2024 MCED Sept Gr 12 P.Sciences P1_MEMO_hlayiso.com_.pdf

Downloaded from hlayiso.com MCED September 2024 Marking Guideline Page 1 of 25 METRO CENTRAL EDUCATION DISTRICT GRADE 12 PHYSICAL SCIENCES: PHYSICS (P1) - MEMORANDUM SEPTEMBER 2024 MARKS: 150 TIME: 3 hours © Copyright Reserved P.T.O

Downloaded from hlayiso.com MCED September 2024 Marking Guideline Page 2 of 25 GRADE 12 MARKING GUIDELINE SEPTEMBER 2024 QUESTION 1 1.1 D  1.2 B  1.3 D  1.4 C  1.5 C  1.6 D  1.7 C  1.8 A  1.9 A  1.10 D  [20] © Copyright Reserved P.T.O

Downloaded from hlayiso.com MCED September 2024 Marking Guideline Page 3 of 25 QUESTION 2 2.1.1 ( ) 2.1.2 FN  FT  Accepted Labels: Fg / Fw / weight / mg / w gravitational force/ force of gravity T FT / Tension / Fstring fk Ffriction / Ff / Friction Ff  N FNormal / Normal / FN Notes: Mark awarded for label and arrow Fg  Do not penalise for length of arrows since drawing is not to scale. Any other additional force(s)  Max 3/4 If force(s) do not make contact with body  Max 3/4 NO MARK awarded for drawing Fg components . (4) 2.1.3 fk = µkN  fk = (0,2)(9,8 x 8)cos30 o  fk = 13,579 (or 13,58) N (3) © Copyright Reserved P.T.O

Downloaded from hlayiso.com MCED September 2024 Marking Guideline Page 4 of 25 2.1.4 POSITIVE MARKING from 2.1.3. 8 kg Block Fnet = ma  T + fk + Fg// = ma T – 13,579 – (8 x 9,8)sin30o  = 0 T = 52,779 N  m2 Block  any one subst. of 0 Fnet = ma Fg – T = ma 9,8m2 – 52,779  = 0 9,8m2 = 52,78 m2 = 5,386 kg OR 5,39 kg  Note: Accept using energy principles for full marks. Note: if using the systems approach, max. 2 marks.[1 mark for formula and 1 mark for final answer only] (5) 2.1.5 INCREASE.  From point Y onwards the 8 kg block is moving on a frictionless surface therefore force acting up the slope is greater / force acting down the slope is less. OR Net force increases since frictional force is absent. OR There is now an unbalanced force acting on the object. OR Acceleration is no longer zero.  (2) 2.2.1 Every body in the universe attracts every other body with a force which is directly proportional to the product of their masses  and inversely proportional to the square of the distance between their centres  (2) 2.2.2 Gm1 m2 F=  r2 (6,67 × 10−11 )(5,98 × 1024 )(6,417 × 1023 )  Fg = (6,38 × 106 + 4,50 × 109 + 3,3895 × 106)²  Fg = 1,2585 x 1019 (or 1,26 x 1019 ) N  (4) [22] © Copyright Reserved P.T.O

Downloaded from hlayiso.com MCED September 2024 Marking Guideline Page 5 of 25 QUESTION 3 3.1 1,8 (s) – t2 .  (1) 3.2 Marking criteria:  Correct formula for a   Correct substitution into formula   Final answer correct WITH direction   Accept gradient method for full marks.  Accept using energy principles for full marks. OPTION 1: (AB) OPTION 2: (AB) UPWARDS AS POSITIVE: DOWNWARDS AS POSITIVE: vf = vi + aΔt  vf = vi + aΔt  29,7 = 0 + a(1,8)  -29,7 = 0 + a(1,8)  a = 16,50 m·s-2 up  a = - 16,50 m·s-2 a = 16,50 m·s-2 up  Do not penalise for a = 16,5 m·s-2 up OPTION 3: ∆p Fnet =  ∆t m(vf − vi ) ma = ∆t (29,7−0) a=  1,8 a = 16,50 m∙s-2 , up  (3) © Copyright Reserved P.T.O

