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15 July, 22:56

Object A has of mass 7.20 kilograms, and object B has a mass of 5.75 kilograms. The two objects move along a straight line toward each other with velocities + 2.00 meters/second and - 1.30 meters/second respectively. What is the total kinetic energy of the objects after the collision, if the collision is perfectly elastic?

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  1. 16 July, 02:19
    0
    Two things:

    1) All collisions conserve momentum

    2) Elastic collisions also conserve kinetic energy.

    So basically the kinetic energy before the collision is the same as the kinetic energy after the collision.

    KE = ½mv²

    For object A:

    KE = ½ (7.20 kg) (2.00 m/s) ² = 14.4 J

    For object B:

    KE = ½ (5.75 kg) (-1.30 m/s) ² = 4.86 J

    total kinetic energy = 14.4 J + 4.86 J = 19.3 J
  2. 16 July, 02:33
    0
    The equation for Kinetic Energy in an Elastic Collision is this:

    1/2 (m₁v₁) + 1/2 (m₂v₂) = 1/2 (7.20kg) (2.00m/s) + 1/2 (5.75kg) (-1.30m/s) =

    3.4625m/s

    I think ... technically the equation above equates to the same thing except measuring the final velocity of the given masses, however you were not given that in the equation, making the first half of the full equation the only viable solution.
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