Physical Sciences Grade 11 2023 QP & Memo

Downloaded from hlayiso.com To MOPANI EAST DISTRICT NATIONAL SENIOR CERTIFICATE MARKS : 100 TIME : 2 Hours This question paper consists of 9 pages including this cover page. Page 1of 9

Downloaded from hlayiso.com Downloaded from Stanmorephysics.com INSTRUCTIONS AND INFORMATIO 10. Write your NAME in the appropriate space on the ANSWER BOOK. This question paper consists of 4 questions. Answer ALL the questions. Number the answers correctly according to the numbering system used in this question paper. 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. Show ALL formulae and substitutions in ALL calculations. Round off your FINAL numerical answers to TWO decimal places. Diagrams are NOT necessarily drawn to scale. Write neatly and legibly. Page 2 of 9

Downloaded from hlayiso.com Downloaded from Stanmorephysics.com QUESTION 1 Four options are provided as possible answers to the following questions. Each question has only one correct answer. Write down ONLY the letter (A-D) next to the question number. 1.1 1.2 1.3 1.4 1.5 UNWU>Y GVNN> QUOUN>Y when? leone Choose-the set of physical quantities which are vectors only. work, energy, weight, distance force, mass, time, power force, displacement, velocity, acceleration force,.distance, speed, acceleration (2) When resolving the vector into horizontal component, which one of the following is relevant? Ry = Rsiné@ for the resultant component Ry = Rcosé6 for the resultant component Rx = Rcosé@ for the resultant component Rx = Rsin@ for the resultant component (2) Which of the following forces is equal but in opposite direction to the normal? Fgt to the surface Fg||to the surface Applied force Frictional force (2) Frictional force is: inverse to normal inverse to applied force proportional to applied force proportional to the normal force (2) A negative charge of 1uC which is free to move is placed a distance 2r from a positive charge of 4 uC as shown below: Which one of the following statements regarding the -1uC charge, when it is at distance r is correct: The electrostatic force experienced by the -1yuC charge will: remain the same be halved be doubled increase four times (2) [10] Page 3 of 9

Downloaded from hlayiso.com oulNaded from Stanmorephysics.com 2.1 2.2 2.2.1 2.2.2 2.3 2.3.1 2.3.2 Fi = 8N 450\ Fe= 6N > oe 7 Three forces Fi, F2 and Fs act at a common point O as shown in the diagram above. The magnitude of vector Fi = 8 N, F2=6 Nand F3=5N. A Define the term resultant of a vector. (2) Use calculations to determine the following: The vertical and the horizontal components of vector F1. (4) The resultant force acting at point O. (9) The structure below shows a system that is in an equilibrium state. The strings with forces T1, T2 and Tz are joined at point O. The weight of the block suspended from T3 is 2 OOON. 70° 2 Ta oO T3 2 000N Explain the meaning of a closed vector diagram. (2) Draw a labelled closed vector diagram of forces showing all the forces (2) acting on the point. Indicate at least one angle in your diagram. [19] Page 4 of 9

Downloaded from hlayiso.com Waliade from Stanmorephysics.com crate of mass 2kg Is pu e right across a rough horizontal surface by a constant force F. This constant force F is applied at an angle of 20° to the horizontal as shown in the diagram below. F 20° 6 kg 3.1 State Newton’s first law of motion. (2) 3.2 Draw a free body diagram, showing al the forces acting on the crate. (4) 3.3 A constant frictional force of 3N acts between the surface and the crate. The coefficient of kinetic friction between the crate and the surface is 0.12. Calculate the magnitude of: 3.3.1 Normal force acting on the crate. (3) 3.3.2 force F. (4) 3.3.3 acceleration of the crate. (4) 3.4 A 5 kg block, resting on a rough horizontal table is connected by a light inextensible string passing over a frictionless pulley, is connected to another block of mass 2 kg hanging vertically as shown in the diagram below: A force of 60 N is applied to the 5 kg block at an angle of 10° with the horizontal, thus making the block to accelerate to the left. 60N if 2kg The coefficient of kinetic friction between the 5 kg block and the surface of the table is 0.5. Ignore the effects of air friction. 3.4.1 Draw the free body diagram, showing ALL the forces acting on the 5 kg (4) block. 3.4.2 State Newton’s Second Law of motion in words. (2) 3.4.3. Calculate the vertical component of the 60 N force. (3) 3.4.4 Calculate the horizontal component of the 60 N force. (3) 3.4.5 Calculate the normal force acting on the 2 kg block. (2) 3.4.6 Calculate the tension force connecting the two blocks. (5) 3.5 A massive rock from outer space is moving towards the Earth. 3.5.1 State Newton’s Law of Universal Gravitation in words. (2) 3.5.2 How does the magnitude of the gravitational force exerted by the Earth (2) on the rock changes as the distance between the rock and the Earth becomes smaller? Choose from INCREASES, DECREASES or REMANS THE SAME. Give a reason for your answer. [40] Page 5 of 9

