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A student placed a magnet on top of a plastic support in a bowl of water - AQA - GCSE Physics - Question 5 - 2020 - Paper 1

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Question 5

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A student placed a magnet on top of a plastic support in a bowl of water. This magnet was fixed in position and above the surface of the water. The student put a se... show full transcript

Worked Solution & Example Answer:A student placed a magnet on top of a plastic support in a bowl of water - AQA - GCSE Physics - Question 5 - 2020 - Paper 1

Step 1

Explain why the floating magnet is repelled and attracted.

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Answer

The north pole of the floating magnet is repelled from the north pole of the fixed magnet, while it is attracted to the south pole of the fixed magnet. This occurs because like poles repel each other and opposite poles attract.

Step 2

What happened to the piece of iron?

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Answer

The piece of iron was attracted to the fixed magnet. This happens because iron is a ferromagnetic material and can become magnetized in the presence of a magnetic field.

Step 3

Describe how to use a compass to plot the magnetic field pattern around a bar magnet.

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Answer

  1. Place the compass near one pole of the bar magnet. Observe the direction the compass needle points.
  2. Mark this position on the paper.
  3. Move the compass to this marked point and observe the needle again, marking the position.
  4. Repeat this process, moving the compass and marking the points along the path around the magnet.
  5. Join the marked points with smooth lines to represent the magnetic field lines, adding arrows from the north pole to the south pole.
  6. Continue this process around the magnet, including areas above and below the bar magnet.

Step 4

Write one letter in each box to show the correct sequence.

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Answer

A - C - B

Step 5

Calculate the elastic potential energy of the spring when the door is unlocked.

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Answer

Using the formula for elastic potential energy: Ee=0.5imeskimesx2E_e = 0.5 imes k imes x^2 where: k = 200 , \text{N/m}, x = 0.040 , \text{m}, we calculate:

Ee=0.5imes200imes(0.040)2E_e = 0.5 imes 200 imes (0.040)^2 Ee=0.5imes200imes0.0016E_e = 0.5 imes 200 imes 0.0016 Ee=0.16JE_e = 0.16 \, \text{J}

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