Physical Sciences P1 MEMO - smartlearner.mobi

1 Physical Science P1 1 September 2015 Preparatory Examination NSC-MEMORANDUM Copyrigth reserved Please turn over Basic Education KwaZulu-Natal Department of Basic Education REPUBLIC OF SOUTH AFRICAREPUBLIC OF SOUTH AFRICAREPUBLIC OF SOUTH AFRICAREPUBLIC OF SOUTH AFRICA UMLAZI DISTRICTUMLAZI DISTRICTUMLAZI DISTRICTUMLAZI DISTRICT MARKS : 150 TIME : 3 Hrs This memorandum consists of 14 pages. Physical Sciences : (PHYSICS) P1 PREPARATORY EXAMINATION SEPTEMBER 2015 MEMORANDUM(AMENDED) NATIONAL SENIOR CERTIFICATE GRADE 12 Physical Science P1 2 September 2015 Preparatory Examination NSC-MEMORANDUM Copyrigth reserved Please turn over QUESTION 1 C C D A C C D D C B [20] Physical Science P1 3 September 2015 Preparatory Examination NSC-MEMORANDUM Copyrigth reserved Please turn over QUESTION 2 When a resultant/net force acts on an object, the object will accelerate in the direction of the force the acceleration is directly proportional to the net force and inversely proportional to the mass of the object.

2 (2) FNormal 6 N Tension (4) Fgravity Fnet = ma 6 = a = 2 (3) (positive marking from ) Fres = Fa + (-Fr) ma = Fa + (-Fr) 4 x 2 = T + (- 6) T = 14 N Option 2 Fnet = T 6 ma = T 6 4 x 2 = 14 N T = 14 N (4) (positive marking from ) Option 1 Fres = Fa + (-Fr) ma = mg + (- T) m2 = m . 9,8 + (- 14) 7,8m = 14 m = 1,79 kg Option 2 Fnet = ma W T = 2m 9,8m T = 2m 7,8m = 14 m = 1,79 kg (4) [17] Physical Science P1 4 September 2015 Preparatory Examination NSC-MEMORANDUM Copyrigth reserved Please turn over QUESTION 3 Option 1 upward as positive vf2 = vi2 + 2a y 0 = 3922 + 2 (-9,8) y y = 7840m The total height reached is 7840 + 7840 = 15 680 m Option 2 - upward as negative vf2 = vi2 + 2a y 0 = (-392)2 + 2 (9,8) y y = 7840m The total height reached is 7840 + 7840 = 15 680 m Option 3 upward as positive vf2 = vi2 + 2a y (- 392)2 = 0 + 2 (-9,8)

3 Y y = 7840m The total height reached is 7840 + 7840 = 15 680 m (5) Option 1 upward as positive vf = vi + a t = 392 + (- 9,8) (80) vf = - The fuel tank is moving down at Option 2 - upward as negative vf = vi + a t = (-392) + (9,8) (80) vf = The fuel tank is moving down at (3) Physical Science P1 5 September 2015 Preparatory Examination NSC-MEMORANDUM Copyrigth reserved Please turn over Option 1 - upward as positive v ( ) 392 40 80 t (s) -392 Option 2 - upward as negative v ( ) 392 40 80 t (s) -392 Criteria for marking the graph Marks Correct shape straight line Graph starts at - 392 m s-1 or + 392 m s-1 Graph passes x axes at 40 s The line is extended beyond 80 s Correct labelling of axes (5) [13] Physical Science P1 6 September 2015 Preparatory Examination NSC-MEMORANDUM Copyrigth reserved Please turn over QUESTION 4 The total linear momentum of a closed system remains constant (is conserved).

4 (2) Option 1 pbefore = pafter mb vbi + mBi vBi = (mb + mB ) vf 0,016 x 360 + 0 = (0,016 + 2,984) vf vf = 1,92 Option 2 pi = pf m1 vi1 + m2 vi2 = (m1 + m2) v (0,016) (360) + (2,984) (0) = 3v vf = 1,92 (4) Option 1 FN = mgCos = 3 x 9,8 x Cos15 = 28,398 N Wfr = 4,67 5,52 = - 0,85 J Ffriction = Wfriction x Cos 180 = - 0,85 0,61 = 1,393 N k = Ffr FN = 1,393 28,398 = 0,049 Option 2 Wfr = 4,67 5,52 = - 0,85 J Wfr = Ffr Cos 180 x - 0,85 = Ff (- 1) (0,61) Ff = 1,393 N Ff = k N 1,393 = k (mgcos ) 1,393 = k (3 x 9,8 cos15 ) k = 0,049 Option 3 Wnc = K + U F x Cos = (Kf - Ki) + (Uf - Ui) F(0,61)(-1)=(0 5,52) +(4,67 0) Ff = 1,393 N Ff = k N 1,393 = k (mgcos ) 1,393 = k (3 x 9,8 cos15 )