Downloaded from hlayiso.com Q . A pouiMaded From Stanmorephysics.CQM a1. 0, are at a distance 4.1.1 4.1.2 4.1.3 4.1.4 4.2. 4.2.1 4.2.2 d apart and are suspended from an insulated rigid wooden bar as shown in the diagram below. The charges on the spheres are -4uC and +8uC respectively. Pp Q Sphere Q experiences an electrostatic force. Define electric field. In which direction will sphere Q move? Write down only TO THE LEFT or TO THE RIGHT. The spheres are now made to touch and then separated. Were electrons REMOVED FROM or TRANSFERRED TO P? Give a reason for your answer. Calculate the distance d between the two spheres if the force experienced by P is 0.8 N. Charges of +4uC and -3uC are placed at a distance of 0.2 m apart ona straight line as shown below. Point P is located 0.1m to the right of the - 3yC charge. CO 0.2m >( \e 0.1m ,p +4uC -3yuC Calculate the: net electric field at point P mass of a +6uC charge placed at P that will experience an acceleration of magnitude 5 x10* m.s?. Page 6 of 9 (2) (1) (2) (4) (5) (3)

Downloaded from hlayiso.com 4. D ownlogde small fF Om teh Sfannorgph Baal 694 carrying a charge of +7 4.3.1 4.3.2 4.3.3 4.3.4 x10° C, is suspended from a horizontal surface by a string of negligible mass. A second sphere B, carrying a charge of -8 x10° C, on an isolated stand, attracts sphere A, so that the string forms an angle of 20° with the vertical. The horizontal distance between the centres of the two is 3cm. Refer to the diagram below: ‘0° C) 3.cm -8x10°C State Coulomb’s law in words. (2) Draw a vector diagram of the force acting on sphere A and indicate at (4) least ONE angle. Calculate the magnitude of the electrostatic force that sphere B exerts (4) on sphere A. Calculate the magnitude of the tension force in the string. (4) [31] TOTAL= 100 MARKS Page 7 of 9

Downloaded from hlayiso.com Downloaded from StannrekeplsiésiGsin@es crave 11 PAPER 1 (PHYSICS) GEGEWENS VIR FISIESE WETENSKAPPE GRAAD 11 VRAESTEL 1 (FISIKA) TABLE 1: PHYSICAL CONSTANTS/TABEL 1: FISIESE KONSTANTES NAME/VAAM SYMBOL/SIMBOOL | _ VALUE/WAARDE Sree btraghonstante G 6,67 x 10°" Nm?kg? ee Re 6,38 x 10° m radi, op olekon e -1,6x 10°C Eeetron mass Me 9,11 x 10°" kg Massa van die Aarde M 5,98 x 10" kg FORCE/KRAG F..=ma w=mg p-Smim My = fem r N f, B= N Page 8 of 9

Downloaded from hlayiso.com Downdess| GO UND ANTSINeRIICOLWE YKLANKCEMLIG v=fa T= 7 nj sin®; =n, sin @, n=< ELECTROSTATICS/ELEKTROSTATIKA pO (k = 9,0 x 10°N-m?-C) Es = KO (k= 9,0x 10°N.m2-C%) | n= e ELECTROMAGNETISM/ELEKTROMAGNETISME e--na® At ®=BAcos@ ELECTRIC CIRCUITS/ELEKTRIESE STROOMBANE \-2 r-Y At | 44,444, R=,+ fh ++ Ron h fs _ aes WwW W=Vq p-— At We=VIAt P=VI W=PRAt P=PR 12 we V*At ye R p= R Page 9 of 9

Downloaded from hlayiso.com Downloaded from Stanmorephysics.com DEPARTMENT OF EDUCATION MOPANI EAST DISTRICT NATIONAL SENIOR CERTIFICATE GRADE 11 MARKS : 100 This memorandum consists of pages including this cover page.

