Year 12 Physics Space Practice Quiz 4 Solution

advertisement
Year 12 Physics Space Practice Quiz 4 Solution
/ 50
Data
rE = 6380km
𝐹 = π‘šπ‘”
𝑣 = 𝑒 + π‘Žπ‘‘
TE = 365.25 days
π‘Ÿ 3 𝐺𝑀
=
𝑇 2 4πœ‹ 2
𝑣𝑦2 = 𝑒𝑦2 + 2π‘Žπ‘¦ Δ𝑦
πΊπ‘š1 π‘š2
π‘Ÿ2
π‘š1 π‘š2
𝐸𝑝 = −𝐺
π‘Ÿ
rMars = 3390 km
Δπ‘₯ = 𝑒π‘₯ 𝑑
𝐹=
mMars = 6.4x1023kg
mMoon = 7.3x1022kg
𝑣 2
𝑙𝑣 = 𝑙0 √ 1 − ( 𝑐 )
𝑑0
𝑑𝑣 =
2
√1 − (𝑣𝑐)
π‘š0
π‘šπ‘£ =
2
√1 − (𝑣𝑐)
1
1
Δ𝑦 = 𝑒𝑦 𝑑 + 2π‘Žπ‘¦π‘‘ 2
MEarth = 6.0x1024kg
𝐹=
π‘šπ‘£ 2
π‘Ÿ
πΈπ‘˜ = 2π‘šπ‘£ 2
G = 6.67x10-11Nm2kg-2
1. Which statement about weight is true? Multiple Choice: 1 mark each
a) The mass of an object will be different on a different planet but the weight will stay the same.
b) Weight is a force that can change with location.
c) The weight of an object (in Newtons) can be calculated by multiplying its mass (in kg) by 9.8.
d) Weight is an imaginary force like centrifugal force.
2. What is the gravitational potential energy of a 900kg object 30km above the surface of Mars? (−𝟏. 𝟏𝟐 ×
𝟏𝟎𝟏𝟎 𝐉) 2 marks
3. What is the minimum amount of work that must be done to move an 8000kg Earth satellite from an orbit
with an altitude of 20 000km to an orbit with an altitude of 29 000km? (πŸ‘. πŸŽπŸ— × πŸπŸŽπŸπŸŽ 𝐉)
3 marks
4. A Low Earth Orbit satellite has a period of 94 minutes. What is its altitude?
(477km)
5. Calculate the acceleration due to gravity 12km above the surface of Mars.
(3.69 m.s-2)
6. Which of the following would increase the range of a projectile?
a) Making the launch angle more vertical.
b) Decreasing the launch speed.
c) Increasing the air resistance of the projectile.
d) Launching from a higher location.
7. Calculate the escape velocity of Mars.
(5.02 km.s-1)
𝑀
π‘Ÿ πΈπ‘Žπ‘Ÿπ‘‘β„Ž
( )
=
2𝐺𝑀
π‘Ÿ
6.0×1024
6380000
2 marks
2 marks
8. Explain why Earth has a greater escape velocity than Mars.
π‘‰π‘’π‘ π‘π‘Žπ‘π‘’ = √
2 marks
2 marks
𝑀
π‘Ÿ
The escape velocity for a planet is proportional to √
𝑀
π‘Ÿ π‘€π‘Žπ‘Ÿπ‘ 
( )
=
6.4×1023
3390000
𝑀
π‘Ÿ πΈπ‘Žπ‘Ÿπ‘‘β„Ž
( )
𝑀
π‘Ÿ π‘€π‘Žπ‘Ÿπ‘ 
>( )
πΈπ‘Žπ‘Ÿπ‘‘β„Ž
π‘€π‘Žπ‘Ÿπ‘ 
Therefore, π‘‰π‘’π‘ π‘π‘Žπ‘π‘’
> π‘‰π‘’π‘ π‘π‘Žπ‘π‘’
9. How does building rockets with multiple stages help rockets get into space?
a) Inertia is proportional to mass so rockets with less mass are easier to stop in an emergency situation.
b) Each stage is a complete engine. Having multiple stages means there is less chance of total failure.
c) Acceleration is inversely proportional to mass. Removing empty fuel canisters increases acceleration.
d) Rocket thrust is zero while the stages are being changed. This gives the astronauts a chance to recover.
