Fsfe Physics 2004

MINISTRY OF EDUCATION FIJI SEVENTH FORM EXAMINATION – 2004 EXAMINER’S REPORT PHYSICS GENERAL COMMENTS PAPER I The format for 2004 FSF Physics Examination paper has been similar to 2003 . The examination paper on the whole had tested knowledge and concepts expected to be learned by seventh form physics students in Fiji, particularly : -familiarity with basic concepts ; -problem solving techniques ; -knowledge of practical situations encountered during practical lessons ; and -coverage of the FSF prescription A total of 1280 candidates had registered to this examination. There is a decrease by 6. 4% in the number of candidates registered for the examination, compared to the year 2003.

No grace marks were given as the examination was a valid one and proof read several times. However, there was an inadequacy in question 7, part (iv) of section C. The value of R was not stated. This inadequacy should not have affected the thinking of the candidates as the answer should have been expressed in terms of R. During the duration of the marking of the scripts, it has been revealed that candidates did not follow instructions given on the paper.

All the instructions were very clearly stated on page 1 of the Question paper andAnswer Book. Individual sections and questions and questions had instructions which candidates had to follow. The setting out of the answers was generally good. Candidates should be well advised to list their information clearly, write down the correct formula and show their substitution to an incorrect calculation (answer), partial marks are awarded for correct substitution and consistent working.

Many candidates have poor understanding of the units especially in momentum, torque and magnetism. Mathematical errors, especially those dealing with algebraic fractions and powers of 10 were quite common. . g. ? + ? /2 eg. 1029 2.

Finally, I wish to call upon the teachers to encourage and motivate their students during the teaching and learning of Physics. Teachers should endeavour to maintain link between physical ideas and relevance to students lives in their teaching by making reference to real life experience and relevance of the concepts in real life. Upon request by seventh form Physics teachers solutions to 2004 examination paper is included in the report. NOTE : Overall statistics for Paper I. ParameterScore Highest Score Lowest Score Median Range IQR Mean 94 5 45 89 31 2. 4 This report is based on statistics from a random sample of raw scores.

The distribution of the raw marks was approximately normal. It showed a more realistic spread, the paper discriminating very well between the able and weak candidates. SECTION A This section contained 30 multiple-choice questions, each worth 1 mark. In general this section was not very well done.

Analysis of scripts show that marks range from 5 – 26, with an average of 25, which is 50%. Several candidates did not make a choice on some of the multiple-choice questions. They should be encouraged to do so.While some questions in this section were challenging for the very able students, there were enough questions in the paper to enable the weaker candidates to score marks.

The designing of questions had taken into account of Blooms Taxanomy. 3. The table given below is taken from a random sample of 100 scripts. The correct answer is marked by an asterisk *.

Q. No. ABCDQ. No.

ABCD 1171973*1627*8587 2764146*17580*69 3867*71818113253*4 4683947*19867*619 514*9631520933*850 6867*8172121213*64 75833*542247*43019 847*7343231031473* 96073*212492120*50 1077053*102533*74713 11667*22526833*581 12545833*27653*3110 1347*104122872267*4 4718867*291065*223 151547*6323042367*6 SECTION B This section contained ten questions and all are compulsory. Each question was worth 3 marks. Analysis shows that this section was difficult to most candidates. The marks ranged from 0 – 28. 5 (out of a possible 30), with an average of 11. 4.

A handful of candidates did not attempt this section at all and as a result, they lowered the mean score of this section. The following remarks relate to individual questions. QUESTION 1 The question was very poorly done. It was very discouraging to see that students at Form 7 level lack the understanding of synthesis the word problems.Generous marks were awarded for correct substitutions and consistent working even if the calculations were incorrect especially in part (b) of this question.

4. The correct way to problem solve this question is : 2m TCOS 60° A B 1m C 100N 100N 280N (a)Taking A as pivot : Clockwise moment = Anti clockwise moment (280N x 1m) + (100N x 1m) + (100N x 2m) = 2mx T COS 60° 580N=T (b)FAX=T sin 60° = 502. 3 N Some students calculated FAY = 19ON. The question had very clearly stated calculate the horizontal component of force excited by the wall at A. QUESTION 2This question was also very poorly done.

