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PHYSICS, M5 2019 VCE 4 MC

The magnitude of the acceleration due to gravity at Earth's surface is \(g\).

Planet \(\text{Y}\) has twice the mass and half the radius of Earth. Both planets are modelled as uniform spheres.

Which one of the following best gives the magnitude of the acceleration due to gravity on the surface of Planet \(\text{Y}\)?

  1. \(\dfrac{1}{2} g\)
  2. \(1g\)
  3. \(4g\)
  4. \(8g\)
Show Answers Only

\(D\)

Show Worked Solution

\(g=\dfrac{GM}{r^2}\)

\(\text{Planet Y:}\ \dfrac{G \times 2M}{(\frac{r}{2})^2}=\dfrac{2GM}{\frac{r^2}{4}}=8 \times \dfrac{GM}{r^2}=8g\)

\(\Rightarrow D\)

Mean mark 58%.

PHYSICS, M5 2020 VCE 2*

Jupiter's moon Ganymede is its largest satellite.

Ganymede has a mass of 1.5 × 10\(^{23}\) kg and a radius of 2.6 × 10\(^6\) m.

Determine the magnitude of Ganymede's surface gravity?   (2 marks)

Show Answers Only

\(1.5\ \text{ms}^{-2}\)

Show Worked Solution
\(F\) \(=\dfrac{GMm}{r^2}\)  
\(mg\) \(=\dfrac{GMm}{r^2}\)  
\(g\) \(=\dfrac{GM}{r^2}\)  
  \(=\dfrac{6.67 \times 10^{-11} \times 1.5 \times 10^{23}}{(2.6 \times 10^6)^2}\)  
  \(=1.5\ \text{ms}^{-2}\)  

PHYSICS, M5 2021 VCE 4*

The planet Phobetor has a mass four times that of Earth. Acceleration due to gravity on the surface of Phobetor is 18 m s\(^{-2}\).

If Earth has a radius \(R\), calculate the radius of Phobetor in terms of \(R\)?   (2 marks)

Show Answers Only

\(\sqrt{2.22} \times R\)

Show Worked Solution

\(g=\dfrac{GM}{R^2}\ \Rightarrow \  R=\sqrt{\dfrac{GM}{g}}\)

\(\text{Phobetor:}\ M → 4M\ \ \text{and}\ \ g → 1.8g\)

\(\text{Radius of Phobetor}\) \(=\sqrt{\dfrac{4GM}{1.8g}}\)  
  \(=\sqrt{\dfrac{2.22GM}{g}}\)  
  \(=\sqrt{2.22} \times \sqrt{\dfrac{GM}{g}}\)  
  \(=\sqrt{2.22} \times R\)  
♦ Mean mark 44%.

PHYSICS, M5 2023 VCE 2

Phobos is a small moon in a circular orbit around Mars at an altitude of 6000 km above the surface of Mars.

The gravitational field strength of Mars at its surface is 3.72 N kg\(^{-1}\). The radius of Mars is 3390 km.

  1. Show that the gravitational field strength 6000 km above the surface of Mars is 0.48 N kg\(^{-1}\).   (2 marks)
  1. Calculate the orbital period of Phobos. Give your answer in seconds.  (3 marks)
  1. Phobos is very slowly getting closer to Mars as it orbits.
  2. Will the orbital period of Phobos become shorter, stay the same or become longer as it orbits closer to Mars? Explain your reasoning.  (2 marks)

Show Answers Only

a.    \(0.48\ \text{N kg}^{-1}\)

b.    \(2.77 \times 10^4\ \text{s}\)

c.    The orbital period will become shorter.

