2025 RVHS H2 Physics P3 Soln Prelim

RVHS JC2 H2 Physics Preliminary Examinations Paper 3 Mark Scheme Questions Answers Marks 1 (a) Systematic errors are constant deviations of the readings of a physical quantity, either consistently higher, or consistently lower, than its true value. Random errors are deviations of readings of a physical quantity, randomly scattered about the mean reading. [B1] [B1] (b) (i) 21.150 cm 20.980 cm Half the smallest division for read-off [B1] (ii) 0.005 cm (half smallest division) [B1] (iii) A = π(D/2)2 = πD2/4 A/A = 2D/D Using D = (0.005 + 0.005) = 0.01 = 2(0.01)/(21.15 – 20.98) = 11.76 ~ 12% (max 2 s.f.) [M1] [A1] Questions Answers Marks 2 (a) Using 𝑣2 = 𝑢2 + 2𝑎𝑠, 𝑢 = 0 or 1 2 𝑚𝑣2 = 𝑚𝑔Δℎ to arrive at 𝑣 = √2 × 9.81 × 1.96 M1 𝑣 = 6.20(12) m s-1 A1 (b) unchanged B1 Normal contact force on ball is vertical B1 (c) 𝑣2 = 𝑢2 + 2𝑎𝑠, 𝑣 = 0 or 1 2 𝑚𝑣2 = 𝑚𝑔Δℎ to obtain rebound speed 𝑣 = √2 × 9.81 × 0.98 = 4.38 (49)m s-1 B1 impulse = change in momentum 𝑚(𝑣𝑓 − 𝑣1) impulse = 0.034(4.3849 − (−6.2012) = 0.36 N s or kg m s−1 A1

Questions Answers Marks 3 (a) The torque of a force about an axis is the product of that force and the perpendicular distance from the line of action of the force to the axis. B1 (b) Ty = Tcos20 = 410cos20 = 385.7 N Tx = Tsin20 = 410sin20 = 140.2 N Sum of forces in horizontal direction = 0, Rx = Tx = 140.23 N Sum of forces in vertical direction = 0, Ry = Ty – 65 – 20 = 300.27 N Resultant force at elbow = (Rx2 + Ry2)0.5 = 331 N tan-1(140.23/300.27) = 25o Direction = 65 o below the positive x-axis M1 A1 C1 / A1 (c) Using Newton’s 2nd law of motion, the large change in momentum and the short time duration of impact will result in greater force. Using Newton’s 3rd law, the force on hand is equal in magnitude and opposite in direction to the force on wooden boards. B1 B1 300.27 140.23 25° 65°

Questions Answers Marks 4 (a) There is no further gain in KE as the acceleration of the metal sphere reaches zero, leading to a constant velocity. The acceleration reaches zero as the viscous (resistive) force increases with velocity of the sphere, until the viscous force equals to the weight of the sphere. [B1] [B1] (b) According to Newton’s second law, this force results in an acceleration a where F = ma The work done by the constant force F is W = F.s = mas But v2 = u2 + 2as → as = 1 2 (v2 − u2) ∴ W = 1 2 mv2 − 1 2 mu2 = final kinetic energy – initial kinetic energy = ΔEk If the object start from rest ( 1 2 mu2 = 0), ∴ W = 1 2 mv2 (derived) [B1] [B1] [B1] [A0] Constant acceleration is necessary for the derivation [B1]

Questions Answers Marks 5 (a) Angular velocity is the rate of change of angular displacement. Unit is rad s–1 [B1] [B1] (b) From forces p.o.v., 𝑚𝑔 + 𝑇 = 𝑚𝑣2 𝐿 Analyzing FBD of string just taut, when the particle reaches C Hence T = 0 N 𝑚𝑔 = 𝑚𝑣2 𝐿 𝑣 = √𝑔𝐿 [M1] [B1] [A0] (c) From energy p.o.v., initial KE = Gain in GPE + final KE 1 2 𝑚𝑉2 = 𝑚𝑔2𝐿 + 1 2 𝑚𝑣2 1 2 𝑚𝑉2 = 𝑚𝑔2𝐿 + 1 2 𝑚𝑔𝐿 𝑉2 = 𝑔4𝐿 + 𝑔𝐿 𝑉2 = 𝑔4𝐿 + 𝑔𝐿 𝑉 = √5𝑔𝐿 = 7.0 [M1] [A1] Questions Answers Marks 6 (a) (i) The electromotive force (e.m.f.) of a source is defined as the amount of energy transferred from non-electrical forms of energy to electrical energy per unit charge as the charge passes through a complete circuit. B1 (ii) Potential difference = [4.5 / (4.5 + 0.5)] x 3 = 2.7 V A1 (b) Voltage across XP = (Lxp/Lxy) x 2.7 = (0.8/1.5) x 2.7 = 1.44 V At balanced point, VEF = VXP = 1.44 V Current = 1.44 / 3 = 0.480 A M1 A1 (c) (i) Lxp decreases Since R = ρL/A, when A decreases, R increases. Vxy increases so Lxp decreases B1 B1 (ii) Since balance length decreases to potential difference Vxp= VEF, at balance length, there are no changes for the ammeter reading. B1

