Tackling Exam

7/29/2019 Tackling Exam

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1AS and A Level Physics Cambridge University Press Tackling the examination

You have done all your revision and now you are in the examination room. Tis is your chance

to show off your knowledge. Keep calm, take a few deep breaths, and try to remember all thethings you have been told.

Ensure you have read the rubric and you know how many questions to answer (on the AS and A2physics papers it is easy you must answer all the questions).Open your paper and read the first question carefully. Make sure you know what you are expectedto do. Note the command words that are used.Do notstart writing until you have carefully read the question, and re-read the subsection you areabout to answer.

When solving problems remember all the things you have been told about setting your work out:show your working, remember to convert to base units, and include units in your answers.

Now print out the question on the next two pages and try to answer it. Do not look at the

sample answer until you have finished.

1AS and A Level Physics Cambridge University Press 1Tackling the examination

You have done all your revision and now you are in the examination room. Tis is your chance

Tackling the examination


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2 AS and A Level Physics Cambridge University Press Tackling the examination

1 Te Young modulus for structural steel is 200 GPa.

a DefineYoung modulus.

[1]

b o measure the Young modulus of structural steel, a sample is drawn into a wire ofradius 0.48 mm. A length of wire of 1.25 m is used in the experiment.

i Calculate the tensile stress on the wire when a stretching force of 54.0 N isapplied to it.

[2]

ii Calculate the extension of the wire.

[3]

iiiSuggest and outline how the extension of the wire, in this experiment, could

be measured.

[2]

Te maximum stress this type of steel can withstand before breaking is 420 GPa.


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c Calculate the maximum strain this steel can withstand.

[2]


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Analysing your answersAnalysing your answersNow look at the answers given by two students, and the comments from the examiner.

Student AStudent A1 Te Young modulus for structural steel is 200 GPa.


[1]

Te student has some idea but the answer is too vague. Tey have tried to explain what is

meant by Young modulus but have got into a bit o a muddle. Compare this with the answergiven by student B. (0/1)



[2]

From this jumble o numbers it appears that the student does seem to recognise that stress isorce divided by area. However, they have totally ignored the act that the radius is measuredin millimetres, so that fnal answer should be in N mm2 not pascals. Te student wouldscore the frst mark or knowing the ormula or stress and attempting to substitute in thequantities. (1/2)

Young mouus is a measue o how much a maeia wil sech when

a oce is apie aking ino accoun is engh an is hickness.

54

(

0.482)

= 746 Pa


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ii Calculate the extens ion of the wire.

[3]

Again there is no explanation o what is going on. It is possible to deduce that the studentis aware o the ormula. Again, they do not convert to base units, but this would not bepenalised twice. Te student then goes straight rom this to the answer. Unortunately theymake a slip in their numbers, but as there is no evidence where they made the error, they loseboth the next two marks. (1/3)

iiiSuggest and outline how the extension of the wire, in this experiment, could

be measured.

[2]

Te student has recognised that the extension is small and come up with a solution,gaining the frst mark. However they do not go on to explain how they would use themicroscope. (1/2)

Te maximum stress this type of steel can withstand before breaking is 420 MPa.

c Calculate the maximum strain this steel can withstand.

[2]

It is good that his frst attempt gave a silly answer the wire stretching to 210 times its

original length! Tis has prompted the student to think about consistency with units. Notethat strain is pure ratio and has no units. (2/2)

200 = 7.46 x1.25

= 0.56 m

Te incease is smal so I wou use a micoscoe.

420 = 210200

= 0.021


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Student BStudent B1 Te Young modulus for structural steel is 200 GPa.


[1]

Tis is a perect answer. It is perectly alright to give a defnition in the orm o anequation. (1/1)



[2]

Tis is an excellent answer. Te student has written down the equation that they needto use and then listed the quantities they need, ensuring that they have converted tobase units. (2/2)

ii Calculate the extension of the wire.

[3]

Another fne answer, although there is an arithmetic error in the fnal answer. However, itis clear that this is just a slip as the student shows a correct manipulation o the ormula sothey will only lose one o the marks. Te correct answer should have been4.7 104 m. (2/3)

Young modulus =stressstrain

stress = forcearea force = 54.0 N

radius = 0.48 mm = 0.48 103 m

= 3 254

(0.48 10 ) = 746 107 Pa

E= 200 GPa = 200 109 Pa length = 1.25 m

200 10

9

=

7746 10 1.25

x

x=

7

9

746 10 1.25200 10

= 5.6 104m

E= stressstrain


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iiiSuggest and outline how the extension of the wire, in this experiment, couldbe measured.

[2]

Tis is an alternative answer. Again, the student has recognised that the extension is verysmall and has come up with a good solution to the problem. Tey give an indication howthey would use the apparatus and their explanation is su cient to gain the second mark. Itwould be even better with a simple diagram. (2/2)

Te more experience you have o doing practical work, the more easily you will be able to

solve problems like this.

