SPM 2013 Physics Paper 1 Answers

2013 SPM Physics Paper 1 Answers

Here are my answers - hopefully they are the same as the examiners': 

1 A	2 A	3 A	4 B	5 C	6 B	7 C	8 B	9 D	10 B
11 A	12 D	13 C	14 A	15 D	16 C	17 C	18 C	19 A	20 B
21 A	22 C	23 A	24 B	25 C	26 A	27 A	28 A	29 D	30 D
31 C	32 D	33D	34 B	35 B	36 B	37 A	38 C	39 C	40 A
41 D	42 B	43 D	44 A	45 A	46 D	47 B	48 E	49 C	50 B

(My answers for SPM 2013 Chemistry Paper 1 can be viewed by clicking here.

My answers for SPM 2013 Modern Maths Paper 1 (1449/1) are here.)

My Comments on Some Physics P1 Questions:

Q 30: The answer is D. More than twice before, 'telescope at normal adjustment' was tested in SPM:
          * In 2005 P2 Q11(a)(iii) - diagram @ pg. 387 of Past-Yr Papers 2005-2012 - D;
          * In 2006 P2 Q9(c)(i) - diagram @ pg. 393 of Past-Yr Papers 2005-2012 - D; and,
          * see also pg. 299 of Oxford-Fajar (2013 Ed.) Fig 5.67 for telescope - Answer D!

Q32: The answer is D. Reason:

"As water waves enter shallower region (at headland first, then only the bay), their speed slows down with the shortening of wavelength (again at headland first, then only the bay) without any change in wave frequency. This refraction of water waves results in the wavefronts  taking the shape of the shoreline with wavelengths (distance between successive wavefronts) at the headland shorter than their corresponding wavelengths at the bay as shown by the diagram. At the bay, wave energy is spread-out over longer wavefronts resulting in calmer sea due to dispersal of wave energy. Answer: D

For many other questions, I have inserted some hotlinks which you can click on to direct you to the relevant pages of this blog where the answers can be found/deduced. For examples, if you click on the hotlink of:

* Q 33 D, you will be directed to the page on "Diffraction" where its Para 6 has the same diagram as in the answer D (but in different orientation);

* Q 32 D, you will be directed to the page on "Refraction of Waves" where its Para 7 shows similar diagram and the answer should be D;

* Q 31 C, you will be directed to the page on "Diffraction" where its Para 2 shows C should be the answer;

* Q 28 A, you will be directed to the page on "Refraction" where its Para 6, pt 1 shows you the answer;

* Q 27 A, you will be directed to the page on "Light" where its Para 2(d)(v)(2) shows you the answer;

* Q 26 A, you will be directed to the page on "Refraction" where its Para 6, pt 1 shows you the answer;

* Q 25 C, you will be directed to the page on "Refraction" where its 2nd Para 6 shows you the answer;
and, so on......

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Need Help In Physics, Chemistry and Maths in or around Petaling Jaya?

Good News: Vacant tuition slots are now up for grab!

(June 18th, 2017):

Hello,

Summer May/June 2017 IGCSE O-level and Edexcel O-level exams were just over. As such, some tuition slots are available for grabs on '1st-come-1st-serve" (3 slots have been taken). A few more slots are available for the following subjects:

* IGCSE: Math 0580 (Angela Loo, Absalom Wong, ... scored A's)
                Math 0606 (Nicole Gan scored A (83%), Kai Xun scored B* (79%) ...)
                International Maths 0607 (Zi Zhun waiting results - Switzerland-bound)
                Physics 0625 (Nicole Gan scored A* (93%), Lim Jin Quan A* (mark NA),...)
                Chemistry 0620; (Nicole Gan scored A* (94%), Lim Jin Quan A* (mark NA),...)

* Edexcel: Maths 4MAO / 4MBO / 4PMO (Previously taught 4MBO / 4PMO at Int'l Sch)
                 Chemistry 4CHO (Just tutored Ms. McLean Maths 4MAO, Chem n Physics)
                 Physics 4PHO (Ms. McLean waiting for results)

(CIE AS/A2: Chemistry 9701 / Physics 9702 can be considered if approached early)

Sms or whatsapp me Douglas at: 011-2120 1608; or,
Email: tutortan1@gmail.com

Thank you.

