Physical Sciences Grade 11 Gauteng March Test 2025 QP & Memo

Testpapers.co.za Ekurhuleni North D6 DISTRICT COMMON TEST 2025 GRADE 11 PHYSICAL SCIENCES March Test MARKS: 100 TIME: 2 hours This question paper consists of 12 pages including 2 data sheets

Testpapers.co.za Downloaden)sezosrichtartnacein lnsaiesTéaiimarch 2025 INSTRUCTIONS AND INFORMATION 1. Write your NAME at the TOP of every ANSWER BOOK. 2. Answer ALL the questions in the ANSWER BOOK. ~~ 3. You may use a no mmable' calculator. tical instruments. 5. YOU ARE ADVISED 4E ATTACHED DATA SHEETS. 6. Number the answe' re according to the numbering system used in this question paper. Bs Give brief motivations, discussions, et cetera where required. 8. Round off your final numerical answers to a minimum of TWO decimal places. 9. Write neatly and legibly

Testpapers.co.za . ich 2025 Question 1 — Multiple choice 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 (1.1-1.5) in the ANSWER BOOK 44 Which one of the following statements is true: (i) A vector is a gust gnitude and direction. (ii) Displaceme ent | id direction of the straight line drawn from the starting of the motion. (iii) Displacement and e are vectors. A (i) and (ii) B (ii) and (iii) C (i) and (iii) D (i), (ii) and (iii) 4.2 Four forces act on a body as shown in the diagram: a 3N 4N 4N 5N v What is the magnitude and direction of the resultant force? 16N south 2N south 8N east 8N south 0a DPY

Testpapers.co.za ; irch 2025 4.3..Three objects X, Y and Z hang from strings knotted together at O, as shown in the diagram. Fy Fe Fy Which vector diagram best represents the relationship of the forces acting on the kni - Fs Fa. 5 F, 1.4. The magnitude of a force A is 60N at an angle of 50° to the vertical as shown in the diagram. Ss so Wa The vertical component of force A is... 45,95N 38,57N 21,43N 14,03N 00OWOD>

Testpapers.co.za ; irch 2025 Which of the following statements correctly explains why a passenger in a car who is not restrained by a seat belt, continues to move forward when the brakes are applied suddenly? A The braking force applied to the car exerts an equal and opposite force on the passenger. B_ The braking force (called inertia) acts on the passenger. C Aresultant forward force acts on the passenger. DA zero resultant force acts on the passenger. 0,7kg What is the magnitude of the force in string P? 15N 42N 27N 35N 0QWD> According to Newton’s Third law, when a rocket is launched vertically upwards, the “action force” is represented by... the gravitational force of the earth on the rocket downwards. the force of the rocket on the exploding gases, downwards. the force of the exploding gas on the rocket downwards. The gravitational force of the rocket on the earth upwards. 0OQODPY

1.8 1.9 1.10 Testpapers.co.za irch 2025 A 2,5 kg object weighs 24,5 N on Earth. What would be the object’s mass (in Kg) on planet Mars if the acceleration due to gravity on Mars is 3,71 m-s. 9,28 7,42 2,5 6,6 2 00ND > The distance bet by the spher« n without chang new electrosta s, Qi and Qzis r. The force exerted Both charges are now doubled een them. The magnitude of the e: 4F Ya F 2F hF 2 0aOoWP,Y The figure below shows the electric field around a point charge q, placed at a distance of 0,3m away from another charge Q. The electric field strength measured is 800N.C™. If the distance of the point Q, the electric field strength v away from 1600N.C* 200N.C* 400N.C*1 800N.C* 2 [10 x 2 = 20] 0OWDD>

