2025 TJC H2 Phy Prelim P2 Solution

2025 H2 Physics P2 Solution 1 (a) The distribution is not uniform with more mass loaded nearer to the left. B1 (b) Resultant force on the boat is zero Resultant moment about any pivot is zero B1 B1 (c) T1 +T2 = 15000 15000(0.75) =T2 (2) T1 = 9400 N T2 = 5600 N C1 A1 A2 (d) Wd = 15000 (30) P = 15000 (30)/12 = 37500 W B1 2 (a) The cube undergoes resonance when the driving frequency of the water wave equals the natural frequency of the oscillating system/cube. There is maximum transfer of energy from the water wave to the cube/ The energy of the system/cube becomes a maximum and the system/cube oscillates with maximum amplitude. B1 B1 (b) v = fλ f = 2.0 Hz Substitute f = 2.0 Hz into l gf 2 1= l = 0.0621 m M1 A1 (c) (i) Increase in wavelength results in decrease in driving frequency, thus driving frequency is not equal to natural frequency, Amplitude decreases. M1 A1 (ii) Increase in mass results in an increase in l, and an decrease in natural frequency. Thus, driving frequency is not equal to natural frequency, Amplitude decreases. M1 A1 (d) Drag force due to water on the cube causes damping. Thus, maximum amplitude of oscillation occurs at a driving frequency 2 Hz, which is smaller than the natural frequency. Since l gf 2 1= , a smaller value of f used in calculation results in a larger value of l calculated. As such, the value determined in (b) is larger than actual measurement. B1 B1

3(a) Using v = fλ v = 3 1 1.4 4.0 10 −   = 350 m s-1 M1 A1 (b)(i) Particle R (As the particle is undergoing SHM, at the amplitude, the instantaneous velocity of the particle is zero). A1 (b)(ii) Particle Q . A1 (b)(iii) Displacement of article Q will be positive next instant. A1 Vertical lines represent equilibrium position of particle along the wave. Particle P is at the centre of compression and particle Q is at the centre of rarefaction | | | | 0.7 1.4 2.1 2.8 position / m displacement / nm 5.00 −5.00 P Q R

(c)(i) Distance travelled by wave in 1 ms = vt = (350)(1 x 10-3) = 0.35 m The graph should have shifted to the left by 0.35 m. Award mark as long as one full wavelength is drawn with displacement at 5.00 nm at initial position. C1 A1 (c)(ii) Phase difference between particle R and S 0.7360 3601.4 x   =  =   180= OR Phase difference between particle Q and S is 90o. C1 A1 displacement / nm t / ms 5.00 −5.00 | | | | | | 2.0 4.0 6.0 8.0 10.0 12.0 Z | | | | 0.7 1.4 2.1 2.8 position / m displacement / nm 5.00 −5.00 P Q R Y

4 (a) 1. Waves must meet  rad out of phase. 2. Waves must have equal amplitude. A1 A1 (b) (i) Wavetrains from S1 and S2 are coherent and superpose at points along YZ. When path difference is an integral multiple of , the waves meet in phase, constructive interference takes place to give a series of maxima. C1 C1 (ii) It decreased to one quarter of the original x (since x = D/a) A1 (iii) The line YZ is not parallel to the slits or the slits not normal to the (incident) microwaves A1 (iv) Place a polariser in front of the transmitter and rotate through 90o OR rotate transmitter/detector through 90o. If this causes minimal/zero signal at some angles, the wave is plane polarized. M1 A1 (c) Distance between two nodes = ½  = speed of detector / frequency of detection = 10 / 1.5 = 6.7 mm Hence, wavelength = 13 mm f = c/ = 3.00 x 108 / 13 x 10-3 = 2.3 x 1010 Hz. C1 C1 A1 (d) (i) White light diffracts after passing through the slits in the grating. For zeroth order maxima, each of the wavelengths in the white light travels the same path length/ zero path difference. The amplitudes add up vectorially to produce a resultant white colour maxima B1 A1 (ii) wavelength  of red light > wavelength blue light ( red > blue ) For waves from any two adjacent slits, path difference is d sin, where d is the separation between the slits. To produce a maxima for 1st order, path difference, dsin  = 1  Hence, maxima for different colors occurs at different angle  , with red light at a larger angle B1 A0 B1 5 (ai) Vrms = 𝑉0 √2 = 170 √2 = 120 V A1 (aii) 𝜔 = 2𝜋/𝑇 = 314 𝑇 = 0.0200 s A1 (b) (i) 𝑉𝑆 𝑉𝑃 = 𝑁𝑆 𝑁𝑃 𝑉𝑆 170 = 3500 2000 VS = 298 V C1

