A toy gun has a spring of 400 N/m. It is squeezed by 5.0 cm behind a ball of 20 g, and released…

(a) $\tfrac12 mv^2 = \tfrac12 kx^2$. We want $v$, so $v = x\sqrt{\dfrac{k}{m}} = 0.050\ \text{m} \times \sqrt{\dfrac{400\ \text{N/m}}{0.020\ \text{kg}}} = 7.07\ \text{m/s}$.

(b) $T = 2\pi\sqrt{\dfrac{0.020\ \text{kg}}{400\ \text{N/m}}} = 0.0444\ \text{s}$. The ball goes from an end to the middle, a quarter of a period: $\dfrac{0.0444\ \text{s}}{4} = 0.0111\ \text{s}$, about $11\ \text{ms}$.