Many students lacked the understanding of 2 – dimensional momentum especially when Pythagoras theorem and/or sine rule is not to be used. A few students were able to calculate the magnitude of B’s final velocity, however, forgot or overlooked the direction of B’s final velocity. To tackle such question, students could have used components of horizontal and vertical momentum and conservation of momentum to determine the magnitude ot B’s final velocity. Students should use their mathematical skills and their understanding on Geometry to determine the direction (? . Majority of the students used vector triangle to solve for VB.

Generous marks were awarded for drawing the correct vector triangle, stating the conservation of momentum marking one of the angles of the triangle as 70° and calculating before collision equals 0. 3 kgm/s 5. The correct way to problem solve this question is using the method : before collision = after collision • x 0. 2kg x 1. 50m/s COS 45° = 0.

2kg x 1ms cos 25° + BX BX = 0. 031 kg m/s • x 0. 2kg x – 1. 50 m/s sin 45° = 0.

2kg x 1 m/s sin 25° + BY BY = - 0. 97 kg m/s B = ? BX? + BY? = 0. 3 kg m/s VB = RB MB VB = 0. 75 m/s ? = tan-1 RBY RBX ? = 84° Method 2 Using cosine rule : : a? = b? + c? - 2bc cos A to calculate RB. VB = RB/MB Direction ; ? is marked on the vector triangle students can use their mathematical skills to calculate ?.

6. QUESTION 3 This question was very well answered by majority of the students. Analysis showed that students had a thorough understanding of the concept of conservation of energy. This question was very similar to Question 3 of 2003 however the formula for I was changed to I = ? mr2.The correct working to derive the expression = ?? 4gh is 3 Loss in Ep = Gain in Ek Ep = Ek trans + Ek rot mgh = ? mv? + ? Iw? = ? mv? = ? (? mr? ) (v/r)? = ? mv? + ? mv? mgh = ? mv? v? = 4gh 3 v = ? 4gh 3 QUESTION 4 This question was generally well done showing that students had good understanding of the concept of thin film interference, however a few students had difficulty in conversions : especially mm to m and nm to m.

The correct working for answering this question is : ? = ? ? x t = ? (360 x 10-9m) 0. 25 x 10-3 m ? = 7. 2 x 10-4 5cm ? = t/5cm t = (7. 2 x 10-4) (0. 05m) t = 3. 6 x 10-5m Another method that could be used to calculate the thickness t is using the formula dx = (n-? ) where n = 1 L d = t 7.

QUESTION 5 This question was very poorly done. The question as a whole was very simple, testing students understanding of the concept on SHM. Analysis of scripts showed that students had difficulty interpreting the reference circle. Teachers and students are advised to take note that reference circle is included in the seventh form Physics prescription.The correct working for answering this question is : (a)y = A sin wt = 20cm sin (2 16s ) (6s) y = 14. 14cm common error : students swapped T and ? (b)Vm = VP cos ? = WA cos ? Vm = 7.

38cm/s Common error : students wrote Vm = WA (c)? max = W? A ? = 3. 08cm/s? common error : stating the correct formula. QUESTION 6 This question was very well answered by many students, however, a few students had difficulty in conversion of area : 2cm? to m?. Most of the students were able to state the correct formula and calculate the required parameters. Consistency marking was done for consistent working.

The correct working for answering this question is : (a)C = ? oA/d = (8. 85 x 10-12 Fm-1) (2 x 10-4 m? ) 5 x 10-3m C = 3. 54 x 10-13 F 8. (b)Q = CV Q = 3. 54 x 10-11C (c)E = V/d E = 2 x 104 Vm-1 QUESTION 7 This question had 2 parts ; (a) and (b) for part (a), students were required to calculate the magnitude of force and state the direction of the force. Many students overlooked the second part of (a) i.

e. the direction. The concept force between two current carrying wires placed at a distance (d) apart, is a repeat of form 6 work and students are expected to know the concept.It was very disappointing to see that students do not know the conventions used for currents into the page and out of page. This concept is introduced at form 5 level physics under the topic electromagnetism. The correct working for answering this question is : (a)F = k I1 I2 l d = (2 x 10-7 NA-2) (3A) (3A) (20 x 10-2m) 0.

1m F = 3. 6 x 10-6 N Direction : Repulsion F F Common error : - students used ? o instead of K. - overlooked direction (b) x region of stronger field between wires 9. QUESTION 8 This question was generally well done.

