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17 December, 06:37

0.05 kg bullet collides and sticks to a 2.5 kg stationary block suspended from a string. the bullet and block swings to a maximum height of 12 cm. what was the initial speed of the bullet?

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  1. 17 December, 06:56
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    First calculate the combine velocity of the bullet and stick, using the formula

    kinetic energy = potential energy

    0.5mv^2 = mgh

    where m is the combined mass

    v is the velocity

    g is the acceleration due to gravity (9.81 m/s^2

    h is the heigth

    0.5 (0.05 + 2.5 kg) v^2 = (9.81 m/s^2) (0.05 + 2.5 kg) (0.12 m)

    solve for v

    v = 1.53 m/s

    using momentum balance

    M = mv

    m1v1 + m2v2 = mv

    where m1 is the mass of the bullet

    v1 is the initial velocity of the bullet

    m2 mass of the stick

    v2 initial velocity of stick

    (0.05 kg) (v1) + (2.5 kg) (0) = (0.05 + 2.5 kg) (1.53 m/s)

    v1 = 78.25 m/s
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