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

Note that the initial momentum is zero so the horizontal component of the person's momentum will be equal in magnitude to the horizontal component of the rock's momentum. Note that the person and the rock will be moving in opposite directions. $m_p~v_p = m_r~v_{r,x}$ $v_p = \frac{m_r~v_{r,x}}{m_p}$ $v_p = \frac{(3.00~kg)(12.0~m/s)~cos(35.0^{\circ})}{70.0~kg}$ $v_p = 0.421~m/s$ The person's speed will be 0.421 m/s.