edexcel science 2011 p3 sample exam

7/30/2019 Edexcel Science 2011 P3 Sample exam

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EXTENSION UNITS PHYSICS

SAMPLE EXAM P3

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FORMULAE

You may find theses formulae useful

The relationship between focal length, object and image distance

current = number of particles per second charge on each particle I = N q

kinetic energy = electronic charge accelerating potential difference KE = mv2

= e V

momentum = mass velocity p = m v

The relationship between temperature and volume for a gas

The relationship between volume and pressure for a gas V1P1 = V2P2

The relationship between the volume, pressure and temperature for a gas

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SAMPLE EXAM P3

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SAMPLE EXAM P3 1 Turn over

Surname Name

AMERICAN ACADEMY LARNACA

YEAR 5 SAMPLE EXAM

Extension Units Physics

Unit P3: Applications of Physics Higher Tier

Time: 1 hour

You must have:

Calculator, ruler

Total Marks

Instructions

Use black ink or ball point pen.

Fill in the boxes at the top of this page with your Surname and Name. Answer all questions.

Answer the questions in the spaces provided

there may be more space than you need.

Information

The total mark for this paper is 60.

The marks for each question are shown in brackets

use this as a guide as to how much time to spend on each question.

Questions labelled with an asterisk (*) are ones where the quality of your written communication

will be assessed

you should take particular care with your spelling, punctuation and grammar,

as well as the clarity of expression, on these questions.

Advice

Read each question carefully before you start to answer it.

Keep an eye on the time.

Try to answer every question.

Check your answers if you have time at the end.


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_______________________________________________________________________________________

1. Liquid nitrogen has important uses in industry for fast cooling and preservation.

(a) Nitrogen has to be cooled to become a liquid.

A factory will cool 200 000 litres of nitrogen at 27 C and normal atmospheric pressure to itsboiling point of196 C and normal atmospheric pressure.

(i) Calculate the volume of nitrogen at its boiling point.

[3 marks]

V

(ii) Liquid nitrogen is about 2900 times denser than gaseous nitrogen at 77 K (it means

that at 77 K, liquid nitrogen will expand 2900 times as it changes to gas).

Show that the factory will obtain about 20 litres of liquid nitrogen.

[1 mark]

(b) The picture below shows what happens when liquid nitrogen is not stored properly.

The tank on the left has exploded because of the high pressure.


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(i) Explain in terms of particles why the pressure of a gas increases as temperature

increases.

[2 marks]

(ii) The tank on the right remained intact because it has a release valve that allows gas

to escape if the pressure exceeds a certain value.

Explain why this will protect the tank from exploding.

[2 marks]

[Total for Question 1 = 8 marks]

_______________________________________________________________________________________


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_______________________________________________________________________________________

2. The picture below shows protons, neutrons and electrons at the same speed, moving through a

region (shaded area) where there is a magnetic field perpendicular to the path of the particles.

(a) Use the properties of the particles to explain why they are deflected in this way.

[3 marks]

(b) When the electrons are at position A, draw arrows to show

(i) the velocity of the electrons, label the arrow V

[1 mark]

(ii) the force acting on the electrons, label the arrow F

[1 mark]

protons

neutrons

electrons

magnetic field

A


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(c) Alpha particles, , are accelerated to a speed of 201 10

3m s

1.

They are smashed onto stationary gold nuclei, .

The nuclei move off at a speed of 8000 m s1

.

Show that the -particles rebound with a speed of 193 103

m s1

.

Hence deduce whether the collision is elastic or inelastic.

[3 marks]


_______________________________________________________________________________________

201 103

m s1

8000 m s1

v1?


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_______________________________________________________________________________________

3. (a) Magnifying glasses are simple convex lenses.

Magnification is achieved by placing the object closer to the lens than the focal point.

(i) Complete the picture below to show where the image will be formed.

[1 mark]

(ii) Magnification is defined as

(you can use similar triangles to understand why).

Find the magnification of an object placed 3 cm away from a lens of focal length

equal to 5 cm.

Find the height of the image if the object has a size of 5 mm.[3 marks]

M

z

f

f


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(b) Human eyes also use lenses.

Large parts of the eye are filled with liquid, the aqueous and vitreous humor.

(i) Name and identify on the diagram the muscles responsible for accommodation.

[1 mark]

(ii) A light ray is moving from the lens into the vitreous humor.

Calculate the angle of refraction as the light ray enters the vitreous humor.

Refractive index of lens = 1.41, Refractive index of vitreous humor = 1.34.

Draw the refracted ray.

[3 marks]

(iii) State the near and far point for an average adult.

[1 mark]

1. near point = 2. far point =

(iv) Explain how a convex lens can correct vision for long-sighted patients.

[1 mark]


_______________________________________________________________________________________

aqueous humor vitreous humor

30

lens, n = 1.41

vitreous humor, n = 1.34


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_______________________________________________________________________________________

4. (a) An X-ray machine has an accelerating voltage of 80 kV.

