NJC 2025 H2 Physics Prelim P1 Ans

[Turn over 9749/01 H2 Physics Multiple Choice Question Number Key Question Number Key Question Number Key 1 C 11 C 21 A 2 D 12 D 22 B 3 B 13 B 23 B 4 A 14 D 24 B 5 B 15 D 25 D 6 C 16 B 26 A 7 B 17 C 27 C 8 B 18 C 28 C 9 C 19 A 29 B 10 B 20 C 30 C Question Number Key Solution 1 C Option A is the approximate mass of one paper clip. Option B is the approximate mass of one coin. Option D is an unreasonable estimate. 2 D Volume = 𝜋 ( 𝑑 2) 2 (ℎ) Density, 𝜌 = mass/volume = 4𝑚 𝜋ℎ𝑑2 ∆𝜌 𝜌 = ∆𝑚 𝑚 + ∆ℎ ℎ + 2 ∆𝑑 𝑑 = 10 + 2(3) + 2 = 18% 3 B 𝑠 = (15.0)(2.00) + 1 2 (−9.81)(2.00)2 = 10.4 m 4 A Vertical component of 9.0 N force = 9.0 sin45° = 6.36 N < weight of crate (19.6 N), so no lifting of crate above the ground. 𝐹net = horizontal component of 9.0 N force − frictional force = 9.0 cos 45° − 2.0 = 4.36 N 𝑎 = 𝐹net 𝑚 = 4.36 2 = 2.2 m s−2 5 B By COLM, 𝑚1𝑢1 + 𝑚2𝑢2 = (𝑚1 + 𝑚2)𝑣 (5.0)(4.0) + (2.0)(−3.0) = (5.0 + 2.0)𝑣 𝑣 = 20.0−6.0 7.0 = 2.0 m s−1 Total final kinetic energy = 1 2 (𝑚1 + 𝑚2)𝑣2 = 1 2 (5.0 + 2.0)(2.0)2 = 14 J

2 Question Number Key Solution 6 C (subscript s denotes stationary train and a denotes accelerating train) 𝐹𝑠 = weight of mass = 1.2 × 9.81 = 11.772 N When train is accelerating, the spring settles at angle to the vertical so that the horizontal component of the tension provides the resultant force for the train to accelerate. 𝐹𝑎 = √11.7722 + (1.2 × 5.0)2 = 13.213 N Since force is proportional to extension, 𝐹𝑠 𝐹𝑎 = 𝑥𝑠 𝑥𝑎 𝑥𝑎 = 𝐹𝑎 𝐹𝑠 (𝑥𝑠) = 13.213 11.772 (2.4) = 2.7 cm 7 B Minimum force needed to lift weight = 900 N Hence minimum torque needed to lift weight = 900 × 0.20 = 180 N m This torque is provided by the couple of forces F on the lever. Minimum force F = 180 / 1.20 = 150 N 8 B Useful power = 0.9 × 𝑚𝑔ℎ = 0.9 × (𝜌𝑉𝑔ℎ 𝑡 ) = 0.9 × (𝑉 𝑡 ) 𝜌𝑔ℎ = 0.9 × (5.7)(1000)(9.81)(30) = 1.5 MW 9 C Frictional force provide the centripetal force 0.2 = 0.01(0.05) 𝜔2 𝜔 = 20 rad s-1 10 B Immediately after launch, spacecraft is still at/near Earth’s surface, so gravitational field strength remains as g. 11 C 𝑔𝐸 = 𝐺𝑀𝐸 𝑟𝐸 2 … (1) 𝑔𝑁 = 𝐺𝑀𝑁 𝑟𝑁 2 … (2) (2)/(1): 𝑟𝑁 𝑟𝐸 = √𝑀𝑁(𝑔𝐸) 𝑀𝐸(𝑔𝑁) = √17 = 4.1 12 D Loss in thermal energy of water = 0.160 × 4200 × 100 = 67200 J Mass of ice melted = 67200 / 336000 = 0.200 kg Total mass of water = 200 + 160 = 360 g 13 B increase in internal energy = 80 + (−100) = −20 J 14 D For the 4 options, ke and total energy are similar. gpe increases linearly with height (mgx). epe decreases as height increases (smaller extension) and quadratic 𝐸𝑒𝑙 = 1 2 𝑘𝑥2

