Gr12 Physical Sciences P1 Solutions Learner Assistance Document 2020

Education KwaZulu-Natal Department of Education REPUBLIC OF SOUTH AFRICA CURRICULUM DIRECTORATE LEARNER ASSISTANCE REVISION DOCUMENT P1 SOLUTIONS GRADE 12 - 2020 PHYSICAL SCIENCES

PHYSICS P1- SOLUTIONS 1. Force diagrams a) P – Normal force , Q – gravitational force b) P – Normal force , Q – gravitational force, R– Applied force, S- kinetic frictional force c) P – Normal force , Q – gravitational force, R– Applied force, S– kinetic frictional force d) P– Tension force Q– gravitational force e) P– gravitational force f) A – Normal force B – Applied force C –Frictional force D–Gravitational force 2. HOMEWORK : FREE BODY DIAGRAMS, 1. 25 kg 15 kg Fnormal 240 N Fnormal FTension FTension Ffriction Ffriction Fweighl Fweighl 2. FTension FTension m2 m1 Fweighl Fweighl

3. 4 kg 8 kg FNormal F FTension FNormal Ffriction Fweighl Fweighl Fnet = ma 1.Take initial motion to the right as being positive 1.1 Rx =Rcos = 100cos600 = 50 N Fnet =ma Rx + (-Ff) = ma 50 – 12 =20a a = 1, 9 ms-2 1.2 Ff = μKFN , FN =mg +Ry =20x9.8+17.321 =213.321N 12 = μK(213.321) μK = 0,0563 1.3 ∆x = vi∆t + ½ a∆t2 = 0 + ½ (1, 9) (10)2 = 95 m 2. Take the initial upward motion of the crate as being positive 2.1 Fnet = ma Fcable + (-Fweight) = ma , constant velocity »»» a=0 Fcable + (-500) (9.8) = (500) (0) Fcable = 500(9.8) =4900 N 2.2 Fnet = ma Fcable + (-Fweight) = ma Fcable + (-500) (9.8) = (500) (2) Fcable = 5900 N

3. Ftension Ftension 3kg 2kg Fweight FWeight Take the downward motion of the 3kg mass piece as being positive 3kg 2kg Fnet = ma Fnet = ma Fweight + (–FT) = ma Fweight + (–FT) = m (-a) (3 x 9.8) – FT = 3a (2 x 9.8) – FT = –2a 29, 4 – FT = 3a 19, 6 – FT = –2a……. (2) FT = 29, 4 – 3a ……. (1) Substitute (1) in (2) 19, 6 – (29, 4 – 3a) =–2a a = 1, 96 ms-2 4. Take the initial upward motion of the lift to be positive FSCALE/FNORMAL FWEIGHT 4.1 Fnet = ma Fscale + (-Fweight) = ma Fscale – (80 x 9.8) = (80) (2) Fscale = 944 N

4.2 Fnet = ma Fscale + (–Fweight) = m (–a) Fscale – (80) (9.8) = (80) (-2) Fscale = 624 N 4.3 FNET =ma Fscale + (–Fweight) = m (–a) Fscale – (80) (9.8)= (80)(0) , constant velocity »»» a=0 Fscale = 784 N 4.4 a=g= 9.8 ms-2 Fnet = ma Fscale + (–Fweight) = m (–a) Fscale – (80) (9.8) = (80) (–9.8) Fscale = 0 N 5. Take the motion down the incline as positive. FNORMAL Fg// FWEIGHT F 200 5.1 Constant velocity »»»  F = Fnet =0 N, a=0 ms-2 FNET =ma Fg// =m.g.sin =50∙9.8∙sin200 Fg// + (–Ff) = m∙(a) = 167,5898 N 167, 5898 + (–Ff) =0 Ff = 167, 59 N, up the slope 5.2 μ=tan =tan 200=0, 3639 OR Ff = μkFN FN = mgcos =50(9, 8cos200) =460,449N 167, 59 = μk∙460,449 μk = 0.3639 6. 6.1 RX = R∙cos = 150∙cos150 =144, 8888 N Fnet =ma RX= ma