Downloaded from hlayiso.com MCED September 2024 Marking Guideline Page 6 of 25 3.3 POSITIVE MARKING FROM 3.2 Marking criteria:  Correct formula with Δy   Correct substitution into formula   Correct substitution into: vf = vi + 2aΔy  2 2  Addition of displacements for max height   Final answer correct  OPTION 1 OPTION 2 UPWARDS AS POSITIVE: DOWNWARDS AS POSITIVE: ∆𝐲: (AB) ∆𝐲: (AB) vf2 = vi2 + 2aΔy  vf2 = vi2 + 2aΔy  (-29,7)2 = (0)2 + 2(-16,50)Δy  (29,7)2 = (0)2 + 2(16,50)Δy  Δy = - 26,73 m Δy = 26,73 m, (upwards) Δy = 26,73 m, upwards ∆𝐲: (BC) ∆𝐲: (BC) vf2 = vi2 + 2aΔy vf2 = vi2 + 2aΔy (0)2 = (-29,7)2 + 2(9,8)Δy  (0)2 = (29,7)2 + 2(-9,8)Δy  Δy = - 45,0046 m Δy = 45,005 m (upwards) Δy = 45,005 m, upwards Max height = 26,73 + 45,0046 Max height = 26,73 + 45,0046 = 71,735 (or 71,73) m  = 71,735 (or 71,73) m  OPTION 3 OPTION 4 UPWARDS AS POSITIVE: DOWNWARDS AS POSITIVE: ∆𝐲: (AB) ∆𝐲: (AB) Δy = viΔt + ½aΔt2  Δy = viΔt + ½aΔt2  Δy = (0)(1,8) + ½(16,5)(1,8)2  Δy = (0)(1,8) + ½(-16,5)(1,8)2  Δy = 26,73 m Δy = - 26,73 m Δy = 26,73 m, upwards ∆𝐲: (BC) ∆𝐲: (BC) vf2 = vi2 + 2aΔy vf2 = vi2 + 2aΔy (0)2 = (29,7)2 + 2(-9,8)Δy  (0)2 = (-29,7)2 + 2(9,8)Δy  Δy = 45,005 m (upwards) Δy = - 45,0046 m Δy = 45,005 m, upwards Max height = 26,73 + 45,0046 = 71,735 (or 71,73) m  Max height = 26,73 + 45,0046 = 71,735 (or 71,73) m  © Copyright Reserved P.T.O

Downloaded from hlayiso.com MCED September 2024 Marking Guideline Page 7 of 25 OPTION 5 OPTION 6 UPWARDS AS POSITIVE: DOWNWARDS AS POSITIVE: ∆𝐲: (AB) ∆𝐲: (AB) v +vf v +vf ∆y = ( i ) ∆t  ∆y = ( i ) ∆t  2 2 (0)+(29,7) (0)+(−29,7) Δy = (1,8)  Δy = (1,8)  2 2 Δy = 26,73 m Δy = - 26,73 m Δy = 26,73 m, upwards ∆𝐲: (BC) ∆𝐲: (BC) vf2 = vi2 + 2aΔy vf2 = vi2 + 2aΔy (0)2 = (29,7)2 + 2(-9,8)Δy  (0)2 = (-29,7)2 + 2(9,8)Δy  Δy = 45,0046 m (upwards) Δy = - 45,0046 m Δy = 45,0046 m, upwards Max height = 26,73 + 45,0046 Max height = 26,73 + 45,0046 = 71,7346 (or 71,73) m  = 71,7346 (or 71,73) m  (5) © Copyright Reserved P.T.O

Downloaded from hlayiso.com MCED September 2024 Marking Guideline Page 8 of 25 Marking criteria:  Correct formula with Δy   Correct substitution into formula  vi +vf  Correct substitution of t into: ∆y = ( ) ∆t  2  Addition for max height   Final answer correct  OPTION 7 OPTION 8 UPWARDS AS POSITIVE: DOWNWARDS AS POSITIVE: ∆𝐲: (AB) ∆𝐲: (AB) v +vf v +vf ∆y = ( i ) ∆t  ∆y = ( i ) ∆t  2 2 (0)+(29,7) (0)+(−29,7) Δy = (1,8)  Δy = (1,8)  2 2 Δy = 26,73 m Δy = - 26,73 m Δy = 26,73 m, upwards ∆𝐭: (BC) ∆𝐭: (BC) vf = vi + aΔt vf = vi + aΔt 0 = (-29,7) + (9,8) Δt 0 = 29,7 + (-9,8) Δt Δt = 3,0306 s Δt = 3,0306 s ∆𝐲: (BC) ∆𝐲: (BC) v +vf ∆y = ( i ) ∆t vi +vf 2 ∆y = ( ) ∆t (−29,7)+(0) 2 Δy = (3,0306)  (29,7)+(0) 2 Δy = (3,0306)  Δy = - 45,0044 m 2 Δy = 45,0044 m Δy = 45,0044 m, upwards Max height = 26,73 +  45,0044 Max height = 26,73 +  45,0046 = 71,7344 (or 71,73) m  = 71,7344 (or 71,73) m  (5) © Copyright Reserved P.T.O

