RVHS H2 Physics P1 Soln

River Valley High School Pg 1 of 4 H2 Physics 9749 JC2 Preliminary Examinations 2023 River Valley High School 2023 JC2 Preliminary Examinations Physics Paper 1 Qn Ans Qn Ans Qn Ans 1 B 11 C 21 A 2 D 12 C 22 A 3 C 13 B 23 D 4 C 14 A 24 A 5 C 15 B 25 B 6 C 16 D 26 D 7 B 17 A 27 B 8 D 18 D 28 B 9 C 19 D 29 A 10 A 20 D 30 D Qn Key guide 1 B Change in vector = final – initial 2 D Option A  2500 W per hour Option B  Based on J1 practical exercise ~ 1.9 m2 Option C  ½(100)(10) 2 = 5000 J Option D  Atmospheric pressure is already 100 KPa 3 C Equation of motion 𝑣𝑣2 = 𝑢𝑢2 + 2𝑎𝑎𝑎𝑎 gives � 𝑣𝑣𝐴𝐴 3 � 2 = 𝑣𝑣𝐴𝐴 2 + 2(−9.81)(0.40) gives 𝑣𝑣𝐴𝐴 = 2.97 = 3.0 m s−1 4 C Let velocity before collision be 𝑣𝑣, and after collision be 𝑉𝑉. Then conservation of momentum gives 𝑚𝑚 𝑣𝑣 + 0 = 2 𝑚𝑚 𝑉𝑉, giving 𝑉𝑉 = 1 2 𝑣𝑣. Initial kinetic energy 𝐸𝐸𝑘𝑘 = 1 2 𝑚𝑚𝑣𝑣2 and after collision, kinetic energy of the system is 1 2 (2𝑚𝑚) � 𝑣𝑣 2 � 2 = 1 4 𝑚𝑚𝑣𝑣2. so change in kinetic energy is 1 4 𝑚𝑚𝑣𝑣2 − 1 2 𝑚𝑚𝑣𝑣2 = − 1 4 𝑚𝑚𝑣𝑣2 = − 𝐸𝐸𝑘𝑘 2 5 C Acceleration along the slope = 30−0 6 = 5 m s−1. Resolve weight along and perpendicular to slope. Along the slope the component of weight is 𝑚𝑚𝑚𝑚𝑎𝑎𝑚𝑚 𝑚𝑚𝑚𝑚 . Hence Newton’s 2nd law gives 𝑚𝑚(𝑚𝑚 𝑎𝑎𝑚𝑚𝑚𝑚𝑚𝑚) = 5 𝑚𝑚, so 𝑚𝑚 = sin−1 � 5 9.81� = 30.6𝑜𝑜 6 C Let W be the weight on the board When the child is not on the board, F(2.0) = W(2.5) The force on the spring F = 1.25W When the child is on the board, F’(2.0) = W(2.5) + (40)(9.81)(5.0) The force on the spring F’ = 1.25W + 981 Hence, the addition force on the spring due to the child is 981 N. The additional compression on the spring = 981/10000 = 9.8 cm 7 B Additional energy stored in spring = 1 2 × 181 × (0.0362 − 0.0212) = 0.07738 J

River Valley High School Pg 2 of 4 H2 Physics 9749 JC2 Preliminary Examinations 2023 loss in GPE = 3.8 × (0.178 − 0.163) = 0.057 J Hence, work done by force = 0.07738 − 0.057 = 0.0204 J Option C distractor if student did not subtract GPE from the EPE 8 D Let driving force by the car be 𝐹𝐹, so Newton’s second law gives 𝐹𝐹 − 200 = 800 × 0.5, giving 𝐹𝐹 = 600 N. Hence power = 𝐹𝐹𝑣𝑣 = 600 × 20 = 12 000 W 9 C Centripetal acceleration = rω2 = r(2π/T)2 Since both r and T are constants, centripetal acceleration does not change over time. 10 A Conservation of energy gives 2400 × 120 = 0.106 × 2260000 + heat loss. Heat loss = 48 440 J. So rate of heat loss = 48440 120 = 404 = 400 W 11 C density of the gas 6.80×1015 1×10−6 × 2.02×10−3 6.02×1023 = 2.282 × 10−5 kg m−3 pressure 𝑝𝑝 = 1 3 × 2.282 × 10−5 × 19002 = 27.456 Pa 12 C equating 𝑚𝑚 = 𝑝𝑝𝑝𝑝 𝑅𝑅𝑅𝑅 at 40 m deep and at the surface, we get (𝜌𝜌𝜌𝜌ℎ+𝑝𝑝𝑎𝑎𝑎𝑎𝑎𝑎)×20 cm3 𝑅𝑅(4+273.15) = 𝑝𝑝𝑎𝑎𝑎𝑎𝑎𝑎×𝑝𝑝 𝑅𝑅(2 0+273.15) , and 𝜌𝜌 𝑚𝑚ℎ = 1 200 × 9.81 × 40 = 4.709 × 105 Pa So 𝑉𝑉 = � 4.709+1.0 1.0 � � 293.15 277.15� × 20 cm3 = 120.8 = 120 cm3 13 B a = g − rω2 Since g can also be expressed as 3 22 44()33ρ π ρπ= = =Gm Gg r Grrr