Higher – ODU Powerpoint Answers

1. a) False.

b) True.

c) True.

d) False.

2. a) Distance travelled = 70km.

b) Average speed = 31.1kmh-1.

c) Displacement = 50km@037º.

d) Average velocity = 22.2kmh-1@037º.

3. a) 7kmh-1.

b) 5kmh-1@127º.

4. A.

5. E.

6. Horizontal distance = 50m.

7.

8.

a) Displacement from A to B = 12.6km@166º.

b) Average velocity = 8.4kmh-1@166º.

c) Average speed = 8.4kmh-1.

9. A.

10. C.

11. D.

12. 3.33ms-2.

13. a) a = -9ms-2.

b) The negative acceleration means that the car is decelerating.

14. Options 1 and 2 are correct.

15.

16.

17. E.

18. s = 261.1m.

19. E.

20. E.

21. a) v = 8ms-1.

b)

c) s = 38m.

22. a) t = 3.1s.

b) sv = 45.9m.

c) sH = 248m.

d) vR = 50ms-1@127º.

23. a) θ = 53º.

b) sv = 19.6m.

c) sH = 60m.

d) vR = 24.7ms-1@143º.

e) The ball leaves and hits the ground with the same vertical velocity.

This means that the ball will be hitting the ground at the same height ‘above

sea level’ that it is hit from.

Therefore vertical displacement equals zero.

24. Options 2 and 3 are correct.

25. a) t = 4s.

b) sH = 160m.

c) vR = 56.3ms-1@135º.

d) Air resistance is assumed to be negligible.

26. a) uH = 18.9ms-1.

uv = 14.8ms-1.

b) t = 1.5s.

c) sv = 11.2m.

d) sH = 61.4m.

27. a) uH = 36.5ms-1.

uv = 22.8ms-1.

. b) t = 4.33s.

c) O -> P = 85m.

P -> Q = 73m and so O -> Q = 158m.

28. a) uH = 10ms-1.

uv = 17.3ms-1.

b) sv = 15.3m.

c) sH = 35.4m.

d) Air resistance is assumed to be negligible.

29. a) t = 0.61s.

b) sv = 1.84m.

c) vR = 6.18ms-1@014º.

30. a) uH = 4.44ms-1.

b) t = 0.65s.

c) h = 3.67m.

31. Upthrust Force = 5500N.

32. a = 3ms-2.

33. Upthrust Force = 32,000N.

34. a = 10.6ms-2.

35. Total Pulling Force = 30N.

36. Resultant Force = 17.5N.

37. a) m = 70kg.

b) a = 0.57ms-2.

c) a = - 0.5ms-2.

38. Force of Friction = 17.6N.

39. Ew = 847J.

40. m = 964kg.

41. a) a = 0.3ms-2.

b) sH = 5.4m.

42. T = 3244N.

43. a) Force of Friction = 165N.

b) Force of Friction = 319N.

44. a) W = 234.6N.

b) a = 3.14ms-2.

c) The student would stop accelerating and would move with a constant speed

when; the frictional forces in total are equal and opposite to the component

of weight acting down the slope.

The greater the speed of the student on the sledge the greater the air

resistance. The air resistance and the frictional force on the snow would

eventually equal the component of weight acting down the slope.

The balanced forces would then make the student and the sledge move

with a constant speed.

45. a) T = 4414N.

b) a = 10.2ms-2.

c) As the capsule rises the angle will reduce from 25º to say 20º.

This would reduce the tension in each chord and also the total vertical

component of tension.

This will then reduce the unbalanced force exerted on the capsule and it

will therefore reduce its acceleration.

46. A.

47. A.

48. a) Total momentum before the collision = Total momentum after the collision

(providing that no external forces are acting)

b) uA = 20ms-1.

c) Kinetic Energy before collision = 312,000J.

Kinetic Energy after collision = 270,000J.

Collision is inelastic as the kinetic energy is not conserved.

49. a) v1 = – 5ms-1

b) B.

50. a) vB = 5.66ms-1.

b) Kinetic Energy before collision = 329,184J.

Kinetic Energy after collision = 44,850J.

Collision is inelastic as the kinetic energy is not conserved.

51. vB = 0.13ms-1.

52. a) v2 = 480ms-1.

b) Kinetic Energy before collision = 172,800J.

Kinetic Energy after collision = 259,200J.

Collision is an explosion as kinetic energy is gained.

53. a) m = 0.06kg.

b)

c) The area under the force-time graph is as measure of:

Impulse or

Change in momentum

54. a) Impulse = 18x10-3Ns.

b) Change in momentum over the first 4s = 15x10-3kgms-1.