5 K = 0,049 (8) [14] Physical Science P1 7 September 2015 Preparatory Examination NSC-MEMORANDUM Copyrigth reserved Please turn over QUESTION 5 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) non conservative (1) Wfriction = Ffriction x Cos x = 3400 x Cos 1800 x 25 Wfriction = - 85000J (3) Wnet = k = mvf 2 - mvi2 = (12000) (20)2 - (12000) (25)2 = - 135 0000 J (4) positive marking from Option 1 Ep + Wfriction = Wnet mgh + 85000 = 1350000 12000 x 9,8 x h = 3 425000 h = 10,76 m Option 2 Wnet = Wg + Wf + WN 135 0000 = ( mgh 85000) + 0 1265000 = (12 000) (9,8) h h = 10,76 m (5)

6 [15] Physical Science P1 8 September 2015 Preparatory Examination NSC-MEMORANDUM Copyrigth reserved Please turn over QUESTION 6 Doppler Effect (1) Zero , no relative motion . (2) fL = v vL fs v vs 3035 = 340 + 0 2500 340 - vs vs = 59,93 (5) Determine the heartbeat of fetus. Determine whether the arteries are narrowed. (2) [10] Physical Science P1 9 September 2015 Preparatory Examination NSC-MEMORANDUM Copyrigth reserved Please turn over QUESTION 7 The electrostatic force experienced per unit positive charge placed at that point. (2) E = kQ r2 = (9 x 109) (4 x 10-9) (5 x 10-2)2 = 1,44 x 104 (4) FE T mg F = mg = (0,08 x 10-3) (9,8) 100 = 7,84 x 10-4 N FQ = 7,84 x 10-4 x tan 10 = 1,38 x 10-4 N (4)

7 Positive marking from and Option 1 E = F q 1,44 x 104 = 1,38 x 10-4 q q = 9,60 x 10-9 C Charge on N is q = 9,60 x 10-9 C Option 2 F = kqQ r2 1,38 x 10-4 = (9 x 109) q (4 x 10-9) (5 x 10-2)2 3,45 x 10-7 = 36q q = 9,60 x 10-9 C Charge on N is q = 9,60 x 10-9 C (4) Physical Science P1 10 September 2015 Preparatory Examination NSC-MEMORANDUM Copyrigth reserved Please turn over -+QP Criteria for marking Marks Correct shape Correct direction Field lines not touching each other or entering the spheres non uniform (1) the electric field is non-uniform.

8 (1) [19] Physical Science P1 11 September 2015 Preparatory Examination NSC-MEMORANDUM Copyrigth reserved Please turn over QUESTION 8 Slip ring . Ensure continuous rotation of the coil. (2) ABCD (1) 2 VVmaxrms= 250= = 35,36 V (3) Criteria for marking Marks Graph starts at 50 V Correct shape Time interval (0,01 ms for 90 rotation) The graph ends at (0,07,0) (4) [10] 0,030,070,05t ( s )0,01-50V ( V )50 Physical Science P1 12 September 2015 Preparatory Examination NSC-MEMORANDUM Copyrigth reserved Please turn over QUESTION 9 V2 = IR = 0,4 x 10 = 4V R1 + 8 = 2,04 = 20 R1 = 12 (5) Rint = Vlost I = 3 0,6 = 5 (3) Option 1 1 = 1 + 1 Rp R1 R2 = 1 + 1 10 20 1 = 3 Rp 20 Rp = 6,67 Rext = 12 + 6,67 = 18,67 emf = I (r + R) = 0,6 (5 + 18,67) = 14,2 V Option 2 V = IRT = 0,6 (12 + 6,67)

9 = 11,2 V emf = V + V1 = 11,2 + 3 = 14,2 V (6)NB: Positive marking from to Physical Science P1 13 September 2015 Preparatory Examination NSC-MEMORANDUM Copyrigth reserved Please turn over P = VI = 3 x 0,6 = 1,8 W (3) Increase, RT increases Imain decreases Vlost decreases Vext increases (Emf is constant) OR the total resistance will increase thus increasing the potential difference . OR from Ohm s Law: resistance is directly proportional to potential difference OR R is directly proportional to V (3) [20] Physical Science P1 14 September 2015 Preparatory Examination NSC-MEMORANDUM Copyrigth reserved Please turn over QUESTION 10 Photoelectric effect (1) particle nature (of light) (1) the frequency of light is higher than the threshold frequency in E to F and (photo) electrons are emitted.

10 (2) OR A to C ; the frequency of incident light is below threshold frequency, therefore no photoelectrons are emitted. frequency OR intensity (1) current (1) Decrease in intensity decreases the current produced by the cell. The number of photoelectrons emitted decreases.. (2) 5,00 x 1014Hz (1) hf = hf0 + Ek (6,63 x 10-34) (6,5 x 1014) = (6,63 x 10-34) (5,00 x 1014) + Ek Ek = 9,8787 x 10-20 J (3) [12] TOTAL: 150