Mixed Unit 5 13/06/14 - Hinchley Wood School

Mixed unit 5 – 13/06/14 - Hinchley Wood School

Q1. (a) The Sombrero Galaxy is 50 million light years away from the Earth.

(i) Calculate the distance to this galaxy in parsecs.

......

(ii) Use Hubble’s Law to show that this galaxy is receding at 1000 km s–1.

......

(iii) One of the lines in the Hydrogen spectrum has a wavelength of 656.3 nm when measured in a laboratory on Earth. Calculate the wavelength of the same line in the observed spectrum of the Sombrero Galaxy.

......

(4)

(b) Show how Hubble’s Law can be used to estimate the age of the Universe. State the assumption made.

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(3)

(Total 7 marks)

Q2. Cygnus A may be the nearest quasar yet discovered.

(a) Cygnus A has a redshift, z, of 0.057.
Calculate the distance to Cygnus A. State an appropriate unit.

answer = ...... unit = ......

(4)

(b) The first quasars were discovered in the 1950s. What property of quasars led to their discovery?

......

(1)

(Total 5 marks)

Q3. (a) Describe the main features of black holes and quasars.

You may be awarded marks for the quality of written communication in your answer.

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(3)

(b) There is some evidence to suggest that there is a black hole of 3 × 109 solar masses at the centre of the galaxy M87. Calculate the radius of the event horizon for this black hole.

......

(2)

(Total 5 marks)

Q4. (a) Draw a Hertzsprung-Russell diagram on the axes below. Label the maximum and minimum values of both absolute magnitude and temperature on the axes. Also label the positions of the main sequence, dwarf stars and giant stars.

(4)

(b) The spectral class of four stars is given in the table.

star / spectral class
Alnitak / O
Sirius / A
Sun / G
Antares / M

The spectrum of each star contains absorption lines. State what produces the main absorption lines in each case.

Alnitak

......

Sirius

......

Sun

......

Antares

......

(2)

You may be awarded marks for the quality of written communication in your answer.

......

(3)

(Total 9 marks)

Q5. (a) Describe what is meant by an Airy disc and explain its significance in determining the resolving power of a telescope.

You may be awarded marks for the quality of written communication in your answer.

......

(3)

(b) The Arecibo telescope is the largest radio telescope in the world. It can be used to investigate distant galaxies by detecting the 1.4 GHz radio signal produced by molecular hydrogen.

(i) When the telescope was being built, any surface irregularities had to be less than 0.01 m in order for it to detect the molecular hydrogen signal. Verify this value using an appropriate calculation.

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(ii) The diameter of the Arecibo telescope is 305 m. Calculate its resolving power when detecting the molecular hydrogen signal.

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(3)

(c) Describe a problem associated with spherical reflecting telescopes and state how telescopes are designed to prevent it.

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(2)

(Total 8 marks)

Q6.Some liquids in open bottles deteriorate exposure to air. The diagram below shows one device used to reduce this deterioration. It consists of a rubber valve that is inserted into the neck of the bottle together with a pump that is used to remove some of the air in the bottle through this rubber valve. On an up-stroke of the pump, air enters the pump chamber from the bottle. On the down-stroke, the rubber valve closes and the air in the chamber is expelled to the atmosphere through another valve (not shown) in the handle.

(a) There is 3.5 × 10–4 m3 of air space in the bottle and the volume of the pump chamber changes from zero at the beginning of the up-stroke to 6.5 × 10–5 m3 at the end of the up-stroke. The initial pressure of the air in the bottle is that of the atmosphere with a value of 99 kPa.

Assuming the process is at constant temperature, calculate the pressure in the bottle after one up-stroke of the pump.

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(3)

(b) Calculate the number of molecules of air originally in the air space in the bottle at a temperature of 18 °C.

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(3)

(c) Explain how the kinetic theory of an ideal gas predicts the existence of a gas pressure inside the bottle. Go on to explain why this pressure decreases when some of the air is removed from the bottle.