(i) Calculate the maximum speed that the electrons develop.

mass of electron = 9.1 1031

kg

charge of electron = 1.6 10

19

C [2 marks]

M

(ii) The X-ray machine is on for 6 seconds.

It produces a current equal 0.02 mA.

Calculate the total number of electrons given out by the cathode.

[2 marks]

Total

(b) I X-rays and stand out.

To view blood vessels for example the patient is injected with iodine.

Then the patient is either observed live or a series of pictures are taken in succession.

It allows doctors to detect conditions like angiomas, which are benign tumours formed by

the cells of blood vessels.

Circle the angioma in the fluorography picture below.

[1 mark]


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(c) Medical X-rays are divided X-rays.

Hard X-rays have a higher energy than soft X-rays.

The division is taken to be at approximately 5 1018

Hz.

(i) Mk H X-rays.

[1 mark]

Penetration of 1 cm muscle tissue for various frequencies

The graph above shows the percentage of X-rays of different frequencies that can

penetrate 1 cm of muscle.

(ii) Explain why a large percentage of penetration is important for the good quality of

the X-ray image.

[1 mark]

0

10

20

30

40

50

60

70

80

90

100

0 5 10 15 20

%

Penetration

Frequency / 1018 Hz


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(iii) T X-rays, filters are used like the ones shown below.

They incorporate thin aluminium foil through which X-rays have to pass after they

are produced.

Explain why using a lead filter would be less practical.

[1 mark]

(iv) The graph below shows the relative amount of X-rays of different frequencies that

reach the patient with and without a filter.

Explain why using a filter is safer to the patient with little effect on image quality.

[2 marks]


_______________________________________________________________________________________

0 5 10 15 20

Amount

of X-rays

Frequency / 1018 Hz

without filter

with 3 mm Aluminium filter


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_______________________________________________________________________________________

5. Methionine is a sulphur-containing essential amino-acid, the structure of which is shown below.

I and rebuilding process.

To study the production or use of methionine in cells, some methionine molecules can undergo a

, Sulphur atom, S-32, is replaced by theradioactive isotope S-35.

(a) S-35 is produced by bombarding Chlorine nuclei, Cl-35, with neutrons.

X

Complete the missing numbers in the nuclear equation above and identify particle X.

[2 marks]

X

(b) Compare the radioactive S-35 with the stable S-32.

(i) Explain what type of decay you expect S-35 to undergo.

[2 marks]

(ii) Complete the equation below, identify particle Y based on your answer in (i) and

explain what happens to the S-35 nucleus when it becomes more stable.

[2 marks]

Oxygen

Nitrogen

CarbonSulphur

Hydrogen

METHIONINE MOLECULE


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*(c) It was always assumed that isotopic labelling with S-35 leaves the cell unaffected.

Now there is evidence that the cell activity is significantly impaired during the process,

to the extent that results may be considered unreliable.

S-35 has a half-life of 87.5 days.

Refer to the properties of the S-35 decay and all the other changes that take place.Refer to the damage that can happen to a cell and explain why some researchers do not

consider the results obtained from S-35 labelling completely reliable.

[6 marks]


_______________________________________________________________________________________


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_______________________________________________________________________________________

6. (a) Optical fibres are becoming more useful in communications.

They are made of an outer glass cladding and a denser inner core.

(i) Calculate the critical angle between the core and the cladding.

[2 marks]

(ii) Describe what will happen if the light ray is incident on the core-cladding boundary

at an angle larger than the critical.

[1 mark]

(iii) If the optical fibres are bent too much, the signal will become weak quickly.

Explain why.

[2 marks]


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SAMPLE EXAM P3 15

*(c) The heart can also be monitored using an electrocardiograph.

The pictures below show a typical arrangement and a normal electrocardiogram

obtained in this way.

Discuss the two ways of monitoring the heart, the one that uses pulse oximeters and the

one that uses electrocardiographs.

Refer to advantages and disadvantages of each.

[6 marks]


_______________________________________________________________________________________

END

TOTAL FOR PAPER = 60 MARKS


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SAMPLE EXAM EXTENSION UNITS PHYSICS MARK SCHEME

P3 MARK SCHEME 16

MARK SCHEME

p. 2 & 3

1. (a) (i) T1 = 27 + 273 = 300 K, T2 = 196 + 273 = 77 K [1 mark]

V1 / T1 = V2 / T2 => 200 000 / 300 = V2 / 77 [1 mark] substitution

V2 = 200 000 77 / 300 = 51 333 litres [1 mark] rearrangement & answer

(ii) 51 333 / 2900 = 17.7 litres [1 mark]

(b) (i) As temperature => KE of particles and speed [1 mark]

=> more frequent / violent collisions with container walls => force etc. [1 mark]

(ii) Gas escapes => fewer gas molecules [1 mark]

=> less collisions with container walls => less force / pressure [1 mark]

p. 4 & 5

2. (a) Neutrons have no charge => not deflected [1 mark]

Protons +ve, electronsve => so deflected in opposite directions [1 mark]

Protons heavier than electrons => protons deflected less [1 mark]

(b) (i)

(ii)

