Limpopo-Physical-Science-Grade-10-September-2023-QP-and-Memo_hlayiso.com_.pdf

Downloaded from hlayiso.com LIMPOPO© PROVINCIAL GOVERNMENT REPUSLIC OF SOUTH AFRICA EDUCATION MOGALAKWENA DISTRICT PHYSICAL SCIENCES saat NATIONAL Sq MARKS: 100 TIME: 2 hours This question paper consists of 11 pages including this one Page 1 of 11

Downloaded from hlayiso.com Aossleaded from Stanmorephysics.com Controlled test term 3 INSTRUCTIONS 1. This question paper consists of 11 pages including the cover page 2. Answet allthe questions in the answer book 3. You are advised to use the attached DATA SHEETS. 4. Round off your final answer to a minimum of TWO decimal places 5. Show all your calculations including formulae where applicable. 6. Candidates may use non-programmable calculators. 7. Write neatly and legibly.

Downloaded from hlayiso.com Aossleaded from Stanmorephysics.com Controlled test term 3 QUESTION 1: MULTIPLE-CHOICE QUESTIONS Four options are provided as possible answers to the following questions. Each question has only ONE correct answer. Write only the letter (A-D) next to the questionpnumber (1.1 - 1.10) in the ANSWER BOOK 1.1. A caristravelling at a speed of 30 m.s“'on a straight road. What would-be the speed of the car in km-h* ? A 30km-h" Bo 130 kmh! C 8.33 kmh! D108 km-h" (2) 1.2 The gravitational potential energy of an object relative to the ground is dependent on the object’s A velocity B Position Cc Change in velocity D Speed (2) 1.3. What volume (in dm?) does 1 gram of hydrogen gas occupy at STP? A 11.2 B 44.8 Cc 22.4 D 56 (2) 1.4 The gradient of a velocity time graph is equivalent to the ... A Acceleration B Position Cc Total distance covered D Displacement (2) 1.10 Which ONE of the vector diagrams below will result in the largest resultant vector? A B c [20]

Downloaded from hlayiso.com Aossleaded from Stanmorephysics.com Controlled test term 3 QUESTION 2 2.1 (2) Define-displacement ice session, a basketball player runs around the basketball court starting He runs to point B, C, D and then returns to point A. B Length = 30 m Cc E a b (| ) dle / NY 3 S £ a A D 2.2.1 Draw the vector diagram the movement of the basketball (2) player for one lab around the court. (Ensure that diagram is fully labelled). Using scale 1cm=5m 2.2.2 Calculate the distance he travelled for one lap (2) 2.3 If it took 19 seconds for the basketball player to complete lab around the court. 2.3.1 Calculate his average speed. (4) [10]

Doenleaded from Stanmorephysics.com Downloaded from hlayiso.com QUESTION 3 3.1. Define empirical formula 3.2 An inorganic substance was analyzed and found to be containing 65.34% of oxygen, 32.65%, of Sulphur and. x amount of hydrogen. 3.2.1 Calculate the percentage of hydrogen element in the substance 3.2.2 Determine the empirical formula for the compound. 3.3. 6.257g of hydrated copper sulphate (CuSO,:nH2O) was heated, and during the heating the mass,was recorded. The mass of the content was decreasing and after a while the mass remained unchanged at 4g. 3.3.1 Give a reason why the mass decreased. 3.3.2 Determine the value of n in the CuSO,4-nH,O 3.4. Define the term molar. mass. 3.5 Calculate the following for Ata(SO4)s : 3.5.1 Its molar mass 3.5.2 Its percentage composition 3.5.3 The number of moles present in 85,5 g Controlled test term 3 (2) (2) (4) (2) (4) (2) (2) (3) (3) [24]

Downloaded from hlayiso.com Aossleaded from Stanmorephysics.com Controlled test term 3 QUESTION 4 4.1 4.2 4.3 44 45 A skateboarder, starting from the top of a ramp 4,5 m above the ground; skates down the ramp, as shown in the diagram below. The mass-of-the skateboarder and his board is 65 kg. Ignore the effects of friction : Define the term gravitational potential energy in words. (2) Calculate the gravitational potential energy of the skater just before he (3) skates down the ramp. State the principle of conservation of mechanical energy in words. (2) Use the principle stated in QUESTION 4.3 to calculate the magnitude (4) of the velocity of the skateboarder when he reaches the ground at point X. Will the skateboarder be able to reach point Y if he were to remainon (5) his skateboard? Write YES or NO and support the answer with a relevant calculation. [16]

