Title 2015 GCE A level H2 Physics 9646 Ans.pdf ✅

2 Paper 2 ( ): optional but good to include { }: tutor’s remarks 1 (a) Spring constant k = Mg/x = (0.45× 9.81)/0.30 = 14.715 Nm-1 = 14.7 Nm-1 . When x = 0.30 m, elastic potential energy = 1⁄2 kx2 = 1⁄2 (14.715)(0.30)2 = 0.6622 J = 0.662 J (b) (i) Between l = 40.0 cm and equilibrium position at l = 60.0 cm, gravitational potential energy decreases while kinetic energy and elastic potential energy increase. Between equilibrium position at l = 60.0 cm and lowest point at l = 80.0 cm, gravitational potential energy and kinetic energy decrease while elastic potential energy continues to increase. At the lowest position, kinetic energy is 0, gravitational potential energy is minimum, and elastic potential energy is maximum. (ii) Gain in kinetic energy = loss in gravitational potential energy - gain in elastic potential energy 1⁄2 Mv2= Mgh - 0.6622 1⁄2 (0.300)v2 = (0.300)(9.81)(0.300) - 0.6622 1⁄2 (0.300)v2 = 0.8829 - 0.6622 = 0.2207 v = 1.21 m s -1 (iii) (The mass has no kinetic energy at the beginning and the end.) Loss in gravitational potential energy = gain in elastic potential energy. Mgx = 1⁄2 kx2 Mg = 1⁄2 kx (0.300)(9.81) = 1⁄2 (14.715)x x = 0.400 m 2 (a) {Use 3 convenient data points to show that the relationship is true} If I ∝ V I = kV Then, k=V/I and k is constant. When V= 6.0V and I = 1.25A, k = 6.0/1.25 = 4.8 V A-1
3 When V= 9.6 V and I = 2 A, k = 9.6/2 = 4.8 V A-1 When V= 12 V and I = 2.5A, k = 12 /2.5 = 4.8 V A-1 Since k is a constant, I ∝ V (b) (i) Resistance is the ratio of potential difference V to current I in (a), R = 4.8 Ω (ii)1. Current I = 12 V/(4.0+ 5.0) Ω = 1.33 Α. 2. Current I = 12 V/(4.8 + 2.7) Ω = 1.6 A (iii) Potential difference between C and the negative terminal (N) of the battery VCN=[6.0 / (4.0 + 5.0)] x 12 = 8.0 V Potential difference between D and the negative terminal (N) of the battery VDN = [2.7 / (4.8 + 2.7)] x 12 = 4.32 V Potential difference between C and D = VCN - VDN = 8.0 - 4.32 = 3.68 V 3 (a) The photon energy must be larger than the work function of metal, which is the minimum energy to cause emission of electron from the surface of metal. If the photon energy is larger than the work function, the excess becomes the maximum kinetic energy of the emitted electrons. (b) (i) When the stopping potential is applied, even the fastest electrons with the maximum kinetic energy will have all their KE transferred into electric potential energy of eVs (and therefore are just brought to rest at C). (ii) The current stays constant, as it is limited by the rate of emission of photoelectrons which is dependent on the intensity of EM radiation (or rate of incidence of photons on M). (c) Photon energy = maximum KE of emitted electrons + work function hf = eVs + Φ = 2.2 eV + 1.8 eV = 4.0 eV = 6.4 × 10−19 J f = 6.4 × 10−19/6.63 × 10-34 = 9.653 × 10 Hz (d) The variation with V of I for electromagnetic radiation of the same frequency but twice the intensity is labelled (d).
4 4 (a) (i) Centripetal acceleration a = Rω2 = R(2π/T)2 = (4.22 x 108 )(2π/1.53 x 105 ) 2 = 0.712 m s-2 (b) (i) The attractive gravitational force exerted by Jupiter on each moon provides the centripetal force required for the circular motion of each moon about the centre of Jupiter. The gravitational field strength at each moon, which is its gravitational force per unit mass, acts towards the centre of Jupiter. centripetal acceleration, which is its centripetal force per unit mass, has the same magnitude as its gravitational force per unit mass, as well as direction also towards the centre of Jupiter. (ii) gIo = GMJupiter / RIo 2 --- (1) gAm = GMJupiter / RAm 2 --- (2) (2)/(1) : RIo / RAm = √( gAm / gIo ) = √( 3.87 / 0.712 )=2.33 5 (a) (i) Force F = qE = (2 × 1.60 × 10-19)(1.5 × 105 ) = 4.8 × 10-14 N (ii) Force F = (1/4πεo)(q1q1)/(r2 ) = 8.99 x 109 (2 x 1.60 × 10−19) 2 /(4.0 × 10-12) 2 = 5.75 x 10-5 N {Examiner: Most candidates calculated correctly. The main errors seen were not squaring the charge or the charge separation, after giving the correct expression for the electrostatic force.} (b) The charge of the two ions A and B have the same magnitude but opposite sign, and they are situated in a uniform electric field. The electric force (by the electric field) on ion A is equal and opposite to that on ion B. The two forces cancel out. The electrostatic force on A due to B, is equal and opposite to the electrostatic force by B on A. These 2 forces also cancel out. {Remember to talk about the forces in a(i) and a(ii) Examiner: The majority of candidates gave answers that described the effect of only one of the types of force that had been calculated in (a)(i) and (ii). There were references to forces cancelling out or where the nature of the force was not identified.} (c) Torque = Fd sin ߠ) = 4.8 x 10 -14)(4.0 x 10-12) sin 60° = 1.66 × 10-25 N m