144, 8888= 30a a = 4, 83 ms-2 6.2 Fnet = ma RX+ Ff = ma 144, 8888 + (–24) =30∙a a = 4, 03ms-2 6.3 FN = Fw – Ry , Ry = Rsin = 100 sin150= 25,882N = (30)(9.8) – 25,882 = 268, 81 N Ff = μkFN 24 = μk( 268, 81) μk = 0,0893 VERTICAL PROJECTILE MOTI0N SOLUTIONS Activity 1 1. 1.1 vf2 = vi2 + 2aΔy (0)2 = (4)2+2(-9, 8) y 0 = 16 + (-19, 6) y -16 = -19, 6y  y = 0, 81 m 1.2 v f = v i + a Δt 0 = (–4) + (9, 8) Δt Δt = 0,41 s ∴ The ball takes 0, 41 s to reach the highest point in its projection (5) 1.3. Time upwards = time downwards ∴ total time in the air is (2) (0, 41) = 0, 82 s 1.4. Total displacement = Δ y = 0 m (1) Displacement is measured in a straight line from the initial position (the thrower’s line from the original to the final position (the thrower’s hand is the initial and final position). Activity 2 2.1 3 - 2, 04= 0, 96 s

2.2 Option 1 Δy = vi t + 12a Δt2 = (10) (3) + 12(–9, 8) (3)2 = 14, 1 m Δy = 14, 1 m below the starting point vf2 = vi2 + 2aΔy 0 = 100 + 2(–9, 8) Δy Δy = 5, 1 m Maximum height above the ground = 5, 1 + 14, 1 = 19, 2 m Option 2 Δy = vi Δt + 12a Δt2 = 0 + 12 (–9, 8) (3 – 1, 02) = –19, 21 m Δy = 19, 21 m (maximum height above the ground) Option 3 vf2 = vi2 + 2aΔy (–19, 4) = 0 + 2 (–9, 8) Δy Δy = 19, 2 m (maximum height above the ground Activity 3 1. 15 m.s-1 (1) 2. Inelastic collision The speed/velocity at which the ball leaves the floor is less / different than that at which it strikes the floor OR The speed/ velocity of the ball changes during the collision. Therefore, the kinetic energy changes/is not conserved. (2) 3 . vf2= vi2 + 2ay (20)2 = (10)2+ 2(9, 8) y ∴ y = 15, 31 m (3) Displacement from floor to maximum height vf2 = vi2 + 2ay (0)2 = (–15)2 + 2(9, 8) y y = –11, 48 m Total displacement = –11, 48 + 15, 3 = 3, 82 m or 3, 83 m Activity 4 4.1 B 4.2 A 4.3 C

MOMENTUM & IMPULSE Activity 1 1.1 Momentum is defined as the product of mass and velocity of an object. 1.2 Impulse is the product of the net force and the time in which the net force acts on an object. 1.3 Elastic collision is defined as collision in which the kinetic energy is conserved. 1.4 Inelastic collision is defined as collision in which the kinetic energy is not conserved. 1.5 Principle of conservation of linear momentum it states that the total linear momentum of an isolated system remains constant. 1.6 An isolated system is the one on which the net external force acting on the system is zero. Activity 2 2.1 Diagram A: m = 150 ÷ 1000 = 0, 15 kg v i = + 18 m·s−1 v f = −12 m·s−1 Diagram B: m = 0, 15 kg → v i = + 18 m·s−1 → v f = 0 m·s−1 Let the direction towards the wall be positive m = 150 g = 0, 15 kg (divide the mass by 1000) v i = 18 m·s−1 pi = mvi = (0, 15) (18) = 2, 7 kg·m·s−1 towards the wall 2.2 m = 150 g = 0, 15 k v f = −12 m·s−1 Pf = mvf = (0, 15) (−12) = 1, 8 kg·m·s−1 away from the wall 2.3 m= 0, 15 kg p= mvf – mvi = (0, 15) (−12 −18) = −4, 5 kg·m·s-1

