Find the maximum KE of a charged Disc

[Hamiltonian]

Homework Statement

A thin disc of Radius R is held closing the opening of a thin hemispherical shell of the same radius. Both the bodies are made of insulating materials and have a uniform surface charge density $\sigma$ and surface mass density. When the system is released what will be the maximum KE of the disc?

Relevant Equations

$V = \frac{K q_1 q_2}{r}$, Conservation of energy&linear momentum

To find the initial potential energy of the system we can assume the disc to be placed inside a hollow sphere of the same radius and $\sigma$, the potential energy inside a charged hollow shell is:
$$V = \frac{\sigma(4\pi R^2)}{4\pi \epsilon_0 R} = \frac{\sigma R}{\epsilon_0}$$
the potential energy of the sphere and disc system will be:
$$U = V(\sigma \pi R^2) = \frac{\sigma^2\pi R^3}{\epsilon_0}$$
$U$ is the potential energy due to the disc and the sphere, but we need to find the potential energy of the disc and hemisphere system, the sphere can be assumed to be made up of two hemispheres hence each contributing to $U/2$
therefore the initial potential energy is:
$$U_0 = \frac{\sigma^2 \pi R^3}{2\epsilon_0}$$
the disc will have maximum KE when it's at $\infty$ so the potential energy of the system $\rightarrow 0$
(assume surface mass density of both hemisphere and disc be $\lambda$)
hence we can conserve energy and linear momentum and solve for the maximum KE of the disc
$$\frac{\sigma^2 \pi R^3}{2\epsilon} = \frac{1}{2}(\lambda \pi R^2){v_d}^2 + \frac{1}{2}(\lambda 4\pi R^2){v_h}^2$$
$$\lambda \pi R^2 v_d = \lambda 4 \pi R^2 v_h \implies v_h = \frac{v_d}{4}$$
where $v_h$ is the velocity of the hemisphere and $v_d$ is the velocity of the disc at $\infty$
solving the above equations gives:
$${v_d}^2 = \frac{4}{5}\frac{\sigma^2 R}{\epsilon_0 \lambda} \implies{KE}_{max} = \frac{2}{5}\frac{\sigma^2 \pi R^3}{\epsilon_0}$$

I don't have the correct answer to this question so can someone check my work and the final answer?

 

[Steve4Physics]

hence we can conserve energy and linear momentum and solve for the maximum KE of the disc
$$\frac{\sigma^2 \pi R^3}{2\epsilon} = \frac{1}{2}(\lambda \pi R^2){v_d}^2 + \frac{1}{2}(\lambda 4\pi R^2){v_h}^2$$
$$\lambda \pi R^2 v_d = \lambda 4 \pi R^2 v_h \implies v_h = \frac{v_d}{4}$$
where $v_h$ is the velocity of the hemisphere and $v_d$ is the velocity of the disc at $\infty$
solving the above equations gives:
$${v_d}^2 = \frac{4}{5}\frac{\sigma^2 R}{\epsilon_0 \lambda} \implies{KE}_{max} = \frac{2}{5}\frac{\sigma^2 \pi R^3}{\epsilon_0}$$

It looks like you have considered the disc and an entire spherical shell moving away from each other.

The disc’s area is $\pi R^2$ and the hemisphere’s is $2\pi R^2$ (not $4\pi R^2$). I.e. the mass of the disc is half the mass of the hemisphere. Conservation of momentum then immediately tells you that $v_h= \frac {v_d}{2}$ in the centre-of-mass frame.

To get the split in kinetic energy you might consider using the handy relationship $KE = \frac{p^2}{2m}$ and using simple proportionality.