8097

November 2000

INTRODUCTION
Course assessment will require candidates to respond to questions which integrate the work studied across the component units as well as to questions which are less structured and set in less familiar and more complex contexts (Advanced Higher Physics, National Course Specification page 19, Second edition - December 1999).
A unit assessment in Advanced Higher Physics contains questions designed to assess achievement of the performance criteria associated with that unit and is not intended to grade or rank the candidates. The course assessment is designed to grade, hence certain questions, or parts of questions, are designed to assess the attainment of the candidate against the Grade descriptions for the award at ‘A’.
Generating evidence for estimates of performance and for appeals
The unit assessments cannot be used on their own to provide an estimate of performance or as evidence for appeals. Course type questions are required for these purposes. These questions must emulate the demands of the external course assessment. This pack provides examples of this type of question. The questions, provided in the specimen question paper, are further sources.
A prelim paper attempted after the completion of two units can provide useful evidence for appeals. Such a paper might reflect the format and duration of the final examination paper and provide useful practice of examination technique. For this paper to be a suitable basis on which to submit an appeal, it must be able to persuade the Principal Assessor that it: • has not been seen by the candidate • offers a suitable level of demand • allows for a demonstration of the criteria associated with the Grade descriptions for the “C” and “A” awards.
The test paper can contain questions which are in the public domain. It is not intended that all staff across Scotland should construct new questions when there are valid past paper questions available from previous years. However the use of a complete paper e.g. that of the previous year or so, is unlikely to prove acceptable to SQA in that there is a high probability that the candidate has seen these questions.
A prelim paper consisting of a variety of questions of suitable demand from a range of sources should be set. The paper should include questions that require candidates to integrate across at least two units. Such a prelim would be supplemented by evidence of attainment in the remaining unit, for example, by a short test of course type questions on that unit.
Although the prelim is a preferred method in many centres, this is not a requirement of SQA and a number of appropriate shorter tests of suitable demand makes a possible alternative.
The questions
Many of the questions in this pack are based on previous CSYS questions. These have been selected and adapted to reflect the demands of the Advanced Higher syllabus.
Physics: Course Questions (Advanced Higher) 1

Solutions and marking scheme
Brief solutions to all questions have been provided at the end of this pack. These solutions are designed to be of assistance to staff in marking student’s scripts.
The general instructions for markers issued by SQA in 1999 should be used as an additional guide for awarding marks. Comments have been made to assist marking and highlight specific points.
For questions involving calculations where two marks are allocated, the following general rule for the award of partial marks should be applied: {½} for the formula, {½} for substitution, {1} for final answer. If the unit in final answer is wrong or missing deduct {½}.
This is denoted in the marking scheme as [standard marking].
Where a question involving a calculation has more than two marks allocated, comment is provided to assist in awarding marks. In some cases alternative methods of responding to the question have been identified.
For questions calling for a description or an explanation a brief solution has been provided. Professional judgement should be applied in giving credit for alternative correct answers.
Credit should be given for correct physics. Wrong physics is always penalised.
It is appropriate to emphasise that in any external examination paper, markers attend a ‘marker’s meeting’ where marking of the individual questions are discussed.
Decisions are then made on any finer points and markers all agree to apply these decisions in their individual marking in order to ensure uniformity across the scripts.
In addition at the ‘standardisation’ the examining team cross mark a selection of scripts from each marker to ensure that there is a uniformity of marking. Any ‘excessive strict’ or ‘excessive lenient’ marking is adjusted accordingly.
Physics: Course Questions (Advanced Higher) 2

MECHANICS
1.
A child of mass 40 kg swings in a horizontal circle of radius 2.0 m by holding the end of a rope attached to a vertical post as shown in the figure below.
The child completes one revolution in 4.0 s (a) Find: (i) the speed of the child (ii) the horizontal component of tension in the rope (iii) the vertical component of tension in the rope (iv) the angle made by the rope with the pole.
8
(b) The child lets go when his feet are 0.50 m above the ground. Calculate how far from the post he lands. State any assumption you made.
4 (12)
Physics: Course Questions (Advanced Higher) 3

2.
A solid wheel and axle is allowed to roll from rest at AA / down an inclined set of rails as shown in Figure 1.
Figure 1 The following information is given: Mass of wheel and axle assembly Height of AA / above BB Distance between AA Diameter of axle / and BB Time to travel from AA / / to BB / / = 1.4 kg = 0.40m
= 2.0 m = 1.0 cm = 30 s (a) Calculate the speed of the solid wheel when it reaches BB / .
(b) Calculate the angular velocity of the solid wheel at BB / .
(c) (i) Write down an expression for the total rotational and translational kinetic energy of the wheel at BB / .
(ii) Find the moment of inertia of the wheel.
(d) The experiment is repeated with a spoked wheel as shown in Figure 2.
The spoked wheel is mounted on an identical axle and has the same mass as the solid wheel. The radius of the spoked wheel is greater than that of the solid wheel.
2 2 3
Figure 2 (i) State and explain how the moment of inertia of the spoked wheel compares with the moment of inertia of the solid wheel.
(ii) State and explain how the time taken for the spoked wheel to reach BB / compares with the time taken for the solid wheel to reach BB / .
4 (11)
Physics: Course Questions (Advanced Higher) 4

3.
(a) State the mathematical expression of the inverse square law of gravitation 1 (b) (i) Show that the speed, v, of a satellite in orbit around the Earth is given by
v
=
GM E r
.
(ii) In the above expression state the meaning of the symbols G, M E and r.
3
(c) A satellite of mass 120 kg is placed in orbit 600 km above the Earth.
(i) Calculate the kinetic energy of the satellite (ii) State the expression for the gravitational potential at a distance r from a mass M.
(iii) Due to the work done by friction the orbit of the satellite decreases to 580 km above the Earth. Calculate the loss of potential energy of the satellite.
(iv) State and explain what happens to the kinetic energy of the satellite when the orbit decreases.
7 (11)
4.
(a) State what is meant by the moment of inertia of a rigid oject about an axis of rotation.
1
(b) The rotor of an electricity generator in a power station can be considered as a uniform cylinder rotating about its axis. The manufacturer quotes the following data for such a rotor: Rotor diameter Rotor mass Moment of inertia of rotor = 3.6 m = 210 × 10 3 kg = 3.4 × 10 5 kg m 2 “Running Speed” = 600 revolutions per minute Time to reach running speed from rest = 30 s (i) Calculate the angular acceleration of the rotor as it is brought from rest to its running speed. You may assume that the acceleration is uniform.
(ii) Calculate the torque exerted on the rotor.
(iii) What is the kinetic energy of the rotor at its “running speed”?
(iv) Due to the motion of the rotor, a nut on the surface of the rotor experiences a force. The mass of the nut is 0.80 kg. The rotor is turning at its designed running speed. Calculate the size of the force experienced by the nut.
(v) Explain why there is a limit to the safe running speed of the rotor.
9 (10)
Physics: Course Questions (Advanced Higher) 5

