X. (5) Copyright reserved

Physideugaoaded from Stanmerephysiss.com September 2025 Test 9 6.3 ted by the balanced equation O2 (g) + 2H20 (g) In-one such — ' UME of gas that forms when ethyne(C2H2) re at reacted. (3) [16] QUESTION 7 (Start on a new page.) The reaction between aluminium carbonate (Al2(COs)s) and hydrochloric acid (HCI) is represented by the balanced e¢ elow._ Ala(COs3)s (s) + 6 aq) + 3CO2 (g) + 3H20 (g) 0,45 dm? of HCi with a concer ol.dm? reacts completely with excess Al2(COs)3_ at standard temperatu pressure (STP). 7.1 Define the term concentration in words. (2) 2 Calculate the: 8s of HCI that reacted. (3) )3 that completely reacted with the 0,45 dm® of (4) 13 Oz) is formed. eld of carbon dioxide (COz) (5) [14] TOTAL: 100 Copyright reserved

Physicalauntyoaded from Stanmer eplay nics .com September 2025 Test 10 DATA FOR PHYSICAL SCIENCES GRADE 10 PHYSICS TABLE 1: PHYSICAL CONSTANTS NAME SYMBOL VALUE Acceleration due to gravity g 9,8 m-s2 Speed of light in a vacuum c 3,0 x 108 m-s* Planck's constant h 6,63 x 10° J-s Charge on electron de -1,6x 107°C Electron mass Me 9,11 x 10°" kg TABLE 2: FORMULAE MOTION Ve = Vv, + aAt v7? = v2 + 2aAx CHEMISTRY TABLE 1: PHYSICAL CONSTANTS NAME SYMBOL VALUE Standard pressure p° 1,013 x 105 Pa Molar gas volume at STP Vin 22,4 dm?-mol* Standard temperature T? 273 K Avogadro's constant Na 6,02 x 1075 mol TABLE 2: FORMULAE m N n=— n=— M Na “Vv _ MV Vin Copyright reserved

Downloadegifresenatasimenephysics.com NSC September 2025 11 TABLE 3: THE PERIODIC TABLE OF ELEMENTS 1 2 3 5 ¥ 9 10 #11 12 13 14 15 16 17 18 (I) (Il) ‘ a) (vy) oY) (V(t) st) Fi Atomic number 2 Le. — wx H He 1 _ 4 3 4 Electronegativity | « Cu {Symbol 5 6 7 8 9 10 2 Li |? Be Elektronegatiwiteit "| — eas Simbool = bt bs 2 OF Ne 7 9 : 11 12 14 16 | 19 20 1 12 t 13 14 15 16 17 18 s Na|* Mg Approximate relative atomic mass 2AQ2 SIS Plt S |3Ce|) Ar 23 24 Benaderde relatiewe atoommassa 27 28 31 32 35,5 40 19 20 21 22| 23 24] 25| 26! 27 28 | 29 30 | 31 32 33 34 | 35 36 s K |2 Cal? Sc[2 Ti |? © Cr |2Mn/|2 Fe |2 Co |2 Ni 2 Cu|2 Zn |2 Ga |2 Ge |S As |S Se |S Br| Kr 39 40 45 43 | 51 52| 55| 56| 59 59 | 63,5, 65 | 70| 73| 75 79 | 80 84 37 38 39 40 | 41 42| 43| 44| 45| 46| 47| #48/ 49/ 50 51 52 53 54 2 Rb/2 Sr & Y [F Zr| Nb|2Mo/2 Tc/8 Ru|& Rh|s Pd 2 Ag |= Cd |= In | Sn |Z Sb|k& Te G I | Xe 86 88 89 1 92 96 4101} 103) 106, 108| 112| 115| 119| 122| 128| 127| 131 55 56 57 72 | 73 74! 75| 76/| 77 78 |- 79 80 | 81 82 | 83 84 | 85 86 5 Cs|2 Ba| La\2 Hf) Ta) W_) Re| Os) Ir| Pt} Au, Hg|/2T@/2Pb\2 Bi |Z Po|G At| Rn 133| 137| 139| 179| 181| 184| 186} 190| 192] 195| 197] 201| 204| 207| 209 87 88 89 n a 3 Fr js be Ac 58 | 59 | 60 | 61 | 62 | 63 | 64 | 65 | 66 | 67 | 68 | 69 | 70 | 71 Ce Pr | Nd |Pm/|Sm| Eu | Gd| Tb | Dy | Ho | Er | Tm | Yb | Lu 140 | 141 | 144 150 | 152 | 157 | 159 | 163 | 165 | 167 | 169 | 173 | 175 90 91 92 | 93 | 94 | 95 | 96 | 97 | 98 | 99 | 100 | 101 | 102 | 103 Th | Pa| U | Np | Pu | Am|Cm_) Bk Cf | Es | Fm/| Md | No | Lr 232 238 Copyright reserved

