2008 Science (Physics) Section A

1. 2008 Science (Physics)Section A Questions, Answers and Comments

2. If you are using topical TYS (2002-2010- version)

3. What did the examiner say?

4. Mass, Weight and Density • Easy chapter to score • Must know definition of mass, weight, gravitational field and gravitational field strength • Must be able to compare mass and weight • Formula : w = mg (note: m is in kg while g, the gravitational field strength depends on the location) • Formula : density = (note: be careful with formula of volume and mass)

5. ANS: Mass = density x volume = 5.5 x 20 = 110 g ANS: mass = 110 g (mass does not change with location)

6. ANS: w = mg = x 1.6 = 0.176 N Mass must be in kg.

7. Electromagnetic Spectrum • EM waves • are all transverse waves; • travel with speed of 3 x 108m/sin vacuum • Application

8. (a) Infrared (b) Microwave (satellite TV is not the same as TV communication which uses radio wave. Radio wave is easily absorbed and cannot make it to the satellite) (c) Gamma rays

9. Kinematics • Define speed and velocity • Formula (if object is decelerating, then v will be smaller than u and a will be negative) Average speed = total distance travelled total time taken (total time would also include the rest time in the journey)

10. Kinematics • acceleration of free-fall for a body near the earth is 10 m/s2. (if air resistance is ignored, all objects will fall at the same acceleration of 10 m/s2 which means that their velocity will increase by 10 m/s every second)

11. Kinematics • Speed-Time Graph • Area under speed – time graph = distance travelled by object

12. Kinematics

13. Time = distance / speed = 600/40 = 15 s (Do not stop the graph at 15 s) a = (v – u) / t t = (v – u) / a = (40 – 0)/8 = 5 s

14. KE = ½ mv2 = ½ (300) (40)2 = 240 000 J Don’t forget to square the velocity *must calculate acceleration first before force can be calculated (since F = ma) a = v - u t = (0 – 40) / 8 = - 5 m/s2 F = ma = 300 x 5 = 1500 N Don’t mix up initial and final velocity

15. Note details in diagram like the kind of charges The high voltage supply is used to charge the metal plate and metal grid to become +ve and – ve respectively

16. The dust particles will gain negative charge from the metal grid and get attracted to the positive metal plate. Answer must consist of the ‘description’ part and ‘explanation’ part. This is because unlike charges attract. Don’t get mixed up with magnetism and use unlike ‘poles’ “Like charges repel” / “unlike charges attract” can be used to explain most electrostatic qn

17. What else you need to know • How to draw electric field of a point charge • Define electric field

18. Don’t forget to add arrows! Accuracy is very important! Only a small range is accepted by examiner. Focal length = 1.8 cm – 1.9 cm

19. Remember: the larger speed is at the top refractive index of glass = speed of light in air speed of light in glass 1.5 = 3 x 108 / v v = 3 x 108 / 1.5 = 2 x 108 m/s Refractive index is always bigger than 1 Many made mistake here!

20. Take note that the solid in the ice pack melts at 1oC, not 0oC

21. The temperature of the solid in the ice pack will remain at 1oC (its melting point) until all of it has melted). Remember: Temperature of a substance does not change during change of state. This is because when the solid in the ice pack melts, it will absorb heat from the surrounding food and cools it down. For solid that does not melt, less heat is absorbed from the food. When an object cools down, it is because it loses heat. There is no such thing as absorbing “coldness”.

22. What else you need to know? • Which processes below absorb energy and which one releases energy to the environment? Boiling melting evaporation freezing condensation

23. Vector quantity has direction but scalar quantity does not have direction. No need to give specific example of scalar or vector. No need to mention similarity that both have magnitude since qn ask for ‘difference ‘ . If qn says ‘compare’, must also state similarity.

24. 5 N 3 N 3 N 5 N 4 N 4 N • Must draw ARROWS !!! • Must label the forces, not the length of the line. • Can check ans by using Pythagoras Theorem. Magnitude = 5 N

25. What else must you take note of for vector questions? • Bring long ruler and protractor for Physics paper • Use their diagram if it is provided in the qn (e.g. 2009 QA3, 2010 QB14) The angle have been measured for you, just continue on the lines drawn. Your line can be longer or shorter than theirs.

26. This button enable the circuit to be close again after it is broken by the circuit breaker. A device that protects an appliance by stopping current larger than its rated value from entering it. (Same as a fuse but it is re-usable) Solenoid will always be magnetised when current pass through it but it will only attract armature if magnetic field is strong enough. Remember: current pass through coil, it does not pass through iron core. Purpose of spring is to pull the armature back to original when it is no longer attracted by solenoid.

27. When current flows in the solenoid, it gets magnetised. • When the current is larger than its rated value, the magnetic field is strong enough to attract the armature. • The springy metal will break contact with the iron armature. • The circuit remains open until the reset button is pressed to push the springy metal to touch the armature again. 4 mark question so you must have 4 or more points. Organise each point as a complete sentence to make it easier for examiner to award marks. • What’s wrong with this answer? • When current flows in the solenoid, it becomes magnetised and attracts the armature……. This point is often not mentioned by students.

28. This will ensure that when the circuit breaker blows, the appliance will be isolated from high potential / will not be ‘live’ / will not be connected to high potential and that will protect the user from electrocution. A very common question, must memorise answer. Check out the ‘Common Questions Asked in Practical Electricity’ in the Wikispaces and ensure that you can answer all of them.

29. This means that the lamp is operating at its rated power. (It is possible that some appliance is operating at a power that is less that its rated value. This is when the voltage supplied is less than the operating voltage.) Parallel circuit hence the voltage across the lamp and resistor is the same. Since they are also parallel to the battery, their voltage will be the same as the emfof the battery P = V I 24 = 12 x I I = 24/12 = 2 A Must state formula

30. V = R I 12 = 3 x I I = 12/3 = 4 A 2 A 4 A R (bulb) = V/I = 12/2 = 6Ω 1/RE = 1/R1 + 1/R2 = 1/6 + 1/3 = 3/6 RE = (3/6)-1 = 2 Ω Total voltage = EMF of battery Effective resistance means total resistance RE = V I = 12 (2 + 4 ) = 2 Ω Total current = current that enters battery Don’t forget to invert the ans to get RE

31. Go on to try other year’s paper!