ASRJC 2025 Prelim H2 P1 MS and Answers

1 9749/01/ASRJC/2025 Prelim [Turn Over Anderson Serangoon Junior College 2025 JC2 H2 Physics P1 Prelim Mark Scheme Paper 1 (30 marks) E – Easy, A – Average, D – Difficult 1 2 3 4 5 6 7 8 9 10 C B B C C A D B B D 11 12 13 14 15 16 17 18 19 20 B B C A B A A D D A 21 22 23 24 25 26 27 28 29 30 D A B A D A D C C C 1 C Typical mass of a small car = 1500 kg Average KE of car = ½ (1500) (302) = 0.68 × 106 = 1 × 106 J (1 s.f.) E 2 B For a fixed resistor, R = constant or ratio of V/I is a constant. The presence of a non-zero y-intercept indicates a systematic error. There is no random error as all points lie on the best fit line E 3 B Using s = ut + 0.5at2, as u = 0 and a = constant, Hence, s  t2 2 2 (0.5 ) 0.25 xT LT xL = = A 4 C Terminal velocity was reached in water as shown by the horizontal line. Area A (distance travelled in air) > Area B (distance travelled in water). Rate of change of velocity in air (gradient of curve) is decreasing. A area A (in air) area B (in water)

2 9749/01/ASRJC/2025 Prelim 5 C Forces must be of the same type, opposite in direction and acts on different body. A 6 A 90 kg min-1 = 1.5 kg s-1 N 305.120)( ==== t mvt mvF A 7 D Since the rod is in equilibrium, Σ = 0 Taking moment about O, T(2Lsin 30o) = 20g(L 2 cos 60o) + 30g(3L 2 cos 60o) T = 269.78 = 270 N D 8 B WD to stretch = ½ × 0.5 × 35 = 9 J Energy recovered = ½ × 0.6 × 20 = 6 J Total WD in 1 cycle = Energy remaining in 1 cycle = 9 – 6 = 3 J Or by counting the no of squares in the area between the two curves Energy remaining = (5 × 0.1) × 6 = 3 J D 9 B Constant force: Acceleration, a = constant Initial speed, u = 0 Therefore v = u + at = at P = Fv = Fat tP  A 20 g 30 g 120o

3 9749/01/ASRJC/2025 Prelim [Turn Over 10 D The vertical component of the acceleration is the centripetal acceleration which is present since the particle is performing circular motion. The horizontal component of the acceleration causes the speed of the object to increase. E 11 B On P, Since weight of mass is 1.0 N, g = weight/mass = 1.0 / 1.0 = 1.0 m s–2 On Q, Mass of 1.0 kg remains unchanged on Q. g = 1.0 x 10 = 10 m s–2 Weight of mass = 1.0 x 10 = 10 N E 12 B 5 1 2 2 2 2 1 3 1 3 1 31.0 10   1.2 500 ms NmPc V Pc c c  − =   = =       = E 13 C For full cycle processes, the net change in internal energy is zero. 560 [( 4.2) (1.0 10 (20.0 5.0) 10 )] 2.7 J U Q W Q Q −  = + = + − +   −  = E 14 A at equilibrium position, x = 0 mm amplitude = 50 mm period = 2.0 s Time shutter remained open t, from x = 0 s to x = 25 mm, = 225 50sin 2.0 t t = 0.17 s D

