Physical Sciences Grade 12 Paper 1 KZN March 2018 Memo

Downloaded from hlayiso.com Education KwaZulu-Natal Department of Education REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE GRADE 12 " PHYSICAL SCIENCES: PHYSICS (P1) * : COMMON TEST , 4 MARCH 2018 : MARKS: 50 TIME : 1 hour This question paper consists of 6 pages and a 1-page data sheet. Copyright Reserved Please Turn Over ———— << M4

Physical Sciences P1 Downloaded from hlayiso.com March 2018 Common Test NSC-GRADE 12 INSTRUCTIONS AND INFORMATION TO CANDIDATES 10. 11. 12. Write your name on the ANSWER BOOK. This question paper consists of FOUR 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 subsections, 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 SHEET. Show ALL formulae and substitutions in ALL calculations. Round off your final numerical answers to a minimum of TWO decimal places. Give brief motivations, discussions, et cetera where required. Write neatly and legibly. Copyright Reserved Please Turn Over M4

Physical Sciences P1 Downloaded from hlayiso.com 3 March 2018 Common Test NSC-GRADE 12 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 question number (1.1 — 1.4) in the ANSWER BOOK, for example 1.5 D. 1.1 M2 1.3 A heavy crate is at rest on a rough horizontal surface. Three different forces, each applied to the crate, attempt to move the crate as shown in the diagram. F2 Fa a a CRATE }#—— > CRATE CRATE I Il Il In each case, the crate is just at the point of moving. The maximum force of static friction between the crate and surface is ... A Greatest in situation (1) B Greatest in situation (II) Cc Greatest in situation (III) D Equal in all three situations. (2) A satellite orbits Earth at a point x, where the gravitational force is a quarter (1) of the gravitational force it experiences on the surface of the Earth. x © Satellite Earth If the radius of the Earth is R, then the height of the satellite ABOVE THE SURFACE of Earth will be ... A 4R B 2R Cc R D eR (2) The net force acting on an object is a measure of the of the object. A Change in kinetic energy B Rate of change of kinetic energy Cc Rate of change of momentum D Change in momentum (2) Copyright Reserved Please Turn Over M4

Downloaded from hlayiso.co NSC-GRADE 12 Physical Sciences P1 1.4 Two balls P and Q are thrown into the air simultaneously from the same height above the ground. P is thrown vertically upwards and Q vertically downwards Mo 8 Common Test with the same initial speed. Neglect air resistance. Take downwards as the positive direction. Which of the following is TRUE for P and Q when they hit the ground? VELOCITY ACCELERATION A Both have the same velocity Both have the same acceleration B P has a greater velocity than Q | P has a negative acceleration; Q has a positive acceleration Cc P has a greater velocity than Q | Both have the same acceleration D Both have the same velocity P has a negative acceleration; Q has a positive acceleration. QUESTION 2 A train carriage A of mass 3000 kg is moving horizontally to the right at 6 m.s~". It collides with another train carriage B of mass 4200 kg that is initially at rest, as shown in the diagram below. 6m.s* —> A The graph below shows the variation with time t¢ of the velocities of the two train carriages before, during and after the collision. 6 \ train|cafriggd BI ° \ v(m.s*) aj t (Ss) “4 o train Fayriage ry 2.1 Define momentum. 22 Determine, by means of appropriate calculations, whether the total momentum of the system is conserved or not after the carriages collided. 2.3. Calculate the average net force exerted by train carriage A on B. (2) [8] (2) (6) (5) [13] Copyright Reserved Please Turn Over M4

Physical Sciences P1 Downloaded from hlayiso.com NSC-GRADE 12 QUESTION 3 March 2018 Common Test A hot-air balloon is rising vertically upwards with a constant speed of 2.5 m.s“'(Figure 1). When the balloon is at a certain height (h) above the ground, an object is accidentally dropped from the balloon. The object reaches the ground 4 s later. Figure 2 shows the velocity time sketch graph (where down is positive) for the vertical motion of the object from the moment it is dropped until it hit the ground. Neglect air resistance. (g = 9,8 m.s*) 3.1 3.2 3.3 3.4 3.5 Copyright Reserved fas ms v(m.s") R ay RRS TS AES St Bh ee sere eee 1 ay” Py ground Figure 1 Figure 2 Define a projectile. What is the magnitude and direction of the velocity of the object at point P. What physical quantity does the gradient of the graph correspond to? What is the significance of point Q on the graph with respect to the motion of the object? USING THE GRAPH ONLY (No equations of Motion), calculate: 3.5.1 The velocity of the object when it hits the ground. 3.5.2 The height, h, from which the object was dropped. MA (4) [16] Please Turn Over

Downloaded from hlayiso.com 6 Physical Sciences P1 March 2018 Common Test NSC-GRADE 12 QUESTION 4 Block A, that is at rest on a horizontal rough surface, is used as an anchor to hold block B, of mass of 60 kg, in the air at a certain height above the ground. The two blocks are connected with rope S that makes an angle of 25° with the vertical. Block B is suspended from the ceiling by cable R that makes an angle of 8 with the ceiling. 60kgl =B Block A experiences a frictional force of magnitude 150 N. The entire system is in equilibrium . Neglect air resistance. 44 Define frictional force. (2) 4.2 What is the magnitude of the resultant force acting on block B? (1) 43 Drawa labelled free-body diagram indicating all the forces acting on block B. (3) 4.4 Calculate the vertical component of the force in cable R. (5) 4.5 Calculate the angle, B, between cable R and the ceiling. (2) [13] TOTAL: [50] Copyright Reserved Please Turn Over M4

