2013 H2 Physics MS EJC Suggested

©EJC 2022 9646/H2 PHYSICS 2013 EUNOIA JUNIOR COLLEGE General Certificate of Education Advanced Level Higher 2 PHYSICS SUGGESTED SOLUTIONS 9646 October/November 2013 Paper 1 Multiple Choice Question Key Question Key 1 A 21 B 2 C 22 B 3 A 23 C 4 D 24 B 5 C 25 A 6 B 26 A 7 A 27 D 8 D 28 A 9 C 29 A 10 A 30 B 11 B 31 A 12 D 32 A 13 A 33 B 14 A 34 B 15 D 35 B 16 C 36 C 17 C 37 B 18 C 38 B 19 B 39 C 20 A 40 D Notes Q17: The mass at the end is 0.4 of the mass at the beginning, so 0.6 of the initial mass escapes. The equation pV=nRT needed to be used throughout. Q27: Most candidates chose option B, 12 ohms, indicating, perhaps that these candidates took into account the increase in resistance either as a result of the increase in length or the decrease in width, but not both.

©EJC 2022 9646/H2 PHYSICS 2013 Paper 2 Structured Questions Qns Marks 1(a) s = ut + ½ at2 = 0 + ½ (1.5)(22) = 3.0 m height = s (sin 40°) = 1.9 m M1 M1 A1 1(b)(i) a = 0 Resultant Force = 0 N = mg = (95)(9.81) = 930 N A1 1(b)(ii) Man has zero acceleration and hence experienced no resultant force. Since there are only 2 forces, the normal contact force must be of the same magnitude as weight. A1 1(c)(i) Three forces: normal contact force, weight, and friction of ground on man. Note: Existence of friction between man and the floor of the cable car was rarely mentioned. Some common errors include erroneous additional quantities such as air resistance and driving force. A1 1(c)(ii) Horizontal frictional force will cause the man to accelerate horizontally. The normal contact force is larger in magnitude than the man’s weight, resulting in a vertical acceleration. The resultant force causes the man to accelerate upwards along the cable. A1 A1 1(d) Distance moved during deceleration, s, s = ½ (v + u)t = ½ (0 + 3.0) (3) = 4.5 total distance moved in direction of motion = 3.0 + (3.0 x 120) + 4.5 = 367.5 m vertical distance travelled = 367.5(sin 40°) = 236 m Gain in potential energy = mgh = (95)(9.81)(236) = 2.2 × 105 J M1 M1 A1 1(e)(i) A1

3 ©EJC 2022 9646/H2 PHYSICS 2013 [Turn over Qns Marks 1(e)(ii) A1 1(e)(iii) A1 2(a) Product of force and the time duration of the impact. Comment: Some answers were unacceptable as definition because the wording given was imprecise, for example, impulse was defined as the force acting for or over a certain period of time. B1 2(b)(i) Magnitude of change in momentum, p = Area of inverted triangle = 1 2 (0.32)(0.50) = 0.080 N s Δ𝑝 = 𝑚(Δ𝑣) → Δ𝑣 = Δ𝑝 𝑚 = 0.080 0.150 = 0.53 ms−1 C1 C1 A1 2(b)(ii)1. Force increases at a uniform rate in the negative direction. Velocity decreases at an increasing rate from 0.267 ms-1 until it comes to a rest at t = 0.50 s B1 B1 2(b)(ii)2. Force increases at a uniform rate in the positive direction. Velocity increases from rest at a decreasing rate until it reaches 0.267 ms-1 in the negative direction. B1 3(a)(i) Resultant force, of constant magnitude, acting on object must point in a direction that is perpendicular to the direction of motion of the object towards a centre. Notes: It is insufficient to mention only the centripetal force without addressing the resultant force. A1

4 ©EJC 2022 9646/H2 PHYSICS 2013 Qns Marks 3(a)(ii) Acceleration is the rate of change of velocity with respect to time. Velocity is vector quantity. Here, it has constant magnitude but changing direction. A1 3(b)(i) Gravitational force acting on object provides the centripetal force necessary for the object to move in a circular motion 𝐹𝐺 = 𝐹𝑐 → 𝐺𝑀𝑚 𝑟2 = 𝑚𝑣2 𝑟 𝑟 = 𝐺𝑀 𝑣2 = (6.67 × 10−11)(6 × 1024) 25002 = 6.4 × 107 m Note: Must mention gravitational force provides the centripetal force. M1 C1 A1 3(b)(ii)1. Potential energy of satellite, U, decreases. 𝑈 = − 𝐺𝑀𝑚 𝑟 . When r is smaller, U becomes more negative, U decreases. A1 3(b)(ii)2. Kinetic energy of the satellite, Ek increases. 𝐸𝑘 = 𝐺𝑀𝑚 2𝑟 = − 1 2 𝑈. When r is smaller, EK increases. A1 4(a) The internal energy of a substance is the sum of the kinetic energy due to the random motion of the molecules and potential energy due to intermolecular forces of attraction Note: Examiners deem “sum” and “random” as key words to the definition. A1 4(b)(i) 𝑝𝑉 = 𝑛𝑅𝑇 = 𝑀𝑡𝑜𝑡𝑎𝑙 𝑚𝑚𝑜𝑙𝑎𝑟 𝑅𝑇 𝑀𝑡𝑜𝑡𝑎𝑙 = 𝑝𝑉𝑚𝑚𝑜𝑙𝑎𝑟 𝑅𝑇 = (105)(0.075)(0.030) (8.314)(25 + 273.15) = 0.091 kg C1 C1 A1 4(b)(ii) The oven is not air-tight but has constant volume. Some air leaves the oven when heated. 𝑝𝑉 = 𝑛𝑅𝑇 → 𝑝𝑉 = 𝑀𝑡𝑜𝑡𝑎𝑙 𝑚𝑚𝑜𝑙𝑎𝑟 𝑅𝑇 → 𝑝𝑉 = 𝜌𝑉 𝑚𝑚𝑜𝑙𝑎𝑟 𝑅𝑇 → 𝑝𝑚𝑚𝑜𝑙𝑎𝑟 𝑅 = 𝜌𝑇 𝜌25 𝜌200 = 200 + 273.15 25 + 273.15 = 1.59 C1 A1

