Answer:
The answer is C > A > B
Step-by-step explanation:
This a conservation of momentum problem, that means that the initial momentum of the system is equal to final momentum of the system.
The mass of the big fish is 5 times greater than the mass of the smaller fish, therefore, to make things simpler,
Let the mass of the big fish be 5 kg and the mass of the small one be 1 kg.
In our case we have:
Momentum before lunch = momentum after lunch.
[tex](v_{1} *m_{1}) + (v_{2} *m_{2}) = (m_{1} *m_{2})* v\\[/tex]
In the first case (A), v1 = 4 m/s ; v2 = 0 m/s
(4 m/s*5kg) + (0 m/s*1 kg) = (5kg + 1 kg)*v
20 + 0 = 6v
V = [tex]\frac{20}{6}[/tex]
V = [tex]\frac{10}{3}[/tex] m/s
In the second case (B), v1 = 4 m/s ; v2 = -1 m/s (v2 is negative because it’s going to the opposite direction of the bigger fish)
(4 m/s*5kg) + (-1 m/s*1 kg) = (5kg + 1 kg)*v
20 - 1 = 6v
V = [tex]\frac{19}{6}[/tex] m/s
In the third case (C), v1 = 5 m/s ; v2 = -2 m/s (v2 is negative because it’s going to the opposite direction of the bigger fish)
(5 m/s*5kg) + (-2 m/s*1 kg) = (5kg + 1 kg)*v
25 - 2 = 6v
V = [tex]\frac{23}{6}[/tex] m/s
Therefore, the greatest speed is the third case and the smallest is the second case. So, C > A > B
