Physical Sciences Grade 11 KZN March Test 2026 Paper & Memo

Downloaded from Stanmorephysics.com KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA PROVINCIAL STANDARDISED ASSESSMENT GRADE 11 MARKS: 100 DURATION: 2 hours This question paper consists of 13 pages and 1 data sheet.

Phyiowshancdewréaom) Stanmorepbgsics.com March 2026 2 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. Stanmorephysics,com 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 SHEET. nfl calculations. Show ALL formulae and she itutiol s to a inirmur of TWO decimal places. Round off your final numerical Give brief motivations, disc et cetera where required. Write neatly and legibly. Copyright reserved

Phyiowshancdewréaom) Stanmorepbgsics.com March 2026 3 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—1.7) in the ANSWER BOOK, for example 1.8 E. 1.1 Consider the vector diagram below: iy: x , ni the fesutant vector of the other three? Which ONE of the following rep) ~ Se. A W = i BX c oY a D Z (2) 1.2 The forces acting on an object are in equilibrium. Consider the following statements: (i) The object could be stationary (ii) The object could be moving at a constant velocity (iii) | The acceleration of the object is zero Which of the above statements is/are true? A (i) only B (ii) only C (i) and (iii) only D (i), (ii) and (iii) (2) Copyright reserved

Phyiowshanceréaom) Stanmorepsbgsics.com March 2026 4 Bs A box is at rest on a horizontal surface. BOX SURFACE Which of the following forces is the reaction force to the WEIGHT of the box? A Normal force acting on the box B Force that Earth exerts on the box C__ Force that the box exerts on Earth D_ Force that the box exerts on the surfack (2) ine by a force F, as shown in the angle 8 with the horizontal 1.4 A block of mass m is pulled up diagram below. The incline The NORMAL force acting on the block is given by ... A mgsin@ B= mgcos@ C mg sin@ — Fcos@ D —_mgcosé - Fsin@ (2) Copyright reserved

Phyiowshancderéaom) Stanmorepbgsics.com March 2026 5 1.5 The acceleration due to gravity on the surface of Earth is g. Which ONE of the following is the gravitational acceleration on another planet which has HALF the mass of Earth and ONE-QUARTER of Earth’s radius? A 2g B 4g C 8g D 16g (2) 1.6 A small sphere, carrying a constant charge is placed r metres from a fixed point, P. The electric field EatP for different values of r. Which of the following graphs correctly shows the relationship between the electric field E at P and r?? A B Cc D = Zz = < wi i ie Ww 2 (m2) 72 (m2) 12 (m2) 2 (m2) (2) Copyright reserved

Physiewshenderfiaom) Stanmorephysics.com March 2026 6 1.7 Two identical conducting spheres carrying charges Q and -2Q are placed r metres apart. The electrostatic force acting on each sphere is F. The spheres are brought into contact and are then separated so that the distance between them remains r metres. Which ONE of the following is the magnitude of the electrostatic force that the spheres exert on each other after they are separated? A F 2 F Bg c F 8 2. D oF [2X7=14] Copyright reserved

Phyiowshanderéaom) Stanmorepbgsics.com March 2026 7 QUESTION 2 (Start on a new page.) A box is lifted vertically upwards at CONSTANT VELOCITY using two light, inextensible strings, P and Q. The strings pass over frictionless pulleys attached to a horizontal ceiling and are attached to the box at point X, as shown in the diagram below.. Stanmorephysics,com e. The tension in string P, which makes an angle of 50° with the ceiling, is 200 N. e String Q makes an angle of 40° with the ceiling. Ignore the effects of air resistance and assume that the box does not rotate while being lifted. Ceiling 200 N 2.1 What is meant by a closed vector diagram? (2) 2.2 Draw a fully labelled closed vector diagram representing all forces acting on the box. Indicate any TWO angles. (3) 2.3 Calculate: 2.3.1 The magnitude of the tension in string Q (3) 2:3.2 The mass of the box (3) [11] Copyright reserved

