A2 Special Relativity

86 minutes

79 marks

Q1.          A particle passes through a detector and 152 ns later hits a target 45.0 m away from the detector.

(i)      Calculate the speed of the particle between the detector and the target.

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(ii)      Calculate the transit time of the particle from the detector to the target, in the frame of reference of the particle.

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Q2.          The Michelson-Morley experiment represented in the diagram was designed to find out if the speed of light depended on its direction relative to the Earth’s motion through space. Interference fringes were seen by the observer.

(a)     (i)      Explain why interference fringes were seen.

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(ii)     The interference fringe pattern did not shift when the apparatus was rotated by 90°. Explain the significance of this null observation.

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(5)

(b)     Einstein postulated that the speed of light in free space is invariant. Explain what is meant by this postulate.

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(Total 7 marks)

 

 

Q3.          (a)     One of the two postulates of Einstein’s theory of special relativity is that physical laws have the same form in all inertial frames of reference.

Explain, with the aid of a suitable example, what is meant by an inertial frame of reference.

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(b)     A certain type of sub-atomic particle has a half-life of 18 ns when at rest. A beam of these particles travelling at a speed of 0.995c is produced in an accelerator.

(i)      Calculate the half-life of these particles in the laboratory frame of reference.

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(ii)     Calculate the time taken by these particles to travel a distance of 108 m in the laboratory at a speed of 0.995c and hence show that the intensity of the beam is

reduced to 25% of its original value over this distance.

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(Total 7 marks)

 

 

Q4.          (i)      Calculate the kinetic energy, in J, of a proton accelerated in a straight line from rest through a potential difference of 1.1 × 109 V.

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(ii)      Show that the mass of a proton at this energy is 2.2 m0, where m0 is the proton rest mass.

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(iii)     Hence calculate the speed of a proton of mass 2.2 m0.

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......................................................................................................................(Total 7 marks)

 

 

Q5.          (a)     In a science fiction film, a space rocket travels away from the Earth at a speed of 0.994 c, where c is the speed of light in free space. A radio message of duration 800 s 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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(b)     A student claims that a twin who travels at a speed close to the speed of light from Earth to a distant star and back would, on return to Earth, be a different age to the twin who stayed on Earth. Discuss whether or not this claim is correct.

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(Total 7 marks)

 

 

Q6.          (a)     Calculate the speed at which a matter particle has a mass equal to 10 times its rest mass.

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(b)     Explain why a matter particle can not travel as fast as a photon in free space even though its kinetic energy can be increased without limit.

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(Total 6 marks)

 

 

Q7.          π mesons, travelling in a straight line at a speed of 0.95 c, pass two detectors 34 m apart, as

shown in the figure below.

(i)      Calculate the time taken, in the frame of reference of the detectors, for a π meson to travel between the two detectors.

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(ii)      π mesons are unstable and decay with a half-life of 18 ns when at rest. Show that approximately 75% of the π mesons passing the first detector decay before they reach the second detector.

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Q8.          (a)     One of the two postulates of Einstein’s theory of special relativity is that the speed of light in free space is invariant.

(i)      Explain what is meant by this postulate.

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(ii)     State and explain the other postulate.

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(b)     A stationary muon has a rest mass of 1.9 × 10–28 kg.

For a muon travelling at a speed of 0.995 c, where c is the speed of light in a vacuum, calculate

(i)      its mass,

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(ii)     its total energy, in J.

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(iii)     its kinetic energy, in J.

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(Total 10 marks)

 

 

Q9.          (a)     One of the two postulates of Einstein’s theory of special relativity is that the speed of light in free space, c, is invariant.

Explain what is meant by this statement.

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(b)     A beam of identical particles moving at a speed of 0.98c is directed along a straight line between two detectors 25 m apart.

The particles are unstable and the intensity of the beam at the second detector is a quarter of the intensity at the first detector.

Calculate the half-life of the particles in their rest frame.

 

 

 

 

 

 

 

answer = ...................................... s(4)

(Total 5 marks)

  

Q10.          In an experiment, a beam of protons moving along a straight line at a constant speed of1.8 × 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 thedetectors.

 

 

 

 

answer = ..................................... m(1)

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

 

 

 

 

 

answer = ..................................... m(2)

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

Calculate the ratio 

 

 

 

 

 

 

 

 

 

answer = .........................................(5)

(Total 8 marks)

  

Q11.          The figure below represents the Michelson-Morley interferometer. Interference fringes are seen by an observer looking through the viewing telescope.

