Physical Science P1 1 September 016 Preparatory Examination Education PHYSICAL SCIENCES: K Z (PHYSICS) l N t l P1 NATIONAL GRADE 1 MARKS : 150 This memorandum consists of 13 pages.
Physical Science P1 September 016 Preparatory Examination QUESTION 1 1.1 A () 1. C () 1.3 B () 1.4 A () 1.5 A or D () 1.6 D () 1.7 B () 1.8 C () 1.9 A () 1.10 D () QUESTION.1 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 net force and inversely proportional to the mass of the object ()..1 For the 7,5 kg mass T - Fg = ma T - (7,5)(9,8) = 0 T = 73,5 N ().. For the 1 kg mass N F f T Fg Fcos30 0 + f - T = ma 0,866 F + μs N - 73,5 = 0 0,866 F + 0,45(1)(9,8) - (0,45)Fsin30 0-73,5 = 0 0,64 F + 5,9-73,5 = 0 F = 3,16 N (accept range: 3,106 3,16) (4)
Physical Science P1 3 September 016 Preparatory Examination.3 F = 6346,07 = Gm1m r 6.,67 x 10-11 x 5,98 x (R + h) 10 4 x 650 R + h = 639170,4 m QUESTION 3 3.1 OPTION 1 h = 639170,4-6,38 x 10 6 = 1170,4 m = 11,7 km (Range: 10,00km 11,7km) UPWARDS POSITIVE v f = v i + aδt 0 = +19 + (-9,8)(Δt) (5) [13] Δt = 1,94 s (3) OPTION DOWNWARDS POSITIVE v f = v i + aδt 0 = -19 + (+9,8)(Δt) Δt = 1,94 s (3) 3. A A accelerates for a greater time than B and the velocity of A is bigger than the velocity of B at x seconds/ A travels a greater downward distance than B/ The gradient of graph A is steeper than the gradient of graph B at 4s/ The mechanical energy of A is greater than the mechanical energy of B but the potential energy of A and B on the ground are the same, therefore the velocity of A must be greater than B () 3.3 OPTION 1 UPWARDS POSITIVE STONE A: Δy = v i Δt + ½aΔt = (19)(4) + ½(-9,8)(4 ) = -,4 m (for any one) STONE B: Δy = v i Δt + ½aΔt -,4 = (0)(4 - t) + ½(-9,8)(4 - t) t = 3,30 s
Physical Science P1 4 September 016 Preparatory Examination (5) OPTION DOWNWARDS POSITIVE STONE A: Δy = v i Δt + ½aΔt = (19)(4) + ½(9,8)(4 ) =,4 m (for any one) STONE B: Δy = v i Δt + ½aΔt,4 = (0)(4 - x) + ½(9,8)(4 - x) x = 3,30 s (5) OPTION 3 UPWARDS POSITIVE STONE A: Δy = v i Δt + ½aΔt = (19)(4) + ½(-9,8)(4 ) = -,4 m (for any one) STONE B: Δy = v i Δt + ½aΔt -,4 = (0)(4 - x) + ½(-9,8)(t B ) t B = 0,7 s 4 - x = 0,7 x = 3,30 s (5) OPTION 4 DOWNWARDS POSITIVE STONE A: Δy = v i Δt + ½aΔt = (19)(4) + ½(=9,8)(4 ) = =,4 m (for any one) STONE B: Δy = v i Δt + ½aΔt,4 = (0)(4 - x) + ½(9,8)(t B ) t B = 0,7 s 4 - x = 0,7 x = 3,30 s (5) OPTION 5 UPWARDS POSITIVE Δy A = Δy B any one (v i Δt + ½aΔt ) A = (v i Δt + ½aΔt ) B (19,4)(4) + ½(-9,8)(4) = (0) + ½(-9,8)(4 - x) x = 3,3 s (5)
Physical Science P1 5 September 016 Preparatory Examination OPTION 6 DOWNWARDS POSITIVE Δy A = Δy B any one (v i Δt + ½aΔt ) A = (v i Δt + ½aΔt ) B (19,4)(4) + ½(9,8)(4) = (0) + ½(9,8)(4 - x) x = 3,3 s (5) OPTION 7 UPWARDS POSITIVE Δy A = Δy B any one (v i Δt + ½aΔt ) A = (v i Δt + ½aΔt ) B (19,4)(4) + ½(-9,8)(4) = (0) + ½(-9,8)(Δt B ) Δt B = 0,7 s 4 - x = 0,7 x = 3,30 s (5) OPTION 8 DOWNWARDS POSITIVE Δy A = Δy B any one (v i Δt + ½aΔt ) A = (v i Δt + ½aΔt ) B (19,4)(4) + ½(9,8)(4) = (0) + ½(9,8)(Δt B ) Δt B = 0,7 s 4 - x = 0,7 x = 3,30 s (5)