∴ p is 4,5 kg·m·s-1 away from the wall 2.4 Diagram A: 𝐹𝑛𝑒𝑡 =𝑃𝑡 𝑀𝑉 − 𝑀𝑉 = 𝑓 𝑡 𝑖 (0,15)(−12) − (0.15)(+18) = 0,1 Force exerted by the wall is 45 N away from the wall 𝑃 Diagram B Fnet = 𝑡 𝑚𝑣𝑓−𝑚𝑣𝑖 = 𝑡 = 0,15(0) - 0,15(18) 0, 1 = -27 N  Force exerted by the wall is 27N away from the wall 2.5 pf = -1,8 kg.m.s-1 away from the wall pi = +2,7 kg.m.s-1 toward the wall p = mvf - mvi = -1, 8 - (+2, 7) = 4, 5 kg.m.s-1 away from the wall -1, 8 kg.m.s-1 +2, 7 kgm.s-1 p = - 4,5 kg.m.s-1 Fnett = p = (0,175) [(–30) – (12)] = –7, 35 N·s Therefore 7, 35 N·s away from the bat (4) 2. Fnet t = –7, 35 = Fnet (0, 05) −7,35  Fnet = 0,05 =-147N Therefore 147 N away from the bat (3) 3. 147 N towards the bat. According to Newton’s Third Law of Motion the force of the bat on the ball is equal to the force of the ball on the bat, but in the opposite direction, (5) Fbat on ball = −Fball on bat

Activity 4 4.1 An isolated system is the one on which the net external force acting on the system is zero. 4.2 m1 = 4 000 kg and m2 = 3 000 kg v1i = +1, 5 m·s−1 v2i = 0 m·s−1 v1f =? v2f = + 2, 8 m·s−1 Σpi=Σpf m1 v1i + m2 v2i = m1 v1f + m2 v2f (4000) (1, 5) + (3000) (0) = (4000) v1f + (3000) (2, 8) 4000 v1f = 6000 – 8400 v1f = 0,6 m·s-1 ∴ 0, 6 m·s-1 to the west 4.3 Σ EKi = ½ m1 v1i 2 + ½ m2 v2i2 = ½ 4000) (0, 6)2+ ½ (3000) (2, 8)2 = 720 + 11760 Σ EK f= ½ m1 v1f2 + ½ m2 v2f2 =½ (4000) (1, 5)2 + ½ (3000) (0)2 = 4500 J 4500 J  12480 J  Collision is inelastic Work, Energy and Power Activity 1 1.1 Work energy and power - The net work done on an object is equal to the change in the object’s kinetic energy. 1.2 Conservative forces - A force for which the work done (in moving an object between two points) is independent of the path taken. 1.3 Non – conservative forces - A force for which the work done (in moving an object between two points) depends on the path taken.

1.4 Isolated system - A system in which the net external force acting on the system is zero. 1.5 Principle of Mechanical energy - The total mechanical energy (sum of gravitational potential energy and kinetic energy) in an isolated system remains constant. 1.7 Power - The rate at which work is done or energy is expended. Activity 2 1.1 B 1.2 D 1.3 D 1.4 A 1.5 B 1.6 B Activity 3 1.1 The total mechanical energy (sum of gravitational potential energy and kinetic energy) in an isolated system remains constant. 1.2 No. 1.3 1.4 N Ff W