Douglas

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Interesting SPM 2013 Question on Half-Life

1) SPM 2013 Physics Paper 1 has an interesting question on half-life calculation.

Q 48 reads:

"25% of Gallium-65 atoms have decayed after 15 minutes. What is the half-life of Gallium-65?

A 30.0 minutes

B 23.4 minutes

C 15.0 minutes

D 7.5 minutes

(Shouldn't there be another choice, say:

E 36+ minutes?)

My Comments (based on info given in the Q48):

Half-life of Gallium-65 based on info given in the Q48:
25% have decayed means 75% have not decayed. And that means, its half-life > 15 minutes. So was the half-life A, B or none of the 4 given choices? Do you not agree that the half life should be calculated as follows:

The 1/4-life Method:

At the end of 15 minutes (1st 1/4-life):   

a) amount decayed = 25% = ¼ of original amount

b) amount undecayed = 75% = ¾ of original amount

At the end of 30 minutes (2nd 1/4-life):  

a) amount decayed   = ¼ x ¾ of original amount = 3/16 of original amount

b) total amount decayed = (¼ + 3/16) of original amount = 7/16 of the original amount

c) total amount undecayed = (1 – 7/16) of original amount = 9/16 of the original amount

To decay to ½ (or 8/16) of original amount:

a)      A further 1/16 is to be decayed

b)      1/16 = [(1/4) x (9/16)] x (4/9)

c)      Time needed = [15 minutes] x (4/9) =  6.67 minutes

Therefore, half-life = (15 + 15 + 6.67) minutes

                               = 36.67minutes

Answer A (30 minutes) cannot be the acceptable answer because in 30 minutes only 7/16 and not 1/2 (or 8/16) of the radioactive substance has decayed. Conceptually, A is a wrong answer. Q48, to me, has no acceptable answers. This is also confirmed by logarithm method of calculating half-life as follows:

Logarithm Method to Calculate Number of Half-Lives:

Let:      n = number of half-lives

            A_o = Original amount of radioactive substance (or original level of radioactivity)

            A_c = Current amount of radioactive substance (or current level of radioactivity)

Then,    (1/2)ⁿ x A_o = A_c           

                        (1/2)ⁿ = A_c/A_o

                        (1/2)ⁿ = 0.75A_o/A_o = 0.75

log both sides:      log (1/2)ⁿ = log (0.75)

 Hence,       n  = log (0.75) ÷ log (1/2) = 0.415037499 half-life

To Calculate the Half-Life, T_1/2

Let       T_1/2 = Half-life time

            T_o = Time as at original amount of radioactive substance

(or original level of radioactivity)

            T_c = Time as at current amount of radioactive substance

(or current level of radioactivity)

            n = number of half-lives

Then,    T_1/2 = (T_c - T_o) ÷ n = 15 minutes ÷ n = 15 min ÷ [log (0.75) ÷ log (1/2)]

            T_1/2 = 36.14131259

            T_1/2 = 36.1 minutes

Based on facts given in Q48, logarithm method shows a half-life 36.1 minutes.

But online searches (please click here or here) show that isotopes Gallium-65  actually have a half-life of 15.2 minutes

So, which is the examiners' preferred answer and why?

Isn't this question both interesting and bewildering?! :)

In conclusion: Overall, except for Q48, Paper 1 is still a good paper - of respectable standard! :)

(10.4.2014: The examiners should have replaced the words "Gallium-65" in Q 48 with the words "isotopes-X". Otherwise, it's like a biology question on man that talks about the man having ovaries, uterus, etc. that makes the candidates wonder whether it is a question about man or something else. The more well-read the students, the more disadvantaged he/she would be - that is ridiculous!)

