Physical Sciences Grade 12 P1 KwaZulu Natal Mock 2019 Memo

Downloaded from hlayiso.com Physical Sciences/P1 NSC – Marking G NATIONAL SENIOR CERTIFICATE PHYSICAL SCIENCES P1 MARKING GUIDELINE PREPARATORY EXAMINATION SEPTEMBER 2019 GRADE 12 MARKS: 150 N.B. This marking guideline consists of 12 pages including this page. Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences/P1 2 Preparatory Examination 2019 NSC – Marking Guideline QUESTION 1 1.1 A (2) 1.2 D (2) 1.3 D (2) 1.4 B (2) 1.5 B (2) 1.6 C (2) 1.7 C (2) 1.8 B (2) 1.9 A (2) 1.10 D (2) [20] QUESTION 2 2.1 When one body exerts a force on a second body, the second body exerts a force of equal magnitude in the opposite direction on the first body.  (2) OR If body A exerts a force on body B, then body B exerts an equal and opposite force on body A NOTE If any of the underlined key words in the correct context is omitted deduct 1 mark. Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences/P1 3 Preparatory Examination 2019 NSC – Marking Guideline 2.2 T Accepted Labels: T FT / Tension / Fcord on m1 w weight / Fg / Gravitational force / Fearth on mA /mg/force of Earth on block. w Criteria  Mark awarded for label and arrow.  Do not penalize for length of arrow since drawing is not to scale  Any other additional force(s): Max.: 1/2  If force(s) do not make contact with dot: Max: ½ (2) 2.3 TAKE CLOCKWISE AS POSITIVE ∆y = vi∆t + ½ a∆t2  0,5 = 0.∆t + ½ a (1,43)2  a = 0,49 m·s-2  Consider mA: Fnet = ma Any one T - mAg = mAa T - (1,9)g = (1,9)(0,49)  T – (1,9)g = 0,931…………. (1) Consider mB: Fnet = ma mBg – T = mBa (2,1)g - T = (2,1)(0,49)  (2,1)g - T = 1,029…………...(2) Solving (1) and (2) : (2,1)g – (1,9)g = 1,96 (simplification) (0, 2)g = 1,96 g = 9,80 m·s-2  (7) [11] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences/P1 4 Preparatory Examination 2019 NSC – Marking Guideline QUESTION 3 3.1 10 m  (1) 3.2 1,2 (s ) (1) 3.3 An object upon which the only force acting is the force of gravity. (2) 3.4 Take downward motion as NEGATIVE. (Other option: take downwards as positive)) v f  v i  a t  0 = vi + (–9,8)(0,6)  vi = 5,88 m.s–1 , upwards  (3) 3.5 Positive marking from QUESTION 3.4 OPTION 1 OPTION 2 y  v i t  21 at 2  ∆U + ∆K = 0 ½ mvi2 + mghi = ½ mvf2 + mghf  = (5,88) (0,6)+ ½ (–9,8) (0,6)2 = 1,764 m ½m(5,88)2+m(9,8)(10)=0 + m(9,8)h Maximum height = 10 + 1,764 h = 11,76 m = 11,76 m OPTION 3 OPTION 4 v f  v i  2ay  2 2  v  vi  y   f  t  (0)2 = (5,88)2 + 2(–9,8) y   2  y = 1,764 m  = ½ (0+5,88) (0,6)  Maximum height = 10 + 1,764 = 1,764 m  = 11,76 m Maximum height = 10 + 1,764 = 11,76 m (4) 3.6 Positive marking from QUESTION 3.4 and 3.5 From maximum height downwards v f  v i  2ay  2 2 = (0)2 + 2 (–9,8) (–11,76) vf = 15,18 m.s–1 OR From the balcony upwards v f  v i  2ay  2 2 = (5,88)2 + 2 (–9,8) (–10)  vf = 15,18 m•s–1 (3) Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences/P1 5 Preparatory Examination 2019 NSC – Marking Guideline 3.7 Positive marking from