Answer to Question #145093 in Physics for Joshua

Question

A 16 kg tuna fish moving horizontally to the right at 6 m/s swallows a 3 kg dalagang bukid that is swimming to the left at 8.5 m/s. What is the speed of the tuna fish immediately after, if the forces exerted on the fish by the water can be neglected?

Answer 1

The total momentum of two fishes befor the impact was:

$$p = m_1v_1 - m_2v_2$$

where $m_1 = 16.6kg$ is the mass of the tuna, $v_1 = 6m/s$ is the speed of the tuna, $m_2 = 3kg$ is the mass of the dalagang bukid and $v_2 = 8.5m/s$ is the speed of the dalagang bukid.

The momentum of two fishes after the impact is:

$$p' = (m_1 + m_2)v'$$

where $v'$ is their speed (tuna's speed, in fact) after the impact.

Let's apply the momentum conservation law:

$$p = p'\\ m_1v_1 - m_2v_2 = (m_1 + m_2)v'\\ v' = \dfrac{m_1v_1 - m_2v_2}{m_1 + m_2}\\ \space \\ v' = \dfrac{16.6\cdot 6 - 3\cdot 8.5}{16.6 + 8.5} \approx 2.95m/s$$

Answer. 2.95 m/s.