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 s (time for 1 complete revolution), which is larger than s. It cannot be a tau particle as it does not live long enough to make the orbits shown in Fig. 4.2. (since it would have decayed to something else before it can even make 1 complete revolution!) [1] [1] Total = 12 5 (a)(i) The nucleus (volume) is very small compared to the atom [1] (ii) The nucleus is charged The mass is concentrated in a small region/volume/core/nucleus [1] [1] (b)(i) Nuclear fusion is the process where two light/small nuclei are combined to produce a heavier/larger nucleus with the release of energy. [1] (ii) [1] (iii)1. X shown at value of A at 56 at the peak of graph [1] 2. Y shown at value of A close to 1 [1] *(iv) energy from 1 nucleus of Z = (1.77  1013) / (6.02  1023) = 2.94  10–11 J {since qn gave “1 mol of Z” releases 1.77 x 1013 J of energy} nucleon number of Z = 93 + 139 + 2 – 1 = 233 From: E released = BE of products (Sr & Xe) – BE of reactants (Z), Thus BE of Z = [(1.25 + 1.81)  10–10] – 2.94  10–11 = 2.77  10–10 J binding energy per nucleon = (2.77  10–10) / (233) = 1.189 x 10-12 J [1] [1] [1]

4 SAJC 2024 Prelim / 8867 [Turn Over = 1.189 x 10-12 J / (106 x 1.6 x 10-19) MeV = 7.43 MeV [1] (c) Current ratio 2 Y to 1 Zr, so initially there was 3 Y. ܰ ܰ௢ = 2 3 = ൬1 2൰ ௡ = ൬1 2൰ ௧ ௧భ/మ = ቀ ଵ ଶቁ ೟ మ.ళ ೏ೌ೤ೞ n = 0.585 ݈݃ ଶ ଷ =݈݃ ଵ ଶ × ௧ ௧భ/మ or t = n x t1/2 = 0.585 x 2.7 days t = 1.58 or 1.6 days [1] [1] [1] (d) ܫ= ܳ ݐ= ݁ܰ ݐ= (9.8 × 10ଵ଴)(1.6 × 10ିଵଽ) 2 × 60 = 1.31 x 10-10 A = 131 pA (3 sf) (where p = prefix called ‘pico’ = 10-12) [1] [1] Total = 16 6 (a)(i) From 0 ms to 45 ms, the rate of increase of the radius R of fireball decreases with time (since gradient becomes gentler). From 45 ms to 60 ms, the rate of increase is constant (since constant gradient). [1] [1] (a)(ii) With a length and width of 12 m, the fireball will take a maximum of 25 ms to travel from one end of the room to the other (or max of 4.5 ms from the middle of the room to one end) leaving very little time (for the fire-fighter) to react. [1] [1] (b)(i)1 When t = 40 ms, lg t = lg 40 = 1.60 When R = 14.6 m, lg R = lg 14.6 = 1.16 (Accept R from 14.4 to 14.6 m) The point is plotted in the graph below (circled). [1]

5 SAJC 2024 Prelim / 8867 [Turn Over (b)(i)2 Correct best fit line drawn (balance number of points above and below the line) [1] (b)(ii) Gradient = (1.245 – 0.925) / (1.80 – 1.00) = 0.40 [1] [1] (b)(iii) From given eqn R n = k t m, n lg R = m lg t + lg k lg R = (m/n) lg t + (1/n) lg k Thus gradient = (m/n) = 0.40 (from (b)(ii) value) The two smallest integers that would give a ratio of 0.40 is n = 5 and m = 2. [1] [1] [1] (c) By taking values of R 5/V for the 5 data points at t = 40 ms, c can be shown to be a constant. Volume V / m3 R / m c = R 5/V c / m2 12.5 × 10−3 14.6 5.31 × 107 10.0 × 10−3 14.0 5.38 × 107 7.5 × 10−3 13.2 5.34 × 107 5.0 × 10−3 12.2 5.41 × 107 2.5 × 10−3 10.6 5.35 × 107 For all 5 sets of readings, the value of c is approximately the same. Hence, c is constant at t = 40 ms (within limits of experimental error). [2] [1] Total = 14

6 SAJC 2024 Prelim / 8867 [Turn Over Section B 7 (a)(i) Ty = 4.8 sin 30o = 2.4 N [1] (ii) Taking moments about the hinge, Sum of CW moments = sum of ACW moments (W x 0.60) + (0.30 x 0.80) + (Ty x 1.2) = (8.2 x 0.50) W = 1.6 N [2] [1] (iii) Hor component of hinge force = Tx = 4.8 cos 30o = 4.2 N [1] (iv) From EPE = ½ Fx 0.32 = ½ (8.2)(x) x = compression = 0.078 m [1] [1] (v)1. Decrease in GPE = mgh = (0.30)(0.090) = 0.027 J [1] 2. By COE, Loss in GPE = gain in KE of block 0.027 = (0.044) – KEA KEA = 0.017 J = ½ mvA2 Thus vA2 = (0.017 x 2)/ (0.30/9.81) vA = 1.1 m s-1 OR KEB = 0.044 J = ½ mvB2 = ½ (0.30/9.81) vB2 Thus vB2 = 2.88 From v2 = uA2 + 2aysy 2.88 = uA2 + 2 (9.81)(0.090) uA = speed at A = 1.1 m s-1 [1] [1] [1] [1] [1] [1] (vi) Resultant force = gravitational force/ weight is vertical only (Hence no resultant force in horizontal direction) [1] (vii) Straight line with positive gradient starting from non-zero value of vY at time tA [1] [1] (b)(i)1. Acceleration = gradient of v-t = (0.30 – 0.12)/ (0.35 – 0.15) = 0.90 m s-2 [1] 2. By PCLM, m1u1 + m2u2 = m1v1 + m2v2 (0.25 x 0.48) + (0.75 x 0.12) = (0.25 v) + (0.75 x 0.30) v = (-)0.060 m s-1 [1] [1]

7 SAJC 2024 Prelim / 8867 [Turn Over OR u1 – u2 = v2 – v1 (si