Find the change in kinetic energy of a 0.650 kg fish leaping to the right at 15.0 m/s that collides inelastically with a 0.950 kg fish leaping to the left at 13.5 m/s

The change in the kinetic energy refers to the work done in displacing a body, thus, the change in the kinetic energy of an object refers to the work done on the object.
The correct formula to use is:
W = Initial kinetic energy - Final kinetic energy;
Where, W = change in kinetic energy
Final kinetic energy and initial kinetic energy = 1/2 MV^2
Initial velocity = 15 m/s
Final velocity  = 13.5 m/s
Initial mass = 0.650 kg
Final mass = 0.950 kg
W = 1/2 [0.650* (15 *15)] - 1/2 [0.950 * (13.5 * 13.5)]
W = 146.25 - 173.13 = 26.88
Therefore, the change in kinetic energy is 26.88 J.
 The negative sign has to be ignored, because change in kinetic energy can not be negative.

Define
m₁ = 0.650 kg the mass of the fish leaping to the right.
u₁ = 15.0 m/s is its velocity.
m₂ = 0.950 kg, the mass of the fish leaping to the left
u₂ = 13.5 m/s is its velocity.

Let v = the mutual velocity after inelastic impact.
For conservation of momentum,
m₁u₁ + m₂(-u₂) = (m₁+m₂)v
0.65*15 - 0.95*13.5 = (0.65 + 0.95)*v
v = - 3.075/1.6
   = - 1.9219 m/s 

The change in KE of the 0.65 kg fish is
(1/2)*(0.65 kg)*(15 + 1.9219)² = 93.06 J

93.1 J

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