1. (a) Hybrid cars produce less pollution. [1A] [1] (b) Using ultra lightweight material to reduce the mass of the hybrid car. [1A] The gasoline engine and the electric motor can work together to produce a large accelerating force. [1A] By Newton's Second Law, a = F/m, the acceleration of the car can be raised. [1A] [3] (c) In a hybrid car, the kinetic energy is partly converted to electrical energy. [1A] In a standard gasoline car, the kinetic energy is mostly converted to heat and sound. [1A] [2] 2. (a) K.E. gain = P.E. loss [1M] = 60 x 10 x (15 - 3.5) = 6900 J [1A] [2] (b) P.E. loss = work done against the resistive force 6900 + 60 x 10 x (3.5 - 0.5) = F x (3.5 - 0.5) [1A + 1M] F = 2900 N [1A] [3] (c) P.E. loss is the same [1A] The distance travelled by the man in the cushion is larger. [1A] The average resistive force by the cushion is smaller. [1A] [1C] [effective Communication] [4] 3. (a) R [1A] (b) (i) Water temperature decreases to the freezing point, 0°C. [1A] Ice at 0°C is still melting. [1A] [2] (ii) Water reaches the room temperature. [1A] [1] 4. (a) (i) Energy required = m c ΔT + m l v [1M + 1M] = (0.02) (4200) (100 - 28) + (0.02) (2.26 x 10 6 ) = 51248 J (~51200J) [1A] [3] (ii) 51248 = (637) (200 - T) [1M]
2009 CE Physics Marking Scheme
Nov 24, 2014
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
1. (a) Hybrid cars produce less pollution. [1A][1]
(b) Using ultra lightweight material to reduce the mass of the hybrid car. [1A]The gasoline engine and the electric motor can work together to produce a large accelerating force. [1A]By Newton's Second Law, a = F/m, the acceleration of the car can be raised. [1A]
[3](c) In a hybrid car, the kinetic energy is partly converted to electrical energy. [1A]In a standard gasoline car, the kinetic energy is mostly converted to heat and sound. [1A]
[2]2. (a) K.E. gain = P.E. loss [1M]
= 60 x 10 x (15 - 3.5)= 6900 J [1A]
[2](b) P.E. loss = work done against the resistive force 6900 + 60 x 10 x (3.5 - 0.5) = F x (3.5 - 0.5) [1A + 1M]
F = 2900 N [1A][3]
(c) P.E. loss is the same [1A]The distance travelled by the man in the cushion is larger. [1A]The average resistive force by the cushion is smaller. [1A][1C] [effective Communication]
[4]3. (a) R [1A]
(b) (i) Water temperature decreases to the freezing point, 0°C. [1A]Ice at 0°C is still melting. [1A]
[2](ii) Water reaches the room temperature. [1A]
[1]4. (a) (i) Energy required = m c ΔT + m lv [1M + 1M] = (0.02) (4200) (100 - 28) + (0.02) (2.26 x 106) = 51248 J (~51200J) [1A]
[3] (ii) 51248 = (637) (200 - T) [1M]
T = 120°C [1A] [2]
(b) To produce 40 g steam, the temperature of the plate will be decreased by
160°C. [1M]It will then below the boiling point of water. [1A]
[2]
5. (a) Because the ray x is diverging from the principal axis. [1A]Therefore, it is a concave lens. [1A]
[2](b) (i)
Extend ray of refracted ray of ray x [1M]Ray passing through the optical centre [1A]Correct image [1A]
[3](b) (ii) 0.58 [1A]
[1](c) Correct position [1M]
[1](d) Correct ray [1M]
[1]6. Connect the two crocodile clips across one nichrome wire. [1A]
Record the ammeter reading. [1A]Repeat the experiment above two procedures using another nichrome wire with different thickness of the same length. [1A]The resistance is smaller if the ammeter reading is larger. [1A][1C] [effective Communication]
[5]
7. (a) Mass of Rn-222 left = [1M]
= 0.625 μg or 6.25 x 10-7g [1A][2]
(b) α [1A][1]
(c) + [1A + 1A]
[2](d) Let x and y be the no. of α particles and β particles respectively
(222-206) = 4x [1M]x = 4 [1A]
(86 - 82) = (2) (4) – y [1M]y = 4 [1A]
[4]8. (a) out of paper [1A]
[1](b) (i)
[1A]
[1](ii) The magnitudes of the magnetic forces are the same because [1A]
they are action-reaction pair. [1A][2]
(c) The directions of the current in X and Y are always opposite. [1A] The forces are always repulsive and X and Y will not attract and repel each other alternatively. [1A]
[2]
9. (a)(i)
Correct labelled axes with units [1A]Correct scale [1A]Correct points [1A]A correct line through the points [1A]
[4](ii) Slope = 1.57 s2m-1 [1A]
[1](iii) By s = 0.5at2, slope of the graph = 2/a [1M] [accept 1.5 – 1.6]
a = 2/slope = 1.27 ms-2 [1A] [accept: 1.25 – 1.33][2]
(b)
1 correct force with correct name all correct [1A]all correct [1A]
[2](c) mg sinθ - f = ma [1M] f = 0.178 x 10 x sin25° - 0.178 x 1.27
f = 0.526 N [1A] [accept: 0.516 – 0.530][2]
10. (a) (i) λ= 9 cm / 2.25= 4 cm [1A]
[1](ii) T = (7 - 3) / 5 = 0.8 s [1M]
f = 1 / T= 1.25 Hz [1A]
[2](iii) v = fλ [1M]
= 1.25 x 4 = 5 cm s-1
s = vt [1M]= 5 x 3= 15 cm [1A]
[3](b) (i) Q [1A]
[1](ii) S [1A]
[1](c)
- Correct waveform [1A]- Correct position of S [1A]
[2]
11. (a)Ultrasound is sound wave of frequency higher than 20 kHz.[1A][1]
(b) Time for tsunami to reach the shore = = 7200 s [1A]Time require for the travel of the signals
= time required for the travel of the signals [1A]Time left for the announcing warning signals = 7200 s – 4.25 = 2h > 1h[1M]The system can meet the requirement.
[3](c) Ultrasound cannot travel in outer space. [1A]
[1](d) The speed of EM wave is higher. [1A]Therefore, Peter’s suggestion more appropriate. [1A]
[2]
12. (a) number of turns = 500 x 100 / 10 = 5000 turns [1A]
[1](b) (i) V = IR [1M]
= 0.1 x 80 = 8 V [1A]
[2](ii) VMN = VMP + VPQ + VON
100 = 8 + VPQ + 8 VPQ = 84 V [1A]
[1](iii) P = VI [1M]
= 84 x 0.1= 8.4 W [1A]
[2](c) Deliver the power though the cable using a higher voltage. [1A]
Use ticker wires for cable. [1A][2]
Paper 2 Answer (percentages)1. C (89)2. A (81)3. C (76)4. A (62)5. A (90)6. B (36)7. A (78)8. C (86)9. C (64)10. B (83)11. D (82)12. D (57)13. D (76)14. C (54)15. A (49)16. C (64)17. A (50)18. B (66)19. C (42)20. B (16)21. D (62)22. D (64)23. D (88)24. B (79)25. B (70)26. D (55)
27. B (33)28. C (51)29. C (57)30. D (56)31. C (55)32. B (54)33. A (54)34. B (60)35. A (69)36. C (41)37. B (56)38. D (50)39. D (47)40. D (38)41. A (31)42. C (70)43. D (66)44. A (54)45. B (65)
Related Documents
