Physical Sciences Grade 12 P1 KwaZulu Natal Mock 2020 QP

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Downloaded from hlayiso.com Downloaded from Stanmorephysics.com NATIONAL SENIOR CERTIFICATE GRADE 12 PHYSICAL SCIENCES P1 (PHYSICS) PREPARATORY EXAMINATIONS SEPTEMBER 2020 MARKING GUIDELINE Time: 3 Hours Marks: 150 This marking guideline has 17 pages.

Downloaded from hlayiso.com Physical Sciences/P1 2 September 2020 Downloaded from Stanmorephysics.com NSC SECTION A QUESTION 1 1.1 B (2) 1.2 D (2) 1.3 D (2) 1.4 D (2) 1.5 D (2) 1.6 A (2) 1.7 D (2) 1.8 C (2) 1.9 C (2) 1.10 A (2) 2 x 10 = [20]

Downloaded from hlayiso.com Physical Sciences/P1 3 September 2020 Downloaded from Stanmorephysics.com NSC SECTION B QUESTION 2 2.1 Normal force is the force or the component of a force which a surface exerts on an object with which it is in contact, and which is perpendicular to the surface.  (2) 2.2 FAy FA FAx N N  Fgx fk  fk  Fgy Fg Accept the following symbols N FN/Normal/Normal force fk Kinetic friction force/f/Ff/fr FA F/Fapplied Fg W/78,4 N Notes • Mark is awarded for label and arrow. • Do not penalise for length of arrows. • Deduct 1 mark for any additional force. • If force(s) do not make contact with body/dot : Max:3/4 • If arrows missing but labels are there: Max:3/4 (4) 2.3 Considering the forces parallel to the plane:: Fnet = ma Fnet = 0  FAx – fk - Fgx = 0 FAcos fk + Fgx 51cos = 1 + 8(9,8)sin30o o  (3)

Downloaded from hlayiso.com Physical Sciences/P1 4 September 2020 Downloaded from Stanmorephysics.com NSC 2.4 POSITIVE MARKING FROM 2.3 Consider the forces perpendicular to the plane: Fnet = ma Fnet = 0  FAy + N – Fgy = 0 FAsin N = Fgy 51sin 37,98o + N = 8 x 9,8cos(30o) ∴ N = 36,51 N (3) 2.5 Increases  (1) [13] QUESTION 3 3.1 3.1.1 UPWARDS IS POSITIVE UPWARDS IS NEGATIVE       (3) 3.1.2 OPTION 1 UPWARDS IS POSITIVE UPWARDS NEGATIVE vf2 = vi2 + 2aΔy vf2 = vi2 + 2aΔy 0 = 122 + 2(–9,8)Δy 0 = (–12)2 + 2(9,8)Δy Δy = 7,35 m Δy = – 7,35m ∴Max height = 50+7,35 ∴Maximum height = 50 +7,35 = 57,35m = 57,35m OPTION 2 POSITIVE MARKING FROM 3.1.1 UPWARDS IS POSITIVE UPWARDS IS NEGATIVE   ]  Δy = 7,35 m Δy = –7,35 m ∴Max height = 50+7,35 ∴Max height = 50+7,35 = 57,35m = 57,35m

Downloaded from hlayiso.com Physical Sciences/P1 5 September 2020 Downloaded from Stanmorephysics.com NSC OPTION 3 POSITIVE MARKING FROM 3.1.1 UPWARDS IS POSITIVE Δy= viΔt + ½aΔt2 Δy= (12)( 1,23) + ½(–9,8)(1,23)2 Δy = 7,35 m ∴Maximum height = 50+7,35 = 57,35m UPWARDS IS NEGATIVE Δy= viΔt + ½aΔt2 Δy= (–12)( 1,23) + ½(9,8)(1,23)2 Δy = –7,35 m ∴Maximum height = 50+7,35 = 57,35m OPTION 4 (K + U)1 = (K + U)2 (½mv2 + mgh)1 = (½mv2 + mgh)2 ½m(12)2 + m(9,8)(50)  = ½m(0)2 + m(9,8)h  h = 57,35 m (4) 3.1.3 OPTION 1 UPWARDS POSITIVE vf = vi + aΔt vf = 12 + (–9,8) (4)  vf = –27,20 m•s-1 ∴ velocity of the ball is 27,20 m•s-1 downwards UPWARDS NEGATIVE vf = vi + aΔt vf = –12 + (9,8) (4)  vf = 27,20 m•s-1 ∴ velocity of the ball is 27,20 m•s-1 downwards OPTION 2 UPWARDS POSITIVE Δy= viΔt + ½aΔt2 Δy= (12)(4) + ½(–9,8)(4)2 Δy = –30,40 m