Phyiowshancderéaom) Stanmorepbgsics.com March 2026 8 QUESTION 3 (Start on a new page.) When three forces, F1, F2 and Fa act on an object at point O as shown in the diagram Salcteea eet moves from rest in a straight line at a bearing of 25°. Th forces re not drawn to scale. = =8N Fo=12N 2 E ge I + * Ss | Define the term resultant vector. ai ; (2) 3.2 Calculate the magnitude of the RESULTANT of: 3.2.1 The horizontal components of F1 and F2 ONLY (2) 3.2.2 The vertical components of F1 and F2 ONLY (2) 3.3 Determine the magnitude of Fs. (6) 3.4 A fourth force F4 now also acts at point O along the vertical axis, so that the object moves due EAST, along the positive x-axis. Determine the magnitude and bearing of Fa. (3) [15] Copyright reserved

Physiewshendecrfiaom) Stanmorephysics.com March 2026 9 QUESTION 4 (Start on a new page.) An investigation is conducted to determine the relationship between the net force acting on an object and its acceleration. The apparatus used in the experiment is shown below. Ticker- timer Trolley with Ticker-. Mass Pieces tape Hanger with mass pieces The trolley moves down a friction-compensated ramp, pulled by mass pieces on a mass hanger. The hanger is attached to the trolley by a light, inextensible string that passes over a frictionless pulley. The net force is increased by moving mass pieces from the trolley to the hanger. 4.1 Explain how a ticker-timer and ticker-tape is used to adjust the ramp in order to compensated for friction. Stanmorephysics,com 4.2 Identify: A:2:1 The dependent variable 4.2.2 A controlled variable 4.3 Write down an investigative question for this experiment. The ticker timer has a frequency of 50 Hz, and the ticker tape shown below (not drawn to scale) is divided into intervals of five dots. Copyright reserved (3) (1) (1) (2)

Physiewshenderfaom) Stanmorephysics.com March 2026 10 4.4 Determine the time elapsed for every 5-dot section of the tape. (3) 4.5 Using the data on the ticker-tape above, calculate the magnitude of the trolley’s acceleration. (4) The results obtained is shown on the graph below: Acceleration vs Net Force a 9 = 2,1 , = ' 2 ! NS ' 5 14 o ' ° H Oo ' <= 07 H ; nt aii 0,49 0,98 \ 1,40 i Net Force (N) 4.6 Write down the mathematical relationship between the net force exerted on the trolley and the acceleration produced. (1) 4.7 Give a reason why the mass pieces added to the hanger were initially placed on the trolley. (2) [17] QUESTION 5 (Start on a new page) A 10 kg block resting on a rough horizontal surface is connected to a 20 kg block, by a light, inextensible string passing over a frictionless pulley. The 20 kg block hangs vertically, as shown in the diagram below. Fa=177N 10 kg 20 kg Copyright reserved

Physiewshenderfaom) Stanmorephysics.com 1 March 2026 A force of 177 N is applied to the left of the 10 kg block, to control the LOWERING of the 20 kg block to the ground. The kinetic frictional force between the 10 kg block and the surface is 11,5 N. 5.1 State Newton’s second law of motion in words. (2) 5.4 Draw a labelled free-body diagram showing all forces acting on the 20 kg block. (2) 5.3 Calculate the magnitude of the: 5.3.1 Acceleration of the 20 kg block (5) 5.3.2 Tension in the string connecting the blocks (2) 5.4 The 177 N force is now applied at an angle 6 to the horizontal as shown in the diagram below. ] +I Fa=177N 20 kg Will the kinetic frictional force on the 10 kg block INCREASE, DECREASE or REMAIN THE SAME? Explain the answer. (3) [14] Copyright reserved

