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

We can use conservation of momentum to solve this question. $m_2~v_2 = m_1~v_1$ $v_2 = \frac{m_1~v_1}{m_2}$ $v_2 = \frac{(15.0~kg)(1.10~m/s)}{19.50~kg}$ $v_2 = 0.846~m/s$ The speed of the large fish after eating the small fish is 0.846 m/s. (b) $K_1 = \frac{1}{2}m_1~v_1^2$ $K_1 = \frac{1}{2}(15.0~kg)(1.10~m/s)^2$ $K_1 = 9.075~J$ $K_2 = \frac{1}{2}m_2~v_2^2$ $K_2 = \frac{1}{2}(19.5~kg)(0.846~m/s)^2$ $K_2 = 6.978~J$ $\Delta K = K_2-K_1$ $\Delta K = 6.978~J - 9.075~J$ $\Delta K = -2.10~J$ The amount of mechanical energy dissipated during this meal was 2.10 J.