Downloaded from hlayiso.com Physical Sciences/P1 6 September 2020 Downloaded from Stanmorephysics.com NSC  –  vf =–27,20 m•s-1 ∴ velocity of the ball is 27,20 m•s-1 downwards UPWARDS NEGATIVE Δy= viΔt + ½aΔt2 Δy= (–12)(4) + ½(9,8)(4)2 Δy = 30,40 m  –  vf =27,20 m•s-1 ∴ velocity of the ball is 27,20 m•s-1 downwards OPTION 3 UPWARDS POSITIVE Δy= viΔt + ½aΔt2 Δy= (12)(4) + ½(–9,8)(4)2 Δy = –30,40 m vf2 = vi2 + 2aΔy vf2 = (12)2 + 2(–9,8)( –30.40)  vf =27,20 m•s-1 ∴ velocity of the ball is 27,20 m•s-1 downwards UPWARDS NEGATIVE Δy= viΔt + ½aΔt2 Δy= (–12)(4) + ½(9,8)(4)2 Δy = 30,40 m vf2 = vi2 + 2aΔy vf2 = (–12)2 + 2(9,8)( 30.40)  vf =27,20 m•s-1 ∴ velocity of the ball is 27,20 m•s-1 downwards (3)

Downloaded from hlayiso.com Physical Sciences/P1 7 September 2020 Downloaded from Stanmorephysics.com NSC 3.1.4 OPTION 1 UPWARDS POSITIVE Δy= viΔt + ½aΔt2 If candidate calculated 30.40 m in Q 3.1.3. above award 2 marks Δy= (12)(4) + ½(–9,8)(4)2 here. Δy = –30,40 m ∴Position above the ground = 50 – 30,40 = 19,60 m UPWARDS NEGATIVE Δy= viΔt + ½aΔt2 Δy= (–12)(4) + ½(9,8)(4)2 Δy = 30,40 m ∴Position above the ground = 50 – 30,40 = 19,60 m OPTION 2 POSITIVE MARKING FROM 3.1.3 UPWARDS POSITIVE  –  Δy = – 30,40 m ∴Position above the ground = 50 – 30,40 = 19,60 m UPWARDS NEGATIVE  –  Δy = 30,40 m ∴Position above the ground = 50 – 30,40 = 19,60 m

Downloaded from hlayiso.com Physical Sciences/P1 8 September 2020 Downloaded from Stanmorephysics.com NSC OPTION 3 POSITIVE MARKING FROM 3.1.3 UPWARDS POSITIVE vf2 = vi2 + 2aΔy (–27,20)2 = (12)2 + 2(–9,8) Δy  Δy = –30,40 ∴Position above the ground = 50 – 30,40 = 19,60 m UPWARDS NEGATIVE vf2 = vi2 + 2aΔy (27,20)2 = (–12)2 + 2(9,8) Δy  Δy = 30,40 ∴Position above the ground = 50 – 30,40 = 19,60 m (4)

Downloaded from hlayiso.com Physical Sciences/P1 9 September 2020 Downloaded from Stanmorephysics.com NSC 3.2  25  Vel oci  ty 12 (m• s-1)  1,23 time(s)  OR Vel oci ty  (m• s-1)  1,23 time(s)  –12   –25 CRITERIA FOR MARKING OF GRAPH Correct shape (the three lines must have same  gradient) Indication of initial velocity (12 m.s–1)  Indication of velocity of bounce (25 m.s–1)  1.23s  First line longer below x-axis than above  (5) If 25 m.s–1 line beyond x-axis then -1 Do not penalise if dotted lines are solid [19]

