NATIONAL SENIOR CERTIFICATE GRADE 12

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1 Physical Sciences/P1 NSC Limpopo DoE September014 NATIONAL SENIOR CERTIFICATE GRADE 1 PHYSICAL SCIENCES: PHYSICS (P1) FISIESE WETENSKAPPE: FISIKA (Vr1) PREPARATORY EXAMINATIONS SEPTEMBER 014 MEMORANDUM This memorandum consists o 10 pages. / Hierdie memorandum bestaan uit 10 bladsye Copyright reserved Please turn over

2 QUESTION/ Vraag B () 1. A () 1.3 A () 1.4 C () 1.5 B () 1.6 C () 1.7 A () 1.8 A () 1.9 A () 1.10 C () [0] QUESTION/ Vraag.1 When a net orce is applied to an object, the object accelerates in the direction o the net orce. The acceleration is directly proportional to the (net) orce and inversely proportional to the mass o the object. // Wanneer 'n resulterende/netto krag op 'n voorwerp inwerk, versnel die voorwerp in die rigting van die krag teen 'n versnelling direk eweredig aan die krag en omgekeerd eweredig aan die massa van die voorwerp. (). N 3 kg 0 N T Consider the 1 kg block: 1 N 1 kg T F net = ma = T + 1a = T 1 T = a equation/vergelyking 1 Consider the 3 kg block: F R = ma = F applied + T + F 3a = 0 T.. equation/ vergelyking Substituting or T: 3a = 0 (a + 1) 4a = 17 a = 4,5 m.s (5) OR/ OF F net = ma (F ) = (M + m) a

3 0 ( + 1) = 4a a = 17/4 = 4,5 ms (5).3. T = a + 1 = 4,5 + 1 = 5,5 N () [1] QUESTION/ VRAAG m 0,8 (s) (1) (1) 3.3 (Take downward motion as NEGATIVE. (Other option: take downwards as positive)) 3.4 v = vi + a t 0 = v i + ( 9,8) (0,4) v i = 3,9 m. s 1, up + marking 3.3 OPTION/OPSIE 1 1 y = v i t + a t = (3,9) (0,4) + ½ ( 9,8) (0,4) = 0,784 m The maximum height = 8 + 0,78 = 8,78 m OPTION/OPSIE + marking 3.3 U + K = 0 ½ mv i + mgh i = ½ mv + mgh ½xm(3,9) + mx9,8x8 = 0 + mx9,8xh h = 8,78 m (½xm(3,9) + 0 = 0 + mx9,8xh h = 0,78 h tot = 8,78 m ) Any one OPTION 3 + marking 3.3 v v a y = i + (0) = (3,9) + ( 9,8) y y = 0,784 m The maximum height = 8 + 0,78 = 8,78 m OPTION 4 v v + i y = t = ½ (3,9 + 0) (0,4) + marking 3.3 = 0,78 m The maximum height = 8 + 0,78 = 8,78 m (4) 3.5 rom maximum height downwards/ vana maksimum hoogte awaarts v v a y = i + = (0) + ( 9,8) ( 8,78) + marking v = 13,1 m.s 1 rom 3.4

4 OR / o rom the balcony upwards/ vana balkon opwaarts v v a y = i + = (0) + ( 9,8) ( 8) v = 13,1 m.s ,9 Initial velocity/aanvanklike snelheid inal velocity/inale snelheid Shape/vorm v (m.s 1 ) 0,4 t (s) 13,1 QUESTION/ VRAAG 4 [15] 4.1 N F Applied/man 100 N Weight/ w/f G /gewig (4) OR/OF N F Applied/man 100 N F Weight/ w/f /gewig

5 4. OPTION 1 Work Done by Gravity = F x Cosθ W G = (10 x 9,8) (5) Cos = 167,59 J 90+0 OPTION Work done = E p =mgh-0 = mg x Sin 0 0 = 10 x 9.8 x 5 x Sin 0 0 = 167,59 J OPTION 1 Σ Fy = 0: N + F G = 0 F N mg Cos 0 0 = 0 F N = (10) (9,8) Cos 0 0 = 9,1 N k = µ F N = (0,4) (9,1) = 36,8 N 180 Work Done by riction = k x Cos θ W = (36,8) (5) Cos = 184 J OPTION F k = µ k N = 0,4 mg Cos θ = 0,4 x 10 x 9,8 Cos 0 0 = 36,8 Work Done by riction = k x Cos θ W = (36,8) (5) Cos = 184 J (6) W F = F x Cosθ =(100) (5) Cos0 0 = 500 N (1) 4.5 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 orce is equal to the change in the object's kinetic energy. () 4.6 W NET = E K + marking 4., 4.3.1, 4.3., 4.4

