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 The values of a at the equator and the pole will increase when r increases. 14 A 𝐹𝐹 = 𝐺𝐺𝐺𝐺𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑅𝑅2 = (6.67 × 10−11)(6.0 × 1024)(60) (4.23 × 107)2 = 13.4 𝑁𝑁 15 B Potential energy vs time, for a situation of oscillating object at extreme position at t = 0 s. 16 D D is always not true because the damping force can be in the same direction as the acceleration of the body. 17 A 18 D Since Q is at maximum displacement, it should have zero speed. To check the movement of P and Q, draw the waveform at a short instant later. From the diagram, we can see that P is moving upwards while R is moving downwards. original waveform waveform after a short instant R P

River Valley High School Pg 3 of 4 H2 Physics 9749 JC2 Preliminary Examinations 2023 19 D 11 9 3 3 1 tan 4.8 630 10sin 0.32 10 9.45 10 m θ λθ − − − = = ×= = × = × yy L b y Since the width of the central bright fringe is 2y1 and the width of each higher order bright fringe is y1, 3 13 3(9.45 10 ) = 28 mm−= = ×yy 20 D Let point O be the point of zero potential. When switch S is closed, resistance between OP will reduce. Using the principle of potential divider, the p.d. between OP also reduces which leads to a lower potential at P (lower positive potential value). Since p.d between OP reduces, it will lead to a larger p.d. between OQ and a lower potential at Q (lower negative potential value). 21 A Using I = Ne/t Number of particles travelling in one second = I/e = 10 x 10-6 / 1.6 x 10-19 = 6.25 x 1013 Since the particles travel 2.5 x 107 m in one second, The number of particles in one centimetre length = 6.25 x 1013 / 2.5 x 109 = 25000 22 A Algebraic sum of potential due to charges at A and B is zero, so 1 4𝜋𝜋𝜀𝜀0 +2.4 × 10−6 7.0 cm + 1 4𝜋𝜋𝜀𝜀0 −2.9 × 10−6 BX cm = 0 giving BX = 8.458 cm Pythagoras theorem gives 𝐿𝐿 = √8.4582 − 7.02 = 4.74 cm 23 D In the setup, VPQ = 3/ (3+1.5+22)12.0 = 1.31 V. Hence, if the test cell is in range of mV, Vpq need to be further reduced, thus, increase 22 Ω to 1 kΩ. 24 A Closer magnetic flux lines means a stronger field to the left. S Q P R2 R1 R3 O

River Valley High School Pg 4 of 4 H2 Physics 9749 JC2 Preliminary Examinations 2023 Flux lines pointing left to right suggests a N pole at the left. N pole of bar magnet will be repel while S pole will be attracted, therefore bar magnet will turn anticlockwise and to the left. 25 B 26 D F = B I L = (B L) I Gradient of F vs I graph = B L = 10 / 0.25 = 40 e.m.f. induced = B L v = (40) (xo ω) = (40) (0.024) (2π 80) = 480 V 27 B mcP t θ∆= => mct P θ∆= For steady d.c 2 mct IR θ∆= as P = I2R For a.c with resistance 2R, 2 2( )2 2 ac mc mct I IRR θθ∆∆= = as 2( )2 2 ac IPR= act =T 28 B Current = power / voltage = 1 x 106 / 66 x 103 = 15.15 A Power loss = I2R = (15.2)2(5) = 1100 W 29 A ℎ𝑓𝑓 = 𝜙𝜙 + 𝐸𝐸𝑘𝑘,𝑚𝑚𝑚𝑚𝑚𝑚 gives ℎ𝑐𝑐 𝜆𝜆 = 𝜙𝜙 + 𝐸𝐸𝑘𝑘,𝑚𝑚𝑚𝑚𝑚𝑚, so 6.63 × 10−34 × 3.0 × 108 250 × 10−9 = 4.0 × 10−19 + 1 2 × 9.11 × 10−31 × 𝑣𝑣𝑚𝑚𝑚𝑚𝑚𝑚2 𝑣𝑣𝑚𝑚𝑚𝑚𝑚𝑚 = 9.3 × 105 m s−1 30 D Rate of decay ) (2ln2ln 2121 AnNtNtN A = = =λ where n and NA are number of moles and Avogadro’s number respectively. Hence 21t nA∝ . The ratio 200 6200 is the smallest, i.e. 0.0323. Option A is 0.0476 Option B is 0.0755 Option C is 0.129