Physical Sciences Grade 11 KZN March Test 2025 Paper & Memo

Testpapers.co.za J _KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE SSSSSSESESSESS, ES TASK Pe Pee eee ear Seseeseenest Seeeeeeseens . = gSeeeeeeen2e2eees TIME: 2 hours MARKS: 100 This question paper consists of 12 pages, TWO data sheets and a special answer sheet for Question 3. Copyright reserved Please turn over

t Testpapers.co.za March 2025 Assessment Task nou INSTRUCTIONS AND INFORMATION TO CANDIDATES 1. Write your name on the ANSWER BOOK. 2. This question paper consists of EIGHT questions. Answer ALL the questions in the ANSWER BOOK. 3. Start EACH question on a NEW page in the ANSWER BOOK. 4. Number the answers correctly according to the numbering system used in this question paper. 5. Leave ONE line between two subsections; for example between QUESTION 2.1 and QUESTION 2:2. 6. You may use a non-programmable calculator. 7. You may use appropriate _ instruments. wnt 8. You are advised to use the attached DATA SHEET. 9. Show ALL formulae and substitutions in ALL calculations. 10. Round off your final numerical answers to a minimum of TWO decimal places. 11. Give brief motivations, discussions, et cetera where required. Copyright Reserved Please Tum Over

Testpapers.co.za March 2025 Assessment Task NSU QUESTION 1 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 — 1.7) in the ANSWER BOOK, for example 1.8E 1.1 The statements below refer to vector and scalar quantities. (i) A vector has magnitude and direction while a scalar has magnitude only. (ii) A scalar quantity can always be added to a vector quantity. (ili) Force is an example of a vector quantity, while the change in position is an example of a scalar quantity. Which of the above statements is/are TRUE? A (i) only B (i) and (ii) only C (i) and (iii) only y D (i), (ii) and (iii) ~ : (2) 1.2 Two forces Fa and Fs act ata point O as shown. > 0 Fe Which ONE of the following forces correctly represents the direction of the resultant of these two forces? wars a3 2 eS Copyright Reserved (2) Please Turn Over

et Testpapers.co.za March 2025 Assessment Task 1.3 A force of magnitude F acts on a block at an angle @ to the horizontal, as shown in the diagram below. The block, initially at rest on a flat, frictionless surface, accelerates uniformly. Angle @ is now halved. How will the NORMAL FORCE and the NET HORIZONTAL FORCE acting on the block change? HORIZONTAL FORCE A Increases B Increases Cc Decreases D Decreases 1.4 Agirl of mass m and a boy c 2m face each other while standing at rest on roller- skates, as shown in the diagram below. The girl pushes the boy to the right with a force F. Ignore all frictional effects. If the girl accelerates to the left at x m-s, what will be the magnitude of the acceleration of the boy immediately after they separate? A xXx B %x C “wx D 2x (2) Copyright Reserved Please Turn Over

Testpapers.co.za March 2025 Assessment Task NSG 1.5 Aman of mass 95 kg stands on a bathroom scale in a stationary lift. The scale registers a reading of 931 N. Which ONE of the followin di the scale is possible if the lift now moves upwards at constant accel ' Less than 95 N. — A B Less than 931N.. C Equal to 931 N. D Greater than 931 N. (2) Copyright Reserved Please Turn Over

P Testpapers.co.za March 2025 Assessment Task NSC 1.6 Two identical metallic spheres, each carrying a charge Q, are placed so that their centres are r metres apart. Which ONE of the following changes, made simultaneously, will DOUBLE the electrostatic force that the charged spheres exert on each other? _~ | Distance between spheres | Magnitude of the charges A r Double the charge on both 2 spheres ua Reduce the charge on one of the B 2 Q spheres to z 4 2 Reduce the charge on one of the v2 | Spheres to ~ D juble the charges on both spheres : (2) 1.7 Asmall test charge q is pI from a point charge, Q. Which ONE of the following graphs correctly sho jonship between the electrostatic force, F, experienced by q and the electric fi , generated by Q when r is constant? A ft B F (N) F (N) > = E (N-C*) E (N-C*) D F (N) > “C1 (2) E (N-C*) [14] Copyright Reserved Please Turn Over