1.4.1 1.4.2 The net work done on an object is equal to the change in the object’s kinetic energy. 1.4.3 A force for which the work done (in moving an object between two points) depends on the path taken. Activity 4 1.1 WFapplied = Fapplied·Δx·cos θ = (220)(10)(cos 0°) = (220)(10)(1) = 2 200 J (3) 1.2. WFnormal = Fnormal· Δx· cos θ = mg· Δx · cos 90°= (50)(9,8)(10)(0) =0J (3) 1.3. WFfriction = Ffriction·Δx · cos θ = (40)(10)(cos 180°) = (40)(10)(−1) = −400 J (3) 1.4. Wnet = ƩW = WFapplied + WFfriction = (2 200) + (−400) = 1 800 J (3) Activity 5 1.1 The net work done on an object is equal to the change in the object’s kinetic energy.(2) T 1.2 W 1.5 WFg = Fg· Δx· cos θ = (6)(9,8)(1.6)(cos 0°)= 94.08 J (3)

DOPPLER EFFECT SOLUTIONS Assignment 1. Doppler Effect is the change in frequency (or pitch) of the sound detected by a listener, because the sound source and the listener have different velocities relative to the medium of sound propagation. 2. v  vL 2.1 f L  fs v  vs 340  0 fL  350  377,78Hz 340  25 340  0 2.2 f L  350  326,03Hz 340  25 340  5 fL  450  456,62 Hz 3. 340  0 340  0 4.1 f L  1200  4080Hz 340  240 340  240 fL  4080  6960Hz 4.2 340  0 340  0 fL  350  377,78Hz 5. 340  25

340  vL 450  400 340  0 vL =42,5 m.s-1 More exercises 1.1 v  vL fL  fs v  vs 1500 fL  (250000) 1500  20 = 253,38 x 103 Hz 1.2 Remains the same 2.1 Doppler Effect 2.2 v  vL fL  fs v  vs 340 90 / 100 fs  ( fs ) 340  vs vS = 37,78 m•s-1 2.3 Smaller than 3.1 Doppler Effect 3.2 v  vL fL  fs v  vs 340 90 / 100 fs  ( fs ) 340  vs vS = 37,78 m•s-1

The detected frequency is independent of the distance between the source and the observer. 4.1 Towards 4.2 v  vL 4.3 f L  fs v  vs 340  0 450  fs 340  20 153000- 9000 = 340fs fs = 423,53 Hz ELECTROSTATICS SOLUTIONS QUESTION ONE 1.1 The electrostatic force between charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them 𝑄+𝑄2 1.2 Qnew = 2 −6𝑥10−6 + 2𝑥10−6 = 2 −6 =−2 𝑥 10 C 𝑄 1.3 nelectrons = 𝑞𝑒 − 2𝑋10−6 = −1,6 𝑋10−19 =1,25x1013 electrons 𝑄1𝑄2 1.4 F = k 𝑟2 2𝑋10−6 .2𝑋10−6 = 9x109 (20𝑋10−2 )2 =9x103 N

𝑄1𝑄2 1.5 1.5 F = k 𝑟2 2𝑋10−6 .4𝑋10−6 =9x109 (30𝑋0−2 )2 =8X10-1 N FR2 =FX2+ FY2 = (9x103)2 + (4X10-1)2 = 9X103 N 8𝑥10−1 Tan ᶿ = 9𝑥103 ᶿ = 5,09x10 QUESTION 2 2.1 The electrostatic force between charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them 𝑄1 𝑄2 2.2 F = k 𝑟2 4𝑥10−6 .2𝑥10−6 =9X109 (50𝑥10−3) 2 2,88X101N 2.3 𝑄1 𝑄2 2.4 F = k 𝑟2 8𝑋10−6 .2𝑋10−6 =9X109 (100𝑋10−3) 2 = 1,44X101N FNET = F1 + F2 1 + 1,44X101 = 2,88 X10 = 1,44X101N 2,44𝑥101 Tan ᶿ = 1,44𝑥101 ᶿ = 5,95x101 QUESTION 3

