3) A rather large fish is about to eat an unsuspecting small fish. The big fish has a mass of 5kg and is swimming at 8 m/s, while the small fish has a mass of 1 kg and is swimming at -4 m/s. What is the
velocity of big fish after lunch?

the velocity of the big fish after the launch is 6 m/s.

Explanation:

Given;

mass of the big fish, m₁ = 5 kg

velocity of the big fish, u₁ = 8 m/s

mass of the small fish, m₂ = 1 kg

velocity of the small fish, u₂ = -4 m/s

Let the final velocity of the big fish after launch = v

Apply the principle of conservation linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

5 x 8   + 1 x (-4) = v(5 + 1)

40 - 4 = 6v

36 = 6v

v = 36/6

v =  6 m/s.

Therefore, the velocity of the big fish after the launch is 6 m/s.

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west