Physical Sciences Grade 11 P1 March Test 2018 Paper & Memo

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Downloaded from hlayiso.com Physical Sciences /P1 2 March 2018 Common Test NSC INSTRUCTIONS AND INFORMATION TO CANDIDATES a Write your name on the ANSWER BOOK. This question paper consists of FIVE 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 Leave ONE line between two subsections, 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. Show ALL formulae and substitutions in ALL calculations. Round off your final numerical answers to a minimum of TWO decimal places. Give brief motivations, discussions, et cetera where required. Copyright Reserved Please Turn Over

Physical Sciences /P 1 March 2018 Common Test Downloaded from hlayiso.com NSC QUESTION 1 : MULTIPLE CHOICE QUESTIONS 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 — 1.4) in the ANSWER BOOK, for example 1.5 D. 1.1. Two forces A and B act at the same point O as shown in the sketch below. When the angle © between the forces changes, the resultant force obtained varies between 12 N and 6 N. What is the possible magnitude of the larger applied force? A B Cc D 3N 18.N 6N ON (2) 1.2 A book of mass 80 g is at rest on a horizontal table. Ground The Reaction force to the book’s weight is ......... A The force exerted by the book on the table. B The force exerted by the table on the book. Cc The force of the Earth on the book D The force of the book on the Earth. (2) Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences /P1 4 March 2018 Common Test NSC 1.3. The velocity-time graph below represents the motion of a body along a frictionless horizontal surface. Which one of the statements below is INCORRECT for this motion? A The acceleration is constant, and is opposite to the direction of motion. B The acceleration is constant, and in the same direction to the motion. Cc A constant net force acts on the body but in the opposite direction to the motion. D A constant applied force acts on the body in the opposite direction to the motion. (2) 1.4 Two objects A and B are released from the same height(h) at the same time. The weight of A is twice that of B. If we ignore the effects of air resistance, we would notice that: Object A strikes the ground first. Object B strikes the ground first. They both strike the ground at the same time. They both strike the ground at the same time, but with different velocities. 0 OD > (2) [8] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences /P1 5 March 2018 Common Test NSC QUESTION TWO Three forces Fi, F2 and Fs act at a point, as shown on the Cartesian plane in the diagram below Yo 10 5 Fy Fs ¢ —+ > -10 5 10s Vv Scale: 1 square unit represents 1 N 2.1. Define the “resultant vector” (2) 2.2. Write down the X-component and Y-component for force Fo. (2) 2.3. Calculate the NET horizontal component for the forces F;, F2 and Fs (1) 2.4 Calculate the NET vertical component for the forces F;, Fz and Fs (1) 2.5 Calculate the magnitude and direction of the resultant force of these three forces. (4) [10] Copyright Reserved Please Turn Over

Physical Sciences PD OWNloaded from hlayiso.com March 2018 Common Test NSC QUESTION THREE A man stands on the scale in a stationary lift as shown in the diagram below. The scale is calibrated in Newtons and reads 882 N. (Image adapted from Fundamentals of Physics Extended -10" edition) 3.1. Calculate the mass of the man. (2) 3.2 Draw and label a free body diagram to show the forces acting on the man. (2) The Lift begins to move. 3.3 Calculate the reading on the scale when the lift is moving: 3.3.1 Downwards at constant velocity of 4 m:s'. (2) 3.3.2 Upwards with a uniform acceleration of 2 m:37 (3) [9] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences /P1 March 2018 Common Test NSC QUESTION FOUR Two wooden blocks of mass 20 kg and 10 kg are placed on a rough horizontal surface. A rope connects the two blocks. When a horizontal force, F, of magnitude 55 N is applied to the blocks as shown, the blocks accelerate to the right at 0,36 ms? 20 kg F=55 N 10 kg }——> The kinetic frictional force between the 20 kg block and the surface is TWICE ‘hat between the 10 kg block and the surface. 4.1. State Newton’s Second law of motion in words. (2) 4.2. Drawa labeled free body diagram to show all the HORIZONTAL FORCES acting on the 10 kg block as it moves. (3) 4.3 By applying Newton’s 2” law of motion separately to each of the blocks, calculate: 4.3.1. The magnitude of the frictional force between the 10 kg block and the surface. (5) 4.3.2 Ux the coefficient of kinetic friction between the blocks and the surface. (3) The rope connecting the two blocks is now cut. 4.4 How will this affect the acceleration of the 10 kg block? (Choose from: INCREASES, DECREASES or REMAINS THE SAME.) Explain. (3) [16] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences /P1 8 March 2018 Common Test NSC QUESTION FIVE 5.1. State Newton's Law of Universal Gravitation in words (2) 5.2 A satellite of mass 225 kg orbits the Earth at an altitude of 6370 km. 7 we Gp Satellite x Nile ¥ va Ripe a ve / $ . % ‘ S \ / oO N f \ \ i \ i \ ! \ ! | | 1 t \ i ‘ i ‘ i ‘ é \ f \ va ‘ ra 5 f , ft ‘ f * “ x a “Bs Fd If the gravitational force of attraction between the Earth and the satellite is 551,33 N, calculate the mass of the Earth. (5) [7] TOTAL MARKS :50 Copyright Reserved Please Turn Over