Downloaded from hlayiso.com 2 Balitaded from Stanmorephysics.com Cw Add Dv Dw 1.2 1.3 14 1.5 QUESTION 2 2.1 together. /V Option 1 Fix = Ficos30° ¥ = 8cos30° =6.93N/ 2.2.1 Fiy = Fisin30° V = 8sin30° =40N/ Option 2 Fix = Fisin60° J = 8sin60° =6.93N/ Fiy = Ficos60° ¥ = 8cos60° =4.0N/ Horizontal vectors Fix = -6.93 N Fox = +6 N Fsx = Fscos60° or = 5cos60° J =2.5N 2.2.2 Frx = -6.93 +6-2.5 = -3.43N = 3.43 N to the left / Vertical forces Fiy = Fasin30° = 8sin30° / =4.0N Fay =ON Fay = Fssin60° or = 5sin60° J =4.33N Fry = 4-4.33 = -0.33 N = 0,33 N downwards J Frx A single vector having the same effect as two or more vectorsacting (2) (2) (2) (2) [10] (2) (4) (9) Fx = Fasin30° = 5sin30° J =2.5N Fay = Fscos30° = 5cos30° JV =4.33N

Downloaded from hlayiso.com Downloaded hysics.com Fr Fr = \/ (Frx)? + (Fry)? Jv =\/ (3.43)? + (0.33)? =3.45N J tan 6 = 0.33 JV 3.43 6 =5.50° So Fr = 3.45 N at the bearing of 264.50° J (270° — 5,50°) 2.3.1 A closed vector diagram is a set of vectors drawn on the cartesian plane using the tail-to-head method and that has a resultant with a magnitude of zero, Jv 2.3.2. th Ts , VJ [19] QUESTION 3 3.1. An object will continue in its atate of rest / uniform motion unless a resultant force acts on it. / / 3.2. Ny Nn ' Fy J ~ OR JF Fx v Fg / wi Fe/wi 3.3.1 Fk = ykN J 3 = (0.12)(N) J N=15N/ 3.3.2 Fnet = ma / N+Fy-W=0 15 + Fsin20°/ —(6)(9.8) / =0 F=128.06N/ 3.3.3 Fnet = ma / Fcos20° - Fy = ma 128.06 cos20° /- 3 = 6aJ/ (2) (4) (3) (4)

Downloaded from hlayiso.com Downlo®titte?om 'Stanmorephysics. com 3.4.1 3.4.2 3.4.3 3.4.4 3.4.5 3.4.6 3.5.1 Fa >TOR Fx« mT Fe/W Fg/W Marking guidelines W/ Weight/ Fg / T/ Tension / Fa/ Applied force / Fx J N/ Normal force / When force is applied on an object it produces acceleration in the (2) direction of the force and this acceleration is directly proportional to the applied force and inversely proportional to the mass of the object. / / Fx = Fcos10° J (3) = 60cos10° / =59.1N/ Fy = Fcos10° V (3) = 60sin10° V =104N/ Fy +N=WJ/ (2) 10.4 + N =5(9.8) N=38.6N/ Fnet = ma J (5) For the 5kg block Fx-T=ma 59.1-T=5a/ T = 59.1 58 .oeeecceeeeees (1) For the 2kg block Fnet = ma T-W=ma T — (2)(9,8) = 2a/ T= 19.6 + 28 oc ceeeeeeeee (2) Equating (1) and (2) 59.1-5a=19.6+2a a=5.64m,s?/ substituting a = 5.64 m,s° in either (1) or (2) gives: T=30.88N/ Each body in the universe attracts every other body with a force that is (2) directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres. //

Downloaded from hlayiso.com 3 Pownloagert From Stanmorephysics.com r2 QUESTION 4 4.1.1. Electric field is a region in space where wherein an electric charge experiences a force. v / 4.1.2 To the left / 4.1.3 Removed / The charge on P is positive / 414 F=kOiQ2 J d2 0,8/ =(9 x109) x( 2 x10°) x (2 x10°9) / d2 d=0,21m/ 4.2.1 E=kO J 2 r E4= (9 x109) x (4.x10°) JV (0.3) = 4x10° N,C* to the left Es = (9 x109) x (3.x10°°) / (0.1)2 = 2.7x10® N,C* to the left Ep = Ea + E3 = (0.4 — 2.9) x10 N.C / = 2,3 x 10° N.C to the left / 4.2.2 F=maJ QE =ma m = (6 x10°)(2,3 x10°)_J/ (5x10?) = 2.76 x10? kg V (2) (2) (1) (2) (4) (5) 4.3.1 The magnitude of the electrostatic force exerted by two charges oneach (2) other is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. > Fg T Marking guidelines 4.3.2 Fe W / Weight/ Fg / T/ Tension J FE/ Electrostatic force / Angle indicated ¥ 4.3.3 F = kQ1Q2V/ r2 (4) (4)

Downloaded from hlayiso.com _ -9 -9 . Downloade ; ephysics.com = 5,60 x107N J 4.3.4 Fg = mg (4) = (2,0 x10%)(9.8) ¥ = 1.96 x10° NV T? = Fg2+ Fe? = (1.96 x10°)2 + (5,6 x10%)2/ = 2.04 x10? NJ [31] TOTAL = 100 MARKS