Physical Sciences Grade 12 P1 Western Cape Mock 2015 Memo

Downloaded from hlayiso.com PHYSICAL SCIENCES/ P1 1 CAPS CAPE WINELANDS EDUCATION DISTRICT CURRICULUM AND ASSESSMENT POLICY STATEMENT GRADE 12 PHYSICAL SCIENCES: PHYSICS (P1) SEPTEMBER 2015 MEMORANDUM MARKS 150 TIME 3 hours Please turn over

Downloaded from hlayiso.com PHYSICAL SCIENCES/ P1 2 SEPTEMBER 2015 CAPS QUESTION 1 1.1 A✓✓ (2) 1.2 A✓✓ (2) 1.3 D ✓✓ (2) 1.4 A ✓✓ (2) 1.5 D ✓✓ (2) 1.6 C ✓✓ (2) 1.7 A ✓✓. (2) 1.8 B ✓✓ (2) 1.9 B ✓✓ (2) 1.10 C ✓✓ (2) [20] QUESTION 2 2.1 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 force and inversely proportional to the mass of the object.✓ (2) 2.2 (3) 2.3 downward = positive (clockwise) Ftension + Fgravitation = Fresultant✓ Ftension + (2)(9,8) = 2(3)✓ (3) Please turn over

Downloaded from hlayiso.com PHYSICAL SCIENCES/ P1 3 SEPTEMBER 2015 CAPS Ftension = -13,6 N Ftension = 13,6 N ✓ upward 2.4 Right = positive (clockwise) Frope + Ffriction = Fresultant✓ 13,6 N + Ffriction =4(3)✓ (3) Ffriction = -1,6 N Magnitude of friction force: 1,6 N ✓ 2.5 The force that the rope exerts on the box✓ and the force that the box exerts on the rope.✓ OR the force that the Earth exerts on the box and the force (2) that the box exerts on the Earth. [13] QUESTION 3 3.1 The product of the resultant/net force acting on an object and the time the resultant/net force acts on the object.✓✓ (2) 3.2 Right = positive F𝛥t = m𝛥v =m(vf - vi)✓ (or other correct form of the equation) F(0,1) = 0,4[ 1,49 - (-6)]✓ (3) F = 29,96 N ✓ 3.3 The total linear momentum of a closed system✓ remains constant (is conserved)✓ (2) 3.4 m1v1i + m2v2i = (m1 + m2) v✓(Right = positive) 50v1i + (0,4)(-6) = 50,4(1,49)✓ v1i = 1,55 m∙s-1✓ (3) 3.5 Total kinetic energy before collision: ½ m1v1i2 +½ m2v2i2 = (0,5)(50)(1,55)2 + (0,5)(0,4)(-6)2✓ =67,26 J✓ Total kinetic energy after collision: ½ (m1 + m2)v2 = (0,5)(50,4)(1,49)2✓ = 55,95 J✓ Ek before≠Ek after✓ (5) inelastic collision. [15] Please turn over

Downloaded from hlayiso.com PHYSICAL SCIENCES/ P1 4 SEPTEMBER 2015 CAPS QUESTION 4 4.1 vf2 = v2i +2 g𝛥y✓ (Upwards = positive) vf2 =(8)2 + 2(-9,8)(2)✓ vf = 4,98 m∙s-1✓ (3) 4.2 Time it took to reach the ceiling: vf = vi + gt✓ 4,98 = 8 + (-9,8)t✓ t = 0,31 s. Therefore: time it took for ball to bounce back: 0,65 - 0,31 = 0,34 s✓ initial velocity of the ball when it bounce back: ∆y = vit + ½ gt2✓ 2 = vi(0,34) + (0,5)(9,8)(0,34)2✓ (6) v = 4,22 m∙s-1✓. 4.3 (5) [14] QUESTION 5 5.1 A force for which the work done in moving an object between two points depends on the path taken/is not independent of the path taken✓✓ (2) 5.2 0J✓ (1) 5.3 Fg// - ( f + F) = 0✓ (Accept other correct symbols) OR/OF F = mg sinθ – fk OR/OF F = mgsinθ – 266 F = [100(9,8) sin 25 o]✓ – 266✓ Please turn over

Downloaded from hlayiso.com PHYSICAL SCIENCES/ P1 5 SEPTEMBER 2015 CAPS 414,167- 266 F = 148,17 N✓ (4) NOTE/LET WEL No mark for diagram 1 mark for use of any of the three formulae 5.4 (6) Please turn over

Downloaded from hlayiso.com PHYSICAL SCIENCES/ P1 6 SEPTEMBER 2015 CAPS [13] QUESTION 6 6.1.1 The change in 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) 6.1.2 increase ✓ (1) 6.1.3 As the police officer move closer to the alarm, he would observe a sound with a shorter wavelength✓ than was originally omitted. Since the wavelength is inversely proportional to the frequency of the wave, the frequency will increase (become more / higher).✓ (2) 6.1.4 𝑣 + 𝑣𝐿 𝑓𝐿 = 𝑣 𝑓𝑠 ✓ (OR Formula as on data sheet) 340 + 40 = ✓(1200)✓ 340 = 1 341, 18 Hz✓ (4) Please turn over