3.1 Electric field is a region of space in which an electric charge experiences a force. 3.2 𝑘𝑄 3.3 E1 = 𝑟 2 3𝑥10−9 = 9X109 (10𝑥10−3)2 = 2,7x105 N.C-1 𝑘𝑄 E2 = 𝑟 2 5𝑥10−9 = 9x109(30𝑥10−3)2 = 5 x104 N.C-1 ENET = E1 + E2 𝑘𝑄 = E1 = 𝑟 2 3𝑥10−9 = 9X109(10𝑥10−3)2 = 5 x 104 + 2,7x105 = 3, 20 x105 N.C-1 𝐹 3.4 E = 𝑒 𝐹 3, 20 x105 = 1,6 𝑥10−19 F = 5, 12 x10-14 N

QUESTION 4 4.1 Because we need to charge the sphere. 𝑘𝑄 4.2 E1 = 𝑟 2 4,8𝑥10−8 = 9X109(5𝑥10−2)2 = 1,73 x105 N.C-1 4.3 4.4 T FE W 𝑄1 𝑄2 4.5 F = k 𝑟2 4,8𝑥10−8 .4𝑥10−8 = 9X109 (5𝑋10−2) 2 = 8,29X10-3N 𝐹 4.6 E = 𝑄 8,29X10−3 E = 4,8 𝑥10−8 F = 1, 73 x105 NC-1 QUESTION 5 5.1 Electric field is a region of space in which an electric charge experiences a force. 𝑄 5.2 ne = 𝑞𝑒 10 𝑥10−6 = 1,6𝑥10−19 = 6, 25 x1013 electrons

𝑘𝑄 5.3 E= 𝑟2 10𝑥10−6 = 9 x109(2𝑥10−2)2 = 2, 25 x108 NC-1 5.4 5.5 The electrostatic force between charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them 𝑄1 𝑄2 5. F = k 𝑟2 10,𝑥10−6 .20𝑥10−6 = 9X109 (3𝑥 −2) 2 = 2, 00 x10-3 N ELECTRIC CIRCUITS SOLUTIONS MULTIPLE CHOICE QUESTIONS 1. A 2. A 3. C 4. B 5. A 6. D 7. C 1 1 1 1.1    Rp R1 R 2 1 1  =  60 60  Rp = 30 Ω  1.2 Rext = 30 + 25 = 55 Ω  Emf = I(R + r)   12 = I (55 + 1, 5)   I = 0, 21 A 

1.3 V = IR  = (0, 21)(30)  = 6, 3 V  QUESTION 2 2.1 1, 5 V  V 2.2 gradient/m = I 0,65 - 1,5  = 1,0 - 0  = - 0,85 Ω  2.3 Internal resistance  2.4 Decreases  When I increase: “Lost volts”/ Ir increases.  Vext = emf – Ir decreases.  QUESTION 3 3.1 12 V  3.2 3.2.1 Option 1 Option 2 V 9,6 emf = IR + Ir I= = =4A 12 = I(2,4) + 2,4  I = 4 A R 2,4 3.2.2 emf = IR + Ir  12 = 9, 6 + 4r   r = 0, 6 Ω  3.3 Option 1 Option 2 emf = I(R + r)  Emf = Vterminal + Ir  12 = 6(R + 0,6)  12 = Vterminal + 6(0,6)  Rext = 1,4 Ω  Vterminal = 8,4 V 1 1 1 V 8,4 = +  I2,4 Ω = = = 3,5 A R R1 R2 R 2,4