55. Change in momentum = 27kgms-1.

56. a) Impulse = 0.35Ns.

b) Impulse = change in momentum = -0.35kgms-1 (upwards)

c) v = 5.94ms-1.

d) v = - 2.75ms-1.

57. a)

b)

The area under each graph graphg is equal and so the Impulse is the

same in each case. Impulse = change in momentum

The soft ball has a greater time of contact, but the average force

exerted on the ground is smaller

The hardball has a smaller time of contact, but the average force

exerted on the ground is greater.

c)

Padding in the helmet will provide a ‘cushioning effect’ on the

ground with the crash helmet.

The persons head will be in contact with the ground for longer, but a

smaller average force will be exerted on the skull.

There will be less damage to the skull.

58. a) v = 1.66ms-1.

b) u1 = 500ms-1.

59. F = 1348N.

60. h = 180km.

61. a) m = 6.38x1023kg.

b) g = 0.81Nkg-1.

62. Asteroid 1 mass = 40kg.

Asteroid 2 mass = 10kg.

63. a) g = W/m

g is the force exerted on a mass of 1kg in the gravitrational field.

b) i) F = GMm/r2 and W = F = mg.

By equating these two equations => GMm/r2 = mg

=> G = mgr2/Mm

=> G = gr2/M

ii) G = 6.67x10-11Nm2kg-2. Therefore QED

64. t’ = 10.4 hours.

65. v = 0.575c

66. a)

Time dilation

Length Contraction

Then by applying the Lorentz Factor into the two equations above:

t’ = tɤ

l’ = l/ɤ

b) v = 0.2c then ɤ = 1.02. => Not appreciable.

v = 0.9c then ɤ= 2.29. => Appreciable.

67. a) v = 1.91x108ms-1.

b) t = 2.58x1012s.

c) t = 1.98x1012s.

68. a) f0 = 826Hz.

b) f0 = 739Hz.

69. The waves or wavefronts are closer together as the train approaches the

person.

As the wavelength decreases then the frequency of sound picked up by the

observer increases.

The waves or wavefronts are further apart as the train passes by the

person.

As the wavelength increases then the frequency of sound picked up by the

observer decreases.

70. a) A => Distant Star. B => Earth.

The Doppler Effect has shown us that the frequency of light will decrease

when the source is moving further away from a stationary observer, and

this means that the wavelength of light will increase.

Red light has a high wavelength and a low frequency and so the light

emitted in this case from the Hydrogen emission spectrum will be shifted

towards the red end of the visible spectrum as the star is continually getting

further away from us, due to the expansion of the universe.

b) i) Z = + 0.18

ii) v = 5.4x107ms-1.

71. a) The blue shift is due to the Sun getting closer to the Earth and emits light

of lower wavelength and higher frequency when viewed by a stationary

observer on Earth.

The red shift is due to the Sun getting further away from the Earth and

emits light of higher wavelength and lower frequency when viewed by a

stationary observer on Earth.

b) Red Shift ratio -> Z is + ve.

Blue Shift ratio -> Z is – ve.

c) The Binary star is blue shifted when moving towards the Earth and red

shifted when it is moving away from the Earth.

The blue shift is due to the star getting closer to the Earth and emits light

of lower wavelength and higher frequency when viewed by a stationary

observer on Earth.

The red shift is due to the star getting further away from the Earth and

emits light of higher wavelength and lower frequency when viewed by a

stationary observer on Earth.

72. a) v = Hod

v = Recessional velocity of a star/galaxy (ms-1)

Hubble’s Constant -> Ho = 2.3x10-18s-1.

d = distance to a star or galaxy (m)

b) Hubble’s Constant Ho = 2.3x10-18s-1.

Red Shift ratio -> Z is +ve.

c) v against d graph for Hubble’s Law

73. a) Hubble’s Law tells us that the universe is continually expanding.

Hubble’s Law can be used to estimate the age of the universe.

b) d = 1.21x1025m.

74. a) To apply the Blackbody Radiator Theory to stars, we assume that although

a star is not a perfect blackbody radiator, it approximates close enough so

that the Blackbody Radiator Theory can be applied.

b) i)

ii) The Blackbody Radiator graph shows that for stars:

Higher temperatures are related to higher thermal energies and thus lower

peak wavelengths.

Hotter stars emit higher thermal energies which will have a greater area

under the curve of an Radiation Irradiance v Wavelength graph.

75. a) The higher the peak wavelength of light emitted from a star the lower the

surface temperature of the star and vice-versa.

b) λp = 4.83x10-7m.

c) i) Hertzsprung- Russell Diagram (H-R Diagram)

ii) The H-R Diagram tells us the following information about a star:

Type of Star

Radius of the Star

Age of the Star