2021 JPJC Prelim H2 Physics P2 Solutions

2021/JPJC/Prelim/9749/02 [Turn over Answers to 2021 JC2 Preliminary Examination Paper 2 (H2 Physics) Suggested Solutions: No. Solution Remarks 1(a)   2 sin 9.81 sin30 4.905 4.9 m s (shown) ag       [1] correct working shown 1(b)    22 2 1 sin30 0.50 1.0 msin30 2 2 4.905 1.0 3.13 m s h d d v u as v v        [1] for correct d [1] correct equation and substitution [1] correct answer 1(c)       2 2 1 2 12.0 3.13sin30 9.81 2 0.50 s horizontal distance 3.13cos30 0.50 1.35 m      s ut at tt t [1] for correct equation and substitution [1] for correct t [1] for correct distance 1(d)(i) In the system of object and trolley, there are no external forces acting on them in the horizontal direction. The only horizontal forces are contact forces (action-reaction pair) acting on each other when object hits the trolley. Hence, total momentum in t he horizontal direction remains constant (or is conserved). [1] for explanation [1] for conclusion 1(d)(ii)         1 1 Applying conservation of momentum, rightward as +ve, 0.35 3.13cos30 1.2 4.0 1.2 0.35 2.48 m s final speed 2.48 m s Direction is to the left f f v v          [1] for correct equation and substitution [1] for correct answer [1] for correct direction 2(a) From the given expression 21 3pc  , we have 21 3 Nmpc V ----- (1) (where m is the mass of 1 molecule) Using the ideal gas equation pV NkT , we have NkTp V ------ (2)

2 2021/JPJC/Prelim/9749/02 No. Solution Remarks Equating (1) and (2), we have 21 3 Nm NkTcVV  21 3 m c kT 213 22m c kT Since k is the Boltzmann constant, the average kinetic energy 21 2 mc of a molecule of mass m, is directly proportional to the thermodynamics temperature T. (shown) [1] for equating (1) and (2) [1] for correct derivation of expression [1] for the correct conclusion 2(b)(i) Total pressure,     total atm 35 5 5 15 1.03 10 9.81 1.01 10 2.526 10 2.53 10 Pa p h g p        [1] for correct method and numerical substitution [1] for correct answer 2(b)(ii) Using pV nRT ,    2 2 1 1water tank 1 21 2 7 3 5 5 53 2.32 10 9.4 10 2.526 10 8.633 10 8.63 10 cm p V pV pVV p         [1] for correct expression and numerical substitution [1] for correct answer 2(c) Pressure at depth of 35 m,     2 atm 35 5 35 1.03 10 9.81 1.01 10 4.547 10 Pa p h g p      Volume of air at depth of 35 m and temperature of 19 °C, 2 2 1 1 21 35m 15m 12 21 21 5 5 5 53 2.526 10 19 273.15 8.633 10 24 273.154.547 10 4.715 10 cm p V pV TT pTVV pT                      [1] for correct method and numerical substitution [1] for correct volume

3 2021/JPJC/Prelim/9749/02 [Turn over No. Solution Remarks Duration of time the air will last for the diver 5 5 4.715 10 45 8.633 10 24.58 24.6 min t     [1] for correct answer 3(a) If path difference of the sound waves along XY is odd-integer of half-wavelength, the waves meet in antiphase and results in destructive interference and hence minimum intensity. [1] [1] 3(b)(i) 22BR 12.00 5.50 13.20 m   [1] 3(b)(ii) R is first order maxima, path difference = 1 BR AR 13.20 12.50 0.70 m     [1] [1] 3(b)(iii) 1 470 0.70 330 m s vf      [1] [1] 3(c) Since 2AI , for ' 3 II , the amplitude of the waves from B is 1 3 A . At Q, the resultant amplitude of the wave interfering destructively is resultant 1 3 A A A  resultant 0.423AA  Hence, the resultant intensity is   2 resultant 0.423 0.179AII [1] amplitude of source B [1] resultant amplitude [1] answer 4(a)(i)  36 12 3.0 A PV   I I I [1] for equation and substitution [1] for correct answer 4(a)(ii)    3.0 20 60 3600 C Qt   I [1] for correct equation and substitution [1] for correct answer

