## University Physics with Modern Physics (14th Edition)

(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.
(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.