YIJC 2023 H2Phy Prelim P3 Solution

2023 YIJC JC2 H2 Preliminary Examination Paper 3 Suggested Solutions 1 (a) The principle of conservation of momentum states that total momentum of the system remains constant provided that there is no net/resultant force exerted on the system B1 B1 (b) (i) Since no net external force, total momentum along AB remains constant. Total initial momentum along AB = Total final momentum along AB 4.0 (4.2 cos θ) + 2.8 (6.0 cos θ) = (4.0 + 2.8)(3.7) 33.6 cos θ = 25.16 θ = 41.5 C1 A1 (ii) Impulse = change in momentum Considering direction perpendicular to line AB Impulse = final momentum – initial momentum = 0 – 4.0 (4.2 sin 41.5°) = − 11.1 N s (accept positive value) C1 A1 (iii) Since, the total final momentum along the direction perpendicular to AB (vertical) is zero, the total initial momentum along the direction perpendicular to AB (vertical) must equal to zero. Since both balls have the same initial momentum (16.8 kg m s−1), the angles have to be the same, so that the total initial momentum along the direction perpendicular to AB equals to zero. B1 B1 (iv) Total KE before collision = ½ (4.0)(4.2)2 + ½ (2.8)(6)2 = 85.7 J Total KE after collision = ½ (4.0 + 2.8)(3.7)2 = 46.5 J Since the KE is not conserved, the collision is inelastic. M1 A1 2 (a) (i) 1. A gas consists of a very large number of molecules. 2. The molecules are in constant, random motion and obey Newton's laws of motion. 3. Collisions between gas molecules are elastic or the collision between gas molecules and the walls of the container are elastic. 4. Intermolecular forces act only during collisions between molecules. The duration of a collision is negligible compared with the time interval between collisions. 5. The volume of the gas molecules themselves is negligible compared with the volume occupied by the gas (this implies that almost all of the gas is empty space). Any 2 of the above B1 B1

2 (ii) As volume compresses, the time between successive collisions of a molecule against the wall decreases or the frequency of the collisions of molecules against the wall increases. For same change in linear momentum, the net force acting on the wall increases, hence increasing pressure B1 B1 (b) (i) pV = nRT For A, 1.5 × 105 × 2.0 × 10–3 = n × 8.31 × 200 n = 0.18 mol A1 (ii) 213 22 rmsE mc kT== .. 27 2 2313 40 1 66 10 1 38 10 20022 rmsc−−    =    crms = 353 ≈ 350 m s–1 M1 A1 (iii) For C to D ΔU = Q + W 3 2 (0.181)(8.31)(400 - 720) = - 50 + W W = - 670 J Work is done by the gas M1 A1 A1 3 (a) The periodic motion of an object about the equilibrium point in which its acceleration is proportional to the displacement of the object from the equilibrium point and always acts in the opposite direction to its displacement. B1 (b) (i) 𝜔 = 2𝜋𝑓 = 2𝜋(12) = 75 rad s−1 C1 (ii) Amplitude x0 = 32 / 2 = 16 mm = 0.016 m 𝑣0 = 𝜔𝑥0 = (75.4)(0.016) = 1.2 m s−1 Obtain zero mark if the first M1 mark is not obtained, ie using the incorrect amplitude. M1 A1 (iii) Maximum contact force at lowest point 𝐹𝑛𝑒𝑡 = 𝑁 − 𝑊 = 𝑚𝜔2𝑥0 𝑁 = 𝑚𝜔2𝑥0 + 𝑚𝑔 = (0.025)(75.42)(0.016) + (0.025)(9.81) = 2.5 N M1 A1

3 (iv) 1 mark each Note: just need to mark and label one of the "1" M2 4 (a) (i) x = 18.0 cm A1 (ii) VA + VB = 0 . ( . ) ( . ) 9 00 3 6 10 04 0 180 4 0 120 Bq   − += qB = −2.4  10−9 C (use of VA = VB giving 2.4  10−9 C scores one mark) C1 C1 A1 (b) By conservation of energy, Total energy at x = 5.0 cm = total energy at x = 27.0 cm e(560) + KEA = e(−600) + KEB (= 0 J) KEA = 1.9  10−16 J ½ mv2 = 1.9  10−16 J Speed of electron = 2.02  107 m s−1 C1 C1 A1 (c) The gradient gives the magnitude of the net E-field strength. Since force experienced equals to charge multiply with net E-field The largest force experienced is at x = 27.0 cm since the gradient is the largest. B1 B1 B1

