Answer
(a) The speed of the large fish after eating the small fish is 0.846 m/s.
(b) The amount of mechanical energy dissipated during this meal was 2.10 J.
Work Step by Step
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.
