NJC 2022 Prelim P3 Guide

[Turn over National Junior College 2022 Preliminary Examination 9749 H2 Physics Paper 3 Suggested Solution Section A 1 (a) EK positive and decreasing, not touching x-axis B1 EP (roughly, but not too far off) symmetrical to EK B1 ET clearly shown (visibly or written in text) on the x-axis B1 (b) (i) Kinetic energy lost equals the gain in gravitational potential energy when object of mass m moves from the surface of Earth to infinity ½ mv2 – 0 = 0 – (–GMm / R) B1 v2 = 2GM / R (ii) gravitational field strength at the surface of planet g = GM / R2 equals the acceleration of free fall at the surface of planet B1 from (i), v2 = 2 (GM / R2) × R M1 v2 = 2gR A0 (c) (i) 3/2 kT = ½ m v2 = mgR C1 T = 2mgR / 3k = (2 × 6.6 x 10-27 × 9.81 × 6.4 × 106) / (3 × 1.38 × 1023) C1 T = 2.0 × 104 K A1 (ii) temperature where all substances have minimum internal energy B1 energy r EK EP ET

2 2 (a) sum of random distribution of kinetic and potential energies of the molecules of the gas B1 no intermolecular forces for ideal gas M1 so no potential energy A1 therefore, internal energy equals to total kinetic energy of the molecules only (b) (i) higher temperature increases r.m.s. speed and change in momentum of each molecule colliding with wall of container M1 increase volume reduces frequency of collision M1 force is product of frequency of collision and change in momentum which is constant when increase in change in momentum is balanced by reduction in frequency of collision A1 (ii) increase in internal energy = 3 2 k(460 − 280) × 1.2 × 6.02 × 1023 (= 2692 J) C1 heat supplied = 1.2 × 0.032 × c × (460 − 280) (= 6.912c) C1 heat supplied = increase in internal energy – work done on gas 6.912c = 2692 + 1300 (correct use of first law) C1 c = 580 J kg–1 K–1 or 578 J kg–1 K–1 A1 (c) constant volume so no work done B1 less thermal energy needed for same temperature rise, so smaller specific heat capacity B1

[Turn over 3 3 (a) series of distinct coloured lines against dark background B1 (b) photon emitted when electron changes from higher to lower energy B1 energy of photon proportional to frequency M1 each line represents photon of specific / distinct energy A1 only transition between discrete energy levels leads to specific energies (c) Δ𝐸 = 1.5 − 0.85 (= 0.65 eV or 1.04 × 10–19 J) C1 wavelength = ℎ𝑐 Δ𝐸 = 6.63×10−34×3×108 1.04×10−19 = 1.91 × 10−6 m M1 wavelength is longer than visible light / infrared, so not visible A1 (d) (i) waves passing slits spread B1 waves from each slit overlap M1 with phase difference 360° / path difference λ A1 (ii) angle of λ1 maximum = tan−1 147−73.5 240 (= 17.027o or 0.29718 rad) C1 sin 𝜃2 = sin 17.027° × 654×10−9 488×10−9 (⇒ 𝜃2 = 23.106o or 0.40328 rad) C1 𝑥−73.5 240 = tan 23.106° M1 x = 175.9 cm A1

4 4 (a) emission of electrons from a metal surface when surface is irradiated with electromagnetic radiation above threshold frequency B1 (b) photon has energy dependent on frequency B1 minimum energy required to remove electron from surface B1 emission if energy of photon greater than minimum / work function / threshold energy B1 so, threshold frequency exist (c) (i) stopping potential is the minimum potential difference to stop the most energetic electrons from reaching the anode B1 (ii) for positive values of V, all the electrons emitted from cathode will reach anode B1 (iii) 𝐸 = 𝜙 + 𝑒𝑉𝑠 ℎ𝑓 = 1.6 × 1.6 × 10−19 + 2.4 × 1.6 × 10−19 C1 𝑓 = 9.7 × 1014 Hz A1 (iv) same Vs B1 saturated current doubled B1 V /V I / A − 2.4