10. Identify the error in the following calculation of the mass of the Sun.
1 mark
2 3
2
3
4πœ‹ π‘Ÿ
4 × πœ‹ × 6380000
𝑀𝑆𝑒𝑛 =
=
= 1.54 × 1017 π‘˜π‘”
2
𝐺𝑇
6.67 × 10−11 × 315576002
The radius used in the calculation is the planetary radius. This is incorrect. The correct radius to use is the
orbital radius.
11. Which statement about the Earth’s orbit is true?
a) Kepler’s law of periods predicts that the Earth has a shorter orbital period than Mars.
b) Kepler’s law of periods predicts that the Mars has a shorter orbital period than Earth.
c) Kepler’s law of periods does not apply to planets revolving around the Sun.
d) The Sun takes 365.25 days to revolve around the Earth.
12. What is the magnitude of the Earth’s gravitational force acting on a 325kg satellite in orbit at an altitude of
15000km?
(285N)
2 marks
13. A 2.1 kg object is swung in a horizontal circular path on a 75cm string. The string has a maximum tension of
8.5N. At what speed will the string break? (1.7 m/s)
1 mark
14. Which statement best describes Isaac Newton's concept of escape velocity?
a) The height of a mountain needed to launch an object into orbit by shooting it from a cannon.
b) The horizontal speed of a projectile needed to launch it into orbit.
c) The vertical component of velocity of projectile needed to reach the edge of the Earth's atmosphere.
d) The velocity of a projectile needed to send a cannon shell to the Moon as foretold by Jules Verne.
15. Which statement about a rocket launched vertically is true?
a) Acceleration increases primarily because air resistance decreases with altitude.
b) Acceleration decreases because the rocket is increasing in mass due to relativistic mass dilation.
c) Rockets can't use full power near the ground for safety reasons so the initial acceleration is low.
d) Acceleration increases because it loses mass and the Earth's gravity decreases with altitude.
16. Which scientific principle best explains the motion of rockets?
a) Conservation of mass.
b) Conservation of energy.
c) Conservation of enthalpy.
d) Conservation of momentum.
17. The mass of the Sun is 1.989x1030kg. Use Kepler's Law of Periods to estimate the Earth-Sun
distance.
(1.496x108km)
2 marks
18. A marble travelling at a speed of 2.9 m/s rolls off a 1.4m high table. What is the range of the
marble? (1.6m). 2 marks
19. Which of the following has the greatest orbital radius?
a) The Moon. (T = 27.32 days)
b) A geostationary satellite (T = 24hours).
c) A Low Earth Orbit Satellite (T = 88 to 127 minutes).
d) The International Space Station. (T = 92.92 minutes)
20. A soccer ball is kicked with a speed of 28m/s at an angle of elevation of 26o. What is the maximum
height of the soccer ball above the launch point? (7.7m). 2 marks
21. Why are solar flares responsible for increased orbital decay of LEO (Low Earth Orbit) satellites?
a) They eject large numbers of charged particles that stick to satellites and increase their mass.
b) They eject large numbers of charged particles that stick to solar panels and decrease their efficiency.
c) They eject large numbers of charged particles that cause fuel loss by making holes in the fuel tank.
d) They eject large numbers of charged particles that increase drag through atmospheric expansion.
22. Which experiment aimed to detect the aether?
a) The Rutherford-Bohr experiment.
b) The Michelson-Morley experiment.
c) The Doppler experiment.
d) The Bose-Einstein experiment.
23. What would be the fate of astronauts if they attempted to re-enter the Earth's atmosphere at too great an
angle (too close to perpendicular)? (Burn up)
1 mark
24. What would be the fate of astronauts if they attempted to re-enter the Earth's atmosphere at too small an
angle (too close to tangential)? (Bounce off / Never return to Earth / Freeze to death).1 mark
25. The average distance between the centre of the Earth and the centre of the Moon is 384400km. What is the
distance from the centre of the Earth of the point of zero nett gravitational attraction along the line joining
the centres of the Earth and Moon? Neglect the gravitational effect of other bodies and the motion of the
Earth and the Moon. (346212km). 2 marks
26. What name is given to a frame of reference that involves no acceleration? (an inertial frame of reference)
1 mark
27. Imagine a person travelling on a train on Earth at the speed of light. Based on our current understanding of
Physics, describe the path of a ball dropped inside the train as observed by the person travelling in the train.
