Advanced Level Physics Paper II Solutions

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Advanced Level Physics Paper II Solutions

The IncredibleCollege

Advanced Level Physics Paper II Solutions

Section I

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1.E

2.E

3.B

4.D

5.C

6.E

7.D

8.C

9.C

10.D

11.A

12.B

13.C

14.B

15.C

16.D

17.E

18.E

19.D

20.D

21.C

22.C

23.C

24.A

25.B

26.C

27.B

28.A

29.D

30.B

31.C

32.C

33.A

34.C

35.C

36.C

37.C

38.B

39.C

40.B

41.D

42.C

43.B

44.E

45.A

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Section II

1.(a)1st law– Everybody continues to be in a state of rest or moves with uniform velocity unless a resultant force acts on it. 1

2nd law – The rate of change of momentum of a body is directly proportional to the resultant force acting on the body and occurs in the direction of this force. 1

3rd law – To every action there is an equal and opposite reaction. Action and reaction are acting on two different bodies. 1

(b) The principle of the conservation of linear momentum states that the momentum of a system is conserved if no external resultant force acts on the system. 1

Suppose an object A of mass m1and velocity u1, collides with another object B of mass m2 and velocity u2, moving in the same direction as shown below.

1.(b)(Continued)

Assume that the final velocities of A and B are respectively v1 and v2.

By Newton’s 2nd law of motion,

force acting on A by B, and½

force acting on B by A, during collision.½

Moreover, by Newton’s third law,

FAB = - FBA½

So m1u1 + m2u2 = m1v1 + m2v2½

i.e. the linear momentum of the system is conserved.

(c)The student is incorrect. ½

There is not a real force to push the object outward.½

A force is required to keep an object moving in a circle of radius r.½

This is provided by the friction between the object and the turntable.½

If the centripetal force required by the object is greater than the maximum friction, the object would maintain its velocity by its inertia ½

and would bemoving in a straight line. So it seems to curve outwards.½

(d)(i)Consider the translational motion of the cylinder,

N – Mg cos  = 0for motion normal to the incline,½

Mg sin  f = Mafor motion along the incline.½

Consider the rotational motion of the cylinder,

(Note : The rotational motion about the centre of mass follows from the Newton’s second law for rotation.)

 = I ½

Neither N nor Mg can cause rotation about the centre of mass,

fr = I ½

where I = Mr2

(ii) From (i),

f r = I 

f r = I½

f = I

1.(d)(ii)(Continued)

Substitute into translation equation of motion, we have

Mg sin  I = Ma½

a =½

Now, I = Mr2

∴a =g sin ½

(iii)If a similar hollow cylinder of the same mass and radius is used, its moment of inertia, I = Mr2, about its centre is larger than the solid one. 1

From (d)(ii), the acceleration of the hollow cylinder down the incline would be smallerthan the previous one 1

since now a =.1

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2.(a)When a motor is switched on, a current passes through the armature and the field coil. A magnetic field is then produced ½

so that the coil carrying current will experience a torque and then rotate. ½

At the same time, an induced e.m.f. is produced ½

in a direction so as to oppose the motion. ½

The dynamic balance between these torque ½

controls the speed of the motor.½

(b)(i)Correct position of coil1

Torque and its direction correctly shown1


2.(b)(Continued)

(ii)

1

Force on one side PQ, F = (BIl)N½

Torque = (BIl)Nb = BI(lb)N = BIAN½

(c)The induced e.m.f. in the coil is the back e.m.f. in the coil and is given by

 = BAN1

(d)(i)When the loading on the motor is increased, the torque required correspondingly increases. ½

From the result in part (b)(ii), = BIAN. Because I, ½

the current flowing in the coil increases ½

and will subsequently increase the p.d. across the armature. In addition, ½

applied voltage = p.d. across the armature + back e.m.f.

This change in the p.d. across the armature will decrease the back e.m.f.½

As a result, the speed of the motor slows down.½

(ii)There will be no difference in both the current and the speed of rotation 1

and the direction of rotation will also be unchanged.1

(iii)The current flowing through the field coil will be decreased ½

and the magnetic field produced will subsequently be reduced. ½

The decrease in the magnetic field will then lead to

a decrease in the back e.m.f. ½

Therefore, more current will flow through the armature. Both these change

in the magnetic field and the current through the armature will affect the

torque directly. However, with the latter being the dominant effect,

the torque will be increased ½

and it enables the armature to rotate faster ½

until the back e.m.f. is again nearly equal to the applied voltage.½

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3.(a)(i)(1)A plane polarized light is one where the electric field intensity vibrates in a fixed plane. ½

An unpolarized light is one where there is no specific plane about which the electric field vibrates. ½

Laser is an example of polarized light ½

while sunlight is an example of unpolarized light.½

(2)Look into the beam of light through a piece of polaroid. If the intensity of light changes as the polaroid is rotated, the light is plane polarized. 1

If not, the light is unpolarized.1

(ii)(1)When a partially polarized light is resolved into x and y-components.