[Turn over 5 5 (a) (i) magnetic flux density decreases with distance from straight wire, so magnetic flux linkage in loop decreases as it falls M1 e.m.f. induced in loop proportional to rate of change of magnetic flux linkage A1 OR loop cuts magnetic flux as it falls (M1) e.m.f. induced proportional to rate of cutting of magnetic flux (A1) (ii) magnetic flux into page through the loop decreasing B1 induced current produces magnetic flux into page to oppose the decrease B1 (right hand grip rule gives) induced current flowing clockwise in loop B1 OR magnetic flux density is higher in upper half then lower half of loop, so rate of flux cut and induced e.m.f. larger in upper half than lower half of loop (B1) direction of induced e.m.f. (drives induced current left to right) is same in both halves, so net e.m.f. (B1) drives current clockwise in loop (B1) (b) induced e.m.f. = − ΔΦ Δ𝑡 = − (30−120)×10−3×𝜋×0.0502 0.040 M1 = 0.0177 V = 18 mV A0

6 6 (a) period = 15 ms C1 frequency = 1 / period = 1 / 0.015 = 67 Hz A1 (b) r.m.s current = 0.75 / √2 C1 = 0.53 A A1 (c) energy = Irms2 R t = 0.532 × 450 × 30 × 10–3 C1 = 3.8 J A1 (d) instantaneous power P dissipated in resistor is P = I2 R where I is instantaneous current and R is resistance of resistor B1 current sill flows in resistor in opposite directions during half-cycles, so heating resistor B1 OR average power <P> dissipated in resistor is <P> = I rms2 R where I rms is root-mean-squared current and R is resistance of resistor (B1) I rms not zero, so heating resistor (B1)

[Turn over 7 Section B 7 (a) progressive: energy transferred along direction of propagation of wave B1 longitudinal: particles along wave oscillate along axis parallel to direction of propagation of waves B1 (b) (i) amplitude = 2.0 × 10–11 m A1 frequency = 500 Hz A1 (ii) maximum speed = 𝜔𝑥0 = (2𝜋 × 500) × 2 × 10−11 M1 = 6.3 × 10−8 m s−1 A0 (iii) maximum kinetic energy = ½ mv02 1 2 × 𝑚 × (6.3 × 10−8)2 = 2.4 × 10−19 C1 𝑚 = 1.2 × 10−4 kg A1 (iv) ellipse / circle M1 with correct axes labels A1 (c) (i) energy transferred from beating wings to the ear / eardrum vibrates at same frequency as beating wing B1 frequency within audible range of person (20 Hz – 20kHz) B1 (ii) intensity ∝ 1 distance2 C1 1.6𝐼 𝐼 = ( 2.0 2.0−𝑥) 2 where I is original intensity C1 x = 0.42 m A1 v / m s–1 y / nm 6.3 x 10–8 0.020 –0.020 –6.3 x 10–8

8 (iii) air molecule travels 4 × 5.0 × 10–9 (= 2.0 × 10–8 m) each cycle C1 total distance covered in 1800 beats = 1800 × 2.0 × 10–8 = 3.6 × 10–5 m A1 (iv) device generate sound at same frequency as buzzing sound B1 of equal / similar amplitude B1 and 180o out-of-phase / anti-phase B1 leads to destructive interference of buzzing sound and the generated sound B1 that cancels the buzzing sound

[Turn over 9 8 (a) (i) 1. mass B1 2. positive electric charge B1 (ii) magnetic force acts on moving charge in magnetic field, force on it may not be due to electric field B1 (b) (i) magnitude of magnetic flux density at wire X = 𝜇0𝐼 2𝜋𝑑 = 4𝜋×10−7×290 2𝜋×0.050 M1 = 1.2 × 10−3 T A0 (ii) 𝐹 = 𝐵𝐼𝐿 ⇒ 𝐹 𝐿 = 𝐵𝐼 force per unit length = 1.2 × 10–3 × 290 C1 = 0.35 N m–1 A1 (c) (i) electron moving perpendicular in magnetic field experience a magnetic force B1 force = 𝐵𝑞𝑣 = 4.64 × 10−3 × 1.6 × 10−19 × 2.9 × 107 M1 = 2.15 × 10−14 = 2.2 × 10−14 N A0 (ii) magnetic field between the wires is not uniform B1 magnetic force / resultant force on electron not constant M1 so, claim is incorrect A0 (iii) curve downwards B1 (d) [Method 1] electric force on electron = 𝑞𝐸 = 1.6 × 10−19 × 13500 0.10 = 2.16 × 10−14 N M1 either force equal to (c)(i), so constant velocity or force ≠ (c)(i), so cannot be constant velocity A1 [Method 2] For electron to move a