Physical Sciences /P1 NSC Downloaded from hlayiso.com March 2018 Common Test DATA FOR PHYSICAL SCIENCES GRADE 11 PAPER 1 (PHYSICS) GEGEWENS VIR FISIESE WETENSKAPPE GRAAD 11 VRAESTEL 1 (FISIKA) TABLE 1: PHYSICAL CONSTANTS/TABEL 1: FISIESE KONSTANTES NAME/NAAM SYMBOL/SIMBOOL VALUE/WAARDE Soni von Aurel Re 6,38 x 10° m TABLE 2: FORMULAE/TABEL 2: FORMULES MOTION/BEWEGING Vv, =v,+aat Ax =v At+fadt? vo =v? +2aAx ae (SY a 2 | FORCE/KRAG Fie = ma w=mg Gm,m 1, F= 2 2 by = i. f, Hy = 7 Copyright Reserved

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Downloaded from hlayiso.com Physical Sciences P1 3 NSC-~(Grade 11) QUESTION THREE 3.1 W =mg 882 = m.9,8 v m =90kg v 3.2 Fecate / Fromei!Fu ¥ Foravity / Fweight Fg ~ Take downward motion as positive 3.3.1 Fer = ma NB: Fecale = m-g Fy +( —Fecate ) = m-a 882— Fecale = 0 ¥ Fecale = 882 N v = 90-98 = 882 Nv Max mark : 1/2 3.3.2 Take upward motion as positive Fret = m-a v (any one} Fscate + (— Fg) = m-(a) scale + (— 882) = 90(2)v Fecale = 1062 N ¥ Copyright reserved Maren 2018 Common Test Physical Sciences P1 4 March 2018 Common Test NSC-(Grade 11) QUESTION FOUR 4.1. Ifa non zero NET force acts on an object, then the object accelerates in the direction of the NET force where the acceleration of the object is directly proportional (2) to the NET force and inversely proportional to the mass of the object.“ (2 or 0) (2) 4.2 10kq block od Fv By eS 1 mark - label and vector (3) 4.3.1 Consider the 10kg mass (2) Fer = ma ¥ (any one] F + (-T) +(-fa) = m-a 55—-T-f = 10(0,36) T=514-f v.. If masses are added , then maximum 3/5 (1) Consider the 20kg mass Fret = m-a T + (-2f) = 20(0,36) T=7.2+2f v... weve) Equating (1) and (2) 51,4-fk =7.2+2f v (1 mark for equating the two equations) (3) fx =14,73 N (9] The Kinetic frictional force on the 10kg block = 14,73N ¥ / fkioxg = 14,73N¥ (5) 43.2 f= pK N Qe — fic = px mg) oa 2(14,73) = ux-(20}(9,8) or bk = 0,15” 4.4 or vy hen the string is cut, the accelerating mass decreases ¥ and since the acceleration is inversely proportional to mass, the acceleration increases. ” 414,73 = pe (109.8) ¥ pe = 0,15 ¥ (3) or the mass of the system decreases” and by Newton's second law, the acceleration increase” (3) [16] Please turn over Please turn over Copyright reserved

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