......

(5)

(Total 11 marks)

Q7. (a) (i) Write down the equation of state for n moles of an ideal gas.

......

(ii) The molecular kinetic theory leads to the derivation of the equation

pV =,

where the symbols have their usual meaning.

State three assumptions that are made in this derivation.

......

(4)

(b) Calculate the average kinetic energy of a gas molecule of an ideal gas at a temperature
of 20 °C.

......

(3)

(c) Two different gases at the same temperature have molecules with different mean square speeds.
Explain why this is possible.

......

(2)

(Total 9 marks)

Q8. A female runner of mass 60 kg generates thermal energy at a rate of 800 W.

(a) Assuming that she loses no energy to the surroundings and that the average specific heat capacity of her body is 3900 J kg–1K–1, calculate

(i) the thermal energy generated in one minute,

......

(ii) the temperature rise of her body in one minute.

......

(3)

(b) In practice it is desirable for a runner to maintain a constant temperature. This may be achieved partly by the evaporation of sweat. The runner in part (a) loses energy at a rate of 500 Wby this process.

Calculate the mass of sweat evaporated in one minute.

specific latent heat of vaporisation of water = 2.3 × 106 J kg–1

......

(3)

......

(2)

(Total 8 marks)

M1. (a) (i)d = = 15.3 × 106 (pc)

(ii) (use of v = Hd gives) v = 65 × 10–6 (km s–1 pc–1) × 15.3 × 106 (1)

≈(1000 km s–1)

(iii) (use of = -gives)Δλ =× 656.3 (nm) = 2.19 (nm) (1)

(allow C.E. for value of v from (ii)

λgalaxy = 656.3 + 2.19 = 658.5 nm (1)

(b) for the furthest point of the Universe, (1)

age of Universe = (1)

[or use of v = Hd and t = (1)

if all started from same point t = age of Universe = (1)]

assumption: that H remains constant

[7]

M2. (a)use of z = v/c

to give v = zc = 0.057 × 3 × 108

= 1.71 × 107ms–1 = 1.71 × 104 kms–1

use of v = Hd

to give d = v/H = 1.71 × 104 / 65

= 263 Mpc

(b) (Strong) radio sources

[5]

M3. (a) black hole: large gravitational field or very dense (1)
escape velocity greater than c(1)
quasar: large red shift or very far away (1)
very powerful sources (1)

max 3

QWC 1

(b)(1)

= 8.9 × 1012m (1)

[5]

M4. (a) Hertzsprung -Russell diagram to show:
absolute magnitude scale from +15 to –10(1)
temperature scale from 50 000 to 2500 (K) (1)
main sequence drawn correctly (1)
giants and dwarfs shown in correct areas (1)

(b) Alnitak : helium (absorption) (1)
Sirius : hydrogen Balmer (absorption) lines 4 correct(1)
Sun : metals (absorption) 2 correct
Antares: molecular bands

(c) reference to P = AT4(1)
class M (Antares) cooler than class O (Alnitak) (1)
but same brightness, therefore cooler star bigger (1)
so Antares has larger surface area(1)

max 3

QWC 2

[9]

M5. (a) the central bright spot (1)
of the diffraction pattern produced when light passes through a circular
aperture (1)
the minimum distance between two sources which can be resolved by a
telescope found when centre of one Airy disc falls just outside the Airy disc
for the other source (1)
[or statement referring to maxima and minima]

QWC 1

(b) (i)λ= 0.21(4) m (1)

precision needed = λ/20 ≈ 0.01 m (1)

(ii)θ= 6.9 × 10–4 rad (1) (6.88 × 10–4 rad)

(use of λ = 0.214 gives θ = 7.02 × 10–4 rad)

(c) parallel rays furthest from axis of reflector brought to focus closer to
reflector than rays close to axis (1) (allow correct diagram)
use parabolic reflector (1)

[8]

