(I) Calculate the Duration of the Message When It Is Received at the Earth

Hinchley Wood School

Q1. (a) In a science fiction film, a space rocket travels away from the Earth at a speed of 0.994c, where c is the speed of light in free space. A radio message of duration 800s is transmitted by the space rocket.

(i) Calculate the duration of the message when it is received at the Earth.

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(ii) Calculate the distance moved by the rocket in the Earth’s frame of reference in the time taken to send the message.

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(4)

(b) A student claims that a twin who travels at a speed close to the speed of light from Earth to adistant star and back would, on return to Earth, be a different age to the twin who stayed onEarth. Discuss whether or not this claim is correct.

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(3)

(Total 7 marks)

Q2. Hertz discovered how to produce and detect radio waves. He measured the wavelength of radio waves produced at a constant frequency using the arrangement shown in the diagram below.

(i) Explain why the strength of the detector signal varied repeatedly between a minimum and a maximum as the detector was moved slowly away from the transmitter along the dotted line.

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

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(ii) Hertz found that a minimum was detected each time the detector was moved a further 1.5m away from the transmitter.
Calculate the frequency of the radiowaves.

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(Total 5 marks)

Q3. A narrow beam of electrons is produced in a vacuum tube using an electron gun, part of which is shown in Figure 1.

Figure 1

(a) (i) State and explain the effect on the beam of electrons of increasing the filament current.

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(2)

(ii) State and explain the effect on the beam of electrons of increasing the anode potential.

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(2)

(b) The beam of electrons is directed at right angles into a uniform magnetic field as shown in Figure 2.

Figure 2

(i) Explain why the electrons move in a circular path at a constant speed in the magnetic field.

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(3)

(ii) When the speed of the electrons in the beam is 7.4 × 106 m s–1 and the magnetic flux density is 0.60 m T, the radius of curvature of the beam is 68 mm.

Use these data to calculate the specific charge of the electron, stating an appropriate unit. Give your answer to an appropriate number of significant figures.

answer = ......

(4)

(iii) Discuss the historical relevance of the value of the specific charge of the electron compared with the specific charge of the H+ ion.

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(2)

(Total 13 marks)

Q4. In an experiment, a beam of protons moving along a straight line at a constant speed of 1.8×108ms–1 took 95 ns to travel between two detectors at a fixed distance d0 apart, as shown in the figure below.

(a) (i) Calculate the distance d0 between the two detectors in the frame of reference of the detectors.

answer = ...... m

(1)

(ii) Calculate the distance between the two detectors in the frame of reference of the protons.

answer = ...... m

(2)

(b) A proton is moving at a speed of 1.8 × 108ms–1

Calculate the ratio

answer = ......

(5)

(Total 8 marks)

Q5. The radioactive isotope of sodium has a half life of 2.6 years. A particular sample of this isotope has an initial activity of 5.5 × 105Bq (disintegrations per second).

(a) Explain what is meant by the random nature of radioactive decay.

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

......

(2)

(b) Use the axes to sketch a graph of the activity of the sample of sodium over a period of 6years.

(2)

(i)the decay constant, in s–1, of , 1 year = 3.15 × 107 s

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(ii)the number of atoms of in the sample initially,

......

(iii)the time taken, in s, for the activity of the sample to fall from 1.0 × 105Bq to 0.75×105Bq.

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(6)

(Total 10 marks)

Q6. (a) A cylinder of fixed volume contains 15 mol of an ideal gas at a pressure of 500kPa and a temperature of 290 K.

(i) Show that the volume of the cylinder is 7.2 × 10–2 m3.

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(ii) Calculate the average kinetic energy of a gas molecule in the cylinder.

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(4)

(b) A quantity of gas is removed from the cylinder and the pressure of the remaining gas falls to 420 kPa. If the temperature of the gas is unchanged, calculate the amount, in mol, of gas remaining in the cylinder.

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(2)

(c) Explain in terms of the kinetic theory why the pressure of the gas in the cylinder falls when gas is removed from the cylinder.

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(4)

(Total 10 marks)

Q7.(a)Describe the changes made inside a nuclear reactor to reduce its power output and explain the process involved.

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(2)

(b) State the main source of the highly radioactive waste from a nuclear reactor.

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(1)

(c) In a nuclear reactor, neutrons are released with high energies. The first few collisions of a neutron with the moderator transfer sufficient energy to excite nuclei of the moderator.

(i)Describe and explain the nature of the radiation that may be emitted from an excited nucleus of the moderator.