5. When a car is moving at a steady speed, it may be assumed that each piston in the engine moves with simple harmonic motion.
(a) A piston has a mass of 0.35 kg and a stroke (twice the amplitude) of 7.6 cm. When the engine is turning at 3000 revolutions per minute, find: (i) the maximum acceleration of the piston (ii) the position or positions where the maximum acceleration occurs (iii) the maximum force exerted on the piston (iv) the maximum speed of the piston (v) the maximum kinetic energy of the piston.
9
(b) Explain, in terms of the physical principles involved, the effects of over revving the engine.
2 (11)
6.
(a) In a particle accelerator, protons can be made to travel at speeds approaching the speed of light. Using the relationship:
m
=   1 −
m
0
v
2
c
2   1 2 (i) determine the speed of a proton, when its relativistic mass is double its rest mass (ii) calculate the relativistic energy of a proton travelling at this speed.
4
(b) In a hydrogen atom, the angular momentum of the electron is quantised.
(i) Calculate the lowest value for the angular momentum of the electron.
(ii) When the electron has its lowest angular momentum, the orbital radius of the electron is 5.3 × 10 -11 m. Calculate the linear momentum of the electron in this case.
(iii) Calculate the de Broglie wavelength of the electron in a hydrogen atom when it has its lowest angular momentum.
6 (10)
Physics: Course Questions (Advanced Higher) 6

7. The figure below shows a Ferris wheel, at a fun fair, which rotates in a vertical plane about a horizontal axis.
Data for the Ferris wheel is given below.
Moment of inertia of the wheel empty Maximum passenger mass per chair Number of chairs Radius of the wheel = 8.8 × 10 5 kg m 2 = 200 kg = 16 = 10 m The Ferris wheel is loaded to a maximum. When rotating at its maximum angular speed, the period of rotation of the wheel is 30 s (a) Calculate the moment of inertia of the fully loaded Ferris wheel.
(b) Calculate the rotational kinetic energy of the loaded wheel at this maximum angular speed.
(c) With the driver motor switched off and no application of the brakes, the system decelerates uniformly to rest from its maximum speed. The wheel takes 20 revolutions to come to rest.
(i) Calculate the angular deceleration of the wheel.
(ii) Calculate the frictional torque which causes this deceleration.
2 2 4
(d) A passenger will experience a force exerted by the seat. When the seat is at the highest position, this force is a minimum.
(i) Explain why this is the case.
(ii) Calculate the value of this force experienced by a 50 kg passenger.
4 (12)
Physics: Course Questions (Advanced Higher) 7

8.
(a) An ice skater changes her angular velocity in a spin by pulling in her outstretched arms (Figure 1(a)) until they are close to her body (Figure 1(b)).
Figure 1(a) Figure 1(b) (i) State what happens to her angular velocity (ii) Explain the change produced, taking into account that her body mass does not alter and that no external torque is applied.
(b) A turntable is in the form of a disc with a moment of inertia about the central vertical axis of 0.024 kg m 2 . The turntable rotates in the horizontal plane on very low friction bearings about its centre. The angular velocity of the turntable is 8.0 rad s -1 .
A small dart of mass 0.15 kg is now dropped from rest just above the turntable. The dart sticks into the turntable at a distance of 0.14 m from its centre as shown in Figure 2.
3
Figure 2 (i) Calculate the moment of inertia of the dart and turntable together.
(ii) Calculate the new angular velocity of the system.
(iii) Calculate: (A) the kinetic energy of the turntable before the dart is dropped (B) the kinetic energy of the turntable and dart after the dart has been dropped.
(iv) Account for the change in kinetic energy.
8 (11)
Physics: Course Questions (Advanced Higher) 8

9. In an experiment to measure the gravitational acceleration g, a mass was suspended from a spiral spring. The resulting extension of the spring was measured. The mass was then pulled down a little further, released, and allowed to oscillate. The periodic time of the oscillations was measured.
This was repeated for a number of different masses.
The graph of extension against mass from the readings obtained is shown below.
160 120 80 40 0 0 0.1
0.2
0.3
Mass/kg
0.4
0.5
Physics: Course Questions (Advanced Higher) 9

(a) From the graph: (i) obtain a value for the spring extension per unit mass (ii) find the uncertainty in this value (iii) what mass would produce an extension of 1 m?
(b) When the mass is displaced from its equilibrium position, the restoring force is given by:
F
= −
ky
where k is the force per unit extension or spring constant measured in N m -1 .
(i) Show that the periodic time T of the oscillation is given by:
T
= 2 π
M k
where M is the mass suspended from the spring.
(ii) The values obtained for the periodic time of the oscillations for different masses are shown below.
Mass M / kg
0.10
0.20
0.30
0.40
0.50
Periodic time T / s
0.38
0.53
0.63
0.74
0.82
5
(A) (B) (C) Plot a straight line graph to verify the relationship between T and M.
Use the gradient of your graph to find a value for k.
From your value for k and your answer to (a) (iii), obtain a value for g, the gravitational acceleration.
8 (12)
Physics: Course Questions (Advanced Higher) 10

