2021 H2 Phy Prelim Paper 2 Solution (forJCs)

2021 JC2 H2 Preliminary Examination Paper 2 Suggested Solutions 1 (a) The principle of conservation of linear momentum states that the total momentum of a system of interacting bodies remain constant provided no external force acts on the system. 1 1 (b) (i) By the principle of conservation of momentum, mXuX + mYuY = mXvX + mYvY 4.0(500) + 28.0(340) = 4.0(220) + 28.0(v) v = 380 m s–1 1 1 (ii) Relative speed of approach = uX – uY = 500 – 340 = 160 m s–1 Relative speed of separation = vY – vX = 380 – 220 = 160 m s–1 Since the relative speeds of approach and separation are the same, the collision is elastic. Or Check whether total final kinetic energy equals the total initial kinetic energy. 1 1 (iii) The forces on the two bodies (or on X and on Y) are equal in magnitude and opposite in direction The duration of impact for both forces is the same and force is change in momentum / time 1 1

2 2 (a) On a banked track, the normal contact force by the road on the car will be tilted to the vertical at an angle equal to the angle of inclination of the track. The horizontal components of both the normal force and the frictional force by the road on the car contributes to the required centripetal force to keep the car moving in a circular path . This increases the centripetal force, which would otherwise be provided by just the frictional force alone. 1 1 (b) (i) [1] – One mark for each correct force. * Allow two normal contact forces and two friction forces between the two wheels and the road. * Minus [1] for not naming any force in full * Minus [1] if centripetal force is drawn into diagram 3 (ii) Horizontally, the resultant force towards the centre of the circle is the centripetal force, Vertically, there is no resultant force, Substituting into (1), 14791.7 = 40v2 – 10392.3 v = 25 or 25.1 m s1 1 1 1 30 Normal force, N Frictional force, F Weight, W

3 3 (a) Internal energy is the sum of microscopic kinetic energy and potential energy of all the molecules in the gas Microscopic potential energy for a real gas is not equal to zero as there is intermolecular bonding between the molecules. So internal energy is not only dependent on temperature. 1 1 (b) p= 1 3 ρ ⟨ c2⟩ = 1 3 M V ⟨ c2⟩ , where M = total mass of the ideal gas V = total volume of the ideal gas 3 2 pV = 1 2 M ⟨ c2⟩ 3 2 nRT = 1 2 M ⟨ c2⟩ = Total KE of gas Since internal energy, U = total KE of ideal gas, as PE of ideal gas = 0  U  T 1 1 1 4 (a) (i) pV = NkT (2.5 × 105) × (3.8 × 10–2) = N × (1.38 × 10–23) × (181 + 273) N = 1.5 × 1024 1 1 (ii) From first law of thermodynamics, increase in internal energy = heat energy supplied + work done on the system Since the process took place at constant volume, work done = 0 Thus, increase in internal energy = 2700 + 0 = 2700 J (Since it is an increase in the internal energy, the change in internal energy is positive 2700 J) 1 1 (b) ∆ EK = 3/2 N k (∆T) 2700 = 3/2 (1.5 × 1024) (1.38 × 10–23) × ∆T ∆T = 87 T = 181 + 273 + 87 = 541 K 1 1 5 (a) (i) Given I = nAve, since I, n, e are the same for cross-sections X and Y, 1 1 (ii) 1 1

