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