10. (a) (i) State what is meant by ‘escape velocity’.
(ii) Show that the expression for the escape velocity from a planet of mass M is given by:
v
= 2
GM r
where the symbols have their usual meaning.
(iii) Calculate the escape velocity from the Moon. The radius of the Moon is 2.6 × 10 6 m.
(b) It is believed that stars which are sufficiently massive may collapse to form a black hole when their nuclear fuel is exhausted at the end of their life. A star with a mass three times the mass of the Sun could collapse to form a black hole of radius less than 10 km.
What is meant by a ‘black hole’?
5 1 (6)
Physics: Course Questions (Advanced Higher) 11

11. A disc is mounted on a bearing as shown in the figure below.
Hanging weight The hanging weight produces a constant torque on the disc. The disc rotates with a uniform angular acceleration, α.
The disc is released from rest and the time taken for five complete rotations while accelerating is measured with a stopclock. The experiment is repeated seven times and the readings are shown below.
Experiment 1 2 3 4 5 6 7 Time t for five rotations / s 9.80
9.42
9.66
9.93
9.92
9.35
9.63
(a) Find the mean time t and the approximate random uncertainty in the mean.
3
(b) The equation for the angular displacement of an object accelerating uniformly from rest is given by θ = 1 2 α (i) Use the mean time t to calculate the angular acceleration α.
(ii) Calculate the uncertainty in α from the uncertainty in t.
4
(c) The moment of inertia I of the disc is given by
I
= (
mgr d
) .
where m is the mass of the hanging weight, g is the acceleration due to gravity, r is the radial arm of the accelerating torque, and the hanging mass is removed.
α
d
is the angular deceleration of the moving disc due to the frictional torque after Calculate the value of I and the uncertainty in I given that:
m
r α
d g
= 0.011 ± 0.001 kg = 0.012 ± 0.002 m = 0.050 ± 0.003 rad s -2 = 9.81 ± 0.01 m s -2 .
4 (11)
Physics: Course Questions (Advanced Higher) 12

ELECTRICAL PHENOMENA
12. (a) Air ceases to be an effective insulator at an electric field strength of 3.0 × 10 6 V m -1 .
(i) Write down an expression for the electric field strength E between parallel plates separated by a distance d when the potential difference between the plates is V.
(ii) Two parallel plates are separated by a distance of 1.5 mm. Calculate the p.d. between the plates when insulation of the air between plates breaks down.
(b) An uncharged spherical conducting sphere is placed in the region between two charged plates as shown in the figure below.
_ +
2
(i) State the nature of the electric field inside the sphere.
(ii) Copy the diagram.
(A) (B) Show the distribution of charge on the sphere.
Sketch the electric field in the region between the plates.
(c) (i) The electrostatic potential V at a distance r from a point charge Q is
V
=
Q
4 πε
o r
.
Write down an expression for the potential at the surface of a conducting sphere of radius R and charge Q.
(ii) Calculate the charge on a conducting sphere of diameter 100 mm when the potential of the sphere is 1.8 kV.
4 3 (9)
Physics: Course Questions (Advanced Higher) 13

13. (a) (i) Draw an arrangement of point charges which gives a point in space where the electrostatic potential is zero and the electric field strength is non zero. Indicate the point concerned.
(ii) Draw an arrangement of point charges which gives a point in space where the electrostatic potential is non zero and the electric field intensity is zero. Indicate the point concerned.
(b) Two protons are separated by a very large distance. They are projected towards each other along the same straight line with equal speeds.
(i) The protons are momentarily at rest at a minimum separation r
min
.
Write down an expression for the electrostatic potential at one proton due to the field of the other.
(ii) Hence write down an expression for the electrostatic potential energy when the separation is r
min
(iii) Given that r
min
is 5.0 × 10 -15 m find the initial speed of the protons, stating any principle that you use in your calculation.
(iv) Calculate the potential difference required to accelerate protons in a linear electrostatic accelerator in order to achieve the speed determined in (b) (iii).
4 7 (11)
14. (a) An electron moving with a constant speed of 2.0 × 10 6 m s -1 enters a region of space where there is a uniform magnetic field. The magnetic induction of the magnetic field is 0.50 mT. The path of the electron makes an angle of 30 ° with the direction of the magnetic field as shown in the figure below.
(i) Explain why the resultant movement of the electron must be helical.
(ii) Calculate the component of the electron’s velocity perpendicular to the field.
(iii) Write down an expression equating the force on the electron with the central force required for circular motion.
(iv) Find the time for the electron to complete one turn of the helical path.
(b) The speed of the electron is doubled. Explain the effect on the time calculated in (a) (iv)?
7 2 (9)
Physics: Course Questions (Advanced Higher) 14

15. (a) The force F on a current carrying conductor in a magnetic field is given by:
F
=
IlB
s in θ .
Derive the relationship for the magnitude of the force acting on a charge q moving with speed v perpendicular to a magnetic field B.
(b) A beam of positively charged particles enters through slit X into a region where there is a uniform magnetic induction B and uniform electric field strength E.
2
Z X Y Each particle has a mass m, charge q and speed v.
Plate Z is made positive with respect to the earthed plate.
(i) State the directions of the electric and magnetic fields when the particles pass through the region undeflected to leave by slit Y.
(ii) Write down an expression equating the forces due to the electric and magnetic fields.
(iii) Derive an expression which shows the relationship between v, E, and B. (iv) The potential difference between the plates is 550 V. The plate separation is 0.12 m.
(A) Calculate the electric field strength between the plates.
(B) The charged particles travelling undeflected between X and Y have a speed of 1.2 × 10 3 m s -1 .
Calculate the magnetic induction of the magnetic field.
7 (9)
Physics: Course Questions (Advanced Higher) 15