Show Worked Solution

a.    \(\text{Find the mass of Mars:}\)

         \(g\) \(=\dfrac{GM}{r^2}\)
  \(M\) \(=\dfrac{gr^2}{G}\)
    \(=\dfrac{3.72 \times (3390 \times 10^3)^2}{6.67 \times 10^{-11}}\)
    \(=6.409 \times 10^{23}\)

 
\(\text{Find the gravitational field strength where}\ r= 9390\ \text{km:}\) 

         \(g\) \(=\dfrac{GM}{r^2}\)
    \(=\dfrac{6.67 \times 10^{-11} \times 6.41 \times 10^{23}}{(9390000)^2}\)
    \(=0.48\ \text{N kg}^{-1}\)

 

b.     \(\dfrac{r^3}{T^2}\) \(=\dfrac{GM}{4\pi^2}\)
  \(T\) \(=\sqrt{\dfrac{4\pi^2 r^3}{GM}}\)
    \(=\sqrt{\dfrac{4\pi^2 (9390000)^3}{6.67 \times 10^{-11} \times 6.409 \times 10^{23}}}\)
    \(=27652\ \text{s}\)
    \(=2.77 \times 10^4\ \text{s}\)

♦ Mean mark (b) 53%. 
COMMENT: Rounded calculations, such as using 6.41 × 10^23 were marked as wrong.

c.    \(T=\sqrt{\dfrac{4\pi^2 r^3}{GM}}\)

\(T \propto \sqrt{r^3}\)

→ \(\text{decreasing the orbital radius will also decrease the orbital period.}\)

PHYSICS, M5 2023 VCE 3 MC

Space scientists want to place a satellite into a circular orbit where the gravitational field strength of Earth is half of its value at Earth's surface.

Which one of the following expressions best represents the altitude of this orbit above Earth's surface, where \(R\) is the radius of Earth?

  1. \(\dfrac{\sqrt{2} R}{2}-R\)
  2. \(\sqrt{2} R\)
  3. \((\sqrt{2} R)-R\)
  4. \(2 R-\sqrt{2} R\)
Show Answers Only

\(C\)

Show Worked Solution

→ For gravitational field strength, \(r \propto \sqrt{\dfrac{1}{g}}\)

→ Therefore if \(g\) → \(\dfrac{g}{2}\), the radius will become \(\sqrt{2}R\).

→ Thus the altitude above the surface of the earth is \(\sqrt{2}R\ -\ R\).

\(\Rightarrow C\)

♦♦ Mean mark 39%.

PHYSICS, M5 2023 HSC 23a

The James Webb Space Telescope (JWST) has a mass of 6.1 × 10³ kg and orbits the Sun at a distance of approximately 1.52 × 10\(^{11}\) m.

The Sun has a mass of 1.99 × 10\(^{30}\) kg.

Calculate the magnitude of gravitational force the Sun exerts on the JWST.  (2 marks)

Show Answers Only

\(F= 35\) \( \text{N}\)

Show Worked Solution

\(F\) \(=\dfrac{GMm}{r^2}\)  
  \(=\dfrac{ 6.67 \times 10^{-11} \times 1.99 \times 10^{30} \times 6.1 \times 10^3}{(1.52 \times 10^{11} )^2}\)  
  \(=35\) \(\text{N}\)  

PHYSICS, M5 2023 HSC 1 MC

The gravitational field strength acting on a spacecraft decreases as its altitude increases.

This is due to a change in the

  1. mass of Earth.
  2. mass of the spacecraft.
  3. density of the atmosphere.
  4. distance of the spacecraft from Earth's centre.
Show Answers Only

\(D\)  

Show Worked Solution

\(g=\dfrac{GM}{r^2}\ \ \Rightarrow\ \ g \propto \dfrac{1}{r^2}\)

\(\therefore\) As r increases, the gravitational field strength decreases

\( \Rightarrow D\)

PHYSICS, M5 EQ-Bank 21

Planet `X` has a mass eight times that of Earth. The acceleration due to gravity on the surface of this planet is half that on the surface of Earth.

