Physical Sciences Grade 12 Paper 1 Limpopo Mock 2022 Memo

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Downloaded from hlayiso.com Downloaded f rom St anmorephysics. com ‘ NATIONAL SENIOR CERTIFICATE GRADE 12 PHYSICAL SCIENCES: PHYSICS (P1) SEPTEMBER 2022 MARKING GUIDELINES MARKS: 150 This marking guidelines consist of 15 pages. Copyright reserved please turn over

Downloaded from hlayiso.com Downloaded Physical f rom St anmorephysics. Sciences/P1 2 com LimpopoDoE/September 2022 Marking Guidelines QUESTION 1 1.1 B ✓✓ (2) 1.2 A ✓✓ (2) 1.3 C ✓✓ (2) 1.4 A ✓✓ (2) 1.5 B ✓✓ (2) 1.6 D ✓✓ (2) 1.7 D ✓✓ (2) 1.8 A ✓✓ (2) 1.9 C ✓✓ (2) 1.10 A ✓✓ (2) [20] Copyright reserved please turn over

Downloaded from hlayiso.com Downloaded Physical f rom St anmorephysics. Sciences/P1 3 com LimpopoDoE/September 2022 Marking Guidelines QUESTION 2 2.1.1 The force that opposes the tendency of motion of a stationary object relative to a surface. ✓✓ (2) 2.1.2 N fs ● Accept the following symbols: N✓ FN/Normal force/Fsurface on crate w fs ✓ f/Ff/frictional force/static frictional force w✓ Fg/mg/weight/gravitational force/FEarth on crate Take uphill as POSITIVE Notes: • Mark is awarded for label and arrow. • Do not penalize for length of arrows • Deduct 1 mark for any additional force. • If force(s) do not make contact with dot/body: 2/3 • If arrows missing: 2/3 (3) 2.1.3 Fnet = ma max Any one ✓ ∴fs + (-mgsin θ) = mamax µs฀ FN - mgsin θ = mamax µs฀ mgcos θ - mgsinθ = mamax max ∴fs - m(9,8)(sin 10°) = mamax …..① max But fs = µ𝑠 N ∴fsmax = µs (mgcos θ)….② Subst. ② into ① : ∴µs (mgcos θ) - m(9,8)(sin 10°) = m(amax) ∴(0,35) m(9,8)( cos 10°)✓-m(9,8)( sin 10°) = m(amax) (÷m) : ∴(0,35)(9,8)( cos 10°) ✓– (9,8)( sin 10°) ✓= amax ∴the maximum acceleration is 1,676 m฀s-2 ✓ (5) Copyright reserved please turn over

Downloaded from hlayiso.com Downloaded Physical f rom St anmorephysics. Sciences/P1 4 com LimpopoDoE/September 2022 Marking Guidelines 2.2.1 Marking criteria -1 mark for each key word/phrase omitted in the correct context. Each body in the universe attracts every other body with a (gravitational) force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. ✓✓ OR: Every particle in the universe attracts every other particle with a (gravitational) force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. ✓✓ (2) 2.2.2 GME gE = 2 = 9,8 m฀s-2 RE GMM gM = 2 ✓ RM M G ( E) 153 = ✓ 𝑅 2 ( 5𝐸 ) 25 GME = 2 153 RE 25 = (9,8) 153 = 1,60 m฀s-2 ✓ (3) (downwards) 2.2.3 Equal to ✓ (1) [16] Copyright reserved please turn over