16. (a) (i) State what is meant by electric field strength E at a point.
(ii) Write down an expression for the force F which acts on a charge q in an electric field of strength E. (b) In an oscilloscope, electrons travelling at a speed of 1.5 × 10 7 m s -1 enter midway between the y-plates as shown in the figure below. The length and separation of the plates are shown. The potential difference between the plates is 40 V.
0.050 m
2
0.020 m (i) Describe the motion of the electrons as they pass between the plates.
(ii) Calculate the vertical displacement y of the electrons as they leave the region between the plates.
(iii) By considering horizontal and vertical components of velocity, or otherwise, calculate the angle θ between the horizontal and the path of the electrons as they emerge from the field between the plates.
7 (9)
17. (a) A solenoid is connected to a d.c. supply as shown in the diagram below.
S The switch S is closed.
(i) Draw a graph to show how the current varies with time.
(ii) Explain the shape of the graph.
(b) A coil of inductance 6.0 H and resistance 4.0 Ω is connected to a 24 V d.c. supply of negligible internal resistance.
Calculate: (i) the maximum current in the circuit (ii) the maximum potential difference across the coil (iii) the maximum rate of change of current in the coil.
3 5 (8)
Physics: Course Questions (Advanced Higher) 16

18. In the apparatus shown in the figure below, a beam of singly charged positive ions enters an evacuated chamber at A. The ions travel between plates P 1 and P 2 before passing through slit S, normal to XY.
The whole apparatus is in a uniform magnetic field of magnetic induction 0.60 T. The potential difference between plates P 1 and P 2 is 3000 V. The distance between P 1 and P 2 is 20 mm. The deflected ions form lines on a photographic plate W.
(a) Show in a sketch, the polarity of plates P 1 and P 2 and the direction of the magnetic field which will make the ions follow the path shown.
(b) (i) Show that ions passing through the slit S have a velocity v given by
v
=
E B
where the symbols have their usual meaning.
(ii) Explain what will happen to ions which have a speed greater than this value.
(c) Calculate the velocity of ions passing through S.
(d) Describe and account for the shape of the path of the ions after passing through S.
(e) Some ions form a trace on W at a point R. R is 190 mm from S.
Calculate the mass of one of these ions.
2 2 2 2 3 (11)
Physics: Course Questions (Advanced Higher) 17

19. A battery of e.m.f. 10V and negligible internal resistance is connected to a data capturing device and a coil of wire as shown in the circuit diagram below.
S data capture device to measure the current When the switch is closed, the data capturing device measures the current at intervals of 0.1 ms.
The current in the circuit is found to vary over the first millisecond as shown in the graph below.
Physics: Course Questions (Advanced Higher) 18

19. continued (a) Switch S is closed. The coil produces a back e.m.f.
(i) State the size of the back e.m.f. at this time.
(ii) Explain why the back e.m.f. opposes the battery e.m.f.
(b) Use the graph to determine: (i) the initial rate of change of current (ii) the self inductance of the coil
2 4
(c) The final steady current in the circuit is 25 mA.
(i) Find the resistance of the coil.
(ii) Calculate the energy stored in the inductor when the current is at its maximum.
4
(d) A coil of negligible resistance is connected to a variable frequency sinusoidal a.c. supply. Sketch a graph showing how the current in the coil varies with frequency. The amplitude of the output voltage from the supply is constant.
1 (11)
20. (a) The magnetic induction B at a distance r from an infinitely long straight wire carrying a current I is given by
B
= µ 2 π 0
I r
Derive the expression for the force per unit length between two long parallel wires carrying currents I
1
and I
2
at a distance r apart.
(b) A long horizontal cable P lying in a North-South direction provides the electricity supply to an industrial process. The cable carries a current of 200 A in the direction from North to South. The vertical component of the Earth’s magnetic field is 4.36 × 10 -5 T.
(i) What is the size and direction of the force per unit length acting on the wire due to the vertical component of the Earth’s magnetic field?
2
(ii) A second identical cable Q is laid parallel to and at a distance of 1.5 m from cable P in the same horizontal plane. Cable Q carries a current of the same size and in the same direction as the current in P.
What is the size and direction of the force per unit length acting on P due to Q?
6 (8)
Physics: Course Questions (Advanced Higher) 19

21. (a) Two point charges Q 1 and Q 2 are separated in air by a distance r.
Write down an expression for the force acting on charge Q 2 .
(b) A negative pion is made up of a d quark and a u antiquark. The electric charge on the d quark is 1/3 e and that on the u antiquark is 2/3 e where e is the charge on an electron.
(i) The distance between the quark-antiquark pair is 1.0 × 10 -15 m.
Calculate the electric force between the quark-antiquark pair in the pion.
(ii) Name the force which holds this quark-antiquark pair together.
(c) The figure below shows a charge q travelling directly towards a fixed stationary charge Q. Charge q is initially travelling with a velocity v.
1 3
+
q v
+
Q
Show that, for a head on collision, the distance of closest approach, r
c
given by is
r c
=
qQ
2 πε 0
mv
2 where m is the mass of charge q and ε 0 is the permittivity of free space.
(d) An alpha particle of mass 6.7 × 10 -27 kg is moving with an initial velocity of 1.0 × 10 7 m s -1 directly towards a fixed stationary gold nucleus. The atomic number of gold is 79.
Find the distance of closest approach of the alpha particle to the gold nucleus.
(e) (i) Explain why the radii of all atomic nuclei are very tiny.
(ii) Name the force associated with beta decay in radioactive nuclei.
2 2 3 (11)
Physics: Course Questions (Advanced Higher) 20

WAVE PHENOMENA
22. (a) (i) Explain what is meant by a stationary wave.
(ii) State the differences between a travelling wave and a stationary wave.
(b) (i) Explain why the relationship
y
=
a
sin 2 π  
ft
−
x
λ   represents a travelling wave.
(ii) A travelling sound wave is represented by the expression
p
= 2 sin π ( 330
t
−
x
) where p is the pressure variation in pascals and the other symbols have their usual meanings.
Find (A) the wave frequency (B) the wavelength (C) the wave speed.
3
23. (a) A source of sound waves of frequency f is moving towards a stationary observer. Show that the frequency detected by the stationary observer is given by
f
′ =
f v v
−
v s
.
where v is the speed of sound in the medium and v s is the speed of the source of sound waves relative to the observer.
(b) A car has a siren emitting a note of frequency 550 Hz. The car is travelling at 45 m s -1 towards a stationary observer.
(i) Calculate the frequency of the sound heard by the observer as the car approaches.
(ii) What frequency will be heard by the observer after the car has passed the observer?
(c) In a gas discharge tube, moving gas atoms emit light after being put into an excited state by passing a current through the gas.
(i) State the effect on the frequency of the light observed by a stationary observer in (A) (B) the direction in which the atom producing the light is moving the direction opposite to that in which the atom producing the light is moving.
(ii) Explain briefly why cooling the discharge tube would have the effect of making the light emitted more monochromatic.
6 (10) 2 4 3 (9)
Physics: Course Questions (Advanced Higher) 21

