# Physical Sciences Grade 12 Western Cape November 2016 Memo > Grade 12 Physical Sciences memorandum from the Western Cape November 2016, 10 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/wp-metro-central-physics-fisika-memo/3dec7cde-4802-45ea-923a-8db7cf978bd0 - Updated: 2026-09-27 - Subject: Physical Sciences - Grade: Grade 12 - Document type: Memorandum - Year: 2016 - Exam period: Western Cape November - Curriculum: CAPS - Pages: 10 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 12](https://www.hlayiso.com/library/grade-12) › [Physical Sciences](https://www.hlayiso.com/library/grade-12/physical-sciences) › [Physical Sciences Grade 12 Western Cape November 2016 Memo](https://www.hlayiso.com/library/wp-metro-central-physics-fisika-memo/3dec7cde-4802-45ea-923a-8db7cf978bd0) ## About this document Physical Sciences · Grade 12 · Western Cape November · 2016. Memorandum, 10 pages. Read online or download the PDF. Read it free online on the page above; download the PDF from there. ## Document text Text extracted from the PDF. Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com Page |1 WCED Metro Central Common Paper Physical Sciences Paper 1 September 2016 MARKING MEMORANDUM QUESTION 1 1.1 D  1.2 D  1.3 C 1.4 A  1.5 B  1.6 B  1.7 A  1.8 C  1.9 D  1.10 C  [20] ### Page 2 Downloaded from hlayiso.com Page |2 QUESTION 2 2.1 ⃗  ⃗⃗ 1 mark per arrow and label FN subtract 1 mark for each of Ty ⃗  the following errors: Ffs  no dot shown Tx  T shown with its components (unless components in ⃗ dashed lines) Fg (4) 2.2 When a resultant (net) force acts on an object, the object will accelerate in the direction of the force. This acceleration is directly proportional to the force and inversely proportional to the mass of the object. 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. (2 or 0) (2) 2.3  ( )  Vertical forces; taking up as positive ⃗ ⃗⃗ ⃗⃗ ⃗⃗   ( )( ) ………(A) (5) 2.4 Horizontal forces; taking left as positive ⃗ ⃗⃗ ⃗  N  ……….. (B) (A) / (B):  Sub into (B) OR Subst into (A) ( ) T sin(48,8o) = 137,06  T = 182,16 N (4) 2.5.1 DECREASES  (1) 2.5.2 From: . The angle ( ) increases , so the vertical component of the tensional force ( ) will increase . OR From: FN + Ty = Fg  increases/ Ty increases  The parcel will not push as hard into the table surface  so the normal force will decrease in magnitude. (2) [18] ### Page 3 Downloaded from hlayiso.com Page |3 QUESTION 3 3.1 ⃗ ⃗ ⃗  ⃗ ( )( ) ⃗ up (3) + ve 3.2  ( )( )  5% energy loss so this represents 95% of the energy after the bounce.  (4) + ve  3.3 (EP + EK )TOP = (EP + EK )BOTTOM  [mgh + ½(0,05)(3)2] top = ( 0 + 0,91)  (all subst) h = 0,685/ (0,05)(9,8) = 1,398 m = (0,4 m)  OR Work Energy Theorem OR Wnc = Ep + Ek (3) 3.4  Position (m)   time (s)  label of both axes (4) [14] ### Page 4 Downloaded from hlayiso.com Page |4 QUESTION 4 4.1 The total linear momentum of an isolated (closed) system remains constant (is conserved). OR In an isolated system  The total linear momentum before collision equals the total linear moment after collision.  (2) 4.2 Linear momentum conservation: Take “towards Orion” as the positive direction: ∑⃗ ∑⃗  ⃗ ⃗ ⃗ ( )( ) ( ) ( ) ( ) 10 mA = 1,8 x 1020 + 7,2 x 1019 mA = 2,52 x 1019 kg Take “towards Orion” as the positive direction: Take “towards Orion” as the positive ∑⃗ ∑⃗  direction: ∑⃗ ∑⃗  ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ ( )( ) ( ) ( ) ( )( ) ( ) ( ) (A) mB = 4 mA - 9 x 1019 (A) Mass conservation: Mass conservation:  (B)  (B) sub (A) into (B): sub (A) into (B): mA + 4 mA - 9 x 1019 = 3,6 x 1019 5 mA = 3,6 x 1019 + 9 x 1019  sub into (B): mA = 2,52 x 1019 kg  (5) 4.3 Take “towards Orion” as the positive direction: ⃗ ⃗  ⃗ (⃗ ⃗) ⃗ ( )( ) /kg·m·s-1 towards Orion  (magnitude + direction) (3) 4.4 3,6 x 1019 - (2,52 x 1019)   sub and ( )( )( ) ( )  sub converted (4) [15] ### Page 5 Downloaded from hlayiso.com Page |5 QUESTION 5 5.1 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 net (resultant) force is equal to the change in the object's kinetic energy.  (2) 5.2  ( )( )( )   OR work due to a conservative forces is equal to negative change in potential energy associated with that conservative force: ( ) ( )( ) ( )  (4) 5.3    ( )(  )  OR  ( ) ( ) ( )( )(( )) ( )((  ) )  (6) 5.4.1 REMAINS THE SAME  (1) 5.4.2 The gravitational force is conservative (non-contact) force, so the work done by the gravitational force will not depend on the path taken.  The starting and ending points are the same. Therefore the work done by the gravitational force will remain the same. (2) [15] ### Page 6 Downloaded from hlayiso.com Page |6 QUESTION 6 6.1.1 The apparent change in frequency in sound heard due to the relative motion between listener and/or source. (2) 6.1.2 ( )   ( )  (335 + vS) = 335 vS = 335 – 0,93 x 335 vS = = 25,22 m·s-1  (4) 6.2.1 Absorption (line spectrum)  (1) 6.2.2 Red-shift  (1) 6.2.3 Away from  (1) [9] ### Page 7 Downloaded from hlayiso.com Page |7 QUESTION 7 7.1 The force of attraction or repulsion between two charges is directly proportional to the product of their charges  and inversely proportional to the square of the distance between them/ their centres. (2)  7.2  (  ) (4) + ve 7.3 ⃗ L on K   ⃗ J on K   Fnet (3) 7.4 MAGNITUDE: ( ) ( )  √ ( )   DIRECTION tan  = 2,  = 63,43° Angle (show tan or other method) / BEARING 206,57O (4) 7.5 The electric field at a point is the electrostatic force experienced per unit positive charge  placed at that point.  (2) + ve 7.6    (3) [18] ### Page 8 Downloaded from hlayiso.com Page |8 QUESTION 8 8.1 emf  (1) 8.2 Load voltage OR external voltage OR terminal voltage  (1) 8.3    (3) 8.4 3Ω  same V over each resistor and I is inversely proportional to R  (3) 8.5      VP = IP RP = (4,5)(1,2)  18 – (0,9 + 5,4) = 5,4 V     R = V/I = 11,7/ 4,5 = 2,6   (7) 8.6 Temperature  (1) 8.7 RP increases when S2 is opened so Rcir increases  so Icir / current strength through ammeter decreases  so Vint (= Ir) decreases (r constant)  so Vext increases (Vext = ε - Vint)  (4) [20] ### Page 9 Downloaded from hlayiso.com Page |9 QUESTION 9 9.1 0,10 s  (1) 9.2 √ √ (2) 9.3.1 (3) 9.3.2 TOO DIM Vrms for bulb = 100 V BUT Vrms of generator – 60 V.  (2) 9.4 I (A) 0,10 0,20 0,05 0,15 Time (s) (2) [10] ### Page 10 Downloaded from hlayiso.com P a g e | 10 QUESTION 10 10.1 Planck’s constant  (1) 10.2 Threshold frequency ( ) is the minimum frequency of light needed to emit (eject) electrons  from the surface of a certain metal / material. (2) 10.3  ( )( ) (3) 10.4.1 The greater brightness would: - increase the number of photoelectrons 10.4.2 - but would have no effect on their kinetic energies / Remain the same (2) 10.5   ( )  OR E = W o + Ek Ek = E - Wo  2 ½ mv = hf - Wo ½ (9,11 x 10 ) v  = (6,63 x 10-34)(5 x 1015) - (9,282 x 10 -19)  -31 2 v = 2,29 x 106 m·s-1  OR Learners can calc the gradient of the graph which = 6,67 x 10-34 and then use above method. 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