RVHS 2025 Prelim (Answers)

2025 J2 H3 Preliminary Examinations Mark Scheme Section A Question Answer Marks 1(a) ( 𝑚 + 2 𝑚 ) 𝑣 𝐶𝑀 = 𝑚𝑣 𝑣 𝐶𝑀 = 𝑣 3 In centre of mass frame (before collision), P: 𝑣 𝑝 = 𝑣 − 𝑣 3 = 2 𝑣 3 Q: 𝑣 𝑄 = 0 − 𝑣 3 = − 𝑣 3 In centre of mass frame (after collision), P: 𝑣 𝑝 = − 2 𝑣 3 Q: 𝑣 𝑄 = 𝑣 3 In lab frame (after collision), P: 𝑣 𝑝 = − 2 𝑣 3 + 𝑣 3 = − 𝑣 3 Q: 𝑣 𝑄 = 𝑣 3 + 𝑣 3 = 2 𝑣 3 Ratio = 2 𝑣 3 / 𝑣 = 2 3 C1 M1 A1 (b) 𝑣 𝑝 = − 𝐾𝐸 𝑄 𝐾𝐸 0 = 1 2 2 𝑚 ( ) 2 𝑣 3 ( ) 2 1 2 𝑚 ( ) 𝑣 ( ) 2 = 8 9 = 89 % M1 A1 (c) After n collisions, 1 9 ( ) 𝑛 ×10× 10 3 = 0 . 10× 10 − 3 𝑛 = 8 . 389 collisions are needed M1 A1 River Valley High School Pg 1 of 12 J2 H3 Physics 9814 Preliminary Examination 2025

Question Answer Marks 2(a)(i) M1 M1 A0 (ii) Acceleration of the particle = . Since acceleration is always directed to B, its direction is opposite to that of the displacement. By comparison to , Since motion is SHM, speed of particle upon arrival at B = M1 M1 M1 A1 (b)(i) Gravitational field strength along AB can be taken to the superposition of field strength due to the original sphere as well as that of a “negative” mass at where the cavity is. At a point r away from centre of cavity A1 M1 River Valley High School Pg 2 of 12 J2 H3 Physics 9814 Preliminary Examination 2025

Since are all constants, the field strength is uniform. (ii) Since the field strength is uniform, the particle will be accelerating at a constant rate. Hence the ratio is 1:1 M1 A1 (c)(i) To determine an expression for the motion, we consider a time interval ∆ t during which the rocket has mass M , travelling with velocity v , and subsequently ejects ∆ M of gas to the surroundings to gain a velocity of ∆v . The change in momentum during time interval ∆ t is therefore As time interval becomes smaller, we get the following. M1 A1 (ii) By rearranging, we obtain the following expression Since M = M 0 + m 0 , and v = 0 at t = 0, K = M1 M1 A1 River Valley High School Pg 3 of 12 J2 H3 Physics 9814 Preliminary Examination 2025

Question Answer Marks 3(a) . where is the potential difference across the capacitor 𝑑𝑤 = 𝑉𝑑𝑞 𝑉The total work done in charging up the capacitor is then given by: 𝑊 = 0 𝑄 ∫ 𝑉𝑑𝑞 𝑊 = 0 𝑄 ∫ 𝑄 𝐶 𝑑𝑞 The electric potential energy stored is equal to the total work done in charging the capacitor, 𝑈 𝑖 = 0 𝑄 ∫ 𝑄 𝐶 𝑑𝑞 = 𝑄 2 2 𝐶 M1 M1 A1 (b)(i) An inductor stores energy in the magnetic field generated by the current. B1 (b)(ii) At a steady-state, the inductors have no effect on the circuit as Thus the current is that of a circuit with a battery and a resistor: 𝐼 = 𝐸 𝑟 = 75 100 = 0 . 75 𝐴 A1 (b)(iii) There are now 2 capacitors in the circuit which are in series. We can find the effective capacitance: 1 𝐶 𝑠 = 1 40 𝑥 10 − 6 + 1 25 𝑥 10 − 6 ∴ μ F 𝐶 𝑠 = 15 . 4 For the 2 inductors in series: L s = 15 +10 = 25 mH Using conservation of energy, the maximum energy that will be stored on the effective capacitance is equal to the maximum energy stored on the effective inductance at the maximum current. 𝑞 2 2 𝐶 𝑠 = 1 2 𝐿 𝑠 𝐼 2 M1 M1 River Valley High School Pg 4 of 12 J2 H3 Physics 9814 Preliminary Examination 2025