Testpapers.co.za a irch 2025 Question 2 The figure below represents a weight W (50N), hanging from a cable, attached to a horizontal beam. A force (F) is applied at 90° to the weight. The angle between the cable and the beam is 25°. Beam 25° 2.1 Using the triangle draw a fully labelled vector diagram (not to scale). agimorep 2.2 Calculate the magnitude of the applied force F. 3 4 2.3 Calculate the tension force of the cable. 3 [10] Question 3 A crate is connected to two ropes as shown in the diagram. Rope A exerts a force of 100N at an angle of 20° to the horizontal. Rope B exerts a force of 120N at an angle of 35° to the horizontal. 3.1. Define the term vector component. 3.2 Calculate the x component for rope A and rope B. FB DN 3.3. Calculate the y component for rope A and rope B. 3.4. Calculate the magnitude of the resultant force of Rope A and Rope B. 5 [15]

Testpapers.co.za . ich 2025 Question 4 44 4.2 The graph was obtained in an investigation of the relationship between force and acceleration, on a surface which is friction-compensated. 1,0 a(ms”*) 4.1.1 Whatis the accelerat V en a force of 2,5 N is applied? 1 4.1.2 What is the relationship between the two quantities plotted? 1 4.1.3 What force must be applied to obtain an acceleration of 1 0,5ms?? 4.1.4 Howwill the graph differ if the same surface was used, but 2 friction is not compensated for? 4.1.5 State Newton's second law of motion in words. 2 Two masses of 1kg and 4kg are attached to a light string AB that passes over a pulley as shown. The 1kg mass rests on a frictionless table. A constant horizontal force P is applied to the 1kg mass to accelerate in the direction of force P. A P kdl > B ) If the tension in string AB is 48N, calculate: 4.2.1. The acceleration of the 4kg mass. 4 4.2.2 The magnitude of force P. 4 {15}

Testpapers.co.za ; irch 2025 Question 5 A 5kg block is being pulled up a rough incline plane of angle 40° to the horizontal by a constant force F. the block moves up the plane at a constant speed. B 5.1 State Newton’s { 2 a 4 5.2 Draw a labelled free m for the 5 kg mass. 4 5.3 The kinetic frictio ce | he block and the inclined surface is 19,78N. Calculate the magnitude of force F. 5 [11] Question 6 Two identical spheres with unequal masses m1 and ma, which are released simultaneously from rest, fall freely from a height h above the surface of the earth. The two spheres reach the ground at the same time. 6.1 State Newton’s Law of Gravitation in words. 2 6.2 Use formulae to show that the gravitational acceleration of any object near the earth’s surface irrespective of its mass, can be given by g=GM r2 Where M = mass of the earth 3 r = radius of the earth 6.3. An object undergoing free fall on the planet Mars, experiences a gravitational acceleration of magnitude 3,8 m.s*. The mass of Mars is 6,4 x 10*°kg. Calculate the radius of the planet Mars in metres. 4 [9]

Testpapers.co.za ich 2025 Question 7 7.1 Charges A, B and C are placed in a straight line, as shown in the diagram below. © ® © Qe 6mm Qa 12mm Qc =12nC +8nC -10nC 2 2 4 4 7.2 Apoint charge Q has an electric field surrounding it. 7.2.1. Explain what is meant by the term electric field strength at a 2 point. 7.2.2 Calculate the magnitude of the electric field strength 0,1m from Q if the charge of Q is -3 x 10°C. 3 7.2.3 If asmall point charge q (+6,4 x 10°'9 C) is placed in the field of Q, calculate the force exerted on q by Q. 3 [20] END OF PAPER TOTAL = 100 Marks 10

Testpapers.co.za irch 2025 DATA FOR PHYSICAL SCIENCES GRADE 12 PAPER 1 (PHYSICS) GEGEWENS VIR FISIESE WETENSKAPPE GRAAD 12 VRAESTEL 1 (FISIKA) TABLE 1: PHYSICAL CONSTANTS/TABEL 1: FISIESE KONSTANTES NAME/NAAM SYMBOL/SIMBOOL VALUE/WAARDE 11

Testpapers.co.za reh 2025 TABLE 2: FORMULAE/TABEL 2: FORMULES MOTION/BEWEGING Vpsvyjraat Ax =v,At+4adt? orlof Ay = v,At+4adt? vy =v, +2adx orlof v,’ =v,’ + 2aay ax= (“Sa or/of Ay -(4 —. Ja FORCE/KRAG Fret = ma ELECTROSTATICS/ELE Aimn-)| 12

Testpapers.co.za Ekurhuleni North D6 DISTRICT COMMON TEST 2025 GRADE 11 PHYSICAL SCIENCES March Test Marking Memorandum P.T.O.

