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22 October, 06:03

An 8.0-kg object moving 4.0 m/s in the positive x direction has a one-dimensional collision with a 2.0-kg object moving 3.0 m/s in the opposite direction. The final velocity of the 8.0-kg object is 2.0 m/s in the positive x direction. What is the total kinetic energy of the two-mass system after the collision?

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  1. 22 October, 09:25
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    Kf = 41 J

    Explanation:

    Theory of collisions

    Linear momentum is a vector magnitude (same direction of the velocity) and its magnitude is calculated like this:

    p=m*v

    Where:

    p : Linear momentum

    m: mass

    v:velocity

    There are 3 cases of collisions : elastic, inelastic and plastic.

    For the three cases the total linear momentum quantity is conserved:

    P₀ = Pf Formula (1)

    P₀ : Initial linear momentum quantity

    Pf : Final linear momentum quantity

    Data

    m₁ = 8.0-kg : mass of object₁

    m₂ = 2.0-kg : mass of object₂

    v₀₁ = 4.0 m/s, to the right : Initial velocity of m₁

    v₀₂ = 3.0 m/s, to the left : Initial velocity of m₂

    vf₁ = 2.0 m/s, to the right : Final velocity of m₁

    Problem development

    We assume that the 2kg object move to the right at the end of the collision, so, the sign of the final speeds (vf₂), we assume vf₂ positive:

    We appy the formula (1):

    P₀ = Pf

    m₁*v₀₁ + m₂*v₀₂ = m₁*vf₁ + m₂*vf₂

    (8) * (4) + (2) * (-3) = (8) * (2) + (2) * vf₂

    32-6 = 16 + (2) * vf₂

    32-6-16 = (2) * vf₂

    10 = (2) * vf₂

    vf₂ = 10 / (2)

    vf₂ = 5 m/s, to the right

    Kinetic energy

    Kinetic energy is that which is due to the movement of bodies and is calculted like this:

    K = (1/2) mv²

    Where:

    K : Kinetic energy (J)

    m: mass (kg)

    v : speed (m/s)

    Total kinetic energy of the two-mass system after the collision (Kf)

    Kf = Kf₁ + Kf₂

    Kf₁ = (1/2) m₁vf₁² = (1/2) (8) (2) ² = 16 J

    Kf₂ = (1/2) m₂vf₂² = (1/2) (2) (5) ² = 25 J

    Kf = 16 J + 25 J

    Kf = 41 J
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