Downloaded from hlayiso.com PHYSICAL SCIENCES/ P1 7 SEPTEMBER 2015 CAPS 6.2.1 An atomic absorption spectrum is formed when certain frequencies of electromagnetic radiation that passes through a medium✓, e.g. a cold gas, is (4) absorbed.✓ An atomic emission spectrum is formed when certain frequencies of electromagnetic radiation are emitted✓ due to an atom's electrons making a transition from a high-energy state to a lower energy state.✓ 6.2.2 The absorbed electromagnetic radiation for the light from Andromeda appear at higher frequencies than the absorbed electromagnetic radiation for light from the Sun.✓ (1) 6.2.3 Blue shift ✓ (1) [15] QUESTION 7 7.1 The magnitude of the electrostatic force exerted by one point charge (Q 1) on another point charge (Q2) 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) 7.2 𝑘𝑄1 𝑄2 F= ✓ 𝑟2 9 ×109 × 𝑄 × 16 𝑥 10−9 7,2 x 10-6 = (0,4)2 ✓ QA = -8 nC.✓ (3) 7.3 Electric field at P due to A 𝑘𝑄 𝐸 = 2✓ 𝑟 9 × 109 × 8 × 10−9 = ✓ (0,3)2 = 800 N∙C-1 l✓ Electric field at P due to B 9 × 109 × 16 × 10−9 E= (0,7)2 ✓ = 293,88 N∙C-1✓ 800 + 293,88 = 1 093,88 N∙C-1✓ (6) 7.4 B TO A✓ (1) 7.5 8𝑛𝐶 + 16𝑛𝐶 2 = 12 nC✓ 4 nC electrons were transferred from B to A✓ (4) 4×10−9 1,6 ×10−19 ✓= 2,5 x 1010 ✓electrons [16] Please turn over

Downloaded from hlayiso.com PHYSICAL SCIENCES/ P1 8 SEPTEMBER 2015 CAPS QUESTION 8 8.1 The potential difference across a conductor is directly proportional to the current in the conductor✓ at constant temperature.✓ (2) 8.2 Negative✓ (1) 8.3 1 1 1 (3) 𝑅 = 𝑅 + 𝑅✓ 1 2 1 1 1 = + ✓ 𝑅 4 2 R = 1,33𝛺✓ 8.4 V = IR✓ 3,9 = I x 1,33✓ I = 2,925 A✓ 𝜀 = 𝐼𝑅 + 𝐼𝑟✓ 6 = 3,9 + 2,925 r✓ (6) r = 0,72𝛺✓ 8.5 2,925 3 ✓= 0,975 A.✓ or V = IR 3,9 = I (4)✓ (2) I = 0,975 A✓ 8.6.1 Increase✓ 8.6.2 Stays the same ✓ (1) 8.7 𝜀 = 𝐼( 𝑅 + 𝑟)✓ 6 = I ( 4 + 0,72)✓ I = 1,27 A✓ (3) [19] QUESTION 9 9.1.1 Generator ✓ (1) 9.1.2 Kinetic/mechanical energy✓ → electrical energy ✓ (2) 9.1.3 B to A ✓ (1) 9.1.4 DC ✓ (1) Please turn over

Downloaded from hlayiso.com PHYSICAL SCIENCES/ P1 9 SEPTEMBER 2015 CAPS 9.1.5 The split ring commutator ensures✓ that the current that passes through to the external circuit is always in the same direction.✓ (2) 9.1.6 Use a coil that consist of more windings ✓ Increase the strengths of the magnets.✓ (2) 9.2.1 The rms value of AC is the DC potential difference which dissipates the same amount of energy as AC✓✓ (2) 𝑉 Vrms = 𝑚𝑎𝑥 2 ✓ 9.2.2 √ 39,45 = ✓ √2 = 27,9 V ✓ (3) 9.3 It can be stepped up or stepped down / is easier to transmit ✓ (1) [15] QUESTION 10 10.1.1 Photoelectric effect ✓. (1) 10.1.2 Wo = hfo✓ = 6,63 x 10-34 x 4,389 x 1014 ✓ = 2,91 x 10-19 J✓ (3) 10.1.3 E = hf = 6,63 x 10-34 x 4,83 x 1014 = 3,2 x 10-19J E = hfo + ½ mv2 ✓ 3,2 x 10-19✓ = 6,63 x 10-34 x 4,39 x 1014 + (0,5)(9,11 x 10-31) v2✓ (4) v = 2,5 x 105 m∙s-1✓ 10.2.1 Increase ✓ (1) 10.2.2 increase✓ (1) [10] TOTAL [150] Please turn over