𝑞 = 𝐼 𝐿 𝑠 𝐶 𝑠 = 0 . 75 25 𝑥 10 − 3 𝑥 15 . 4 𝑥 10 − 6 = 4.7 x 10 -4 C A1 Question Answer Marks 4(a)(i) M1 M1 A1 (a)(ii) M1 A1 (a)(iii) If the string has mass, the total inertia of the system would be greater and the net force would be exerted on the total mass of the system. The acceleration will therefore be And hence acceleration will be smaller. M0 A1 (b)(i) M1 M1 A1 River Valley High School Pg 5 of 12 J2 H3 Physics 9814 Preliminary Examination 2025

(b)(ii) As α → g , the tension will tends to zero. B1 (c) Consider a point where the system is moving at constant speed For the hanging mass, For mass on the slope, M1 M1 M1 M1 A1 Question Answer Marks 5(a)(i) let change in intensity be , then we are given or ∆ 𝐼 ∆ 𝐼 ∆ 𝑥 ∝ − 𝐼 ∆ 𝐼 ∆ 𝑥 = − 𝑘𝐼 ; − ve sign is introduced because energy is lost B1 rearranging to form differential equation, 𝐼 0 𝐼 ∫ 1 𝐼 𝑑𝐼 = − 𝑘 0 𝑥 ∫ 𝑑𝑥 B1: forming d.e. B1: with correct limits B2 solving to get 𝐼 = 𝐼 0 𝑒 − 𝑘𝑥 A1 (ii) , or ∆ β = 5 = 10 log 𝑙𝑜𝑔 𝐼 𝑓 𝐼 𝑖 log 𝑙𝑜𝑔 𝐼 𝑓 𝐼 𝑖 = 1 2 giving 𝐼 𝑓 𝐼 𝑖 = 10 M1 Using or 𝐼 ∝ 1 𝑟 2 𝐼 𝑓 𝐼 𝑖 = 𝑟 𝑖 𝑟 𝑓 ( ) 2 = 10 B1 giving m 𝑟 𝑓 = 1 10 1 4 ×1 . 2 = 0 . 67 A1 5(b) path difference = π 𝑟 − 2 𝑟 B1 We want smallest radius that produce minimum intensity, phase difference = π B1 gives ∆ ϕ = ∆ 𝑥 λ ×2π π = π 𝑟 − 2 𝑟 λ ×2π M1 River Valley High School Pg 6 of 12 J2 H3 Physics 9814 Preliminary Examination 2025

cm 𝑟 = λ 2 ( π − 2 ) = 40 . 0 2 ( π − 2 ) = 17 . 5 A1 5(c)(i) ω 1 𝑡 + ω 2 𝑡 = 2 1 2 ω 1 − ω 2 ( ) ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ cos 𝑐𝑜𝑠 1 2 ω 1 + ω 2 ( ) ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ω ' = 1 2 ( ω 1 − ω 2 ) ω = 1 2 ( ω 1 + ω 2 ) A1 (ii) ω≫ω' M1 since ω ' = 1 2 ( ω 1 − ω 2 ) 𝑓 ' = 1 2 ( 𝑓 1 − 𝑓 2 ) or 2 𝑓 ' = 𝑓 1 − 𝑓 2 B1 Hear a louder (4 times intensity) sound, since intensity amplitude 2 ∝ B1 with frequency = 𝑓 1 − 𝑓 2 B1 River Valley High School Pg 7 of 12 J2 H3 Physics 9814 Preliminary Examination 2025

