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 = 7.5 cm [C1] (ecf) Draw a vertical line 7.5 cm from plate A [M1] (c) The electron will stop before the equipotential line as the horizontal component of velocity in the direction of the field would be lower. [A1] (d)(i) ( ) ( ) y y yy xx F ma qE a a m s s a t t ts s u t m 27 19 4 12 2 2 2 12 2 8 68 1.67 10 1.6 10 2.0 10 1.92 10 1 2 110.50 10 1.92 1022 5.1 10 3.5 10 5.1 10 0.179 −− − − − − ==  =    = =  =  = = =    = (d)(ii) a proton = qE / m a alpha = 2qE / 4m = ½ a proton = 9.6 x 1011 m s-2 Same vertical displacement for both proton and alpha [C1] t alpha = 7.2 x 10-8 s A 10.0 cm electron v B P (b)(ii) [C1] [A1] [C1] [A1] (ecf) [C2] [A1] [C1]

Using 𝑠𝑥 = 𝑢𝑥𝑡, since ux is the same for both, (𝑠𝑥)𝑎𝑙𝑝ℎ𝑎 = 0.252 𝑚 therefore alpha particle will exit the plates [A1] 4 (a) The heating effect is due to power dissipation in the coil which is dependent on the root-mean-square current and not on the average current. [C1] The root mean square current is the value of the steady direct current that would dissipate heat at the same rate as the alternating current in a given resistor and it is non-zero. Hence, there is heating effect in the coil. [C1] (b) (i) 2 26 6 3 15 16 240 20 10 20 10 1.39 10240 10 ss pp s loss VN VN V k kV P I R W = =  = =  =   =   (ii) 𝑃𝑙𝑜𝑠𝑠 = ( 20×106 16×103) 2 × 20 × 10 = 3.13 × 108𝑊 𝐸𝑛𝑒𝑟𝑔𝑦 𝑠𝑎𝑣𝑒𝑑 = (3.13 × 108 − 1.39 × 106) × 24 ÷ 1000 = 7.48 × 106𝑘𝑤ℎ 𝐴𝑚𝑜𝑢𝑛𝑡 𝑠𝑎𝑣𝑒𝑑 = 7.48 × 106 × 0.10 = $7.48 × 105 5 (a) oRWcos14 8500==  oR 8500 cos14= Horizontal component of R = oR sin14 = o o 8500 sin14cos14 = 2100 N B1 B1 (b) Horizontal component of R provides the centripetal force for circular motion.  vm r 2 2100=  v 28500 21009.81 150  =  v = 19 m s-1 M1 A1 (c) There is a greater centripetal force to be provided at higher velocity. Hence, there is a horizontal component of frictional force acting on the car besides the reaction force R, providing the centripetal force required for its circular motion The car will tend to slide up the slope. M1 A1 [M1] [A1]

6 (a) (i) -When an electron transits from a higher energy level to lower energy level, energy is released as electromagnetic wave or in the form of a photon, of energy hf . The difference in energy between 2 levels, E is equal to the energy of the photon/ or E = hf. -since energy level is discrete, the difference in energy is also discrete . Hence, only certain discrete frequencies exist, hence line spectra obtained. B1 B1 (ii) Energy has to be supplied to the electron to bring it to infinity/ to remove the electron, where PE is zero without a change in KE OR The electron is bound to the nucleus. Total energy of the system/atom is negative. Hence work has to be done to remove the electron from the atom. B1 (ii) Using 𝐸𝑛 = − 27.9 𝑛2 n / eVnE 1 −27.9 2 −6.98 3 −3.10 4 −1.74 (iii) Using ∆𝐸 = 𝐸𝑛 − (−27.9) = 27.9 − 27.9 𝑛2 , where E represents the remaining energy of the colliding electron. n / eVE 1 → 2 20.92 1 → 3 24.80 1 → 4 26.16 1 → 5 26.78 Calculation of ΔE for 1 → 4 and 1 → 5 Electron does not have enough energy to excite to n = 5 Hence highest energy level reach is n = 4 M1 A1 (iv) Using hcE =  hc E =  n 4 3 2 1 E/ eV -1.74 -3.10 -6.98 -27.9 All E values correct – B1 Relative spacing between energy levels correct – B1 Diagram must be fully labelled with n and E values – B1

Shortest wavelength 34 8 19 41 6.63 10 3.00 10 47.5 nm26.16 1.60 10 hc E − − →   = = =  C1 A1 (b) (i) Maximum energy of electron is equal to energy of photon with shortest wavelength. max min hcE = [C1] Accelerating potential 34 8 19 10 min 6.63 10 3.00 10 1.60 10 2.00 10 6220 V hcV e − −−   ==    = Note: min must be read to ½ small square, otherwise minus 1 mark C1 A1 (ii) -Bombarding high energy electrons knock out the inner shell electrons of the target atoms. Hence, electrons transit from higher orbital shells to the vacant inner shells -These transitions result in the emission of (X -ray) photons whose energies are given by the difference in the energy levels = hc  , resulting in sharp peaks at specific wavelengths of 4.00 x 10-10 m, 6.60 x 10-10 m and 9.95 x 10-10 m. B1 B1 7 (a) (i) 1 By Newton’s 3rd law, the rocket exerts a force on the gases so the gases exert an equal and opposite force on the rocket. By Newton’s 2nd law, this (net) force on the rocket will cause it to accelerate. B1 B1 2 Total momentum of rocket and gas as a system remains c