Downloaded from hlayiso.com Aossleaded from Stanmorephysics.com Controlled test term 3 QUESTION 5 5.1 5.2 5.3 5.4 5.5 A car accelerates from rest at 15 m.s~? for 2 seconds on a horizontal road. Definethe term acceleration. (2) Caléulate the: 5.2.1 Distance covered by the car. (3) 5.2.2 Velocity of the car. (3) While travelling at a constant velocity of 108 km. h~* , the driver of a car notices a sign warning motorists to keep a safe 2-second following distance. At that instant the car is 80 m behind a truck that is travelling at a constant velocity of 90 km.h7}. 108 km:h" > = ot | 80m a= \ EEE EEE Explain the meaning of a safe 2-second following distance. (2) Calculate the safe 2-second following distance behind the truck. (6) Calculate how long it will take the motorist to get to a safe 2-second (5) following distance behind the truck. [21]

Downloaded from hlayiso.com Aossleaded from Stanmorephysics.com Controlled test term 3 QUESTION 6 The velocity versus time graph for a racing car moving eastwards, is show?f below. = N E 7 2 E > time (s) s 6.1 Write down the initial velocity of the car. (2) 6.2 Write down the speed of the car at time t= 10s (2) 6.3 Describe the motion of the car for the section labelled CD (2) 6.4 Support the answer to QUESTION 4.3 above by calculating the (4) acceleration for section CD 6.5 Without any calculation, compare the magnitude of the acceleration (2) of the car in part DE with that of part CD of the journey. Write only GREATER THAN, LESS THAN or EQUAL TO. Give a reason for the answer. 6.6 Determine the total displacement for the motion of the car. (7) [19] Total [100]

Downloaded from hlayiso.com Aossleaded from Stanmorephysics.com Controlled test term 3 DATA FOR PHYSICAL SCIENCES GRADE 10 PAPER 1 (PHYSICS) GEGEWENS VIR FISIESE WETENSKAPPE GRAAD 10 VRAESTEL 1 (FISIKA) TABLE 1: PHYSICAL CONSTANTSI/TABEL 1: FISIESE KONSTANTES elelall f SYMBOL/SIMBOOL | VALUE/WAARDE Spoed van'ig inn vakuum ¢ 3.0 x 10° ms" eine Me 9,11x 107" kg TABLE 2: FORMULAE/TABEL 2: FORMULES MOTION/BEWEGING Vv; =Vj + at Ax=v,At+ Laat? Vy +V v,7 -v? +2aAx AX — = Lat WORK, ENERGY AND POWERIARBEID, ENERGIE EN DRYWING U=mgh or/of E,=mgh k=4 mv? or/of E, =4mv? Em =Ex+E,. or/of Ey=K+U WAVES, SOUND AND LIGHT/GOLWE, KLANK EN LIG v=far 7-1 f c E=hf or/of E= ne ELECTROSTATICS/ELEKTROSTATIKA ne 2 ELECTRIC CIRCUITS/ELEKTRIESE STROOMBANE Q=IAt SS he es R, =R, +R, +... V=—

Downloaded from hlayiso.com Aossleaded from Stanmorephysics.com Controlled test term 3 DATA FOR PHYSICAL SCIENCES GRADE 10 PAPER 2 (CHEMISTRY) GEGEWENS VIR FISIESE WETENSKAPPE GRAAD 10 oar VRAESTEL 2 (CHEMIE) TABLE ‘yang SICAL CONSTANTS/TABEL 1; FISIESE KONSTANTES AMEINAAM SYMBOL/SIMBOOL | VALUE/WAARDE TABLE 2: FORMULAE/TABEL 2: FORMULES m n= Na m Vv --_O! n=— or/of c Ve