If Earth has a radius of 1, what is the radius of Planet `X` ?   (3 marks)

Show Answers Only

`4`

Show Worked Solution

Radius of Earth:

`F_(g)` `=(GM_em)/(r_e)^2`  
`g` `=(F)/(m)`  
  `=(GM_e)/(r_e)^2`  
`:.r_(e)` `=sqrt((GM_e)/(g))`  

 
Radius of planet `X`:

`(g)/(2)` `=(G xxM_x)/(r_x)^2`  
  `=(G xx8M_e)/(r_x)^2`  
`:.r_(x)` `=sqrt((16GM_e)/(g))`  
  `=4r_e`  

 
`:.` Since Earth has a radius of 1 `=>\ r_(x)=4`

PHYSICS, M5 2015 HSC 19 MC

An astronaut working outside a spacecraft in orbit around Earth is not attached to it.

Why does the astronaut NOT drift away from the spacecraft?

  1. The force of gravity acting on the astronaut and spacecraft is negligible.
  2. The spacecraft and the astronaut are in orbit around the Sun with the Earth.
  3. The forces due to gravity acting on both the astronaut and the spacecraft are the same.
  4. The accelerations of the astronaut and the spacecraft are inversely proportional to their respective masses.
Show Answers Only

`D`

Show Worked Solution

By Elimination:

→ Gravity is the force keeping the spacecraft and the astronaut in orbit, so it is not negligible (eliminate A).

→ Earths rotation around the sun is not relevant to the motion of the astronaut and the spacecraft relative to Earth (eliminate B).

→ Using  `F=(GMm)/(r^2)`, the force due to gravity acting on both is proportional to their respective masses. Since the spacecraft is significantly heavier, it will experience a greater force due to gravity (eliminate C).

→ The accelerations, `g=(F)/(m)` of the spacecraft and the astronaut are inversely proportional to their respective masses. As they both experience a force, `F`, due to gravity proportional to their masses, their accelerations are the same. Therefore, the astronaut will not drift from the spacecraft.

`=>D`


♦♦♦ Mean mark 15%.

PHYSICS, M5 2015 HSC 11 MC

Which of the following diagrams correctly represents the force(s) acting on a satellite in a stable circular orbit around Earth?
 

 

Show Answers Only

`D`

Show Worked Solution

→ In a stable, circular orbit the satellite requires no propulsion force to keep it in orbit. This is due to it undergoing uniform circular motion with gravity acting as its centripetal force.

→ A force of gravity acts on the satellite, with direction towards Earth’s centre of mass. 

→ No ‘reaction force’ acts on the satellite. This is because any ‘reaction’ to the gravitational force exerted on the satellite by the Earth is an equal and opposite force exerted on the Earth by the satellite (Newton’s Third Law).

→ i.e. The reaction force acts on the Earth, not the satellite.

`=>D`


♦♦♦ Mean mark 29%.

PHYSICS, M7 2016 HSC 28

The following diagram shows the acceleration of a rocket during the first stage of its launch.
 

Explain the acceleration of the rocket with reference to the law of conservation of momentum.   (5 marks)

Show Answers Only

→ The combustion of fuel and expulsion of gases causes propulsion of the rocket. The rocket exerts a force on these gases in order to expel them backwards.

→ An equal and opposite force is exerted back on the rocket by the gases (Newton’s Third Law).

→ As the rocket and gases form a closed system, the law of conservation of momentum applies, meaning the momentum of the rocket is equal in magnitude and opposite in direction to the momentum of the fuel.

→ The momentum of the rocket is given by  `p=mv`. Since the force (and hence momentum) of the fuel is constant, the velocity of the rocket must increase as its mass decreases due to the expulsion of its fuel.

→ The rocket will accelerate. Also, using  `F=ma`, as the force acting on the rocket due to the gases is constant, the acceleration of the rocket must increase as its mass decreases.

Show Worked Solution

→ The combustion of fuel and expulsion of gases causes propulsion of the rocket. The rocket exerts a force on these gases in order to expel them backwards.

→ An equal and opposite force is exerted back on the rocket by the gases (Newton’s Third Law).

→ As the rocket and gases form a closed system, the law of conservation of momentum applies, meaning the momentum of the rocket is equal in magnitude and opposite in direction to the momentum of the fuel.

→ The momentum of the rocket is given by  `p=mv`. Since the force (and hence momentum) of the fuel is constant, the velocity of the rocket must increase as its mass decreases due to the expulsion of its fuel.

