Physical Sciences Grade 12 P1 Limpopo November 2016 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 12 PHYSICAL SCIENCES: PHYSICS (P1) PREPARATORY EXAMINATIONS SEPTEMBER 2016 MEMORANDUM MARKS: 150 TIME: 3 hours This memorandum consists of 10 pages. Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 Limpopo DoE September 2016 NSC QUESTION 1 1.1 D (2) 1.2 A (2) 1.3 C (2) 1.4 B (2) 1.5 C (2) 1.6 A (2) 1.7 D (2) 1.8 D (2) 1.9 B (2) 1.10 A (2) [20] QUESTION 2 2.1 2.1.1 When a net force (Fnet) is applied to an object (of mass, m) it accelerates in the direction of the (net) force. The acceleration (a) is directly proportional to the (net) force and inversely proportional to the mass of the object. (2 or 0) OR The net force acting on an object is equal to the rate of change of momentum of the object (in the direction of the force). (2 or 0) (2) 2.1.2  N / FN / Force of surface on trolley  F/Applied force  F / Ff / Friction  w / Fg / weight / gravitational force (4) (Accept the components of Fg INSTEAD of Fg but not both Fg and the components. No arrows = ¾; forces not touching dots = 3/4)    2.1.3 (a) Ff = μkFN = μk (mg Cos 30 ) = 0,2 (33,95) = 6,79 N o (3) (b) Positive marking from 2.1.3 a  Fg// = mgSin 300 = (4)(9,8)sin 30 = 19,6 N Fnet = maR = F + Ff + Fg//    (4) (0,43) = F + (–6,79) + (–19,6) F = 28,11 N  (5) 2

Downloaded from hlayiso.com Physical Sciences/P1 Limpopo DoE September 2016 NSC   2.2 = = 18944,34 N  (4)  [18] QUESTION 3 Any sign convention is accepted, i.e. learners may also work with upwards as positive 3.1   3.1.1 9,8 (m.s–2) – object is falling through the air. (2)  3.1.2 tx = (0,5 + 0,5) = 1 s (1) 3.2 The area under the graph represents the displacement.   displacement = area = ½ base x height = ½ x (–0,5) x 3,9 = –0,98 m height = 0,98 m  (3) 3.3 3.3.1 From the graph, downward motion as POSITIVE.  v f = vi + a ∆t  = –4,9 + (+9,8) (1,7) = 11,76 m.s–1  (3) 3.3.2 Upwards as - Upwards as + ∆y = v i ∆t + 21 a∆t 2 ∆y = v i ∆t + 21 a∆t 2     = (–4,9) (1,7) + ½ (+9,8) (1,7)2 = (4,9) (1,7) + ½ (-9,8) (1,7)2 = 5,83 m  = - 5,83 Δ y = 5,83 m  3

Downloaded from hlayiso.com Physical Sciences/P1 Limpopo DoE September 2016 NSC OR Upwards as - Upwards as +  ∆y = v i ∆t + 21 a∆t 2  ∆y = v i ∆t + 21 a∆t 2 = (- 4,9) (0,7) + ½ (+9,8) (0,7)2  = (4,9) (0,7) + ½ (- 9,8) (0,7)2  = 5,83 m  = - 5,83 Δ y = 5,83 m  (3) 3.4 a (m.s–2)   t (s) (2) [14] 4

Downloaded from hlayiso.com Physical Sciences/P1 Limpopo DoE September 2016 NSC QUESTION 4  4.1 N / FN / Normal Force   0,86 N / f / Ff / friction 0,3 N / Fapp / Applied Force  w / Fg / Weight / Gravitational force (4) 4.2 Fnet = Fapplied + Ff = 0,30 + (–0,86) = –0,56 N = 0,56 N opposite direction of motion/ to the left    (3)  4.3 Work Energy theorem states that, 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)  4.4 W NET = ∆ EK Fnet ∆x Cos ϴ = ½ mvf2 – ½ mvi2   0,56 ∆x Cos 180 = ½ (0,8) (0,2)2 – ½ (0,8) (1,2)2 (0,56) x (∆x) x (–1) = 0,016 – 0,576 ∆x = 1 m  OR W NC = ∆ EK + ∆ EP  f ∆x Cos 180 + FAPPL ∆x Cos 0  = ½ mvf2 – ½ mvi2 + 0  0,86. ∆x Cos 180 + 0,3. ∆x Cos 0 = ½ (0,8) (0,22) – ½ (0,8) 1,22 (-0,86 +0,3). ∆x = 0,016 – 0,576 ∆x = 1 m  (4)   4.5 Friction OR applied force (note: symbols not accepted) (1) [14] 5

Downloaded from hlayiso.com Physical Sciences/P1 Limpopo DoE September 2016 NSC QUESTION 5 5.1 Impulse is the product of the resultant/net force acting on an object and the time the resultant/net force acts on the object.  (2 or 0) 5.2 5.2.1 OPTION 1 OPTION 2 Take direction towards left as Take direction towards right as positive positive Σpi = Σpf  Σpi = Σpf  0 = Mvcannon + mvball Any one 0 = Mvcannon + mvball Any one     0 = (1250) v + 1,25 (–80) 0 = (1250) v + 1,25 (80)  v = 100 / 1250 = 0,08 m.s–1, left  v = 100 / 1250 = – 0,08 m.s–1 v = 0,08 m.s–1, left  5.3.2 (5) OPTION 1 OPTION 2 Take direction towards left as Take direction towards right as positive positive   Fnet∆t = m∆v = mvf – mvi  Fnet∆t = m∆v = mvf – mv i F (1,0)= (1250) ((0)–(0,08)) Fnet (1,0)= (1250) ((0) – (–0,08)) F = –100 N Fnet = 100 N  Fnet = 100 N  OR v f = vi + a ∆t 0 = 0,08 + a (1,0)  a = – 0,08 Fnet = m x a  = 1250 x (–0,08) = –100 N = 100 N (4) [11] 6

