MI 23M1PrelimAS (H2 PHYSICS P2)

[Turn over 2023 Prelim Exams (Solutions) PU3 H2 Physics 9749 Paper 2 Qn Answers Marks 1 a Since the units of all the terms are the same, the equation is homogenous. units of v2, u2, and 2as presentation and conclusion B1 B1 b 𝑚𝑎𝑠𝑠 = (𝑙𝑒𝑛𝑔𝑡ℎ × 𝑤𝑖𝑑𝑡ℎ × ℎ𝑒𝑖𝑔ℎ𝑡) × 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 = (1.99 × 1.95 × 2.02) × 1.24 = 9.7199 = 9.71 𝑔 𝐵𝑦 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑖𝑛𝑔 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 / 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑢𝑛𝑐𝑒𝑟𝑡𝑎𝑖𝑛𝑡𝑦: 𝛥𝑚 𝑚 = 𝛥𝑙 𝑙 + 𝛥𝑤 𝑤 + 𝛥ℎ ℎ + 𝛥𝜌 𝜌 = 0.01 + 0.01 + 0.01 + 0.05 1.24 = 0.070323𝛥𝑚 = 0.070323 × 9.7199[∗ 𝑀1] = 0.68353 = 0.7 𝑔 𝑚𝑎𝑠𝑠 = (9.7 ± 0.7) 𝑔 Alternatively, max-mean method: mmax = (1.99 x 1.01)(1.95 x 1.01)(2.02x1.01)(1.29) = 10.418 g 𝛥𝑚 = 𝑚𝑚𝑎𝑥 − 𝑚𝑚𝑒𝑎𝑛 = 10.418 − 9.7199 = 0.7 𝑔 C1 C1 C1 A1 (C1) (C1)

2 2 a Use of mgh = ½ m v2 and makes h subject h = 14.7 or 15 m A1 bi Calculate the final vertical velocity at C (using v = 0 + at=9.81x1.6) v = 15.7 or 16 m s-1 C1 A1 bii Straight line of positive gradient) Starting at 0 ms-1 and ends at 16 ms-1 at 1.6 s. A1 biii Use of pythagoras' theorem: resultant v2 = 15.72 + 172 v = 23 or 23.1 m s-1 M1 c slope: smaller change in vertical component of velocity/ smaller change in vertical component of momentum by Newton’s second law, the force experienced = rate of change of momentum is less, so less risk of injury M1 A1 3 ai Archimedes’ Principle states that the upthrust on a body completely or partially submerged in a fluid is equal in magnitude and opposite in direction to the weight of the fluid the body displaces. B1 aii Pressure increases with depth in a fluid. When an object is submerged in a fluid, bottom is at a greater depth than at the top, hence, the pressure is greater at the bottom than at the top of the object. The difference in pressure results in a net upward force acting on the object, which is upthrust. B1 B1 b C1 C1 A1 c A1

3 [Turn over 4 (a) 1 mark for correct direction of arrows 1 mark for similar length of arrows in vertical direction B1 B1 (b) Since frictional force provides for the centripetal force, A1 (c) (i) For a surface which is banked, the horizontal component of the normal reaction is an additional source of the centripetal force. Since Centripetal Force = , an increase in the centripetal force would allow the rider to turn the corner at a higher speed without slipping, provide the mass of cyclist and bicycle and radius of turn remain constant. B1 B1 (ii) Considering forces acting on the rider/bicycle in the horizontal direction: ---------------------- (1) For the equilibrium in the vertical direction: --------------------(2) Solving (1) & (2), 🡪max velocity during turning v = 15.7 m s-1 C1 C1 A1 reaction force weight normal contact force friction weight

4 5 a The rocket experiences equal and opposite gravitational forces exerted by Earth and Moon. Hence there is zero net force as the forces cancel out. B1 b 1. Moon does not have an atmosphere hence there is no need to supply energy to do work against air resistance. 2. The gain in gravitational potential energy from Moon to P is much lower than that from Earth to P hence less kinetic energy is required for the rocket to move from Moon to P. B1 B1 c C1 A1 di The gravitational potential at a point is the work done per unit mass in bringing a point mass from infinity to that point. B1 dii C1 C1 A1

