Pure Physics DBQ Package (answers)

CHS Data-based Questions Skills Package – Suggested Answers 1 1. CGS: Mechanics (+ Thermal Physics) [GRILS/Practical Planning] Qn Part Suggested Answers Marks Remarks 9 (a) 𝑄 = 𝑚𝑐𝛥𝜃 Q = 0.0010 × 4200 × 1.0 = 4.2 J A1 (b) Work done against weight = GPE gained = mgh = 50 × 10 × 201 = 100500 J = 1.0 × 105 J (2 s.f.) M1 A1 Accept answer to 2 or 3 s.f. (c) Min amount of energy from carbohydrates = 60% × 100500 = 60300 J Min amount of calories provided = 60300 / 4.2 = 14357 Min mass of carbohydrates needed = 14357 / 4000 = 3.59 g M1 A1 Allow ECF from (b) (d) Use calibrated electronic balance to measure his mass m. Use metre rule to measure the height of one step, h, of the staircase. Count the total number of steps, n, that he has to run up. The total height that he runs up to will be H = nh. Using a stopwatch, record the time taken t for him to reach H. His average power can then be calculated from Pave = mgH / t. Repeat the above steps at least 2 more times. Take the mean of Pave for a more accurate answer. B1 B1 B1 B1 1 m for stating the 3 main measuring instruments and their corresponding physical quantities 1 m for stating the formula for average power Pave 1 m for describing a clear and sensible procedure 1 m for repeating and taking the mean of Pave (e) t = E / P = 100500 / 48 = 2.1 × 103 s A1 Allow ECF from (b)

CHS Data-based Questions Skills Package – Suggested Answers 2 2. HCI: Mechanics [GRILS] Qn Part Suggested Answers Marks Remarks 9 (a) 𝐾𝐸 = 1 2 𝑚𝑣2 = 0.5 (0.025) (56)2 = 39.2 = 39 J M1 A1 (b)(i) v = u + at 0 = 56 + a(0.020) a = −2800 m/s2 deceleration = 2800 m/s2 M1 A1 (ii) 𝐹𝑛𝑒𝑡 = 𝑚𝑎 f = (0.025)(2800) f = 70 N M1 A1 (c) Work done = fd = (70)(0.12) = 8.4 J M1 A1 (d) Loss in KE = WD against friction + Gain in GPE 39.2 = 8.4 + mgh 39.2 = 8.4 + (0.025 + 8.0) (10) h h = 0.3838 = 0.38 m (theoretical value) However, experimental value = 0.292 m, hence there is a discrepancy M1 Aa1

CHS Data-based Questions Skills Package – Suggested Answers 3 3. Nan Chiau: Mechanics (+ Thermal Physics + Magnetism and Electricity) [GRILS/PRC] Qn Part Suggested Answers Marks Remarks 10 (a) 80 − 50 = 30 m A1 (b) Any possible reason: -To protect the wind turbine from damage due to its blades being ripped off / wear and tear / prevent turbulence or prevent the generator being burnt out or catastrophic failure -Stop working during hurricane -Current too high causing fuse to melt and open circuit so not working B1 Note for student: “Suggest” question so you need to come up with an answer that may be based on or even beyond what is covered in the syllabus. (c)(i) 𝐸 = 0.5 𝜋 × 502 × 143 × 1.25 = 13469578.5 W = 13.5 MW (3 s.f.) A1 (ii) Efficiency = useful power / input power × 100% = ( 3034 / 13470 ) × 100% = 22.5% (3 s.f.) M1 A1 (iii) Any possible reason: -Due to the limited number of blades mounted, not all the KE of the wind is converted to electrical energy. -In any mechanical system due to friction among moving parts or resistance in the electrical components, some of the electrical energy is wasted as heating effect or thermal energy on the wiring and electrical components. -Work is done against air resistance. -Some energy is converted into sound energy. B1 (d) -Diagram showing cyclic arrows in anti-clockwise direction of movement of hot air rising from land and movement of cold air towards the land. -Due to higher specific heat capacity of water than land B1 each, max 2 marks

CHS Data-based Questions Skills Package – Suggested Answers 4 -The warm air over the land expands, becomes less dense and rises, creating a partial vacuum. -The heavier, denser, cooler air over the water flows in to take its place, creating sea breeze. -Higher air pressure above sea and lower air pressure above land causes air flow from sea to land. -Explanation of convection current formed. (e) Lightning conductors mounted at the highest point, reducing the probability of lightning striking due to the point action of the sharp points which produces positively charged ions below a highly negatively charged cloud. OR Earth wire connected to the metallic part of the wind tower or a network of strike termination devices (lightning rods); A network of conductors to move electrical energy from the strike termination devices toward earth; a network for ground terminations (ground rods), direct excess charges to the ground/earth. B1 B1 B1 B1

CHS Data-based Questions Skills Package – Suggested Answers 5 4. NYGH: Waves [PRC/AUSGAI] Qn Part Suggested Answers Marks Remarks 13 E (a) Light from a wider range of directions can enter the light collector with less reflection, so more light can be captured B1 (b)(i) 55° and 0° A1 (ii) Light ray laterally displaced for flat light collector Light ray passes through undeviated for hemispherical collector B1 B1 1 m to be deducted from total mark for missing direction on rays (c)(i) Incident angle and angle of reflection are equal Reflections constructed until ray exits the tube B1 B1 (ii) The greater the number of reflections, the smaller the amount of light transmitted. At each reflection, some light is absorbed and some light is refracted. OR Longer distance travelled when there are more reflections, hencegreater loss of light to surroundings. B1 B1 (d) Light hits the acrylic film at different angles of incidence due to the irregular surface. Hence the light is refracted out from the film at different angles. B1 B1

CHS Data-based Questions Skills Package – Suggested Answers 6 13 O (a) Less easily refracted / no significant loss of energy over long distances B1 (b)(i) 170 ms A1 (ii) Time taken to receive reflected signal is long for vacant lots as the distance travelled by signal is longer. Amplitude of signal is smaller as there is greater energy loss for longer distance travelled for vacant lots B1 B1 (iii) Distance = speed x time = 340 m/s × (20 × 10-3 s) / 2 = 3.4 m M1 A1 Note for student: 20 ms is the time difference between empty and occupied lot; so we need to divide by 2 because signal is reflected from top of vehicle (travelled twice the distance) (iv) Reference value remains the same as frequency does not affect speed of the wave. B1 (v) Ultrasound travels faster in liquid Time taken between transmission and reception of signals is shorter. OR The amplitude is smaller as there is greater energy loss in liquid than in gas. B1 B1 (c) Presence of wind will carry away the wave and hence the wave cannot be detected by the receiver. B1

CHS Data-based Questions Skills Package – Suggested Answers 7 5. VS: Waves [AUSGAI/GRILS] Qn Part Suggested Answers Marks Remarks 10 (a)(i) 50 cm A1 (ii) v = fλ = 2.0 (50) = 100 cm/s d = st = 100 (0.06) = 6.0 cm M1 A1 (iii) Arrow pointing down B1 (iv) −9.0 cm A1 Accept −8.0 cm to −11.0 cm Allow ECF from (a)(ii) Note for student: To estimate the displacement of P, draw the wave after it has travelled for 0.060 s (b)(i) To ensure that no air is trapped between transmitter and skin, otherwise nearly all the transmitted pulse will be reflected at the surface of the skin. B1 (ii) The cross at the rightmost tip of the tumour B