Technical Sciences Grade 12 IEB Paper 1 2019 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE EXAMINATION NOVEMBER 2019 TECHNICAL SCIENCES: PAPER I MARKING GUIDELINES Time: 3 hours 150 marks These marking guidelines are prepared for use by examiners and sub-examiners, all of whom are required to attend a standardisation meeting to ensure that the guidelines are consistently interpreted and applied in the marking of candidates' scripts. The IEB will not enter into any discussions or correspondence about any marking guidelines. It is acknowledged that there may be different views about some matters of emphasis or detail in the guidelines. It is also recognised that, without the benefit of attendance at a standardisation meeting, there may be different interpretations of the application of the marking guidelines. IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: TECHNICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 2 of 10 QUESTION 1 MULTIPLE-CHOICE QUESTIONS 1.1 D 1.2 D 1.3 D 1.4 A 1.5 A 1.6 A 1.7 B 1.8 C 1.9 A 1.10 C QUESTION 2 2.1 FN / N FN / N Fapp Fy T T Fx Fg / W Fg / W FORCE DESCRIPTION MARKS FN / N Normal force Fapp / Applied force / Fx and Fy components Fg / W Weight T Tension in the rope 2.2 When a net force is applied to an object of mass, m, it accelerates the object in the direction of the net force. Or When a net force, Fnet, is applied to an object of mass, m, it accelerates in the direction of the net force. The acceleration, a, is directly proportional to the net force and inversely proportional to the mass. Or (In terms of momentum) The net (or resultant) force acting on an object is equal to the rate of change of momentum of the object in the direction of the net force. IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: TECHNICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 3 of 10 2.3 (to the right is +) 5 kg trolley 10kg trolley Fnet = ma Fnet = ma Fx + ( −T ) =ma T = ma 50cos 30°− T = 5a 1 T = 10a 2 −T = 5a − 43,3 T = 10a Or −10a =5a − 43,3 = 0 15a − 43,3 −15a −= 43,3 a = 2,89 mis −2 to the Right a = 2,89 mis −2 2.4 Positive marking from Question 2.3 −T = 5a − 43,3 −T = ( 5 × 2,89 ) − 43,3 T = 28,85 N Or T = 10a T= 10 × 2,89 T = 28,9 N 2.5 The rate at which velocity changes. 2.6 If resistance is taken into account Fnet will decrease. According to Newton's second law Fnet ∝ a Thus acceleration will also decrease. IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: TECHNICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 4 of 10 QUESTION 3 3.1 t = 1,5 s 3.2 Displacement per time interval remains constant, thus velocity is constant. 3.3 The total linear momentum of an isolated system remains constant in magnitude and direction. 3.4 Σ pi = Σ pf (to the west is +) ∆v mx = m x v f + my v f v= + yv i ∆t 2,88 − 1,38 ( 0,25 + 0,75 )( 0 ) = ( 0,25 )( −5 ) + ( 0,75 ) v f v= 0,6 − 0,3 v f = −1,67mis −1 v = 5 mis −1  1.67mis −1 to the east ∴v f = 3.5 Impulse is the product of the net force acting on an object and the time the net force acts on the object. 3.6 Positive marking from Question 3.4 Fnet i∆ =t my v f − m y v i Fnet i0,3= ( 0,75 )( −1,67 ) − ( 0,75 )( 0 ) Fnet = −4,18 N ∴ Fnet = 4,18 N east 3.7 3.7.1 An elastic collision is a collision where momentum and kinetic energy are conserved. Or Elastic collision: Σ pi = Σ pf and ΣEK before = ΣEK after 3.7.2 An inelastic collision is a collision where only momentum is conserved. Or Inelastic collision: Σ pi = Σ pf and ΣEK before ≠ ΣEK after IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: TECHNICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 5 of 10 QUESTION 4 4.1 Energy is the ability to do work. 1 4.2 = E k m ⋅v 2 2 1 Ek = ( 8 ) (102 ) 2 Ek = 400 J 4.3 fk = µk FN =fk mk ( m ⋅ g ) =fk ( 0,15 )( 8 ⋅ 9,8 ) fk = 11,56 N 4.4 Positive marking from Question 4.3 WfK= fK ⋅ ∆X ⋅ cosθ W= fK 11,56 ⋅ 6 ⋅ cos180° WfK = −70,56 J IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: TECHNICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 6 of 10 QUESTION 5 5.1 Power is the rate at which work is done (or energy is expended). 