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2 Significant Errors in Malaysian Form 5 Physics and Chemistry Textbooks (KBSM)

On 10/9/2013, I wrote to Ministry of Education on the above matter; and I am glad to say that the Ministry has agreed with my findings. And, what is heartening is that the Ministry replied within 8 days - Excellent job to all concerned  at the Ministry of Education. The lady who handled my call - to get the correct email to send my mail - was very professional. I therefore urge all educated parents to be vigilant and proactive about errors they detect in their kid's textbooks -  do write to the Ministry for actions because we don't want our kids to study the wrong things and give the  correct "wrong" answers in their public exams.  With that, may I wish all school children sitting for PMR and SPM very soon: Good Luck and Best Wishes!!! (P/s: Only 2 significant errors for the many pages that I have gone through - really not bad!) Here are my letter and the Ministry's reply: My letter:

"10/9/2013

To: Whom It May Concern

From: A Concerned Parent

Re: Significant Errors in Malaysian Form 5 Physics and Chemistry Textbooks (KBSM)

In good faith, I write to inform about what I think are significant errors in the above-mentioned Form 5 Science Stream textbooks.

If the Ministry agrees that they are indeed serious errors, I sincerely hope that "Corrections Circulars" can be sent out urgently to notify all relevant parties - particularly to the Form 5 Science students who will be sitting for their SPM exams. very soon. The errors:

1. Form 5 Physics Textbook (KBSM):

• At page 13: the 2nd paragraph on refraction: "By comparing the angle of incidence, i with angle of refraction, r in Figure 1.19, you will find that when waves travel from a denser medium to a less dense medium, they are refracted towards the normal." The correct underlined wordings should be "away from normal" and not "towards the normal" . This is a very serious fundamental error!
• 
  

2. Form 5 Chemistry Textbook (KBSM):

• At page 188: The chemical formula for "lauryl alcohol" was given twice there as CH₃(CH₂)₉CH₂OH. The correct chemical formula for "lauryl alcohol" should be: CH₃(CH₂)₁₀CH₂OH (error highlighted in red).

I would be grateful for your follow-up actions.

Thank you.

Sincerely,

Douglas Tan

(Handphone: 011-2328 xxxx)

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Ministry of Education's Reply:

"Rough translation of the Ministry's reply (please see below for the original):

Mr. Douglas Tan.

Sir,

1. With respect, I refer to your email...

2. Your statement about the factual errors has been referred to ...(of the Ministry) for scrutiny/investigation/study.

3. Upon our scrutiny/investigation/study, we found that there is basis to your statement about the factual errors.

4. We will carry-out immediate remedial and follow-up actions on the matter.

5. We thank you for your pro-activeness and working together on the matter.

Thank you."
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The original:

Ramlan Abu Talib <ramlan@moe.gov.my>	8:51 AM (1 hour ago)
to me, Pengarah, Paridin, Che, Krishnan, Syed, Sim, Wan

Malay

English

Translate message

Turn off for: Malay

En. Douglas Tan,

Tuan,

1.     Saya dengan hormatnya merujuk kepada e-mel tuan kepada Bahagian Buku Teks, Kementerian Pendidikan Malaysia,  bertarikh 10 September 2013 berhubung perkara di atas.

2.     Penyataan tuan terhadap kesilapan fakta dalam buku teks Physics Form 5 dan Chemistry Porm 5 KBSM telah dirujuk kepada Sektor Penerbitan Sekolah Menengah untuk penelitian.

3.     Setelah dibuat penelitian , pihak kami mendapati bahawa penyataan pihak tuan terhadap hal ini mempunyai asas.

4.     Pihak kami akan melaksanakan tindakan penambahbaikan dan tindakan susulan terhadap hal ini dengan segera.

5.     Pihak kami mengucapkan terima kasih di atas keprihatinan dan kerjasama tuan terhadap hal ini.

Sekian, terima kasih.

       ..Ramlan...