(The ball would fall towards the floor at 9.8m/s2 just like it would for any person not moving relative to
the event such as a person dropping a ball while standing still. The speed of the observer relative to the
ball is what is important.) 1 mark
28. How much work must be done to accelerate an object with a rest mass of 130kg from 0.9c to 0.99c? 2 marks
𝒇
𝟏
𝟏
𝐖 = πš«π‘¬π’Œ = π‘¬π’Œ − π‘¬π’Šπ’Œ = πŸπ’ŽπŸŽ.πŸ—πŸ—π’„ (𝟎. πŸ—πŸ—π’„)𝟐 − πŸπ’ŽπŸŽ.πŸ—π’„ (𝟎. πŸ—π’„)𝟐
𝟏
= 𝟐(
πŸπŸ‘πŸŽ
√𝟏 −
𝟎. πŸ—πŸ—πŸ
πŸπŸ‘πŸŽ
𝟏
) (𝟎. πŸ—πŸ—π’„)𝟐 − 𝟐 (
) (𝟎. πŸ—π’„)𝟐 = πŸ‘ × πŸπŸŽπŸπŸ— 𝑱
√𝟏 − 𝟎. πŸ—πŸ
29. A spacecraft travels from Earth to a distant star at a speed of 0.8c. The journey takes 6.91 years according to
on-board clocks. What is the distance to the star as measured by an observer on Earth?
2 marks
π’•πŸŽ = πŸ”. πŸ—πŸ π’šπ’†π’‚π’“π’”
𝒗 = 𝟎. πŸ–π’„
π’πŸŽ =?
𝒗=
π’•πŸŽ
π’πŸŽ = 𝒗 × π’•π’— = 𝒗 ×
√𝟏 −
π’πŸŽ
𝒕𝒗
π’•πŸŽ
𝒕𝒗 =
𝟐
√𝟏 − (𝒗𝒄)
= 𝟎. πŸ–π’„ ×
𝟐
(𝒗𝒄)
πŸ”. πŸ—πŸ π’šπ’†π’‚π’“π’”
√𝟏 −
= πŸ—. 𝟐𝟏 π’π’š = πŸ–. πŸ•πŸ × πŸπŸŽπŸπŸ” π’Ž
(𝟎. πŸ–)𝟐
30. An alien race measures time in ligendas. A spacecraft travels from the alien's home planet to a distant star at
a speed of 0.95c. The journey takes 23.4 ligendas according to on-board clocks. What is the distance of the
journey as measured by an observer on the alien's planet? Hint: Measure distance in light ligendas. 2 marks
π’•πŸŽ = πŸπŸ‘. πŸ’ π’π’Šπ’ˆπ’†π’π’…π’‚π’”
𝒗 = 𝟎. πŸ—πŸ“π’„
π’πŸŽ =?
π’πŸŽ
𝒕𝒗
𝒗=
π’πŸŽ = 𝒗 × π’•π’— = 𝒗 ×
π’•πŸŽ
√𝟏 −
𝒕𝒗 =
π’•πŸŽ
𝟐
√𝟏 − (𝒗𝒄)
= 𝟎. πŸ—πŸ“π’„ ×
πŸπŸ‘. πŸ’ π’π’Šπ’ˆπ’†π’π’…π’‚π’”
𝟐
(𝒗𝒄)
√𝟏 − (𝟎. πŸ—πŸ“)𝟐
= πŸ•πŸ. 𝟐 π’π’Šπ’ˆπ’‰π’• π’π’Šπ’ˆπ’†π’π’…π’‚π’”
31. A subatomic particle created in the upper atmosphere travels a distance of 435m in its own frame of
reference and 1027m in an Earth frame of reference before decaying. How fast is it travelling?
