2021 Prelim H2Phy P3 Ans

1 2021 DHS H2 Physics Prelim Paper 3 Suggested Solutions Section A 1 (a) change in velocity / time (taken) or rate of change of velocity B1 (b) (i) vX = (24 / 1.5) = 16 m s–1 A1 (ii) tan 28° = vY / vX or vX = v cos 28° and vY = v sin 28° M1 vY = 16 tan 28° or vY = 16 × (sin 28° / cos 28°) so vY = 8.5 m s–1 A0 (iii) v = u + at C1 t = (0 - 8.5) / (-9.81) = 0.87 s A1 (iv) straight line from positive vY at t = 0 to negative vY at t =1.5 s M1 line starts at (0, 8.5) and crosses t-axis at (0.87, 0) and does not go beyond t = 1.5 s. A1 (c) v2 = u2 + 2as 0 = 8.52 + 2(–9.81)s or (s = ut + ½at2) s = 8.5×0.87 + ½ × (–9.81) × 0.872 or (s = vt – ½at2) s = 0 – ½ × (–9.81)×0.872 or (s = ½ (u + v)t or area under graph) s = 0.5 × 8.5 × 0.87 C1 s = 3.7 m A1

2 (d) acceleration (of freefall) is unchanged / not dependent on mass, and so no effect (on maximum height) OR explanation in terms of energy: (initial) KE mass, (Δ)KE = (Δ)PE, (max) PE mass, and so∝ ∝ no effect (on maximum height) B1 2 (a) Force experienced by a mass in a gravitational field B1 (b) (i) total downward force on the moving pulley = 960 + (2.4 x 9.81) = 984 N C1 As the moving pulley is moving upwards at a constant speed, net force = 0 2T cos 11.5o = 984 C1 T = 502 N A1 (ii) Cosine of angle (with vertical) decreases OR angle of rope (with vertical) increase M1 Hence tension increases A1 (c) (i) rope / lower pulley has to be lifted up OR load has kinetic energy (any one) B1 (ii) force applied is less than weight of load B1 3 (a) pV / T = constant T = (6.5 x 106 x 30 x 300) / (1.1 x 105 x 540) C1 = 985 K A1 (b) (i) The increase in the internal energy of a system is equal to the B1 sum of the thermal energy supplied to the system and work done on the system. B1 (ii) no thermal energy supplied to system, thus increase in internal energy = work done on system B1 so U increases B1 U increases so kinetic energy of atoms increases and hence T increases B1

3 (c) KE ∝ T, hence v ∝ , and so C1 ratio = A1 4 (a) (i) 0.050 m A1 (ii) C1 C1 T = 0.75 s A1 (b) one point labelled P where ellipse crosses displacement axis B1 5 (a) When two or more waves of the same kind meet at a point at the same time,B1 the displacement of the resultant wave is the vector sum of the displacements of the individual waves at that point at that time. B1 (b) (i) A1 (ii)The distance of the upper loudspeaker from the man is C1 C1 This means that the waves interfere destructively and the man will hence detect a minimum of sound intensity.

4 A1 (iii)1. increase the frequency of the sound B1 2. increase the separation between the two loudspeakers B1 6 (a) The electric force per unit positive charge experienced by a small stationary test charge placed at that point. B1 (b)(i) to the right / from the left / from A to B / in the same direction as electron velocity B1 (ii) v2 = u2 + 2as a = (1.5 × 107)2/ (2 × 2.0 × 10–2) C1 Other alternative calculations for the C1 mark: e.g. a = 1.5×107 / 2.67×10–9 e.g. a = [(1.5×107 × 2.67×10–9) – 2.0×10–2] × [2 / (2.67×10–9)2] e.g. a = (2.0×10–2 × 2) / (2.67×10–9)2 = 5.6 × 1015 m s–2 A1 (iii) E = F / Q = (9.11 × 10–31 × 5.6 × 1015) / 1.6 × 10–19 C1 = 3.2 × 104 V m–1 A1 (c) straight line with negative gradient starting at an intercept on the v-axis and ending at an intercept on the t-axis. B1 7 (a)Angular velocity is the rate of change of angular displacement of a radius joining the body to the axis of rotation. B1

5 (b)(i) (ii) magnitude of the centripetal force about P is the same M1 A1 (iii) x = 0.4d C1 Consider S2, gravitational force provides centripetal force (c)(i) The work done per unit mass in bringing a small test mass B1 from infinity to that point. B1 (ii) curve from r to 4 r, with gradient of decreasing magnitude and line showing potential is negative throughout B1 line passing through (2r, -0.5) and (4r, -0.25) B1

6 Section B 8 (a)(i) A photon is a discrete packet (quantum) of energy of electromagnetic radiation. B1 (ii) The work function energy of a metal is the minimum energy of photon to cause emission of electron from the surface of a metal. B1 (b)(i) 1. f0 correctly labelled B1 2. parallel line with same gradient and higher horizontal intercept B1 (ii) Planck constant B1 (iii) Max KE corresponds to electrons emitted from surface Energy is required to bring electron to surface B1 (iv) intensity determines number of photons arriving per unit time B1 Hence determines number of electrons emitted per unit time, and not energy. B1 (c)(i) 13.60 eV B1 (ii)1. correct energy levels B1 correct direction of arrows B1 2. energy in Joules = 1.90 x 1.6 x 10-19 = 3.04 x 10-19 C1 E = hc/  = (6.63 x 10-34)( 3 x 108) / (3.04 x 10-19) = 6.54 x 10-7 m A1 Visible light region (red) B1 (d)(i)Wavelength that is associated with a particle that is moving. B1 (ii)  = h/mv C1 v = (6.63 x 10-34) / (9.11 x 10-31)(1.2 x 10-10) C1 = 6.1 x 106 m s-1 A1

7 (iii) wavelength is about the separation of atoms in a crystal B1 can be used in electron diffraction B1 9 (a)(i) The average time taken for the initial number of nuclei (or activity) of B1 a particular radioactive nuclide to reduce to half its original value. B1 (ii) Decay is not affected by external or environmental factors (such as pressure, temperature etc) B1 (iii) A1 (iv) (v) Loss of mass = 241.0568229 – (237.0481673 + 4.0026032) C1 = 6.0236 × 10−3 u Energy = (0.0060524 u x 1.66 x 10−27 x (3.00 x108)2 C1 = 9.0423 x 10−13 J = 5.65 MeV A1 (b) For nuclei having high nucleon numbers, the binding energy per nucleon decreases with larger nucleon numbers. B1 When two such nuclei fuse together, they will produce a daughter nucleus

8 which has an even larger nucleon number and smaller binding energy per nucleon. B1 The total binding energy of the products is less than that of the initial nuclei, hence there is an increase in the total mass of the system, and energy has to be supplied for such a reaction to take place. B1 (c) (i) B1 B1 (ii) 1. Initially alpha particle must have some kinetic energy. B1 2. ~ THE END ~