Plk Vicwood K.T. Chong Sixth Form College

05 AL Physics/Essay Marking Scheme/P.1

PLK VICWOOD K.T. CHONG SIXTH FORM COLLEGE 05’ AL Physics: Essay Marking Scheme

1. (a) The acceleration (or net force acting) is towards a fixed point (equilibrium position) and its ½ magnitude is directly proportional to the displacement from the fixed point. ½ or Net (restoring) force = -kx, where k is the force constant, x is the displacement from equilibrium position.

Simple harmonic motion is isochronous, i.e. the period is independent of its amplitude. 1 2

(b) (i)

Q  l

When the bob is at P where arc OP = x and OQP = , theT forces on the bob are tension T and weight mg (or using diagram). ½ P O m Resolving mg radially and tangentially we have themg unbalanced sin  restoring force mg sin  towards O. The equation of motion is given by  mg cos  1 mg -mg sin  = ma, a = acceleration ½

Since the bob’s displacement x = l and sin  ~  when  is small, the equation becomes 1 x -mg =  mg = ma ½ l g g  a =  x = -2x (where 2 = ) 1 l l The motion is thus simple harmonic (a  -x) if the oscillations are of small amplitude (i.e.   10). 2 l The period T = = 2 ½ 5½  g

or  = I˙˙ mgl sin  = ml 2˙˙ mgl  = ml 2˙˙ (sin   ) g ˙˙   = 0 l g 2 l  2 = and T = = 2 l  g 05 AL Physics/Essay Marking Scheme/P.2

(b) (ii) Set up a simple pendulum using a long string (say, at least 0.8 m) clamped firmly at one end using a stand. ½

Use a stop watch to measure the period T (by timing, say, 20T) ½ for different values of length l of the string ½ and plot a graph of l against T2. ½ A straight line which passes through the origin should be obtained, and its slope equals 1 g . ½ 42

g l = T 2 42  g = 42  slope of the graph ½

Precautions (1) at least 10 oscillations to be timed (for l = 1.0 m) (2) an angle of swing less than 10. (3) oscillation should be in a vertical plane (4) the length l to be measured to the centre of the bob ANY 1½ (5) the oscillations to be counted as the bob passes a fixed point, say, THREE the equilibrium position O. (i.e. to make sure that each count @½ corresponds to a complete oscillation) (6) check the support to make sure that l is the same when the bob swings to both sides (7) draught-free enclosure 5½

(c) (i) Since g is much smaller on the moon, the period T would be much longer for the same 1 pendulum length l. ½ 1½

(ii) No oscillation can be set up (or circular motion) since the pendulum experiences ½ ‘weightlessness’ in a spacecraft orbiting round the earth (Qr Both the spacecraft and the 1 pendulum are in a state of ‘free fall’). 1½

2. (a) (i) Light can be plane-polarized as it comprises transverse electric and magnetic fields at right angles to each other and to the direction of propagation. 1 Sound wave is longitudinal, i.e. the oscillation of particles is along the direction of propagation, thus cannot be not plane-polarized. 1 2

(ii) (I) Use a piece of Polaroid to view a lamp. When the Polaroid is rotated about an axis 1 perpendicular to itself, the brightness does not vary, implying that the light from the 1 lamp is not plane-polarized. 2

(II) A narrow beam of unpolarized light from a slide projector is sent through a tank of ½ water in which scattering particles are obtained by adding one drop of milk. ½

Polarization of the scattered light is detected by rotating a piece of Polaroid in either 1 side positions A or B. The scattered light along PA must be polarized in a direction parallel to PB while 1 05 AL Physics/Essay Marking Scheme/P.3

that along PB must be polarized in a direction parallel to PA.

(iii) The ‘glare’ reflected from water surface is (completely or partially) plane-polarized. 1

Polaroid discs, suitably orientated, are used in sunglasses as ‘filters’ so as to reduce the ‘glare’ effectively. 1 Since a polaroid allows light of certain polarization to pass through o r selectively absorb in a certain direction. 1 3

(b) In the testing of optical ‘flats’, the plate under test is made to form an air wedge with a standard plane glass surface. ½

Any uneven parts of the surface, corresponding to slight difference in path difference, 1 which require more grinding will appear as irregularities in what should be parallel, equally spaced, straight set of fringes. ½ or In the blooming of lens, a lens is being coated with a film of transparent material to a ½ thickness of one quarter of a wavelength of light in the film.