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Downloaded from Stanmorephysics.com QUESTION 14 (2) (2) (2) (2) (2) (2) [12] QUESTION 2 2.1.1. OPTION 1 distance speed = time 60¥ v= — 30 at At=2sv iafimorephysics.com OPTION 2 4 AX= v,At+ panty Ax= ( OPTION 3 Ve + Vi v 5 at 30 +30 1 60v = (30)At + 5(0)AteY 60v= ( Jaw At=2sv At=2sv (4) 2.1.2 POSITIVE MARKING FROM 2.1.1 OPTION 1 v= it Ax =— v 22 z Ax = 44 m V east (Accept: Right)” OPTION 2 4 AX = viAt + paar? il = (22)(2) + 3 )(2)°v Ax = 44 m V east (Accept: Right)” Copyright reserved

Downloaded from Stanmorephysics.com 2.1.3. The rate of change of position/ The total displacement over the total time.v OPTION 1 [POSITIVE MARKING FROM 2.1.1] Time taken for the buck to move from B to C: 2.1.4 distance speed = time to move from B toC =6-2¥Y =4s OPTION 2 POSITIVE MARKII 44v =88m v= 33 m:s" east V Copyright reserved (2) (6)

Downloaded from Stanmorephysics.com 2.2 scale and direction ark — F3 drawn right from the head of F2 with the correct scale and direction. ark — Final answer (160 N). Can include on diagram or on its own. ark — F1 drawn with correct scale and direction ark — F2 drawn from the head of F1 with correct ark — Fr drawn with correct direction and scale. F3=160N v¥ QUESTION 3 3.1 The cyclist changed direction / The direction changed from west to east. V 3.2 we Ax OR Vv At _ 60-0V ~ 30-04% = 2m:s' west Vv 3.3 Om-s2/Zero/OV 3.4 The graph is a straight line OR Graph has a constant gradient. v The velocity is constant. ¥ 3.5 yee X2-— X% 60-X we = eel 60 - 70 X=35m Distance = 60 + (60 — 35) v =85mv Copyright reserved (5) [20] (1) (4) (1) (2) (4) [12]

Downloaded from Stanmorephysics.com QUESTION 4 4.1 The total path length travelled. vv 4.2 1 OR Ax= viAt + at? _ AK vit 2 1 2 = 32(3) + 2 (3) v a= Ax (3) Ax = 96 mv 4.3 2(-2,5)(Ax) | 2 ee At) = 12,8's ¥ At =12,8sv 44 POSITIVE MARKING FROM 4.2 AND 4.3 1 OR Ax= v,At+ aat? Him 1 2 =5Y a v 5v (15,8) + 5 (0)(15.8) = 5v(15,8) ¥ Ax=79m Ax =79m Distance = 500 — 96 - 204,8 -79 v =120,2 mv Copyright reserved 1 Ax=vjAt+ =aAty AtY 204,8 Vv = 32At+ 2(2,5)aby (2) (2) (4) (4) [12]

Downloaded from Stanmorephysics.com QUESTION 5 5.1 The amount of substance having same number of particles as there are (2) atoms in 12 g carbon -12. OR That quantity of substance that contains 6,02x10?5 elementary particles. v ¥ able but mark according to the mark allocation] (4) 5.2.2 POS! KCIOs : KCI 1:1 (OR2:2) n(KCl) = 0,25 mol ¥ N n(KCl) = 5¥ N 0,25 = ———_v 6,02x107° Sv (4) 5.2.3 2 58 n(O2) = 0,375 mol ¥ Vv = —Vv n(02) V, m Vv =—vV 0,375 224 V(O2) = 8,4 dm? Y (Accept 8,51 dm’) (4) [14] Copyright reserved

Downloaded from Stanmorephysics.com QUESTION 6 6.1.1 6.2 6.3 m-(O) = 1000 — (265,31 + 353,74) v = 380,95 g ¥ POSITIVE MARKING FROM 6.1.1 (2) Simplest Ratio Simplest Ratio (Modified) 10 x M(H,0) M(X2CO3.10H20) % H,0 = 62,937V = M(X) = 23 g.mot' | Xis Nav OPTION 1 V(CO,)= =x 90 OR 2x90 v = 60 cm? C2H2 : CO2 1 :2 (OR 2:4) V(CoHa)= 5% 60 ¥ V(C2H2) = 30 cm3 ¥ OPTION 3 6 gas moles > 90 cm? 1 gas mole = = v = 15 cm$ V(C2H2) = 2x 15 ¥ = 30 cmv Copyright reserved Empirical Formula: K2Cr2O7 ¥ Y (all modified simplest ratios) (6) (5) OPTION 2 V(H20)= 4x90 OR 2x90 v = 30 cm? C2H2 : H2O 1 : 1 (OR 2:2) V(C2H2)= V(H20) ¥ V(C2H2) = 30 cm3 ¥ (3) [16]

Downloaded from Stanmorephysics.com QUESTION 7 7.1. The-number of moles of solute Y per cubic decimetre of solution. ¥ OR The number of moles of a substance V per unit volume (of a solvent) ¥ [Calculation can be in a single table but mark according to the mark allocation] : 1 (or6:3) Ga TOTAL: Copyright reserved (5) [14] 100