Physical Sciences Grade 11 IEB June 2019 Memo

Downloaded from hlayiso.com JUNE EXAMINATION - 2019 MEMORANDUM: GRADE 11 PHYSICAL SCIENCE [Total marks = 130] SECTION A: MULTIPLE CHOICE 11 B 1.2 D 13 D 14 D « 15° OD (s) Question 2: 2.1.1 Rate of change of position or displacementY wv (2) 2.1.2 Distance = Area under graph = % bh v = 29,8 v (3) = (17, 5)W),Y Ss .ve3.4im sty Vew x: 2.1.3. Take motion north as positive — Os But * a (ty) uw = “se Mi 0-341 2 2 . a= Y= S55 36 v=-0,24ms* ie 0,24ms south” azojzums = 8) Soret, 2.1.4. When a net force, Fnet, is applied to an object of mass, m, it accelerates in the 22.1 direction of the net force. v The acceleration, a, is directly proportional to the net force and inversely proportional to the mass. ” (2) Take motion north as positive Fret = mav , = (20)(-0,24) Vv = - 4,87 Nie. 4,87 N south” 7 (4) or = yrt+2cas)s —~ Page 1 of 7

Downloaded from hlayiso.com 222. Se LL} 4 batk™ t- - UO) + 6 (4. 2)(.3)" — bose = Os = 3, Of an 1e 3,08 ‘q ~ (2) | | | | . = -€ 65+ aleay L- Ka Gelms* 12 Ablos ~ “P- | | | | | = AWA0 SSN le FOSSM up ms oy 4 ™\ e aa 3 22 S L(s) S . a aves “| Shape “ Page 2 of 7 ° (c)

Downloaded from hlayiso.com Question 3: 3.1 Fn Fu = normal force Fis™® = maximum static frictional force ¥ Fig F, = gravitational force (-1 if relative magnitudes are incorrect) Fy _ Fygl = 1382 V 3.20 Ue= FN (240)(9,8)c0s(36°)¥ = 0.73Y i(02,%1 3.3.1. Greater than ¥ 3.3.2 Fp, =U, FN ~ (H, is constant) oF oc Fry Inclined plane: Fu = mgcos® “Fu <img vv Horizontal surface: Fu=mg ‘. Fu is greater on the horizontal surface «Fy, is greater on the horizontal surface Question 4: Newton’s Laws of Motion (3) (4) 4.4.1 Fy = normal force “ fs Fx = kinetic frictional force Fy = gravitational force Fa Fr Fa = applied force ” Fu = Fy + Fay Fr Le Fax> Fad (-1 if relative magnitudes are incorrect: Page 3 of 7 (4) C23

Downloaded from hlayiso.com 4.1.2. Take motion to the left as positive Frat = ma ¥% a 7 or ley. 480 costs?) ¥— Fix = (200)(0,9) Be 3, BS Fe = 200 CA) oF = 203.35 N ¥ (6) 21 Reforma LEFT Rotem Rg. GoTess ~ See Oexae-T zO be a T- Ounaé 2O,4a —-—Ts O6n ~ 5,e¢ - © 73,42 Te S88 “Oba T= Oe 5,55 Oba = ha tB Ie a aiionse (s) 4.2.2 When object A exerts a force on object B, object B simultaneously exerts an oppositely directed force of equal magnitude on object A. vv (2) 4 4.2.3. Force of the 0,5 kg mass dost on the pan. @) Cis 3 Question 5: 5.1 Weight is the gravitational force that the earth exerts on an object (on or near its surface) while mass is the quantity of matter in a body. ¥ (2) 5.2 Graph — on answer sheet Heading y-axis title and unit x-axis title and unit scale (plotted points > 1/2 graph paper) “ plotted points (accurate and visible to within half a small square) v line of best fit ~ (6) Page 4 of 7

5.4 5.6 3,0 2,5 weight (N) nh ro) = an Downloaded from hlayiso.com a Graph to show weight vs mass 0 0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 mass (kg) . Gradient = AylAx ¥ = (values must be from LOBF on graph) ¥ Gradient = 3,76 N-kg* or 3,76 m-s7 (accept 3,56 — 3,96) Vv {3) w=mg v Therefore g = gradient g =3,76ms? “ (2) mars (1) _ 6M _, 6.7x107")M g= 3 v 40,58 = (4,19 10%)? v M=1,23 x 10% kg Vv (4) [18] Page 5 of 7

Downloaded from hlayiso.com Question 6 . ee We CRW 6.1.1 oe NeeRy =z 1S Chie =1,5x0,16” _ . = 0.24 mol = 0,24 mel (3) 6.1.2 n=m/iM¥ = 15/100V =0,15 moiv” HNO3:CaCOz = 2:1. HNOsis the limiting reagent” (4 62 n=vVinv = 3,36/22,4¥ / = 0,15 mol Cl2 produced” n(NaCh) reacted = 0,15 x 2 = 0,3 mol” ‘ m (NaCl) reacted =nxMv = 0,3'% 58,5¥ = 17,55 g¥ % purity = 17,55/20 x 100 = 87,8 %¥ (8) : [15] Question 7: 7.1. NHav 7.2. KBrv 7.30 Znv 74 Chav Ne 75 Nev 76 SiOav 6] Question 8: 8.1.1 An intramolecular bond occurs between atoms within moleculesv ¥ (2) 8.1.2 8.2.1 8.2.2 8.3 An intermolecular force is a weak force of attraction between molecules, ions or atoms of noble gasesv (2) London Forces” (1) Pure Covalent Bondy” (2) Although fodine and bromine both have London intermolecular forcesv’, lodine has more electrons Wand therefore seis up larger tempory dipoles’. More energy is needed v’to overcome the stronger Loncon forces in lodine. (4) Page 6 of 7

8.4 8.5.1 8.5.2 Downloaded from hlayiso.com A large amount of energy is needed to break the Many” Strong v electrostatic forces “in a crystal lattice. pop een leas nok (3) ion-dipole: o ( ) ammonia ~ Both ammonia and water have strong hydrogen bonds. The IMF's need to be of similar strength” for a substance to dissolve in another (3) {18] [Total marks = 130] Page 7 of 7