Downloaded from hlayiso.com Physical Sciences/P1 10 September 2020 Downloaded from Stanmorephysics.com NSC QUESTION 4 4.1 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. (2) 4.2 4.2.1 (K + U)1 = (K + U)2  2 2 (½mv + mgh)1 = (½mv + mgh)2 (½(1,005)v2 + 0) = 0 + (1,005)(9,8)(0,5)  v1 = 3,13 m•s-1 ∴The velocity of the block – bullet system immediately after collision is 3,13 m•s-1 (3) 4.2.2 POSITVE MARKING FROM 4.2.1 Σpi = Σpf  m1v1i + m2v2i = (m1+ m2)vf (0,005)v1i + (1)(0)  = (0,005+ 1)(3,13)  v1i = 629,13 m•s-1  (4) 4.3 Less than (1) [10] QUESTION 5 5.1.1 Total mechanical energy of an isolated system remains constant (2) 5.1.2 No, (1) 5.1.3 OPTION 1 (K + U)P = (K + U)Q – Wf OR  Wnc = ΔEp + ΔEk (½mv2 + mgh)P = (½mv2 + mgh)Q – W f [½(1200)(0,8)2 + (1200)(9,8)(1,8)]P = [½(1200)(4)2 + 0 – Wf]Q Wf = –11952 J Wf = Ff Δx cos 180  –11952 = –Ff 2,2 Ff =5432,73 N

Downloaded from hlayiso.com Physical Sciences/P1 11 September 2020 Downloaded from Stanmorephysics.com NSC OPTION 2 Wnet = ΔK Wf + W g = ½mvf2 - ½mvi2  Wf + mgh = ½m(vf2- vi2) Wf + mgh = ½mvf2 - ½mvi2 Wf + 1200(9,8) (2,2) cos324,90o = ½(1200)(4)2 - ½(1200)(0,8)2  (accept 35.1o) Wf = –11952 J Wf = Ff Δx cos 180  –11952 = –Ff 2,2 Ff =5432,73 N (5) 5.2 5.2.1 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.  Note: -1 mark for each key word/phrase omitted in the correct context. (2) 5.2.2 W = ΔK net  2 2 Wf + W g + W N = ½mvf – ½mvi  8000cos180oΔx + 0 + 0 = ½(1000)(0) – ½(1000)(352)  Δx = 76,56 m (4) 5.2.3 W = ΔK net Wf + W g + W N = ½mvf2 – ½mvi2 8000cos180o(30) + 0 + 0 = ½(1000)vf2 – ½(1000)(352)  vf = 27,29m•s-1 (3) 5.2.4 INCREASE On a rainy day the road surface would have less frictional force needed for the braking.  (2) [19]

Downloaded from hlayiso.com Physical Sciences/P1 12 September 2020 Downloaded from Stanmorephysics.com NSC QUESTION 6 6.1 Doppler Effect.  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.  OR Doppler Effect.  It is the change in the observed frequency of a sound wave when the source of sound is moving relative to the listener.  (3) 6.2 Towards (1) 6.3 v  vL fL  fs  v  vs    OR    fs = 800 Hz (range: 800 – 833.33 Hz) v = fsλ 340 = 800λ λ = 0,425 m (range: 0,425 – 0,408) (7) 6.4 Greater than.  The reflected waves are moving toward the ambulance.  (2) 6.5 ● blood flow meter  ●echocardiograms.  ●ultrasound technology ●monitor pregnancies ●examine soft tissue injuries. ( any two) (2) [15]

