EJC 2020 J1H2 Promo Mark Scheme

© EJC 2020 9749/01/J1H2PROMO/2020 EUNOIA JUNIOR COLLEGE JC1 PROMOTIONAL EXAMINATIONS 2020 General Certificate of Education Advanced Level Higher 2 PHYSICS Multiple-Choice, Structured Questions and Practical MARK SCHEME w/ EXAMINERS’ COMMENTS Maximum Mark 9749 October 2020 Paper 1 Solution 1 A 𝑈𝑛𝑖𝑡𝑠 𝑜𝑓 𝐸 = 𝑘𝑔 𝑚2 𝑠−2 𝑈𝑛𝑖𝑡𝑠 𝑜𝑓 𝑐 = 𝑚𝑠−1 For equation to homogeneous, units of (𝛼𝑐2)2 and (𝛽𝑐)2 must be equal to units of 𝐸2 (𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝐸)2 = (𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝛼)2(𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝑐)4 (𝑘𝑔 𝑚2 𝑠−2)2 = [𝛼]2(𝑚2 𝑠−1)4 [𝛼] = 𝑘𝑔 Therefore, 𝛼 represents mass. (𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝐸)2 = (𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝛽)2(𝑢𝑛𝑖𝑡𝑠 𝑜𝑓 𝑐)2 (𝑘𝑔 2 𝑠−2)2 = [𝛽]2(𝑚 𝑠−1)2 [𝛽] = 𝑘𝑔 𝑚 𝑠−1 Therefore, 𝛽 represents momentum. 2 B 𝜌 = 𝑀 𝑉 = 𝑀 𝑙𝑏𝑡 Δ𝜌 𝜌 = Δ𝑀 𝑀 + Δ𝑙 𝑙 + Δ𝑏 𝑏 + Δ𝑡 𝑡 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑢𝑛𝑐𝑒𝑟𝑡𝑎𝑖𝑛𝑡𝑦 𝑜𝑓 𝜌 = 1 + 1 + 2 + ( 0.4 11.2 × 100%) = 7.6% Refer to EJC 2020 Measurement tutorial Qn 7 for deeper understanding on handling percentage uncertainty when “stacking” is involved. 3 B B will result in the voltmeter giving readings that are larger than the true value. 4 D Option C is actually the velocity-time graph of the car. Note that the gradient at t = 0 for the displacement-time graph should be zero as the car starts from rest. Given that the area under the acceleration-time graph cancels out one another, the final velocity of the car is zero as change in velocity is zero. Hence gradient of displacement-time graph is zero at the end. Lastly, when acceleration is 0 between t and 2t, the car is travelling at constant speed. Hence gradient of displacement-time graph is constant.

2 © EJC 2020 9749/J1H2PROMO/2020 5 B As both the hammer and spanner are released from the same vertical height and have the same initial vertical velocity relative to the first floor they will land in the first floor at the same time as they experience the same acceleration. 6 C Solution: the 12 N force will be exerted equally o n all springs as they are being connected in series. Using Hooke’s Law, F = kx, xA = 12/2 = 6 m xB = 12/3 = 4 m xC = 12/6 = 2 m Therefore answer is 3 : 2 : 1. 7 D Pressure at point A = 𝜌𝑤𝑎𝑡𝑒𝑟𝑔ℎ + 𝑃𝑎𝑡𝑚 = (1000)(9.81)(ℎ) + 𝑃𝑎𝑡𝑚 = 9810ℎ + 𝑃𝑎𝑡𝑚 Pressure at point B = 𝜌𝐻𝑔𝑔ℎ𝐻𝑔 + 𝜌𝑜𝑖𝑙𝑔ℎ𝑜𝑖𝑙 + 𝑃𝑎𝑡𝑚 = (13600)(9.81)(0.05) + (800)(9.81)(ℎ − 0.05) + 𝑃𝑎𝑡𝑚 = 6278.4 + 7848ℎ + 𝑃𝑎𝑡𝑚 Pressure at the same depth are equal at equilibrium. 6278.4 + 7848ℎ + 𝑃𝑎𝑡𝑚 = 9810ℎ + 𝑃𝑎𝑡𝑚 ℎ = 3.20 𝑚 8 A Team C and F alone gives a resultant force of 300 N along OF Team A and D gives a resultant force of 300 along OD Team B and E gives a resultant force of 300 N along OE Drawing the vector diagram gives resultant force of all teams to be 600 N along OE. Question can also be d one by resolving resultant forces due to CF and AD along the direction OE.