Section B Question Answer Marks 6(a)(i) Consider a ring of mass dm, radius r and thickness dr . The mass of the ring is , where is mass per unit length 𝑑𝑚 = λ 𝑑𝑥 = 𝑀 𝐿 𝑑𝑥 λ Since , 𝑟 2 = 𝑥 2 𝐼 𝑟𝑜𝑑 𝑎𝑡 𝑐𝑒𝑛𝑡𝑟𝑒 = − 𝐿 /2 𝐿 /2 ∫ 𝑟 2 𝑑𝑚 = − 𝐿 /2 𝐿 /2 ∫ 𝑥 2 𝑀 𝐿 𝑑𝑥 = 𝑀 𝐿 𝑥 3 3 ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ − 𝐿 /2 𝐿 /2 = 1 12 𝑀 𝐿 2 Using parallel axis theorem, 𝐼 𝑟𝑜𝑑 𝑎𝑡 𝑒𝑛𝑑 = 1 12 𝑀 𝐿 2 + 𝑀 𝐿 2 ( ) 2 = 1 3 𝑀 𝐿 2 C1 M1 A1 (a)(ii) There are no unbalanced torques about the pivot so angular momentum is conserved. But the pivot exerts an unbalanced horizontal external force on the system, so the linear momentum is not conserved. B1 B1 (a)(iii) 𝑚 2 𝑣 0 𝑑 = − 𝑚 2 𝑣𝑑 + 1 3 𝑚 1 𝐿 2 ω ( 3 . 00 ) ( 10 . 0 ) ( 1 . 50 ) = − ( 3 . 00 ) ( 6 . 00 ) ( 1 . 50 ) + 1 3 ( 9 . 17 ) ( 2 . 00 ) 2 ω rad/s ω = 5 . 88 M1 A1 (b)(i) 60° = π 3 𝑟𝑎𝑑 average angular velocity = = 1.05 rad s − 1 π 3 𝑟𝑎𝑑 1 𝑠𝑒𝑐𝑜𝑛𝑑 A1 (b)(ii) 𝐼 = 1 3 𝑚 𝑎𝑟𝑚 𝐿 𝑎𝑟𝑚 2 + 1 3 𝑚 𝑙𝑒𝑔 𝐿 𝑙𝑒𝑔 2 𝐼 = 1 3 0 . 13 ( ) 75 ( ) 0 . 70 ( ) 2 + 1 3 0 . 37 ( ) 75 ( ) 0 . 90 ( ) 2 = 9 . 08 𝑘𝑔 𝑚 2 Rotational KE = 1 2 𝐼 ω 2 = 1 2 9 . 08 ( ) 1 . 05 ( ) 2 = 5 . 0 𝐽 M1 A1 (b)(iii) Translation KE = 1 2 𝑚 𝑣 2 = 1 2 75 ( ) 1 . 4 ( ) 2 = 72 . 34 𝐽Total KE = 72 . 35 + 5 . 0 = 77 . 34 𝐽 percentage of his kinetic energy is due to the rotation = 5 . 0 77 . 34 = 6 . 46% M1 A1 River Valley High School Pg 8 of 12 J2 H3 Physics 9814 Preliminary Examination 2025

(c)(i) Balancing vertical forces, 𝐹 𝑟𝑜𝑑 𝑐𝑜𝑠 θ = 𝑓 + 𝑤 + 𝐹 𝐹 𝑟𝑜𝑑 𝑠𝑖𝑛 θ = 𝑛 Since , 𝑓 = µ 𝑘 𝑛 𝐹 𝑟𝑜𝑑 𝑐𝑜𝑠 θ = µ 𝑘 𝑛 + 𝑤 + 𝐹