4 5 (a) (i) Total charge = nALe A1 (ii) Current I = total charge / time = nALe / t = nAve M1 A0 (b) For each cross-section X and Y, v = I / nAe Since IX = IY (series connection) and nX = nY (same material) average drift speed of free electrons at cross-section Y average drift speed of free electrons at cross-section X = X Y A A / ( . ) /  = 2 2 4 0 69 4 d d = 2.1 C1 A1 (c) (i) Resistance = L A =   2 4 L d = ( )( ) ( ) . . −− − 72 23 4 5 1 x 10 50 x 10 0 36 x 10 RAB = 2.5  M1 A0 (ii) 1. LCD = 2LAB, dCD = ½ dAB, CD = AB (same material) Hence, RCD = 8RAB = 20  IAB = E / (RAB + R) = 6.0 / (2.5 + 2.5) = 1.2 A ICD = E / RCD = 6.0 / 20 = 0.3 A Hence, total current supplied by E = 1.2 + 0.3 = 1.5 A C1 C1 A1 2. Potential drop from A to M = IAB(0.5RAB) = 1.2(1.25) = 1.5 V Potential drop from C to N = ICD(0.5 RCD) = 0.3(10) = 3.0 V Hence, p.d. between M and N = 1.5 V M1 A1

5 6 (a) 238 234 4 92 90 2U Th + → Correct notation for alpha particle Total A number conserved and total Z number conserved M1 A1 (b) (i) Initial k.e. of the alpha particle = 25  (5  103)  (5.2  10−18) = 6.5  10−13 J Assumption: (1) All the initial kinetic energy of the alpha particle was used to produce ion - pairs. (2) No energy was used to excite molecules that the alpha particle encountered. M1 B1 (b) (ii) 1. ½ (4  1.66  10−27) v2 = 6.5  10−13  v = 1.399  107 ≈ 1.4  107 m s-1 M1 A1 (b) (i) 2. By conservation of momentum, 0 = mαvα + mThvTh 0 = (4u × 1.399 × 107) + (234u × vTh) vTh = – 2.39  105 ≈ 2.4  105 m s-1 (speed) M1 A1 (c) (i)  = 12 ln2 t  = 0.693 24.1 24 3600 = 3.328  10−7 ≈ 3.3  10−7 s−1 M1 A1 (c) (ii) A = N A r mAN M =    = ( 73.33 10 − )  ( 235 6.02 10234 ) = 4.28  1015 ≈ 4.3  1015 Bq M1 A1 7 (a) (i) 1. The frequency of the source is the number of oscillations per unit time. B1 2. nλ A1

6 (ii) Either Or distance nv time t == f oscillation per unit time so fλ is the distance per unit time M1 Since nf t= , v = fλ distance per unit time is v so v = fλ A1 (b) (i) The incident wave from the loudspeaker and the reflected wave from the hard surface superpose / interfere / overlap. Since the two waves have the same speed and frequency, a stationary sound wave is formed B1 B1 A0 (ii) The positions of the three nodes and two antinodes are where the intensity are zero and maximum respectively. B1 (iii) Node to node = ½ λ = 34 cm v = fλ 330 = f × 0.68 f = 485 ≈ 490 Hz C1 A1 (iv) When x = 20 cm, Intensity = 0.925 I When x = 40 cm, Intensity = 0.25 I Since I = kA2 2 20 2 40 0.925 0.25 x x AI IA = = = 20 40 1.92x x A A = = = C1 C1 A1 (c) (i) 1. Each wavelength (of the white light) travels the same path difference or are in phase when reaching the zero order. All the wavelengths (of the white light) constructively interfere, producing a maximum. Hence the diffraction pattern has white light at the zero order B1 B1 A0 2. Since the wavelength of red light is longer than that of the blue light, the first order of red light is diffracted by a larger angle th