1 1 1 I tail lamps = 6 – 3,5 = 2,5 A =   1,4 2,4 R  R = 3,36 Ω V 8,4 Rtail lamps = =  = 3,36 Ω Each tail lamp I 2,5 : R = 1,68 Ω  Rtail lamp = 1,68 Ω  Option 3 Option 4 V = IR  For parallel combination: 12 = (6)R  I1 + I 2 = 6 A Rext = 2 Ω V V ∴ + =6 2,4 R taillamps ∴Rparallel = 2 – 0,6 = 1,4 Ω 1 1 1 1 1 8,4 ( + ) = 6 = +  2,4 R taillamps R R1 R2 1 1 1 =   ∴ Rtail lamps = 3,36 1,4 2,4 R  R = 3,36 Ω Rtail lamp = 1,68 Ω  Each tail lamp R = 1,68 Ω  3.4 Increases  Resistance increases, current decreases  Ir (lost volts) decreases  QUESTION 4 4.1 The current in a conductor is directly proportional to the potential difference  across its ends at constant temperature.  OR The ratio of potential difference to current is constant at constant temperature.  4.2 1 1 1 1 1 4.2.1   =    Rp = 0, 7 Ω  R p R1 R 2 1,4 1,4 OR R1R 2 1,4  1,4 Rp = = = 0, 7 Ω  R1  R 2 1,4  1,4

4.2.2 OPTION 1: V = IR emf = I(R + r)  = (15)(0,7)  12 = I(0,7 + 0, 1)  = 10,5 V  I = 15 A OPTION 2: emf = I(R + r) R= I= =  = 15 12 = Vexternal + (15)(0,1)  A Vexternal = 12 – (15)(0,1) 0,7 =  = 10,5 V  ∴V = 10,5 V  V”lost” = Ir = (15)(0,1) = 1,5 V Vexternal = 12 – 1,5 V = 10,5 V OPTION 3 Voltage divides 0,7: 0,1 / 7:1 Iheadlight = = 7,5 A ∴Vheadlight = x 12  V = IR = (7,5)(1,4) = 10,5 V = 10,5 V  = 11,83 V  4.2.3 OPTION 1 OPTION 2 OPTION 3 P=  I (light) = 7, 5 A I (light) = 7, 5 A =  P = VI  P = I2R  = 78, 75 W  = (10, 5) (7, 5)  = (7, 5)2(1, 4)  = 78,75 W  = 78,75 W 

4.3 Decreases  (Effective/ total) resistance decreases.  - (Total) current increases.  “Lost volts” / Vinternal / Ir increases, thus potential difference / V (across headlights) decreases. V2 P= decreases. R QUESTION 5 5.1 9V Potential difference measured when: switch is open / no current flows / circuit is open/no work done is in external circuit  5.2 Emf = IR + Ir  9 = Vext + (3)(0,3)  Vext = 8,1 V Vext = I(R1 + R2) 8,1 = 3(3R)  R1 = 0,9  5.3 Decreases  5.4 Increases  Resistance decreases.  Current increases.  Ir increases. QUESTION 6 6.1 Any two: Temperature  Cross sectional area (thickness) of material  Length 6.2 Conductor Q  For the same potential difference,  wire Q has a higher current than wire P.  Therefore wire Q has a lower resistance than wire P. 

OR Conductor Q The gradient of the graph for wire Q is bigger than that for wire P.  I V Gradient = is bigger, thus = R is smaller.  V I QUESTION 7 7.1 Vint = 45 – 43, 5 = 1, 5 V  V 1,5 I= = =3A R 0,5 V12 Ω = IR12 Ω = 3 x 12 = 36 V  V// = 43, 5 – 36 = 7, 5 V (If only V// = 7, 5 V: 2 marks) V 7,5 I = // = = 0,75 A R 10 7.2 IR = 3 – 0, 75 = 2, 25 A  V 7,5 R = // = = 3, 33   I 2,25 7.3 Increases  The total resistance increases,  Therefore the current decreases therefore Vinternal decrease  therefore reading on V increases. QUESTION 8 8.1 Rheostat A Resistor V

