Answer
(a) $v_{max} = \sqrt{\frac{kx_0^2 + mv_0^2}{m}}$
(b)$x_{max} = \sqrt{\frac{kx_0^2 + mv_0^2}{k}}$
Work Step by Step
$Energy = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2$
(a) The maximum speed will occur when the potential energy in the spring is 0 and all the energy in the system is kinetic energy:
$\frac{1}{2}mv_{max}^2 = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2$
$v_{max} = \sqrt{\frac{kx_0^2 + mv_0^2}{m}}$
(a) The maximum stretch will occur when the kinetic energy in the spring is 0 and all the energy in the system is potential energy:
$\frac{1}{2}kx_{max}^2 = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2$
$x_{max} = \sqrt{\frac{kx_0^2 + mv_0^2}{k}}$