Majority of the students were able to correctly answer this question.This question had tested the concept of magnetic field of a soleroid and in particular the hand note used to determine the direction of magnetic field, the effect on magnetic field strength when (i) diameter is halved (ii) the length is reduced by half, while keeping the number of turns constant and calculating magnetic field strength. The answer(s) for this question is as follows :- (a) Use R. H Grip Rule in reverse manner for soleroids. (b)(i)half d ; no effect B = 0.

5T (ii)B ? haL l ; ? doubles B = 2 x 0. 5T = 1 T (c)B = ? o NI l B = 5 x 10-3 T 10. QUESTION 9Overall, this question was very poorly answered. Approximately about 90% of the students do not know how to derive an expression for the resonant frequency and define the term reactance. It is a pity to see that students hide behind the formulae and use symbols (notations) for defining a term. Analysis reveals that students haven’t learnt the concept, they have however claimed the formula for resonant Frequency.

The answer(s) for this question is/are as follows :- (a)At resonanceVl = Vc I X L = I Xc 2? f L = 1 4? f C f? = 1 4?? L C 1 f= ? 4?? L C 1 f= 2? ? L C b)Reactance is defined as : the ratio of voltage to current for a capacitor or inductor in an a. c circuit. QUESTION 10 This question was also very poorly done. Students failed to comprehend magnetic field at an angle of 30° from the vertical shown on the diagram, and as a result were unable to correctly calculate magnetic flux.

The other fact picked up was that students had difficulty in conversions; cm to m, ms to s. Majority of students could not calculate area of the square loop given the dimension in centimeters Faradays law is used to calculate the induced emf in the loop. Many students used B l to calculate emf.The correct working for answering this question is : O = BA cos ? ; A = l? = (4 x 10-2m)? = 1.

6 x 10-3 m? 11. O = D. ST (1. 6 x 10-3 m? ) cos 30° = 6.

93 x 10-4 Wb ? = ? O ? t = 6. 93 x 10-4 Wb 20 x 10-3 s ? = 3. 46 x 10-2 V Common error : Rounding off. Some students wrote 4 x 10-2V; as a result, they were penalized. SECTION C This section of the paper contained seven questions, out of which, any four questions were to be answered.

Each question was worth 10 marks. The following table presents some statistics on this section. PARAMETERSRAW SCORES HIGHEST SCORE LOWEST SCORE RANGE MEAN SCORE 39. 5 0 9. 5 16 The table shown below indicates the mean score per question.

Question No. 1234567 Mean Score6. 53. 54. 5 This section was fairly done. A very many candidates who did not do well in Section A and B managed to pull through this section.

A few candidates were able to pass the overall paper due to this section. The following remarks relate to individual questions : QUESTION 1 This question was very popular, and generally very well answered by majority of the students. Analysis of scripts portrayed the following : Highest : 10 Lowest : 0 Range : 10 Mean Score : 6. 5 % attempting this question : 85 12.

Specific Comments (a)All the parts (i) – (iv) were done satisfactorily. The correct answers for these were : (i)= Fr = 0. 08 Nm (ii)? = T/I = 0. 27 rad s? (iii)? W = ? t = 0. 54 rad/s (iv)L = IW = 0.

16 kg m? /s (b)This question was generally well done, however, many students failed to indicate the direction of the force in (i). Some students had difficulty in re-arranging equations. Many were able to clearly state qB? = M?? /r, but cannot find r. H is advisable that weaker students substitute first and then solve.

A few students substituted incorrect mass (9. 11 x 10-31 kg. Mass, of proton = 1. 673 x 10-27kg.

This value is given in the physical data on page 3 of the answer book. (i)F = qB? F = 2. 88 x 10-12N direction : to the left (ii)r = r = 0. 52m (c)This question was fairly done. It has been observed that students are very good at questions involving calculations, but when it comes to explanation, they cannot explain the concept tested.

The marking criteria for parts (i) and (ii) were such that though students could not phase the sentence properly, key words such as out of phase and destructive interference for (i) and absorb other colours and allow red light to pass for (ii) were looked for while marking.The correct answer for this question was : (i)Light waves arrive out of phase at B and interfere destructively forming a dark band. (ii)Filter will absorb other wavelength of lights and will only allow red light to pass through. (iii)? = dx/L = 6 x 10-7m 13. QUESTION 2 In general, many candidates did not choose to answer this question.

Analysis of scripts portray the following Highest Score:8. 5 Lowest Score:0. 5 Range:8 Mean Score:3. 5 % attempting this question :