6 W G + W + W N + W F = ½ mv ½ mv i ( 168) + ( 184) + (0) + (500) = ½ (10) v ½ (10) ((1,5) v = 5,64 m.s 1 (4) QUESTION/ VRAAG 5 [0] 5.1 Inelastic Energy lost due to sound and heat or The two locomotives move together Onelasties Lokomotiewe beweeg saam na die botsing () 5. (take right a positive) Σ p = Σ [accept: m 1 v i + m v i = m 1 v 1 + m v /p beore = p ater ] beore p ater (6000)(4) + (5000) (v B ) = ( ) v B = v B = 1,8 m.s 1 to the right (6) Other option: take right as negative [8] QUESTION/ VRAAG Blood low meter / Doppler low meter / bloedvloeimeter/dopplervloeimeter (1) Higher pitch /hoër toonhoogte (1) 6.. Lower pitch / laer toonhoogte (1) v ± vl v + v 6.3 Use either L = s or L = L S v ± v s v L = x L = 1 963,6 Hz (5) 6.4 labels t [11] QUESTION/ VRAAG Electric ield at a point is the (electrostatic) orce experienced per unit positive charge placed at that point. // Die elektriese veld by 'n punt is die elektrostatiese krag wat per eenheidspositiewe-lading wat by daardie punt geplaas is ondervind word () 7. Curved electric ield line important. on outside A B Geboë elektriese veld en lyne belangrik op buitekant

7 1 mark: shape o ield between and outside 1 mark direction kq 7.3 E1 = r = 9 x 10 9 (15 x 10 9 ) ( x 10 ) = N.C 1 to the let/ links () E kq = r = 9 x 10 9 (Q x ) (8 x 10 ) = 1,4065 x 10 1 x Q x to the let / links E NET = E 1 + E 3,943 x 10 5 = ,4065 x10 1 Q X Q X = +4,04 x 10 8 C (7) 7.4 decrease / aneem (1) [1] QUESTION/VRAAG OR V P = R 13,5 = (18) R R = 4 Ω P = VI I R = P/V = 13,5/18 = 0,75 A V = IR R = V/I R = 18/0,75 = 4 Ω

8 8. V R = I 4 = 18 I I = 0,75 A R = V p I 1 1 = 18 I 1 I 1 = 1,5 A I Total = 1,5 + 0,75 =,5 A = I 10 Ω (5) OR R P = 8Ω V = IR P 18 = I (8) I =,5 A (5) 8.3 Internal resistance is the opposition to the ollow o charge within a cell (or an ammeter) / Interne weerstand is die teenstand teen die vloei van stroom binne-in n sel (o n ammeter) () 8.4 When the switch is closed, the pd across the 10 Ω resistor is V 10 = IR 10 =,5 x 10 =,5 V 8.5 The pd across the external resistors is given by: V 1 = V ext = V P + V 10 = 18 +,5 = 40,5 V When the switch is open the total pd (em) is: Em = V total = 45,9 V The lost volts is: V L = V Total V ext = 45,9 40,5 = 5,4 V V L r = I = 5,4,5 =,4 Ω (5)

9 OR V TOT = E = I TOT (R ext + R int ) 45,9 = (,5) (R P r) = (,5) ( r) R P = 8Ω r =,4 Ω 8.6 Increase / Toeneem (1) QUESTION/ VRAAG Ohm's Law states that: the potential dierence across a conductor is directly proportional to the current in the conductor at constant temperature./ Die potensiaalverskil oor 'n geleier is direk eweredig aan die stroom in die geleier by konstante temperatuur () 9.3 Inverse o resistance or 1/R / omgekeerde van weerstand (1) [19] 9. Y The inverse o the gradient o graph Y is greater Gradient o graph Y smaller. Thus 1/R smaller. Hence, R is greater. Y; Y se gradient kleiner dus is die omgekeerde van R kleiner, dus R groter () 9.4 R X can be ound by inding the gradient o the graph or X gradient = I V m = 0,4 0, (or other correct values rom graph) 8 4 (o ander korrekte waardes vana graiek) = 0,05 Rx = 1 0,05 = 0 Ω (4) [9] QUESTION/VRAAG Mechanical energy is converted to electrical energy. / meganiese energie na elektriese energie (1) AC generator / WS generator (1) (slip rings/ sleepringe) (1) anti-clockwise / anti-kloksgewys (1) The (rate o) change in magnetic lux/ magnetic ield linkage is at a maximum // die tempo van verandering van magnetise luks is maksimum. ()

10 10.3 V rms Vmax 330 = = = 33,36 V P rms = V rms I rms = 33,36 x 15 = 3500,5 W (5) [11] QUESTION/VRAAG Cathode / katode (1) 11. Threshold requency /dumpelrekwensie (1) 11.3 Work unction is the minimum quantity o energy which is required to remove an electron rom the surace o a given solid, usually a metal. die minimum energie benodig om 'n elektron uit die oppervlak van 'n metaal vry te stel. () 11.4 E= W 0 = h 0 = (6,63 x ) (5 x ) = 3,315 X J h = W 0 + mv (6,63 x ) ( 1 ) = 3,315 x x =,15 x Hz (5) 11.6 same / dieselde (1) [13] Grand total: 150