4 2021/JPJC/Prelim/9749/02 No. Solution Remarks 4(a)(iii)    36 20 60 43200 J E Pt   [1] for equation, substitution and correct answer 4(a)(iv) 12 3.0 4.0 VR I   [1] for equation, substitution and correct answer 4(b)(i) 11 34.0 1.33 12 9.0 A1.33 eff eff R R   I [1] for correct Reff [1] for correct current 4(b)(ii)1. There is no change to the potential difference across the lamps since they are all connected in parallel to an ideal battery. With the same potential difference and resistance, the power dissipated and brightness remain unchanged. [1] correct explanation [1] correct conclusion 4(b)(ii)2. Due to the internal resistance of the battery, the potential difference across the lamps decreases . Hence, the brightness of the lamps decreases. [1] p.d. across lamps decreases [1] brightness decreases 5(a)   7 0 4 3 2 4 4 4004 10 3.8 1.6 3.8 10 1.19381 10 T 0.04080 3.8 10 2 1.2 10 Wb Bn NBA                      I [1] [1] for correct substitution 5(b) 4 4 mean e.m.f. 2 1.2 10 0.30 8.0 10 V t       [1] for correct substitution [1] for correct answer

5 2021/JPJC/Prelim/9749/02 [Turn over No. Solution Remarks 5(c) [1] E = 0 V from t = 0 s to 0.5 s and 0.8 s to 1.4 s [1] E = 8.0 × 10-4 V from t = 0.5 s to 0.8 s [1] E = −2.0 × 10-4 V from t = 1.4 s to 2.0 s 5(d) The iron core increases the magnetic flux density , resulting in a larger rate of change of flux linkage. Hence, the mean e.m.f. induced in coil Y is larger. [1] for correct answer 6(a)(i) Wave nature contradicts: Since energy is arriving in continuous manner, a certain time is needed for the electron to gather enough energy before it is ejected. Particulate nature supports: Energy of the incident photon will be instantaneously transferred to the absorbing electron in a one-one interaction. [1] [1] 6(a)(ii) Work function of a surface is the minimum energy required to liberate an electron from the surface. [1] 6(a)(iii) By Conservation of energy, Kinetic energy of electron = Energy of photon – Work done to emit an electron from the surface. But electrons can be ejected from different layers of metal, hence work done can vary, resulting in different KE. When the work done is minimum, KE will be maximum. [1] using conservation of energy or Einstein’s photoelectric equation [1] for any correct explanation. 6(b) Energy of photon, 350 nm 34 8 9 19 6.63 10 3.00 10 350 10 5.6829 10 J 3.55 eV hc            [2] for any correct calculations

6 2021/JPJC/Prelim/9749/02 No. Solution Remarks Energy of photon, 700 nm 34 8 9 19 6.63 10 3.00 10 700 10 2.8414 10 J 1.78 eV hc            To emit an electron, energy of photon must be greater than work function. So electrons will be emitted from potassium metal using light of 350 nm. [1] for correct conclusion 7(a)(i) Energy density is the amount of energy that can be released per unit mass of fuel. [1] for correct answer 7(a)(ii) Rate of consumption of natural gas 6 6 1 total electrical power output efficiency of power stations energy density of natural gas 12600 10 0.27 56 10 12600 0.27 56 833 830 kg s         [1] for correct method and numerical substitution [1] for correct answer 7(b)(i) Solar intensity incident on Earth   26 211 2 radiant power of the Sun, area of perpendicular surface, 3.90 10 4 1.50 10 1379 1400 W m P A        I [1] for correct expression and numerical substitution [1] for correct ans