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Downloaded from hlayiso.com MCED September 2024 Marking Guideline Page 10 of 25 3.4 POSITIVE MARKING FROM 3.3 Marking criteria:  Formula with Δt   Correct substitution into formula Δt   Addition of total time   Final answer correct  OPTION 1: OPTION 2: UPWARDS POSITIVE DOWNWARDS POSITIVE ∆𝐭: (BD) ∆𝐭: (BD) y = vit + ½ at2  y = vit + ½ at2  - 26,73 = (29,7)t + ½ (-9,8)t2  26,73 = (-29,7)t + ½ (9,8)t2  t = 6,8568 s t = 6,8568 s Value of t2 Value of t2 t2 = 1,8 + 6,8568  t2 = 1,8 + 6,8568  = 8,6568 (or 8,66) s  = 8,6568 (or 8,66) s  OPTION 3: OPTION 4: UPWARDS POSITIVE DOWNWARDS POSITIVE ∆𝐭: (CD) ∆𝐭: (CD) y = vit + ½ at2  y = vit + ½ at2  - 71,7346 = (0)t + ½ (-9,8)t2  71,7346 = (0)t + ½ (9,8)t2  t = 3,82618 s t = 3,82618 s Value of t2 Value of t2 t2 = 1,8 + 3,0306 + 3,82618  t2 = 1,8 + 3,0306 + 3,82618  = 8,65678 (or 8,66) s  = 8,65678 (or 8,66) s  © Copyright Reserved P.T.O

Downloaded from hlayiso.com MCED September 2024 Marking Guideline Page 11 of 25 OPTION 5 OPTION 6 UPWARDS AS POSITIVE: DOWNWARDS AS POSITIVE: 𝐯𝐟 : (BD) 𝐯𝐟 : (BD) vf 2 = vi2 + 2aΔy vf 2 = vi2 + 2aΔy vf 2 = (29,7)2 + 2(-9,8)(-26,73) vf 2 = (-29,7)2 + 2(9,8)(26,73) vf = - 37,4966 m.s-1 vf = 37,4966 m.s-1 ∆𝐭: (BD) ∆𝐭: (BD) vf = vi + aΔt  vf = vi + aΔt  (-37,4966) = (29,7) + (-9,8)Δt  (37,4966) = (-29,7) + (9,8)Δt  Δt = 6,8568 s s Δt = 6,8568 s s Value of t2 Value of t2 t2 = 1,8 + 6,8568  t2 = 1,8 + 6,8568  = 8,6568 (or 8,66) s  = 8,6568 (or 8,66) s  OPTION 7 OPTION 8 UPWARDS AS POSITIVE: DOWNWARDS AS POSITIVE: 𝐯𝐟 : (CD) 𝐯𝐟 : (CD) vf 2 = vi2 + 2aΔy vf 2 = vi2 + 2aΔy vf 2 = (0)2 + 2(-9,8)(-71,7346) vf 2 = (-0)2 + 2(9,8)( 71,7346) vf = - 37,4966 m.s-1 vf = 37,4966 m.s-1 ∆𝐭: (CD) ∆𝐭: (CD) vf = vi + aΔt  vf = vi + aΔt  (-37,4966) = (0) + (-9,8)Δt  (37,4966) = (-0) + (9,8)Δt  Δt = 3,82618 s Δt = 3,82618 s Value of t2 Value of t2 t2 = 1,8 + 3,0306 + 3,82618  t2 = 1,8 + 3,0306 + 3,82618  = 8,65678 (or 8,66) s  = 8,65678 (or 8,66) s  (4) © Copyright Reserved P.T.O