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(2)

(ii)The subsequent collisions of a neutron with the moderator are elastic.

Describe what happens to the neutrons as a result of these subsequent collisions with the moderator.

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(2)

(Total 7 marks)

Q8.A bicycle and its rider have a total mass of 95 kg. The bicycle is travelling along a horizontal road at a constant speed of 8.0 m s–1.

(a) Calculate the kinetic energy of the bicycle and rider.

......

(2)

(b) The brakes are applied until the bicycle and rider come to rest. During braking, 60% of the kinetic energy of the bicycle and rider is converted to thermal energy in the brake blocks. The brake blocks have a total mass of 0.12 kg and the material from which they are made has a specific heat capacity of 1200 J kg–1 K–1.

(i) Calculate the maximum rise in temperature of the brake blocks.

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(ii) State an assumption you have made in part (b)(i).

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(4)

(Total 6 marks)

M1. (a) (i)t0 = 800 (s) (1)
(use of t = t0gives)t = 800(1 – 0.9942)–1/2(1)

= 7300 s (1)

(ii)distance (= 0.994ct = 0.994 × 3 × 108 × 7300)
= 2.2 × 1012m (1)(2.18 × 1012m)
(allow C.E. for value of t from (i))

(b)space twin’s travel time = proper time (or t0) (1)

time on Earth, t = t0(1)

t > t0
[or time for traveller slows down compared with Earth twin] (1)
space twin ages less than Earth twin (1)
travelling in non-inertial frame of reference (1)

max 3

[7]

M2. (i)reflected waves and incident waves form a stationary/standing
wave pattern or interfere/reinforce/cancel (1)
nodes formed where signal is a minimum (1)

(ii)λ/2 = 1.5 (m) [or λ = 3 (m)]
[or nodes formed at half–wavelength separation] (1)

(use of c = fλ gives)(1)

= 100 MHz (1)

[5]

M3. (a) (i) The number of electrons (per second) in the beam will increase (1)
because the filament will become hotter and will emit more
electrons (per 2 second) (1)

(ii)the speed (or kinetic energy) of the electrons will increase (1)

because the electrons (from the filament) are attracted towards
the anode with a greater acceleration (or force) (1)

(or gain more kinetic energy in crossing a greater pd)

(b) (i) (magnetic) force on each electron in the beam is perpendicular
to velocity (1)

no work is done on each electron by (magnetic) force so ke
(or speed) is constant (1)

magnitude of (magnetic) force is constant because speed
is constant (1)

(magnetic) force is always perpendicular to velocity so
is centripetal (1)

max 3

(ii)rearrangingr = gives (1)

= 1.81 × 1011(1) C kg–1(1)

for correct answer to 2 sf(1)

(iii)specific charge for the electron ≈ 2000 × specific charge of H+(1)

(accept = and accept any value between 1800 and 2000)

which was the largest known specific charge before the specific
charge of the electron was determined/measured (1)

(or which could be due to a much greater charge or a much smaller
mass of the electron)

[13]

M4. (a) (i)d0 =(speed × time = 1.8 × 108 × 95 × 10–9) = 17(.1) m

(ii)d (= d0 (1 – v2/c2)½)

= 17.1 × (1 – (1.8 × 108/3.0 × 108)2))½

= 14 m (or 13.7 m or 13.68 m)

t = t0 (1 –v2/c2)–½

95 = t0 × (1 – (1.8 × 108/3.0 × 108)2)–½ gives t0 = 76 ns

d = vt0 = 1.8 × 108 × 76 × 10–9 = 14 m (or 13.7 m or 13.68 m)

(b)m (= m0 (1 –v2/c2)–½)

= 1.67(3) × 10–27 × (1 – (1.8 × 108/3.0 × 108)2)–½)

= 2.09 × 10–27 kg

kinetic energy = (m–m0) c2

or correct calculation of E = mc2 (= 1.88 × 10–10 J)

orcorrect calculation of E0 = m0c2 (= 1.50 × 10–10 J)

= 0.25 (allow 0.245 to 0.255 or ¼ or 1:4)

[8]

M5. (a) (use of ‘isotope’ instead of ‘nucleus’ not accepted)
there is equal probability of any nucleus decaying,
it cannot be known which particular nucleus will decay next,
it cannot be known at what time a particular nucleus will decay,
the rate of decay is unaffected by the surrounding conditions,
it is only possible to estimate the proportion of nuclei decaying
in the next time interval
any two statements (2)