4 (iii) When the cross-sectional area is smaller, its resistance per unit length is larger. With the same current through entire wire, the voltage per unit length of the damaged part is larger. 1 (b) (i) I1 + I3 = I2 1 (ii) p.d. across BJ of wire changes so there is a difference between the p.d across wire points BJ and p.d. across cell E 1 1 (iii) 1 1 1 6 (a) Faraday’s law of electromagnetic induction states that the magnitude of the induced e.m.f. E is directly proportional to the rate of change of the magnetic flux linkage. 1 (b) (i) Rate of increase in area of loop ABDCA = [3-(-5)]  0.15 = 1.2 m2 s-1 Induced e.m.f. = dϕ dt = (0.20)(1.2) = 0.24 V Alternatively,  = BlvAB + BlvCD = 0.20(0.15)(3) + 0.20(0.15)(5) = 0.24 V 1 1 1 (ii) [The magnetic flux  = BA.] Since the area of the loop is increasing, the magnetic flux is increasing. By Lenz’s law, the induced current in the loop will create a magnetic field that is acting out of the paper to oppose the increasing flux. By the right hand grip rule, the induced current will flow in the anti- clockwise direction of ABDCA (Cannot use Fleming’s Right Hand Rule as an explanation in this case as question specifies using Lenz’s law.) 1 1 (iii) By Fleming’s Left Hand rule, the induced current will lead to a magnetic force that is in opposite direction to the velocity of the rods. This will cause the rod to slow down. 1

5 In order to keep the speed constant such that the net force is zero, an external force (equal and opposite) is required to counter the magnetic force. 1 7 (a) 1. The existence of a threshold (minimum) frequency below which no photoelectrons are emitted no matter how intense the EM radiation is. 1 2. Above the threshold frequency, the maximum kinetic energy of the photoelectron (stopping potential) increases with the frequency and is independent of the intensity. 1 3. There is no appreciable time delay between the incident EM radiation and the emission of photoelectrons even for very low EM intensities. 1 (b) (i) Work function energy refers to the minimum energy required to liberate an electron (escape) from the metal surface. 1 (ii) Photon energy = = = 5.23  10-19 = 3.27 eV 1 1 (iii) Sodium and calcium (both must be listed) (allow ecf) The incident photon has energy greater than their work function energies 1 1 (c) From de Broglie’s equation, = 1.7410-27 N s From Newton’s 2nd law (rate of photons incident on metal)(change in momentum for one photon) = (7.6 × 1014)(1.7410-27) = 1.3 10-12 N 1 1 1

6 8 (a)  radians 1 (b) (i) Award 1 mark for each correct path. 2 (b) (ii) Resultant displacement/amplitude is the vector sum of the resultant displacement/amplitude of the waves. (It is “considering”, hence, do not need to be too strict on the words) The two waves reaching the receiver have different amplitude/energy/ power/intensity. Because (any one of the possible reasons) 1. the distance travelled by the two waves are different. 2. some energy is absorbed upon reflection. Hence no complete cancellation / resultant displacement/amplitude is not zero. 1 1 1 (c) (i) Measure the distance of the two ends of plate M And adjust (make sure) plate M such that the two ends are equidistant from the marked line. OR Use a set square to align a ruler to be perpendicular to the reference line. Move the set square along the ruler to align plate M at the new position. 1 1

7 (ii) Recognising the path difference Path difference = 2 z− y 1 mark to recognise the path difference = 2 √ x2 +( y 2 ) 2 − y 1 mark to be able to write the equation in terms of the hypothenuse of the triangle = √ 4 x2 +( y)2 − y Since there is a phase change at the surface, for destructive interference Path difference = n 1 mark to explain why path difference is equated to n 1 1 1 z z y 2

8 (d) (i) Gmax line is drawn through bottom of n=3 error bar and through top of n=11 error bar. Gradient Gmax = 0.0312 to 0.0327 1 1 (ii) Gmin line is drawn through top of n=5 error bar and through bottom of n=13 error bar. Gradient Gmin = 0.0242 to 0.0257 1 1 (iii) average  = Gmax+Gmin 2 Allow ECF 1 (iv) It is in band C. (allow ecf) 1

9 (v)  = Gmax  ave = 0.36  102 Percentage uncertainty in  = ∆ ❑ ×100 = 12.6 % Allow ECF 1 1 (e) y = kx 4.5 = k1 (2.9) 22.3 = k2 (5.8) 55.5 = k3 (9.0) 99.5 = k4 (12.0) k1 = 1.5 k2 = 3.8 k3 = 6.1 k4 = 8.29 Conclusion: Since the k values are significantly different, y is not proportional to x. (1 mark for showing at least two k values with correct calculation 1 mark for conclusion) 1 1 (f) Because the wave reflecting from plate