Two solid marbles A and B with a mass of 3.00 kg and 6.50 kg respectively have an elastic collision in one dimension. Before collision solid

Question

Two solid marbles A and B with a mass of 3.00 kg and 6.50 kg respectively have an elastic collision in one dimension. Before collision solid marble A (3.00 kg) was at rest and the other solid marble (6.50 kg) had a speed of 3.50 m/s. Calculate the magnitudes of velocities of two solid marbles vA and vB after collision.

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Amity 4 years 2021-08-12T22:17:35+00:00 1 Answers 18 views 0

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  1. Edana
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    2021-08-12T22:19:15+00:00
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    Answer:

    va = 4.79 m/s

    vb = 1.29 m/s

    Explanation:

    Momentum is conserved:

    m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

    (3.00) (0) + (6.50) (3.50) = (3.00) v₁ + (6.50) v₂

    22.75 = 3v₁ + 6.5v₂

    For an elastic collision, kinetic energy is conserved.

    ½ m₁u₁² + ½ m₂u₂² = ½ m₁v₁² + ½ m₂v₂²

    m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²

    (3.00) (0)² + (6.50) (3.50)² = (3.00) v₁² + (6.50) v₂²

    79.625 = 3v₁² + 6.5v₂²

    Two equations, two variables.  Solve with substitution:

    22.75 = 3v₁ + 6.5v₂

    22.75 − 3v₁ = 6.5v₂

    v₂ = (22.75 − 3v₁) / 6.5

    79.625 = 3v₁² + 6.5v₂²

    79.625 = 3v₁² + 6.5 ((22.75 − 3v₁) / 6.5)²

    79.625 = 3v₁² + (22.75 − 3v₁)² / 6.5

    517.5625 = 19.5v₁² + (22.75 − 3v₁)²

    517.5625 = 19.5v₁² + 517.5625 − 136.5v₁ + 9v₁²

    0 = 28.5v₁² − 136.5v₁

    0 = v₁ (28.5v₁ − 136.5)

    v₁ = 0 or 4.79

    We know v₁ isn’t 0, so v₁ = 4.79 m/s.

    Solving for v₂:

    v₂ = (22.75 − 3v₁) / 6.5

    v₂ = 1.29 m/s

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