1 CJC JC2 H2 Prelim 2011 Paper 2 Solutions 1 (a) (i) Force on air

= rate of change of momentum of air

= change of momentum of air per unit time

= change in velocity x mass flow per unit time

= (v – 0) x (ρ.πr2vt)/t

= ρ.πr2v2

(ii) By Newton’s 3rd Law, Force on fan by air = Force on air by fan = ρ.πr2v2

(iii) As the fan blades rotate, they exert a force on the air. By Newton’s Third Law, the air exerts an equal and opposite force on the axle of the fan.

(b) (i) No. When the fan blades rotate, momentum is transferred to the air, which in turn transfers momentum to the sail, which is connected to the whole trolley. At the same time, due to Newton’s Third Law, the gain in momentum of the air causes the axle of the fan to gain a momentum that is equal in magnitude but in the opposite direction. Since the fan is also fixed on the trolley, the NET force on the trolley is zero. Thus the trolley will not move and his idea won’t work.

(ii) remove the sail reduce the mass of the trolley increase blade area place fan on the floor (any two plausible methods)

2 (a) (i) Simple harmonic motion is defined as the motion of an object whose acceleration a is

directly proportional to its displacement x from a fixed point (equilibrium position) and is always directed towards that fixed point.

(ii) Since acceleration is always directed towards the equilibrium position, hence: When the object is moving towards the equilibrium position, its acceleration and velocity are in the same direction. When the object is moving away from the equilibrium position, its acceleration and velocity are in opposite direction.

(b) (i) As em wave acts as an external oscillator that provides energy to ions continuously, and the frequency of oscillation of ions is equal to the driving frequency of em wave, ions are undergoing forced oscillations.

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3. (a) 1 V = 1J C-1

Emf of 0.15 V means that 0.15 joules of energy is converted from other forms into electrical energy in driving per coulomb of charge round the electrical circuit.

(b) 2 x 0.25 Ω = 0.50 Ω

(c) Resistance of 1 line of cells = 0.50 Ω Resistance of 2 lines of cells connected in parallel: 1/R = 2 x 1/0.50 R = 0.50 / 2 = 0.25 Ω

(d) (i) resistance of the electroplaques in 1 line = 5000 x 0.25 resistance of the eel R (140 electroplaques connected in parallel): 1/R = 140 x 1/(5000 x 0.25) R = 5000/140 x 0.25 = 8.9286 Ω = 8.93 Ω

(ii) Total circuit resistance = resistance of eel + resistance of surrounding water = 8.9286 + 800 Current = Voltage / resistance = (5000 x 0.15) / (8.9286 + 800) = 0.92715 = 0.927 A

(iii)

3 By potential divider principle, 800/ (8.9286 + 800) x (5000 x 0.15) = 742 V OR current through each line of electroplaques in the eel’s body = 0.92715 / 140 = 6.6225 x 10-3 A (5000 x 0.15) – (6.62 x 10-3)(5000 x 0.25) = 742 V

- (iv) current through each line of electroplaques in the eel’s body

= 0.92715 / 140 = 6.6225 x 10-3 A The current through each line of electroplques is too small (approximately by about two orders smaller than the current through the surrounding water) to harm the eel.

4 (a) Stimulated emission leads to a deluge of photons released (1 leads to 2 leads to 4 etc)

A photon that is incident on an atom is not absorbed, but instead triggers another photon to be released. This causes the number of photons to double. Each of these photons then goes on to trigger other atoms to release photons in the same way. Hence the number of photons keeps doubling.

(b) The single slit is to ensure that the two sources are coherent The laser is already coherent and hence a single slit before the double slit is not necessary.

(c) (i) A dopant is introduced with a lower valence number than the intrinsic semiconductor. The dopant hence introduces a hole just above the valence band, which allows an electron to transit easily to this hole, leaving a hole behind in the valence band which increases the number of holes available for conduction.