Testpapers.co.za irch 2025 Question 1 — Multiple choice ana” ; i a dale ; 4.3—D vv ; 14 ByYv 2 15 Dvv ; 16 Cvv 2 17) AVY : 18 Cvv 19 Avv . 110 BYY . [10x 2=20] Question 2 a4 ‘ WwW T v v F 2.2 Tan 25°v = 50v oF Tan Gah Be E 50 a F = 107,23NV F = 107,23NY 23 T2=W2 + Fv or Sin 25°v = 50 v 3 = (50)? + (107,23)? v T T = 118,31N v T= 118,31N v [10] P.T.O.

Testpapers.co.za ich 2025 Question 3 3.1 The components of a vector are vectors that together have the same 2 effect as the original vector. vv 3.2. RopeA Rope B 4 Cos 20° = X Cos 35° = X 100 v 120 v X=93,97N ¥ X = 98,30N v 3.3. RopeA y . 4 sin 20° =X | =X 100 , 120 v X =34,20N = 68,83N v 3.4. X component =- 8,30 = 4,33N v Y component = -34 =34,63N v F: Y P 2= x2 + Y2 aay F 2 = (4,33)? + (34,63)? v X 5 4 rp 34,89N v [15] Question 4 4.1.1 1ms?v 1 4.1.2 a Fnetv 1 4.1.3 1,25N ¥ 1 4.1.4 It will not pass through the origin. vv 2 4.1.5 An object will accelerate in the direction of a net force. The 2 acceleration is directly proportional the net force and inversely proportional to the mass. vv 4.2.1 T-—W=Fnet (ma) ¥ 48 — (4)(9,8) v =4av a=2,2m.s? v 4 P.T.O.

Testpapers.co.za . ich 2025 4.2.2 Fnet=mav P-T =ma P—48v = (1)(2,2)v 4 P=50,2NY [15] Question 5 5.1 A body will remain in its state of rest or continue in its state of uniform 2 motion, unless a cted upomby an unbalance force vv 5.2 ms 5.3 Fnet =ma ¥ Fnet = 0 F -fk - Fgll=ma F —19,78 v - (5)(9,8)sin 40° v = Ov F- 19,78 -31,50=0 5 F = 51,28N v (11] Question 6 6.1 Every particle in the universe will attract every other particle with a force 2 that is directly proportional to the product of their masses and is inversely proportional to the square of the distance between them. v ¥ 6.2 W=mg...1 Fg=GmM ....2 v 3 r (1)=(2) mg=GinM vv g=GM re (ie 63 g=GM v 4 r 3,8v = (6,67 x 10°") (6,4 x 1029) v r r= 3351668,87m v [9] P.T.O.

Testpapers.co.za irch 2025 Question 7 7.1.1, The force between two charges at rest is directly proportional to the prroduct of the charges and inversely proportional to the square of the distance between them. v v 2 2 4 7.14 F=KQiQ2 =(9 x 10°)(8 x 10°)(10 x 10°) v = 0,005N v re (0,012)? v Fnet = + 0,024-—0,005 =0,019N left v 4 7.2.1 Force exerted per unit positive charge. vv 2 7.2.2 E=KQv = (9 x 10°)(3 x10%) v = 2700N.Ctv re (0,1)? 3 7.2.3 F=Eqv= (2700)(6,4 x 107%) v= 1,73 x 10°SNY = THE END TOTAL = 100 Marks P.T.O.