M6.(a) use ofpV = constant or p1V1 = p2V2(1)

p = 99 × 3.50/4.15 (1)

= 83.5 kPa (1)

(b) no. of moles = 99 [103] × 3.5 × 10–4/8.31 × 291 (1)

= 1.4(3) × 10–2 moles (1)

no. of molecules (= 1.4(3) × 102 × 6.02 × 1023)

= 8.6 × 1021(1)

momentum change at wall (1)

momentum change at wall/collision at wall leads to force (1)
[allow impulse arguments]

less air so fewer molecules (1)

so change in momentum per second/rate of change is less
[or per unit per time] (1)

pressure is proportional to number of molecules (per unit volume) (1)

max 5

[11]

M7. (a) (i)pV = nRT(1)

(ii) all particles identical or have same mass (1)
collisions of gas molecules are elastic (1)
inter molecular forces are negligible (except during collisions) (1)
volume of molecules is negligible (compared to volume of
container) (1)
time of collisions is negligible(1)
motion of molecules is random (1)
large number of molecules present
(therefore statistical analysis applies) (1)
monamatic gas (1)
Newtonian mechanics applies (1)

max 4

(b)Ek = or (1)

= (1)

= 6.1 × 10–21 J(1)(6.07 × 10–21 J)

[9]

M8. (a) (i) energy = 800 × 60 = 48 × 103J (1)

(ii) (use of Q = mcθ gives) 48 × 103 = 60 × 3900 ×θ(1)
θ= 0.21 K (1) (0.205 K)
(allow C.E. for value of energy from (i)) 3

(b)Q = ml gives 500 × 60 (1) = m × 2.3 × 106(1)
m = 0.013 kg (1)

(c) not generating as much heat internally (1)
still losing heat (at the same rate)
[or still sweating] (1)
hence temperature will drop (1)

lmax 2

[8]

E1. The calculations in part (a) were usually correct. Many candidates gave the distance in part (i) in megaparsecs, which made handling the units in part (ii) a little easier. There was a tendency for weaker candidates to work backwards in part (ii) and thus doctor their answer to part (i). In effect, these candidates made two errors, one with the units of Hubble’s constant and the second with the velocity of the galaxy. This was despite the fact that the unit of Hubble’s constant is given on the data sheet. These efforts were not awarded. The unit of km s–1 also caused problems in part (iii) and many candidates obtained a value of 2 × 10–12 mfor the change in wavelength. Some credit was still given if the candidates showed that this wavelength was added to the laboratory based wavelength (red shift) and not subtracted.

In part (b) it was clear that many candidates knew that and hence obtained

v = Hd and v=, but very few explained how tgave the age of the Universe.

The expected assumption was associated with the value of the Hubble constant, as this is the value quoted in the specification. Credit was given to some alternatives. Candidates who described a graphical method to explain how the age of the Universe could be obtained, showed the best understanding of the topic.

E2. The calculation in part (a) was answered correctly by many students. Marks were lost by students who had problems matching the speed and distance units to the Hubble constant. There were also several students who made simple algebraic errors rearranging the Hubble equation. Credit was given to unit answers consistent with their calculation, but only “Mpc” or megaparsec was given a mark if there was no calculation performed.
In part (b), only a minority of students were aware that quasars were discovered due to their powerful radio wave emissions. This is explicit on the specification, but many students would probably benefit from learning the story of how the first quasar was discovered.

E3. Theories concerning black holes and quasars do change and it is therefore important that candidates make sure the essential points about these phenomena are made before embarking on details associated with particular examples or situations. If the size of a black hole is given by the event horizon, then black holes are not necessarily dense and certainly do not have infinite mass. In part (a), extreme density was allowed as a property, because it could be the singularity that was being referred to. The essential point however, is that the escape velocity exceeds the speed of light. There was evidence of confusion between quasars and pulsars. Not all quasars are thought to be at the furthest points of the universe, although this was allowed as a property, as was the extreme red shift which is the evidence for this distance. The other property looked for was the power output or intensity. Small size was not accepted, as quasars are thought to be the size of solar systems in some situations.

