The system consists of the ball ($m$) and the spring gun ($M$). Since the horizontal surface is frictionless, there are no external horizontal forces, so the system's momentum is conserved.

At maximum compression, the ball and the gun move together with a common final velocity, $V$. Conservation of momentum:

$$p_i = p_f$$ $$mv = (m+M)V$$ $$V = \frac{m}{m+M}v$$

The total mechanical energy of the system is also conserved. The initial kinetic energy of the ball is converted into the final kinetic energy of the combined system and the elastic potential energy ($E_p$) of the spring. Conservation of energy:

$$K_i = K_f + E_p$$ $$\frac{1}{2}mv^2 = \frac{1}{2}(m+M)V^2 + E_p$$

[Q1] Solve for the elastic potential energy $E_p$:

$$E_p = \frac{1}{2}mv^2 - \frac{1}{2}(m+M)V^2$$

Substitute the expression for $V$:

$$E_p = \frac{1}{2}mv^2 - \frac{1}{2}(m+M)\left(\frac{m}{m+M}v\right)^2$$ $$E_p = \frac{1}{2}mv^2 - \frac{1}{2}\frac{m^2 v^2}{m+M} = \frac{1}{2}mv^2\left(1 - \frac{m}{m+M}\right)$$ $$E_p = \frac{1}{2}mv^2\left(\frac{M}{m+M}\right)$$

[Q2] The ratio of the potential energy to the initial kinetic energy ($K_i = \frac{1}{2}mv^2$) is:

$$\frac{E_p}{K_i} = \frac{\frac{1}{2}mv^2\left(\frac{M}{m+M}\right)}{\frac{1}{2}mv^2} = \frac{M}{m+M}$$