Physiewshendecrfaom) Stanmorepbysics.com March 2026 12 QUESTION 6 (Start on a new page) A rocket, with its fuel, has a mass of 20 000 kg on the surface of Earth. The rocket is launched vertically upwards and its mass decreases uniformly as the fuel burns. At a height of 6,4 x 10° m above the surface of Earth, the rocket’s mass DECREASES by 5000 kg. 6.1 State Newton’s law of universal gravitation in words. (2) 6.2 Calculate the weight of the rocket on the surface of Earth. (2) 6.3 Determine the percentage reduction in the rocket’s weight when it is 6,4 x 10° m above the surface of Earth. (5) The rocket’s mass decreased by 25% as it ascends from the ground to 6,4 x 10° m above the surface of Earth 6.4 Explain why the percentage reduction in the rocket’s weight is considerably larger than the percentage reduction in the rocket’s mass. (2) _d [11] QUESTION 7 (Start on a new page) A metal sphere P, which was initially neutral, is given a charge of — 8 uC. TA Determine the number of electrons added to P to obtain a charge of — 8 UC. (3) Sphere R, carrying a charge of 5 uC, is placed 20 cm to the right of sphere P along a straight line. Point x is located 5. emto the right of sphere R, as shown in the diagram below. 20 cm 5cm \ Qr=-8 uC | iP R Xx Qr=5 uC 72 Define the term electric field at a point in words. (2) 7.3 Calculate the net electric field at point x. (5) Copyright reserved

Physiewshenderfiaom) Stanmorephysics.com March 2026 13 A third sphere S, of unknown charge, is placed 10 cm to the right of sphere R, as indicated in the diagram below. 10 cm The net electrostatic force acting on sphere P due to spheres R and S is 6,6 N to the right. 74 Determine the: 7.41 Electrostatic force that sphere S exerts on sphere P (5) 74.2 Charge on sphere (3) [18] Copyright reserved

Ph Pewsdeadadrdeom Stanmorapdiysics.com March 2026 14 DATA FOR PHYSICAL SCIENCES GRADE 11 PAPER 1 (PHYSICS) TABLE 1: PHYSICAL CONSTANTS NAME SYMBOL VALUE Acceleration due to gravity g 9,8 m.s? Universal gravitation constant G 6,67 x 10°71 N-m?-kg2 Radius.of the Earth Re 6,4. x 108m Mass of the Earth Me 5,98 x 104 kg Coulombs constant k 9 x 109 N-m2-C2 Charge on electron qe -1,6 x 10°79C Electron mass Me 9,11 x 10°31 kg TABLE 2: FORMULAE MOTION Vy = Vj + aAt v2 = v2 + 2aAx FORCE Fret =ma w=mg f= u,N fk = UN _ Gmym2 _M FS r g= Gy ELECTROSTATICS kQ,Q, _ kQ F= 2 E= z WwW F Vem E=-— q q Q Q n=— or n=— e co Copyright reserved

KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE MARKS: 100 DURATION: 2 hours Copyright reserved

Downloaded from Stanmorephysics.com QUESTION 1: 14 Byy (2) 1.2 Dvv (2) 1.3 Cyv (2) 1.4 Byv (2) 1.5 Cvv (2) 1.6 Dvv (2) 1.7 Cvv (2) [14] QUESTION 2 2.1 A closed vector diagram is a representation of several vectors and the resultant is zerov so that the last vector ends where the first vector starts. V (2) 2.2 OPTION 1 OPTION 2 Ww Marking Criteria: e Three forces correctly drawn and labelled in a right-angled triangle. ¥ e Direction of forces correctly shown. ¥ ¢ Any two angles correctly shown on triangle. V (3) 2.3.1 OPTION 1 OPTION 2 OPTION 3 T T sin50° sin 40° ea UE e = _O = tan 50 ce tan 40 iz Th i, 200 T, sin 50° sin 40° eo v= Vv eo v= 2 v= tan 50 Te tan 40° v 500 v 200 Ts Ta = 167,82 N ¥ To = 167,82 N ¥ To = 167,82 N ¥ (3) Copyright reserved

