Suggested Answers to 2015 A Level H2 Physics 9646 PAPER 1 and 2

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Physics Paper 1 MCQ

1) A 2) D 3) C 4) A 5) C

6) A 7) D 8) C 9) A 10) C

11) B 12) A 13) D 14) A 15) A

16) A 17) C 18) A 19) C 20) B

21) C 22) D 23) B 24) B 25) A

26) B 27) B 28) B 29) D 30) B

31) B 32) D 33) B 34) C 35) C

36) C 37) C 38) A 39) D 40) B

Physics Paper 2

1. a) EPE = 0.662J

bi) The system has only GPE before the 300g mass is released. After releasing, part of the GPE is converted to KE and to EPE. At the lowest point, once the mass comes to a stop, all of the GPE will have been converted to EPE.

ii) v = 1.21 m/s

iii) 0.40m

2a) Select 2 points from the graph: (6.0, 1.25) and (12.0, 2.50)

(V – 12.0)/(I – 2.50) = (12.0 – 6.0) / (2.50 – 1.25)

V – 12.0 = (6.0/1.25) (I – 2.50)

V = 4.81I

V is proportional to I

bi) 4.8Ω

ii) 1.3 A , 1.6A

iii) 6.35 V

3a) Incident photons must have energy greater than or equal to the work function of M to emit one electron. Energy of photon is related to the frequency of the electromagnetic radiation by E = hf

bi) Free electrons at the surface are emitted with maximum KE but electrons of the outer shell require more energy to escape thus they will be emitted with less KE. A minimum potential difference is present to stop the photoelectrons with the maximum KE from reaching the collector in order to reduce the current to zero.

ii)It is limited by rate of emission of photoelectrons, which is dependent on the intensity.

c) 9.45 x 10^14 Hz

d) Starts from -2.2v and follows the same shape of the original. The value of y-intercept is doubled due to the doubled intensity.

4a) 0.712 m/s^2

bi) Since the 2 moons move in circular orbits around Jupiter’s center of mass, thus the centripetal force is also the gravitational force due to Jupiter, which acts towards Jupiter’s center of mass.

ii) 2.33

5ai) 4.80 x 10^-14 N

ii) -2.30 x 10^-16 N

b) Electrostatic force on A due to B and B due to A are internal forces. They have the same magnitude but different in direction. Thus the resultant force action on the molecule is zero.

c) 1.6628 x 10^-25 Nm

6a) 128, 54

b) 1510 s

7ai) s = -3.0

ii) -3.170 , -0.693

iii) Plot the point and draw the best fit line.

iv) 3.10

v) There is a linear relationship between ln(y1/m) and ln(l/m) with a gradient of 3.1 and with a y- intercept of -1.02.

vi) 3

b) k = 2.70 x 10^-10

ci) Acceleration is directly proportional to the displacement. The negative sign indicates that the acceleration is opposite to the displacement and is always directed to the position of zero displacement.

ii) 0.560

Apparatus.

Filament lamp, power supply, beaker, thermometer, Stirrer, Stopwatch, Voltmeter, Ammeter, Rheostat, Switch, Weighing machine, Waterproof tape.

Draw a diagram with a filament lamp immersed in water, with the thermometer and stirrer inside the beaker. Draw the circuit of the lamp with connecting ammeter and voltmeter and the power source.

Procedure.

Measure mass of water in beaker, m.
Insulate the beaker.
Connect filament lamp with circuit, wrap the metal cap with the waterproof tape.
Set up apparatus as shown.
Record initial temperature.
Close the circuit and start the stopwatch.
When the temperature hits 70ºC, stop the stopwatch and record the readings of the voltmeter and ammeter.
Repeat steps 5 to 7 after replacing the hot water in the beaker with the same amount of water at room temperature. Obtain different potential differences by changing the resistance of the rheostat.

Working

1 – (Thermal energy/electrical energy output)
Relationship between I and t is given by η = aV^b , where a and b are constants. The equation can be written as lg η = b lg V + lg a
Plot a graph of lg η against lg V and obtain a and b

Accuracy

Minimise heat loss to surroundings by using the lid and lagging.
Conduct preliminary experiments to decide on the suitable values of resistance and voltage supply.
Ensure experiment is not conducted over long periods of time so that efficiency of the lamp is not affected by the thermal energy released.

Safety