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24 January, 03:32

A 25,000-kg train car moving at 2.50 m/s collides with and connects to a train car of equal mass moving in the same direction at 1.00 m/s. How much does the kinetic energy of the system decrease during the collision?

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  1. 24 January, 03:40
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    14062.5 J

    Explanation:

    From the law of conservation of momentum,

    Total momentum before collision = Total momentum after collision.

    V = (m₁u₁ + m₂u₂) / (m₁+m₂) ... 1

    Where V = common velocity after collision

    Given: m₁ = m₂ = 25000 kg, u₁ = 2.5 m/s, u₂ = 1 m/s

    Substitute into equation 1

    V = [25000 (2.5) + 25000 (1) ] / (25000+25000)

    V = (62500+25000) / 50000

    V = 87500/50000

    V = 1.75 m/s.

    Note: The collision is an inelastic collision as such there is lost in kinetic energy of the system.

    Total Kinetic energy before collision = kinetic energy of the first train car + kinetic energy of the second train car

    E₁ = 1/2m₁u₁² + 1/2m₂u₂² ... Equation 2

    Where E₁ = Total kinetic energy of the body before collision, m₁ and m₂ = mass of the first train car and second train car respectively. u₁ and u₂ = initial velocity of the first train car and second train car respectively.

    Given: m₁ = m₂ = 25000 kg, u₁ = 2.5 m/s, u₂ = 1 m/s

    Substitute into equation 2

    E₁ = 1/2 (25000) (2.5) ² + 1/2 (25000) (1.0) ²

    E₁ = 12500 (6.25) + 12500

    E₁ = 78125+12500

    E₁ = 90625 J.

    Also

    E₂ = 1/2V² (m₁+m₂) ... Equation 3

    Where E₂ = total kinetic energy of the system after collision, V = common velocity, m₁ and m₂ = mass of the first and second train car respectively.

    Given: V = 1.75 m/s, m₁ = m₂ = 25000 kg

    Substitute into equation 3

    E₂ = 1/2 (1.75) ² (25000+25000)

    E₂ = 1/2 (3.0625) (50000)

    E₂ = (3.0625) (25000)

    E₂ = 76562.5 J.

    Lost in kinetic Energy of the system = E₁ - E₂ = 90625 - 76562.5

    Lost in kinetic energy of the system = 14062.5 J
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