Te maximum stress this type of steel can withstand before breaking is 420 MPa.c Calculate the maximum strain this steel can withstand.

[2]

Another perect answer! (2/2)

Student A scored 5/10. Tis should be enough for a grade E, perhaps even a grade D.Student B scored 9/10. An excellent effort. If this standard is maintained through the paper

the student would certainly score an A*.

Now go through your answer to the question and see what you think you might have scored.Te lessons to learn from this exercise are:read the question properly taking note of any command words;show your working under the pressure of examinations it is easy to make arithmetic slips;convert quantities to their base units to avoid errors with power of ten;learn definitions carefully.

Te exension is vey smal so he suen wou hang he wie om he ceiing

an atach a venie scae o i. He wou hang a secon wie nex o i wih a main

scae. As he oae he wie he cou ea of om he venie.

sess = 420 MPa = 420 106 Pa

E= 200 GPa = 200 109 Pa

E= sesssain

sain =

6

9420 10 Pa 200 10 Pa

= 0.021


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Two more to tryTwo more to tryNow try these two questions, but this time try and work to a time scale. Tere are 60 marksfor Paper 2 and the time allowed to complete the paper is 60 minutes about 1 minute per

mark. Te first of these questions is for 11 marks and the second is for 7 marks, so if you cancomplete the questions in under about 20 minutes you are on track to complete the paper.

2 Figure 1 shows the principles of an experiment a student set up to demonstrateinterference of light.

Figure 1

light

1.60 mm

2.50m

screen

Te light comes from a coherent, monochromatic source.

a i Explain what is meant by acoherentsource of light.

[1]

ii Explain what is meant by amonochromatic source of light.

[1]

b On the screen a series of bright and dark fringes are seen. Explain how the fringesare formed.

[3]

c Figure 2 shows the fringes as seen on the screen.

Figure 2

bright fringes

dark fringes9.20 mm


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Calculate the wavelength of the light.

Wavelength = m [3]

d Te student now replaces the light source Awith a white light source.State and explain how the pattern on the screen would change.

[3]

3 Figure 3 shows a potential divider circuit.

Figure 3

X

Y

2.0k

3.0k

Te battery has an e.m.f. of 6.0 V and its internal resistance is negligible.a Explain what is meant by the term e.m.f.

[1]


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b Show that the potential difference across the 3.0 k resistor is 3.6 V.

[2]

c When a voltmeter is connected across XY it only reads3.2 V.i Suggest why the reading is less than the calculated value and deduce the resistance

of the voltmeter.

[1]

ii Calculate the resistance of the voltmeter.

esisance= k[3]


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[1]

Te student has a bit more idea here, but this is really not enough or A level. We are lookingor the idea o a single requency or wavelength. (0/1)

b On the screen a series of bright and dark fringes are seen.Explain how the fringes are formed.

[3]

Tere is some idea here that it is phase dierences between the waves rom the two slits thatproduces intererence, but the explanation is not linked to the dark and bright ringes thatare observed. (1/3)

c Figure 2 shows the fringes as seen on the screen.

Calculate the wavelength of the light.

[3]

Te student has orgotten that the 9.2 mm measures the distance across 10 ringes and hasnot included this in their answer. Tis is a serious mistake, betraying a lack o true under-standing. One mark has been given as compensation because it is elt that the studentdisplays some knowledge o the principle, and the rest o the calculation is good. (1/3)

Students answersStudents answersQuestion 2Question 2Student AStudent A2 a i Explain what is meant by acoherentsource of light.

[1]

Te idea o coherence is not easy, but this student has not got any idea at all. (0/1)

Ligh o a singe coou

Te waves om he wo ae eihe in se o ou o se causing hem o

ad u o subac

=2.5

1.6 1039.2 103

= 5.888 103 m

Ligh o which is consan bighness


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[3]

Te student recognises that coloured ringes will be seen, but the comment that white lightis made up o colours does not really start to explain why coloured ringes are seen. We arelooking or an understanding that the colours have dierent wavelengths, and thereore themaxima and minima o the colours are in dierent places. (1/3)

Student BStudent B2 a i Explain what is meant by acoherentsource of light.

[1]

Tis almost gets there. Te student recognises that there is something constant aboutthe phase, but is unable to ully express this. We are looking or an answer where thestudent clearly explains that there is no (or a constant) phase dierence across the wholewaveront. (0/1)


[1]

Perect! (1/1)

b On the screen a series of bright and dark fringes are seen.Explain how the fringes are formed.

[3]

An excellent answer. I particularly like the comment regarding the waves travel dierentdistances rom the two slits. (3/3)

Cooue inges wil now be seen as whie igh is mae u o

ifeen coous.