Physical Science P1 6 September 016 Preparatory Examination 3.4 OPTION 1 UPWARDS POSITIVE v (m.s -1 ) 19,00 A 0 1,94 3,30 4,00 B t (s) CRITERIA For the same graph: v = 19 m.s -1 at t = 0 / v = 0 m.s -1 at t = 1,94 s The other graph starts at time = 3,30 s and v = 0 m.s -1 Both graphs are straight lines and parallel to each other Both graphs are correctly labeled with A to the left of B (5)
Physical Science P1 7 September 016 Preparatory Examination OPTION DOWNWARDS POSITIVE (maximum 4 out of 5) v (m.s -1 ) A B 0 1,94 3,30 4,00 t (s) -19,00 CRITERIA For the same graph: v = -19 m.s -1 at t = 0 / v = 0 m.s -1 at t = 1,94 s The other graph starts at time = 3,30 s and v = 0 m.s -1 Both graphs are straight lines and parallel to each other Both graphs are correctly labeled with A to the left of B Minus 1 mark at the end QUESTION 4 4.1 Inelastic, the kinetic energy is not conserved. () 4. Total p before = Total p after (m 1 v 1i )bullet + (m v i )block = (m 1 v 1f )bullet + (m v f )block (m 1 )(300) + 0 = (m 1 )(10) + (m )(1,8) m = 100 m 1 (or m 1 = 0,01 m ) (4) [15] 4.3 E ki(bullet) + E ki(block) - 75,76 = E kf(bullet) + E kf(block) ½ mv i + ½ mv i - 75,76 = ½ mv f + ½ mv f ½m 1(300) + 0-75,76 = ½ m 1(10) + ½ m (1,8) 45 000m 1-75,76 = 700 m 1 + 1,6 m 45 000m 1-75,76 = 700 m 1 + 1,6 (100 m 1) m 1 = 0,0 kg For subtracting from the sum of the initial kinetic energies m = kg (5)
Physical Science P1 8 September 016 Preparatory Examination QUESTION 5 5.1 Net work done on an object is equal to the change in the kinetic energy of the object. F () Normal /N 5. F Engine -1 for additional forces Direction and label must be correct [11] F Friction /f/f k F Gravity /Fg (4) 5.3 fk = µ kn = (0,017) (1500)(9,8)(cos 10,1 0 ) = 45,94 N (3) 5.4 sin 10,1 = 59/AB AB = 33,85 m W net = ΔEk W Fe + W N + W Ff + W Fg + W Fgp = 0 F e x cos θ + 0 + F f x cos θ + 0 + F gp x cos θ = 0 F e (33,85)(cos0 0 ) + (45,94) (33,85)(cos180 0 ) + mgsin θ x cos θ = 0 33,85 F e - 81861,13 + (1500)(9,8)(sin,10,1 0 )( 33,85)(cos180 0 ) = 0 33,85 F e = 949158,63 F e = 851,61 N P = F.v = (851,61)(10) = 8516,10 W (Accept: 8 539,33 W) (7) W nc = ΔEp + ΔEk W f + W Fe = mgδh + 0 F f x cos θ + F e x cos θ = (1500)(9,8)(59) (45,94)( 33,85)(-1) + 33,85 F e = 867300 F e = 851,6 N P = F.v = (851,6)(10) = 8516,0 W (7) [16] OR
Physical Science P1 9 September 016 Preparatory Examination QUESTION 6 6.1 The apparent/observed (change in the) frequency (or pitch) of the sound detected by a listener because the sound source and the listener have different velocities relative to each other. (ALL or NOTHING marking) () 6. v = f x λ Positive marking 340 = f x 0,7 from 6. to 6.4. f = 47, Hz (3) 6.3 Towards. The frequency of the sound waves heard by the traffic official is greater than the frequency of the sound waves emitted by the