Downloaded from hlayiso.com Downloaded Physical f rom St anmorephysics. Sciences/P1 5 com LimpopoDoE/September 2022 Marking Guidelines QUESTION 3 3.1 Motion during which the only force acting on an object is the gravitational force.✓✓ (2) 3.2 OPTION 1: UPWARDS POSITIVE: For stone X: For stone Y: 1 2 1 2 ∆y = vi ∆t + a(∆t) ✓ ∆y = vi ∆t + a(∆t) 2 2 1 2 49 1 0 = (3v)(10) + (-9,8) (10) ✓ 0 = ( )∆t + (-9,8) (∆t)2 ✓ 2 3 2 0 = 30v – 490 49 0 = ∆t ( -4,9∆t) 490 = 30v 3 49 ∆t = 0 s or ∆t = 3,333 s v = ( 3 ) m฀s-1 ∆t = 3,33 s (3,333 s) ✓ (4) DOWNWARDS POSITIVE: For stone X: For stone Y: 1 2 1 2 ∆y = vi ∆t + a(∆t) ✓ ∆y = vi ∆t + 2a(∆t) 2 1 49 1 0 = (3v)(10) + (9,8) (10)2 ✓ 0 = (− )∆t + 2(9,8) (∆t)2 ✓ 2 3 0 = 30฀v + 490 49 0 = ∆t (− +4,9฀∆t) -490 = 30฀v 3 49 ∆t = 0 s or ∆t = 3,333 s v = (− 3 ) m฀s-1 ∆t = 3,33 s (3,333 s) ✓ OPTION 2: UPWARDS POSITIVE: For stone Y: 49 For stone X: vBi = 3 m฀s-1 or 16.333 m฀s-1 Consider upward motion: Consider upward motion: vf = vi + a∆t✓ vf = vi + a∆t 0 = vi + (-9,8)(5)✓ 0 = 16,333 + (9,8)∆t✓ vi = 49 m฀s-1 ∆t = 1,666(1,67 s) ∆t(total) = 2 x 1,666 = 3,33 s Copyright reserved please turn over

Downloaded from hlayiso.com Downloaded Physical f rom St anmorephysics. Sciences/P1 6 com LimpopoDoE/September 2022 Marking Guidelines DOWNWARDS POSITIVE: For stone X: For stone Y: 49 Consider upward motion: vBi = - 3 m฀s-1 or -16.333 m฀s-1 vf = vi + a∆t✓ Consider upward motion: 0 = vi + (9,8)(5)✓ vf = vi + a∆t vi = -49 m฀s-1 0 = -16,333 + (9,8)∆t✓ ∆t = 1,666(1,67 s) ∆t(total) = 2 x 1,666 = 3,33 s 3.3 POSITIVE MARKING FROM QUESTION 3.2: OPTION 1: UPWARDS POSITIVE: For stone Y For stone X 1 2 1 2 ∆y = vi ∆t + a(∆t) ✓ ∆y = vi ∆t + a(∆t) 2 2 5 2 49 5 1 2 1 = (49)(5) ✓+ 2(-9,8)(5) ✓ H=( ) (3) + 2(-9,8)(3) ✓ 3 245 245 245 = = 9( )✓ = 9H H=( )m 2 18 18 DOWNWARDS POSITIVE: For stone Y For stone X 1 2 1 2 ∆y = vi ∆t + 2a(∆t) ✓ ∆y = vi ∆t + a(∆t) 2 49 5 1 5 2 1 2 H = (− ) (3) + 2(9,8)(3) ✓ = (-49)(5) ✓+ 2(9,8)(5) ✓ 3 245 245 245 H = (− )m =- = 9(− )✓ = 9H 18 2 18 OPTION 2: For stone X UPWARDS POSITIVE: 1 ∆y = vi∆t + a∆t2 2 1 For stone Y = (16.333)(1,666) + (9,8)(1,666)2✓ 2 1 = 13,61 m = H ∆y = vi∆t + a∆t2✓ 2 13,61 X 9 = 122,49 m = 9H(122,5 m)✓ 1 = (49)(5) ✓+ (-9,8)(5)2✓ 2 = 122,5 m Copyright reserved please turn over

