MJC_H1_PHY_P2_Answers

Meridian Junior College Solutions to H1 Physics Prelim Paper 2 JC2 Preliminary Examinations 2016 Proposed solutions to H1 Physics Prelim Paper 2 1 (a) (i) Average speed = 0.200 / 0.06428 = 3.11139 m s−1 [C1 for value of v] Max = 0.202 / 0.06427 = 3.14299 m s −1 Min = 0.198 / 0.06429 = 3.07979 m s−1 ∆v = (3.14299−3.07979) / 2 = 0.0316 m s−1 [C1 for value of ∆v] OR: 0.002 0.00001 0.010160.200 0.06428 vdt vdt     ∆v = 0.01016 × 3.11139 = 0.0316 m s−1 [C1 for value of ∆v] Final answer = (3.11 ± 0.03) m s−1 [A1 for ∆v expressed to 1 s.f., v expressed to 2 d.p. (same number of d.p. as ∆v] (ii) Idea 1: “Average” vs “instantaneous” [B1] There is an increase in speed as the card is falling through the light gate, so the measured speed is only an average value (based on total distance / total time) and this is not equal to the instantaneous value. Idea 2: “Speed” vs “velocity” [B1] The light gate only detects the duration of the obstruction, and thus cannot detect direction of travel (which direction the card is moving). (iii) Card does not fall straight (e.g. rotates) and this affects the effective length of the card that obstructs the infrared beam. (b) (i) Obtain acceleration from gradient of graph = 0.33 0.10 9.20.025   [M1,A1] (ii) The gradient [A1] of the graph gives a value for acceleration that is lower than the expected (accepted) value for acceleration. [M1] (Y-intercept of the graph is not a systematic error, because it just shows that the initial speed of the light gate is 0.10 m s-1 when it first crosses / triggers the light gate. Does not have to be corrected as the y-intercept is not required to obtain the value of acceleration.)

Meridian Junior College Solutions to H1 Prelim Paper 2 JC2 Preliminary Examinations 2016 2 2 (a) Horizontal line from t = 0 s to t = 6.0 s, at 310 kJ [B1] Diagonal line from (6.0 , 310) to (7.4, 250) followed by horizontal line to the t = 9.0 s [B1] (b) From the Fig 2.2, t = 3.0 s to t = 4.2 s, the increase in gravitational potential energy is 130 60 70 kJ 70000 70000 [M1]300 9.81 23.785 23.8 m [A1] GPE mgh h        The height, and therefore the diameter, is 23.8 m (c) (i) time start: 6.0s time end: 7.4 s [B1 – both must be correct] Due to the dissipative force of friction, kinetic energy is lost [B1] However, gravitational potential energy is constant [B1] since height remains the same. Kinetic energy Potential energy energy / kJ 0 50 100 150 200 250 300 350 0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 time / s

Meridian Junior College Solutions to H1 Prelim Paper 2 JC2 Preliminary Examinations 2016 3 (ii)  3 -1 loss of KErate of KE dissipation duration in zone A 250 190 10 [M1]1.4 42 857 42.9 kW or kJ s [A1]    (ii