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

To determine the final velocity of the second particle, we can use the principle of energy conservation. The kinetic energy (KE) gained by a charged particle when accelerated through a potential difference (V) is given by:

$KE = QV$

For the first particle:

• Mass = $m$, Charge = $Q$, Potential difference = $V$, Final velocity = $u$
• The kinetic energy is:

$KE_1 = QV = \frac{1}{2}mu^2$

For the second particle:

• Mass = $\frac{m}{2}$, Charge = $2Q$, Potential difference = $2V$, Final velocity = $v$
• The kinetic energy is:

$KE_2 = 2Q(2V) = 4QV$

Equating the kinetic energy for the second particle:

$4QV = \frac{1}{2} \left(\frac{m}{2}\right)v^2$

Simplifying this, we find:

$4QV = \frac{mv^2}{4}$

Multiplying both sides by 4 gives us:

$16QV = mv^2$

This allows us to express the final velocity $v$ as follows:

$v = \sqrt{\frac{16QV}{m}}$

We already know from the first particle that:

$u = \sqrt{\frac{2QV}{m}}$

Squaring both sides of this equation:

$u^2 = \frac{2QV}{m}$

Now, substituting $\frac{QV}{m}$ from this equation back into our previous result, we can express it in terms of $u$:

$v = \sqrt{8} \cdot u = 2\sqrt{2}u$

Thus, the final velocity of the second particle is:

$v = 2\sqrt{2}u$