Downloaded from hlayiso.com MCED September 2024 Marking Guideline Page 12 of 25 3.5 POSITIVE MARKING FROM 3.3 and 3.4 UP AS POSITIVE: Criteria:  velocity of the rocket when the fuel runs out (29,7 m.s-1)   29,7  time t2 when rocket crashes on the ground   straight line and positive gradient for first 1,8 s   straight line and less steep negative gradient v (m·s-1) from 1,8 s to t2    t1 t2 0 1,8 8,6568  Δt (s) (4) -37,4966 DOWN AS POSITIVE: 37,4966 v (m·s-1) t1 t2 0 1,8 8,6568  Δt (s) Criteria:  velocity of the rocket when the fuel runs out   (29,7 m∙s-1)   time t2 when rocket crashes on the ground  - 29,7  straight line and negative gradient for first 1,8 s    straight line and less steep positive gradient from 1,8 s to t2  (4) [17] © Copyright Reserved P.T.O

Downloaded from hlayiso.com MCED September 2024 Marking Guideline Page 13 of 25 QUESTION 4 4.1 In an isolated system  the total mechanical energy remains constant/is conserved  OR In an isolated system The sum of gravitational potential energy and kinetic energy remains constant  (2) 4.2 (Ep + Ek) Before = (Ep + Ek) After 1 1 (mgh + 2 mv2) Before = (mgh + 2 mv2)After  1 1 (3)(9,8)(1,2)  + 2 (3)(0)2 = (3)(9,8)(0) + 2 (3)v2  v = 4,8497 or 4,85 m∙s-1  Accept W NC for full marks. (4) 4.3 In an isolated system  total linear momentum remains constant /is conserved.  OR In an isolated system the total linear momentum before collision equals the total momentum after collision.   [1 mark] NOTE: NO marks to be awarded if closed system is used. (2) 4.4 POSITIVE MARKING FROM 4.2 Right + ∑ p before = ∑ pafter  mpvpi + mbvbi = mpvpf + mpvbf (3)(4,8497)  + (1)(0) = (3)(1,8) + (1)vbf  vbf = 9,1491 (9,15) m·s-1  (right) (4) 4.5 POSITIVE MARKING FROM 4.4 (Ep + Ek) Before = (Ep + Ek) After 1 1 (mgh + 2 mv2) Before = (mgh + 2 mv2)After 1 1 (1)(9,8)(0) + 2 (1)(9,1491)2  = (1)(9,8)h  + 2 (1)(0)2 h = 4,27 m  Accept W NC for full marks Do NOT accept equations of motion. (3) [15] © Copyright Reserved P.T.O

Downloaded from hlayiso.com MCED September 2024 Marking Guideline Page 14 of 25 QUESTION 5 5.1 The net work done on an object by a force is equal to the change in the object's kinetic energy.  OR The work done on an object by a resultant / net force is equal to the change in the object's kinetic energy.         (2) 5.2 Ff  Accepted Labels: FN  Fg / Fw / weight / mg / w gravitational force/ force of gravity fk Ffriction / Ff / Friction N FNormal / Normal / FN Notes: Fg  Mark awarded for label and arrow Do not penalise for length of arrows since drawing is not to scale. Any other additional force(s)  Max 2/3 If force(s) do not make contact with body  Max 2/3 (3) 5.3 OPTION 1 Wnet = (WFnormal) + Wfriction + W Fgravity  W net = (0) + µkN.x.cosθ + mg.x.cosθ Wnet = 0 + (0,42)(850)(9.8)(cos 30°) (200)(cos180°)  + (850)(9,8)(200)(cos 60°)  W net = (- 605 975,2955) + 833 000 W net = 227 024, 7045 (or 227 024, 7) J  OR Wnet = (WFnormal) + Wfriction + W Fgravity  W net = (0) + µkN.x.cosθ + mg.x.cosθ Wnet = 0 + (0,42)(850)(9.8)(cos 30°) (200)(cos180°)  + (850)(9,8)Sin300(200)(cos 0°)  W net = (- 605 975,2955) + 833 000 W net = 227 024, 7045 (227 024, 7) J  (Check the range) © Copyright Reserved P.T.O