→ The rocket will accelerate. Also, using  `F=ma`, as the force acting on the rocket due to the gases is constant, the acceleration of the rocket must increase as its mass decreases.


♦ Mean mark 52%.

PHYSICS, M5 2016 HSC 25

Two teams carried out independent experiments with the purpose of investigating Newton's Law of Universal Gravitation. Each team used the same procedure to accurately measure the gravitational force acting between two spherical masses over a range of distances.

The following graphs show the data collected by each team.
 

 

  1. Compare qualitatively the relationship between force and distance in the graphs.   (2 marks)
  1. Assess the appropriateness of Team A's data and Team B's data in achieving the purpose of the experiments.   (3 marks)
Show Answers Only

a.    Force vs Distance relationship:

→ Both graphs show that as distance between the centres of masses increases, force increases.

→ Team A’s data shows that the force between masses is inversely proportional to the square of the distance between them.

→ Team B’s data shows that force decreases at a constant rate.
 

b.   Data appropriateness:

→ Team A’s data set uses a good range of distances however it is inappropriate as it incorporates too few measurements for a valid relationship to be determined.

→ Team B’s data set is inappropriate as it does not incorporate a sufficient range of distances to determine a valid relationship although it has a sufficient number of measurements.

Show Worked Solution

a.    Force vs Distance relationship:

→ Both graphs show that as distance between the centres of masses increases, force increases.

→ Team A’s data shows that the force between masses is inversely proportional to the square of the distance between them.

→ Team B’s data shows that force decreases at a constant rate.
 

b.   Data appropriateness:

→ Team A’s data set uses a good range of distances however it is inappropriate as it incorporates too few measurements for a valid relationship to be determined.

→ Team B’s data set is inappropriate as it does not incorporate a sufficient range of distances to determine a valid relationship although it has a sufficient number of measurements.


Mean mark (b) 56%.

PHYSICS, M5 2016 HSC 21

  1. Why does orbital decay occur more rapidly for satellites in a low-Earth orbit than for satellites in other orbits?   (2 marks)
  1. Calculate the magnitude of the gravitational force that acts on a 50 kg satellite when it is 8000 km from Earth's centre.   (3 marks)
Show Answers Only

a.    Satellites in a low-Earth orbit:

→ Encounter more atmospheric drag than satellites in higher orbits due to their lower altitude.

→ This reduces their orbital velocity which decreases orbital radius which in turn causes orbital decay.
 

b.   312.7 N

Show Worked Solution

a.    Satellites in a low-Earth orbit:

→ Encounter more atmospheric drag than satellites in higher orbits due to their lower altitude.

→ This reduces their orbital velocity which decreases orbital radius which in turn causes orbital decay.


Mean mark (a) 59%.
b.    `F` `=(GMm)/(r^(2))`
    `=(6.67 xx10^(-11)xx6xx10^(24)xx50)/((8000 xx10^(3))^(2))`
    `=312.7\ text{N}`

PHYSICS, M5 2016 HSC 17 MC

A projectile was launched horizontally inside a lift in a building. The diagram shows the path of the projectile when the lift was stationary.
 

The projectile was launched again with the same velocity. At this time, the lift was slowing down as it approached the top floor of the building.

Which diagram correctly shows the new path of the projectile (dotted line) relative to the path created in the stationary lift (solid line)?
 

 

Show Answers Only

`C`

Show Worked Solution

→ Due to the motion of the elevator, the downwards acceleration of the ball is less than when the elevator is stationary.

→ The new path that the ball takes will appear wider relative to its path in the stationary elevator.

`=>C`


♦ Mean mark 46%.

PHYSICS, M5 2018 HSC 21

  1. Compare the force of gravity exerted on the moon by Earth with the force of gravity exerted on Earth by the moon.   (2 marks)
  1. The acceleration due to gravity on the moon is `1.6 \ text{m s}^(-2)` and on Earth it is `9.8 \ text{m s}^(-2)`. Quantitatively compare the mass and weight of a 70 kg person on the moon and on Earth.   (2 marks)
Show Answers Only

a.   Using Newton’s Third Law:

→ The force of gravity of the Earth on the moon is equal in magnitude and opposite in direction to the force of gravity exerted on Earth by the moon.
 

b.   Comparison of mass:

→ The mass of the person on both Earth and the moon is 70 kg.
 

