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24 April, 21:10

Two objects, X and Y, are held at rest on a horizontal frictionless surface and a spring is compressed between them. The mass of X is 2/5 times the mass of Y. Immediately after the spring is released, X has a kinetic energy of 50 J and Y has a kinetic energy of:

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  1. 24 April, 22:13
    +1
    20J

    Explanation:

    Using conservation law of momentum;

    since the bodies were at rest, their initial momentum is zero

    0 = M1Vx + M2Vy

    - M1Vx = M2Vy where Vx is the final velocity of x after the spring has been release and Vy is final velocity of y and M1 and M2 are the masses of x and y

    also M1 = 2/5 M2

    substitute M1 into the the equation above

    -2/5 M2Vx = M2Vy

    cancel M2 on both side

    -2/5Vx = Vy

    comparing the kinetic energy of both x and y

    for x K. E = 1/2 M1 Vx²

    and y K. E = 1/2M2 Vy²

    substitute for M1 = 2/5 M2

    K. Ex = 1/2 * 2/5 M2 Vx²

    divide K. Ex / K. Ey = (1/2 * 2/5 M2 Vx²) / 1/2 M2 Vy²

    cancel the common terms

    K. Ex / K. Ey = (2/5 Vx²) / Vy²

    substitute - 2/5Vx for Vy

    (2/5 Vx²) / (-2/5 Vx) ² = (2/5 Vx²) / (4/25 Vx²)

    cancel Vx²

    (2/5) / (4/25) = 2/5 : 4/25 = 2/5 * 25/4 = 5/2

    the ratio of x and y kinetic energy is 5:2

    since the kinetic energy of x is 50

    50 : 20 = 5 : 2 if 10 is used to divide both sides

    the kinetic energy of y = 20 J
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