Since the horizontal surface is smooth, the mechanical energy of the mass-spring system is conserved. Let the initial state be when the ball is at displacement $x_1$ and released from rest ($v_1=0$), and the final state be when the ball is at displacement $x_2$ with velocity $v_2$.

The conservation of mechanical energy is expressed as:

$$E_1 = E_2$$ $$\frac{1}{2}mv_1^2 + \frac{1}{2}kx_1^2 = \frac{1}{2}mv_2^2 + \frac{1}{2}kx_2^2$$

Since $v_1 = 0$:

$$\frac{1}{2}kx_1^2 = \frac{1}{2}mv_2^2 + \frac{1}{2}kx_2^2$$ $$mv_2^2 = k(x_1^2 - x_2^2)$$ $$v_2 = \sqrt{\frac{k}{m}(x_1^2 - x_2^2)}$$

Substitute the given values in SI units ($m=0.1$ kg, $x_1=0.1$ m, $x_2=0.06$ m):

$$v_2 = \sqrt{\frac{360}{0.1}((0.1)^2 - (0.06)^2)}$$ $$v_2 = \sqrt{3600(0.01 - 0.0036)} = \sqrt{3600(0.0064)}$$ $$v_2 = \sqrt{23.04} = 4.8 \text{ m/s}$$