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]    (iii) From the Fig 2.2, the carriage enters zone A with a velocity of 250 kJ  2 -1 1 250 0002 2 250 000 [M1]300 40.824 40.8 m s mv v    (iv) Taking the rate of kinetic energy dissipation as the “power” of friction 42 857 [M1]40.824 1049.7 1050 N [A1] PF v PF v     3 (a) 1 111 1 0 10 [M1]33 13.3 [A1] XY RRR R RRR       (b) (i) 3.0 [M1]13.3 0.40 0.219 A [A1] total VI R   (ii)    34 5 3 22 3 3 Since the resistors , and are identical and in parallel, 0.219Current through 0.073 [C1]3 0.073 10 0.0532 W [A1] Alternatively; use potential divider to find p.d. across R ; use RR R R PI R VP     2 R

Meridian Junior College Solutions to H1 Prelim Paper 2 JC2 Preliminary Examinations 2016 4 (c) [M1] Correct orientation (Not accepting just arrows, students should at least draw the triangle to indicate direction of allowed current flow) [A1] Correct symbol ( , circle optional, with lines to connect to the rest of the circuit) 4 (a) Emission line spectrum consists of discrete bright coloured lines on a dark background. [B1] Absorption spectrum consists of dark lines against a continuous white light spectrum. [B1] (b) Summary of key marking points: Thermal excitation / electrical discharge (B1) Photons emitted due to de-excitation of atoms (B1) Each line corresponds to one particular photon energy of specific frequencies/ wavelengths. (B1) Extended version: Gases such as hydrogen or neon can be placed in a discharge tube at low pressure. A voltage (several kilo-volts) is applied between metal electrodes in the tube which is large enough to produce an electric current in the gas. The gas becomes excited by the collisions with the electrons passing through the tube, from cathode to anode of the discharge tube. [B1] The excited gas atoms are unstable. When the gas atoms undergoes a transition to a lower energy level, the excess energy is emitted as electromagnetic radiation (photon) with a specific frequency. [B1] R2 R1 R3 R4 R5 X Y 0.40 3.0 V

Meridian Junior College Solutions to H1 Prelim Paper 2 JC2 Preliminary Examinations 2016 5 The frequency f of the emission line is dependent on the difference between the high and low energy levels, E = hf. Due to the discrete energy levels, only certain high-to-low energy level transitions are possible within the atom, therefore only certain frequency lines are present in the spectrum. [B1] (c) No two gases give the same exact line spectrum. [B1] Hence by comparing the line spectrum of the given sample with that of known elements, we can identify the elements in that sample. [B1] (d) (i) 13.6 eV [B1] (ii) Kinetic energy of electrons = 2.00 x 10-18 / 1.60 x 10-19 = 12.5 eV hence the highest state that the atom can be excited to n=3. [B1] Note: The typical excitation processes that we would expect would be: From n=1 to n=2, From n=1 to n=3. When the atom de-excites, the following transitions may be observed From n =3 to n=2 From n =2 to n=1 From n =3 to n=1 [B1 for all three correct transitions – the second B1 mark will not be given if excitation instead of de-excitations were mentioned.] 5 (a) (i) 1. Magnitude = 40.0 N Angle = 0° 2. Horizontal, assuming Y hor pointing to left: 100 sin60° − Yhor = 0 or 100 sin60° = Y hor Vertical, assuming Y vert pointing upward: 100 cos60° + Yvert − 40.0 = 0 or 100 cos60° + Y vert = 40.0 Yhor = 86.6 N (i.e. 86.6 N in leftward direction) Yvert = −10.0 N (i.e. 10.0 N in the downward direction)

Meridian Junior College Solutions to H1 Prelim Paper 2 JC2 Preliminary Examinations 2016 6 [M1 for horizontal forces & vertical forces] Solving,  = 90° + 6.6° = 96.6° [A1 for angle] Using Pythagoras’ Theorem, Y = 87.2 N [A1 for magnitude] [Alternative methods using vector triangles are also acceptable.] (ii) Horizontal component of X is not zero. [M1] To keep forces in equilibrium, there must be a (non-zero) horizontal component of Y [A1]. Thus rope B cannot be parallel to the weight of S. (b) (i) The initial total momentum of the two carts is zero because the two carts have the same mass and have the same speed but travel in opposite directions. [For learning only:] In equation form (let mass of each cart be M): M(10) + M(−10) = Mv1 + Mv2 0 = v1+ v2 This means that after the collision, the total momentum of the carts must be zero. Since the two carts have the same mass, the velocities of the two carts will be equal in magnitude but in opposite directions. [B1] (ii) Since the mass, initial speeds, final speeds, of the two carts are the same, the two carts must have lost the same amount of kinetic energy. [B1, for appropriate comparison of masses] W