Sing Yin Secondary School

2001 Physics

CE Mock Exam Marking Scheme

Section A

1.a.P – A,Q – C,R – B.(2A)

(one correct 1 mark, all correct 2 marks)

b.Cost = (1.5kW)(40/60)hr  $0.8 = $0.8(2M+1A)

  1. The loudness of Note 1 is smaller than that of Note 2 because its amplitude is smaller than that of Note 2.

The pitch of Note 1 is lower than that of Note 2 because its frequency is smaller than that of Note 2. (4M+1C)

Input 1 / Input 2 / Output
0 / 0 / 0
0 / 1 / 1
1 / 0 / 1
1 / 1 / 1

3.a.The truth table of an OR gate

(2 M)

  1. A – switch,A – thermistor,

B – resistor,B – rheostat,

C – thermistor,ORC – switch,(3A)

D – rheostat.D – resistor.

(Grouping correct (switch with resistor and thermistor with rheostat), 1 mark.

Two correct positions, 2 marks,

All correct, 3 marks,

Grouping not correct, NO marks)

4.a.average speed = 0.014/0.2 = 0.07ms-1(0.5M+0.5A)

b.velocity increases to maximum then decreases.(0.5A+0.5A)

accelerates then decelerates.(0.5A+0.5A)

5.a.speed along path = (3/4)(2)(1.2)/2 = 2.827ms-1(0.5M+0.5A)

b.velocity = ((1.2)2+(1.2)2)1/2 / 2 = 0.8485ms-1(0.5M+0.5A)

6.a.1.45 sinC = 1.15 sin90o => C = 52.48o(1M+1A)

b.sin = 1.45sin(90o-52.48o) =>  = 62.03o(1M+1A)

c.

(1M)

d.increases.

7.a.Convex. Because the image is real.(1A+1A)

  1. long sight(1A+0.5A+0.5A)

Section B

8.a.green house effect/trapped thermal energy(2M)

b.Scale, title + axis labels, data points, curve(1M+1M+1A+1A)

c.specific heat capacity = [(4)(60)(2000)]/[(1.5)(75-55)]=16kJkg-1K-1(1M+1A)

d.75oC

  1. specific latent heat of vaporization = (3)(60)(2000)/(1.5-1.3)=1.8MJkg-1 (1M+1A)
  2. energy saved = (16000)(1.5)(55-25) = 720kJ(1M+1A)
  3. insulation, stirring(1M+1M+1C)

9.a.Ep = (46)(10)(0.5) = 230J(0.5A+0.5M)

b.i.FA(1.8) = (46)(10)(1.8-1.1) => FA=178.9N(2M+1A)

ii.FB = 46(10)-178.9 = 281.1N(1M+1A)

iii.WA=(178.9)(0.5) = 89.45J(1M+1A)

iv.

(1A+1A)

v.B may be stronger.(1M)

c.i.F = (46)(0.3) = 13.8N(0.5M+0.5A)

ii.No. A horizontal component is needed for acceleration.(1M+1C)

d.i.Zero. Forces perpendicular to displacement.(1M)

ii.Yes. For muscle contraction.(1M)

10.a.i.To the right, (towards the cell)(1A)

  1. (1)increases(1A)

(2)unchanged (or decreases)(1A)

(3)increases(1A)

(4)unchanged (or direction reversed)(1A)

b.i.Clockwise(1A)

ii.P(1A)

iii.By Lenz’s Law, the induced current would flow in a direction opposing the changing of magnetic flux which cause the current. Therefore, the rod experiences a backward force and finally comes to stop. (3M+1A)

OR

By Fleming’s left hand rule, the magnetic force acting of the rod PQ is towards left. The rod decelerates and comes to stop.

iv.It becomes the internal energy of the iron rod. (or electrical energy)(1A)

  1. (1/2)mv2 = mCT

T = 4.1710-5oC(2M+1A)

11.a.i. = 10cm, A = 5cm(2A)

  1. minimum v = 25cms-1

minimum f = v/ = 2.5 Hz(2A)

  1. P,T are in phase.(1A)

Q,S; P,R; R,T are antiphase.(any two, 2A)

  1. P and R are instantaneously at rest

Q is moving upwards(3 A)

b.i.(1) Straight line pattern(1A)

(2) direction of displacement reversed(1A)

  1. The frequency of both wave is the same.

The wave in the string is a transverse standing wave while the one in the air is a longitudinal traveling wave. (2M+1C)

12.a.So that the activity of the source is constant throughout the experiment.(2M)

b.To produce a fine beam of radiation.(2M)

  1. The reading drop a lot when blocked with a sheet of paper, so  radiation is present. The reading does not drop much when the radiation is further blocked by an aluminum block, so  radiation is not present. The reading drops when further blocked by a lead block, so  radiation is present.

As a conclusion,  and  is emitted by the source.(4M+1C)

d.correct diagram(2M)

Done in a cloud-chamber (or other detectors)(1M)

Magnetic field have to be strong (for observable  deflection)(1M)

e.To produce ions around the tip in order to reduce the chance of lightning/ enhance the discharging process. (2M)