2. DOW EG ARE MARKING FRUNCE SR (BORGP FON 3) OPTION 1 OPTION 2 OPTION 3 - Tp Tp sin 50° = W cos 40° = WwW W= [1.2 Tae 200 200 i °o = ___ Oo - ae is 2 2 sin 50) W cos 40 Ww W= |200°+ 167,82°v W = 261,08 N W = 261,08 N W = 261,08 N W=mg W.= mg W=mg 261,08 = m(9,8) ¥ 261,08 = m(9,8) v 261,08 = m(9,8) “ m = 26,64 kgv m = 26,64 kgv m = 26,64 kgv QUESTION 3 3:1 A single vector that has the same effect as two or more vectors added together. vv 3.2.1 Fx = 12cos(30°) — 8cos(30°) ¥ {Note: sin(60°) could replace cos(30°) | *"in either substitution} =3,46N V (East) ' | 3.2.2 Fy = 12sin(30°) + 8sin(30°) ~ y d {Note: cos(60°) could replace sin(30°) ore; in either substitution} = 10 N v (North) 3.3 POSITIVE MARKING FROM Q3.2.1 AND Q3.2.2 Fax = Fscos(70°) OR = Fsx=Fssin(20°) ¥ = 0,34F3 East Fay = Fssin(70°) OR Fsx=Fscos(20°) ¥ = 0,94F3 South Fx = 0,34F3 + 3,46 V East Fy = 10 - 0,94F3 V North 0,34F3 + 3,46 10 - 0,94F3 0,34F3+3,46 | OR 10 - 0,94F Ola oF}- tan(25°) = ~7970,94F; tan(65°)= 9 3aF,+3.46 Fs= 1,55 Nv Copyright reserved (3) [11] (2) (2) (2)

Downloaded from Stanmorephysics.com 3.4 POSITIVE MARKING FROM QUESTION 3.3 F4 = 10 - 0,94Fs = 10 -0,94(1,55) v = 8,54 N V ata bearing of 180° ¥ QUESTION 4 441 Ticker tape is attached to the trolley, and threaded through the ticker timer. ” The trolley is gently pushed down the ramp without any mass pieces attached.v The angle of the incline is adjusted until the dots on the ticker-tape are equally spaced. ¥ 4.2.1 Acceleration’ 4.2.2 Mass (of the system) / angle of the incline (any one) ¥ 4.3 What is relationship between the resultant force acting on the trolley and the acceleration it produces?v ij 44 1 a = / =0,02s crenif Time for 5 dot interval = 0,02 x 5“ =0,1sv 4.5 OPTION 1 OPTION 2 OPTION 3 _ Ax _ Ax _ Ax v" it v" it "it 0,007 0,007 0,021 = Vv = = ——V/ re OO Or OA = 0,07 ms = 0,07 m:s* = 0,21 ms? 0,035 0,021 0,035 = v = —_—v = OF tama" 0 = 0,35 m:s* = 0,21 mst = 0,35 m-s* _ Av _ Av _ dv a it * at a" it - 0,35 - 0,07 | J 0,21 - 0,07 | - 0,35 - 0,21 if 0,2 0,1 0,1 =1,4ms?v =1,4ms?v =1,4m-s2v 46 Acceleration is directly proportional v'to the net force (OR a a Fnet) 4.7 It ensures that the mass of the system remains constant Vv Copyright reserved (3) [15] (3) (1) (1) (2) (3) (4) (1) (2) [17]

