CJC H1 PHY P2 QP and SOL

1 CATHOLIC JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATIONS Higher 1 CANDIDATE NAME CLASS 2T INDEX NUMBER PHYSICS 8866/02 Paper 2 28 August 2015 2 hours Additional Materials: Answer Paper READ THESE INSTRUCTIONS FIRST Write your index number and name on all the work you hand in. Write in dark blue or black pen on both sides of the paper. [PILOT FRIXION ERASABLE PENS ARE NOT ALLOWED] You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Section A Answer all questions. Section B Answer any two questions. Circle the 2 questions that you answered in the table below. At the end of the examination, fasten all work securely together. The number of marks is given in brackets [ ] at the end of each question or part of the question. This document consists of 24 printed pages [Turn over] FOR EXAMINER’S USE SECTION A (40 MARKS) 1 /7 2 / 8 3 /7 4 /8 5 /3 6 /7 SECTION B (40 MARKS) 7 / 20 8 / 20 9 / 20 TOTAL /80

2 PHYSICS DATA: speed of light in free space, c = 3.00 x 108 m s-1 permeability of free space, o = 4 x 10-7 H m-1 elementary charge, e = 1.60 x 10-19 C the Planck constant, h = 6.63 x 10-34 J s unified atomic mass constant, u = 1.66 x 10-27 kg rest mass of electron, me = 9.11 x 10-31 kg rest mass of proton, mp = 1.67 x 10-27 kg acceleration of free fall, g = 9.81 m s-2 PHYSICS FORMULAE: uniformly accelerated motion, s = u t + 2 1 a t2 v2 = u2 + 2 a s work done on / by a gas, W = p V hydrostatic pressure p = gh resistors in series, R = R1 + R2 + ... resistors in parallel, R 1 = ... 21 11 RR

3 SECTION A (40 marks) Answer all questions in Section A. 1 A car of mass 1380 kg, travelling at 31.1 m s -1, is brought to rest by applying the brakes. The average braking force is estimated to be 1.38 x 104 N. Calculate (a) the initial kinetic energy of the car, kinetic energy = ………………………..J [1] K.E. = ½ m v2 = ½ (1380)(31.1)2 = 667400 J A1 (b) the average deceleration of the car, deceleration = ……………………… m s-2 [1] F = ma  a = F/m = 1.38 x 104 /1380 = 10.0 m s-2 A1 (c) the distance travelled before it comes to rest. braking force = …………………………N [2] Work done = KE loss Fd = KE d = KE loss / F = 667400/1.38 x 104 = 48.2 m M1 A1 (d) Suggest whether the answer in (c) is an over-estimation or under-estimation. [2] In practice, air resistance and rolling friction of the road are presence. The total decelerating force is larger and hence the distance travel will be shorter. The value is an overestimation. B1 B1 2 (a) State what is meant by the equilibrium of a body. [2] It does not accelerate linearly, velocity is constant It does not change in rotational speed, angular velocity is constant B1 B1 (b) Fig. 2 shows a girl supported by two ropes. She is in equilibrium. She has a weight of 392 N.

4 (i) Calculate the tension T1 and T2 in the ropes. tension T1 = ………………N tension T2 = ………………N [4] T1 cos 50 = T2 cos 40 0.643 T1 = 0.766 T2 ………………..(1) T1 sin 50 + T2 sin 40 = 392 0.766T1 + 0.643T2 = 392………..(2) Solve (1) and (2) T1 = 244 N T2 = 205 N M1 M1 A1 A1 (ii) The girl is pulled vertically downwards so the ropes stretch. She is then released. Explain why the method you used in (i) could not be used to determine the tensions in the ropes immediately after she is released without additional information. [2]  the girl is not in equilibrium and the resultant force is not zero  The downwards force is unknown & the angle of inclination of T1 and T2 are unknown. B1 B1 Fig.2

5 3 A long -jumper leaps off the starting block at a speed 8.6 m s -1 at an angle  to the horizontal and lands on level pit. (a) Explain why the longer-jumper needs to have an upwards component of velocity at take-off, as well as forward velocity component to reach a good horizontal distance. [2] the upwards component gives him airborne time t The forwards component ux gives him forward distance travelled because x = ux t B1 B1 (b) (i) Suppose that the angle  = 35o, calculate the time to reach the maximum height and the horizontal distance of the long jumper. In your calculations, you should neglect the presence of air resistance. time = ……….……….s horizontal distance =……..…………..m [4] Vertical motion without air resistance Using “v = u + at” 0 = 8.6sin35 + (-9.81)t t = 0.5028 = 0.50 s M1 A1 airborne time = 2 x 0.5028 = 1.006 s horizontal distance = 7.6 cos35 x 1.006 = 6.26 = 6.3 m M1 A1 (ii) Why does his horizontal distance is less than the answer to (b)(i) when air resistance is taken into consideration. [1] Air resistance opposes the motion, So the airborne time will decreases The horizontal component of the velocity also decreases with time Since horizontal distance = horizontal velocity of velocity x airborne time The horizontal range is smaller B1 4 A glass tube of Helium gas atoms are excited when a potential difference is applied across it. When the emitted light is viewed through a spectrometer, three emission lines of blue, green and yellow colours are observed. Fig. 4.1 shows the spectral lines, together with the associated photon wavelengths of each colour. Fig. 4.1 Light from the gas is incident on the surface of a metal plate X. The electrons liberated from the plate are attracted to the anode as shown in Fig. 4.2

6 Fig 4.2 The experiment is then repeated using two other metal plates Y and Z of different work function energies. The table below shows the work function energies of the different plates. Plate Work Function Energy / eV X 1.58 Y 2.42 Z 3.17 (a) What is meant by the term work function energy of a metal? [1] Work function energy is the minimum energy required to eject an electron from a metal surface in the photoelectric effect. B1 (b) The figure below shows the variation of current I in the circuit with applied potential difference V between the metal plate and anode when the blue light from the gas is allowed to incident on plate X. Fig. 4.3 V / V I / μA VS

7 (i) Calculate the stopping potential Vs [3] 𝐸 = ∅ + 𝑒𝑉𝑠 4.450 x 10 − 19 = (1.58)(1.60 𝑥 10−19) + (1.60 𝑥 10−19)𝑉𝑠 𝑉𝑠 = 1.2 𝑉 Blue light E = hf = hc /  = 6.63 x 10-34 x 3 x 108 / 447 x 10-9 = 4.450 x 10-19 J X plate Work function = (1.58 )(1.60 𝑥 10−19) 𝐽 A1 M1 A1 (ii) On Fig. 4.3, sketch the variation of current I with applied potential difference V when light from the gas is incident on plate Y. Explain your answers. [3] (ii) On Fig. 4.3, sketch the variation of current I with applied potential difference V when light from the gas is incident on plate Y. Explain your answers. [4] Smaller stopping potential M1 – Smaller saturated current The work function energy of plate Y is higher than plate X. Same photon energy Smaller maximum kinetic energy of the emitted electrons Smaller stopping potential The maximum current will also decrease as  The number of photons incident is the same,  The photon energy is the same  the work function energy is higher  less electrons can be liberated current is smaller B1 B1 B1 B1 (iii) State with a reason, w hich of the plates used for the photoelectric effect experiments will a zero reading