½ 05 AL Physics/Essay Marking Scheme/P.4

The light reflected from the top (ray 1) and bottom (ray 2) surfaces of the coating/film 1 then interfere destructively as ray 2 has to travel twice the thickness of the film (path difference = 2  ¼ = ½), the amount of light reflected can be significantly reduced. ½

(The refractive index of the film is less than that of glass and so both rays 1 and 2 suffer a 180 phase change at a rare-dense boundary. The net effect of the phase 5 changes on the path difference is thus zero.)

or Measurement of wavelengths ½

Explanation should include: the origin of path difference 1 the rays that interfere ½ the kind of interference or pattern results ½

(Accept any other reasonable answers.)

3. (a) (i) The electromotive force of the battery is the energy transferred to unit charge (or work done per unit charge) from chemical energy of the battery when the charge passes through the battery. 1

The potential difference between points AB is the amount of electrical energy changed to other forms of energy when unit positive charge passes from A to B in the external 1 circuit.

Since the battery has internal resistance, some of the electrical energy given to unit charge is dissipated within the battery and only part of it is delivered to the load in the external circuit across AB. 1 3

R mA 1

Since the internal resistance of the voltmeter RV is comparable to R, the milliammeter should be connected in series with R. In fact, the voltmeter reading V is the sum of

p.d.’s (VR + VmA) across both R and the milliammeter, which is greater than the actual ½ p.d. across R.

The calculated value of R is greater than its actual value since 1 V V V V = R mA = R  mA where I is the reading of the milliammeter. ½ 3 I I I

(b) Maim assumptions: The range of intermolecular forces (both attractive and repulsive) is small compared with the ½ average distance between molecules. Also, there are a large number of molecules even in a small volume. ½

Suppose there are N molecules each of mass m in a cubical box of side l. ½ Consider one molecule moving with velocity c, which can be resolved into components u, v ½

and w in the directions Ox, Oy and Oz respectively. 05 AL Physics/Essay Marking Scheme/P.5

Consider motion along Ox, the molecule with momentum mu strikes the shaded wall X and ½ rebounds with same speed in an opposite direction. ½ Thus its change of momentum (due to the impulse provided by the wall) is

mu – (–mu) = 2mu ½

(b) If the molecule travels to the wall opposite X and rebounds back to X again without striking any other molecule on the way, the time interval between successive collisions of the ½ molecule and wall X is 2l/u, it follows ½ 2mu mu2 rate of change of momentum at X= = ½ 2l / u l mu2 By Newton’s second law, force on molecule (by wall) = ½ l mu2 By Newton's third law, force on wall by molecule = ½ l The total force on wall X due to impacts by all N molecules is given by m force on X = (u2  u2  ...  u2 ) ½ l 1 2 N

where u1, u2, ... are the different Ox components of the velocities of molecules 1,2, … Hence, pressure p on wall X of area l2 is given by m 2 2 2 p = (u1  u2  ...  uN ) ½ l3 2 If u represents the mean value of the squares of all the velocity components in the Ox direction, then u2  u2  ...  u2 u2 = 1 2 N ½ N m 2  p = (Nu ) ______(*) l3

For any molecule, c2 = u2 + v2 + w2 ½

c2 = u2  v2  w2

However, since N is large and the molecules move randomly, all the mean values of u2, v2 and ½ w2 should be equal. Thus

c2 = 3u2 ½

2 Nmc 3 (*) becomes p =  c2 (N, m, l are constants) 1 10 3l3 4. (a) (i) The induced e,m.f. in a coil is directly proportional to the rate of change of flux-linkage 1 d through it, i.e.   (N) . 1 dt (ii) When a piece of metal moves in a magnetic field (or is exposed to a changing magnetic 1 05 AL Physics/Essay Marking Scheme/P.6

field), e.m.f. is induced in the metal, thus inducing currents. ½

Electromagnetic damping of moving-coil meters enables the coil to take up its deflected ½ position quickly without overshooting and oscillating about its final reading. ½ The coil is wound on a metal frame in which large eddy currents are induced and that ½ the k,e. of the coil is transferred to electrical energy of the eddy currents. This causes opposition to the coil’s motion as it cuts across the radial magnetic field of ½ the permanent magnet. ½ 4

(b) (i)

The primary and secondary coils are wound on the two arms of a complete, laminated soft iron core such that the turn ratio NP:NS is 55:3. ½ 2