24. (a) (i) State the conditions for two light beams to be coherent.
(ii) The conditions for coherence are usually more difficult to satisfy for light than for sound.
Explain why this is so.
(b) Interference fringes may be formed by division of amplitude when light from an extended source is incident normally on a thin wedge shaped film.
(i) State what is meant by ‘division of amplitude’.
(ii) There is no coherence between light waves originating from different points on an extended source. Explain why an extended source can be used to produce interference fringes.
(iii) Show that the distance ∆x between fringes formed by a thin film whose thickness increases by an amount D over a length L is given by ∆x = λ
L
2
nD
where λ is the wavelength of the light in a vacuum (or air) and n is the refractive index of the medium of the film.
(c) Two optically flat glass plates of length 125 mm are separated at one end by a hair as shown in the figure below. This forms a thin air wedge.
2 5
hair Light of wavelength 589 nm illuminates the plates from above.
At the line of contact of the glass plates there is a dark fringe.
A further 55 dark fringes are counted in a distance of 50.3 mm.
(i) Explain why there is a dark fringe at the point of contact of the glass plates.
(ii) Calculate the thickness of the hair.
(iii) State what would happen to the fringes if the space between the glass plates was filled with water, rather than air.
5 (12)
Physics: Course Questions (Advanced Higher) 22

25. White light is reflected normally from one part of a soap film with surfaces A and B as shown in Figure 1. The refractive index of the soap film is 1.38.
Figure 1 The light is then focussed on to the slit of a spectrometer as shown in Figure 2.
Figure 2 The first order spectrum observed has a dark band with a minimum intensity at an angle θ corresponding to a wavelength (in air) of 578 nm.
(a) (i) State what is meant by the optical path difference between waves reflected from the two surfaces A and B of the film.
(ii) Explain how the absence of light at 578 nm can be accounted for by interference effects in the thin film.
(iii) Calculate the minimum thickness of soap film consistent with a dark band at 578 nm.
(b) White light is reflected normally from a different part of the film and the position of the dark band in the first order spectrum is found to be closer to the red end of the spectrum.
Explain what information this gives about the difference in film thickness between the two positions used for reflection.
(e) After the conclusion of the experiment it was noticed that a black band formed across the film at the top of the film and increased in width until it extended over most of the film just before the film broke.
Explain this observation.
5 2 2 (9)
Physics: Course Questions (Advanced Higher) 23

26. The figure below shows a non reflective coating for a lens. This coating produces destructive interference of the reflected beams at normal incidence.
(a) Derive a relationship between the minimum thickness d of the coating, the wavelength λ of the light in air and the refractive index n of this coating.
(b) Magnesium fluoride is used as a non reflective coating for a lens. The refractive index of magnesium fluoride for a wavelength of 550 nm is 1.38. Calculate the thickness of magnesium fluoride required to produce destructive interference for light of wavelength 550 nm.
2 2
(c) Explain what effect the coating will have on the intensity of transmitted light.
2
(d) The lens is illuminated with white light and viewed in the reflected light.
Describe and explain the appearance of the lens
2 (8)
27. When unpolarised light is incident on the surface of water at the Brewster angle, the reflected light is polarised.
(a) Explain what is meant by ‘light is polarised’.
(b) Explain why sound is not polarised by reflection.
1 1
(c) Calculate Brewster’s angle for monochromatic light reflected from water.
The refractive index of water is 1.33 for this light.
(d) Suggest why polarising sunglasses might be of use to a trout fisherman.
2 1 (5)
Physics: Course Questions (Advanced Higher) 24

28.
A source of monochromatic light is placed behind two narrow slits separated by 0.25 mm. A light detector, 2.5 m in front of the slits, is moved from X to Y as shown in the figure below.
As the detector moves from X to Y, a series of maximum and minimum readings is observed.
The distance of the detector from X for each maximum reading is recorded in Table I.
Table 1 Order of maximum reading 0 th 1 st 2 nd 3 rd 4 th 5 th Distance from X/mm 0 6.9
14.1
21.0
27.9
35.0
(a) (i) Show that the fringe spacing ∆x for an experimental arrangement of this type is given by the expression ∆x = λ
D d
where the symbols have their usual meaning.
(ii) Calculate the wavelength of light used in the experiment.
(b) The experiment is repeated using a different colour of light and the results are shown in Table 2.
Table 2 Order of maximum reading 0 th 1 st 2 nd 3 rd 4 th 5 th Distance from X/mm 0 6.0
12.0
17.9
24.0
30.1
4
(i) State how the wavelength of this light compares with that used in the original experiment.
(ii) Suggest one way in which the experimental arrangement could be changed so that the separation of the maximum readings returned to those in Table I.
2 (6)
Physics: Course Questions (Advanced Higher) 25