Ramlan bin Haji Abu Talib

b.p Pengarah

Ketua Unit Khidmat Selia

Sektor Penyeliaan dan Penyelidikan

Bahagian Buku Teks, Kementerian Pelajaran Malaysia

Aras 1-3, Blok E-15, Parcel E

62604 PUTRAJAYA

(03-88844484) ) (019-6644435) (Fax-03-88844489)"


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Approximate Wavelength Formula, λ = ax/D

Approximate Formula for Wavelength (λ = ax/D) from Interference Pattern:

• Where, λ = wavelength

                   a = distance between the 2 coherent wave sources
             
                   x = distance between 2 adjacent antinodal lines
                         (radiating ripples or interference fringes or points of loudness)

                   D = perpendicular distance between the parallel lines
                          where a and x are  measured

• Empirical evidence from experiment shows:

• that x is directly proportional to λ

• that x is inversely proportional to a

• therefore, mathematically, x = kλ/a, where k is found to be = D

• Thus, x = Dλ/a

• And, changing the subject of formula, we have: λ = ax/D

• Application 1: The "Approximate Wavelength Formula, λ = ax/D" can be used to find the wavelength of waves from their interference patterns as a, x and D can be determined from the interference pattern. 

• Application 2: From the same formula λ = ax/D or x = Dλ/a (same formula with different subjects)
• The interference pattern of red, green or blue light can all be explained by the "Approximate Wavelength Formula, x = Dλ/a" 

                        x _red > x_blue

                     

• The distance between adjacent fringes x of red light is bigger than that of green or blue light because the wavelength of red light (λ _red) is longer as compared to that of green light (λ_green) or blue light.

• Since  λ _red > λ_green > λ_blue, from  x = Dλ/a, therefore, x _red > x_green > x_blue.

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Need Help In Physics, Chemistry and Maths in or around Petaling Jaya?

Good News: Vacant tuition slots are now up for grab!

(May 30th, 2016): Summer May/June 2016 IGCSE O-level and CIE AS/A2 exams are ending. As such, some tuition slots are available for grabs on '1st-come-1st-serve". The tuition slots are for:

* IGCSE: Math 0580
                Math 0606
                Physics 0625
                Chemistry 0620

(Edexcel Maths and Physics; SPM Maths, Physics and Chemistry - hourly rates: min. RM80/hr 1-to-1 or RM50/hr/student for group of 2 to 4)

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WhatsApp me Douglas at: 018-3722 482 or, 
email: tutortan1@gmail.com

Thank you.

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Reflection of Light

1.      Law of Reflection:

• ·                       The angle of incidence, Ө_i = The angle of reflection, Ө_r 
• ·                        The incident ray, the reflected ray and the normal lie in the same plane. 

                  

·         The normal is the line drawn at right angle to the mirror where the ray strikes

·         The angle of deviation of a ray is the angle between what would have been the direction of the original ray and the actual direction of the deviated ray.)

   

2.      Image Formed by Reflection:

a.    Plane Mirror

                         i.          What is a plane mirror?

A plane mirror is a flat, smooth and reflective surface which reflects in uniform direction light falling on it.

                        ii.          Characteristics of image formed by reflection in a plane mirror:

1.    Virtual – A virtual image cannot be formed on a screen;

2.    Same size as the object (congruent);

3.    Upright;

4.    Laterally inverted – Left appears as right and right appears as left;

       

5.    The image is as far behind the mirror as the object is in front;

6.    A line joining the image and the object would be:

a.    at right angle to the mirror (perpendicular); and,

b.   bisected by the mirror;

7.    The image would move:

a.         the same distance and at the same speed as the object moves - towards or away from the mirror, following and depending on the object;

b.        twice the distance or twice the speed of the mirror as the mirror moves - - towards or away from the object, following and depending on the mirror;

c.         or rotate twice the angle of rotation of the mirror about a point in the mirror in clockwise or anticlockwise direction, following and depending on the mirror.

                      iii.          Applications of Reflection in Plane Mirror.

1.        Household mirror - We need plane mirror half our height to fully view our full height as follows:

              

2.        Dance studio

3.        Optician’s eyesight testing room

4.        Mirror periscope

5.        Anti-parallax mirror (in Ammeters, Voltmeters, etc.)

6.        Laterally-inverted spelling of the word “AMBULANCE” at the front part of ambulance so that driver of the front car when viewing through his rear view mirror would read it as “AMBULANCE” and gives way accordingly.

7.        Kaleidoscope

8.        Decoration to create feeling of spaciousness – in living halls, hotel lobby, lift, etc.

                      iv.          Ray Diagrams Showing Image Formation by Reflection in Plane Mirror

    

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