Downloaded from hlayiso.com Downloaded Physical f rom St anmorephysics. Sciences/P1 7 com LimpopoDoE/September 2022 Marking Guidelines DOWNWARDS POSITIVE: For stone X 1 ∆y = vi∆t + a∆t2 For stone Y 2 1 1 = (-16.333)(1,666) + (9,8)(1,666)2✓ ∆y = vi∆t + a∆t2✓ 2 2 = - 13,61 m = H 1 = (-49)(5) ✓+ (9,8)(5)2✓ -13,61 X 9 = -122,49 m = 9H(122,5 m)✓ 2 = -122,5 m (5) 3.4 POSITIVE MARKING FROM QUESTION 3.2: 49 49 Velocity (m฀s-1 ) Velocity (m฀s-1 ) 0 5 10 0 5 10 Time (s) Time (s) -49 -49 Marking criteria: • Correct shape (Should intersect t-axis) ✓ • Final velocity and initial velocity shown ✓ • 5 s shown for maximum height ✓ (3) [14] Copyright reserved please turn over

Downloaded from hlayiso.com Downloaded Physical f rom St anmorephysics. Sciences/P1 8 com LimpopoDoE/September 2022 Marking Guidelines QUESTION 4 4.1 The product of an object’s mass and its velocity. ✓✓ (2 or 0) (2) 4.2 The total (linear) momentum of an isolated system remains constant (is conserved). ✓✓ (2) 4.3 OPTION 1 OPTION 2 ∑ pi = ∑ pf Any one✓ ∆pA = - ∆pB Any one✓ mA vAi +mB vBi = mA vAf + mB vBf mA (vAf − vAi ) = mB (vBf - vBi ) mA (4) + mB (-1) ✓= mA (1) + mB (3)✓ mA (1-4) ✓ = -mA (3-(−1)) ✓ 4mA - mA = 3mB + mB mA (-3) = −mB (4) 3mA = 4mB ✓ mA (-3) = mB (4) ✓ mA 4 mA -4 4 mB = 3 = = mB -3 3 mA : mB = 4:3 mA : mB = 4:3 Marking criteria: • Formula • Right hand substitution into formula • Left hand substitution into formula • This step: mA (-3) = mB (4) (4) ✓ 4.4 pi ✓ pf ∆p ✓ (3) Criteria mark • Large initial momentum in the same direction as final momentum ✓ • Small final momentum in the same direction as initial momentum ✓ • Change in momentum in the opposite direction ✓ [11] Copyright reserved please turn over

Downloaded from hlayiso.com Downloaded Physical f rom St anmorephysics. Sciences/P1 9 com LimpopoDoE/September 2022 Marking Guidelines QUESTION 5 5.1 The net work done on an object is equal to the change in the object’s kinetic energy. ✓✓ OR: The work done on an object by the net force is equal to the change in the object’s kinetic energy. ✓✓ (2) OPTION 1: 5.2 Wnet = ∆EK OR Wf + Ww + WN = ∆EK Any one✓ 1 f∆x฀cos θ + mg∆x(cos θ) + 0 = 2m(vf2 − v2i ) 1 2 2 (45)(∆x)( cos 180°) + (10)(9,8)(∆x)( cos 125°) ✓+ 0 = 2(10)(0 - 8,84 )✓ -45฀∆x – 56,2105฀∆x = -390,728 ∆x = 3,86 m x = 3,86 m✓ (4) NB: The work done by the gravitational force W w can also be calculated as follows: OR: Ww = -∆Ep Ww = mgsinθ∆xcos฀ = (10)(9,8)(sin35°)xcos180° = mg(hi – hf) = (-56,2105x)J =(10)(9,8)(0 – xsin35°) = (-56,2105x)J OPTION 2: Wnc = ∆Ep + ∆Ek 1 = mg(hf – hi) + 2m(vf2 − v2i ) Any one✓ 1 Wf = mg(hf – hi) + 2m(vf2 − v2i ) 1 f∆xcos θ = mg(hf – hi) + 2m(vf2 − v2i ) 1 (45)xcos 180° = (10)(9,8)(xsin35° - 0)✓ +2 (10)(02 - 8,842 )✓ -45x = 56,2105x – 390,728 x = 3,86 m✓ 5.3 N f ● Accepted labels N✓ FN /Normal force/Fnormal f✓ f/Ff / frictional force/ fs w✓ Fg/mg/weight/gravitational force w Copyright reserved please turn over

