a 0.300kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.80 m/s when it makes a head on collision with a stationary 0.150kg glider.
(a) Find the magnitude and direction of the final velocity of each glider in the collision is elastic.
(b) Find the final kinetic energy of each glider

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  1. and your question is?
    you have the conservation of momentum, and the energy conservation. You have two unknowns, and two equations.
    find in momentum one velocity in terms of the other, then put that into the second equation and do the algebra. It will be a quadratic.

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  2. M1 = 0.30kg, V1 = 0.8 m/s.
    M2 = 0.150kg, V2 = 0.

    M1*V1 + M2*V2 = M1*V3 + M2*V4.
    0.3*0.8 + 0.15*0 = 0.3*V3 + 0.15*V4,
    Eq1: 0.3*V3 + 0.15*V4 = 0.24.

    Conservation of KE Eq:
    V3 = (V1(M1-M2) + 2M2*V2)/(M1+M2).
    V3 = (0.8(0.3-0.15) + 0.30*0)/(0.3+0.15) = 0.267 m/s. = Final velocity of M1.

    In Eq1, replace V3 with 0.267 and solve for V4:
    0.3*0.267 + 0.15*V4 = 0.24.
    0.15*V4 = 0.16,
    V4 = 1.07 m/s. = Velocity of M2.

    Direction: Both gliders are moving to the right.

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  3. b. KE1 = 0.5M1*V3^2 = 0.5*0.3*0.267^2 = 0.011 J. = Kinetic energy of M1.

    KE2 = 0.5M2*V4^2 = 0.5*0.15*1.07^2 = 0.086 J. = Kinetic energy of M2.

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