(a)     Explain why the interference fringes shift their position if the distance from either of the two mirrors to the semi-silvered block is changed.

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(b)     Michelson and Morley predicted that the interference fringes would shift when the apparatus was rotated through 90°. When they tested their prediction, no such fringe shift was observed.

(i)      Why was it predicted that a shift of the fringes would be observed?

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(ii)     What conclusion was drawn from the observation that the fringes did not shift?

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(Total 6 marks)

 

 

Q12.(a)    Calculate the speed of a particle at which its mass is twice its rest mass.

 

 

 

 

speed ..................................... m s−1

(2)

(b)     Use the axes below to show how the mass m of a particle changes from its rest mass moas its speed v increases from zero.

          Mark and label on the graph the point P where the mass of the particle is twice its rest mass.

(3)

(c)     By considering the relationship between the energy of a particle and its mass, explain why the theory of special relativity does not allow a matter particle to travel as fast as light.

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(Total 7 marks)

 

M1.          (i)        = 2.96 × 108 m s–1 (1)2

(ii)      t = 152 ns (1)

t0   = 152  (1)

= 25 ns (1)2

QWC 2[4]

 

 

M2.          (a)     (i)      two beams (or rays) reach the observer  (1)interference takes place between the two beams  (1)bright fringe formed if/where (optical) path difference =whole number of wavelengths(or two beams in phase)[or dark fringe formed if/where (optical) path difference =whole number + 0.5 wavelengths](or two beams out of phase by 180 °C/ π/2 /½ cycle)  (1)

(ii)     rotation by 90° realigns beams relative to direction of Earth’smotion  (1)no shift means no change in optical path difference betweenthe two beams  (1)( ) time taken by light to travel to each mirror unchangedby rotation  (1)distance to mirrors is unchanged by rotation  (1)( ) no shift means that the speed of light is unaffected[or disproves other theory]  (1)

max 5

(b)     the speed of light does not depend on the motion of thelight source  (1) or that of the observer  (1)

2[7]

 

 

M3.          (a)     Newton’s laws obeyed in an inertial frame [or inertial frames move at constant velocity relative to each other] (1)

suitable example (e.g. object moving at constant velocity) (1)2

(b)     (i)      (use of t = t0  gives)          t0 = 18 (ns) (1)

t = 18 × 10–9   (1)

  = 1.8 × 10–7 s (1)

(ii)     time taken   = 3.6 × 10–7 s (1)

time taken = 2 half-lives, which is time to decrease to 25%intensity (1)

[alternative scheme: (use of l = l0 gives) l0 = 108 (m)

l = 108  = 10.8 m (1)

time taken   = 3.6 × 10–8 s

= 2 half-lives, which is time to decrease to 25% intensity (1)]5

[7]

 

 

M4.          (i)      Ek (= eV) (= 1.6 × 10–19 × 1.1 × 109)= 1.8 × 10–10 (J) (1)   (1.76 × 10–10 (J))

(ii)      (use of E = mc2 gives)        Δm =   = 2.0 × 10–27 (kg) (1)

=   (1)

(allow C.E. for value of Ek from (i), but not 3rd mark)

 m = m0 + Δm (1)       (= 2.2 m0)

(iii)     (use of m = m0  gives)    2.2m0 =   (1)

v =   (1)

   = 2.7 × 108 m s–1 (1)[7]

 

 

M5.          (a)     (i)      t0 = 800 (s) (1)

(use of t = t0  gives)   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))

4

(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]

 

 

M6.          (a)     10m0 = m0  (1)

gives   = 1 – 0.01 = 0.99 (1)

v (= 0.995c) = 2.98(5) × 108 m s–1 (1)

3

(b)     m = m0  (1)

m → infinity as v → c (1)

[or m increases as v increases]

Ek(= mc2 –m0c2) → infinity as v → c (1)

v = c would require infinite Ek (or mass) which is (physically)

impossible (1)Max 3

[6]

 

 

M7.          (i)      time taken  s (1)

(ii)      use of t =  where t0 = 18 ns

and t is the half-life in the detectors’ frame of reference (1)

s (1)

time taken for Π meson to pass from one detector to the other= 2 half-lives (approx) (in the detectors' frame of reference) (1)2 half-lives correspond to a reduction to 25%,so 75% of the Π mesons passing the first detectordo not reach the second detector (1)

alternatives for first 3 marks in (ii)

1. use of t = , where t0 = 18 ns

= 57.6(ns)journey time in detector frame (= 2t) = 2 × 57.6ns (  2 half-lives)