5 ©EJC 2022 9646/H2 PHYSICS 2013 [Turn over Qns Marks 5(a) Notes: Must use ruler to construct the field lines. Field lines must touch the plates and have even spacing between them to demonstrate a constant field. A1 5(b)(i) 𝐹 = 𝑄𝐸 = 𝑄 (Δ𝑉 𝑑 ) = (1.6 × 10−19) ( 24 12 × 10−3) = 3.2 × 10−16 N C1 A1 5(b)(ii) 𝑊 = 𝐹𝑑 = (3.2 × 10−16)(12 × 10−3) = 3.8 × 10−18 J A1 5(b)(iii) 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝐾𝐸 = 𝐸 = 1 2 𝑚𝑣𝑖 2 = 1 2 × 9.11 × 10−31 × (4.5 × 106)2 = 9.22 × 10−18𝐽 Electron slows down as it moves from A to B due to repulsion 𝐹𝑖𝑛𝑎𝑙 𝐾𝐸 = 9.22 × 10−18 − 3.8 × 10−18𝐸 = 5.38 × 10−18𝐽 𝑠𝑝𝑒𝑒𝑑 = √2𝐸𝐾 𝑚 = √2 × 5.38 × 10−18 9.11 × 10−31 = 3.44 × 106 M1 M1 A1 6(a) Emission of electrons from a cold metal surface when electromagnetic radiation of sufficiently high frequency falls on it. B1

6 ©EJC 2022 9646/H2 PHYSICS 2013 Qns Marks 6(b) Electrons near the surface of the metal need to be supplied with a minimum amount of energy to overcome work-function energy before they can be removed from the surface. Photons must transfer this minimum amount of energy to these electrons for them to be removed. So these photons must possess this minimum amount of energy. Photon energy is given by product of Planck constant and frequency E = hf. Photons with this minimum amount of energy must have a minimum frequency, hence photons must have frequency above this threshold frequency for the photoelectric effect to take place. Note: Candidates generally quoted the Einstein photoelectric equation but did not always mention that hf was the energy of a photon. Many answers omitted the term photons or work-function energy. B1 B1 B1 7(a) -190 oC to 10 oC, R increases linearly with θ. R reaches a peak of 2080 Ω at 15 oC. 15 oC to 40 oC, R decreases at an increasing rate with θ. 40 oC to 100 oC, R decreases linearly with θ. 100 oC to 200 oC, R decreases at a decreasing rate with θ. B1 B1 7(b) when R = 1780 , θ = 50°C, so 1780 × 50 = 89 000 when R = 240 , θ =150°C, so 240 × 150 = 36 000 since the product of Rθ is not the same, R is not inversely proportional to θ M1 A1 7(c)(i) A1

7 ©EJC 2022 9646/H2 PHYSICS 2013 [Turn over Qns Marks 7(c)(ii) A1 7(d)(i) 𝑅 = 𝐴𝑒 𝐸𝑔 2𝑘𝑇 ln 𝑅 = 𝐸𝑔 2𝑘 (𝑇−1) + ln 𝐴 (Linearization) The proposal is true if a graph of ln(R) against T-1 is linear. The graph of ln(R) against T-1 shown in Fig. 7.3 is linear for θ above 100°C. This supports the proposal. M1 A1

8 ©EJC 2022 9646/H2 PHYSICS 2013 Qns Marks 7(d)(ii)1. Gradient = Δ𝑦 Δ𝑥 = 6.10−5.00 (2.500−2.250)×10−3 = 4400 𝐸𝑔 = 2𝑘(4400) = 2(1.38 × 10−23)(4400) = 1.21 × 10−19 J = 0.76 eV M1 M1 A1 7(d)(ii)2. 5.00 = ln 𝐴 + 4400(2.250 × 10−3) → 𝐴 = 0.0074 Ω A1 7(e) n-type semiconductor is doped with impurity that has donor energy level just below conduction band. There is a much greater increase in mobile charge carriers (electrons) in conduction band and hence lower resistance in a n-type semiconductor as compared to an intrinsic