Downloaded from hlayiso.com Physical Sciences P1 March 2018 Common Test NSC-GRADE 12 DATA FOR PHYSICAL SCIENCES (PHYSICS) GRADE 12 GEGEWENS VIR FISIESE WETENSKAPPE (FISIKA) GRAAD 12 TABLE 1: PHYSICAL CONSTANTS/TABEL 1: FISIESE KONSTANTES NAME/NAAM SYMBOL/SIMBOOL VALUE/WAARDE Acceleration due to gravity e2 Swaartekragversnelling g opts Universal gravitational constant le 6,67 x 1071 N.m2.kg2 TABLE 2: FORMULAE/TABEL 2: FORMULES MOTION/BEWEGING vp =Vitaat Ax = v,At+ 4aAt® orf/of Ay =v,At+ $adt? 5G va +V Vv, +V; vp? =vi2 + 2adx orfof v,? =v, +2ady_ | AX = (aes) At or/ofAy = (SS Ja 12 K=E, =. mv 2 FORCE/KRAG Fret = Ma p=my Fie Mt =Ap=mv , my; Fy =mg Fe Sm,m, r £0 = y,N f= LN Copyright Reserved M4

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Downloaded from hlayiso.com education ) Department: Education PROVINCE OF KWAZULU-NATAL : PHYSICAL SCIENCES P14 ‘ | MEMORANDUM : COMMON TEST | : MARCH 2018 ; | NATIONAL SENIOR CERTIFICATE N.B. This memorandum consists of 4 pages including this page. Copyright Reserved Please Turn Over bh

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Downloaded from hlayiso.com Physical Sciences 2 March 2018 Common Test NSC — Memorandum (GRD 12) PHYSICAL SCIENCES PAPER ONE QUESTION 1 11. Cv (2) 12 Cvv (2) 13. Cvv (2) 14 Avv (2) [8] QUESTION 2 24 The product of mass and velocity of the object. vv (2) 2.2 total Poefore = Mavat Mave — — = (3000)(6) + (4200)(0) v = 18000 kg.m.s" ,to the right ~ anyone vy total pafter = Ma vat Meve =(3000)(-1) + (4200)(5) ~ = 18000 kg.m.s" ,to the right” Total Pvetore = total pater Momentum is conserved v (6) [ m(v,- v,) / __m(v,- vj) 2.3 Fret = aA “ OR Fret ~ At _ 4200v(6 -0) ¥ _ 3000¥(-1 - 6)” ~ av - 24 = 10500N tothe right ¥ = -10500 N (to the right) Fg = 10500 N to the righty (5) - ~ [13] QUESTION 3 3.1 An object upon which the only force acting is the force of gravity. v ¥ (2) 3.2. 2.5m.s! Vupwards ¥ (2) 3.3 (Gravitational) acceleration “ (1) 3.4. Maximum height reached by the object. v ” (2) Copyright Reserved Please Turn Over

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Downloaded from hlayiso.com Physical Sciences =] March 2018 Common Test NSC — Memorandum (GRD 12) 3.5 If equations of motion are used for the 3.5.1 and 3.5.2 subtract 2 marks for each question. de V4 wef bab” va 3.5.1 gradient = —— _ + )(« ) At 2-2 the) y 98 = oe = LET toa NK v Vv, +25 . (> | 98 =—R8 numerator and denominator ‘ ‘ 4-0 v Vz = 36,70 m.s'v (downwards) ¥ Note: If point R and Q are used the answer is 36,75 m.s! (5) Positive marking from 3.5.1 Ue 3.5.2 Height = Area ofA, +AreaofA, ¥ BN " Ye wat = (A) x (0,25) x (2.5) + (4) x (4 - 0,25) x (36,7) ¥¥ Af wot sale Ke = 68,69 m% a > ste) + 4/9 (6 ] iv OR GPs botiy a (»/ Height = Area of trapezium of the v — t graph. ¥ = %4(2,5 + 36,7)(3,5) vv = 68,60mY” (4) [16] QUESTION 4 4.1 The force that opposes the motion of an object Vand acts parallel with the surface. v (2) 42 ONY (Accept 0 / Zero) (1) 4.3 F, R Fs w [ Notes : Accepted Labels Mark | w weight /Fe/ Fy 4 FR Tension force in cable / Tz iv Fs Tension force in rope / Ts v | Any additional force: deduct 1 mark (maximum %) Lines must touch object otherwise (maximum % ) Subtract one mark if arrows are not shown (maximum % ) (3) Copyright Reserved Please Turn Over

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Downloaded from hlayiso. Physical Sciences 4 NSC — Memorandum (GRD 12) com March 2018 Common Test 44 2Fx=0 tan 25°= Fs.sin 25° - 150 = 0 W Fs= 354.93 N Anyone FE = OR SV _ ik-o ———= Then, 2Fy = Fry + Fsy=mg Fey = Fry + (354.93).cos 25° Y= (60)(9.8) ” = Fry = 266.32 N ¥ a Positive marking from 4.4 45 266,32 tan B= 59 B = 60,612 v OR W = 588N S = 354,93 N R=? R? = W? + S?— 2WSCos 25° R? = 588? + 354,932 — 2(588)(354,93)Cos 25° ¥ R = 305,66 305,66Sin (90° - B) = 354,93Sin25° B= 60,61°V Copyright Reserved 150 v sw = 321,68 Ov 60 x 9,8 Y- 321,68” 266,32N v nas (.)) (2) [13] TOTAL MARKS: 50

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