(c) m1u1 + m2u2 = m1v1 + m2v2Take mass of = 4, mass of Au = 197

4 201 000 + 0 = 4 v1 + 197 8000 [1 mark] formula & substitution

=> v1 = (804 000 1 576 000) / 4 = 193 000 m s1

[1 mark] rearrangement

KEbef= (4) (201 000)2

= 8.0802 1010

ignore unit (mass not in kg anyway)

KEaft = (4) (193 000)2

+ (197) (8000)2

= 8.0802 1010

KEbef= KEaft => elastic [1 mark]

A

V, tangent to path [1 mark]

F, towards centre /

perpendicular to path [1 mark]


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P3 MARK SCHEME 17

p. 6 & 7

3. (a) (i)

[1 mark]

(ii) 1 /f= 1 / u + 1 / v=> 1 / 5 = 1 / 3 + 1 / v=> 1 / v= 2 / 15 => v= 7.5 cm [1 mark]

M = 7.5 / 3 = 2.5 [1 mark]size = 5 2.5 = 12.5 mm [1 mark]

(b) (i)

[1 mark]

(ii) n1 sin 1 = n2 sin 2 => 1.41 sin 30 = 1.34 sin 2 [1 mark]

2 = sin1

(1.41 sin 30 / 1.34) = 31.7 [1 mark]

(iii) 1. 25 cm; 2. infinity [1 mark]

(iv) L / E

=> convex lens helps bend the light rays more [1 mark]

f

f

image

ciliary

muscles

30

away from normal

[1 mark]


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P3 MARK SCHEME 18

p. 8, 9 & 10

4. (a) (i) e V= KE= mv2

=> 1.6 1019

80 000 = 9.1 1031

v2

[1 mark] substitution

v2

= 2.8 1016

=> v= 1.68 108

m s1

[1 mark] rearrangement & answer

(ii) I = N e => 0.02 10

3

= N 1.6 10

19

=> N = 1.25 10

14

electrons / sec [1 mark]Total number = 6 N = 6 1.25 1014

= 7.5 1014

electrons [1 mark]

(b)

[1 mark]

(c) (i) f => energy

> f< 5 1018

Hz; f> 5 1018

Hz [1 mark]

(ii) Film / Detector of X-rays found behind patient (or similar) [1 mark]

(iii) Lead absorbs too much radiation => little left for detection (or similar) [1 mark]

(iv) Soft X-rays mostly filtered that do not penetrate tissue anyway [1 mark]

Soft X-rays are absorbed so increase dosage for patient [1 mark]


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P3 MARK SCHEME 19

p. 11 & 12

5. (a)

X

[1 mark]

X = proton [1 mark]

(b) (i) S-35 has too many neutrons compared to stable S-32 [1 mark]> -decay [1 mark]

(ii)

-

[1 mark]

Y = electron, 1 neutron changes to a proton and an electron [1 mark]

*(c) S-35 undergoes -decay, releasing electrons

Electrons are ionising, especially since the source is now in the cell

-particles can ionise DNA, causing mutation

-particles can ionise other molecules in the cell that will change their normal function

These changes and/or energy released may kill cellS changes to Cl => this changes the properties of methionine

Long half-life means there is time to observe how methionine behaves in cell

but also too much time for damage to cell to be severe

Level Marks

1 1 2 Refers to ionisation / mutation and other effects of radioactivity

to cells

2 3 4 Connects problems of radioactivity specifically to properties of

-particles and the half-life of S-35

3 5 6 Refers to specific changes due to methionine and relates to possibility

of unreliable results during research


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P3 MARK SCHEME 20

p. 13, 14& 15

6. (a) (i) n1 sin 1 = n2 sin 2 => 1.62 sin c = 1.52 sin 90 [1 mark] substitution

c = sin1

(1.52 / 1.62) = 69.8 = 70 [1 mark] rearrangement & answer

(ii) Total internal reflection (or similar) [1 mark]

(iii) Angle of incidence becomes < than critical angle [1 mark]

Some light will be transmitted through the cladding

=> less will be left for reflection => signal weakens [1 mark]

(b) Oxygenated blood will absorb more infrared than red

=> detectors will detect more red than infrared [1 mark]

*(c) ECG based on electrical signals produced during muscle activity

ECG uses conductivity of tissue that transmits electricity to skin

Oximetry based on absorption and transmission properties of Hb and oxyHb

Advantages of oximetry

Easy method, minimum / small / cheap

apparatus

Gives extra information (besides pulse rate)

of amount of oxygen in blood

Disadvantages of oximetry

Provides only simple pulse rate, no details of

heart cycle

Advantages of ECG

Gives excessive detail on heart cycle

Provides more than PQRST cycle giving

detailed information on function and

working of the heart

Disadvantages of ECG

Not easy / simple method

Requires more specialized knowledge or

setting up by a specialist

Level Marks

1 1 2 Refers to the general principles of working of the ECG and

pulse oximetry

2 3 4 Refers to and explains one advantage of ECG or oximetry

3 5 6 Refers to and explains a second advantage of ECG or oximetry

edexcel science 2011 p3 sample exam

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