Downloaded from hlayiso.com Physical Sciences/P1 Limpopo DoE September 2016 NSC QUESTION 6 6.1 6.1.1 The Doppler Effect is the change in the observed 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 or 0) OR The change in the (observed) frequency when there is relative motion between the source and the observer.  (2 or 0) 6.1.2 away (1) v ± vL 6.1.3 fL = fs  v ± vs  340 − 0  88 = x 90 340 + v s vs = 7,73 m.s–1 (5) (note: it is not necessary to show the zero) 6.2 6.2.1 Red Shift occurs when absorption lines are shifted towards smaller  frequencies ( or larger wavelengths) (which is the red end of the spectrum)(2) 6.2.2 absorption (spectrum)  (1)   6.2.3 F. fastest galaxy/highest velocity (2) [14] QUESTION 7 7.1  direction Q1 Q2  pattern  field lines do not touch (3) 7

Downloaded from hlayiso.com Physical Sciences/P1 Limpopo DoE September 2016 NSC 7.2 The magnitude of the electrostatic force exerted by one point charge (Q 1) on another point charge (Q 2) 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) kQ 1Q 2 7.3 F=  r2 3 9 x 10 9 x Q x Q  4 x10 N = - (6 x 10 -3 ) 2 Q = (-) 4 x 10–9 C  (4) kQ 7.4 E=  r2  4,44 x 105 = 9 x 109 (4 x 10–9) (r)2  r = 9 x 10–3 m   d (or r) = 9 x 10–3 m – 6 x 10–3 m = 3 x 10–3 m = 3 mm (5) [14] QUESTION 8  8.1 Emf is the total potential difference across an electric circuit when the switch is open.  OR   Emf is the energy supplied per coulomb of charge/unit charge moving through the battery (2) 8.2  8.2.1 Internal Resistance // lost volts (used up by internal r) (1)  8.2.2 0 (V) / zero (1) 8

Downloaded from hlayiso.com Physical Sciences/P1 Limpopo DoE September 2016 NSC 8.2.3  V (or I = ) R   I= = 0,9A Rp = 12 Ω  Rext = 12 + 43 = 55 Ω   Emf = I (R + r) = 0,9 (55 + 5) = 0,9 x 60 = 54 V  (7) 8.2.4 Positive marking from Q 8.2.3  Therefore, VP = I x RP = 0,9 x 12 = 10,8 V (OR VP = 49,5 – V43 = 49,5 – (0,9 x 43) = 10,8 V)   V 10,8 I 30Ω = 30Ω = = 0,36 A R 30 (3)   8.3 Decrease.  Total resistance in the circuit increases. Current decreases. Hence, power decreases. (3) [17] QUESTION 9   9.1 Internal resistance in the opposition to the flow of charge in a cell/ an ammeter (in an electric circuit.) (2 or 0) 9.2  9.2.1 3 V (1)   9.2.2 ‘lost’ volts = 3,0 – 2,0 = 1 V OR 1 V  (2) 9

Downloaded from hlayiso.com Physical Sciences/P1 Limpopo DoE September 2016 NSC 9.2.3 r can be found by finding the gradient of the graph gradient = ∆ I ∆V  = 0,4 – 0,6  1–0 = –0,2 1 = – 0,2 Ω–1 Rint = 5 Ω  (other correct values from the graph can be used for the calculation) (3) [8] QUESTION 10 10.1 10.1.1 (split – ring) commutator  (1) 10.1.2 coil (1) Vmax  10.2 10.2.1 Vrms = 2  Vmax  200 = Vmax = 282,84 V (3) 2   10.2.2 P = V2rms Therefore, R = 2402 = 26,18 Ω R 2200 During load shedding:  P = V2rms = 2002  = 1527,88 W (4) R 26,18 OR   P = V2rms Therefore, R = 2402 = 26,18 Ω R 2200 V=IR 200 = I (26,18) I = 7,64 A   P = VI = 200 (7,64) = 1527,88 W (4) [9] 10

Downloaded from hlayiso.com Physical Sciences/P1 Limpopo DoE September 2016 NSC QUESTION 11    11.1 E = hf = hc = 6,63 x 10–34 x 3 x 108 = 3,9 x 10–19 J (4) λ 510 x 10–9  OR c=fxλ 3 x 108 = f x 510 x 10–9  f = 5,88 x 10 14  E = hf = 6,63 x 10–34 x 5,88 x 10 14 = 3,9 x 10 -19 J  (4) 11.2 Positive marking from 11.1  Caesium.Its work function is less than the energy of a photon of green light. (2) 11.3 Positive marking from 11.1 and 11.2  E = hf = W 0 + Ek Ek = E – W 0  = 3,9 x 10–19 – 3.36 x 10–19 = 5,4 x 10 –20 J  (3) 11.4 11.4.1 remain the same  (1) 11.4.2 remain the same  (1) 11.4.3 remain the same  (1) [12] Grand Total: 150 11