2. use of t =   where t = 119 ns

       = journey time in detector frame

=37nsjourney time in rest frame = 2 × 18 ns (2 half-lives)[5]

 

 

M8.          (a)     (i)      speed of light in free space independent of motion of source (1)

and of motion of observer (1)

(ii)     laws of physics have the same form in all inertial frames (1)

inertial frame is one in which Newton’s 1st law of motion isobeyed (1)

laws of physics unchanged in coordinate transformation (1)

from one inertial frame to another (1)max 4

(b)     (i)      m (= m0 (1 – v2/c2)–1/2)   = 1.9 × 10–28 × (1 – 0.9952)–1/2(kg) (1)

                                             = 1.9 × 10–27 kg (1)

(ii)     E (= mc2)  = 1.9 × 10–27 × (3.0 × 108)2 (1)

                          = 1.7 × 10 10 J (1)

(iii)     EK (= E – m0c2)   = 1.7 × 10–10 (1.9 × 10–28 × (3.0 × 108)2) (1)

                                   = 1.5 × 10–10J (1)6

[10]

 

 

M9.          (a)     c is the same, regardless of the speed of the light source or theobserver (1)1

(b)     distance between detectors in rest frame of particles(= 25 × (1 – 0.982)1/2) = 5.0 m (1)

time taken in rest frame of particles   = 1.7 × 10–8 s (1)

time taken to decrease by ¼ = 2 half lives (1)

half life (= 1.7 × 10–8/2) = 8.5 × 10–9 s (1)

[alternatively

time taken in rest frame of detectors   = 8.5 × 10–8 stime taken in rest frame of particles(= 8.5 × 10–8 × (1 – 0.982)1/2) =1.7 × 10–8 s)]

4[5]

  

M10.          (a)     (i)      d0 =(speed × time = 1.8 × 108 × 95 × 10–9) = 17(.1) m 1

(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)

or

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)2

(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)

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

 =   

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

5[8]

  

M11.          (a)     bright (or dark) fringe is seen where the two beams are in phase(or out of phase by 180°) 

changing the distance to either mirror changes the path (or phase) difference(between the two beams) so fringes shift 

2

(b)     (i)      speed of light was thought to depend on the speed of the light source(or the speed of the observer)   (or on the motion of the Earth(through the aether))

distance travelled by each beam unchanged (by rotation) 

time difference between the two beams would change on rotation 

phase difference would therefore change (so fringes would shift) 3

(ii)     speed of light is independent of the speed (or motion) of the light source(or the observer) 

(or ‘aether’ hypothesis incorrect (owtte)) or absolute motion does not exist)1

[6]

 

M12.(a)

(Rearranging gives)

Accept either answer.2

(b)     curve starts at v=0, m = mo and rises smoothly  2nd mark; ecf from a if plotted correctly

curve passes through 2mo at v = 0.87 c (± 0.03c or in 2nd half of x-scale div containing 0.87c)  

3rd mark; There must be visible white space between the curve and the v = c line; also, the curve must reach 7mo at least.

curve is asymptotic at v = c ( and does not cross or touch v =c or curve back )  3

(c)     Energy = mc2 so (as v -> c ) energy of particle increases as mass increases  Alternative scheme for 1 mark only; mass infinite at v = c which is (physically) impossible  

mass -> infinity as v -> c so energy -> infinity which is (physically) impossible  

[OR for one mark only

force = ma so force increases as mass increases

Mass -> infinity as v->c so force -> infinity which is (physically ) impossible  ]2

[7] 

E1.          Most candidates were able to score the first mark in part (i), but in part (ii), many candidates thought 152 ns was the proper time and proceeded to obtain a larger value for the transit time. Candidates who realised that the proper time needed to be calculated, generally went on the score full marks. A small minority of candidates set up the length contraction expression and then calculated the transit time correctly.

 

E2.          Although most candidates knew, in part (a)(i), that the fringes were due to interference between two beams, few candidates mentioned that the two beams reached the observer. Many candidates were confused about the general conditions for a bright fringe or a dark fringe and often referred to phase difference in terms of wavelength, or gave the path difference for a dark fringe as half a wavelength instead of an odd number of half wavelengths.

In part (a)(ii), most candidates were aware that the ether theory was abandoned as a result of the Michelson-Morley experiment, but very few were able to explain in adequate terms, either why a fringe shift was predicted using the theory or why such a shift was not observed. Few candidates mentioned that the beams were realigned relative to the Earth’s direction of motion or that the time taken by light to travel along each path and the distance travelled was unchanged, when the apparatus was rotated.