Solutions and notes to assist marking
Marks are shown bold inside curly brackets e.g. {2} or {
1 / 2
}.
Where necessary guidance in the award of partial marks is shown.
MECHANICS
1. (a) (i) 3.1 m s -1 {2} [standard marking] (ii) 197 N (allow 192 N) {2} [standard marking] (iii) 392 N {2} [standard marking] (iv) 26.7 ° {{2} [standard marking] (b) horizontal distance 0.99 m {2} [standard marking] distance from post (using Pythagoras) 2.2 m {1} assumption - initial movement at a tangent {1} 2. (a) 0.13 m s -1 {2} (b) 26 rad s -1 {2} (c) (i) Total E k = ½ mv 2 {
1 / 2
} + ½Iω 2 {
1 / 2
} (ii) I = 0.016 kg m 2 {2} [standard marking] (d) (i) Will be greater {1}, since mass on average further from the axis. {1} (ii) Longer {1} since more rotational E k , less translational E k at the bottom so average speed smaller. {1}
Gm m
1 2 3. (a)
F
=
r
2 {1} (b)
mv
2
r
=
r
2 {
1 / 2
} + {
1 / 2
}
v
2 =
GM E r
[ {
1 / 2
} for evidence that v =
GM E
is obtained]
r
Where G is the universal gravitational constant, M E is the mass of the Earth and r is the distance between the satellite and the centre of the Earth {
1 / 2
}.
(c) (i) E k = ½ mv 2 {
1 / 2
} E
k
120 × − 11 × × 6 24 {
1 / 2
} + {
1 / 2
} + {
1 / 2
} for correct data in numerator and denominator E
k
= 3.4 × 10 9 J {1} [deduct {
1 / 2
}if unit wrong or missing] (ii) E
p
= −
GM
{{1}
r
(iii) 2.0 × 10 7 J {2} [no mark for formula, since above used] [ {
1 / 2
} + {
1 / 2
} for correct substitutions for GMm + both r’s] [ {1} for final answer, deduct {
1 / 2
} if wrong or missing unit] (iv) E k increases {1},since v increases as r decreases {1} Physics: Course Questions (Advanced Higher) 26

4. (a) Moment of inertia is a measure of the reluctance of a body to accelerate rotationally in response to a torque {
1 / 2
}, takes account of distribution of mass as well as mass. {
1 / 2
} (b) (i) 2.1 rad s -2 {2} [standard marking] (ii) 7.1 × 10 5 N m {2} [standard marking] (iii) 6.7 × 10 8 J {2} [standard marking] (iv) 5.7 × 10 3 N {2} [standard marking] (v) Nut might shear {1} 5. (a) (i) 3.75 × 10 3 m s -2 {2} [standard marking] at the extreme positions {1} (iii) 1.3 × 10 3 N {2} [standard marking] (iv) 11.9 m s -1 {2} [standard marking] (v) 24.9 J {2} [standard marking] (b) Increases the max acceleration {1}, increases the max force on piston {1} 6. (a) (i) 2.6 x 10 8 m s -1 {2} [formula given, {
1 / 2
}+ {
1 / 2
}for correct substitutions] (ii) 3.0 × 10 -10 J {2} [standard marking] (b) (i) 1.1 × 10 -34 kg m 2 s -1 ( or Js) {2} [standard marking] (ii) 2.0 × 10 -24 kg m s -1 {2} [standard marking] (iii) 3.3 × 10 -10 m {2} [standard marking] 7. (a) 1.2 × 10 6 kg m 2 {2} [standard marking] (b) 2.6 × 10 4 J {2} [standard marking] (i) 1.7 × 10 -4 rad s -2 {2}[standard marking] (ii) 204 Nm {2} [standard marking] (c) (i) At the top of the motion the force upwards from the seat has to equal the weight less the central force required for circular motion. {1} At the bottom the force upwards from the seat has to equal the weight plus the central force. {1} (ii) 468 N {2} [standard marking] 8. (a) (i) Increases {1} (ii) Mass brought in towards the axis {
1 / 2
}, moment of inertia decreases {
1 / 2
} angular momentum remains unchanged {
1 / 2
}, hence angular velocity increases {
1 / 2
} (b) (i) 0.027 kg m 2 {2} [standard marking] (ii) 7.1 rad s -1 {2} [standard marking] (iii) (A) 0.77 J {2} [standard marking] (B) 0.68 J {1} [same calculation as (A) hence 1 mark for answer only] (iv) Lateral friction and slight lateral movement produce heat [allow: ‘heat produced due to friction between dart and disc’] {1} Physics: Course Questions (Advanced Higher) 27

9.
(a) (i) 360 mm kg -1 {2} (allow ± 10) (ii) ± 20 mm kg -1 {2} (allow between ± 15 and ± 25 ) (iii) 2.8 kg {1} (b) (i) F = -ky and F = ma hence a = −
k M
y {
1 / 2
} a = - ω 2
y T
(ii) (A) (B) (C) = {
1 / 2
} ω 2 = −
k
{
1 / 2
} [notice that a is acceleration]
M
2 ω π = 2 π
M k
[ {
1 / 2
} for evidence that
T
= 2 π
M k
is obtained] plot T
2
against M with: both axes correctly labelled {
1 / 2
} points plotted correctly {1} best straight line drawn {
1 / 2
} 29.7 N m -1 {2} 10.6 m s -2 {2} [for answers to B and C allow appropriate range] [answers to B and C should be given to two significant figures, above answers are given to three significant figures to assist marking] 10. (a) (i) Minimum velocity required at the surface of a planet to remove an object to infinity. {1} (ii) To escape total energy E k + E p (at infinity) = 0 hence total energy on surface of planet E k + E p (surface) = 0 {
1 / 2
} ½mv 2 {
1 / 2
} +
GMm r
{
1 / 2
} = 0
v
2 = GM [{
1 / 2
} for evidence that v =
r
(iii) 1.9 × 10 3 m s -1 {2} [standard marking] 2
GM r
is obtained] (b) The escape velocity is greater than the speed of light so nothing including light can escape from the star. {1}
OR
The collapsed star has a very large gravitational field. Light is affected by gravitational fields and does not escape. {1} 11. (a) 9.67 s {2} [standard marking] uncertainty = max number of min readings {
1 / 2
} = = ± 0.08 s {
1 / 2
} (b) (i) 0.67 rad s -2 {2} [standard marking] (ii) % uncertainty in α is 0.8% {
1 / 2
} % uncertainty in t is 1.6% {
1 / 2
} of 0.67 = 0.01 rad s -2 {1} (c) I = 1.8 x 10 -3 kg m 2 {2} [formula given but give {1}for substitutions] percentage uncertainty in denominator = 1.4% {
1 / 2
} percentage uncertainty in numerator = 19% {
1 / 2
} overall uncertainty = 19% {
1 / 2
} = ± 0.3 x 10 -3 kg m 2 {
1 / 2
} Physics: Course Questions (Advanced Higher) 28

