Prelim SAJC (H1) P2 2024 Ans

1 SAJC 2024 Prelim / 8867 [Turn Over SAJC JC2 H1 Physics Prelim 2024 Paper 2 Mark Scheme Section A 1 (a)(i)1 Intercept on graph/ line does not pass through origin [1] 2 Scatter of readings about best fit line [1] (ii) Correction for zero error (0.05 A) explained Use of V and corrected I values from graph Resistance = V/ I = 22.2 Ω (e.g. from 4.0 / 0.18, where 0.18 is the corrected I value) [1] [1] [1] (b) R = V / I = 6.8 / 0.64 = 10.625 %R = %V + % I = (0.1/6.8) x 100 + (0.01/0.64) x 100 = 3.033 % R = 0.0303 x 10.625 = 0.32 Ω = 0.3 Ω (to 1 s.f.) R = 10.6  0.3 Ω [1] [1] [1] Total = 8 2 (a) The gravitational force of attraction between two point masses is proportional to the product of their masses and inversely proportional to the square of their separation. [1] (b) At 8200 km from centre of Earth, F = 6.0 N F = GMm / r2 6.0 = (GM x 1) / (8200 x 103)2 M1 = 6.0486 x 1024 kg At 16000 km from centre of Earth, F = 1.8 N F = GMm / r2 1.8 = (GM x 1) / (16000 x 103)2 M2 = 6.9085 x 1024 kg Average mass M = 6.48 x 1024 kg. [1] [1] [1] Total = 4 3 (a) time spent between plates: t = L/ v = 10 x 10-2 / 6.5 x 105 = 1.54 x 10-7 s [1] (b) = 1.20 x 1011 m s-2 [1]

2 SAJC 2024 Prelim / 8867 [Turn Over Direction of acceleration toward the right (lower potential, metal plate B) [1] (c) In the direction from plate A to B, (where there is a constant acceleration due to the uniform e-field), = 0 + (1.20 x 1011)(1.54 x 10-7) = 18480 = 1.85 x 104 m s-1 = (184802) + (6.5 x 105)2 = 650262 = 6.5 x 105 [1] [1], ecf (d) {no need to draw the red path, for illustration purposes only} (dotted path not required for answer) [1] Total = 6 4 (a)(i) Using Fleming Left hand rule , the magnetic force will always be perpendicular to the direction of the charged particle’s velocity The magnetic force provides for the centripetal force. [1] [1] (ii) Negative [1] (iii) Magnetic force provides the centripetal force, Bqv = q = q α (since m, B, v are constants) so = = 2 [1] [1] [1] (b)(i) As particle X collides with the gas particles, its magnitude of velocity v will gradually decrease. Since r = and only v is decreasing with B, charge q and mass m constant, So r (radius of circular motion), will be decreasing, as depicted by the spiral nature in the diagram. [1] [1]

3 SAJC 2024 Prelim / 8867 [Turn Over (ii)1. v = rω, where ω = , v = r T = 2π() Using r = , T = 2π() T = [1] [1] 2. For the tau particle, m = 3000(9.11×10-31) = 2.733 × 10-27 kg, B = 1.0 T, q = 1.6 × 10-19 C, T = = = 1.07 × 10-7