3 © EJC 2020 9749/J1H2PROMO/2020 9 B Taking 3M and 2M as one system, Resultant force upslope, 5ma = F – 5mgsin30 a = F/(5m) –gsin30 Considering FBD for 2M: FAonB – 2mgsin30 = 2ma FAonB = 2m(a + gsin30) Sub a into FAonB, FAonB = 2m(F/(5m) –gsin30 + gsin30) = 2F/5 10 C Mass of air swept down per second = density of air x volume of air per second = 1.02 x cross sectional area of air x velocity of air = 1.02 x π x (5.0/2)2 x (18) = 360.5 kg per second Force on air by blade = ∆(mv)/ ∆t = 360.5(18-0) = 6490 N From Newton’s Third Law, magnitude of force on blade by air = magnitude of force on air by blade = 6490 N 11 C Change in momentum = area under the graph from 4 to 8 seconds = ½ (2+4) (12) = 36 Ns 36 = 3.0 (vf – 12) vf = 24 m s-1

4 © EJC 2020 9749/J1H2PROMO/2020 12 C Using Conservation of linear momentum 21 21 2Mv+(M)(-2v)= (2M)v Mv 0= 2v v ---(1)   Using relative speed of approach = relative speed of separation 21 21 v 2v= v v v= v v ---(2)   Solving (1) & (2) 2 1 vv v 2v   2 2 2 k 11total E after collison = (2M)v (M)(2v) 3Mv22  13 B Application of Newton’s 3rd Law. A is a violation of Newton’s 3rd Law. For B, If the forces are in equilibrium, then crate will not accelerate. D will result in 0 resultant force on the crate → 0 acceleration. 14 A Graph given is extension, e vs force applied F. Hence the elastic potential energy stored in spring is the shaded area X. 15 B Output energy = change in GPE = (400)(9.81)(1200) = 4710 kJ Input energy = 4710/(0.8) = 5890 kJ Wasted energy = 5890 – 4710 = 1180 kJ Wasted power = 1180/[2(60)] = 9.83 kW 16 C vr  . Since the Earth rotates about its axis at constant , v is proportional to r.

5 © EJC 2020 9749/J1H2PROMO/2020 17 D At the top, the direction of linear momentum is to the left. GPE is max, KE is constant. At the bottom, the direction of linear momentum is to the right. GPE is min, KE is constant. Total mechanical energy = GPE + KE. 18 B + : 2 2 2 (2 ) 3 2(2 ) net GM G M GMg R R R        Option A: Found the difference between gX and gY. Did not consider that g is a vector and here gX and gY are in the same direction, thus should be added instead. Option C: Careless. Used 2R2 instead of (2R)2. Option D: Used the formula for potential  instead of for g. 19 C 1 pdE dFm dr dr r       Option A: Underestimated the gradient. Option B: Incorrectly used g r  .which is applicable only for a radial field. Here the g-field is the vector sum of the g-field due to the moon and the g-field due to the planet. The net field is not radial. Option D: Overestimated the gradient. 20 A 10[( 60) ( 20)] 400 MJW GPE m         Option B: Did not multiply by m to find GPE. Option C: Did not multiply by m to find GPE. Did not calculate change as final value subtract initial value. Option D: Calculated the work done by the gravitational force instead, which is equal to the negative of the change in gravitational potential energy.

6 © EJC 2020 9749/J1H2PROMO/2020 21 D Option A: 2 T   . Since T = 24 h is constant,  is constant. Option B: 23 22 1GMm mr rr     . Since  is constant, r is constant. Thus 2ar  is constant. Option C: 23 22 1GMm mr rr     . Since  is constant, r is constant. Option D: p GMmEm r  since r is constant. Thus GPE depends on the mass m of the satellite. 22 D 𝑣𝑚𝑎𝑥 = 𝜔𝑥0 = 5 𝑚𝑠−1 𝑥0 = 7 𝑚 Therefore, 𝜔 = 5 7 = 0.7143 rad s-1 And 𝑇 = 2𝜋 𝜔 = 8.8 𝑠 23 C Equation, taking right to be positive: 𝑥 = 50 cos 2𝜋 3 𝑡 When displacement is 28 mm while moving to the right, 28 = 50 cos 2𝜋 3 𝑡 2𝜋 3 𝑡 = 2𝜋 − cos−1 (28 50) 𝑡 = 2.5 𝑠 x/mm 28 t /s Moving to the right (second time) Moving to the left (first time)

7 © EJC 2020 9749/J1H2PROMO/2020 24 D After 1st polariser, by Malus’ Law, 𝐼1 = 𝐼0 cos2 40° After 2nd polariser, by Malus’ Law, 𝐼2 = 𝐼1 cos2 70° Therefore, 𝐼2 = cos2 40° × cos2 70° × 𝐼0 = 0.0686𝐼0 25 B P is moving to the right, but it is not at maximum speed. Q is moving to the left and since it is at equilibrium, it is at maximum speed. R is momentarily at rest, since it is at its amplitude. Its acceleration is at its maximum. S is moving to the right but its acceleration is zero at the equilibrium point. 26 C Wavelength is from compression to compression. 100 m spans across 5 wavelengths. Therefore, 𝜆 = 100 ÷ 5 = 20 m Using 𝑣 = 𝑓𝜆, 1200 = 𝑓(20) Therefore, frequency = 60 Hz 27 C The sources are out of phase by π rad. This is because the path difference S2O – S1O is zero but the waves are in antiphase at O, resulting in destructive interference and a minimum detected. Low signal X Path difference = 2λ High signal Path difference = 3λ/2 Low signal Path difference = λ High Signal Path difference = λ/2 Low signal O Path difference = 0

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