QWC 2

(b)continuous curve starting at 5.5 × 105Bq
plus correct 1st half-life (2.6 yrs, 2.75 × 105Bq(1)
correct 2nd half-life (5.2 years, 1.4 × 105Bq) (1)
(allow C.E. for incorrect 1st half-life)

=8.5 × 10–9 (s–1) (1)

(8.46 × 10–9 (s–1))

(ii) (use of = –λN gives) N = (1)

= 6.5 × 1013 (atoms) (1)

(allow C.E, for value of λ from (i))

(iii) (use of N = N0e–λt and AN gives)

= (1)

= 3.4 × 107 (s) (1)

(allow C.E. for value of λ from (i))

[10]

M6. (a) (i)p V = nR T(1)
V = (1) (gives V = 7.2 × 10–2m3)

(ii) (use of Ek = kTgives) Ek = × 1.38 × 10–23 × 290 (1)

= 6.0 × 10–21 (J) (1)

(b) (use of pV = nRT gives) (1)

[or use pn]

n = 13 moles (1) (12.5 moles)

(c)pressure is due to molecular bombardment [or moving molecules] (1)
when gas is removed there are fewer molecules in the cylinder
[or density decreases] (1)

(rate of) bombardment decreases (1)
molecules exert forces on wall (1)

is constant (1)

[orpV= Nm (c2)(1)

V and m constant (1)

(c2)constant since T constant (1)

pN(1)]

[orp = p(c2)(1)

explanation of ρ decreasing (1)

(c2)constant since T constant (1)

p (c2)ρ(1)]

max 4

[10]

M7.(a)insert control rods (further) into the nuclear core / reactor

a change must be implied for 2 marks
marks by use of (further) or (more)

allow answers that discuss shut down as well as power reduction

which will absorb (more) neutrons (reducing further fission reactions)

If a statement is made that is wrong but not asked for limit the score to 1 mark (e.g. wrong reference to moderator)

(b) fission fragments / daughter products or spent / used fuel / uranium rods (allow) plutonium (produced from U-238)

not uranium on its own

(c)(i) (electromagnetic radiation is emitted)

A reference to α or β loses this first mark

as the energy gaps are large (in a nucleus) as the nucleus de-excites down discrete energy levels to allow the nucleus to get to the ground level / state mark for reason

2nd mark must imply energy levels or states

(ii)momentum / kinetic energy is transferred (to the moderator atoms)
or
a neutron slows down / loses kinetic energy (with each collision)

(eventually) reaching speeds associated with thermal random motion or reaches speeds which can cause fission (owtte)

[7]

M8. (a) (use of Ek = ½mv2 gives) Ek =× 95 × 8.02 (1)

= 3040 J (1)

(b) (i)ΔQ = 0.60 × 3040 = 1824 (J) (1)
(allow C.E. for Ek from (a))
(use of ΔQ = mcΔθ gives) 1824 = 0.12 × 1200 Δθ (1)
Δθ= 13 K (1)(12.7 K)
(allow C.E. for ΔQ)

(ii)no heat is lost to the surroundings (1)

[6]

E1. Many correct calculations were given in part (a). Some candidates used the duration of 800 s for the time t in the expression, rather than t0, and hence obtained a value for t0, which was not required. Of the candidates who did use the duration of 800 s correctly some failed to make progress as a result of careless arithmetical errors. In part (ii), most candidates knew how to calculate the required distance although some candidates, having obtained the correct distance, then applied the relativistic length formula to obtain their final answer.

In part (b) most candidates realised that the space twin aged less because time ran slower for the space traveller. Many candidates were able to use the time dilation equation to explain why this is so, although very few candidates realised that the space twin was in a non-inertial frame of reference for part of the journey.

E2. In part (i), the majority of candidates realised that the reflected waves and direct waves formed a stationary wave pattern and that the signal was a minimum at the nodes. Some candidates referred only to interference, without demonstrating any awareness that stationary waves were formed. In part (ii), a significant number of candidates thought that the wavelength was 1.5 m although most candidates knew how to calculate the frequency of the radio waves from the wavelength.

E3. In (a) (i) most candidates understood the heating effect of an electric current, realising the beam would become more intense due to the release of more electrons, but struggled for a precise and rigorous explanation. Many thought the electrons were emitted with increased speed and a significant number of candidates did not explain the effect on the beam, as asked in the question, and simply said more electrons were emitted. In (a) (ii) most candidates identified that the electrons would be travelling at greater speed but again struggled for a complete explanation.

