NJC 2025 H2 Physics Prelim P1 Sol

[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 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"!"#"(ℎ) Density, 𝜌 = mass/volume =#$%&!! ∆((=∆$$ + ∆&& + 2∆!! = 10 + 2(3) + 2 = 18% 3 B 𝑠=(15.0)(2.00)+12(−9.81)(2.00)"=10.4 m 4 A Vertical component of 9.0 N force = 9.0 sin45° = 6.36 N < weight of crate (19.6 N), so no lifting of crate above the ground. 𝐹)*+=horizontal component of 9.0 N force−frictional force =9.0cos45°−2.0=4.36 N 𝑎=,"#$$=#../"= 2.2 m s−2 5 B By COLM, 𝑚0𝑢0+𝑚"𝑢"=(𝑚0+𝑚")𝑣 (5.0)(4.0)+(2.0)(−3.0)=(5.0+2.0)𝑣 𝑣="1.12/.13.1= 2.0 m s−1 Total final kinetic energy = 0"(𝑚0+𝑚")𝑣"=0"(5.0+2.0)(2.0)"= 14 J

2 Question Number Key Solution 6 C (subscript s denotes stationary train and a denotes accelerating train) 𝐹4=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. 𝐹5=P11.772"+(1.2×5.0)"= 13.213 N Since force is proportional to extension, ,%,&=6%6& 𝑥5=,&,%(𝑥4)=0.."0.00.33"(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×T𝜌𝑉𝑔ℎ𝑡W=0.9×T𝑉𝑡W𝜌𝑔ℎ =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 𝑔7=𝐺𝑀7𝑟7"…(1) 𝑔8=𝐺𝑀8𝑟8"…(2) (2)/(1): 𝑟8𝑟7=_𝑀8(𝑔7)𝑀7(𝑔8)=√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 𝐸9:=0"𝑘𝑥"

[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=𝐼1cos"𝜃 and 𝐼∝𝐴" that 𝑘𝐴"=𝑘(𝐴1)"cos"𝜃 So, 𝐴=𝐴1cos𝜃 Alternatively, resolve amplitude to the plane of polarisation of filter = 𝐸1cos𝜃 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 0.001400sin𝜃"=2(567×102;)⇒𝜃"=27.00° 0.001400sin𝜃.=3(567×102;)⇒𝜃.=42.87° The angle between is 42.87 – 27.00 = 15.87° 19 A From Coulomb’s law, 𝐹∝0<! 𝐹=𝐹=T200600W"=19 𝐹==19×180=20 𝜇N 20 C Loss of kinetic energy = Gain in electric potential energy 9.0×1020.=79×2×(1.6×1020;)"4𝜋𝜀>𝑟 r = 4.0 × 10–14 m 21 A 𝐼=𝐴𝑛𝑞𝑣⇒𝑣=𝐼𝐴𝑛𝑞 𝑣=1..1[0.1×(01'()!](D.E×01!))(0./1×01'*+)= 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=2𝜋𝑇=2𝜋150=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 28 C 𝑝=ℎ𝜆=6.63×102.#2.0×1020" =3.32×102"" kg m s20 𝐸F=𝑝"2𝑚=(3.32×102"")"2×9.11×102.0=6.0×1020# 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 𝑁1 After time t, 88,=0. which is smaller than 0" (one half-life) and larger than 0# (two half-lives).