Downloaded from hlayiso.com MCED September 2024 Marking Guideline Page 15 of 25 OPTION 2 Wnet =  EK  (5) (W Fnormal) + Wfriction + WFgravity =  EK (0) + µkN.x.cosθ + mg.x.cosθ =  EK 0 + (0,42)(850)(9.8)(cos 30°)(200)(cos180°) + (850)(9,8)(200)(cos 60°)  =  EK - 605 975,2955 + 833 000 =  EK W net =  EK = 227 024, 7045 (or 227 024, 7) J  OPTION 3 Wnet = (WFnormal) + Wfriction + W Fgravity  W net = (0) + µkN.x.cosθ + mg.x.cosθ Wnet = (0) + (0,42)(850)(9.8)(cos 30°) (200)(cos180°)  + (850)(9,8)(100)(cos 0°)  W net = (- 605 975,2955) + 833 000 W net = 227 024, 7045 (or 227 024, 7) J  OPTION 4 W net =  EK  (5) (WFnormal) + Wfriction + W Fgravity =  EK (0) + µkN.x.cosθ + mg.x.cosθ =  EK 0 + (0,42)(850)(9.8)(cos 30°)(200)(cos180°) + (850)(9,8)(100)(cos 0°)  =  EK - 605 975,2955 + 833 000 =  EK W net =  EK = 227 024, 7045 (or 227 024, 7) J  OPTION 5 Fnet = Fg // - Fk Fnet = (850)(9,8)(sin 30°) – (0,42)(850)(9,8)(cos 30°)  Fnet = 1135,123522 N W net = Fnet.x.cosθ  W net = (1135,123522)  (200)( cos0°)  W net = 227 024, 7045 (or 227 024, 7) J  (5) © Copyright Reserved P.T.O

Downloaded from hlayiso.com MCED September 2024 Marking Guideline Page 16 of 25 5.4  EK from B to C = EKf - EKi - 108 950 = EKf – 227 024, 7045  EKf at C = 118 074, 7045 OPTION 1 Wnet =  EK  W Fnormal + WFgravity + Wfriction + W Fbrakes =  EK 0 + 0 + µkN.x.cosθ + Fbrakes.x.cosθ = EKf at D - EKi at C 0 + 0 + (0,42)(850)(9.8) (50)(cos180°)  + Fbrakes.(50)(cos 180°)  = 0 – 118 074, 7045  Fbrakes = - 1137,106 N  magnitude of Fbrakes = 1137,106 N  OPTION 2 Wnc =  EP +  EK  Wfriction + W Fbrakes = (EP at D – EP at C) + (EKf at D - EKi at C) µkN.x.cosθ + Fbrakes.x.cosθ = (mghD – mghC) + (EKf at D - EKi at C) (0,42)(850)(9.8)(50)(cos180°) + Fbrakes.(50)(cos 180°) = (0 – 0) + (0 – 118 074, 7045)  Fbrakes = - 1137,106 N  magnitude of Fbrakes = 1137,106 N  © Copyright Reserved P.T.O

Downloaded from hlayiso.com MCED September 2024 Marking Guideline Page 17 of 25 OPTION 3 and 4  EK from B to C = EKf - EKi - 108 950 = EKf – 227 024, 7045  EKf at C = 118 074, 7045 J 1 EK at C = 2 𝑚𝑣 2 1 118 074, 7045= 2 (850)𝑣 2 𝑣𝑎𝑡 𝐶 = 16,6680 m∙s-1 OPTION 3 Wnet =  EK  1 W Fnormal + WFgravity + Wfriction + W Fbrakes = 2 𝑚(𝑣𝑓2 − 𝑣𝑖2 ) 1 0 + 0 + µkN.x.cosθ + Fbrakes.x.cosθ = 2 𝑚(𝑣𝑓2 − 𝑣𝑖2 ) 1 (0,42)(850)(9.8) (50)(cos180°)  + Fbrakes.(50)(cos 180°)  = (850)(02 − 16,66802 ) 2 Fbrakes = - 1137,106 N  magnitude of Fbrakes = 1137,106 N  OPTION 4 Wnc =  EP +  EK  1 Wfriction + W Fbrakes = (EP at D – EP at C) + 2 𝑚(𝑣𝑓2 − 𝑣𝑖2 ) 1 µkN.x.cosθ + Fbrakes.x.cosθ = (mghD – mghC) + 2 𝑚(𝑣𝑓2 − 𝑣𝑖2 ) 1 (0,42)(850)(9.8)(50)(cos180°) + Fbrakes.(50)(cos 180°) = (0 – 0) + (850)(02 − 16,66802 )  2 Fbrakes = - 1137,106 N  magnitude of Fbrakes = 1137,106 N  (7) [17] © Copyright Reserved P.T.O