Downloaded from Stanmorephysics.com QUESTION 5 51 When a resultant/net force acts on an object, the object will accelerate in the direction of the force. This acceleration is directly proportional to the (net) forcevY and inversely proportional to the mass of the object. (2) NOTE: If the first sentence is not stated (for correct context), maximum a 5.2 (2) T/ Tension ¥ Note: subtract 1 mark for additional forces v W / Weight | 5.3.1 10kg block 20 kg block Fret = ma 7 By 5 | Fnet = ma THC Fe) +(eh}ema fF — W +(-T)=ma T-177-11,5V¥=10av (20 x 9,8)-T = 20a Vv T -— 188,5 = 10a (eqn 1) 196 —T = 20a (eqn 2) (eqn 1) + (eqn 2): 196 — 188,5 = 30a 7,5 =30a a =0,25m-s? v (5) 5.3.2 POSITIVE MARKING FROM QUESTION 5.3.1 Substitute in eqn (1) or eqn(2) OPTION 1 OPTION 2 T - 188,5 = 10a 196 —-T = 20a T — 188,5 = 10(0,25) v 196 — T = 20(0,25) v T=191NV T=191NV (2) 5.4 Decrease. ¥ The vertical component of the applied force will cause the normal force to decrease. ¥ The kinetic frictional force is directly proportional to the normal force + v Any one The kinetic frictional force decreases as the normal force decreases (when uk is constant) (3) [14] Copyright reserved

Downloaded from Stanmorephysics.com QUESTION 6 6.1 Every body in the universe attracts every other body with a gravitational force that is directly proportional to the product of their massesv and inversely proportional to the square of the distance between their centres. Vv 6.2 W = mg = 20000 x 9,8 ¥ (= 196 000 N ¥ 6.3 POSITIVE MARKING FROM QUESTION 6.2 OPTION 1 OPTION 2 OPTION 3 G GM =! ~ ae g= =v Snew= ined P r’ 6,67%10'''x5,98%10""x15000 v | 6,67x10'''x5,98x10"" = 2,45 ms? (6,4x10°+6,4%10°) v (6,4x10°+6,4%10°) = 36 517,27N = 12,4345 ms? im WwW =mg iy W=mg ~ | = 15000 x 2,4345¥ =| =15000 x 2,45v = 36 517,5N = 36 750 N j \ =} J 196 000 - 36517,27 9, ; tN ele ff % Loss in weight 796 000 «100 = 81,34% v 6.4 e The rocket is further away from the centre of the Earth OR above the Copyright reserved surface of the earth, therefore gravitational acceleration decreases. ¥ Since BOTH the gravitational acceleration and mass decreasev, there is a greater percentage reduction in weight. OR The gravitational force is directly proportional to the product of the masses and inversely proportional to the square of the distance between the centres. ¥ Since BOTH the mass of the rocket decreases and the distance between the rocket and centre of the earth increases, “there is a greater percentage reduction in weight. (2) (2) (5) (2) [11]

Downloaded from Stanmorephysics.com QUESTION 7 71 Fa, Ge 8x10° a ACCEPT : -8x 10° 7 4,6x 10° ~ 4,6 10° 7.2 7.3 7.4.1 7.4.2 = 5x 10" electrons ¥ The electrostatic force experienced per unit positive charge placed at the point. vv kQ Ep 9 -6 _ 9*10°x Bea 2 0,25 = 1,152 x 108 N-C" left kQ Bor _ 9x10°x 5 x10% 0,052 = 1,8 x 107 N-C" right —Vv v Enet = Er — Ep (Right +) OR Enet = Ep — Er (Left +) =1,8x 107 — 1,152 eo . = 1,152 x 10°-1,8 x 107 v = 1,6 85 x 10” N-C* right ¥ = -1,6 85 x 107 N-C* = 1,6 85 x 10’ N-C* right ¥ kQRQp 7 RP = 9 x10°x 5 x10x 8 x10 v 7 0,22 v =9N right Fret = Fre + Fsp 6,6 =9 + Fspv Fsp =-2,4N =2)4N lefty POSITIVE MARKING FROM 7.4.1 kQ,;Q Fp = — a4 a x10°x Q,x 8 x10° v 0,37 Magnitude of Qs= 3x10®C. ~ Qs=-3x10°CV TOTAL: Copyright reserved (3) (2) (5) (5) (3) [18] 100