[Turn over 3 Question Number Key Solution 15 D Lower amplitude throughout. Peak shifts to a slightly lower frequency with more damping 16 B Obtain from Malus’s law 𝐼 = 𝐼0 cos2 𝜃 and 𝐼 ∝ 𝐴2 that 𝑘𝐴2 = 𝑘(𝐴0)2 cos2 𝜃 So, 𝐴 = 𝐴0 cos 𝜃 Alternatively, resolve amplitude to the plane of polarisation of filter = 𝐸0 cos 𝜃 17 C The narrower the slit, the higher the amount of spreading The longer the wavelength, the higher the amount of spreading. 18 C dsin𝜃 = n𝜆 0.001 400 sin 𝜃2 = 2(567 × 10−9) ⇒ 𝜃2 = 27.00° 0.001 400 sin 𝜃3 = 3(567 × 10−9) ⇒ 𝜃3 = 42.87° The angle between is 42.87 – 27.00 = 15.87° 19 A From Coulomb’s law, 𝐹 ∝ 1 𝑟2 𝐹′ 𝐹 = (200 600) 2 = 1 9 𝐹′ = 1 9 × 180 = 20 𝜇N 20 C Loss of kinetic energy = Gain in electric potential energy 9.0 × 10−13 = 79 × 2 × (1.6 × 10−19)2 4𝜋𝜀𝑜𝑟 r = 4.0 × 10–14 m 21 A 𝐼 = 𝐴𝑛𝑞𝑣 ⇒ 𝑣 = 𝐼 𝐴𝑛𝑞 𝑣 = 0.30 [1.0×(10−3)2](8.5×1028)(1.60×10−19) = 2.2 × 10−5 m s−1 22 B When XJ is 0.50 m, RXJ = 20 / 100 × 50 = 10 Ω V of lamp = 1.5 × (10/20) = 0.75 V P = V2/R P’/P = V’2/V2 P’ = (0.75/1.5)2 P = 0.25P 23 B F = BIL F + 3.6 × 10–3 = B(I + 4)(0.15) F + 3.6 × 10-3 = F + 0.6B B = 0.006 T

4 Question Number Key Solution 24 B Option A: Force on each isotope is same, FB = Bqv (same value of force) Option B: The isotopes have different masses. Mv2 / r = Bqv r = Mv / Bq Option C: Uniform circular motion in magnetic field, so speed and k.e. remains constant for each isotope. Option D: Acceleration ma = Bqv a = Bqv/m (the isotope has different mass) 25 D 𝜔 = 2𝜋 𝑇 = 2𝜋 1 50 = 100𝜋 Maximum magnetic flux linkage = NBA Max Induced emf = 𝜔NBA = (100𝜋)(200)(0.20)[𝜋(0.1)2] = 395 V 26 A I2peak = 10 A2 Pmean = ½ Peak = ½ I2peak R I2dc (R) = ½ (10) R I = 2.23 A 27 C 𝜙 = ℎ𝑐 𝜆0 (This wavelength just causes photoemission of electrons at zero kinetic energy) 𝜆0 = 6.63 × 10−34 × 3.00 × 108 2.3 × 1.60 × 10−19 = 5.4 × 10−7 m 28 C 𝑝 = ℎ 𝜆 = 6.63 × 10−34 2.0 × 10−12 = 3.32 × 10−22 kg m s−1 𝐸𝑘 = 𝑝2 2𝑚 = (3.32 × 10−22)2 2 × 9.11 × 10−31 = 6.0 × 10−14 J 29 B The sequence of decay is not important. Deduce the total change in the proton and neutron number after the series of decay. decay proton neutron 𝛼𝛽𝛽 0 –4 30 C Total number of antimony nuclei at t = 0 is 𝑁0 After time t, 𝑁 𝑁0 = 1 3 which is smaller than 1 2 (one half-life) and larger than 1 4 (two half-lives).