The x and y-components of the electric field have different amplitudes of vibration. 1

The amplitude of one is stronger than the other.1

(2)According to the diagram below, the reflected beam will be completely plane polarized when it is at right angle to the refracted beam. 1

1

From Snell's law,

sin i=n sinrwhere n is the refractive index of glass/water.

sin i=n sin(900i) =n cosi

Thereforetan i = n1

(b)(i)(1)By comparing the line spectrum of light from a distant star with light from a stationary laboratory source, the bright lines are displaced towards the red end of the spectrum from the stationary source.

This phenomenon is call red shift.1

3.(b)(i)(Continued)

(2)Red shift means that wavelengths of light from distant stars are longer than that due to stationary sources. 1

By Doppler effect, it means that the stars must be moving away from the earth, in other words, the universe is expanding. 1

(ii)

1

The wavelength of light sent from A,

1=½

The wavelength of light sent from B,

2=½

Change in wavelength, =12

=

=

=1

By calculating  and using grating experiment (d sin= n), c = speed of light in vacuum and  = wavelength of light observed, rotational speed of the sun can be found. 1

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4.(a)(i)Assume that the attraction between gas molecules are negligible and the collision between gas molecules and the container wall are completely elastic. 1

Impulse exerted by the gas molecule on the container wall 1

Assume that the time of collision is negligible compared with the time between successive collisions. ½

Time between successive collisions = ½

4.(a)(i)(Continued)

Average force exerted by the gas molecule on the container wall is

1

Total pressure exerted by the gas on the container wall is

where i = 1, 2, 3……1

Assume that the total volume of gas molecules is negligible compared with the volume occupied by the gas, i.e. the volume of the container. ½

Therefore, ½

(ii)By the ideal gas equation, ,½

the average k.e. of each gas molecule = ½

where k is the Boltzmann’s constant and

(b)The behaviour of a real gas differs from that of an ideal gas because intermolecular attraction force exists between adjacent gas molecules. 1

The intermolecular attraction force depends on both the number of gas molecules, N, hitting on the container wall and the number of gas molecules, N, hindering those gas molecules hitting on the container wall. 2

However, both N and N.1

Thus, the actual pressure should be modified as .

In addition, a gas molecule has a finite size of volume so that the actual free space occupied by the gas should be reduced from V to . 1

4.(Continued)

(c)The first law of thermodynamics states that the increase in the internal energy of an object is given by the sum of work done on the object and the heat absorbed by the object. 1

Process A to B is an isothermal compression, i.e. both the temperature and the internal energy of the gas are unchanged. 1

As work is done on the gas in compression, heat must be transferred from the gas to the surroundings to keep the temperature of the gas constant. 1

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  1. (a)

Correct diagram1½

Correct labels1

½

The current drops indicate that an atom can only absorb a certain quantum of energy. This implies that the energy levels of an atom are quantized. 1

5.(Continued)

(b)(i)Hydrogen Spectrum

The hydrogen spectrum is an emission line spectrum in the visible range characterised by the energy levels of the hydrogen atom. ½

When a hydrogen atom is excited from the ground state to a higher energy level, it becomes unstable and eventually falls back to one of its lower energy levels. 1

The excess energy E is emitted as electromagnetic radiation of wavelength where E = energy difference between energy levels. ½

X-ray Spectrum

The X-ray spectrum is composed of 2 parts – the line spectrum and the continuous spectrum with a minimum wavelength. ½

In an X-ray tube, energetic electrons bombard the metal target and may eject an electron from the innermost shell (K-shell) of an atom. ½

The excited atom is unstable and an electron from, say, the L-shell may move into the vacancy. ½

The excess energy E is emitted as electromagnetic radiation of very short wavelength as E is very large for transitions of electrons between inner shells. 1

On the other hand, some bombarding electrons lose energy as they decelerates when hitting the metal target and emit the energy in the form of radiation. 1

Since most of the electrons usually have more than one encounter with the metal atoms before losing all their energy, several X-ray photons with different energy are emitted, which contribute to the continuous spectrum. 1

The minimum wavelength of the continuous spectrum corresponds to the bombarding electrons losing energy in a single encounter, which is bounded by the maximum kinetic energy of the electron. ½

(ii)The Sun’s continuous spectrum is crossed by many hundred of dark lines known as Fraunhofer lines indicating that there are certain elements, in vapour form, in the Sun’s chromosphere. ½

The vapours are cooler than the central hot portion, i.e. the photosphere of the Sun. ½

They absorb their own characteristic wavelengths from the Sun’s continuous spectrum and re-radiate them in all directions. ½

Therefore, those electromagnetic radiations of characteristic wavelengths appear as dark lines in a bright background. ½

5.(Continued)

(c)(i)The word LASER means Light Amplification by Stimulated Emission of Radiation. 1

(ii)- Monochromatic

- Coherent

- Intensive & collimated

Any two of the above1

(iii)- Medical use, e.g. fusing a detached retina back into place

- Holography

- Industrial use, e.g. cutting or drilling holes in solid materials

- Laser disc player or compact disc player

- Telecommunications

Any two of the above1

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