2 marks
𝒍𝒗 = πŸ’πŸ‘πŸ“π’Ž
π’πŸŽ = πŸπŸŽπŸπŸ•π’Ž
𝒗 =?
𝒗 𝟐
𝒍𝒗 = π’πŸŽ √𝟏 − ( 𝒄 )
𝒍𝒗 𝟐
𝒗 𝟐
𝟏 − (𝒄) = ( )
π’πŸŽ
𝒍𝒗
𝒗 𝟐
= √𝟏 − ( 𝒄 )
π’πŸŽ
𝒍𝒗 𝟐
𝒗 𝟐
(𝒄) = 𝟏 − ( )
π’πŸŽ
𝒍𝒗 𝟐
𝒗
( 𝒄 ) = √𝟏 − ( )
π’πŸŽ
πŸ’πŸ‘πŸ“ 𝟐
𝒗
( 𝒄 ) = √𝟏 − (
) = 𝟎. πŸ—πŸŽπŸ”
𝒗 = 𝟎. πŸ—πŸŽπŸ”π’„ = 𝟐. πŸ•πŸ × πŸπŸŽπŸ– π’Ž ⋅ 𝒔−𝟏
πŸπŸŽπŸπŸ•
32. A blue car is twice as long as a red car when they are at rest. The cars pass a stationary policeman at the
same time and the policeman determines that the two cars have the same length. The red car is travelling at
0.6c. How fast is the blue car travelling?
3 marks
𝒍𝒃𝒍𝒖𝒆
= 𝟐 × π’π’“π’†π’…
𝟎
𝟎
√𝟏 − (
𝒍𝒓𝒆𝒅
= 𝒍𝒓𝒆𝒅
𝒗
𝟎
𝒍𝒃𝒍𝒖𝒆
𝒗
𝒍𝒃𝒍𝒖𝒆
𝟎
𝒃𝒍𝒖𝒆
𝒗
= √𝟏 − (
𝒄
𝟐
)
𝟐
𝒗𝒓𝒆𝒅
)
𝒄
𝒍𝒃𝒍𝒖𝒆
= 𝒍𝒓𝒆𝒅
𝒗
𝒗
√𝟏 − (
𝒍𝒃𝒍𝒖𝒆
= 𝒍𝒃𝒍𝒖𝒆
𝒗
𝟎
𝒍𝒓𝒆𝒅
𝒗
= √𝟏 − (
𝟐 × π’π’“π’†π’…
𝟎
𝟏
√𝟏
𝟐
𝟏 𝟐
𝒗𝒓𝒆𝒅 = 𝟎. πŸ”π’„
𝟐
𝒗𝒃𝒍𝒖𝒆
)
)
𝒄
𝒍𝒓𝒆𝒅
𝒗
𝟐
𝒗𝒃𝒍𝒖𝒆
)
𝒄
𝟐
𝒗𝒃𝒍𝒖𝒆
)
𝒄
− (𝟎.πŸ”)𝟐 = √𝟏 − (
( 𝟐) (𝟏 − (𝟎.πŸ”)𝟐 ) = (𝟏 − (
𝒗𝒃𝒍𝒖𝒆 =?
𝒍𝒓𝒆𝒅
𝟎
𝟏√
𝟏−
𝟐
(
= √𝟏 − (
𝟐
𝒗𝒓𝒆𝒅
)
𝒄
= √𝟏 − (
𝟐
𝒗𝒃𝒍𝒖𝒆
)
𝒄
𝒗𝒃𝒍𝒖𝒆
𝒄
𝟏 𝟐
= √𝟏 − ( 𝟐) (𝟏 − (𝟎.πŸ”)𝟐 )
𝒗𝒃𝒍𝒖𝒆 = 𝟎. πŸ—πŸπŸ•π’„ = 𝟐. πŸ•πŸ“ × πŸπŸŽπŸ– π’Ž ⋅ 𝒔−𝟏
𝟐
𝒗𝒓𝒆𝒅
)
𝒄
𝟐
𝒗𝒃𝒍𝒖𝒆
)
𝒄
Download