#### Answer

(a) The magnitude of the stone's velocity is 25.8 m/s and it is moving at an angle of $35.5^{\circ}$ below the +x-axis (where the bullet's initial direction of motion is assumed to be the +x-axis).
(b) The collision is not perfectly elastic.

#### Work Step by Step

(a) Let the bullet's initial velocity be in the +x direction, and let the bullet's final velocity be in the +y-direction. We can find the horizontal component of the stone's velocity.
$m_sv_x = m_bv_0$
$v_x = \frac{m_bv_0}{m_s}$
$v_x = \frac{(0.00600~kg)(350~m/s)}{0.100~kg}$
$v_x = 21.0~m/s$
We can find the vertical component of the stone's velocity.
$m_bv_f+m_sv_y = 0$
$v_y = \frac{-m_bv_f}{m_s}$
$v_y = \frac{-(0.00600~kg)(250~m/s)}{0.100~kg}$
$v_y = -15.0~m/s$
We can find the magnitude of the stone's velocity.
$v = \sqrt{(21.0~m/s)^2+(-15.0~m/s)^2}$
$v = 25.8~m/s$
We can find the angle $\theta$ below the +x-axis.
$tan(\theta) = \frac{15.0}{21.0}$
$\theta = arctan(\frac{15.0}{21.0})$
$\theta = 35.5^{\circ}$
The magnitude of the stone's velocity is 25.8 m/s and it is moving at an angle of $35.5^{\circ}$ below the +x-axis.
(b) $K_1 = \frac{1}{2}m_bv_0^2$
$K_1 = \frac{1}{2}(0.00600~kg)(350~m/s)^2$
$K_1 = 367.5~J$
$K_2 = \frac{1}{2}m_bv_f^2 + \frac{1}{2}m_sv^2$
$K_2 = \frac{1}{2}(0.00600~kg)(250~m/s)^2 + \frac{1}{2}(0.100~kg)(25.8~m/s)^2$
$K_2 = 220.8~J$
Since the final amount of kinetic energy is less than the initial amount of kinetic energy, the collision is not perfectly elastic.