Angular momentum $L = mvr_{\perp}$. The line equation is $X - Y + 4 = 0$.
The perpendicular distance $r_{\perp}$ from the origin $(0,0)$ to the line is $\frac{|0 - 0 + 4|}{\sqrt{1^2 + (-1)^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.
Thus, $L = 5 \times (3\sqrt{2}) \times (2\sqrt{2}) = 60 \text{ units}$.

By mechanical energy conservation, kinetic energy converts to spring potential energy: $\frac{3}{4}mv^2 = \frac{1}{2}kx^2$. Substituting the values gives $x = 0.6\text{ m}$.

The acceleration of a rolling body depends on its moment of inertia ratio $I/mR^2$. The solid cylinder has a smaller moment of inertia ratio ($1/2$) compared to the hollow cylinder ($1$), giving it a higher acceleration and causing it to reach the bottom first.

Static friction provides the necessary torque for rolling without slipping, converting translational kinetic energy into rotational kinetic energy without dissipating mechanical energy.

Using conservation of energy, potential energy equals total kinetic energy: $Mgh = \frac{3}{4}Mv^2$. Solving for velocity gives $v = \sqrt{\frac{4}{3}gh}$.

Acceleration of the centre of mass is given by $a = frac{F}{m} + frac{tau}{I}R_{text{eff}}$. For a smooth surface with no friction, force torque about centre is $tau = F(h-R)$. Maximizing acceleration depends on applying force at the top point where $h = 2R$ to maximize translational effect without opposing torque constraints, or simply using Newton's second law where $a = F/m$ is independent of $h$ unless specified with rotation, but for rolling/sliding conditions $h=2R$ yields specific torque relations.

Total kinetic energy of the rolling disc is $K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{3}{4}mv^2$. Substituting $m = 100\text{ kg}$ and $v = 0.2\text{ m/s}$ gives $K = 3\text{ J}$. Work required to stop it is equal to its total kinetic energy, which is $3\text{ J}$.

For a solid sphere, $K_t = \frac{1}{2}mv^2$ and $K_r = \frac{1}{5}mv^2$. The total kinetic energy is $K_t + K_r = \frac{7}{10}mv^2$. The ratio $K_t : (K_t + K_r)$ evaluates to $5 : 7$.

By conservation of angular momentum, the final common angular velocity is $\omega = \frac{\omega_1 + \omega_2}{2}$. The loss in rotational kinetic energy is $Delta E = E_i - E_f = \frac{1}{2}I\omega_1^2 + \frac{1}{2}I\omega_2^2 - 2 \cdot \left(\frac{1}{2}I\omega^2\right)$, which simplifies to $\frac{1}{8}I(\omega_1-\omega_2)^2$.

Acceleration of a rolling body on an inclined plane is given by $$a = \frac{g sin\theta}{1 + I/(mR^2)}$$. Since the acceleration is independent of mass and depends only on the geometry (moment of inertia factor), the sphere has a smaller inertia factor than the disk, meaning the sphere has a greater acceleration and reaches the bottom first.