TJC 2022 Prelim P3 Guide

TJC 2022 H2 Physics Prelim Solutions P3 1 (ai) Internal energy is determined by the state of the system and it can be expresses as the sum of a random distribution of kinetic energy associated with the molecules of the system. [B1] (ii) pV = NKT (1.05  105)  (2.9  10−4) = N  (1.38  10−23)  (303) N = 7.282  1021 = 7.28  1021 [M1] [A1] (iii) ( ) ( ) 23 21 21 3 2 3 1.38 10 3032 6.272 10 6.27 10 J KE kT − −− = =    =  =  [A1] (b) (i) ( )( ) ( ) 21 23 3 2 3 7.282 1 0 1.38 10 357 3032 8.14 J U Nk T −  =  =    − = Allow e.c.f. from aii [M1] [A1] (ii) Since gas A undergoes an adiabatic compression, Q = 0. From the 1st law of thermodynamics, U = Q + W 8.14 = 0 + W W = 8.14 J [A1] (iii) (For gas A, constantPV T = . 454 52.1 101.05 10 2.9 10 1.71 10 Pa303 357 A A p p −−    =  =  For gas B, PV = constant. (1.05  105)  (2.9  10−4) = pB  (2.1  10−4)  pB = 1.45  105 Pa) 2.9 2.1 1.71 1.45 pressure / 105 Pa volume / 10−4 m3 gas B gas A 1.05 [B1] graphs start at same point, with A above B and A steeper than B [B1] arrows from initial to final state. [B1] Initial P, V indicated

2 (a) The acceleration of the particle is proportional to its displacement from the equilibrium position and is always directed towards that position. B1 B1 (b) At y = 0.000 m, TE = KE + GPE + EPE = 0 + 0 + 0.445 = 0.445 J -Accept answer between 0.440 J and 0.445 J. C1 A1 (c) Straight line passing through the origin and (0.393, 0.160). 1. At y = 0.160 m, TE = KE + GPE + EPE 0.445 = 0 + GPE + 0.049  GPE = 0.396 J Acceptable value for the GPE at y = 0.160 m ranges from 0.390 J to 0.396 J. 2. Students who draw curves get zero mark. B1 B1 (d) Method 1: Consider (0.393, 0.160) on GPE line, mg (0.160) = 0.393 m = 0.250 kg Method 2: At y = 0.080 m, the equilibrium position of the SHM, extension e of spring is 0.040 m. (See Fig. 4.1). There is no net force on m, thus ke = mg (61.4)(0.040) = m (9.81)  m = 0.250 kg C2 A0 (e) Method 1: 2 2 2 max 11 22 ooKE mv m x ==  22120.196 (0.250)( ) (0.080)2 T = T = 0.401 s Method 2: 𝜔2 = 𝑘 𝑚 ⇒ (2𝜋 𝑇 ) 2 = 𝑘 𝑚 𝑇 = 2𝜋√ 𝑚 𝑘 = 2𝜋√0.25 61.4 = 0.401 s Method 3: At y = 0.000 m, the extension of spring is maximum, emax = 0.120 m. The mass is at its amplitude position in its SHM, so displacement is xo = 0.080 m. (See Fig. 4.1). +: netF ma=  max oke mg ma−= 2 max () oke mg m x −= 2 (61.4)(0.12) (0.25)(9.81) (0.25)( (0.08))−= 2 0.401 sT  == C1 A1

3 (a) A A VF qE q d V V V V V 16 191.12×10 1.6×10 0.10 0 70.0 70.0 V −− == =  = − = = (b)(i) P P q V mv V V V 2 19 31 6 2 Gain in PE Loss in K.E. of electron 1 2 11.6 10 9.11 10 (4.30 10 )2 70 52.6 V 17.4 V −− = = −   =     − =− = (b)(ii) E =V/d = 70/10 = 7.0 V cm-1 for V = 52.6 V, d = 52.6/7.0 =