In (b) (i) most candidates scored one mark for identifying that the force on the electrons acted at right angles to their direction of motion. More able candidates identified the fact that the force does no work on the electrons and hence their speed remains constant. Very few candidates could go on to say that the force is always at right angles to their direction of motion and hence is centripetal.

The calculation in (b) (ii) should have been a straightforward exercise but less than a quarter of the candidates scored all four marks. The necessary equation is on the data sheet or candidates could equate the centripetal force to the force on an electron in a magnetic field. However many candidates did not know what was meant by specific charge and simply worked out the charge on an electron. A significant number of candidates used values of e or m or both from the data sheet in arriving at their value of e/m. This approach received no marks. The significant figure and unit marks were independent. Although the unit of specific charge was widely known (on the Data and Formulae Booklet) a large number of candidates gave their answer to three significant figures even though all the data was presented to two significant figures.

In (b) (iii) the candidates that had worked out e in (ii) went on to talk about quantisation of charge and Millikan’s experiment for which they gained no credit. This question is a crucial part of the turning points option and as such was poorly answered. Very few candidates could clearly say that the specific charge of an electron was much greater than that of the H ion which could have been due to either a much greater charge or a much smaller mass of the electron.

E4. Most candidates scored the mark in (a)(i) and many went on to successfully complete (a)(ii).

However, careless arithmetic errors were not uncommon in (a)(ii) and some candidates used an incorrect formula.

Some excellent answers were seen in part (b) which demonstrated a very good understanding of the topic and a first-rate grasp of algebra. The more able candidates usually worked through several lines of algebra and successfully reduced the ratio to an expression in terms of v/c only then completed the calculation in a single line. Most candidates started with calculations of either rest energy or relativistic mass or the total energy. Many candidates demonstrated they knew how to calculate the relativistic mass of the proton at the given speed. However, many candidates did not gain more than two marks because they attempted to calculate the kinetic energy using ½ mv2 rather than mc2–moc2, even when they had calculated the correct values of mc2 and moc2.

E5. The question on nuclear instability performed well on the whole and many candidates gained high marks. Answers to part (a) however, were weak, with most candidates achieving only one of the two available marks. This was disappointing since the examiners had recognised as many as five possible marking points. The reason for not gaining the two marks was either because the candidates elaborated unnecessarily on what was essentially one point or because of an imprecise use of language. Many candidates used ‘randomly’ in the context of their answer, without really explaining its meaning. Since the same word was part of the subject of the question, credit was not awarded for such answers. Most candidates gained credit by referring to the uncertainty in knowing which nucleus would decay or when. The examiners did not accept statements which referred to an isotope rather than a nucleus.

The graph in part (b) was usually drawn correctly with three points on it: the first at time zero and the other two at each subsequent half-life.

The calculation in part (c) was well done by most candidates, with many scoring full or almost full marks. A significant number however, were unable to progress beyond the

exponential decay equation in part (iii) and many had in .

E6. There were significant variations between candidates. Most were able to correctly apply the equation of state for an ideal gas but were less confident when it came to calculating in part (a) the average kinetic energy of a gas molecule in the cylinder. It was evident that a significant number of candidates did not know the appropriate formula. In part (c) explaining, in terms of the kinetic theory, why the pressure of the gas fell was well answered. The only common error was assuming that the kinetic energy of molecules fell even though the temperature remained constant.

E7.Most candidates were fully aware of the function of the control rods in absorbing excess neutrons and scored well in part (a). Some candidates said too much by explaining the role of the control rods to absorb neutrons and the moderator to slow neutrons down but then did not make it clear which reduced the power. The weaker candidates talked about control rods controlling the reactions without any further explanation.

Part (b) was answered well by most but it was common to give the answer fuel rods rather than spent fuel rods.

In (c)(i) the most common answer was ‘gamma rays’ but very few then went on to discuss energy levels. Some of those that did then spoilt their answer by referring to changing electron levels.

Part (c)(ii) was a very good question to distinguish between the weak and strong candidates. The weaker candidates focussed on the wording in question concerning elastic collisions. They interpreted this to mean the neutrons maintain their kinetic energy or momentum during all subsequent collisions.

E8. The last question in the paper was generally done well and candidates across the ability range were able to perform the various calculations successfully. A minority were confused by the need to use only 60% of the kinetic energy, but because marks were awarded for consequential errors, this did not prove to be too much of a penalty.

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