Downloaded from hlayiso.com Pyswsleadiad from Stanmorephysics.com Controlled test term 3 TABLE 3: THE PERIODIC TABLE OF ELEMENTS TABEL 3: DIE PERIODIEKE TABEL VAN ELEMENTE 2 3 4 . 6 7 s . J Ww u ard ary “ aL} 16 vw 1% wy - om oO ~~ WH jw Atomic number i. KEY/SLEUTEL Atcomgetal : In uA He 29 4 + Electronegativity Symbol sy] e]_7] 8]. 0| © 2 Li 2 Be Elektronegatwitcte | «* CU |*“eimboo! Sepiclisn|gols Fl] Ne 3 me n 12 4 16 12 Zz 4 T 3 ry 15 16 bid 8 a epcocinnis sohive seriic mane sae sils Pls sigcel| ar Pemoerse omer enoreemaes Aer Shin f eS 6 Se) a 939 as Bele xsezsscel-r-/> 3 = Se ee oe wee eee oboe eh obeeeeeeel © s i= c|° Tif V2 Cri? Mn/® Fe |® Co|® Ni |* Cu/* Zn|* Ga |? Ge|S As|5 Se |S Br} Kr as | 51) sa] 55/56 |" 50] 50| 63.5] 66| 70 | 73| 75 | 70| s0| ea 3] ae] a0] at] aa] 43 | aa] a8) 40] a7 | a8] 00 | 80 | 81 | a2] | ee S$ 2 SrlS Y¥ | Z2r| Nb/= Mol? Tels Rujd Rhis Pd |e Ag|= Cd/= In |S Sn|2 Sbi5 Te/S 1 | Xe we! oo | o1| 92, 96 101] 103] 106 100) 142) 148| 119| 122] 120| 427|_ 109 $6 37 72 7 4 7 76 7 73 7 30 1 82 a3 a4 as a6 3 3 Ba} Laj> Hf) Ta) W| Re| Os) Ir| Pt] Au| Hg|> Te |2 Pb|= Bi |S Pols At} Rn sez s98|" 170) set, se4] 195] 190] 49a] 196] 187 _zor]” 204” a07|" 209 S Fr js Ra) Ac a Ce Pr} Nd/|Pm/| Sm) Eu | Gd| Tb Ho | Er | Tm) Yb | Lu 140 141 146 150 152 157 159 163 165 167 163 173 175 a a Th | Pa| U | Np| Pu| Am/Cm) Bk | Cf | Es | Fm| Md | No | Lr 22 218

Downloaded from hlayiso.com Downloaded from Stanmorephysics.com PROVINCIAL GOVERNMENT REPUSLIC OF SOUTH AFRICA EDUCATION MOGALAKWENA DISTRICT MEMORUNDOM PHYSICAL SCIENCES NATIONAL*SENIOR CERTIFICATE TERM 3:°CONTROLLED TEST 08 SEPTEMBER 2023 GRADE 10 MARKS: 100 TIME: 2 hours — This question paper consists of 7 pages including this one Page 1 of 7

Downloaded from hlayiso.com Downloaded from Stanmorephysics.com QUESTION 1 114. Dv¥v 1.2. Bv¥v 1.3. Cvv 14. Avv 15. Bvy, [10] QUESTION 2.1. The difference in position (space ) VY (2) 2.2.1 B 30m Cc Marking Criteria Y lines drawn to scale £ A Y 4 arrows shown, touching each other a <4 | ¥ correct shape of drawing A 30m D () 2.2.2 Distance=AB+BC+CD+DA =(15)+(30)+(15)+(30) =90mv¥ (2) distance time 2.3.1 Avarage Speed = _ 90mv ~ 1938¥ = 4.74m.s7!¥ [10] Page 2 of 7

Downloaded from hlayiso.com Downloaded from Stanmorephysics.com QUESTION 3 3.1 The simplest whole number ratio of atoms in a compound. VY 3.2.1 %H= 100-65,31%-32,65%=2,04% 3.2.2 H Ss fe} Mass (g) 2104 32.65 65.31 M (g/mol) Jl 32 16 Mole=m/M—|-2!04 1.02 4.08 vv Ratio 2 1 4 v Empirical formula H,S0,% 3.3 3.3.1 water changes into gas V and leaves the (system)” 3.3.2 CuSO, H,0 Mass (g) 4 2.257 v M (g/mol) 159.50 18 Mole=m/M 0.0251 0.125 vv Ratio 1 Sv 3.4. The mass of one of a substance measured in g.mol'vv 3.5.1. M( At2(SOa)s ) = 2(27) +3(32) +12(16) = 342 g.mol'vv __2(27) 3.5.2 % Al = ~100 = 15, 79%Y % SF = 362) % S = 342 x100 = 28,07%V % S = EO «100 = 56,14%V 3.5.3 POSITIVE MARKING FROM m Afa(SO. =~ v n( Al(SO4)s ) = _ 85.5 ~ 342 = 0, 25 mol ¥ (3) [24] Page 3 of 7