Criteria for circuit diagram Mark Battery connected to the resistor as shown – correct symbols used.  Rheostat connected in series with resistor – correct symbols used.  Ammeter connected in series so that it measures the current  through resistor – correct symbols used. Voltmeter connected in parallel across resistor – correct symbols  used. 8.2 Temperature  8.3 B V The ratio is greater than that of A.  I QUESTION 9 1 1 1 1 1 9.1   =   R r1 r2 4 16 ∴ R = 3,2 Ω Reffective = 3,2 Ω + 2 Ω + 0,8 Ω =6Ω 9.2 Option 1: Option 2: V = IR  emf = I(R + r)  12 = I(6)  12 = I(5,2 + 0,8)  I=2A I=2A 9.3 Vparallel = IR  = (2) (3, 2)  = 6, 4 V 6,4 V8Ω = = 3, 2 V  2 QUESTION 10

10.1 Option 1 1 1 1 1 1   = +   R = 6, 47 Ω R e r1 r2 9 23 Rtot = 6, 47 + 2 + 0, 2 = 8, 67 Ω V 12 I = = 1, 38 A  R 8,67 Option 2 1 1 1 1 1   = +   R = 6, 47 Ω R e r1 r2 9 23 Rext = 6, 47 + 2 = 8, 47 Ω Emf = I(R + r)  12 = I (8, 47 + 0, 2)  I = 1, 38 A  10.2 Decreases  (Effective) resistance of circuit decreases  (No current through 15 Ω and 8 Ω resistances) Current (I) increases  Ir (lost volts) increases  Vexternal decreases QUESTION 11 11.1 The current through a conductor is directly proportional to the potential difference across its ends at constant temperature.  11.2 Equal  2 A divides equally at T (and since IM = 1 A it follows that IN = 1 A)  11.3 emf = IR + Ir  17 = 14 + Ir  Ir = 3 V Vlost 3 r= =  = 1, 5 Ω  I 2 11.4 VN = IRN = (1) (2) = 2 V  11.5 VY = 14 – 2 = 12 V  VY = IRY  12 = (2) RY   RY = 6 Ω 

ELECTRODYNAMICS SOLUTIONS QUESTION 1 1.1 Q/split ring commutator/commutator Q (1) 1.2 Replace Q/split ring commutator with slip rings. (1) 1.3 AC can be stepped-up at power stations to reduce energy loss during transmission. (2) 1.4.1 1.4.2

QUESTION 2 2.1 2.1.1 slip rings (1) 2.1.2 brush (es) (1) 2.2 Maintains electrical contact with the slip rings. OR To take current out/in of the coil. (1) 2.3

Mechanical /kinetic energy to electrical energy. (1) 2.4 1½ (1) 2.5 2.6 Parallel to (1) 2.7 QUESTION 3 3.1 3.1.1 A: coil / rotor / armature

B: brushes C: commutator OR Split-ring (commutator) (3) 3.1.2 ANY ONE: Takes current into the coil. Maintains contact with the commutator. (1) 3.1.3 DC motor (1) 3.1.4 Due to the motor effect. OR There is an interaction between the external magnetic field and the magnetic field produced by the current in the conductor. (2) 3.2.1 3.2.2 0, 04 s 3.2.3

QUESTION 4 4.1 Electromagnetic induction. (1) 4.2 Rotate the coil faster/Increase the number of coils/ Increase the strength of the magnetic field. (1) 4.3 Slip rings (1) 4.4.1 It is the value of the voltage in a DC circuit that will have the same heating effect as an AC circuit. Accept: The time averaged voltage of an AC. 4.4.2

V, Vims = eS v V2 - 339,45 Vv v2 Vims = 240,03 V¥ QUESTION 5 5.1.1 OPTION 4 V? ms Pato Y 2 34 | /2 100v =v R R= 578 OV fo} v v OPTION 2, Mase = 340 1 - 549,04 v2 v2 V7 ms Paw oY 2 240,04 v Vims = 100v = R=5780Vv 5.1.2 (3) [8]

5.2 Can be stepped up or down/ can be transmitted with less power loss. (1) QUESTION 6 6.1.1 Move the bar magnet very quickly up and down inside the coil (2) 6.1.2 Electromagnetic induction (1) 6.1.3 Commutator/ split rings (1) 6.2.1 6.2.2