5 [Turn over 6 ai Total resistance of the 4 parallel resistors = (1/2R + 1/2R)-1 = R. I = E/(R + r). B1 A1 aii Power in the 4 parallel resistors = I2 R, where I = E/(R+r) Power in the battery = IE 𝑝𝑜𝑤𝑒𝑟 𝑖𝑛 𝑒𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑟𝑒𝑠𝑖𝑠𝑡𝑜𝑟𝑠 𝑝𝑜𝑤𝑒𝑟 𝑏𝑦 𝑏𝑎𝑡𝑡𝑒𝑟 = 𝐼2𝑅 𝐼𝐸 or 𝐼2𝑅 𝐼2(𝑅+𝑟) = 𝐼 𝑅 𝐸 = 𝐸 𝑅 𝑅 + 𝑟 1 𝐸 = 𝑅 𝑅 + 𝑟 M1 A0 aiii Having 4 resistors in a network shown in Fig. 6.1, each resistor only dissipates ¼ of the power compared to a single resistor. Hence one of the resistors in Fig. 6.1 is less likely to exceed the power rating. B1 bi There will be no deflection in the galvanometer when the p.d across JY = 1.5 V. Let the balance length (between J and Y) be L. Then 𝑉𝐽𝑌 = 𝐿 ×1.50 (1.50+0.50) × 4.0 = 1.5 L = 0.500 m. C1 A1 bii When 4.7  connected with the 1.5 V cell, the terminal p.d across the 4.7  is 𝑉 = 4.7 4.7+0.50 × 1.5 = 1.356 𝑉 There will null deflection when VJY = 1.356 V. Then 𝑉𝐽𝑌 = 𝐿′ ×1.50 (1.50+0.50) × 4.0 = 1.356 L’ = 0.452 m. Alternative working: 𝐿′ = 1.356 3 × 1.0 = 0.452 𝑚 C1 C1 A1

6 7 a {Similarity} The discrete lines of both absorption and emission spectrum occur at same frequencies, {Difference} the absorption has dark lines against a continuous spectrum whereas the emission spectrum has coloured lines against a black background. B1 B1 bi E = hf = ℎ𝑐 𝜆 = (6.63×10−34)×(3.0×108) 656×10−9 ×(1.6×10−19) = 1.90 eV M1 A1 bii Energy level labelled -1.50 eV and nearer to -0.84 eV A1 c 0.92 eV Nothing happens to the lithium atom (ie. it stays in the -5.02 eV level) 0.92 – (5.02 – 4.53) = 0.43 eV Atom will be excited to the -4.53 eV level A1 B1 A1 B1 di When the highly energetic electrons knock out the electrons in the inner shell of the atoms, leaving a vacancy. Electrons in the next higher energy levels transit down to the vacancy and X-ray photons are produced with energy equal to the energy difference between the 2 energy levels. B1 B1 dii 𝑉 = ℎ𝑐 𝑒𝜆𝑜 = (6.63 𝑥 10−34)(3.0 𝑥 108) (1.6 𝑥 10−19)(1.2 × 10−11) = 1.04 × 105 𝑉 = 104 kV C1 A1 diii Same characteristic wavelengths, lower threshold wavelength, higher intensity A1

7 [Turn over 8 (a) There are 3 positively charged particles Applying Fleming’s Left Hand Rule, there are 3 lines curving to the left. A1 M1 (b)(i) Compared to other tracks, the radius of the track in Fig. 8.3 is the smallest. The mass of electron is small. Its momentum and the radius will be small. Hence the proposal is possible. M1 A1 (b)(ii) The electron loses energy as it travels through the chamber. Its momentum and hence its radius decreases, resulting in a spiral. B1 B1 (c)(i) Conservation of charge gives (Charge of K–) + (proton charge) = Total charge after collision (-e) + (+e) = (+e) + (–e) + (+e) + q4 ⇒ q4 = –e C1 A1 (c)(ii) Particle 3 has the lowest momentum because r is smallest. A1 (d)(i) It causes little or no ionization in the target atoms, so it leaves no track. B1 (d)(ii) They are oppositely charged, as the paths curve in opposite directions, implying that the magnetic force points in opposite directions, OR, because they are formed from a neutral particle. B1 B1 (e)(i) 1. (e)(i) 2. particle px / 10-20 N s py / 10-20 N s pz / 10-20 N s E / 10-12 J K– 438.05 -13.24 0.81 1317.12 p 0.00 0.00 0.00 150.13 sum 438.05 -13.24 0.81 1467.25 A1

8 particle px / 10-20 N s py / 10-20 N s pz / 10-20 N s E / 10-12 J K– 79.03 1.48 11.95 252.50 Ω– 7.98 -0.60 2.07 33.38 Ω + 2.02 -6.52 -1.21 30.51 P 80.46 6.85 -3.76 285.22 K0 189.10 -8.69 -13.07 574.78 sum 358.59 -7.48 -4.02 1176.39 A1 (e)(ii)1 px = (438.05 - 358.59) x 10−20 = +79.46 x 10−20 N s py = (-13.24 + 7.48) x 10−20 = -5.76 x 10−20 N s pz = (0.81 + 4.02) x 10−20 = +4.83 x 10−20 N s E = (1467.25 - 1176.36) x 10−12 = 290.86 x 10−12 J A1 A1 A1 A1 (e)(ii)2 = 1.835 x 10-27 kg C1 A1