Pl Westpopers.co-za March 2025 Assessment Task NSC QUESTION 2 Three forces, P, Q and R, of magnitudes 400 N, 250 N and 300 N respectively, act at a point Oin the directions shown in the diagram below. The forces are NOT drawn to scale. t P=400N Q=250N SS ~ R=300N = < 2.1 Define the term resultant vector. 2.2 Calculate the vector sum of the: 2.2.1 HORIZONTAL components of the forces acting at O 2.2.2 VERTICAL components of the forces acting at O 2.3 Using the answers to QUESTIONS 2.2.1 and 2.2.2, draw a labelled vector diagram. Show the resultant force on this diagram. Your diagram need NOT be drawn to scale. 2.4 Hence, calculate the magnitude and direction of the resultant force acting at point O. Copyright Reserved (2) (2) (2) (3) (5) [14] Please Turn Over

P Testpapers.co.za March 2025 Assessment Task NOU QUESTION 3 Grade 11 learners conduct an experiment to investigate three co-planar forces at equilibrium, acting at a point. Strings P, Q and R are joined by means of a knot at point O, and masses are attached at the end of each string, as shown in the diagram below. 3.1. What is the magnitude of the net force acting on the knot (point O)? (1) 3.2 Calculate the magnitude of the force in string P. (2) 3.3 The data obtained from their experiment was plotted on the attached graph sheet. 3.3.1 You must answer this question using the special answer sheet at the end of this question paper. Determine graphically, using the parallelogram law, the magnitude of the (5) resultant of the forces acting in strings P and Q. Use the scale: 10 mm represents 0,1 N. 3.3.2 What is the magnitude of X? (3) [11] Copyright Reserved Please Turn Over

F Testpapers.co.za March 2025 Assessment Task now QUESTION 4 A small caravan of mass 750 kg is being towed by a car of mass 1500 kg by means of a tow-bar, as shown in the diagram below. Caravan - 750 kg Car - 1500 kg When towing the caravan on a straight horizontal road, the car accelerates uniformly from rest, reaching a speed of 30 ms in 25 s. A constant frictional force of 200 N acts on the caravan, while the tension in the towbar i is T. Ignore the mass of the tow-bar and the rotational effects of the wheels. 4.1 Define the term kinetic frictional force. 4.2 Drawa labelled free body diagram showing all the forces acting on the caravan. 4.3 Calculate the magnitude of T, the tension in the tow-bar immediately after the car starts moving from rest. 4.4 After 25 s, the car travels ata CONSTANT VELOCITY. Write down the magnitude of each of the following while the car travels at constant velocity: 4.4.1 The net force acting on the caravan 4.4.2 The frictional force acting on the caravan 4.4.3 The tension in the tow-bar (2) (4) (5) (1) (1) (1) [14] Copyright Reserved Please Turn Over

Pl Testpapers.co.za March 2025 Assessment Task QUESTION 5 Block A, initially at rest on a rough horizontal surface, is connected to block B by a light inextensible string that passes over a frictionless pulley. The mass of block A is X kg, while that of block B is 12 kg. When a force F of magnitude 125 N, acting parallel to the surface, is applied as shown in the diagram below, block A moves to THE LEFT at CONSTANT VELOCITY. F=125N T 5.1. State Newtons Second law of motion in words. (2) 5.2 If the coefficient of kinetic friction (ux) between block A and the surface is 0,25, calculate: 5.2.1 The tension, T, in the string (3) 5.2.2 The magnitude of the kinetic frictional force acting on block A (3) 5.2.3 X, the mass of block A (3) 5.3. The magnitude of force F is now doubled. How will the acceleration of the blocks change? Choose from INCREASES, DECREASES or REMAINS THE SAME. Give a reason for the answer. (2) [13] Copyright Reserved Please Turn Over

ld Testpapers.co.za March 2025 Assessment Task NOU QUESTION 6 A block of mass 35 kg is prevented from sliding down a rough inclined plane by means of a light, inextensible cable that is parallel to the incline and attached to a vertical pole, as shown in the diagram below. i The incline makes an angle of 20° with the horizontal. 6.1 Define the term static friction in words. (2) 6.2. Calculate the magnitude of the component of the weight of the block that acts parallel to the incline. (2) The cable breaks, and the block begins to slide down the incline. 6.3 If the coefficient of static friction is 0,3, show that the magnitude of the acceleration of the block CANNOT be less than 0,589 m:s*. (5) [9] QUESTION 7 The mass of the sun is 1.99 x 10°? kilograms and its distance from the centre of Earth is 150 million kilometers. 7.1 State Newton's law of universal gravitation in words. (2) 7.2. Calculate the gravitational force that the Sun exerts on Earth. (5) [7] Copyright Reserved Please Turn Over

