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Three particles are travelling in the same plane with velocities as shown in the vector diagram - AQA - A-Level Physics - Question 10 - 2021 - Paper 2

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Question 10

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Three particles are travelling in the same plane with velocities as shown in the vector diagram. What is the root mean square speed of the particles?

Worked Solution & Example Answer:Three particles are travelling in the same plane with velocities as shown in the vector diagram - AQA - A-Level Physics - Question 10 - 2021 - Paper 2

Step 1

Calculate the Squared Velocities

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Answer

First, we need to find the square of the velocities for each particle:

  • For the first particle, the velocity is 2m s12 \, \text{m s}^{-1}, so its squared velocity is (2)2=4m2s2(2)^2 = 4 \, \text{m}^2 \text{s}^{-2}.
  • For the second particle, the velocity is 4m s14 \, \text{m s}^{-1}, so its squared velocity is (4)2=16m2s2(4)^2 = 16 \, \text{m}^2 \text{s}^{-2}.
  • For the third particle, the velocity is 6m s16 \, \text{m s}^{-1}, so its squared velocity is (6)2=36m2s2(6)^2 = 36 \, \text{m}^2 \text{s}^{-2}.

Step 2

Sum the Squared Velocities

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Answer

Now sum the squared velocities: 4+16+36=56m2s24 + 16 + 36 = 56 \, \text{m}^2 \text{s}^{-2}

Step 3

Calculate the Root Mean Square Speed

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Answer

To find the root mean square speed, calculate the mean of the squared velocities and then take the square root:

  1. Calculate the mean: Mean=563=56318.67m2s2\text{Mean} = \frac{56}{3} = \frac{56}{3} \approx 18.67 \, \text{m}^2 \text{s}^{-2}

  2. Take the square root: RMS Speed=18.674.32m s1\text{RMS Speed} = \sqrt{18.67} \approx 4.32 \, \text{m s}^{-1}

Thus, rounding gives approximately 4.3m s14.3 \, \text{m s}^{-1}, which matches answer choice A.

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