Downloaded from hlayiso.com MCED September 2024 Marking Guideline Page 18 of 25 QUESTION 6 6.1 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.  OR The apparent change in frequency/pitch/wavelength  of the sound detected by the listener due to the relative motion between listener and source.  (2) 6.2 Moving Towards v±vL fL = f  v±vs s V 253 380 = V−20 fs  (253 380)(v−20) fs = v  (Converting) Moving Away v±vL fL = f v±vs s V 246 710 = V+20 fs  (246 710)(v+20) fs = v Equate (253 380)(v−20) (246 710)(v+20) = v v v = 1 499, 52 m·s-1  (6) 6.3 REMAINS THE SAME  Frequency emitted by the source does not depend on the speed of the source. OR Frequency of the source is the same.       (2) [10] © Copyright Reserved P.T.O

Downloaded from hlayiso.com MCED September 2024 Marking Guideline Page 19 of 25 QUESTION 7 7.1 S  (1) 7.2 Criteria for sketch Marks Correct direction of field lines &  minimum of 8 lines per charge Correct shape of the electric field lines between charges and on the outside of the  charges. No field lines crossing each other. Field lines must touch the charge, but not go  inside the charge. Note: If learner draws field pattern of two opposite charges: 0 ⁄ 3 If only one charge is drawn, max: 1 ⁄ 3 for direction. (3) 7.3 Q1 + Q 2 Q1 + Q 2 Qnew = Qnew = 2 2 6×10−6 + 2×10−6 6+ 2 Qnew =  Qnew =  2 2 ∴ Qs = + 4 × 10−6 C  ∴ Qs = 4 μC  (2) 7.4 Positive marking from 7.3. Charge S Charge R (a) ∆Q = 4 × 10−6 − 2 × 10−6 (a) ∆Q = 4 × 10−6 − 6 × 10−6 = + 2 × 10−6 C = − 2 × 10−6 C Q Q (b) n =  (b) n =  e e 2×10−6 − 2×10−6 n = 1,6 ×10−19  n = − 1,6 ×10−19  n = 1,25 × 1013 electrons  n = 1,25 × 1013 electrons  (3) © Copyright Reserved P.T.O

Downloaded from hlayiso.com MCED September 2024 Marking Guideline Page 20 of 25 7.5.1 The electric field at a point is the electrostatic force experienced per unit positive charge placed at that point.         (2) 7.5.2 POSITIVE MARKING FROM 7.3 Marking Criteria: Enet = ER on X + ES on X Formula kQR kQS  First Substitution Enet = +  [Any formula] Second Substitution r2 r2 Addition (9x109 )(4x10−6 ) (9x109 )(4x10−6 ) Final answer with direction = (10x10−3 )2 +  (2x10−3 )2  = 3,6 x 10 8 + 9 x 10 9 Enet = 9,36 x 10 9 N∙C-1 right  kQ ER on X = r2R  [Any formula] (9x109 )(4x10−6 ) = (10x10−3 )2  = 3,6 x 108 N∙C-1 kQ ES on X = r2R (9x109 )(4x10−6 ) = (2x10−3 )2  = 9 x 10 9 N∙C-1 Enet = (3,6 x 108) +  (9 x 109) Enet = 9,36 x 10 9 N∙C-1 right  (5) [16] © Copyright Reserved P.T.O

Downloaded from hlayiso.com MCED September 2024 Marking Guideline Page 21 of 25 QUESTION 8 8.1.1 12 V  (1) 8.1.2 12 – 8 =4V (1) 8.1.3 Option 1 ΔI gradient =  ΔV 1 0−6 − =  [or any suitable co-ordinates] 𝑟 12−0 1 6 −𝑟 = - 12 r = 2Ω  Option 2 1 r = − gradient  1 r = − ΔI  ΔV 1 r = − 0−6  [or any suitable co-ordinates] 12−0 r = 2Ω  Option 3 Vint = Ir [lose formula mark] 4  = 2𝑟  [or any suitable co-ordinates] r= 2Ω  [max 3 marks] (4) 8.2.1 The maximum (total) energy  provided by a battery per unit (positive) charge / per coulomb passing through it.  OR The maximum (total) work done  by a battery per unit (positive) charge passing through it.  (2) 8.2.2 19,125 V  (1) © Copyright Reserved P.T.O

