A boy is fishing in a stationary (unmoored) boat. Being ecologically conscious, he throws back the 2.0 kg fish he has caught. He gives the fish an initial horizontal velocity of 5.0 m/s relative to the water. The mass of the boy and the boat together is 45 kg. The fish is a largemouth bass, not a red herring.

    a) What is the velocity v_b of the boy and boat after the fish is thrown?

    b) How much energy did the boy have to expend in throwing the fish?

Solution:

a) Use conservation of momentum (in the x-direction, where x=horizontal) as there are no external forces on the boat-fish-boy system in the x-direction. Initial momentum = 0, since everything starts off at rest.

       p_initial = p_final

       0 = m₁v₁ + m₂v₂

thus

     v₂  =  - [ m₁v₁]/m₂  

          =   - (2.0 kg)(5 m/s)/45 kg   =   -0.22 m/s

Note that the minus sign gives us the direction, i.e. the boat & boy recoil opposite to the direction of the fish.

b) The energy expended (or work done) is the change in total kinetic energy K. Initial kinetic energy is zero, so

K = ½ m₁v₁² + ½ m₂v₂²

       = ½ (2.0 kg) (5 m/s)²  +  ½ (45 kg) (-0.22 m/s)²  

       = 26.1 J

(Note: Dont't forget the second term, i.e. the fact that the boy's throw had to give energy to both the fish and the boat)