A1 (ii) 𝑉𝑆 = 𝐼𝑆𝑅 298 = 𝐼𝑆(130) 𝐼𝑆 = 2.288 A 𝐼𝑃 𝐼𝑆 = 𝑁𝑆 𝑁𝑃 = 3500 2000 IP = 4.00 A C1 A1 b iii 𝑉𝑆 = 𝐼𝑆𝑅 298 = 𝐼𝑆(130) 𝐼𝑆 = 2.288 A 𝐼𝑃 𝐼𝑆 = 𝑁𝑆 𝑁𝑃 = 3500 2000 IP = 4.00 A C1 A1 Power loss due to Induced Eddy currents Hysteresis los Any possible causes B1 (c) 6 (a) (i) From Fig. 6.2, the current at t = 0.070 s is 1.500 A. I − − == = =    = 7 0 2 3000 200000.15 4 10 20000 1.500 3.77 10 T n Bn M1 M1 A0 I/A t/s 0 0.020 0.040 2.29 × 2 = 4.58 -2.29

(ii) Using points (0.070, 1.500) and (3.80, 0.550) on the line in Fig. 5.2, d 0.550 1.500 3.0645 3.06d 0.380 0.070t −= =− =−− I According to Faraday’s law, s s0 s0 47 22 dde.m.f. ( ) dd d ()d d d 1500 2.5 10 4 10 20000 ( 3.0645) 2.8882 10 2.89 10 V cs c c N B Att N A n t N A n t   −− −− =− =− =− =− =−       − =  =  I I C1 C1 A1 (iii) The current Is in the solenoid flows from the positive terminal of the battery. By right-hand-grip rule, the magnetic flux density Bs produced points to the left. Since Is decreases, Bs also decreases so by Lenz’s law, the induced magnetic flux density Bc in the coil must point to the left. By right-hand-grip rule, the induced current Ic in the coil flows from left to right through the galvanometer. B1 B1 B1 7 (a) gravitational force exerted by Sun on Earth provides the centripetal force m r ω2 = GMm / r2 r3 = GM × (T / 2π)2 r3 = 6.67 × 10-11 × 2.0 × 1030 × (365 × 24 × 3600 / 2 π )2 1 AU = r = 1.498 x 1011 m = 1.50 × 1011 m B1 C1 C1 A1 solenoid coil A Bs Bc Ic Is

(b)(i) The mass of the Sun is much bigger compared to the total mass of all the planets, and so their gravitational influence is small. B1 (b)(ii) To escape to infinity, Initial KE of object at Jupiter ≥ gain in GPE from Jupiter to infinity (1/2)mv2 ≥ 0 – (-GMm / r) (1/2) v2 ≥ GM / r v2 ≥ (2 × 6.67 × 10-11 × 2.0 × 1030) / (5.2 × 1.50 × 1011) vmin = 1.85 × 104 m s-1. C1 C1 A1 (b)(iii) According to Fig 7.3, the speed of Voyager 2 at Jupiter on 9 July 1979 was 10.5 km s-1 = 1.05 × 104 m s-1, which is less than the escape speed of 1.85 × 104 m s-1 needed at that distance from the Sun. Hence, it was not travelling fast enough to escape to infinity. B1 (b)(iv) From Fig. 3, the speed of Voyager 2 increased from 10.5 km s-1 to 28.0 km s-1 during its interaction with Jupiter. Hence, its gain in momentum = Δp = pf – pi = mvf – mvi = 773 (28.0 – 10.5) × 103 = 1.35 × 107 kg m s-1 Velocity must be read to ½ square (1 dp) C1 A1 (b)(v) Voyager 2 gains kinetic energy/momentum from Jupiter’s orbital kinetic energy about the Sun. In a gravity assist, the spacecraft exchanges momentum/energy with the moving planet, the craft gains a tiny amount of the planet’s orbital energy C1 (c)(i) Half-life, t1/2 = 87.74 years A = A0 exp(-λt) or