5.2 Fg= m ⋅ g = Fg 1500 ⋅ 9,8 Fg =14700 N 5.3 Positive marking from Question 5.2 P= F ⋅ v P =14700 2 P = 24900 W 5.4 Positive marking from Question 5.3 1 hp = 746W x hp = 29400W 29400 × 1 x= 746 x = 39,41 hp (if only the answer is given, award 2 marks) 5.5 E p = mgh E p = 1500 ⋅ 9,8 ⋅ 7,5 E p =110250 J 5.6 The total mechanical energy (sum of gravitational potential energy and kinetic energy) in an isolated system remains constant. 5.7 Positive marking from Question 5.5 Mechanical energy is conserved (EP + EK )B =(EP + EK )A 1 1 mgh + mv i 2 =mgh + mv f 2 2 2 (110250 + 0 )= (1500 ⋅ 9,8 ⋅ 6 ) +  1500 ⋅ v f 2  1 2  −1 =v f 5,42m ⋅ s IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: TECHNICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 7 of 10 QUESTION 6 6.1 Within the limit of elasticity, stress is directly proportional to the strain. σ 6.2 K= ε 5 × 106 K= 5 × 10−4 K = 1× 1010 Pa The most appropriate material would be wood. ∆L 6.3 ε= L ∆L 5 × 10 −4 = 0,12 ∆L = 6 × 10 −5 m 6.4 Stress is internal restoring force per unit area of a body. F 6.5 σ= A 9800 5 × 106 = A A 1,96 × 10 −3 m 2 = QUESTION 7 7.1 Viscosity is the property of the fluid to oppose relative motion between the two adjacent layers. 7.2 20W–30 7.3 It's a (multi-grade) winter oil At cold temperatures (0°) the oil's viscosity is the same as a SAE 10 At 100 °C the viscosity of the oil is the same as a SAE 30 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: TECHNICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 8 of 10 QUESTION 8 8.1 In a continuous liquid at equilibrium, the pressure applied at a point is transmitted equally to the other parts of the liquid. 8.2 A = πr 2 2  11,8 × 10 −2  A=π   2  A = 0,19 m 2 8.3 Positive marking from Question 8.2 F1 F2 = A1 A2 319 F −4 = 2 3,02 × 10 0.19 F2 = 200695,36 N QUESTION 9 9.1 Capacitance of a capacitor is the amount of charge it can store per volt. 9.2 • The charge held depends on the applied voltage across the plates. • The capacitance increases as the total area of the opposing surfaces of the plates increases. • The capacitance increases as the distance between the plates decreases. • The capacitance depends upon the dielectric material (Any 3 of the factors) kε 0 A 9.3 C= d 1⋅ 8,85 × 10 −12 ⋅ ( 0,02 × 0,02 ) C= Both equations 0,003 −12 = C 1,18 × 10 F Q C= V 6 × 10 −11 1,18 × 10 −12 = V V = 50,85V IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: TECHNICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 9 of 10 QUESTION 10 10.1 AMMETER. 10.2 The potential difference across a conductor is directly proportional to the current in the conductor at constant temperature. 10.3 V1 = 4,5V 10.4 V = IR 4,5 = I ⋅6 I = 0,75 A 10.5 The rate at which electrical energy is converted in an electric circuit. V2 10.6 P= R 4,52 P= 6 P = 3,38 W Positive marking from Question 10.4 P = VI P = 4,5 × 0,75 P = 3,38 W Positive marking from Question 10.4 P = I2R P = 0,752 × 6 P = 3,38 W IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: TECHNICAL SCIENCES: PAPER I – MARKING GUIDELINES Page 10 of 10 QUESTION 11 11.1 11.1.1 AC Generator 11.1.2 The principle of electromagnetic induction 11.1.3 • More windings on the coil • Stronger magnets • Greater rate of change of the magnetic flux / increase rotation speed (Any one of the three factors) 11.1.4 Slip rings 11.2 11.2.1 AB NS VS 11.2.2 = NP VP NS 4,5 = 1000 240 NS = 18,75 windings Total: 150 marks IEB Copyright © 2019