Physical Sciences Grade 12 P1 KwaZulu Natal Mock 2018 Memo

Downloaded from hlayiso.com   Physical Sciences (P1) 1 Preparatory Examination September 2018 NSC – Marking Guideline NATIONAL SENIOR CERTIFICATE GRADE 12     PHYSICAL SCIENCES P1 MARKING GUIDELINE PREPARATORY EXAMINATION SEPTEMBER 2018 This marking guideline consists of 11 pages. Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences (P1) 2 Preparatory Examination September 2018 NSC – Marking Guideline QUESTION 1 1.1 Cüü 1.2 Cüü 1.3 Aüü 1.4 Büü 1.5 Düü 1.6 Cüü 1.7 Aüü 1.8 Düü 1.9 Cüü 1.10 Cüü 10x 2 = [20] QUESTION 2 2.1 When a resultant/net force acts on an object, the object will accelerate in the direction of the force at an acceleration directly proportional to the forceü and inversely proportional to the mass of the object. ü OR The resultant/net force acting on an object is equal to the rate of change of momentum of the object in the direction of the resultant/net force. üü (2) 2.2 Notes : Accepted labels Mark w Weight / Fg / FG ü T Tension in string ü f Frictional force ü N Normal force ü (4) 2.3 Box Object Fnet = maü Fnet = ma T + (-f) = ma w + (-T) = ma T – 6 = 9a………..(1) ü (2.5)(9,8) – T = 2.5a………(2) ü Substituting (1) into (2): -6 + (2.5)(9.8) = 11.5aü a = 1.61 m.s-2ü (5) Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences (P1) 3 Preparatory Examination September 2018 NSC – Marking Guideline M1M2 2.4 F=G P r2 But FE&M = FS&M P (net zero gravitational force) (!.!"×!"!!! )(!,!"×!"!! )(!,!!×!"!" ) 1,997×10!"   P =   P !! r = 2,21 x 1011m P (5) [16] QUESTION 3 3.1 An object upon which the only force acting is the force of gravity. ü ü (2) 3.2 OPTION 1 Upward is positive Ball Y Ball X Δy1 = vi Δt + ½ a Δt2ü Δy2 = vi Δt + ½ a Δt2 = 20t + ½(-9.8) Δt2 -(50- Δy1) = (0)( Δt) + ½(-9.8) Δt2 = 20t – 4.9 t2……….(1)  ü Δy1 = -4.9t2 + 50………..(2)  ü Now, 20t – 4.9 t2 = -4.9t2 + 50 ü 20t = 50 t = 2.5 sü (5) Upward is negative Ball Y Ball X Δy1 = vi Δt + ½ a Δt2ü Δy2 = vi Δt + ½ a Δt2 -Δy1 = -20t + ½(9.8) Δt2 (50 – Δy1) = (0)( Δt) + ½(9.8) Δt2 Δy1 = 20t - 4.9 t2……….(1)  ü Δy1 = -4.9t2 + 50………..(2)  ü Solving (1) and (2): 20t = 50ü t = 2.5 sü OPTION 2 The balls are approaching each other (relative velocity increases) Δx (vx + vy) = ü Δt 50 (0ü + 20ü)   = ü (5) Δt Δt = 2.5 s ü Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences (P1) 4 Preparatory Examination September 2018 NSC – Marking Guideline 3.3 OPTION 1 OPTION 2 Δy = vi Δt + ½ a Δt2 ü Δy = vi Δt + ½ a Δt2ü = (20)(2.5) ü + ½(-9.8)(2.5)2ü = (0)(2.5) + ½(-9.8)(2.5)2ü = 19.375 m = -30.625 h = 19.375 mü h = 50 –ü 30.625 = 19.375 mü OPTION 3 Ball X vf = vi + aΔt = 0 + (-9.8)(2.5) ü vf = -24.50 m.s-1 Em at top = Em at meeting point (mgh + ½mv2)at top = mgh + ½mv2)at meeting pointü m(9.8)(50) + 0 = m(9.8)Δy + ½m(-24.50)2ü Δy = 19.375 m h = 19.375 mü (4) 3.4 Mark positively from 3.2 and 3.3 OPTION 1 (downward is positive) Criteria • Shapeü • Starts at 50 mü • Ends at 2.5 sü • Ends at 19.375mü OPTION 2 (downward is negative) Criteria • Shapeü • Starts at -50 mü • Ends at 2.5 sü • Ends at -19.375mü (4) Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences (P1) 5 Preparatory Examination September 2018 NSC – Marking Guideline OPTION 3 Can shift the lower graph to start at the origin, and the rest will remain the same. (4) [15] QUESTION 4 4.1 The product of mass and velocity of the object. üü (2) 4.2 0 kg.m.s-1 (Accept 0 / Zero) ü (1) 4.3 To the right is positive To the right is negative p = mvü p = mvü = (1)(10) ü = (1)(-10) ü = 10 kg.m.s-1 to the rightü = -10 kg.m.s-1 = 10 kg.m.s-1 (to the right)ü (3) 4.4 To the right is positive To the right is negative Σpbefore = Σpafter ü Σpbefore = Σpafter ü 0ü = 100vgb ü+ 10ü 0ü = 100vgb ü+ (-10)ü vfi = -0.1 m.s-1 vfi = 0.1 m.s-1 to the leftü vfi = 0.1 m.s-1to the leftü (5) [11] QUESTION 5 5.1 The work done by a net force on an object is equal to üthe change in the kinetic energy of the object. ü OR Net work done on an object is equal to üthe change in the kinetic energyof the object. ü (2) 5.2 Fg = gravitational force or weightü FN / N= normal forceü Ff / f = frictional forceü   (3) NB : Ignore the relative lengths the forces Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences (P1) 6 Preparatory Examination September 2018 NSC – Marking Guideline 5.3 OPTION 1 Wnet = FnetΔx cosθü = Wf + WFg// = (190)(10)(cos180°)ü + (50)(9.8)(sin30°)(10)(cos0°)ü = -1900 + 2450 = 550 Jü (4) OPTION 2 Fnet = Fg// - Ff = (50)(9.8)(sin 30º) – 190 ü = 55 N Fnet = ma 55 = 50a OPTION 3 a = 1.10 m.s-2 Wnet = Wf + WFg// ü vf2 = vi2 + 2aΔx = (50)(9,8)(10)cos60 ü -1900 ü = 550 J ü = 22 + 2(1.10)(10) ü vf = 5.10 m.s-1 Wnet = ½mvf2 - ½mvi2 ü = ½(50)(5.10)2 -½(50)(2)2 = 550 Jü 5.4 POSITIVE MARKING FROM QUESTION 5.3 OPTION 1 Wnet = ΔK ü OPTION 2 = ½ mvf2 – ½ mvi2 550ü= ½ (50)(vf2 – 22)ü Wnc =ΔK +ΔUü vf = 5.10 m.s-1 ü -1900 = ½(50)(v2 – 22) + (50)(9.8)(hY-hX) -1900ü = ½(50)(v2 – 22) + (50)(9.8)(-10sin30º) ü v = 5.10 m.s-1ü (4) [13] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences (P1) 7 Preparatory Examination September 2018 NSC – Marking Guideline QUESTION 6 6.1 14 x 10−4 s PP (2) 6.2 POSITIVE MARKING FROM QUESTION 6.1 1 f = P T 1 = P 14 x 10 − 4 = 714,29 Hz P (3) 6.3 It 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. PP OR It is the change in the observed frequency of a sound wave when the source of sound is moving relative to the listener.PP (2) 6.4 POSITIVE MARKING FROM QUESTION 6.1 !!!  ! f! = f P !!!! ! !"# (714.29)P = P(600)P !"#!!! v! =  54.40  m.s-­‐1  P (accept 54.28 to 54.40  m.s-­‐1) (5) [12] QUESTION 7 !!   !  !! 7.1 Qnew = ! !×!"!! P!(!!×!"!! )P = ! = -1.5×10!!  CP (3) 7.2.1 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 chargesP and inversely proportional to the square of the distance (r) between them. P OR The force of attraction or repulsion between two point charges is directly proportional to the product of the charges ü and inversely proportional to the square of the distance between them. ü (2) Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences (P1) 8 Preparatory Examination September 2018 NSC – Marking Guideline 7.2.2 Criteria for marking Marks the diagram Correct shape ü   Correct direction ü   Field lines not touching each ü   other or entering the spheres (3) 7.2.3 Tx = (0.07)cos60ºP = 0.035 NP (2) !!!   !! 7.2.4 F =   P !! (!×!"! )(!.!×!"!! )(!.!×!"!! ) 0.035P = P !! r = 0.012  m x = 0.012  mP OR ! !"   E =   ! !.!"# either 𝐸 = !×!"!! P = 729166.67 N.C-1 P   ! 𝐸 = 𝑘 !!  !.!×!"!! 729166.67=  𝑘 P !! r = 0.012 mP (4) [15] QUESTION 8 8.1 The potential difference across the ends of the conductor is directly proportional to the current flowing to the conductor at constant temperature. üü (2) Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences (P1) 9 Preparatory Examination September 2018 NSC – Marking Guideline 8.2.1 OPTION 1 OPTION 2 ! ! R !Ω   = ! P R !Ω   = ! P ! ! 8= 8= !.! !.! V = 2.4  V  V = 2.4  V ! !×! R !Ω   = R !     =   = 2.67  ΩP ! !!! !.! ! 4   = ! R           = ! !.! I     = 0.6  AP 2.67 = P ! I!"!   = (0.3 + 0.6)P = 0.9  AP I         = 0.9  AP OPTION 3 Current through 4 Ω = 2 x 0,3 P   = 0,6A P   ∴ A1 = (0,6 + 0,3) P= 0,9A P (4) 8.2.2 Positive marking from Question 8.2.2 ! R !   = P ! !".! = P !.! = 17.33  ΩP (3) 8.2.3 Positive marking from Question 8.2.2 !×! R ! =   !!! = 2.67  ΩP R ! = (2.67 + 17.33)  P = 20  ΩP (3) 8.2.4 Positive marking from Questions 8.2.1 and 8.2.3 𝑒𝑚𝑓 = 𝐼 𝑅 + 𝑟 P 19P = 0.9 20 + r P 𝑟             = 1.11 ΩP (4) 8.3.1 DecreaseP (1) 8.3.2 Total resistance of the circuit decreases,  P therefore current increases.  P More volts will be lost,  P and Vext decreases (Vext = ε – Ir)P (4) [21] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences (P1) 10 Preparatory Examination September 2018 NSC – Marking Guideline QUESTION 9 9.1 Slip ringsP (1) 9.2 Criteria • Starts at 50 VP • Ends at 50 VP • Ends at 0.08X10-3 sP • 2 cyclesP (4) 9.3 Increase the rotation speed of the coil. PP (2) I max 9.4.1 I rms = P 2 6.27 I rms = P 2 I rms = 4.43 AP (3) 9.4.2 OPTION 1   OPTION 2 V ! R = max PP 𝐼!"# =   !"# P I max ! Vmax V P 52.6 = P 4.43 = rms P 6.27 52.6 Vmax = 329.80V P Vrms = 233.018V Vmax Vrms = P 2 Vmax 233.018 = P 2 Vmax = 329.54V P   (5) [15] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences (P1) 11 Preparatory Examination September 2018 NSC – Marking Guideline QUESTION 10 10.1 The energy of the photons of red light is greaterP than the work function of the metal in the photocell.P OR The frequency of red light is higherP than the threshold/cut-off frequency of the metal in the photocell. P (2) 10.2.1 IncreaseP (1) 10.2.2  Stays the same P Since intensity is unchangedP, same number of photons reach cathode. P Therefore number of photo electrons emitted is unchanged. P (4) 10.3 E = Wo + Ek (max) P                     c h = hfo + ½mv2                               λ   6,63 x 10 -34 × 3 × 10 8 P  6,63 x 10-34(f ) P + ½(9,11 x 10-31)(5,06 x 105)2P   = o       4,5 × 10 −7 P     fo = 6,67 x 1014 Hz P (6)         [13] GRAND TOTAL: 150     Copyright Reserved Please Turn Over