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24 March, 15:13

A 2.00-kg object traveling east at 20.0 m/s collides with a 3.00-kg object traveling west at 10.0 m/s.

After the collision, the 2.00-kg object has a velocity 5.00 m/s to the west. How much kinetic energy

was lost during the collision?

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Answers (2)
  1. 24 March, 15:45
    0
    Momentum = mass x velocity

    Before collision

    Momentum 1 = 2 kg x 20 m / s = 40 kg x m/s

    Momentum 2 = 3 kg x - 10m/s = - 30 kg x m/s

    After collision

    Momentum 1 = 2 kg x - 5 m/s = - 10 m/s

    Momentum 2 = 3 kg x V2 = 3V2

    Total momentum before = total momentum after

    40 + - 30 = - 10 + 3V2

    V2 = 6.67 m/s

    Total kinetic energy before

    = (1/2) [ 2 kg * 20 m/s * 2 + 3 kg * (-10 m/s) * 2 ]

    = 550 J

    Total kinetic energy after

    = (1/2) [ 2 kg * ( - 5 m/s) * 2 + 3 kg * 6.67 m/s * 2 ]

    = 91.73 J

    Total kinetic energy lost during collision

    = 550 J - 91.73 J

    = 458.27 J
  2. 24 March, 16:08
    0
    Kinetic energy before:

    E = 0.5mv^2 + 0.5mv^2

    = 0.5*2*20^2 + 0.5*3 * (-10) ^2

    = 400 + 150

    = 550J

    Kinetic energy after:

    E = 0.5mv^2

    = 0.5*2 * (-5) ^2

    = 25J

    Kinetic energy lost during the collision:

    E = 550 - 25

    = 525J
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