Suggested Answers to Paper 3 of Preliminary H2 Physics Exam 2010

Section A

1a) X is the resultant of the weights of all parts of the carZ is the resultant of the normal force on the car and frictional force on the tyres.

1b) 1. Zx 2000 = 1200 (8) Zx = 11600 NZy = 1200 (9.81) = 11772 NThus, Z = (116002 + 117722) = 16.5 kN

2. Angle = tan1 (11772/11600) = 45.4

2ai)

2aii)

2bi)

2bii)

Consider a body of mass m lifted by a distance h above its initial position at constant velocity by an upward force.

F = mg W = (F) h Ep = mgh

Gravitational field strength is not constant if object is moved over long distances.

The change in volume/potential energy when ethanol vaporizes is much greater than the change in volume/potential energy during melting. OR There is additional work done against the atmospheric pressure during vaporization.

Q = thermal energy to raise temp to boiling point + thermal energy required to vaporize the ethanol

Q = mc∆ + m Lv = ρ V (c ∆ + Lv) = (0.79 g cm-3) (1.0 cm3) [(2.4 Jg-1 K-1) (78-20 K) + 840 Jg-1] = 774 J

3ai) At the top, N + N = mg

To remain in contact with track, N > 0

> mg

> 9.81

Vmin = 13.3 m s-1

3aii) By COME, Ek(bottom) + Ep(bottom) = Ek(top) + Ep(top) + Elost

Vbottom = 29.8 m s1

3aiii) At bottom,

N = 14.8 × 103 N

3b) There must be a non-zero net force pointing towards the centre of loop to change the direction of car.

4a) Description and explanation of any one of the following observations:- no photoelectrons emitted if frequency is below a certain minimum- maximum kinetic energy of photoelectrons is independent of the intensity of the

incident light- graph of KEmax / stopping potential versus frequency of incident light is a

straight line- photoelectric emission is almost instantaneous, even if intensity of light is very

low

4b) Long lifetime t is large E is small energy of level is well-defined

4c) High energy barrier width shorter transmission coefficient greater shorter average waiting time before tunnelling shorter half-life

5a) Semiconductor: At higher temperature, more charge carriers are released as many electrons have enough energy to overcome band gap. Resistance wire: its free electrons encounters more frequent collisions with lattice reducing its ease to flow through.

5b) Given sufficient energy, the electrons in the valence band will be able to transit up to conduction band.

The electrons in the CB and holes in the VB both contribute to the flow of current when a voltage is applied.

5c) When the P type and N type semiconductors are brought together, the electrons will diffuse into the P type and the holes diffuse into the N type.Positive and negative core ions set up internal electric field that would prevent further diffusion.Hence depletion region is formed.

5d)

Section B

6a) The acceleration is directly proportional to its displacement and it is directed towards a fixed point.

6bi) X = 0.025 cos 2 t6bii) Vmax = xo

= 2(0.025) = 0.157 m s-1

6biii) Vmax = xo = 2(0.020) = 0.126 m s-1

6biv) time taken for wave to go from P to Q = d/v = 5.0/2.5 = 2.0s From the graph, points that are 2.0 s apart are in phase. Hence phase difference is zero.

6bv) 1 mark for reduction in amplitude as d increases 1 mark for correct number of cycles (i.e 2) for d = 5 cm .

6ci)wavelength Angle (n=1) Angle (n=2)

400 nm 13.89 28.69750 nm 26.74 64.16 12.85 35.47

is greater than

6cii) At n=3, the 400 nm light is at an angle of 46.1 degrees. It is before n =2 of 750 nm at an angle of 64.2 degrees. Hence overlap occurs since n=3 occur at a smaller angle than n=2.

6ciii) No overlap Brighter

p n

6civ) White light

7ai) Electric field strength is the electric force per unit charge on a positive test charge placed at that point. The SI unit is N C1 or V m1

Electric field strength = V / d

7bi)

7bii) |E| = (4.2 – 2.0)/(30 x 103) = 73 N C1

7biii) loss in KE = gain in PE½ mev1

2 = e(2.2)v1 = 8.8 x 105 m s1

7biv) 1. Even though the distance has been halved, as long as the p.d. remains the same, the velocity v1 will remain the same .

2. As plate A is now at a higher potential than plate B, the electron emitted from plate B will always be attracted to plate A. Thus the minimum velocity required will be zero.

7bv) ½ me (1.1 x 106 cos θ)2 = 1.6.x 1019(2.2)

cos θ = 0.7992θ = 36.90.

7c) (i) Negative charge

(ii) Electric force upwards = gravitational force downwardsqE= mgq/m = g/E

= 9.81/9.67 = 1.01448 = 1.01 C kg1

(iii) If the mass is 72 x 1019 kg, the amount of charge on the particle =72 x 1019 x 1.001448=7.3 x 1018 C

30 mm

θβ2

Plate A 4.2 V

Plate B 2.0 V

β1

v1

v2

This is 7.3 x 10-18 / 1.6 x 10-19 = 45.6 times the charge of an electron This is not valid as charge is quantized in sizes of 1.6 x 10-19 C.

8a) It is a nuclear reaction that involves the splitting of a heavy nucleus into smaller parts.Neutrons and gamma rays are usually produced OR energy is usually released in the process.

8bi) There are 36 protons.There are (92 – 36 =) 56 neutrons.

8bii) mass defect = (235.044 + 1.009) – (91.910 + 140.916 + 3×1.009) = 0.2 u

Energy released = 0.2 (1.66 × 10-27) (3 × 108)2

= 2.99 × 10-11 J

8biii) rate of mass of U used that contribute to useful output = 0.23 × (3.5 × 10-3) = 8.05 × 10-4 kg s-1

useful power output =

= 6.17 × 1010 W

8biv) The energy is usually released as radiation and kinetic energy of fragments.

8ci) It is the time taken for the number of undecayed nuclei to decay by half.

8cii) Since the mass no of the particle is 234 – 230 = 4, it is most likely alpha particle / helium nucleus.

8ciii) decay constant =

= 2.89 × 10-6 year-1

8civ) N = 5.5 × 1026 e-(2.89 × 10-6) (87000)

= 4.28 × 1026 A = (2.89 × 10-6) (4.28 × 1026) = 1.24 × 1021 year-1

8cv) - Used as radioactive tracers in either medicine, agriculture or pipe leakage- Used to monitor thickness of papers or sheets produced in factory- Used of ionizing radiation in radiotherapy to treat cancer/tumor- Used in smoke detectors to trigger alarm- Used in carbon-dating to measure age of material