Comparison of weight:

→ The weight of the person on Earth is given by  `W_e=mg=70 xx9.8=686\ text{N.}`

→ The weight of the person on the moon is given by  `W_m=mg=70xx1.6=112\ text{N.}`

→ The persons weight on Earth is greater than it is on the moon.

Show Worked Solution

a.   Using Newton’s Third Law:

→ The force of gravity of the Earth on the moon is equal in magnitude and opposite in direction to the force of gravity exerted on Earth by the moon.


♦♦ Mean mark (a) 33%.
 

b.   Comparison of mass:

→ The mass of the person on both Earth and the moon is 70 kg.
 

Comparison of weight:

→ The weight of the person on Earth is given by  `W_e=mg=70 xx9.8=686\ text{N.}`

→ The weight of the person on the moon is given by  `W_m=mg=70xx1.6=112\ text{N.}`

→ The persons weight on Earth is greater than it is on the moon.

PHYSICS, M5 2018 HSC 7 MC

A planet `X` has twice the mass and twice the radius of Earth.

What is the magnitude of the gravitational acceleration close to the surface of planet `X`?

  1. `(1)/(2)  g`
  2. `1\ g`
  3. `2\ g`
  4. `4\ g`
Show Answers Only

`A`

Show Worked Solution

Suppose Earth has a mass of `M` and a radius of `r`:

`F=mg_1=(GMm)/(r^2)\ \ => \ \ g_1=(GM)/(r^2)`
 

If `X` has a mass of `2M` and a radius of `2r`:

`g_2` `=(2GM)/((2r)^2)`  
  `=(2GM)/(4r^2)`  
  `=(GM)/(2r^2)`  
  `=(g_1)/(2)`  

 
`=>A`


Mean mark 53%.

PHYSICS, M5 2018 HSC 1 MC

A satellite orbits Earth as shown.

Which diagram correctly shows the direction of the satellite's acceleration?
 

 

 

Show Answers Only

`A`

Show Worked Solution

The direction of acceleration is the same as the direction of force acting on the satellite, which is towards the centre of mass of the Earth.

`=> A`


Mean mark 64%.
COMMENT: A surprising one third of students answered incorrectly.

PHYSICS, M6 2020 HSC 33

A strong magnet of mass 0.04 kg falls 0.78 m under the action of gravity from position `X` above a hollow copper cylinder. It then travels a distance of 0.20 m through the cylinder from `Y` to `Z` before falling freely again.
 

The magnet takes 0.5 seconds to pass through the cylinder. The displacement-time graph of the magnet is shown.
 

Analyse the motion of the magnet by applying the law of conservation of energy.

Your analysis should refer to gravity and the copper cylinder, and include both qualitative and quantitative information.   (9 marks)

Show Answers Only

During the first 0.4 seconds:

→  The magnet is accelerating downwards at 9.8 m s–2 due to gravity.

→  The magnet’s gravitational potential energy is being converted into kinetic energy consistent with the law of conservation of energy.

→  Quantitatively:

`Delta E_(k)` `= Delta U`  
  `=mg Delta h`  
  `=0.04 xx9.8 xx0.78`  
  `=0.30576` J  

 

→  Hence 0.30576 J of the magnet’s gravitational potential energy is converted into kinetic energy as it falls under gravity.

 
As magnet reaches the copper cylinder:

→  Its downwards motion causes the formation of induced currents in the cylinder which produce a magnetic field that opposes the magnet’s motion (Lenz’s law).

→  This causes the magnet to decelerate to 0.4 m s–1 and lose kinetic energy. 