QUESTION 3.4 and 3.6   initial velocity  final velocity 5,88  Shape (Line below x – axis longer) v   Time to reach max height (m.s–1)  t (s) 0,6  (4) [18] -15,18 QUESTION 4 4.1 The total linear momentum in a closed/isolated system remains constant / is conserved. (2) NOTE If any of the underlined key words in the correct context is omitted deduct 1 mark. 4.2 Right as positive Σpi = Σpf  (mvi)1 + (mvi)2 = (mvf)1 + (mvf)2 (5500)v + (2000)(-30) = (5500)(6) + (2000)(10)  v = 20,55 m·s-1  Left as positive Σpi = Σpf  (mvi)1 + (mvi)2 = (mvf)1 + (mvf)2 (5500)v + (2000)(30) = (5500)(-6) + (2000)(-10)  vi = -20,5455 m·s-1 magnitude of velocity = 20,55 m∙s-1  (4) 4.3 Fnet Δt = mvf – mvi  Fnet (0,2) = (2000)(10) – (2000)(-30)  Fnet = 400 000 N Fnet = 400 000 N to the left  OR Fnet Δt = mvf – mvi  Fnet (0,2) = (5500)(6) – (5500)(20,5455)  Fnet = -400 001,25 N Fnet = 400 001,25 N to the left  (4) [10] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences/P1 6 Preparatory Examination 2019 NSC – Marking Guideline QUESTION 5 5.1 The total mechanical energy in an isolated (closed) systemremains constant (is conserved).  (2) NOTE If any of the underlined key words in the correct context is omitted deduct 1 mark. 5.2 OPTION 1 Emech at P = Emech at Q  (mgh + ½ mv2)P = (mgh + ½ mv2)Q 4[(9,8)(3) + ½(0)2]  = 4[(9,8)(1,25) + ½ v2]  v = 5,86 m·s-1  (4) OPTION 2 Emech at P = Emech at Q  (mgh + ½ mv2)P = (mgh + ½ mv2)Q 4[(9,8)(1,75) + ½ (0)2]  = 4[(9,8)(0) + ½ v2]  v = 5,86 m∙s-1  (4) 5.3.1 The net/total work done on an object 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) NOTE If any of the underlined key words in the correct context is omitted deduct 1 mark. If the word “work” is omitted then 0 marks. Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences/P1 7 Preparatory Examination 2019 NSC – Marking Guideline 5.3.2       (3) 5.3.3 Wnet = Δ EK   Wnet = 0 Wf + W g = 0 fΔx cosƟ + mgΔxcosƟ = 0 (15)(X)cos 180° + (4)(9,8)(1,25)cos 0° = 0 X = 3,267 m  (5) 5.4 REMAIN THE SAME.  (1) [17] QUESTION 6 6.1 Doppler Effect.  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 An (apparent) change in observed/detected frequency (pitch), as a result of the relative motion between a source and an observer  (listener). (3) NOTE If any of the underlined key words in the correct context is omitted deduct 1 mark. 6.2 (5) 6.3 Increase  (1) [9] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences/P1 8 Preparatory Examination 2019 NSC – Marking Guideline QUESTION 7 7.1 The magnitude of the electrostatic force exerted by one point charge (Q1) on another point charge (Q2 ) is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance (r) between them.  (2) NOTE If any of the underlined key words in the correct context is omitted deduct 1 mark. 7.2 (3)    7.3 Fg = mg = (0,004)(9,8)  = 0,04 N Frepulsion = Fg × tan 5º = 0,04 × tan 5º  = 3,43 ×10-3 N  (4) 7.4 Positive Marking from 7.3      (4) Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences/P1 9 Preparatory Examination 2019 NSC – Marking Guideline 7.5 Criteria  Shape (pattern)  Direction of field lines   Field lines not touching each other  If field lines are not touching the spheres: Max 2/3 (3) 7.6 Q1 + Q2 Qnew = 2 (1×10−6 )+(9×10−6 ) =  2 = +5× 10 C −6 (2) 7.7 A to B  (1) 7.8 Positive marking from 7.6 Qnew − Q1 n= e (5×10−6 )−(1×10−6 ) =  1,6×10−19 = 2,5× 1013 (electrons)  (2) [21] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences/P1 10 Preparatory Examination 2019 NSC – Marking Guideline QUESTION 8 8.1 OPTION 1 OPTION 2 V2 P = VI P=  13,5 = (18)I  R I = 0,75 A 182 13, 5 =  V = IR R 18 = (0,75)R R = 24  R = 24  (3) 8.2 Positive marking from 8.1 OPTION 1 OPTION 2 1 1 1 V = IR = +  18 = I(24)  Rp R1 R2 I = 0,75A 1 1 1 = +  Rp 12 24 Vp = I12R Rp = 8  18 =I12(12)  I12 = 1,5 A V = IR  Itot = 0,75+1,5 18 = I(8)  = 2,25 A  I = 2, 25A  (5) 8.3 Internal resistance is the opposition to the flow of charge within a cell/battery.  (2) 8.4 Positive marking from 8.2 V = IR  = (2,25)(10)  = 22,5 V  (3) 8.5 OPTION 1 OPTION 2  = I(R + r)  Vext = Vp + V10 45,9 = 2,25 (10+8 + r) = 18 + 22,5 r = 2,40  = 40,5 V Vlost = 45,9 – 40,5 = 5,40 V Vlost = Ir  5,4 = (2,25)r  r = 2,40  (5) 8.6 Increase  (1) [19] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences/P1 11 Preparatory Examination 2019 NSC – Marking Guideline QUESTION 9 9.1 Electromagnetic induction  (1) 9.2 The rms value of the AC is the direct current which dissipates the same amount of energy as AC.  (2) NOTE If any of the underlined key words in the correct context is omitted deduct 1 mark. 9.3 V1  (1) V2 9.4 V1 = OR V2 = √2 V1 (1) √2 V2 Vmax 9.5 V1 = or Vrms =  √2 √2 V2 220 =  √2 V2 = 311,13 V (3) 9.6 OPTION 1 OPTION 2 1 1 Imax Pave = Vmax · Imax  Pave = Vmax ·  2 √2 √2 1 1200 = (311,13) · Imax  (√2)(1200) = (220) · Imax  2 Imax = 7,71 A  Imax = 7,71 A  (3) OPTION 3 OPTION 4 Vrms 220 Pave = Vrms · Irms R= =  = 40,33 Ω Irms 5,455 Vmax 311,13 1200 = 220 · Irms  Imax = = = 7,72 A  R 40,33 Irms = 5,46 A But Imax = √2 · Irms  = (√2)(5,455) = 7,71 A  OPTION 5 (3) (Vrms )2 Pave = R (220)2 R=  = 40,33 Ω 1200 Vmax 311,13 Imax = = = 7,72 A  (3) R 40,33 Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences/P1 12 Preparatory Examination 2019 NSC – Marking Guideline 9.7 ANYONE • Easier to generate and transmit from place to place. • Lesser energy loss in transmission. • Voltage can be easily changed by stepping it up or down. (1) [12] QUESTION 10 10.1 Cathode  (1) 10.2 Threshold frequency  (1) 10.3 The minimum energy that an electron in the metal needs to be emitted from the metal surface.  (2) NOTE If any of the underlined key words in the correct context is omitted deduct 1 mark. 10.4 W0 = hf0  = (6,63 x 10–34) (5 x 1014)  = 3,32 X 10–19 J  (3) 10.5 Positive marking from 10.4 1 hf  W0  mv 2 2    (6,63 x 10–34) (f1) = 3,32 x 10–19 + 11 x 10–19 f1 = 2,15 x 1015 Hz  (5) 10.6 Remain the same  (1) [13] TOTAL MARKS: 150 Copyright Reserved Please Turn Over