A particle of mass $m$ and charge $Q$ is accelerated from rest through a potential difference $V$. The final velocity of the particle is $u$. A second particle of mass $rac{m}{2}$ and charge $2Q$ is accelerated from rest through a potential difference $2V$. What is the final velocity of the second particle?

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A particle of mass $m$ and charge $Q$ is accelerated from rest through a potential difference $V$ - AQA - A-Level Physics - Question 15 - 2022 - Paper 2

Question 15

A particle of mass $m$ and charge $Q$ is accelerated from rest through a potential difference $V$. The final velocity of the particle is $u$. A second particle of m... show full transcript

Worked Solution & Example Answer:A particle of mass $m$ and charge $Q$ is accelerated from rest through a potential difference $V$ - AQA - A-Level Physics - Question 15 - 2022 - Paper 2

Step 1

Calculate the final velocity of the first particle

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Answer

Using the principle of energy conservation, the kinetic energy gained by the first particle can be expressed as:

$KE = QV = \frac{1}{2} mv^2$

Substituting the final velocity $u$ :

$QV = \frac{1}{2} mu^2$

From this, we can express the potential difference:

$V = \frac{mu^2}{2Q}$

Step 2

Calculate the final velocity of the second particle

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Answer

For the second particle with mass $rac{m}{2}$ and charge $2Q$ , the potential difference is $2V$ . The kinetic energy is:

$KE = 2QV = \frac{1}{2} \left( \frac{m}{2} \right) v^2$

Substituting for $V$ from the first particle:

$2\left(\frac{mu^2}{2Q}\right) = \frac{1}{4} mv^2$

Rearranging gives:

$\frac{mu^2}{Q} = \frac{1}{4} mv^2$

Cancelling $m$ and solving for $v$ :

$v^2 = 4u^2$

Thus, the final velocity of the second particle is:

$v = 2u$

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