(ii) (I) The d.c. current in the primary coil will set up a constant magnetic field in the soft 1 iron core. Since there is no change in magnetic flux through the secondary, no e.m.f. would 1 be induced in it. 2

(II) When a sinusoidal input voltage is applied to an ideal transformer, the flux  in the iron core at time t due to the current in the primary coil is also sinusoidal. 1

The e.m.f. S induced in the secondary coil (NS turns) by the same flux  in the ½ d core is also sinusoidal with the same frequency as  and V since derived ½ P dt from  has the same frequency as . d d  = (N ) = N  2 S dt S S dt

(c) An a.c. is fed to a flat, circular coil as shown. The magnetic field produced is detected by an ½ axial search coil held vertically at the centre of the coil. The location/plane of the search coil should be kept fixed relative to the coil. A CRO ½ indicates the induced e.m.f. picked up by the search coil, which is proportional to the ½ magnetic field due to the flat coil. (Accept also using Hall probe and d.c.) ½ 05 AL Physics/Essay Marking Scheme/P.7

The current fed to it is increased in steps, say, 1 A and the corresponding induced e.m.f.’s ½ measured by the CRO are recorded. ½

A graph of induced e.m. f. against the current fed to the coil is plotted and a straight line graph ½ should be obtained, implying the directly proportional relationship. ½ 5

5. (a) (i)

A photocell consisting of a curved cathode and an anode is connected to a battery via a sensitive ammeter. The variable p.d. across the photocell is negative and the photoelectric current is indicated by the ammeter.

For observation (2), the minimum negative p.d. or stopping potential -Vs indicates the ½ maximum kinetic energy of the ejected electrons since

KEmax = eVs ½

The intensity of the same monochromatic light is varied, the corresponding stopping ½ potentials are obtained by adjusting the p.d. of the battery until the ammeter reading just ½ becomes zero.

Experimental result shows that the stopping potentials remain the same (i.e. independent ½ of intensity). 4

(ii) (I) Observation (2) : The emission speeds / kinetic energies should range from zero to a maximum due to electrons having a range of possible kinetic energies inside ½ the metal. Those electrons with the highest kinetic energy are emitted with the maximum speeds.

According to wave theory, we would expect a certain number of photoelectrons to ½ be ejected with greater speeds when the radiation intensity increases but ½ 05 AL Physics/Essay Marking Scheme/P.8

observation (2) gives a different result.

Obervation (3) : According to wave theory, radiation energy is spread over the wavefront and the amount incident on anyone electron would be extremely ½ small.

So for any frequency of radiation, we would expect electrons to be ejected if ½ sufficient time is allowed for them to gather enough energy to escape but ½ observation (3) gives a different result. 3

(II) Einstein assumed that electromagnetic radiations are emitted and absorbed in whole numbers of quanta, called photons. ½ A photon of energy hi causes an electron to be emitted if hf > , where  is the work function or energy supplied to a metal for liberating an electron from metal ½ surface. Thus the maximum kinetic energy (with a maximum speed vmax) of the emitted ½ electron is given by 1 hf -  = mv2 2 max Also, since  is constant for a particular metal, there is a minimum frequency,

the threshold frequency f0, below which no photoemission is possible. ½ (Observation (3)) 1 We then have, hf =  and h(f - f ) = mv2 ½ + ½ 0 0 2 max It follows that the maximum kinetic energy of the ejected electrons depends solely on the frequency I of the incident radiation, increase in the intensity only ½ increases the number of photons and thus electrons ejected. (Observation (2)) ½ 4

(b) Laser operates on stimulated emission of radiation in which, when an inverted population ½ arises, a photon of exactly the correct energy approaches an excited atom, an electron in a ½ higher energy level may be induced to fall to a lower level and emit another photon. ½

A pumping radiation of frequency (E3  E1) / h raises electrons from level 1 to level 3, from which they fall by spontaneous emission to level 2. An inverted population can arise between levels 2 and 1 if electrons remain long enough in level 2.

The spontaneous emission of photon due to an electronic fall from level 2 to level 1 may subsequently trigger the stimulated emission of more photons from other atoms, having the same phase, frequency and direction of travel as the stimulating photon which is itself unaffected. Same frequency – monochromatic 1 Same phase – coherent 1 05 AL Physics/Essay Marking Scheme/P.9

Same direction (and same phase) – intense 1 5