→  Finding the magnets speed before entering the copper cylinder:

`E_(k)` `=(1)/(2)mv^2`  
`0.30576` `=(1)/(2)xx 0.04xx v^2`  
`v` `=3.91`  m s–1  

 
→  Quantifying the kinetic energy loss:

`Delta E_(k)` `=(1)/(2)mv^2 – (1)/(2)m u^2`  
  `=(1)/(2)xx 0.04xx 3.91^2 – (1)/(2)xx 0.04xx 0.4^2`  
  `=0.30256` J  

 
→  Applying the law of conservation of energy shows that 0.30256 J of the magnet’s kinetic energy is being converted to heat energy within the cylinder as the magnet decelerates.

→  Finally, as the magnet passes through the copper cylinder, its gravitational potential energy decreases while its velocity remains constant.

→  Quantifying the decrease in gravitational potential energy:

`Delta U` `=mg Delta h`  
  `=0.04xx 9.8xx 0.2`  
  `=0.0784`  

 

→  Hence, 0.0784 J of the magnet’s gravitational potential energy is converted into heat energy in the cylinder, consistent with the law of conservation of energy.

Show Worked Solution

During the first 0.4 seconds:

→  The magnet is accelerating downwards at 9.8 m s–2 due to gravity.

→  The magnet’s gravitational potential energy is being converted into kinetic energy consistent with the law of conservation of energy.

→  Quantitatively:

`Delta E_(k)` `= Delta U`  
  `=mg Delta h`  
  `=0.04 xx9.8 xx0.78`  
  `=0.30576` J  

 

→  Hence 0.30576 J of the magnet’s gravitational potential energy is converted into kinetic energy as it falls under gravity.

 
As magnet reaches the copper cylinder:

→  Its downwards motion causes the formation of induced currents in the cylinder which produce a magnetic field that opposes the magnet’s motion (Lenz’s law).

→  This causes the magnet to decelerate to 0.4 m s–1 and lose kinetic energy. 

→  Finding the magnets speed before entering the copper cylinder:

`E_(k)` `=(1)/(2)mv^2`  
`0.30576` `=(1)/(2)xx 0.04xx v^2`  
`v` `=3.91` m s–1  

 
→  Quantifying the kinetic energy loss:

`Delta E_(k)` `=(1)/(2)mv^2 – (1)/(2)m u^2`  
  `=(1)/(2)xx 0.04xx 3.91^2 – (1)/(2)xx 0.04xx 0.4^2`  
  `=0.30256` J  

 
→  Applying the law of conservation of energy shows that 0.30256 J of the magnet’s kinetic energy is being converted to heat energy within the cylinder as the magnet decelerates.

→  Finally, as the magnet passes through the copper cylinder, its gravitational potential energy decreases while its velocity remains constant.

→  Quantifying the decrease in gravitational potential energy:

`Delta U` `=mg Delta h`  
  `=0.04xx 9.8xx 0.2`  
  `=0.0784`  

 

→  Hence, 0.0784 J of the magnet’s gravitational potential energy is converted into heat energy in the cylinder, consistent with the law of conservation of energy.

♦ Mean mark 51%.

PHYSICS M5 2022 HSC 35

A capsule travels around the International Space Station (ISS) in a circular path of radius 200 m as shown.
 


 

Analyse this system to test the hypothesis below.  (5 marks)

The uniform circular motion of the capsule around the ISS can be accounted for in terms of the gravitational force between the capsule and the ISS.

Show Answers Only

Find the gravitational force between the capsule and the ISS:

`F` `=(GMm)/(r^2)`  
  `=(6.67 xx10^-11 xx4.2 xx10^5 xx 1.2 xx10^4)/(200^2)`  
  `=8.4 xx10^(-6)  text{N}`  

 
Find the centripetal force required to keep the capsule in its orbit:

`F_(c)` `=(mv^2)/(r)`  
  `=(1.2 xx10^4 xx 0.233^2)/(200)`  
  `=3.26  text{N}`  

 

→ The gravitational force is not sufficient to provide the necessary centripetal force to keep the capsule in orbit around the ISS.

→ The hypothesis is invalid.