Downloaded from hlayiso.com MCED September 2024 Marking Guideline Page 22 of 25 8.2.3 OPTION 1 Vint = Ir  6,5 = I(0,5)  Icircuit = 13 A  OPTION 2 Ɛ = Vext + Ir  19,125 = (19,125 – 6,5) + I(0,5)  Icircuit = 13 A  (3) 8.2.4 OPTION 2 OPTION 1 1 1 1 1 1 1 (a) = +  (a) = +  Rp R1 R2 Rp R1 R2 1 1 1 1 = +  = +  5 2,5 5 2,5 = 0,6 Ω = 0,6 Ω RA1 = 1,6667 + 10 = 11,6667 Ω RA1 = 1,6667 + 10 = 11,6667 Ω (b) Vp = 19,125 – 6,5 (b) Vp = 19,125 – 6,5  = 12,625 V = 12,625 V V 10 (c) R=IP (c) V10Ω = 11,6667 × 12,625 A1 12,625 = 10,8214 V 11,6667 =  V IA1 (d) R10Ω = I10Ω A1 IA1 = 1,08 A  10,8214 10 =  IA1 IA1 = 1,08 A  (4) © Copyright Reserved P.T.O

Downloaded from hlayiso.com MCED September 2024 Marking Guideline Page 23 of 25 8.2.5 POSITIVE MARKING FROM 8.2.3 AND 8.2.4 OPTION 1 OPTION 2 (a) Icircuit = IA1 + IA2 (a) Ɛ = I(R + r) 13 = 1,08 + IA2  19,125 = 13(R + 0,5)  IA2 = 11,92 A Rext = 0,97115 Ω V 1 1 1 (b) R=IP (b) = + A2 Rext R1 R2 12,625 = 11,92  1 1 1 = +  0,97115 11,6667 R R = 1,06 Ω  R = 1,06 Ω  OPTION 3 V (a) RT = I T T 19,125 =  13 RT = 1,47115 Ω (b) Rext = RT – r Rext = 1,47115 – 0,5 Rext = 0,97115 Ω 1 1 1 (c) = + Rext R1 R2 1 1 1 = +  0,97115 11,6667 R R = 1,06 Ω  (3) 8.2.6 Decrease   Total external resistance of the circuit decreases.   current in the circuit increases   Vint (= Ir) increases From: emf = Vext + Vint  emf remains constant  Therefore V (= Vext) decreases (4) [23] © Copyright Reserved P.T.O

Downloaded from hlayiso.com MCED September 2024 Marking Guideline Page 24 of 25 QUESTION 9 9.1 Marking criteria If any of the underlined key words/phrases in the correct context are omitted: - 1 mark per word/phrase. The process whereby electrons are emitted / ejected from a metal surface when light of suitable frequency is incident on that surface. (2) 9.2 5,4 x 10-7  (m)          (1) 9.3 POSITIVE MARKING FROM 9.2 OPTION 1 E = W o + Ekmax  ℎ𝑐 ℎ𝑐 𝜆 = 𝜆 + ½ mev2 0 (6,63 x 10−34 )(3 x 108 ) (6,63 x 10−34 )(3 x 108 )  =  + ½ (9,11x10-31)v2  (3,2 x 10−7 ) (5,4 x 10−7 ) v2 = 5,55937 x 1011 v = 745 611,829 m∙s-1 / 7,4561 x 105 m∙s-1  ℎ𝑐 OPTION 2 (a) E=𝜆 (6,63 x 10−34 )(3 x 108 ) =  (3,2 x 10−7 ) = 6,2156 x 10-19 J ℎ𝑐 (b) W0 = 𝜆 0 (6,63 x 10−34 )(3 x 108 ) =  (5,4 x 10−7 ) = 3,6833 x 10-19 J (c) E = W o + Ekmax  (6,2156 x 10-19) = (3,6833 x 10-19) + Ekmax  Ekmax = 2,5323 x 10-19 J (c) Ekmax = ½ mv2 2,5323 x 10-19 = ½ (9,11x10-31)v2  v = 745 611,677 m∙s-1 / 7,456 x 105 m∙s-1  (5) © Copyright Reserved P.T.O

Downloaded from hlayiso.com MCED September 2024 Marking Guideline Page 25 of 25 9.4 REMAIN THE SAME   Intensity only affects number of photons incident onto the metal per second (and therefore the number of photoelectrons emitted from a metal surface per second) OR Intensity does not affect the kinetic energy of photoelectrons. OR Only frequency affects the kinetic energy of photoelectrons.     (2)  TOTAL: 150 © Copyright Reserved P.T.O