RVHS H2 Physics P3 Soln

River Valley High School Pg 1 of 9 JC 2 H2 Physics 9749 Preliminary Examinations 2024 RVHS JC2 H2 Physics Prelim Examination 3 Mark Scheme 1 (a) (i) time = 0.43 / 1.1 = 0.39(1) s A1 (ii) 𝑠 = 𝑢𝑡 + 1 2 𝑎𝑡2 = 1 2 (9.81)(0.39)2 C1 = 0.75(0) m A1 (iii) vertical velocity: 𝑣 = 𝑢 + 𝑎𝑡 = (9.81)(0.39) = 3.8259 m s-1 M1 𝜃 = 𝑡𝑎𝑛−1 𝑣𝑦 𝑣𝑥 = 𝑡𝑎𝑛−1 3.8259 1.1 = 74(.0)° A1 (iv) 1. Horizontal line at a non-zero value of a. B1 2. Curved line from origin with increasing gradient B1 (b) acceleration of free fall is unchanged / not dependent on mass and so no effect (on time taken) A1 2 (a) 12 and TT : down 3T : up B1 (b) By Principle of Moments, Taking moment about the pivot at the base of wire 1, sum of clockwise moments = sum of anti-clockwise moments ( )( ) ( ) ( ) ( )( ) 3 3 3 14 5 10 0 14 5 350 10 9 8110 0 49 14 5 10 0 8 N p p . . Tmg . mg . .. . T . − = = = = M1 A1 (c) P is in vertical translational equilibrium, R adds additional downward force. 3T is only upward force and so must provide additional tension, more likely to snap. B1 All 3 must be correct.

River Valley High School Pg 2 of 9 JC 2 H2 Physics 9749 Preliminary Examinations 2024 If method involves principle of moments, reference to a pivot must be made known before a mark can be awarded together with the reasoning. 3 (a) loss in GPE = gain in elastic PE 2 22 1 2 (25)(0.050) 0.028959 kg2 2(9.81)(0.110) 0.029 kg(shown) mgh kx kxm gh = = = = = M1 (b) When the spring first compresses, the magnitude of the force from the spring is less than weight. Hence, there is still a downward resultant force that causes the marble to continue accelerating. B1 B1 (c) Max speed of marble happens when force from spring = weight 22 (0.028959)(9.81) 0.01136 m25 11 (25)(0.01136) 0.001614 J22 (0.060 0.01136) (0.028959)(9.81)(0.060 0.01136) 0.02027 J 0.02027 0.001614 0.0 kx mg mgx k gain in EPE kx loss in GPE mg gain in KE loss in GPE gain in EPE = = = = = = = =+ = + = =− =− = 19 J (2sf) M1 M1 M1 A1 4 (a) When a charged particle is travelling in a magnetic field, it experiences a magnetic force that is always perpendicular to its velocity (and the magnetic field lines). Since the charged particle is travelling perpendicular to the uniform magnetic field, and that the resultant force only consists of the magnetic force, the particle travels in a circular path. B1 B1 (b) centripetal force provided by magnetic force M1 M1

River Valley High School Pg 3 of 9 JC 2 H2 Physics 9749 Preliminary Examinations 2024 2 2 2 2 2 mr Bqv vm Bq r m Bq Bq m Bq Tm mT Bq       = =  = = = = A1 (b) magnetic force provides for centripetal force 2 26 5 19 (4.5 10 )(4.8 10 ) 0.60(0.15) 2 4.8 10 mvBqv r mvq Br C − − = ==   = M1 A1 5 (a) (i) Hypothetical gas obeys equation of state pV = nRT (perfectly at all pressures, temperature s and volume) B1 (ii) Mean-square-speed (of atoms / molecules) B1 (iii) 21 3pc = 𝜌 = 𝑁𝑚 𝑉 with N explained (m = mass of a molecule) Or 𝜌 = 𝑀 𝑉 (M = mass of a gas) B1 𝑝𝑉 = 1 3 𝑁𝑚〈𝑐2〉 B1 𝑝𝑉 = 𝑁𝑘𝑇 with p, V, T explained B1 So mean kinetic energy 〈𝐸𝑘〉 = 1 2 𝑚〈𝑐2〉 = 3 2 𝑘𝑇 B1 (b) (i) Internal energy U of a system is sum of a random distribution of kinetic and potential energies associated with the molecules of a system. B1 (ii) (in ideal gas) no intermolecular forces, hence no potential energy B1 Internal energy is (solely) kinetic energy (of particles) Since mean (translational) kinetic energy is proportional to thermodynamic temperature of the gas, the internal energy is directly proportional as well. B1