hooter/ Observed frequency is greater than 433,64 Hz () v ± v L 6.4 fl = fs v ± v s 47, = 340 x 433,64 340 v s v s = 7,78 m.s -1 The speed limit is 100 km.h -1 (7,78 m.s -1 ), while the speed of the car is 100 km.h -1 (7,78 m.s -1 ).The car is not exceeding the speed limit (6) 6.5 less than (1) [14] QUESTION 7 7.1 The (magnitude of the electrostatic) force exerted by one point charge on another point charge is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. () 7. T/tension F E /electrostatic force Fg/weight/force of gravity (3) 7.3 kq1q F = r 9 x F = 10 9 x 40 x = 5,76 x 10-3 N 10 0,05-9 x 40 x 10-9
Physical Science P1 10 September 016 Preparatory Examination Fg = F E tan 60 0 m x 9,8 = (5,76 x 10-3 )(tan 60 0 ) m = 1,0 x 10-3 kg OR Fg = F E /tan 30 0 m x 9,8 = (5,76 x 10-3 )/(tan 30 0 ) m = 1,0 x 10-3 kg OR QUESTION 8 F E = T.cos60 1 mg = T.sin60 F E /cos60 = mg/sin60 m = 1,0 10-3 kg (6) [11] 8.1 8. The electric field at a point is the electrostatic force experienced per unit positive charge/ +1C placed at that point. (ALL or NOTHING marking) () 8.3 E P = kq 1 r Any one = 9 x 10 9 x 3 x 10-9 equation (0,01 - y) right E Q = kq 1 r () = 9 x 10 9 x 7 x 10-9 (y) 9 x 10 9 x 3 x 10-9 = 9 x 10 9 x 7 x 10-9 left (0,01 - y) (y) y = 0,009 m (6) [10]
Physical Science P1 11 September 016 Preparatory Examination QUESTION 9 9.1 R 1 = V I 4,5 I 3A 1,5 = I = I R = 3-1, = 1,8 A V R3 = V R = IR R 3 = = 1,8 x 10 = 18 V V I R 3 = 18 1, = 15 Ω (7) 9. OPTION 1 emf = V + V p + Ir 4 = 4,5 + 18 + (3)(r) r = 0,5 Ω (5) OPTION R P = 10(15) (10 + 15) = 6 Ω έ = I (R + r) 4 = 3 [ (6 + 1,5) + r ] r = 0,5 Ω OPTION 3 POSITIVE MARKING FROM 9.1 R tot = 4/3 = 8Ώ R ext = 6 + 1,5 = 7,5Ώ r = R tot - R ext = 8-7,5 = 0,5Ώ (5) 9.3.1 increases (1) - 9.3. The total resistance of the circuit increases.
Physical Science P1 1 September 016 Preparatory Examination The circuit current decreases. Therefore Ir decreases V 1 = έ - Ir Since έ remains constant and Ir decreases V 1 increases (4) [17] QUESTION 10 10.1 M to G (1) W 10..1 P = t P = 9,45 x 10 700 6 = 131,5 W (3) 10.. V V max rms = V rms = 311.13 = 0,00 V Positive marking from 10..1 P ave = V rms I rms 131,5 = 0,00 x I rms I rms = 5,97 A I I max rms = I 5,97= max Imax = 8,44 A (6) 10.3 8,44 Current (A) 0,04s t (s)
Physical Science P1 13 September 016 Preparatory Examination Criteria Two full cycles with correct shape Showing the maximum current Showing the time 0,04 s for two cycles Marks (3) [13] QUESTION 11 11.1 photoelectric effect (1) 11. the minimum energy that an electron in a metal needs to be emitted from the metal surface. (ALL or NOTHING marking) () 11.3 E = W o + Ek c 1 Any one formula h = W 0 + mv max λ 8-34 3 x 10 1-31 5 6,63 x 10 = W -9 0 + (9,11x 10 )(3,51x 10 ) 487 x 10 Accept: 4,08 10-19 J for the mark. W 0 = 3,5 x 10-19 J (4) 11.4.1 increases, the number of photoelectrons emitted per second increases. () 11.4. remains the same. (1) [10] TOTAL MARKS: [150]