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12 March, 15:47

Two clowns are launched from the same spring-loaded circus cannon with the spring compressed the same distance each time. Clown A has a 40-kg mass; clown B a 60-kg mass. The relation between their kinetic energies at the instant of launch is

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  1. 12 March, 17:50
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    the kinetic energy of clown A is 0.444 times the kinetic energy of clown B.

    Explanation:

    Let the spring constant of the spring is k.

    For clown A:

    m = 40 kg

    let the extension in the spring is y.

    So, the spring force, F = k y

    m g = k y

    40 x g = k x y

    y = 40 x g / k ... (1)

    For clown B:

    m' = 60 kg

    Let the extension in the spring is y'.

    So, the spring force, F' = k y'

    m' g = k y'

    y' = 60 x g / k ... (2)

    Kinetic energy for A, K = 1/2 ky^2

    Kinetic energy for B, K' = 1/2 ky'^2

    So, K/K' = y^2/y'^2 K / K' = (40 x 40) / (60 x 60) (from equation (1) and (2))

    K / K' = 0.444

    K = 0.444 K'

    So the kinetic energy of clown A is 0.444 times the kinetic energy of clown B.
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