Downloaded from hlayiso.com Downloaded Physical f rom St anmorephysics. Sciences/P1 10 com LimpopoDoE/September 2022 Marking Guidelines Notes: • Mark is awarded for label and arrow. • Do not penalize for length of arrows • Deduct 1 mark for any additional force. • If force(s) do not make contact with dot/body: 2/3 • If arrows missing: 2/3 (3) 5.4 OPTION 1: OPTION 2: fs = wⅡ = mgsin θ✓ µs = tanθ = tan35° =0,7002 =(10)(9,8)(sin 35°)✓ fs = µsN = µsmgcosθ (ANY ONE)✓ fs = 56,21 N ✓ =(0,7002)(10)(9.8)(cos35°)✓ =56,21 N✓ (3) [12] QUESTION 6 6.1 The apparent 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) 6.2 v = fs λ ✓ 340 = (300)λ✓ λ = 1,13 m ✓ (3) 6.3 ANY ONE: • To monitor the heartbeat of a foetus (unborn baby).✓ • To measure the rate of blood flow.✓ (1) 6.4.1 OPTION 1: OPTION 2: v±vL v- vS fL = ( ) fs ✓ λB = ✓ v±v S fs 340 340-30 ✓ = (340-30) (300) ✓ = 300✓ = 329,032 Hz = 1,03 m✓ v = fL λ 340 = (329,032)λ ✓ (4) λ = 1,033 m ✓ Copyright reserved please turn over

Downloaded from hlayiso.com Downloaded Physical f rom St anmorephysics. Sciences/P1 11 com LimpopoDoE/September 2022 Marking Guidelines 6.4.2 OPTION 1: OPTION 2: v±vL V- VS fL = ( ) fs λA = ( )✓ v±v S fs 340 ✓ 340+30 = (340+30) (300)✓ =( 300 )✓ = 275,676 HZ = 1,23 m ✓ v=fλ (340) = (275,676) λ 340 λ= (3) 275,676 = 1,23 m ✓ 6.5 Less than✓ (1) [14] QUESTION 7 7.1.1 Distance (between the point charges)/medium/air✓ (1) 7.1.2 The electrostatic force is directly proportional to the product of charges.✓ (1) 7.1.3 ∆F NOTE: accept any gradient = 2 ∆Q value from the graph (4-3)X 1012 = 2 ✓ ∆Q 1 X1012 = 1✓ -2 = 1 X 1012 N฀C ✓ (3) 7.1.4 KQ1 Q2 F= ✓ r2 F k 2 = r2 Q 9 x 109 1 X 1012 ✓= ✓ r2 r2 = 9 x 10-3 r = 0,09487 m (0,095 m)✓ KQ2 3 NOTE: If F = is used, then maximum: r2 4 (4) Copyright reserved please turn over