Show Worked Solution

Find the gravitational force between the capsule and the ISS:

`F` `=(GMm)/(r^2)`  
  `=(6.67 xx10^-11 xx4.2 xx10^5 xx 1.2 xx10^4)/(200^2)`  
  `=8.4 xx10^(-6)  text{N}`  

 
Find the centripetal force required to keep the capsule in its orbit:

`F_(c)` `=(mv^2)/(r)`  
  `=(1.2 xx10^4 xx 0.233^2)/(200)`  
  `=3.26  text{N}`  

 

→ The gravitational force is not sufficient to provide the necessary centripetal force to keep the capsule in orbit around the ISS.

→ The hypothesis is invalid.

PHYSICS M6 2022 HSC 18 MC

A charged oil droplet was observed between metal plates, as shown.
 


 

While the switch was open, the oil droplet moved downwards at a constant speed. After the switch was closed, the oil droplet moved upwards at the same constant speed.

Assume that the only three forces that may act on the oil droplet are the force of gravity, the force due to the electric field and the frictional force between the air and the oil droplet. The magnitudes of these forces are `F_G` (due to gravity), `F_E` (due to the electric field) and `F_F` (due to the frictional force).

Which row of the table shows all the forces affecting the motion of the oil droplet in the direction indicated, and the relationship between these forces?
 

Show Answers Only

`D`

Show Worked Solution

→ For the droplet to move at a constant speed, the net force acting on it must be zero.

→ For the downwards motion, this means the downwards gravitational force is equal in magnitude to the upwards frictional force (i.e. `F_(G)=F_(F)`).

→ For the upwards motion, this means the upwards electric force is equal in magnitude to the sum of the downwards frictional and gravitational forces (i.e. `F_(E)=F_(G)+F_(F)`).

`=>D`


♦♦♦ Mean mark 26%.

PHYSICS, M6 2021 HSC 27

A student is considering how to levitate a thin metal rod in a strong magnetic field of 1.2 T. The current flowing through the rod will be 2.3 A.

  1. Use a labelled diagram to show a suitable orientation of the current and the magnetic field to achieve this outcome. Include relevant forces in your diagram.   (3 marks)
     
     
  1. Explain why the maximum mass per unit length of the rod cannot exceed 0.282 kg m\(^{-1}\). Support your answer with a calculation.   (3 marks)
Show Answers Only

a.


 

b.   The upwards magnetic force → \(F=IlB\, \sin \theta\)

Levitation will occur when

\(W\) \(=F_B\)  
\(mg\) \(=IlB\, \sin \theta\)  
\(\dfrac{m}{l}\) \(=\dfrac{IB\, \sin \theta}{g}\)  
  \(=\dfrac{2.3 \times 1.2 \times \sin\, 90^{\circ}}{9.8}\)  
  \(=0.282\ \text{kg m}^{-1}\)  

 
If the mass per unit length exceeds this, the rod would not levitate.

Show Worked Solution

a.

Note: a different alignment with the magnetic force coming out of the page and the current moving left is also correct.
 

b.   The upwards magnetic force → \(F=IlB\, \sin \theta\)

Levitation will occur when

\(W\) \(=F_B\)  
\(mg\) \(=IlB\, \sin \theta\)  
\(\dfrac{m}{l}\) \(=\dfrac{IB\, \sin \theta}{g}\)  
  \(=\dfrac{2.3 \times 1.2 \times \sin\, 90^{\circ}}{9.8}\)  
  \(=0.282\ \text{kg m}^{-1}\)  

 
If the mass per unit length exceeds this, the rod would not levitate.

PHYSICS, M5 2021 HSC 14 MC

Which of the following statements correctly describes the gravitational interaction between the Earth and the Moon?

  1. The Earth accelerates towards the Moon.
  2. The net force acting on the Earth is zero.
  3. The Moon and Earth experience equal and opposite accelerations.
  4. The force acting on the Moon is smaller than the force acting on the Earth.
Show Answers Only

`A`

Show Worked Solution

According to Newton’s Third Law, the earth and moon experience equal forces in opposite directions. This causes them to accelerate towards each other.

`=>A`


♦ Mean mark 19%.