2012 O LEVEL PHYSICS PAPER 2 ANSWERS. The area of the shaded region is the distance travelled by the car from t = 10 s to t = 16 s.

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1 2012 O LEVEL PHYSICS PAPER 2 ANSWERS SECTION A 1 (a) Acceleration, a = (v-u)/t = (25-20)/4 = 1.25ms -2 The area of the shaded region is the distance travelled by the car from t = 10 s to t = 16 s. (c) (i) Average speed of the car is the total distance travelled by the car divided by the total time taken by the car to travel that distance. The area under the graph for each second, between t = 6 s to t = 16 s, is larger than that between t = 0 s to t = 6 s and t = 16 s to t = 20 s. 2 (a) Work done by the force of swimmer s legs against the wall is converted to kinetic energy of the swimmer. K.E = (0.80) 2 = 19.2 J (c) (i) As the swimmer moves forward, she uses her limbs to push the water backwards. The product of the force of the swimmer on water and the displacement of the water is the work done by the swimmer on the water. The work done by the swimmer on the water is converted to kinetic energy of the swimmer and to do work against the drag forces which is eventually lost as heat and sound energies to the surroundings. 3 (a) Electromagnetic waves do not require a medium for propagation. (i) Visible light and infra-red radiation The wavelength of the radiation increases while the frequency is reduced. The energy released by the lamp per unit time is reduced. (c) (i) Heat energy is transferred from the filament to the metal support and glass support. The nitrogen gas around the filament is heated up by the filament and rises, setting up a convection current within the bulb. As the lamp heats up, convection currents are set up in the 1 P a g e T u i t i o n P h y s i c s. c o m

2 surrounding air, transferring heat away from the lamp into the atmosphere. 4 (a) Density of cold air = (3/2) = g/cm 3 (i) The molecules collide against the walls of the piston. The force exerted on the walls per unit surface area is the pressure of the air in the cylinder. As the volume decreases, the rate of collisions of the molecules against the walls increases. However, the temperature also reduces; hence the kinetic energy of the molecules is decreases. As a result, the force exerted on the walls remains the same. As area remains the same, so does the pressure. (c) As the temperature falls, the atmospheric pressure is larger than the pressure of the air in the piston. Hence the piston moves inwards due to the unequal pressures. 5 (a) 95 s (c) This is because the temperature of the liquid is larger than the temperature of the environment. Temperature remains constant during phase change. The thermal energy lost does not result in a loss of kinetic energy but a loss of potential energy (intermolecular bonds are formed/ strengthened). (d) 2 P a g e T u i t i o n P h y s i c s. c o m

3 6 (a) The image is formed on the same side of the lens as the object. (c) 1. Image size will be smaller. 2. Image distance from the lens will be smaller. 7 (a) Reffective = 1/((1/60)+(1/40)) = 24 Ω I = V/R = 6.0/24 = 0.25 A (c) (i) The e.m.f of 6.0V is shared equally between the 30 Ω resistors along the branch containing points A and B (potential divider method). (iii) VCD = 6.0 (30/40) = 4.5 V VAC = = 1.5 V 8 (a) (i) When switch is closed, current flows in the coil. By Fleming s left hand rule, the arms of the coil which are in a magnetic field will experience a force, causing the coil to turn clockwise when viewed from the side of the commutator. This is due to the presence of the split ring commutator which reverses the direction of the current in each arm of the coil after it has turned 180 ; hence the force on each arm is also reversed. (i) H should be drawn exactly on one of the peaks. 1. The peaks of the graph are higher. 2. Each cycle has a shorter period. 3 P a g e T u i t i o n P h y s i c s. c o m

4 SECTION B 9 (a) (i) In both deep and shallow waters, the speed of the waves is the same when their wavelengths are 10 m. Also, their speed increases as wavelength increases. The speed of the waves in deep waters increases at a faster rate as the wavelength increases and it increases continually as wavelength increases from 10m to 600m. The speed of the waves in shallow water increases more slowly and stops increasing when the wavelength increases till 400m. (i) Considering the waves with highest speeds, Time taken = distance/speed = 2,000,000/30.6 = s (iii) f = v/λ = 30.6/600 = Hz Total time taken by slower wave = (4 3600)+( ) = s Speed of wave = 2,000,000/79800 = 25.1 ms -1 From the table, the wavelength is 40 0m. 10. (a) (i) (iv) 1. The ship will oscillate up and down with a frequency of Hz. 2. The ship will be unstable due to the larger frequency of the oscillations. 12 V R V A 1. The student will shift the probe (as shown in the diagram) upwards, increasing the potential difference across the lamp. 2. As current increases, the lamp heats up, resulting in an increase of resistance of the lamp. As a result, ohm s law (V = IR) is not obeyed. 4 P a g e T u i t i o n P h y s i c s. c o m

5 (i) Q = It = = 48 C V = P / I = E / (t I) = 390 / ( ) = 8.13 V Either 11 (a) This is because wire A is directly connected to the switch which is a safety measure. When the switch is off, the filament bulb and other parts of the lamp near the bulb will not be at a high potential. (i) In the event of a short circuit where a large current flows through the circuit, the fuse will melt and an open circuit will be achieved, stopping the large current from flowing. This action of the fuse helps to protect the device from overheating as well as protect the persons operating the device from getting electric shocks. This is due to the presence of the Earth wire, wire C. When wire A touches the case, it creates a short circuit, resulting in the fuse of the plug being blown and an open circuit is attained. (c) (i) R = V 2 / P = (230) 2 / 100 = 529 Ω R1 / R2 = P2 / P1 = 60 / 100 = 0.60 (d) This is because the double insulation is also a safety measure which increases the circuit protection and reduces the risk of electric shocks significantly. It is in place of the Earth wire requirement. Or 11 (a) (i) 1. Bring the charged rod near to the metal plate. 2. Connect the lead to the metal plate. 3. Remove the lead from the metal plate with the charged rod still nearby. 4. Finally, take the charged rod away from the metal plate. 1. When the charged rod is brought near to the metal plate, electrons will flow towards the rod. Hence the left side of 5 P a g e T u i t i o n P h y s i c s. c o m

6 the metal plate is negatively charged while the right side is positively charged. 2. When the earthed lead is connected to the metal plate, electrons will flow from the earth to neutralise the positively charged end of the metal plate. 3. After taking the charged rod away, the excess negative charges on the metal plate will redistribute themselves uniformly on the metal plate. (i) When light strikes a region, the surface becomes conducting and electrons will flow from the earth to the region to neutralise the positive charges in the region. 1. The toner is positively charged, and it repels the positively charged regions on the drum. 2. Electrostatic induction occurs in the conducting regions as the toner is sprayed and approaches the regions. This causes the toner to be attracted towards the regions. ** End ** 6 P a g e T u i t i o n P h y s i c s. c o m

# Ans Workings / Remarks

# Ans Workings / Remarks 1 A The positive zero error is +0.03 mm. The reading is 1.84 mm. Thus, final reading = 1.84 (+0.03) = 1.81 mm 2 A We can use parallelogram of forces to determine the final resultant