Downloaded from hlayiso.com Downloaded Physical f rom St anmorephysics. Sciences/P1 12 com LimpopoDoE/September 2022 Marking Guidelines 7.2.1 A region of space in which an electric charge experiences a force. ✓✓ (2) 7.2.2 kQ E= ✓ r2 Enet,p = 0 kQ1 kQ2 ✓= r2 r2 (9 X 10 )(8 X 10-6 ) 9 (9 X 109 )(2 X 10-6 ) = ✓ (0,4-d)2 d2 ACCEPT: If 10-6 is omitted since it appears on both sides. d2 (2 X 10-6 ) (0,4-d)2 = -6 (8 X 10 ) = 0,25 d = 0,5 0,4-d d = 0,1333 m ∴The distance is 0,1333 m ✓ (4) 7.2.3 OPTION 1: OPTION 2: Q1 + Q2 Q1 + Q2 Qnew = Qnew = 2 2 8 X 10-6 +2 X 10-6 8 X 10-6 +2 X 10-6 = ✓ = ✓ 2 2 = 5 𝑋 10−6 C = 5 𝑋 10−6 C Q Q n= ✓ n= ✓ e e 5 𝑋 10−6 − 8 𝑋 10−6 5 X 10-6 - 2 X 10-6 n= ✓ n= ✓ −1,6 X 10−19 -1,6 X 10-19 n = 1,875 x 1013 electrons ✓ n = 1,875 x 1013 electrons ✓ (4) [19] Copyright reserved please turn over

Downloaded from hlayiso.com Downloaded Physical f rom St anmorephysics. Sciences/P1 13 com LimpopoDoE/September 2022 Marking Guidelines QUESTION 8 8.1 Emf✓ (1) 8.2.1 Vlost = Ir 24 – 21,6✓ = I(2) I = 1,2 A Itotal = 1,2 A V12 Ω = IR = (1,2)(12)✓ = 14,4 V V10 Ω = 21,6 – 14,4 = 7,2 V V 7,2 I10 Ω = R ✓= ✓= 0,72 A✓ 10 (5) IA = 0,72 A 8.2.2 IX = 1,2✓ – 0,72 = 0,48 A V 7,2 RX = ✓= ✓ I 0,48 = 1,50 Ω✓ (4) 8.3 POSITIVE MARKING FROM QUESTION 8.2.1 P = I2R✓ = (1,2)2✓ (12)✓ = 17,28 W✓ (4) 8.4 Decreases ✓ • Total resistance in the circuit increases. ✓ I • Current in the circuit decreases (I R). ✓ • P = I 2R; when R is constant, P decreases ✓ (4) [18] Copyright reserved please turn over

Downloaded from hlayiso.com Downloaded Physical f rom St anmorephysics. Sciences/P1 14 com LimpopoDoE/September 2022 Marking Guidelines QUESTION 9 9.1.1 From electrical energy to mechanical energy. ✓ (1) 9.1.2 Clockwise. ✓ (2) 9.1.3 (carbon) brush✓✓ (1) 9.1.4 It reverses the direction of the current in the coil after each half-cycle. ✓ (1) 9.1.5 Increases. ✓ The current increases✓ (2) 9.2.1 The rms potential difference is the AC potential difference which dissipates/produces the same amount of energy as an equivalent DC potential difference✓✓ ACCEPT: The rms voltage is the DC potential difference which dissipates/produces the (2) same amount of energy as the equivalent AC potential difference✓✓ 9.2.2 Vrms Vrms = ✓ √2 Vrms 200 = ✓ √2 Vrms = (200)(√2) Vrms = 282,8427 V✓ (3) [12] Copyright reserved please turn over

Downloaded from hlayiso.com Downloaded Physical f rom St anmorephysics. Sciences/P1 15 com LimpopoDoE/September 2022 Marking Guidelines QUESTION 10 10.1.1 The process whereby electrons are ejected from a metal surface when light of suitable frequency is incident on that surface. ✓✓ (2) 10.1.2 The minimum energy that an electron in the metal needs to be emitted from the metal surface.✓✓ (2) 10.1.3 (a) Frequency (of the incident light). ✓ (1) (b) Frequency (of the incident light).✓ (1) 10.1.4 E = hf hc E =λ ✓ (6,63 X 10-34 )(3 X 108 ) ✓ ∴E = -9 (450 X 10 ) ✓ E = 4,42 X 10−19 J ✓ Since photon energy is less than the work function of the metal, so, no emission occurs. ✓ (5) 10.2. Increasing energy 1880 nm ✓ 486 nm ✓ 656 nm ✓ [14] TOTAL: 150 Copyright reserved