F Testpapers.co.za March 2025 Assessment Task NOU QUESTION 8 Two small charged conducting spheres, P and Q, on insulated stands, carry charges of - 4 nC and +12 nC respectively. The centres of the spheres are 25 mm apart, as indicated in the diagram below. X is the midpoint of the line joining the centres of the spheres. 8.1 Define electric field at a point. | (2) - 8.2 Calculate the magnitude of the: 8.2.1 Resultant electric field at point X (5) 8.2.2 Force that an electron will experience at point X (3) The two spheres are brought into contact with each other, and then separated and moved back to their original positions. 8.3 Determine the new charge on each sphere. (2) 8.4 Calculate the amount of charge transferred upon contact. (2) 8.5 Calculate the electrostatic force that the two spheres exert on each other after they made contact with each other and were separated. (4) [18] TOTAL MARKS: [100] Copyright Reserved Please Turn Over

PI Testpapers.co.za March 2025 Assessment Task DATA SHEET PHYSICS: TABLE 1: PHYSICAL CONSTANTS NAME SYMBOL VALUE Acceleration due to gravity g 9,8 m-s? Universal gravitational constant G 6,67 x 10" Nem?-kg? Radius of the Earth R 64x10 °m Mass of the Earth - at. 5,98 x 10% kg Charge on an electron 4 -1,6 x 1049C TABLE 2: FORMULAE MOTION Ve = Vj tat Ax =v, At+Jadt? or Ay =v,At+taat? V7 =v? +2aAx or v,2 =v, +2aAy ax -(“ ola or ay -(“ 2) FORCE Fret =Ma p=mv f" = HN fi =H,N Fret = Ap Ap= mv, - mv, Wemg M M Fou or F-G™ye g= Ge or g= Gr Copyright Reserved Please Turn Over

Testpapers.co.za March 2025 Assessment Task oiwYy ELECTROSTATICS n=— or n=— e a Copyright Reserved Please Turn Over

PF Testpapers.co.za March 2025 Assessment Task NSC DETACH THIS PAGE AND HAND IN WITH YOUR ANSWER BOOKLET. BR ERIK IK IR IK 9K 9K 9K 9K 9K CIC IC IC ICC IC IC IC IK IKK IK IK IK IK IKK IKK KIC I NAME OF LEARNER : GRADE: 11 Question 3.3.1 Copyright Reserved

Testpapers.co.za KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE GRADE 11 NB: This memorandum consists of 8 pages. Copyright reserved Please turn over

Testpapers.co.za March 2025 Common Test QUESTION ONE 11. Avy 12 Avv 13° By¥v 14 Byv 15 Dvv 16. -BYvY 17 Byv 7x2= [14] QUESTION TWO 5 m 2.1 Itis a single vector epresent a number of vectors/ forces 2.2.1 2.2.2 acting on an object in e nd direction. vv (2 or 0) (2) OPTION 1: © Rx = (250¢0830°) +(400 cos70°) + (-300cos50°)v’= +160,48NV OPTION 2 : © Rx = (250 sin60°) +(400 sin20°) + (-300sin40°) v= 160 48 NV (2) OPTION 1: © Ry = 400 sin70° +250 sin30° + (- 300 sin50°)V= +271,06N v OPTION 2; © Ry = 400 cos20° +250 cos 60° + (- 300 cos 40°)V=271,06 NV (2) Copyright reserved Please turn over

Testpapers.co.za March 2025 Common Test 2.3 Positive marking from v = Ry=271,06 N = RyF271,06 N ve, i v ERK = 160,48 Ne DRx = 160,48 N OR vy = Rx = 160,48 NY > Ry=2’ Markir ict diagram Criteria ark allocation = Rx drawn correctly 1 mark (3) > Ry drawn correctly 1 mark Resultant force drawn correctly | 1 mark 2.4 POSITIVE MARKING FROM Q2.2 Rner? = Rx? + Ry? = (160,48)? + (271,06)2v Rnet = 315,00 NY 6 = 59,37° North of EastY OR at a bearing of 30,63° OR 30,63° East of North (5) [14] Copyright reserved Please turn over

