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4 August, 23:40

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

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  1. 4 August, 23:48
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    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
  2. 5 August, 01:40
    0
    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.
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