RVHS H2 Physics P2 Soln

River Valley High School Page 1 of 9 J2 H2 Physics 9749 Preliminary Examinations RVHS JC2 H2 Physics Preliminary Examinations Paper 2 Mark Scheme Questions Answers Marks 1 (a) Impulse is defined as product of (average) force exerted on wall and time of impact B1 (b) Δ𝑝𝑝 = 0.058 × (−34 − 34) M1 = −3.944 kg m s−1 A1 (c) impulse = change in momentum = area under F – t graph M1 𝐹𝐹𝑚𝑚𝑚𝑚𝑚𝑚 × 4 1000 = 3.944 giving 𝐹𝐹𝑚𝑚𝑚𝑚𝑚𝑚 = 986 = 990 N (2 s.f.) A1 (d) From t = 0 to t = 2 ms: decrease at increasing rate. From t = 4 ms to t = 6 ms: decrease at decreasing rate. B1 From t = 2 ms to t = 4 ms: decrease at constant rate. B1 after t = 6 ms, velocity at −34 m s-1 B1 Example Questions Answers Marks 2 (a) molecules has component of velocity in 3 directions, hence 𝑐𝑐2 = 𝑐𝑐𝑚𝑚2 + 𝑐𝑐𝑦𝑦2 + 𝑐𝑐𝑧𝑧2 ------- (1) M1 their motions are random, so averaging gives 〈𝑐𝑐𝑚𝑚2〉 = 〈𝑐𝑐𝑦𝑦2〉 = 〈𝑐𝑐𝑧𝑧2〉 --------- (2) M1 Hence substitute the result of (2) into (1), then 〈𝑐𝑐2〉 = 3〈𝑐𝑐𝑚𝑚2〉 A1 velocity / m s-1 0 2 4 6 t / ms 34 −34

River Valley High School Page 2 of 9 J2 H2 Physics 9749 Preliminary Examinations Hence for N molecules, we have 𝑝𝑝𝑝𝑝 = 1 3 𝑁𝑁𝑁𝑁〈𝑐𝑐2〉 A0 (b) (i) pressure 𝑝𝑝𝑝𝑝 = 𝑛𝑛𝑛𝑛𝑛𝑛 or 𝑝𝑝 = 𝑛𝑛𝑛𝑛𝑛𝑛 𝐿𝐿3 = 3×8.31×(20.0+273.15) 0.2003 M1 so force on each sides 𝐹𝐹 = 𝑝𝑝𝑝𝑝 = 3×8.31×(20.0+273.15) 0.2003 × 0.2002 M1 𝐹𝐹 = 36 500 N A1 (ii) particles would attract one another B1 so average force on wall will decrease B1 3 (a) (i) 𝜔𝜔 = 2𝜋𝜋𝜋𝜋 [M1] → 𝜋𝜋 = 1 2𝜋𝜋 � 10.5 0.450 = 0.76879 ~ 0.77 Hz [A1] (ii) From 50 cm to 80 cm position, a change of 0.300 m: 𝛥𝛥𝛥𝛥𝛥𝛥𝛥𝛥 = 𝑀𝑀𝑀𝑀(𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑛𝑛𝑒𝑒 𝑒𝑒𝑒𝑒 𝑛𝑛) = 0 .450 × 9.81 × 0.300 = 1.32435 J [M1] W hen mass was initially held at 50 cm position, spring was already experiencing EPE, as unstretched length of spring is 32 (or 32.5 cm), therefore initial EPE is due to e = 18 cm. When mass ends up at 80 cm position, spring would have stretched by e = 48 cm. 𝛥𝛥𝛥𝛥𝛥𝛥𝛥𝛥 = 1 2 𝑘𝑘(𝑒𝑒22 − 𝑒𝑒12) = 1 2 (10.5)(0.4802 − 0.1802) = 1.0395 J [M1] So, at the position of 80 cm, the KE present in the mass would be: 𝛥𝛥𝛥𝛥𝛥𝛥 = 1.32435 − 1.0395 = 0.28485 J [C1] 0.28485 = 1 2 𝑁𝑁𝑣𝑣2 → 𝑣𝑣 = 1.12517 ~ 1.13 m s−1 [A1]

River Valley High School Page 3 of 9 J2 H2 Physics 9749 Preliminary Examinations When considering energy conversion for a vertical spring-mass system, there are GPE, EPE and KE. This is at a position close to the maximum displacement. (b) (i), (ii) Fig. 3.3 Labels – Displacement of +/- 24.0 cm, 2 cycles of period 1.3 s (for Line W) [B1] When spring-mass is at equilibrium, reading off the graph, the length of the spring should be 74.0 cm. Given the start length of the spring is 50.0 cm, therefore the displacement of free oscillations will be 24.0 cm. Line W – Accept +ve/-ve Cosine graph [B1] Line X – Check-by-eye decreasing displacement over time, 2 cycles of consistent larger period than Line W [B1] W 1.3 s 2.6 s X 24.0 cm − 24.0 cm displacement time