(ii) In reverse bias, the external electric field strengthens the internal electric field due to the immobile ions in either side of the p-n junction. This drives the holes towards the p side and the electrons towards the n side, causing the depletion region to widen. A depletion region is a region which has no mobile charge carriers. Hence current cannot flow in reverse biased.

(iii) As the External E field increases, the electrons that are bound in the lattice structure may break free and get channelled into an observable current. The electrons that are suddenly released can also collide into other bound electrons,

4 creating the “avalanche” of electrons as seen in the breakdown voltage.

5 (a) Magnetic flux density at a point is the force acting per unit current, on per unit length of

wire lying at right angles to the magnetic field.

(b) Allows a uniform torque to be produced throughout the armature’s rotation.

At all positions, F = BIL sin 90

(c) For each turn of coil, change in magnetic flux = B x [Area swept in one revolution by each 100mm-side] x [2 sides] = 0.5 T x 2πr(100 mm) x 2 [Area swept = cylindrical surface area] = 0.5 T x 2π ( 90 mm / 2) (100 mm) x 2 = 0.028274 Wb For 250 turns of coil, total change in magnetic flux linkage = 250 x 0.028274 Wb = 7.0686 Wb = 7.07 Wb

(d) As the windings start to rotate, the magnetic flux through the armature windings change and hence there is an induced e.m.f. in the windings. By Lenz’s law, this induced e.m.f. will be generated in a direction that opposes the e.m.f. supplied by the source.

(e) Induced emf = rate of change of flux linkage = N x B x [rate of change of area swept by one 100mm-side] x [2 sides] = 250 x 0.5 T x 2πr(100 mm) (frequency) x 2 = 250 x 0.5 T x 2π(90 mm / 2)(100 mm) (1200 / 60) x 2 = 141.37 V = 141 V OR Using answer from (c), Induced emf = rate of change of flux linkage = 7.0686 Wb x frequency

5 = 7.0686 Wb x 1200 rpm = 7.0686 Wb x (1200 / 60) =141.37 V = 141 V OR Induced emf = NBLv x 2 sides = NBL(rω) x 2 sides = 250 x 0.5 T x 100 mm x (90 mm / 2)(1200 x 2π / 60) x 2 =141.37 V = 141 V

(f) (i) As the armature’s rotational speed increases, the rate of change of magnetic flux linkage of the armature windings increases. This causes the magnitude of the back e.m.f. to increase in accordance to Faraday’s law. The current drops gradually as the back e.m.f. increases.

(ii) The current becomes steady as the rotational speed of the coil becomes constant . This means that the magnitude of the back e.m.f. is now constant and thus the net e.m.f. and current is now steady.

(g) I = (240 – back emf at steady state) / 0.8Ω = (240 – 141.37) / 0.8 = 123.29 A = 123 A

(h) Force on one side of each turn of coil, F = BIL sin 90° = (0.5 T) (123.29 A) (length of coil cutting field lines) sin 90° = (0.5 T) (123.29 A) (100 mm) = 6.1644 N Total Torque = N x F x perpendicular distance = 250 x 6.1644 x (90 mm) = 137 Nm

(i) There will be energy is loss in the resistor through heating whenever the circuit is in use, not just at the start of the motor. Supplementary notes: To limit the current at start-up, a current limiter is used. An example is the NTC thermister. At start-up, its temperature is still low and hence its resistance is high, thus reducing the inrush current. As time progresses, its temperature rises due to the current, reducing its resistance by 10 to 50 times, thus reducing the power dissipated in it. The NTC thermistor thus provides protection from undesirably high inrush currents, while its resistance remains negligibly low during continuous operation.

Source: http://www.epcos.com/web/generator/Web/Sections/ProductCatalog/ProtectionDevices/InrushCurrentLimiters/NTCInrushCurrentLimiters/PDF/PDF__Applicationnotes,property=Data__en.pdf;/PDF_Applicationnotes.pdf

6

Q6 Planning Question

Aim: To investigate how the electrical power consumed by a small electric motor varies with the final constant speed as a mass is raised through the vertical height.