F Testpapers.co.za March 2025 Common Test QUESTION 3 3.1 ON/zero ¥ (1) 3.2 Fe =mg = (0,056 x 9,8) = 0,55 NV (2) 3. 3.1 Ri 0,62 N / 60,2 mm MARKING RUBRIC : FORCE DIAGRAM Criteria Mark allocation Fe and Fa drawn correctly . 2x 1 =2 marks Parallelogram completed correctly 1 mark Resultant force drawn correctly 1 mark Final answer: Resultant force 1 mark (correct conversion using scale) (5) Accept (0,58N — 0,64 N) Copyright reserved Please turn over

Testpapers.co.za POSITIVE MARKING FROM QUESTION 3.3.1 3.3.2 Wx=mg 0,62Y =X x9.8V X = 0,063 kg or 63gY QUESTION 4 4.1 It is the force that opposes the motion of a moving object V relative to a surface. ¥ NOTE: Only award the second mark if the context is correct. 4.2 7 N/Normal force/Frnv Kinetic friction fk ‘% @ TTensio” W/Weight v 43 we=vtadt % 30 = 0 +a(25)¥ a=1,2m-s2 Fret = m T + (-fk) =m T = 200 = 750(1,2) ¥ T=1100NY ¥ (any one) 4.4.1 ON /zeroY 4.4.2 200 NY 4.4.3 200 NY Copyright reserved March 2025 Common Test (3) [11] (2) (4) (5) (1) (1) (1) [14] Please turn over

Testpapers.co.za March 2025 Common Test QUESTION 5 54 When a net/resultant force acts on an object, the object accelerates in the direction of the resultant force. This acceleration is directly proportional to the force V and inversely proportional to the mass of the object.” (2) 5.2.1 Fnet =ma v (any one) T + (-We) = ma T = We T-(12 x 9,8)=12(0) OR T = 12x9.8v T = 117,60 NY (3) 5.2.2 POSITIVE MARKING FROM QUESTION 5.2.1 Fret = ma F + (-fx) +(-T) = 0 F = fk+T * 125 + (- fx) +(-117,60) fk = 7,40 NY 5.2.3 POSITIVE MARKING FROM QUESTION 5.2.2 fk = UkNY 7,40 = (0,25)(9,8X) ¥ X = 3,02 kg¥ (3) 53 IncreasesY , acceleration increases since the net/resultant force increases. Accept: a « Fnet¥ (2) = fx + 117,60 (3) [13] Copyright reserved Please turn over

Testpapers.co.za March 2025 Common Test QUESTION 6 6.1 It is the force that opposes the tendency of motion of a stationary object relative to a surface. ¥ NOTE: Only award the second mark if the context is correct. (2) 6.2 Wi=mg sin@ = (35)(9,8)(sin 20°) ¥ = 117,31 NV (2) 6.3 Pe UsN ¥ = 0,3 x (35 x 9,8 x cos 20°) ¥ = 96,69 N fm > fv i Wy fk = Wr- Fret ¥ Sie ’ fi > Wi - Fret “ql } 96,69 > 117,31 -35av | a > 0,589 m-s? 4 (5) QUESTION 7 7.1 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. V NOTE: Marks must only be awarded if the definition is in context with the gravitational law) (2) LZ = Smima, F= 2 v = (6,67 x 10°1')(1,99 x 10°°)(5,98 x 1024) v (150 x 109v)2v =3,53x102N VW NOTE: If the conversion is incorrect, max 3). ; Award a mark for squaring the incorrectly converted mark. (5) [7] Copyright reserved Please turn over

Testpapers.co.za March 2025 Common Test QUESTION 8 8.1 The force experienced by a unit positive charge placed at the point. VW (2) 8.2.1 kQ kQ a Ea= —-9x109x 4x10 5 3 ~T(0,0125)2 _ — 9%10°%12%10" = 230 400 N-C* (0,0125)2 (2,30 x 105 N-C') = 691 200 N-C* (6,91 x 10° N:C") Ener = 230 400 ¥ 691 200 = 9,22 x 10° N-C*1/ 921 600 N-C-1v So NOTE: If an incorr € if all other substitutic , award the second and third marks for the final answer if it is computed correctly. This also 8.5 (5) 8.2.2 POSITIVE MARKING F=Eqv = (9,22 X105)(1 = 1,47x1073N (3) 8.3 Q= Qa Q _-44+12y ~ 2 =4nC v (2) x=-8NC Yow v=enc (2) 8.5 JE M. 8.3 ACCEPT: 2,30 x 10“ N repulsion (4) [18] TOTAL MARKS: 100 Copyright reserved Please turn over