River Valley High School Page 4 of 9 J2 H2 Physics 9749 Preliminary Examinations 4 (a) Interference is the superposing of two or more waves to produce regions of maxima and minima in space, according to the principle of superposition. B1 (b)(i) Using dsinθ = nλ, let θ = 90o and choose the largest λ in order for a full spectrum will be observed, 9 5 1 sin90 (700 10 )3 10 4.8 n n −= ×× = Hence, the maximum order for a full spectrum to be observed is 4. M1 A1 (b)(ii) Considering the first order diffraction angle for 700 nm wavelength, 9 5 1 sin 1 (700 10 )3 10 12.1o θ θ −= ×× = Considering the second order diffraction angle for 400 nm wavelength, 9 5 1 sin 2(400 10 )3 10 13.9o θ θ −= ×× = Since the second order diffraction angle for 400 nm wavelength is larger than the first order diffraction angle for 700 nm wavelength, the spectra do not overlap. C1 C1 A1

River Valley High School Page 5 of 9 J2 H2 Physics 9749 Preliminary Examinations (c) White light consists of visible light of different wavelengths. There is no diffraction of light at the central region so this region remains white. In diffraction of waves, waves with larger wavelength will have larger diffracted angles. Light sources with lower wavelength do not diffract as much and do not achieve overlapping of light sources at the edge to produce white light. B1 B1 Qn Answer Mark 5 (a)(i) Resistance of copper, 𝑛𝑛𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = (1.60 × 10−8)(3.0) 𝜋𝜋 (0.60 × 10−3 + 1.78 × 10−5)2 − 𝜋𝜋 (0.60 × 10−3)2 = 0.705 Ω Effective resistance = � 1 0.236 + 1 0.705� −1 = 0.176 Ω (show intermediate value of higher sf) = 0.18 Ω M1 A1 (a)(ii) As current increases, power dissipated increases. Heat is generated and hence equilibrium temperature increases. Hence, the lattice ions vibrate more vigorously, hindering the flow of ‘charge carriers’, therefore increasing the resistance. B1 B1

River Valley High School Page 6 of 9 J2 H2 Physics 9749 Preliminary Examinations (b)(i) Using Vs/Vp = Ns/Np, (max Vs)/240√2 = 1/10 max Vs = 33.9 V M1 A1 (b)(ii) period = 0.02 s maximum power 2 2 0 0 (33.9) 6380 W0.18 VP R= = = 1 mark for correct shape 1 mark for correct labelling peak power and period B2 6380 Power / W 0.02 0.01 0.03 0.04 time / s 0

River Valley High School Page 7 of 9 J2 H2 Physics 9749 Preliminary Examinations (b)(iii) 2(33.9) 4(0.18) 1600 WP = = OR 2 2 0 0 (33.9) 6380 W0.18 VP R= = = Mean power for sinusodial graph = 0 1 2 P Since it is half-wave rectified, mean power for half-wave rectified graph = 0 1 4 P = 1 (6380) 16004 = W If students use graphical method, deduct 1 mark if students did not provide clear steps on how Vrms and mean P are detemined. There should be clear steps leading to the final answer. C1 M1 A1 C1 M1 A1 (c) A voltage of 24 V steady direct current voltage has the same value as the rms secondary voltage of 24 V of the alternating current. The rms voltage of the half -way rectified voltage is lower (16.9 V) . Hence, the power increases when 24 V direct current voltage is used. B1 Questions Answers Marks 6 (a) Electric field strength at a point is defined by the force per unit charge. Not acceptable Force on a unit charge. [ Unit is N] whereas for force per unit charge, the unit is N C −1] B1 acting on a small positive stationary test charge at that point. B1 (b) (i) electric field strength inside parallel plate = Δ𝑉𝑉 Δ𝑚𝑚 = 1500−0 0.015 = 1.0 × 105 V m−1 M1 Since charge is stationary electric force balances the weight so 𝑞𝑞𝛥𝛥 = 𝑁𝑁𝑀𝑀 o r 𝑁𝑁 × 1.6 × 10−19 × 1.0 × 105 = 4.90 × 10−15 × 9.81 M1 𝑁𝑁 = 3 (given as integer) A1 (ii) initial electric force have the same magnitude as weight. M1

River Valley High School Page 8 of 9 J2 H2 Physics 9749 Preliminary Examinations Now the direction is same as the weight. So net force on charge is twice the weight. 2g A1 downwards B1 7 (a) Faraday's law of electromagnetic induction states that the e.m.f. induced in a conductor is directly proportional to the rate of change of magnetic flux linkage. Lenz's law states that the direction of the induced e.m.f. is such that it may produce an effect that opposes the change causing it. (b) (i) Flux linkage = NBA [M1] = 85 x (𝜋𝜋x 10-3x 2.8)x (π x (1.6 x 10-2)2) = 6.0 x 10-4 Wb [A1] (ii) Flux change = ∆Φ = 2 x 6.00 x 10-4 [M1] Induced e.m.f. = ∆Φ/∆t = 0.00401 V ~ 4.0 mV [A1] (iii) 0 v for t = 0 to 0.3, t = 0.6 to 1.0 and t > 1.6 t [