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

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

Answer: 93.1 J

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.