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Small water drops leave a tap with zero velocity at intervals of 0.20 s - AQA - A-Level Physics - Question 21 - 2020 - Paper 1

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Small water drops leave a tap with zero velocity at intervals of 0.20 s. They then fall freely 0.80 m to reach a horizontal surface. How far has a drop fallen when ... show full transcript

Worked Solution & Example Answer:Small water drops leave a tap with zero velocity at intervals of 0.20 s - AQA - A-Level Physics - Question 21 - 2020 - Paper 1

Step 1

Calculate the time taken for the drop to hit the surface

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Answer

Using the formula for free fall, we can calculate the time taken for the drop to fall 0.80 m:

d = rac{1}{2} g t^2

where:

  • dd is the distance fallen (0.80 m)
  • gg is the acceleration due to gravity (approximately 9.81 m/s²)
  • tt is the time in seconds.

Rearranging for tt, we get:

t = ext{sqrt} rac{2d}{g} = ext{sqrt} rac{2 imes 0.80}{9.81} \approx 0.40 ext{ s}

Step 2

Determine the distance fallen by subsequent drops

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Answer

Since the drops are released every 0.20 s, there will be multiple drops in the time taken for the first drop to hit the surface:

  1. First drop: Falls for 0.40 s Using the formula: d_1 = rac{1}{2} g t^2 = rac{1}{2} imes 9.81 imes (0.40)^2 = 0.78 ext{ m}

    Correction for time intervals: The previous drop hits the surface after that time, so for 0.20 s time interval: 2. Second drop: Falls for 0.20 s d_2 = rac{1}{2} g (0.20)^2 = rac{1}{2} imes 9.81 imes (0.20)^2 \approx 0.20 ext{ m}

Therefore, the drop has fallen 0.20 m when the previous drop hits the surface.

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