Downloaded from hlayiso.com Downloaded from Stanmorephysics.com QUESTION 4 4.1. The energy an object has because of its position in the gravitational field relative to some reference point ” (2) 4.2 Ep=mhgy¥ = (65) AO.8)+(4.5)¥ =2 866,5Jv (3) 4.3 The net/total mechanical energy (sum of kinetic and gravitational potential energy) in an isolated/closed system” remains constant/ is conserved” (2) 4.4 (Ep + Ex)topibo = (Ep + Ex)boromion mgh + 0 =mgh + Yay" (65)(9.8)(4.5)”= 0 + (65 Vv" v v=939ms' ¥ OR/OF (Ep + Ex)topine = (Ep + Ex) pottom/onder mgh +0 =m pacr an A 2 866,5 v =0+ A(65)\- ¥ v=9,39ms" Vv 4.5 OPTION 1 (Ep + Ex)top = (Ep + Ek)bottom mgh + 0 =mgh +%mv? v (65)(9,8)h ¥ +0=0 +%x65x (9,39) Vv 637h = 2 865, 6 h =4,49m NovY:h =4,49m<6mv OPTION 2 Ep aty = mghyv = (65)(9,8)(6) =3 822 Jv Emech < Ep aty ¥ therefore he will not reach point YY“ [16] Page 4 of 7

Downloaded from hlayiso.com Downloaded from Stanmorephysics.com QUESTION 5 5.1The rate ¥ of change of velocity v’. (2) 5.2 5.2.1 AX= vA ty , Ax= 0( 5(15)2°v Ax = 30 m¥ (3) 5.2.2 positive marking from question 5.2.1 OPTION 4/OPSIE 1 OPTION 2/OPSIE 2 vy2= Vi2+2aAxY Vz V j+adt v= 07 ¥+2(15)(30)¥ = OV + 15x2v Vv; = 30 m:s" to the righty /regs vy = 30 m-s"' to the rightv /regs Accept: To the right/East/In the direction of motion (3) 5.3 When following a car, a motorist should keep a safe distance such that it takes more than 2s v’to reach the same position v as the car in front. OR The car will need 2 s to stop in an emergency and not hit the car in front. vw (2) 5.4 Convert 90 km-h" into m-s/Skakel 90 km-h’’ om na m-s* 90km_ 90x 10° i Ee sl¥ th 36007 25s OPTION 4/OPSIE 1: OPTION 2/OPSIE 2: 1 _ (Vit Mi J Ax= vate 5aAty ax=( : )at ar 25 + 25 = AX = (25)(2)¥+ 5(02v ax=( 5 ) vay kay Ax = 50 my Ax = 50 mY a (6) Page 5 of 7

Downloaded from hlayiso.com Downloaded from Stanmorephysics.com 5.6 108km_ 108 x 10° Th 3600 eed/Verskil in spoed: 30 — 25 =5m-s* el 30 m (80 — 50) at 5 m-s"'to be at a 2 second distance behind fore: distance = (v) (t) ¥=30ms'! Difference ifisp Car has td the truck. / 30 = (5) (t( t=6s VV (5) [21] QUESTION 6 (2) 6.1 30 m/s ¥¥ 6.2 40m/s vv (2) 6.3 The speed decreases v uniformly (from 40 m/s to 0 m/s )¥ OR The car slows downy and finally stops¥” 6.4 guh¥y _ (0)-40 ¥ ~ 25-20 ¥ =-8m-s*v R 6.5 Equal toY , Same gradient Vv (2) Page 6 of 7

Downloaded from hlayiso.com Downloaded from Stanmorephysics.com 6.6 OPTION 1/OPSIE 1 Displacement = Area under the v-t graph” Verplasing = Opperviakte onder v-t grafiek =(Atrapezium + Arectangie/reghoek + Atriangle 1/driehoek 1) — Atriangie 2/driehoek 2 (40+30)(5)” +(15x40) “+ % (5 x40)“ —['% (2,5 x 20)]¥ 50 mv east/oosY OR/OF Displa = Area under the v-t graphy Verplasing = Opperviakte onder v-t grafiek = (Atrapeziumitrapesium + Arectangie'regnoek + Atriangle/anienoek)- Atriangle/dnenoek = % (20+15)(10)” + (30 x 20)” + % (5 x40) Y— %4 (2,5 x 20)” = 850 mv east/oosv (7) [19] Total 100 Page 7 of 7