Ligh which is al in hase wih ise

Ligh o a singe equency

Waves om he wo sis ave ifeen isances, i he waves ae in hase when

hey each he sceen hee is a bigh inge, i hey ae exacy ou o hase hen a

ak inge is seen.


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c Figure 2 shows the fringes as seen on the screen.Calculate the wavelength of the light.

[3]

Te student made a silly error, orgetting to change the 9.2 mm to metres. Tey got it rightor the 1.6. Only one mark is lost, as this is clearly an arithmetic error rather than a lack ounderstanding. (2/3)


[3]

Te student recognises that coloured ringes will be seen and that the dierent colourshave dierent wavelengths. However, the fnal step in the argument is missing that themaxima/minima are in dierent places because the longer the wavelength, the greater thepath dierence required or the waves to get exactly (180) out o step. (2/3)

Question 3Question 3Student AStudent A3 a Explain what is meant by the term e.m.f.

[1]

Although this is sometimes accepted at lower levels it is not enough at A level, where an

explanation in terms o the work done in the complete circuit is needed. (0/1)

Te voage acoss he eminas when hee is no curen owing.

=

31.6 10 9.2

10 2.5

= 5.89 104m

Cooue inges wil now be seen as he ifeen coous have ifeen

waveenghs an wil ineee in ifeen aces.


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[2]

Te student has gone through the correct process, but has not recognised that the resistorshave a resistance in the kilohm range, so the current is not 1.2 A, but 1.2 mA. Also, thestudent has not completed the fnal calculation. (1/2)

c When a voltmeter is connected across XY it only reads3.2 V.i Suggest why the reading is less than the calculated value and deduce the resistance of

the voltmeter.

[1]

Te student recognises that the all in potential is to do with the resistance o the voltmeter;their explanation is rather muddled but there is just enough to gain the mark. (1/1)

ii Calculate the resistance of the voltmeter.

6 V

R

I1

I2I

V

2 k

3 k

Te curen = 6.05.0

= 1.2 A

Poenia ifeence = 3.0 1.2

Because he vomee has esisance so a curen asses hough i changing he

oa esisance in he cicui


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[3]

Te student starts o sensibly by drawing out the circuit and labelling the currents andresistances. However, they make the basic mistake o assuming that the current through the2 k resistor is still 1.2 mA, with the inevitable result that the current through the 3 kresistor is also 1.2 mA and the current through the voltmeter is zero. o sort this out they

really need to use Kirchho s laws. (0/3)

Student BStudent B3 a Explain what is meant by the term e.m.f.

[1]

An excellent answer. Te second sentence is not really needed, but it shows that the studentreally understands the concept. (1/1)


[2]

Once more a perect answer. (2/2)

Te oa wok one in he cicui by each couomb o eeciciy. Tis incues he

wok one in iving he curen hough he batey ise.

Te curen = 6.0/5.0 k = 1.2 mA

Poenia ifeence = 3.0 k 1.2 mA = 3.6 V

I=I1+I

2

1.2 =I1+I

2

6 = 3I1 + 2 1.2

I1= 1.2 A

I2= 0


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c When a voltmeter is connected across XY it only reads3.2 V.i Suggest why the reading is less than the calculated value and deduce the resistance

of the voltmeter.

[1]

Te student recognises that the all in potential is to do with the resistance o the voltmeter.Tis explanation is o a high standard. (1/1)

ii Calculate the resistance of the voltmeter

arithmetic error

[3]

error carried orward

Te student starts o sensibly by drawing out the circuit and labelling the currents andresistances. Tey correctly calculate the current through the 2 k resistor. Tey go on tofnd the current through the 3 k resistor, but then the student makes an arithmetic error(It should have been 1.067 A.) Te student then uses the incorrect answer correctly (apartrom losing a minus sign!) to fnd the voltmeters resistance. Te correct fnal answer shouldhave been 9.6 k. (2/3)

Because he vomee has esisance so he esisance in he owe am o he

cicui is now he combine esisance o he 3 k esiso an he vomee.

I=I1+I

2

In he oo conaining he wo esisos:

6 = 3.2 + 2I

I= 2.82

= 1.4 A

oenia o acoss 3 k esiso = 6 2.8 = 3.2 V

so curen = 3.23

= 1.67 A

soI2= 1.4 1.67 = 0.27 A

So esisance o vomee = 3.22.7

= 1.19 k


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Summary of students performanceSummary of students performanceIn question 2 student A scores 3/11 and in question 3 only 2/7. Tis sort of level is below thatof a grade E. aken with question 1, student A might just scrape an E.

Student B, however, scores 8/11 and 6/7, an excellent performance keeping well on track foran A grade.

ConclusionConclusionTese examples should help you see what the examiners are looking for and shoulddemonstrate some of the common mistakes that are made. You need to learn your workthoroughly, but also you must learn to understand.

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