4 9749/01/ASRJC/2025 Prelim 15 B Air molecules undergo simple harmonic motion. As intensity is proportional to square of amplitude, amplitude of vibration of air molecules increases. Since vo = xo = (2πf)xo, maximum speed increases. Speed of wave travel of sound depends on the medium (density, temperature etc), not intensity. A 16 A ( )path differencephase difference 2 phase difference betwee n sourceswavelength =  + As light intensity at P is zero, the waves from R and S meet at P with phase difference of nπ, where n = 1, 3, 5, 7.... A: phase difference at P = 2π + π = 3π (correct) B: phase difference at P = ½ × 2π + π = 2π (incorrect) C: phase difference at P = 2π + ½π = 5/2 π (incorrect) D: phase difference at P = 2π + 0 = 2π (incorrect) A 17 A d sin  = n  For the same d and , 3X = 2 red 3X = 2 × 720 × 10−9 X = 480 nm A 18 D F = qE ma = qE a = qE/m  Since q, E, m are constants, a is constant v2 = u2 + 2ax u = 0, v2 = 2ax KE = ½ m v2 = ½ m(2ax)  KE is proportional to x. A 19 D From definition, WV= Q From definition, W PtV= Q t = I , and VR= I , hence 22 2 2 VWP V = R R QR= = =II E 20 A When S is closed, effective external resistance decreases, hence current (ammeter reading) increases, and cause voltmeter reading to decrease as p.d across internal resistance increases. A

5 9749/01/ASRJC/2025 Prelim [Turn Over 21 D ou 4 24 tVE = + ou 2 3 tV E = A 22 A Potential gradient k = 0.50 1.50.50 0.50 0.7810.96 xy xy VV   += = = E = k lXZ = (0.781)(0.64)= 0.50 V D 23 B A 24 A By Fleming’s Left Hand Rule, force is directed out of the page. As  tends to 0 , force becomes maximum , as wire becomes perpendicular to flux density (or as  tends to 90 , force becomes minimum ). Hence the force is a cosine function. F = BIL where B and L are perpendicular to each other = BI (2) (L/2 cos ) = BIL cos  A 25 D The wire segment can be split into 2 portions, the triangular segment and a rectangular segment. The magnitude of the rate of change of magnetic flux when triangle is entering or leaving the field, is increasing as the rate of change of area in field increases. Induced e.m.f. and hence induced current is increasing. At the rectangular segment, rate of change of flux is constant, thus induced emf and induced current is a constant. A P

6 9749/01/ASRJC/2025 Prelim 26 A o oo III 23 22 22 ==+ )()( A 27 D Frequency of violet light f = 7 × 1014Hz (or wavelength of violet light  = 400 nm) Estimated energy of a photon of violet light, E = hf = (6.63 × 10−34) (7 × 1014) = 5 × 10−19 J A 28 C Based on HUP, x p ≥ h (1.00 x 10-20) (p) ≥ 6.63 × 10−34 p ≥ 6.63 × 10−14 Percentage change in uncertainty of momentum 1 14 14 4 4.00 10 100%6 4.00 1 6. 3 10 0 − − − −=   = 66% D 29 C Since half-life of sample (33 years) is much greater than duration of decay (2 days), the activity of the sample is approximately constant during the duration of decay. From A = -dN/dt The number of decayed nuclei (dN) = A(dt) = 4.0 x 105 (2 x 24 x 3600) = 6.9 x 1010 OR 𝐴 = 𝑁 𝐴𝑂 = [𝑙𝑛2 𝑡0.5 ] 𝑁𝑂 4 × 105 = 𝑁𝑂𝑙𝑛2 33 × 365 × 24 × 60 × 60 𝑁𝑂 = 6.00558 × 1014 𝑁 = 𝑁𝑂𝑒−𝑡 𝑁 = 𝑁𝑂𝑒 −[𝑙𝑛2 𝑡0.5 ]𝑡 = (6.00558 × 1014)𝑒−[ 𝑙𝑛2 33×365×24×60×60](2×24×60×60) = 6.004889 × 1014 Thus, 𝑁𝑂 − 𝑁 = 6.91160 × 1010 = 6.9 × 1010 (2 s. f. ) A

7 9749/01/ASRJC/2025 Prelim [Turn Over 30 C 𝑋𝑏 𝑎 → 𝑌𝑏 𝑎−4 + 𝐻𝑒2 4 + 2 𝑒−1 0 Between X and Y, the number of neutrons differ, and the number of protons are the same. Hence, nucleus Y is an isotope of nucleus X. E