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