Define the problem Independent variable : constant speed of mass raised by motor Dependent variable: power consumed by motor Control variable: mass of the mass

1 Set up the apparatus as shown.

2 Turn on the switch and allow the mass to be pulled up by the motor by coiling the tread around the motor shaft.

3 Set the resistance of the variable resistor to a suitable value such that the motor will pull the mass up at a suitable speed.

4 When the mass reaches the position A, start the time using the light gate 1 at position A connected to the timer.

5 Record the time t1 when the mass reaches the position B and time t2 when the mass reaches the position C using light gate 2 and light 3 which are connected to the timer.

6 Record the voltmeter reading V and the ammeter reading I when the mass is between B and C.

7 Calculate the power consumed by the motor using P = I V.

8 Measure the distance D1 between A and B and D2 between B and C using a metre rule. Ensure that the distance between A and B, and between B and C are large enough to reduce random error in measuring the time interval. The metre rule is placed close to the mass and observations are made at eye level to reduce parallax error.

9 Calculate the speed of the mass using v1 = D1 / t1 and v2 = D2/t2

Variable resistor A

voltmeter

Ammeter

Low voltage power supply

switch

V

motor

Motor shaft

C t2

D2

B t1

D1

A t =0

Tall bench Light gates L1

L2

L3

Digital

timer

7

10 If v1 is equal to v2 then the mass is traveling at a constant speed. If not raise the height of A and B or change the voltage applied to the motor until a constant speed is obtained between A and C.

11 Calculate the average speed v = (v1 + v2) / 2

12 Repeat steps 3 to 11 for different values of the current flowing in the motor by adjusting the variable resistor/voltage supply.

13 The mass of the weight used is kept constant. (control variable)

14 Plot a graph of power against speed to determine how the power consumed by the motor affects the speed of the mass.

Details

15 Measure I and V only when the speed is constant when the mass is between A and C

16 Use of photogates and digital timer to take the time duration to minimise human error as the time interval is small

17 The mass should not swing around as it is raised.

Safety precautions

18 Voltage used must not be higher than the operating voltage of the motor to prevent the motor from overheating and damaging.

19 Ensure that the mass does not hit the shaft of the motor. The motor must be turned off immediately after it has passed position C.

20 The motor must be secured onto the bench using clamps or belts.

Q6 Suggested Mark Scheme marks Define the problem:

Identify IV, DV and at least one CV 1

Workable arrangement and procedure

• Comprehensive statements of procedures

• How power supply is connected to the motor

• How the load is driven by the motor

• Suitable graph to be plotted to show relation

2

Method in measuring speed Method 1 :

• measurement of time interval using stop watch

• measurement of distance using ruler

• use speed = distance / time Method 2:

• use of video footage

• method to determine speed from the footage Method 3:

• use of motion sensor and data logger with suitable arrangement of the device

2

Method to ensure constant speed:

• check speed at two different locations

• check speed on data logger display

1

Arrangement to change speed:

• Use of different powers by changing the voltage/current: o Use of variable resistor o Use of variable supply voltage

1

Method in measuring power consumed by motor

• Clear and well labeled circuit diagram with correct connection of voltmeter/ammeter/ motor/supply unit/variable resistors

• measure I using ammeter through motor

• measure V using voltmeter across terminal of motor

• use of P = VI

2

Method in keeping the control variable/s constant

• mass of the weight used to be kept constant

1

8

Safety

• Voltage applied must be below operating voltage of the motor

• Ensure the motor do not get overheated

• Start with low power to avoid load from accelerating too fast

1

Details

• Measure I and V when speed is constant; when readings are steady

• Use of photogates and digital timer: with explanation how they are used

• Use large distance to reduce random error when use v = d / t to determine speed

• Ensure pulley rotate freely

• Repeat experiment and find average speed AND average power: reduce random error

Max 3 marks