Figure 2 shows a yo-yo made of two discs separated by a cylindrical axle - AQA - A-Level Physics - Question 2 - 2021 - Paper 6

To calculate $v$, we can utilize the principle of conservation of energy. Initially, the potential energy (PE) of the yo-yo is converted into kinetic energy (KE) as it falls:

1. The initial potential energy when the yo-yo is at height $h = 0.50 ext{ m}$ is: $PE = mgh = (9.2 \times 10^{-2} ext{ kg})(9.81 ext{ m/s}^2)(0.50 ext{ m})$

2. The kinetic energy when it has fallen this distance is comprised of both translational and rotational kinetic energy: $KE = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2$ where $I$ is the moment of inertia, and we know that: $\omega = \frac{v}{r}$

3. Substituting for $I$ and $\\omega$ gives: $KE = \frac{1}{2} mv^2 + \frac{1}{2} (8.6 \times 10^5 ext{ kg m}^2) \left(\frac{v}{5.0 \times 10^{-3}}\right)^2$

4. Setting potential energy equal to total kinetic energy: $mgh = \frac{1}{2} mv^2 + \frac{1}{2} (8.6 \times 10^5) \left(\frac{v}{5.0 \times 10^{-3}}\right)^2$

5. By solving this equation for $v$, we find: $v^2 = \frac{2mgh}{m + \frac{I}{r^2}}$

After plugging in the numbers and solving, we will arrive at the value of $v = 1.5 ext{ m/s}$.