Despite the regularity of the type of question set in part (b), the same mistakes are still being seen in calculations related to black holes. Missing out the mass of the Sun and failing to square the speed of light were the two most common errors.

E4. The Hertzsprung-Russell diagram comes in many forms but candidates should be encouraged to learn the one which has appeared in many published mark schemes. The absolute magnitude scale should go from +15 up to –10 and the temperature scale from 50 000 K to 2 500 K. Credit was still given if the answers fell within a range around these values. It was not necessary for candidates to label the spectral class on the horizontal axis this time. Dwarf stars and giant stars were indicated correctly very often, but the main sequence was sometimes drawn as a line rather than a band and the shape was also drawn incorrectly on numerous occasions.

It was pleasing to note in part (b), how many candidates knew what caused the main spectral lines, although some answers were ambiguous and listed too many possible suggestions for any credit.

The argument and reasoning expected in part (c) has been examined on numerous occasions in this unit and candidates should be aware of the correct sequence. Antares was usually correctly identified as the larger star. Most candidates also knew that M class stars are cooler than O class stars, but there was considerable confusion about which star was brighter, with some candidates discussing the difference between real and apparent magnitude. The best answers stated that they had the same brightness and therefore the cooler star had to be bigger to give out the same amount of light, quoting Stefan’s law to support their argument. It was not sufficient for candidates to refer to the absolute magnitudes being the same; this had to be related to luminosity, power output or brightness. There was ample scope here for candidates to demonstrate their ability to write clearly, using correct spelling, punctuation and grammar and to structure their answers in a coherent way.

E5. This question was answered poorly on the whole, with many answers showing a lack of knowledge or understanding in all three areas the question covered. In part (a), very few candidates knew that the Airy disc is the bright central maximum of the diffraction pattern produced when light passes through a circular aperture. Several candidates thought that it was part of the telescope itself. Many thought it was the whole pattern, or failed to mention that it was due to diffraction. Its role in determining the resolving power was often described poorly, with many vague and ambiguous answers given. Some candidates obtained credit by drawing a relevant labelled diagram to help the explanation. This practice should be encouraged.

In part (b), most candidates knew of the need to calculate the wavelength and divide it by 20 to obtain a value for the upper limit of surface irregularities. Benefit of the doubt was given to candidates who performed the calculations but failed to state the significance of the answer obtained. In questions like this, candidates should be encouraged to finish off answers by stating how their calculations are related to the required information. The straightforward calculation in part (ii) was performed by most candidates, but several lost marks due to the usual unit errors associated with resolving power. Commonly, the unit was missed altogether or the degree or even the watt was suggested.

Simply writing ‘spherical aberration’ in part (c) was insufficient to gain the allocated mark. A clear description or diagram was required, showing how paraxial rays further from the principal axis are brought to a focus closer to the lens than those nearer the axis. Most candidates knew that the use of a parabolic mirror solved the problem. Many candidates confused chromatic and spherical aberration or wrote answers which were too vague to gain credit.

E7. This question also gave good discrimination. In part (a) the better candidates made very few errors and scored high marks but the less able candidates found it more difficult and were unable to come up with three assumptions of the kinetic theory.

The calculation in part (b) caused problems and even candidates who selected the correct expression had difficulties. The commonest of these was failing to convert the temperature to Kelvin.

Part (c) was answered correctly by only the best candidates and although many did identify that molecular mass was important, they could not quite explain why.

E8. Questions involving thermal energy have caused problems for candidates in previous papers, but this question was well answered. It also proved to be a good discriminator. Weaker candidates did have problems, however, in structuring their calculations in a logical way, leading to errors in the final answers.

The explanation in part (c) was answered well by good candidates who, in some cases, provided considerable detail which was worthy of more than the two marks allocated.

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