ELECTRICAL PHENOMENA
12. (a) (i)
E
=
V d
{
1 / 2
} (ii) 4.5 kV {1
1 / 2
} [{
1 / 2
} for substitutions {1} for answer] (b) (i) Zero (ii) (A) (c) (i) (B)
V
= correct charges on the outside surface {1} direction of lines shown correctly {1}
Q
lines terminate at 90 o to plates and to surface of sphere {1} 4 πε 0
R
{1} (ii) 1.0 nC {2} [formula given above [{
1 / 2
} + [{
1 / 2
} for substitutions/data] 13. (a) (i) Arrangement {1} [e.g. one positive, one negative] Position {1} [e.g. mid way between equal + and - charges] (ii) Arrangement {1} [e.g. two positive charges] (b) (i) Position {1} [e.g. mid way between equal positive charges]
V q
πε 0
r
min {1} (ii) E
p
=
q
2 4 πε 0
r
min {1} (iii) Change in E p = change in E k OR all initial E
k
converted to E
p
{
1 / 2
} [Correct data substitutions for q {
1 / 2
}, for ε
0
{
1 / 2
}, and for m {
1 / 2
}] 5.2 × 10 6 m s -1 {1} for answer (note: 2 × 1 2
mv
2 =
q
2 4 πε 0
r
min ) (iv) 1.4 × 10 5 V {2} [standard marking] 14. (a) (i) Component of initial velocity parallel to the field remains unchanged. {1} Component of velocity perpendicular to field results in circular motion {1} (ii) 1.0 × 10 6 m s -1 {1} (iii)
Bqv
sin θ = ( sin θ ) 2
r
{1} [OR Bqv ⊥
= mv
⊥ 2 /r ] (iv) 7.15 × 10 -8 s {3} [T = 2 π
r v
{
1 / 2
} and v =
r Bq
{
1 / 2
} hence T = 2 π
m qB
{
1 / 2
}] 15. (a) I =
Nqv
{
1 / 2
} and F BNqvl force per unit charge = {
1
BNqvl Nl
/ 2
= }
Bqv
{1} [OR other equivalent derivation] (b) (i) Electric field from top plate to bottom {1} Magnetic field at right angles {
1 / 2
}, into the paper{
1 / 2
} (ii) Bqv =
Eq
{1} (iii) v = (iv) (A) (B)
E B
{1} 4.6 × 10 3 V m -1 {2} [standard marking] 3.8 T {1}[no partial mark for formula or substitutions, but allow credit if wrong answer for (A) correctly used here] Physics: Course Questions (Advanced Higher) 29

16. (a) (i) The force per unit charge at the point (acting on a small test charge) {1} (ii) F =
Eq
{1} (b) (i) Parabola {1} (ii) horizontally t = (length of plates)/v
H
= 0.05/(1.5 × 10 7 ) {
1 / 2
} vertically E = V/d {
1 / 2
} a = Eq/m {
1 / 2
} s = ½ at 2 {
1 / 2
} 2.0 mm {1} (iii) vertical velocity =
t
{1} tan θ = v v
H
{
1 / 2
} substitutions {
1 / 2
} θ = 4.5
o {1} 17. (a) (i) Shape of graph correct {
1 / 2
} Axes labelled correctly {
1 / 2
} (ii) Increasing current produces back e.m.f. {
1 / 2
} Initial back e.m.f. equal to supply voltage {
1 / 2
} Hence initial rate of current increase is limited {
1 / 2
} Rate of current increase gets less as I rises and voltage across resistor increases. {
1 / 2
} (b) (i) 6 A {2} [standard marking] (ii) 24 V {1} (iii) 4 A s -1 {2} [standard marking] 18. (a) P 1 positive, P 2 negative {1} B field out of paper {1} (b) (i)
Bqv
=
Eq
{1} [hence answer given] (ii) Deflected (towards X) hence do not pass through S. {1} (c) E = V/d {
1 / 2
}; for substitutions {
1 / 2
} 2.5 × 10 5 m s -1 {1} (d) Circular path {
1 / 2
} Force on ion due to motion through B field is constant {
1 / 2
}and perpendicular to direction of motion {
1 / 2
}and acts as a central force. {
1 / 2
} (e) 3.6 × 10 -26 kg {2} [standard marking] 19. (a) (i) 10 V {1} (ii) Induced e.m.f. opposes the change of current, the battery must supply energy in overcoming the back e.m.f. {1} (b) (i) 5 A s -1 {2} [readings from graph at t = 0 and calculation of gradient] (ii) 2 H {2} [standard marking] (c) (i) 400 Ω {2} [standard marking] (ii) 6.25 × 10 -4 J {2} [standard marking] (d) Axes labelled {
1 / 2
} Shape correct {
1 / 2
} 20. (a) B at I
2
due to I
1
= µ force on length l carrying current I
2
= BI l force per unit length = µ 2 π
r
{
1 / 2
} 2 π
r
{
1 / 2
} {
1 /
(b) (i) 8.72 × 10 -3 N {2}, East {1} (ii) 5.3 × 10 -3 N (or N m -1 ) {2} towards Q {1}
2
} = µ
I I l
0 1 2 2 π
r
{
1 / 2
} Physics: Course Questions (Advanced Higher) 30

