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 magn