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

We can use conservation of momentum to find block B's horizontal speed $v_{Bx}$ after the collision. $m_Bv_{Bx} + m_Av_{A2} = m_Av_{A1}$ $v_{Bx} = \frac{m_Av_{A1}-m_Av_{A2}}{m_B}$ $v_{Bx} = \frac{(2.00~kg)(8.00~m/s)-(2.00~kg)(-2.00~m/s)}{4.00~kg}$ $v_{Bx} = 5.00~m/s$ We can find the vertical speed just before block B strikes the floor. $v_{By}^2 = v_{y0}^2+2ay = 0 +2ay$ $v_{By} = \sqrt{2ay} = \sqrt{(2)(9.80~m/s^2)(2.60~m)}$ $v_{By} = 7.14~m/s$ We can find the speed of block B $v_B$ just before it strikes the floor. $v_B = \sqrt{(5.00)^2+(7.14)^2}$ $v_B = 8.72~m/s$ The speed of block B just before it strikes the floor is 8.72 m/s