Questions Answers Marks 7 (a) (i) Longest wavelength → lowest energy photon Therefore emission transition must be between two closest levels, 5 to 4 Do not accept 4 to 5 [B1] (ii) 10 [B1] (iii) Converting 3.6 eV into joules, 5.76  10-19 J is available from the bombarding electrons 5.76 is larger than 5.12, but smaller than 5.81, so maximum absorption of energy will be up to 5.12 Therefore, observed transitions will be between Level 3 and Level 1, i.e. 3 [M1] [M1] [A1] (b) For transitions to be visible to human eye, consider extreme values of 400 nm and 700 nm 400 nm: Using E=hc/λ = (6.63  10-34)(3.00  108)/(400  10-9) = 4.9725  10-19 J 700 nm: (6.63  10-34)(3.00  108)/(700  10-9) = 2.8414  10-19 J From energy levels in Fig. 7.1, only 1 (one) visible line (from Level 1 to Level 2) [M1] [A1]

(c) [B1] (d) Accelerated incident electron ‘collides’ with an electron from the innermost shell (step 1) kicking it out of the K-shell (step 2). The atom is excited due to the vacancy in the K-shell. An electron from the M-shells transits to the vacancy in the K-shell (step 3), emitting a photon in the x-ray frequency, corresponding to the peaks [B1] [B1] Alternate: The high energy incident electrons colliding with the metal target give the innermost shell electrons sufficient energy to be removed from the metal or transited to higher energy levels. The peaks correspond to the characteristic X -rays that are emitted when M-shell electrons electrons transit to the lower energy shells. [B1] [B1] (e) With higher accelerating voltage there will be greater number of electrons that can be removed and accelerated from the filament. Also, generally electrons with higher kinetic energy undergoing deceleration should be able to produce more photons. Hence the higher intensity of the broad spectra with higher accelerating voltage [B1] [B1] (f) The peak frequencies are dependent on the energy transition of the other higher energy shell electrons to the lower energy shells, in which discrete quanta of energy are emitted, depending on energy level structure of the metal. This is independent of applied voltage. [B1] [B1]

Questions Answers Marks 8 (a) beta (minus) decay B1 charge / proton number need to be conserved so with increase in proton number and a particle of negative charge must be emitted. Alternative Also accept nuclear equation B1 (b) (i) Use to kill cancer cells (by implanting it near the tissues) M1 half-life is short so duration of powerful radiation will not be too long or beta particles do not travel far in tissue, so surrounding cell will not be damaged. A1 (ii) 1. Using 𝐴 = 𝜆𝑁, 64 000 = ln 2 2.69×24×60×60 𝑁 to get 𝑁 = 2.14(60) B1 𝑁 = 6.02 × 1023 198 × mass in gram 𝑚 = 7.06 × 10−12 g A1 2. 𝐴 = 64 000 𝑒−ln 2 2.69×13.5 = 64 000 × 0.03085 M1 𝐴 = 1.97 kBq A1

Section B Questions Answers Mar ks 9 (a) (i) No net heat transfer between A and B B1 A and B have same temperature B1 (ii) gradient of graph = 1 𝑐 M1 𝑐 = 1 gradient = 1 4/16 = 4 kJ kg-1 K-1 A1 (iii) 𝑚𝐴𝑐𝐴Δ𝜃𝐴 = 𝑚𝐵𝑐𝐵Δ𝜃𝐵 gives 𝑐𝐴 = 1.5 5.0 × 20 60 × 4000 M1 𝑐𝐴 = 400 J kg-1 K-1 A1 (iv) straight line through origin, 10 times as steep. B1 (b) (i) Δu : increase in internal energy q : heat supplied to the system w : work done on the system M1: definition of quantity A1: direction of change (underlined words) M1 A1

Δu q w process 1 (adiabatic) positive zero positive process 2 (constant volume) negative negative zero process 3 (isothermal) zero positive negative (ii) first row correct B1 (iii) second row correct B1 Volume remains uncha