21. (a) F =
Q Q
1 2 4 πε 0
r
2 {1} (b) (i) correct substitution for charge (1.6 × 10 -19 ) {
1 / 2
} correct substitution for ε
0
or 1 /4 πε o {
1 / 2
} 51 N {1} (ii) Strong force {1} (c) Change in E k = change in E p {
1 / 2
} 1 2
mv
2 + 0 =
qQ
4 πε 0
r c
− 0 {1} {
1 / 2
} for evidence that final expression is obtained (d) (i) 1.1 × 10 -13 m {2}[formula given, {
1 / 2
} + {
1 / 2
} for substitutions] (e) (i) The strong force which holds the nucleons together has a very short range (about 10 -14 m) {1} (ii) Weak force {1}
WAVE PHENOMENA
22. (a) (i) A wave formed by two waves of same wavelength and frequency {
1 / 2
}, travelling in opposite directions. {
1 / 2
} (ii)
Travelling wave
Energy is transported Phase difference between adjacent oscillating points. {
1 / 2
{ }
1 / 2
} Adjacent points have same amplitude (1D wave).
Stationary wave
Energy is stored {
1 / 2
} No phase difference between adjacent oscillating points (between nodes).
Adjacent points oscillate with different amplitude. {
1 / 2
} (Use professional judgement. - {
1 / 2
} for each relevant independent difference) (b) (i) For a displacement y to be constant
ft x
λ must remain constant but as t increases x must increase {1} (and the equation represents a {1} travelling wave) [Use professional judgement with respect to detail] (ii) (A) 165 Hz {1} (B) 2 m {1} (C) 330 m s -1 {2} [standard marking] 23. (a) λ
S
× 1
f
{1}
f v
′ =
f v
−
v f S
{
1 / 2
} hence
f
′ =
v
−
fv v s
{
1 / 2
} (i) 634 Hz {2}[no partial mark for formula since given ] { {
1 / 2
} for v s and {
1 / 2
} for other substitutions] (ii) 486 Hz {2} [standard marking] (b) (i) (A) Will increase (very slightly) {1} (B) Will decrease (very slightly) {1} (ii) Cooling would reduce the speed of the atoms. {1} Physics: Course Questions (Advanced Higher) 31

24. (a) (i) They must have the same wavelength {
1 / 2
} and have a constant phase relationship. {
1 / 2
} - i.e. in general they must be from the source and the same part of the source, and have a very small path difference.
(laser light a special case) (ii) Two light sources cannot maintain a constant phase relationship (whereas two loudspeakers can be driven from the same signal generator).
{
1 / 2
} Light wavetrains are short (whereas continuous waves from a loudspeaker can be very long) {
1 / 2
} (use professional judgement for alternatives) (b) (i) A wave is split into two waves by partial reflection {1} and partial transmission {1} at a surface.
(ii) Each point on the source produces two waves by division of amplitude which are recombined having a path difference determined only by the film thickness (and refractive index). {1} (iii) for a dark fringe, m λ = 2
ny
{
1 / 2
} for the next dark fringe, (
m
+ 1 ) λ = 2 + ∆
xD L
  {1} and λ = 2 ∆xDn
L
{
1 / 2
} hence ∆
x
= 2 λ
L nD
[as given] (c) (i) Phase change of π at the lower surface. {1} (ii) 4.03 × 10 -2 mm {2}[no partial mark for formula since given] [ {
1 / 2
} + {
1 / 2
} for correct substitutions instead] (iii) Spacing would decrease. {1} 25. (a) The equivalent path difference in a vacuum (or air) for two waves, which would produce the same phase difference between the waves. {1}
OR
optical path = geometrical path difference × refractive index {
1 / 2
} the optical path difference is the difference between the two optical paths {
1 / 2
} (b) Light reflected from the two surfaces has an optical path difference equal to a whole number of wavelengths. {1} This produces destructive interference {
1 / 2
} since there is a phase change for one of the reflections. {
1 / 2
} (c) 209 nm {2} [standard marking] (d) The film is thicker {1}, since the wavelength is larger being ‘closer to the red end’ {
1 / 2
} and
d
= λ 2
n
hence d is larger. {
1 / 2
} (e) When the film is very thin and about to break {
1 / 2
}, light reflected from both surfaces has nearly zero optical path difference {
1 / 2
}, and since there is a phase change of π for light reflected from one surface {
1 / 2
}, destructive interference occurs {
1 / 2
}.
Physics: Course Questions (Advanced Higher) 32

26. (a) Optical path difference = 2nd {
1 / 2
} For destructive interference 2
nd
= λ 2 {
1 / 2
} since there is a phase change at both surfaces {
1 / 2
} Hence
d
= λ 4
n
{
1 / 2
} (b) 100 nm {2}[no partial marks for formula since given] [allow {
1 / 2
} + {
1 / 2
} for correct substitutions] (c) Intensity of transmitted light is increased. {1} Since less light is reflected and energy is conserved. {1} (OR constructive interference between transmitted rays) (d) Destructive interference occurs at 550 nm (at this one wavelength) {
1 / 2
} At larger (red) and smaller (blue) wavelengths interference is incomplete. {
1 / 2
} The combination of some reflected red and blue gives purple hue. {1} 27. (a) Electric (or magnetic) field vector oscillates in a single fixed plane. {1} (b) Sound is a longitudinal wave. {1} (c) 53 ° {2} [standard marking] (d) Reflected light is partially polarised. {
1 / 2
} Polarising sunglasses can cut this out allowing trout beneath the surface to be seen. {
1 / 2
} 28. (a) (i) For the first fringe path difference ≈ dsin θ ≈
d
∆ {
D x
1 / 2
} for θ small {
1 / 2
} Since first bright fringe, path difference = λ {
1 / 2
} Hence λ =
d D x
1 / 2
} (ii) 7.0 × 10 -7 m {2}}[no partial marks for formula since given] [allow {
1 / 2
} + {
1 / 2
} for correct substitutions] (b) (i) ∆x smaller hence λ smaller {1} (ii) increase D (or decrease d) {1} Physics: Course Questions (Advanced Higher) 33

DATA Common Physical quantities
Radius of Earth Mass of Earth Mass of Moon Mass of neutron Mass of proton Refractive index
QUANTITY
Gravitational acceleration Mean radius of Moon orbit Universal constant of gravitation Speed of light in vacuum Speed of sound in air Mass of electron Charge on electron Planck’s constant Permittivity of free space Permeability of free space of water of perspex G m n m p h ε 0 µ 0 n n c v m e e
SYMBOL
g R E M E M M 9.8 m s -2
VALUE
6.4 x 10 6 m 6.0 x 10 24 kg 7.3 x 10 22 kg 3.84 x 10 8 m 6.67 x 10 11 m 3 kg -1 s -2 3.0 x 10 8 m s -1 3.4 x 10 2 m s -1 9.11 x 10 -31 kg -1.60 x 10 -19 C 1.675 x 10 -27 kg 1.673 x 10 -27 kg 6.63 x 10 -34 J s 8.85 x 10 -12 F m -1 4 π x 10 -7 H m -1 1.33
1.5
The solutions to the questions use the data values given above.
Physics: Course Questions (Advanced Higher) 34