River Valley High School Pg 4 of 9 JC 2 H2 Physics 9749 Preliminary Examinations 2024 (c) Increase in internal energy / J Heat supplied to gas / J Work done on gas / J A to B 1200 0 1200 B to C -1350 -1350 0 C to D -600 0 -600 D to A 750 750 0 1 if WBC and WDA = 0 1 if ∆UDA calculated correctly. 1 if WAB and WCD, QBC and QDA calculated correctly 6 (a) (i) PSYV and QRXW B1 (ii) electrons moving in magnetic field deflected towards face QRXW / electrons accumulate on face QRXW M1 So face PSYV is more positive A1 (b) (i) Arrow point up the page B1 (ii) 𝐸𝑞 = 𝐵𝑞𝑣 𝑣 = 𝐸 𝐵 = 12 × 103 930 × 10−6 C1 = 1.3 × 107 𝑚 𝑠−1 A1 (iii) 𝐵𝑞𝑣 = 𝑚 𝑣2 𝑟⁄ 𝑞 𝑚⁄ = (1.3 × 107) (7.9 × 10−2⁄ × 930 × 10−6) C1 = 1.8 × 1011 𝐶 𝑘𝑔−1 A1 7 (a) (i) A1 (ii) A1 P / W t / s T 2T V / V t / s T 2T

River Valley High School Pg 5 of 9 JC 2 H2 Physics 9749 Preliminary Examinations 2024 (iii) 2 2 max max max 5.0 12.5 W2.0 12.5 3.1 W44 average VP R PP = = = = = = M1 A1 (b) (i) When an alternating current source is connected to the primary coil, there would be a changing magnetic flux produced. The iron-core strengthens and links the flux through the secondary coil. In the secondary coil, since the magnetic flux through it changes all the time, there would be an e.m.f. induced (according to Faraday’s Law). B1 B1 (ii) emf induced in secondary coil = 230 / 20 = 11.5 V current in secondary coil = V / R = 11.5 / 7.0 = 1.6429 A current in primary coil = 1.6429 / 20 = 0.082 A M1 A1 8 (a) The cut-off wavelength corresponds to the most energetic photon that can be produced. That happens when all the kinetic energy of an accelerated electron is lost in a single collision/interaction with the target atom in producing one photon. A1 (b) When electrons striking the metal target interact with the crystal lattice, forces experienced by the electrons cause them to be accelerated, decelerated or deflected. When this occurs, their kinetic energies are lost through the emission of Bremsstrahlung (or “braking radiation”), which are photons of a range of energies which can lie in the X-ray region. Since the magnitude of the ‘deceleration’ experienced by the incident electrons is different for all and is not discrete , the wavelengths of the emitted photons have a continuous distribution so the Bremsstrahlung produces a continuous spectrum of electromagnetic radiation. OR Bremsstrahlung radiation, which is emitted when high energy external electrons coming close to the nucleus decelerate, accelerate or deflect. This energy lost in terms of photons can be any amount of energy less than the maximum kinetic energy of the electrons, therefore forming continuous spectra. A1 A1

River Valley High School Pg 6 of 9 JC 2 H2 Physics 9749 Preliminary Examinations 2024 (c) (i) area = 0.200 m x 0.300 m = 0.0600 m2 Accept a range of 0.0400 m2 ≤ area ≤ 0.200 m2 A1 (ii) Area of a grain: no of grains = 66 1010 0600.0 −−  = 6.00×1010 no of photons = 6.00×1010 ×10 = 6.00×1011 energy = 6.00×1011 × 10-15 = 6.00 x 10-4 J Accept a range of 4.00×10−4 J ≤ energy ≤ 2.00×10−3 J M1 A1 Section B 9 (a) (i) 1. Diffraction refers to the bending or spreading out of waves when they travel through a small opening or when they pass round a small obstacle. B1 2. Interference refers to the superposing of two or more coherent waves to produce regions of maxima and minima in space, according to the principle of superposition B1 3. Coherence refers to having a constant phase difference (and same frequency) (between waves/sources/particles). B1 (ii) Any two of the following: 1. The waves must overlap to produce regions of maxima and minima. 2. The sources must be coherent. 3. The waves must have the same amplitude or approximately the same amplitude. 4. The waves must be unpolarised or with the same p