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

We can consider the system of both buckets. Let $m_1$ be the 4.0-kg bucket. Let $m_2$ be the 12.0-kg bucket. $K_2+U_2=K_1+U_1$ $\frac{1}{2}(m_1+m_2)v^2+m_1gh= 0 + m_2gh$ $v^2= \frac{(m_2-m_1)~2gh}{m_1+m_2}$ $v= \sqrt{\frac{(m_2-m_1)~2gh}{m_1+m_2}}$ $v= \sqrt{\frac{(12.0~kg-4.0~kg)(2)(9.80~m/s^2)(2.00~m)}{12.0~kg+4.0~kg}}$ $v = 4.43~m/s$ The speed of the bucket is 4.43 m/s when it strikes the floor.