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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 moved

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Answer

The floating magnet is influenced by the magnetic field of the fixed magnet. Since the north pole of the floating magnet is near the north pole of the fixed magnet, they repel each other. This repulsion causes the floating magnet to move away from the fixed magnet.

Step 2

What happened to the piece of iron?

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Answer

When the piece of iron was placed near the fixed magnet, it was attracted to the fixed magnet. This occurs because the iron becomes magnetized and behaves like a magnet, adhering to the fixed magnet.

Step 3

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

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Answer

To plot the magnetic field around a bar magnet using a compass:

  1. Place the compass near one pole of the bar magnet (north or south).
  2. Observe the direction in which the compass needle points, which indicates the direction of the magnetic field.
  3. Mark the position of the compass on a piece of paper.
  4. Move the compass to this marked point and repeat the process, moving around the magnet until you have several points marked.
  5. Connect the points with a smooth line to represent the magnetic field lines.
  6. Add arrows to indicate the direction of the field, pointing from the north pole to the south pole of the 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

The elastic potential energy can be calculated using the formula:

Ep=0.5imeskimesx2E_p = 0.5 imes k imes x^2

where:

  • EpE_p is the elastic potential energy,
  • kk is the spring constant (200 N/m), and
  • xx is the extension (0.040 m).

Substituting the values: Ep=0.5imes200imes(0.040)2=0.16extJE_p = 0.5 imes 200 imes (0.040)^2 = 0.16 ext{ J}

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