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17 March, 01:26

A ball of mass 0.440 kg moving east (+x direction) with a speed of 3.80 m/s collides head-on with a 0.220-kg ball at rest. If the collision is perfectly elastic, what will be the speed and direction of each ball after the collision?

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  1. 17 March, 02:24
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    The speed and direction of each ball after the collision is 1.27 m/s to East direction and 5.07 m/s to East direction.

    Explanation:

    given information

    m₁ = 0.440 kg

    v₁ = 3.80 m/s

    m₂ = 0.220 kg

    v₂ = 0

    collision is perfectly elastic

    v₁ - v₂ = - (v₁' - v₂')

    v₁ = - (v₁' - v₂')

    v₂' = v₁ + v₁'

    according to momentum conservation energy

    m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'

    m₁v₁ = m₁v₁' + m₂ (v₁ + v₁')

    m₁v₁ = m₁v₁' + m₂v₁ + m₂v₁'

    m₁v₁ - m₂v₁ = m₁v₁' + m₂v₁'

    v₁ (m₁ - m₂) = (m₁ + m₂) v₁'

    v₁' = (m₁ - m₂) v₁ / (m₁ + m₂)

    = (0.440 - 0.220) (3.8) / (0.440 + 0.220)

    = 1.27 m/s to East direction

    v₂' = v₁ + v₁'

    = 3.8 + 1.27

    = 5.07 m/s to East direction
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