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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\ \text{J}`
  • Hence 0.30576 joules 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\ \text{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\ \text{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\ \text{J}`
  • Hence 0.30576 joules 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\ \text{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\ \text{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%.