OPTION 1 V, Vv = __ max rms V2 220 = Vina V2 Vinax = 311,13 V Vee _ 33113 Y max me R 40,33 =7,71 AY OR/OF Vinax Pp = _/max'max ave 2 1200,1 = (B13 Mae wen Imax =7,71A OPTION 2 P, erage = Vimslime Y average rms~rms 1200.1 = (220) ims” lims = 5,455 A Imax = V2 (5,455) =7,71AY — (7,715A) OPTION 3 P.verage = TR v 1200,1 = ms(40,33)” lims = 5,455 A Imax = V2 lims = /2 (5,455) = 7.71 AY QUESTION 7

7.1 7.1.1 North pole (1) 7.1.2 Q to P (1) 7.2.1 OR 7.2.2

QUESTION 8 8.1.1DC-generator Uses split ring/commutator (2) 8.1.2

8.2.1 OPTION 1 Vv — Vina OR/OF m2 V, —Vmo___ = 349 < 949,416 Pave = Vins Tins v ™s V2 “> , Prive, =Vins Tins, goo= 0.) v v V2 800 = Iims _ (240,416) ¥ aan Ay lms =3,03AY Inms. = 3,33 AY 8.2.2 Pave =Vins Tims for the kettle 2000 = = UY lot = (8,32 + 3,33) ¥ =11,65AV

PHOTOELECTRIC EFFECT SOLUTIONS QUESTION 1 1.1 It is the minimum energy that an electron in the metal needs to be emitted from the metal surface (2) 1.2 Frequency/Intensity (1) 1.3 The minimum frequency required to remove an electron from the surface of the metal. (2) 1.4 1.5 The photocurrent is directly proportional to the intensity of the incident light. QUESTION 2 2.1.1 The minimum frequency (of a photon/light) needed to emit electrons from (the surface of) a metal. (substance) OR The frequency (of a photon/light) needed to emit electrons from (the surface of) a metal. (substance) with zero kinetic energy. (2) 2.1.2 Silver Threshold/cutoff frequency (of Ag) is higher OR To eject electrons with the same kinetic energy from each metal, light of a higher frequency/energy is required for silver. Since E = Wo + Ek(max) (and Ek is constant),

the higher the frequency/energy of the photon/light required, the greater is the work function/Wo. (3) 2.1.3 Planck’s constant (1) 2.1.4 Sodium (1) 2.2.1 2.2.2 7, 09 x 1015 (electrons per second) OR Same number as that calculated in Question 10.2.1 above 2.2.1 (1) [13] QUESTION 3 3.1 It is the process whereby electrons are ejected from a metal surface (2)

3.2 Graph of/Grafiek van Eximaximaks) VS A Eximax) X (10 “*? J) >|a 3.3.1 OPTION 1 4 = 46x10%m" v fezolv = (3 x 10°)(1,6 x 10°) ¥ = 48x10" Hz OPTION 2 By extrapolation: y-intercept = -W. W. = hf, ¥ 3,2 x 10°°v = (6,63 x 10™)f, ¥ fo = 4,8 x 10"4 Hzv (4) (4)

3.3.2 OR QUESTION 4 4.1 photoelectric effect 𝑐 1 4.2 h𝜆 = Wo + 2mv2 (6, 63 x10-34)(3 x 108) ÷ 200 x 10-9 = 8, 00 x10-19 + 0, 5 x 9, 11 x10-31 x v2 v = 653454, 89 m.s-1 QUESTION 5 5.1 The minimum frequency (of a photon/light) needed to emit electrons from (the surface of) a metal (substance). 5.2

5.3 Smaller (less) than (1) 5.4 The wavelength/frequency/energy of the incident light (photon/hf) is constant . Since the speed is larger, the kinetic energy is larger the work function/W 0 /threshold frequency smaller. (3)