Question

A 20-kg fish swimming at 1 m/s swallows a 5-kg fish swimming toward it at a velocity that bring both to a halt. How fast was the smaller fish moving prior to being eaten?

Answer 1

Answer:

Swallowed fish speed=4 m/s

Explanation:

Given

m₁(fish mass)=20 kg

v₁(fish speed)=1 m/s

m₂(Swallowed fish mass)=5 kg

To find

v₂(Swallowed fish speed)

Solution

From law of conservation of momentum

$m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v\\ (20kg)(1 m/s)+(5 kg)v_{2}=(20kg+5kg)0m/s\\20kg.m/s+5kgv_{2}=0\\v_{2}=(20kg.m/s)/5kg\\v_{2}=4m/s$

As I neglected the negative sign because it only shows that swallowed fish is moving in opposite direction