There were some very good explanations of Einstein’s postulate in part (b) and many candidates scored both marks. Some candidates made irrelevant references to frames of reference.

 

E3.          Very few candidates referred to Newton’s laws or constant velocity in part (a). Some candidates referred to examples in terms of constant speed without mentioning direction. Most examples that gained credit referred to a simple example such as a free object inside an aeroplane.

In part (b)(i), most candidates carried out the calculation correctly. Some, however, inserted 18 ns incorrectly into the time dilation equation. In part (ii), most candidates who scored well, correctly used the distance and speed to calculate the time taken and then compared their answer with the half-life in the laboratory frame of reference. There were candidates who used the length contraction formula correctly and then made a comparison with the distance calculated using an incorrect value of time from part (i).

 

E4.          Most candidates were able to calculate the kinetic energy of the proton correctly in part (i). Many gave a correct and concise calculation in part (ii), although some candidates produced lengthy calculations as a result of converting into units of atomic mass then into MeV then into joules. In part (iii), a significant number of candidates were unable to calculate the speed after correctly identifying the principles involved in the calculation.

 

E5.          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.

 

E6.          Many candidates gained full marks for part (a). Those who failed to gain full marks often made careless algebraic errors even although they were aware of the principle of the calculation. Some candidates incorrectly considered the mass as nine times the rest mass.

In part (b), most candidates were aware that the mass of a particle increases as its speed approaches the speed of light, but very few related their statement to the relativistic mass equation. The kinetic energy of a matter particle was not considered correctly by many candidates, and they often used the non-relativistic expression to justify their statements concerning kinetic energy. Although some candidates did realise that the kinetic energy or the

mass would be infinite at the speed of light, relatively few stated that infinite mass or kinetic energy is impossible. Very few candidates related the two aspects of the explanation to each other.

 

E7.          In part (i), many candidates obtained the correct answer, although some lost the mark by using the answer in a further calculation of the time taken in the meson’s rest frame. In part (ii), most candidates knew that the given figure of 75% meant they needed to show that the mesons took two half-lives to travel from detector 1 to detector 2. However, only a minority of candidates were able to give a clear, correct solution. The majority of candidates who failed to make progress in part (ii) used the time dilation equation incorrectly, either as a result of algebraic errors or as a result of using their answer from part (i) as the proper time t0. Those candidates who chose to use the length contraction equation instead of the time dilation equation were mostly able to correctly complete the calculation.

 

E9.          In (a) many candidates were able to explain the statement by referring to the relative motion of the source or observer. Some candidates had learnt standard responses such as the speed of light is the same in all frames of reference which was insufficient to score the mark here.

In (b) there was a full range of answers. There were some blanks, there were some pages of numbers and equations getting nowhere and there were some excellent well argued four mark answers. However, the most common response was a two mark answer that found the time in the rest frame of the detectors and knew that this was two half-lives. Many candidates were unable to substitute the correct values or use the relevant equations with sufficient care.

  

E10.          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 mo c2.

  

E11.          In part (a) although few students stated the condition for the formation of a bright or dark fringe, many students did know that the shift of the interference pattern occurs because the path difference or the phase difference changed when the distance is changed.

In part (b) many students did appreciate that the speed of light was thought by scientists to be affected either by the motion of the Earth or by an ‘ether wind’. However, only a minority of students appreciated the distances travelled by each beam was unchanged or that the time difference between the two beams changed on rotation. Students often referred to a change in the time taken by the beams rather than a change in the time difference on rotation. Many students lost a mark as they did not refer to the rotation causing a change of the phase difference or a change in the optical path difference.

In part (c) most students appreciated the observation that the fringes did not shift led to the conclusion that absolute motion does not exist.

 

 

E12.(a)    Many candidates obtained both marks with a clear calculation. A significant number of candidates inserted the relative mass values correctly into the relativistic mass equation but were then unable to rearrange the equation correctly as a result of basic algebraic errors.

(b)     The graph proved to be a formidable challenge for many candidates. A significant number of candidates started the graph at the origin and many candidates failed either to plot the 2mo point at 0.866 c or to label it as instructed. Most candidates realised that the graph was a curve with an increasing positive gradient and that it did not cross the v = c line.

(c)     Few candidates were able to gain both marks. Many realised that mass and energy increases as v increases but failed to justify this statement by quoting the equation E = mc2. Some candidates did recognise the mass increased without limit as v tends to c but then stated that this (increase) was impossible. A significant number of candidates did recognise that the relativistic mass equation gives infinite mass at v = c although some candidates failed to state that this is physically impossible.