Downloaded from hlayiso.com Physical Sciences/P1 13 September 2020 Downloaded from Stanmorephysics.com NSC QUESTION 7 7.1.1 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: -1 mark for each key word/phrase omitted in the correct context. 7.1.2 Accept: Density around R maybe greater than Q Q R Marking Criteria Correct Shape  Field lines do not cross  Direction of field lines correct  (3) T 7.1.3 FE  FG  (3) 7.1.4 CONSIDERING THE HORIZONTAL FORCES: Fnet = 0 Tsin = Fe  Tsin =   Tsin = 0,216 …………………………………… 1 CONSIDERING THE VERTICAL FORCES: Fnet = 0 Tcos = mg Tcos = 8 x 10-2 (9,8)  ………………………. 2

Downloaded from hlayiso.com Physical Sciences/P1 14 September 2020 Downloaded from Stanmorephysics.com NSC 1 2 tan = 0,26/(8 x 10-2 (9,8)) = 15,40o OPTION 2  F=  = 0.216N Fg = mg = 8 x 10-2 x 9,8 = 0,784 N tan = F/Fg = 0,216/0,784 = 15,40o (6) 7.1.5 POSITIVE MARKING FROM 7.2 Tcos = 8 x 10-2 (9,8) o Tcos = 8 x 10-2 (9,8)  T = 0,81 N (3) 7.2.1 Electric field is a region of space in which an electric charge experiences a force.  (2) (accept : force experienced per unit positive charged when placed at that point) 7.2.2   OR ( E1 = E2)  2E1 =

Downloaded from hlayiso.com Physical Sciences/P1 15 September 2020 Downloaded from Stanmorephysics.com NSC 2 = 2 = = 4q1 = q2 4 x 0,5 x 10-6 = q2 q2 = 2 x 10-6 C (4) [21] QUESTION 8 8.1 The potential difference across a conductor is directly proportional to the current in the conductor at constant temperature  NOTE If constant temperature is omitted -1 mark (2) 8.2 8.2.1 Total Resistance = Rseries + Rparallel Total Resistance = 4 + ( )-1 Total Resistance = 11,20 I=  I=  I = 1,07A (5)

Downloaded from hlayiso.com Physical Sciences/P1 16 September 2020 Downloaded from Stanmorephysics.com NSC 8.2.2 POSITIVE MARKING FROM 8.2.1 V4Ω = IR V4Ω = 1,07 x 4 V4Ω = 4,28 V V// = 12 - V4Ω V// = 12 – 4,28 V// = 7,72 V I10Ω = I10Ω =  I10Ω = 0,43 A P = I2R V10Ω = IR V10Ω = 0,43 x 10 P = (0,43)2 (10)  V10Ω = 4,3 V P = 1,85 W  P=  P = VI P = 4,3 X 0,43 P=  P = 1,85 W  P = 1,85 W  (5) 8.2.3 POSITIVE MARKING FROM 8.2.1 W = V.IΔt V2 W t R 122 W = I2Rt  x 30 11.2  385,71J = 12 x 1.07 x 30 = 1,072x 11,2 x 30 = 385,20 J = 384,69 J (3) 8.3 Decreases If the 12 Ohm resistor stops working, the total resistance would increase. Resistance is inversely proportional to the current. (3) [18]

Downloaded from hlayiso.com Physical Sciences/P1 17 September 2020 Downloaded from Stanmorephysics.com NSC QUESTION 9 9.1 AC (generator)  (1) 9.2 Slip rings (1) 9.3 The AC potential difference/voltage which dissipates the same amount of energy as DC.  OR (The rms value of AC is) the DC potential difference/voltage which dissipates the same amount of energy as AC.  (2) 9.4 V Max V RMS  2  V RMS 2  1  VRMS = 0,71 V (3) 9.5 0,04 s. (1) 9.6 Positive marking from Q 9.4 PAve  V RMS I RMS  I P Ave  V RMS ( 2 ) Max 2 P Ave  0,71( 2 ) PAve = 1,00 W  (3) 9.7 Any two: (i) easier to generate and transmit from place to place  (ii) easier to convert from a.c. to d.c. than the reverse  (iii) voltage can be easily changed by stepping it up or down (iv) high frequency used in a.c. make it more suitable for electric motors (2) [13] TOTAL: 150